Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}... | \begin{align*}
\sum \frac{a^3}{b+c+d} &\geq \frac{4(a^3+b^3+c^3+d^3)}{3(a+b+c+d)}\\
&\geq \frac{(a+c)^3+(b+d)^3}{3(a+b+c+d)}\\
&=\frac{1}{3}((a+c)^2+(b+d)^2-(a+c)(b+d))\\
&\geq \frac{(a+c)(b+d)}{3}\\
&=\frac{1}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $a+b+c \geq \frac{3+\sqrt{7}}{2}$ for $a,b,c \in [0, \frac32]$ with $a^2+b^2+c^2+abc=4$
Let $a,b,c \in \left[0;\frac{3}{2}\right]$ and $a^2+b^2+c^2+abc=4$. Prove that
$$a+b+c \geq \dfrac{3+\sqrt{7}}{2}.$$
Source: This is a math problem that my teacher gave me $3$ months ago (the submission deadline has expired)... | It suffices to prove that
$$2a + 2b + 2c - 3 \ge \sqrt 7.$$
Using $(a + b + c)^2 \ge a^2 + b^2 + c^2$ and $(a + b + c)^3/27 \ge abc$, we have
$(a + b + c)^2 + (a + b + c)^3/27 \ge 4$
which results in $a + b + c > 3/2$ or
$2a + 2b + 2c - 3 > 0$.
Thus, it suffices to prove that
$$(2a + 2b + 2c - 3)^2 - 7 \ge 0.$$
We have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Factor $1-64m^6$ under $\mathbb{Q}$ Factor $1-64m^6$ under $\mathbb{Q}$. We can employ the well known formula: $a^3-b^3 = (a-b)\cdot(a^2+ab+b^2)$.
$1-64m^6 = (1)^3-(4m^2)^3 = (1-4m^2)(1+4m^2+16m^4)=(1+2m)(1-2m)(1+4m^2+16m^4)$.
However, the answer is provided as $(1+2m)(1-2m)(1-2m+4m^2)\cdot(1+2m+4m^2)$.
I cannot unders... | There is a simpler approach. We could first factor $1 - 64m^6$ as a difference of squares, then factoring the resulting sum of cubes and difference of cubes.
\begin{align*}
1 - 64m^6 & = (1 + 8m^3)(1 - 8m^3)\\
& = (1 + 2m)(1 - 2m + 4m^2)(1 - 2m)(1 + 2m + 4m^2)
\end{align*}
As for your method of first factori... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+... | Write given expression as $A+B$ where $$A = \frac{ab^2}{a+c} +\frac{bc^2}{b+a} +\frac{ca^2}{c+b}$$
and
$$B= \frac{2}{a+c} + \frac{2}{b+a} + \frac{2}{c+b}$$
By AM-GM inequality we have
$$ A\geq \frac{3abc}{\sqrt[3]{(a+c)(c+b)(b+a)}}$$
and $$ B\geq \frac{6}{\sqrt[3]{(a+c)(c+b)(b+a)}}$$
Further again by AM-GM we have $$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
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Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (... | $$r^4+r^2+1=(r^2+1)^2-r^2=(r^2+r+1)(r^2-r+1)$$
Observe that if $f(n)=n^2+n+1, f(r-1)=(r-1)^2-(r-1)+1=r^2-r+1$
If $g(n)=an+b, g(n-1)=a(n-1)+b=an+b-a$
$$\text{Let }\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(r)}{f(r)}-\dfrac{g(r-1)}{f(r-1)}$$ $$\text{so that the partial sum becomes }\sum_{r=1}^n\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
sum of 100 terms of logarithmic expression
Calculate value of
$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$
My try :: $\displaystyle x^4+4y^4$
$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$
$\display... | $\ln (16241)$. Work out the first few cases, use $\ln a + \ln b = \ln ab$ and then prove by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Simplifying the surd $\sqrt{3 - \sqrt{8}}$ Recently I was solving a math problem and I came across the following ( Only part of the problem ):
$$\sqrt{3 - \sqrt{8}}$$
Here's is what I did to simplify the above:
$$(a-b)^2=a^2+b^2-2ab$$
$$\sqrt{1+2-2\sqrt{2}}$$
$$\sqrt{(1-\sqrt{2})^2}$$
$$=1-\sqrt{2}$$
When you expand $(... | We have $\sqrt {3-\sqrt 8 }=\sqrt {3-2\sqrt 2 }.$ We ask whether there exist rational $a,b$ such that that $$3-2\sqrt 2=(a+b\sqrt 2)^2=(a^2+2b^2)+2ab\sqrt 2.$$ Since $a,b\in\Bbb Q$, this requires $$ (\bullet) \quad a^2+2/a^2=3 \land -2ab=2.$$ So $b=-1/a$ from the 2nd equation of $(\bullet).$ So $3=a^2+2b^2=a^2+2/a^2.$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this, do I have to continue calculating with rational fractions? $$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$$
I tried to use different substitutions such as $t=\cos(x)$, $t=\sin(x)$, $t=\tan(x)$, and after expressing $\sin$ and $\cos$ through $\tan(\frac{x}{2})$, I've got $ \int -4 \frac{t^2dt}{(1+t^2)^... | We have
$$(\sin x+2\cos x)(2\cos x-\sin x)=4-5\sin^2x=5\cos^2x-1$$
Therefore
$${\sin^2x\over \sin x+2\cos x} ={\sin^2x(2\cos x-\sin x)\over (\sin x+2\cos x)(2\cos x-\sin x)}\\ =
2{\sin^2x\over 4-5\sin^2x}\cos x-{1-\cos^2x\over 5\cos^2x -1}\sin x$$
Hence substituting $s=\sin x$ and $t=\cos x,$ respectively, gives
$$\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$ (from MIT Integration Bee 2023) This is from the final round of MIT Integration Bee 2023.
$$\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$$
The g... | From @Archisman Panigrahi,
\begin{eqnarray}
I&=&\frac{1}{2} \int_0^{2\pi} \left(\frac{\cos{x}\cos{15x}}{\cos{3x}\cos{5x}}\right)^2 dx= \frac{1}{2} \int_0^{2\pi} \left(\frac{\cos{14x}+\cos{16x}}{\cos{2x}+\cos{8x}}\right)^2 dx\\
&=&\frac{1}{4} \int_0^{4\pi} \left(\frac{\cos{7x}+\cos{8x}}{\cos{x}+\cos{4x}}\right)^2 dx=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Using Maclaurin series to solve $y'=xe^x$. I am trying to solve a differential equation using the Maclaurin series. The differential equation is
$$y'=xe^x$$
My solution 1:
$xe^x$ expressed in sigma notation is: $\sum_{k=0}^\infty\frac{x^{k+1}}{k!}$ and $y'=\sum_{k=0}^\infty(k+1)a_{k+1}x^k$. Equating them will lead to
$... | Method 1:
LHS of the series is given by
$$\begin{aligned} \sum_{k=0}^\infty (k+1)a_{k+1}x^k &= a_1 + \sum_{k=1}^{\infty}(k+1)a_{k+1}x^{k} \\
&= a_1 + \sum_{k=0}^{\infty}(k+2)a_{k+2}x^{k+1}\end{aligned}$$
Note that we have replaced $k$ by $k+1$ in the second equality.
Equality of power series implies that $a_1 = 0$ and
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove that $\frac a{a+2b}+\frac bc+\frac{c^2}{\left(c+2a\right)^2 }\ge1$
(sqing) Let $a$, $b$, $c>0$. Prove that $\dfrac a{a+2b}+\dfrac bc+\dfrac{c^2}{\left(c+2a\right)^2 }\ge1$. Source (unsolved)
In this inequality, $b$ appears the least frequent, so I want to deal with it. I took $a$, $c$ as parameters and derived ... | Using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&\frac{a}{a + 2b} + \frac{b}{c} + \frac{c^2}{(c + 2a)^2} - 1\\
\ge{}& \frac{(a + b)^2}{a(a + 2b) + bc} + \frac{c^2}{(c + 2a)^2} - 1 \\
={}& \frac{(2ab - ac + bc)^2}{(a^2 + 2ab + bc)(c + 2a)^2}\\
\ge{}& 0
\end{align*}
with equality if $a:b:c = 2:1:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find out limit of the following question let, $f(x) =x^\frac{1}{3}$ be a diffrentiable function on $ (0, \infty).$
Given that
$$\frac{f(3+h) -f(3)}{h}=f'(3+\theta(h)h)$$
Then find out
$\lim_{h\to 0+} \theta(h) =? $
Since, $f$ is diffrentiable at $3$ , I think limit must be $0$ as it tends to $f'(3)$
Please help me
| You can obtain explicitly that
$$\theta(h)=\frac{1}{h}\left(\left(3\frac{\sqrt[3]{3+h}-\sqrt[3]{3}}{h}\right)^{-3/2}-3\right)$$
Now, by noticing that $h=(\sqrt[3]{3+h}-\sqrt[3]{3})(\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9})$, compute the desired limit as
$$\lim_{h\to0^+}\theta(h)=\lim_{h\to0^+}\frac{1}{h}\left(\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate: $\int\frac{3e^{2x}-2e^x}{e^{2x}+2e^x-8}dx$ Given $\int\frac{3e^{2x}-2e^x}{e^{2x}+2e^x-8}dx$
My attempt- Substitute $u=e^x \Rightarrow du=dx.e^x$
The integral becomes
$$\int \frac{3u-2}{u^2+2u-8}du $$
Now, the numerator is a linear function and the denominator is a quadratic function. We can rewrite the numera... | No, it is not correct, since$$\int\frac1{(u+1)^2-3^2}=\frac1{\color{red}6}\log\left(\frac{u-2}{u+4}\right).$$So, your approach should lead to$$\log\left(\frac{\left(e^{2x}+2e^x-8\right)^{\frac32}}{\left(\frac{e^x-2}{e^x+4}\right)^{\frac5{\color{red}6}}}\right)+C.$$But there is a simpler approach: using the fact that$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Formula for $\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ b} \ln \left(a^2+b^2+a b \sqrt{2}\right) $ In my post, I had proved that $$
\int_0^{\infty} \frac{\ln \left(x^2+a^2\right)}{b^2+x^2} d x=\frac{ \pi}{b} \ln (a+b) \tag*{(*)}
$$
To go further, I guess that
$$\int_0^{\infty} \frac{\ln ... | An option using a semicircular contour in the upper half-plane: due to its even integrand, if $a,\,b>0$ your integral is$$\begin{align}\Re\int_{\Bbb R}\frac{\ln(x^2-a^2+aix\sqrt{2})}{b^2+x^2}dx=\Re\left[2\pi i\lim_{z\to bi}\frac{\ln(z^2-a^2+aiz\sqrt{2})}{z+bi}\right]=\frac{\pi}{b}\ln(a^2+b^2+ab\sqrt{2}).\end{align}$$Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation $\cos\frac{2x}{3}+\sqrt3\sin\frac{2x}{3}=\frac{8}{3+\cos{4x}}$ Solve the equation $$\cos\dfrac{2x}{3}+\sqrt3\sin\dfrac{2x}{3}=\dfrac{8}{3+\cos{4x}}$$
Let's divide both sides of the equation by $2$ to get $$\dfrac12\cos\dfrac{2x}{3}+\dfrac{\sqrt3}{2}\sin\dfrac{2x}{3}=\dfrac{4}{3+\cos{4x}}\\\iff \sin\l... | To begin with, I would recommend you to notice that
\begin{align*}
-1 \leq \cos(4x) \leq 1 & \Longleftrightarrow 2 \leq \cos(4x) + 3 \leq 4\\\\
& \Longleftrightarrow \frac{1}{4} \leq \frac{1}{\cos(4x) + 3} \leq \frac{1}{2}\\\\
& \Longleftrightarrow 1 \leq \frac{4}{\cos(4x) + 3} \leq 2
\end{align*}
Hence the LHS equals ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is $\csc x = a+bi$ defined? A question in my assignment asks to find which equation can have its roots $\sec^2 x, \csc^2 x$ (no restrictions in $x$ were given).
The equations were quadratic ($x^2+bx+c$) and the one which satisfies $-b=c$ can be the required equations.
There were two which satisfies this,
$x^2-3x+3=0, ... | If you want to solve the equation $$\csc (x+i y)=a+i\,b$$ expand first the cosecant, go to hyperbolic functions, use the conjugate, separate the real and imaginary parts to end with
$$\frac{\sin (x) \cosh (y)}{\cos ^2(x) \sinh ^2(y)+\sin
^2(x) \cosh ^2(y)}=a \tag 1$$
$$\frac{\cos (x) \sinh (y)}{\cos ^2(x) \sinh ^2(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the coordinates in an isosceles triangle if the triangle it self is in positive axis A at $(45,10)$, B at $(10,20)$, $AB=AC$ and angle $C=20$ degree find the coordinates of $C$.suggest the formula so i can write code in Perl.
| We have $A(45,10),B(10,20),C(x_c,y_c)$.
$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$
$C=B=20^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}=\pi /9$ rad
$A=180^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}-40^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}=140^{%
%TCIMAC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How closely can we estimate $\sum_{i=0}^n \sqrt{i}$ By looking at an integral and bounding the error?
| This was an interesting challenge, to try and come up with an estimate for this sum in an elementary way.
The estimate I got was
$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$
for some constant $C$ which appears to be close to $-0.207$ (which I guess will be $\zeta(-1/2)$).
(By $a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/5676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 3
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Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ Are there known sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ for $k > 1$ subject to the constraint $\sum_{p \mid k} \frac{1}{p+1} < 1$? (The factor of +1 in the denominator and the constraint of the sum of inverse "shifte... | My answer on MO:
If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments here), for any set of $n$ positive integers {$a_{1}, \dots, a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| Begin with the Law of Cosines to compute $\cot^2\frac{A}{2}$ in terms of the lengths of the sides of the triangle:
$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$
$$\Rightarrow \begin{cases}
\cos^2{\frac{A}{2}}=\frac{1+\cos{A}}{2}=\frac{2bc+b^2+c^2-a^2}{4bc} = \frac{(b+c)^2-a^2}{4bc} \\
\sin^2{\frac{A}{2}}=\frac{1-\cos{A}}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$ I am quite a baffled now, I am not getting by how it can be written that :
\begin{align*}
&\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ \\
&= \log_{10} \tan... | Yes it is valid.
$\log_{10} \tan 40^{\circ} + \log_{10} \tan 50^{\circ} = \log_{10} \tan 40^{\circ} + \log_{10} \cot 40^{\circ} = \log_{10} 1 = 0$
Similarly combine 41 and 49, 42 and 48 etc.
The product on the left side is $0$ as $\log_{10} \tan 45^{\circ} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/10868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How would I create a rotation matrix that rotates X by a, Y by b, and Z by c? How would I create a rotation matrix that rotates X by a, Y by b, and Z by c?
I need to formulas, unless you're using the ardor3d api's functions/methods.
Matrix is set up like this
xx, xy, xz,
yx, yy, yz,
zx, zy, zz
A Quaternion is fine too... | The rotation matrices around the x, y, and z axes, respectively, are
$$R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}$$
$$R_y(\phi) = \begin{pmatrix} \cos \phi & 0 & \sin \phi \\ 0 & 1 & 0 \\ - \sin \phi & 0 & \cos \phi \end{pmatrix}$$
$$R_z(\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/13272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$ $$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go abo... | It might not be elliptic after all... (unless I have made some mistake)
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
Let $\displaystyle u = x -\frac{1}{x}$.
Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.
Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\fra... | {
"language": "en",
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"source": "stackexchange",
"question_score": "44",
"answer_count": 4,
"answer_id": 1
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How to factorize $4x^4+12x^{10/3} y^{2/3}+ \dots $? Does anyone know how to factorize the following expression:
$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$
?
| It is possible, but likely messy.
First set $\displaystyle z = \left(\frac{x}{y}\right)^{2/3}$ and divide your expression by $\displaystyle y^4$.
We get
$$4z^6 + 12 z^5 + 33 z^4 + 46 z^3 + 33 z^2 + 12 z + 4$$
Now divide this by $\displaystyle z^3$ and set $\displaystyle t = z + 1/z$.
We get
$$4(z^3 + 1/z^3) + 12(z^2 + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Partial fraction integration So I want to find all antiderivaties of
$\frac{x}{x^3-1}$
Since the denominator is of a lesser degree than the numerator, partial fraction is to be used instead of long division.
I've started by doing:
$\frac{x}{x^3-1} = \frac{A}{x^2+x+1} + \frac{B}{x-1}$
Hence,
$x = A(x-1)+B(x^2+x+1)$
Then... | Like most techniques of integration, the ideal of partial fractions is to reduce a difficult integral to integrals that are, if not easy, at least doable. The three types of integrals that show up when doing partial fractions are:
*
*$\displaystyle \int\frac{1}{(ax+b)^n}\,dx$, with $a,b$ constants, $a\neq 0$, and $n... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Singular matrix Suppose I have a singular matrix given by
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14}\\
a_{12} & a_{11} & a_{14} & a_{13}\\
a_{31} & a_{32} & a_{33} & a_{34}\\
a_{32} & a_{31} & a_{34} & a_{33}
\end{pmatrix}$$
Which is a homogeneous system in $(x_1, x_2, x_3, x_4)^T$ where there is a va... | Yes. We have $\det A=\det B\cdot \det C$. There are some different ways to see this; here is one:
Your matrix $A$ can be written as the block matrix $\left(\begin{array}{cc} X&Y\\ U&W\end{array}\right)$, where $X$, $Y$, $U$, $W$ are the following $2\times 2$ matrices:
$X=\left(\begin{array}{cc} a_{11}&a_{12}\\ a_{12}&a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How $m^4+4 = 0 \Rightarrow m = 1 \pm i,-1\pm i$? This is given in my module as a part of a problem's solution:
$$m^4 + 4 = 0 $$ $$\Rightarrow m = 1 \pm i,-1\pm i$$
I am not getting how this conversion is taking place,could somebody explain?
| Here $m = (-4)^{1/4}$.
Now $-4 = 4(\cos(\theta) + i\sin(\theta))$ in polar form where $\theta = (2n-1)\pi$.
By de Moivre's theorem
$m = (-4)^{1/4} = 4^{1/4}(\cos(\frac{\theta}{4}) + i\sin(\frac{\theta}{4}))$
For $n = 0$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$,
For $n = ... | {
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"source": "stackexchange",
"question_score": "4",
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Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$ $\displaystyle\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}\,\mathrm{d}x}$
I used partial fractions, solved $A = 2, C = 3$.
$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$
\begin{align*}
&8x^4+15x^3+16x^2+22x+4\\
... | Given what you have, what happens when you set $x = -1$?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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The inverse of a certain tricky function What is the explicit form of the inverse of the function $f:\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow\mathbb{Z}^+$ where $$f(i,j)=\frac{(i+j-2)(i+j-1)}{2}+i?$$
| Since your function seems to be Cantor's pairing function $p(x,y) = \frac{(x+y)(x+y+1)}{2} + y$ applied to $x= j-2, y = i$, and since the inverse of the pairing function is $p^{-1}(z) = (\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + 3\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2}-z,z-\frac{\lfloor \frac{\sqrt{8z+1}-1}... | {
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"source": "stackexchange",
"question_score": "5",
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Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they... | What your textbook means is $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \iff \left(x \equiv 1 \pmod{5} \right)$$
This is what you checked by plugging in the different cases for $x$.
By taking $x \equiv 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \equiv 0 \pmod{5}$ and hence $$ \left(x \equiv 1 \pmod{5} \right) ... | {
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Given a matrix $A$ find a matrix $C$ such that $C^3$=$A$ This is a question I had on a test, we were told not to use brute-force and figure out a smart way to solve the problem.
We have a matrix $A =$
$\displaystyle\begin{bmatrix}
2 & 3\\
3 & 2
\end{bmatrix}$. Find a matrix $C$ such that $C^{3}$=$A$.
What is the 'smar... | One thing that jumps out at me is that this matrix has two eigenvalues. Find them: det
$
(\begin{bmatrix}
2-\lambda & 3\\
3 & 2-\lambda
\end{bmatrix})$=
$(2-\lambda)^{2}-9$=$-5-4\lambda+\lambda^{2}$. Factoring this you get $(\lambda+1)(\lambda-5)=0$, so your eigenvalues are -1 and 5. Now, find bases for their respecti... | {
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"source": "stackexchange",
"question_score": "6",
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\z... | By induction we can easily prove that for any nonnegative real numbers $a_k$
$$1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j-\sum_{1\le i<j<k\le n}a_ia_ja_k\le\prod_{k=1}^n(1-a_k)\le1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j$$
Taking $a_k=\frac{x^2}{k^2\pi^2}$,we get
$$1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\f... | {
"language": "en",
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"source": "stackexchange",
"question_score": "119",
"answer_count": 15,
"answer_id": 1
} |
Summing the power series $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\prod\limits_{k=1}^n\frac{2k-1}{2k} $ I'd like to determine the function corresponding to the following power series:
$$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1},
$$
where $|x|... | Consider the following series,
\begin{align}
\arcsin x = \sum_{n \geq 0} \frac{(2n)!}{2^{2n} (n!)^{2}} \frac{x^{2n+1}}{2n+1} = \sum_{n \geq 0} \frac{(2n-1)!!}{(2n)!!} \frac{x^{2n+1}}{2n+1},
\end{align}
which converges for $|x| \leq 1$ and where $\frac{(2n-1)!!}{(2n)!!} = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/30343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 1
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$x^3+48=y^4$ does not have integer (?) solutions How does one find all positive integer solutions to the equation $x^3+48=y^4$?
| I realize this is a massive revive, but here is a solution without any powerful theorems about elliptic curves.
We have $x^3 + 64 = y^4 + 2^4$. It is easy to show all (odd) primes dividing the RHS are $1 \pmod{8}$. Now, looking modulo $16$ we deduce either $x \equiv 1 \pmod{16}$ and $y$ is odd or $x,y$ are both even. F... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Calculating prime numbers I have a program which determines if a number is prime or not. The basic algorithm simply checks for the number being divisible by 1 and itself.
My question is, is there an upper limit to checking numbers for divisibility without a remainder? If yes, can you explain why?
My "gut" feeling... | I will try to explain by using an example.
Let's check if $37$ is a prime number.
*
*It's obviously not divisible by $2$,
*it's not divisible by $3$ (because $3+7=10$ which is not divisible by $3$),
*and it's not divisible by five (because the last digit is not $0$ or $5$).
These are all the possible candidate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/35072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at nex... | HINT $\rm\:\ \ mod\ 9:\ (x+3)^3\equiv\:x^3\ \Rightarrow\ f(n+1)\: \equiv\ f(n)\ $ so, by induction, $\rm\ f(n)\equiv\: f(1)\equiv\: 0 $
E.g. a specific inductive step: $\rm\: mod\ 9\::\ \ 3^3\:\equiv\: 0^3\ \Rightarrow\ 1^3+ 2^3+3^3\ \equiv\ 0^3+1^3+2^3\ \equiv\ 0\:.$
NOTE $\ $ The period $3$ repeating behavior is ill... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Proof of inequality I have problems with proving inequality :
$${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$
where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.
Thanks.
| Consider the polynomial $(x-a)(x-b)(x-c)$. Multiplied out, this is
$$
\begin{eqnarray}
(x-a)(x-b)(x-c)
&=&
x^3-(a+b+c)x^2+(ab+ac+bc)x-abc
\\
&=&
x^3-10x^2+(ab+ac+bc)x-abc\;.
\end{eqnarray}
$$
We also have
$$10^2=(a+b+c)^2=(a^2+b^2+c^2) + 2(ab+ac+bc)\;,$$
$$ab+ac+bc=\frac{100-(a^2+b^2+c^2)}{2}\;,$$
and thus
$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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To sum $1+2+3+\cdots$ to $-\frac1{12}$ $$\sum_{n=1}^\infty\frac1{n^s}$$
only converges to $\zeta(s)$ if $\text{Re}(s)>1$.
Why should analytically continuing to $\zeta(-1)$ give the right answer?
| Today I stumbled on a totally elementary derivation of this that does not rely on regulators or any reference to the Riemann Zeta Function. We start by looking at the geometric series
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ... $$
Now instead of computing the taylor series of this function we can to compute its $\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "449",
"answer_count": 18,
"answer_id": 16
} |
Understanding algebraic manipulation If $a+b+c \neq 0 $ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of:
$$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$
What confusing me here, is the not so obvious hint which is given with the problem,which says that tha... | Or just add and subtract $a^2$, $b^2$, $c^2$ to the numerators of the three fractions respectively.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sqrt[x]{y} = y^{1 / x}$? Need help with algebra of exponentiation I need help with algebra of exponentiation.
$$\begin{align*}
\sqrt{x^2} &= (x^2)^{1/2} &\qquad &\text{(since }\sqrt[x]{y}=y^{1/x}\text{)}\\
&= x^{2(1/2)} &&\text{(since }(x^y)^z = x^{yz}\text{)}\\
&= x^{(1/2)2} &&\text{(since }xy=yx\text{)}\\
&= ... | The equality
$$
(x^2 )^{1/2} = x^{2(1/2)}
$$
holds only for $x \geq 0$. If $x < 0$, then
$$
(x^2 )^{1/2} = |x| > 0,
$$
but
$$
x^{2(1/2)} = x < 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/51517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a,b,c>0 \text{ s.t. }a+b+c=1 \implies \sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ How can we show that the assumption $a,b,c>0$ and $a+b+c=1$ implies
$$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$
| First we claim that $\sqrt{ab+c} \geq \sqrt{ab}+c$.
Proof of the claim:
$$\begin{eqnarray*}
\sqrt{ab}+c
&=& \sqrt{ab + 2c\sqrt{ab} + c^2}
\\ &\leq& \sqrt{ab + (a+b) c + c^2}
\\ &=& \sqrt{ab + c(a+b+c)} = \sqrt{ab+c}.
\end{eqnarray*}
$$
The only inequality used here is the AM-GM inequality: $\sqrt{ab} \leq \frac... | {
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On calculating the minor determinant of adjoint matrix Question:
$A$ is an $n$ order square matrix, and $B=A^{(*)}$ is the adjoint matrix of $A$ (i.e. $AB=BA=\det A\cdot I_n$).
Show that
$$\det B
\begin{pmatrix}
1 & 2 & \cdots & k \\
1 & 2 & \cdots & k
\end{pmatrix}
=
(\det A)^{k-1}\cdot
\det A
\begin{pmatrix}
... | It is actually more related to this. Maybe it is better to express it by partitioning.
Given an "invertible" square matrix $A$,
$$
A = \left(\begin{array}{c|c}A_1 &A_2\\\hline A_3 &A_4\end{array}\right), A^{-1} = \left(\begin{array}{c|c}\hat A_1 &\hat A_2\\\hline \hat A_3 &\hat A_4\end{array}\right)
$$
and by steal... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $N= (323232 \cdots 50 \text{ digits})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?
If $N= (\underbrace{323232 \cdots}_{50 \text{ digits}})_9$ (i.e in base $9$) then how to
find the remainder when this $N$ is divided by $8$?
I am looking for a "fast" approach that could be... | The same way we find the remainder of dividing a number by $9$ when the number is written in base 10: add the digits.
Why?
Because remember that writing, say, $(38571)_9$ "really" means
$$1 + 7\times 9 + 5\times 9^2 + 8\times 9^3 + 3\times 9^4.$$
When you divide $9$ by $8$, the remainder is $1$; when you divide $9^2$ b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also l... | We begin by writing $k^3$ in a more clever fashion: $k^3 = k(k-1)(k-2) + 3k^2 - 2k$ :
$$\sum_{k=0}^n k^3 = \sum_{k=0}^n k(k-1)(k-2) + 3k^2 -2k$$
Distributing the summation and adjusting our indices we obtain: $$ \sum_{k=3}^n k(k-1)(k-2) + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$
Notice, $$k(k-1)(k-2) = \frac {k!}{(k-... | {
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"url": "https://math.stackexchange.com/questions/61482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "184",
"answer_count": 28,
"answer_id": 10
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Proving a series for the Watson Triple Integrals? It's long known that the first Watson triple integral evaluates to,
$$I_1 = \frac{1}{\pi^3} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dx \, dy \, dz}{1-\cos x \cos y \cos z} = \frac{\Gamma^4\left(\tfrac{1}{4}\right)}{4\pi^3} = 1.3932039\dots$$
Apparently, it i... | The sum
$$
\frac{4}{3} \sum_{n \ge 0} \frac{(4 n)!}{(n!)^4} z^4 = \frac{4}{3} \sum_{n \ge 0} \frac{ (\frac{1}{4})_n (\frac{3}{4})_n (\frac{1}{2})_n }{ (1)_n (1)_n} \frac{(256 z)^n}{n!} = \frac{4}{3} {}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| 256 z \right... | {
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"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| We can try by induction, that is
*
*base case: $1^3=1^2$
*induction step: assume $1^3 + 2^3 +...+ n^3 = (1 + 2 +...+ n)^2=\frac{n^2(n+1)^2}{4}$ then
$$1^3 + 2^3 +...+ n^3+(n+1)^3=\frac{n^2(n+1)^2}{4}+(n+1)^3=\\=(n+1)^2\left(\frac{n^2}{4}+(n+1)\right)=(n+1)^2\frac{n^2+4n+4}{4}=\frac{(n+1)^2(n+2)^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/62171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
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The square of an integer has form $3n$ or $3n+1$ Prove that if $n\in\mathbb Z$, then $n^2$ is of the form $3q$ or $3q+1$ for some $q\in\mathbb Z$
I would like to show that 3q+2 is = 3q+1 thus $n^2$ can be of the form of 3q or 3q+1.
Case one
$(3k)^2=(3k)(3k)=9k^2=3(3k^2)$ and is still of the form $3q$
When $q=3k^2$
$3q... | When working by cases, you need to go through all three of them. So you should square $3q$ and find $(3q)^2=9q^2=3k$ for $k=3q$, then square $3q+1$ and $3q+2$ and find they equal $3k+1$ for some $k$. Under case two, you need an equal sign between the two lines. You skipped squaring $3q+2$. You also need parentheses... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Properties of Fermat primes Fermat primes 17 and 257 appear a lot in the prime composition of numbers of the form $a^{2^n}+1$. For example, $11^8+1$ is divisible by 17 and $11^{32}+1$ is divisible by 257.
I have verified the following statement for 3, 5, 17 and 257 but can't prove it. I would be grateful if someone cou... | This sort of rephrases Henning's answer. (That one is bottom up, this one is top down, which I like more personally)
Fermat's little theorem says that $x^{2^{2^k}}-1$ has $2^{2^k}$ solution mod $p$. Now $x^{2^{2^k}}-1 = \left (x^{2^{2^{k-1}}} - 1 \right)\left(x^{2^{2^{k-1}}} + 1 \right)$, each factor having at most $2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/64529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Trouble with elementary number theory I'm trying to solve the following two problems:
Show $n^{12}\equiv 1\pmod{72}$ when $(72,n)=1$
and
Compute ${7^8}^9\pmod{100}$
For the first, I saw that $n^{24}=n^{\varphi(72)}\equiv1\pmod{72}$ by Euler's theorem, but this leaves us with $n^{12}n^{12}\equiv 1\pmod{72}$. This co... | $(1)$ Using Carmichael function, $$\lambda(72)=\text{lcm}(\lambda(8),\lambda(9))=\text{lcm}(2,6)=6$$
$$\implies n^6\equiv1\pmod{72}\text{ if }(n,72)=1 $$
$(2)$ Observe that $\displaystyle7^2=49=50-1\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod{100}$
$\displaystyle\implies7^n\equiv1\pmod{100}$ if $4|n$
Now, $8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/71601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Matrix multiplication: is C(AB) the same as (CA)B? I would like to show that $(\mathbf{A} \mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}$, where $\mathbf{A}$ and $\mathbf{B}$ are $N \times N$ square matrices.
I think that this can be done as follows:
First, note that $(\mathbf{A}\mathbf{B})^{-1} (\mathbf{A} \mathbf... | As mentioned in the comments on my question, it is possible to demonstrate to yourself the property of associativity by switching to the index representation of matrices. I did this to verify to myself what was said above.
To show that $\mathbf{C}(\mathbf{A}\mathbf{B}) = (\mathbf{C} \mathbf{A}) \mathbf{B}$, note that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/71674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Beta Function -- finding a lower bound based on parameters I would like to show that
$$ 1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq \frac{1}{c+1}.$$
for all $c \geq 2$.
I have plotted it out for $c$ up through 200, and it seems to hold.
Does anyone have any tips on if this can be formally shown? I imagine an e... | The definition of the beta function is
$$Beta\left(c+1,\frac{1}{c}\right) = \int_0^1 t^c (1-t)^{1/c-1} dt.$$
Since $c \geq 2$ and $t \in [0,1]$, $t^c \leq t$. Thus (applying integration by parts to the first integral), we have
$$Beta\left(c+1,\frac{1}{c}\right) \leq \int_0^1 t (1-t)^{1/c-1} dt = \left. -c t (1-t)^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Number of solutions - please check my solution?
Determine, in accordance to $k$, how many solutions does the given
system of equations have:
$$ \begin{cases}kx+(k+1)y=k-1\\4x+(k+4)y=k\end{cases} $$
And check, for which values of $k$ this system has exactly one solution lying within the boundaries of a triang... | First, let's get back to the mainstream terminology: We have a linear system $Ax=b$ with the data
$$
\pmatrix{k &k+1\\4 &k+4}\pmatrix{x\\y}=\pmatrix{k-1\\k}
$$
Now if $A$ is invertible then there exists a unique solution which is given by
$$
\pmatrix{x\\y} = \pmatrix{k &k+1\\4 &k+4}^{-1}\pmatrix{k-1\\k} = \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
$n! \leq \left( \frac{n+1}{2} \right)^n$ via induction I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.
This is where I am stuck:
$$\left( \frac{n+2}{2} \right)^{n+1}
\geq \dots \geq
=2 \left( \frac{n+1}{2} \right)^{n+1}
= \left( \frac{n+1}{2} \right)^n(n+1)
\geq n!(n+1)
= (n+1)! $$
I appro... | Assuming $n! \le \left( \frac{n+1}{2} \right)^n$ is true, carry the induction step
$$
(n+1) n!\leq (n+1) \left(\frac{n+1}{2}\right)^n =2
\left(\frac{n+1}{2}\right)^{n+1} \stackrel{?}{\leq} \left(\frac{n+2}{2}\right)^{n+1}
$$
But the last inequality is just
$$
2 \le \left( \frac{n+2}{n+1} \right)^{n+1} = \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/76130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
Which base of numerical system have $\frac 15 = 0.33333\ldots$? Which base of numerical system have $\frac{1}{5} = 0.33333\ldots$?
I need assistance in solving this one.
| If we are working in base $b$ (we must have $b\gt3$), then $0.3333\ldots$ is
$$0.3333\ldots = \frac{3}{b} + \frac{3}{b^2} + \frac{3}{b^3}+\cdots$$
Since
$$\sum_{n=1}^{\infty}\frac{3}{b^n} = \frac{3}{b}\sum_{n=0}^{\infty}\frac{1}{b^n} = \frac{3}{b}\left(\frac{1}{1-\frac{1}{b}}\right) =\frac{3}{b-1},$$
then...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Midpoint-Convex and Continuous Implies Convex Given that $$f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,$$
how can I show that $f$ is convex.
Thanks.
Edit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.
|
I would like to properly show by induction that if $ m\in\{0,1,2,\ldots,2^{n}-1\} $ then
$$
f\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)\le \frac{m}{2^n}f(x)
+\Big(1-\frac{m}{2^n}\Big)f(y).
$$
and then the result will follows by contuinity since $$ m=\lfloor2^nt\rfloor \in\{0,1,2,\ldots,2^{n}\} ~~~and~~\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/83383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 6,
"answer_id": 1
} |
A question about series $$\sum\frac{1}{2^n} = 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$
$$\sum \frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$
Why is the first one convergent and the second divergent? First has $$\lim_ {n\to \infty}\ \frac{1}{2^n}$$ and the second $$\lim_{n\to \infty}\ \frac{1}{n}$$ ... | The fact that the terms go to zero is a necessary condition for convergence, but it is not sufficient. These two series show that it is not enough for the terms to get small in order for the series to converge, you need the partial sums to converge.
Remember that the series is really the limit of the partial sums:
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/83437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.
| The statement has a natural geometric interpretation when $a, b, c, d \geq 0$. A circle $x^2 + y^2 = r$ and a curve $x^3 + y^3 = s$ for $x, y > 0$ are symmetric about the line $x = y$ and if they intersect, they either intersect at a point $(x,x)$ or two symmetric points $(x,y)$ and $(y,x)$ for $x \neq y$. Thus if $a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/84855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Limit of geometric mean of N radii of an ellipse Is this equation correct?
$$\lim_{N \to \infty} \prod_{n=1}^N (a^2\cos^2 (2\pi n/N)+b^2\sin^2(2\pi n/N))^{1/N}=ab$$
If so, why?
| This is not an answer, but simply a verification of the integral in Srivatsan's answer.
Using $\cos^2(y)=\frac{1+\cos(2y)}{2}$ and $\sin^2(y)=\frac{1-\cos(2y)}{2}$,
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y\tag{1}
$$
becomes
$$
\log\left(\frac{a^2+b^2}{2}\right)+\frac{1}{2\pi}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/84987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Count the number of ways of four distinct numbers showing up when six dice are rolled Suppose we roll six fair dice, how many ways can four distinct numbers show up?
| I'll assume that you mean six-sided dice.
There are $\binom64$ ways of choosing the $4$ distinct numbers. They can either appear $3,1,1,1$ times or $2,2,1,1$ times. In the first case, there are $4$ choices of the number appearing thrice and $6\cdot5\cdot4$ choices for the positions. In the second case, there are $6$ ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion? Is there a way to evaluate this limit:
$$\lim_{x \to 0} \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$
without using de l'Hôpital rule and series expansion?
Thank you,
| This is similar to the accepted answer by user Pacciu, but avoids the use of "asymptotics" and $\approx$ symbols as these are totally unnecessary and perhaps can be confusing to a beginner. Rules of limits combined with the following standard limits is sufficient: $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\lim_{x \to 0}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A version of Hardy's inequality involving reciprocals. How can one prove for any sequence of positive numbers $a_n, n\ge1,$ we have
$$\sum_{n=1}^\infty \frac{n}{a_1+a_2+a_3+\cdots+a_n}\le 2\sum_{n=1}^\infty \frac{1}{a_n}$$
Added later:
Apparently, this is a version of Hardy's inequality. The above is the case $p=-1$. ... | There is also the following N.Sedrakyan's proof.
By C-S
$$2\sum_{k=1}^n\frac{1}{a_k}>2\sum_{k=1}^n\frac{k^2}{a_k}\left(\frac{1}{k^2}-\frac{1}{(n+1)^2}\right)=2\sum_{k=1}^n\frac{k^2}{a_k}\sum_{m=k}^n\left(\frac{1}{m^2}-\frac{1}{(m+1)^2}\right)=$$
$$=2\sum_{k=1}^n\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\sum_{m=1}^k\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/98483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?
I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$
So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. I... | Calculate!
$(1+i)^0=1$, so $(1-i)^0=1$, so $(1+i)^0+(1-i)^0=2$.
$(1+i)^1=1+i$, so $(1-i)^1=1-i$, so $(1+i)^1+(1-i)^1=2$ .
$(1+i)^2=2i$, so $(1-i)^2=-2i$, so $(1+i)^2+(1-i)^2=0$.
$(1+i)^3=-2+2i$, so $(1-i)^3=-2-2i$, so $(1+i)^3+(1-i)^3=-4$.
$(1+i)^4=-4$, so $(1-i)^4=-4$, so $(1+i)^4+(1-i)^4=-8$.
Now the game starts a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 2
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Write the equations of the system for which $Ax=b$ is compatible $A =\left[ {\begin{array}{cc}
1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&a \\ -2&1&1&b \\ 1&1&-2&c \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&a \\ 0&-3&3&b+2a \... | What you did in the Update just shows that $A{\bf x}=\bf0$ has one (independent) solution. That is the dimension of the null space of $A$ is 1.
So $A{\bf x}=\bf b$ has two independent solutions. That is, the dimension of the column space of $A$ is 2. You have an echelon form of $A$. What you need to do is find a bas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/100421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Circle locus, how to satisfy the equation.
$A(-3,1), B(0,-5), P(X,Y)$
If $|AP| = 2|BP|$ prove that $x$ and $y$ satisfy the equation:
\begin{aligned}
\ x^2+y^2-2x+14y+30 =0
\end{aligned}
I get as far as determining the co-ordinates like so
\begin{aligned}
\ \sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2}
\end{aligne... | If we start from the equation
$$
\sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2},
$$
and square both sides, we obtain the equivalent equation
$$x^2+6x+9+y^2-2y+1=4x^2+4y^2+40y+100,$$
which simplifies to
$$3x^2+3y^2-6x+42y+90=0,$$
which is equivalent to
$$x^2+y^2-2x+14y+30=0.$$
Comment: Draw the line segment that joins... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/101389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$ Q. Prove
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$$
My try: expand and got:
$$\frac{5^{2x-2}+2(5^{x^2-1})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}$$
Now what? I find my pre-calculus skills esp with Indices, Logarithms & Trigo lacking ...... | Factor as:
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{(5^{x-1})^2(1+5^2)^2}{(5^{x-1})^2(1+25^2)}$$
Then simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/102260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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A question about finding the time difference Two people $A,~B$ are participating a running race.Initially of course they are both at rest.They then proceed with constant acceleration. $A$ covers the last $\dfrac{1}{4}$ of the distance in $3$ second while $B$ covers the last $\dfrac{1}{3}$ of his distance in $4$ second.... | Let $a$ and $b$ be the accelerations of $A$ and $B$, respectively, and let $d$ be the length of the race. Let $x_A(t)$ and $x_B(t)$ be the distances covered by $A$ and $B$, respectively, at time $t$ after the start of the race; $x_A(t)=\frac12at^2$, and $x_B(t)=\frac12bt^2$. $A$ reaches the $3/4$ mark at time $t_1$ giv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Writing a Polar Equation for the Graph of an Implicit Cartesian Equation
If $(x^2+y^2)^3=4x^2y^2,$ then $r=\sin 2\theta$ for some $\theta$.
Using $r^2=x^2+y^2, x=r\cos\theta,y=r\sin\theta$, it's easy to get $r^2=\sin^22\theta$.
But I don't know what to do next, since $r$ could be negative in $r=\sin2\theta.$
Actually... | Avoiding the issue entirely:
$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$
$$r^3 = 2(r\sin\theta)(r\cos\theta)$$
$$x = r\cos\theta$$
$$y = r\sin\theta$$
$$r^3 =2xy$$
$$r = (x^2 + y^2)^{\frac 12}$$
$$(x^2+y^2)^{\frac 32} =2xy$$
$$(x^2+y^2)^3=4x^2y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Simplify _Elementary Calculus_ section 1.6 problem 25 Let's try this again. We're still on problem 25 in section 1.6 of Elementary Calculus.
$$\frac{3-\sqrt{c+2}}{c-7}$$
My first thought is (again) to multiply by $3+\sqrt{c+2}$:
$$=\frac{(3-\sqrt{c+2})(3+\sqrt{c+2})}{(c-7)(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{(c-7)(3+\sq... | Everything is fine, apart from the fact that you don't think it is fine.
First, a small piece of advice. Don't multiply unless you have to. At a certain stage you had
$$\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})},$$
which you expanded (but certainly did not simplify) to
$$=\frac{9-(c+2)}{3c+c\sqrt{c+2}-7\sqrt{c+2}-21}.$$
This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving integral equation with Laplace's Transform. I'm trying to prove the following
$$\int\limits_0^\infty {\frac{{\cos tu}}{{{u^2} + 1}}\log udu} = - \frac{\pi }{2}\int\limits_0^\infty {\frac{{\sin tu}}{{{u^2} + 1}}du} $$
The original problem suggested the use of the Laplace Transform, thus I have
$$\int\limits_... | $$- \frac{\pi }{2}\int\limits_0^\infty {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$
Now by partial fractions you can separate and get:
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$\lim _{(x,y) \to (0,0)}$ $\frac{(x^3-y^2)^2}{x^4+y^4}$- two methods different conclusion. I tried to use couple of inequalities in this way:
$0\leq\left|\frac{(x^3-y^2)^2}{x^4+y^4}\right| \leq \frac{(x^3)^2}{x^4}=x^2$, so now I can use the squeeze theorem and conclude that this limit is 0, but when I choose a route s... | This can be done simply:
Take a path with $x=0$ and $y\rightarrow0$; then ${(x^3-y^2)^2\over x^4+y^4}={y^4\over y^4}=1$.
Take a path with $y=0$ and $x\rightarrow0$; then ${(x^3-y^2)^2\over x^4+y^4}={x^6\over x^4}=x^2\rightarrow0$.
So $ \lim\limits_{(x,y)\rightarrow(0,0)} { (x^3-y^2)^2 \over x^4+y^4}$ does not exist.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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reducing $x^4+1$ in $\mathbb{Z}_p[x]$
Possible Duplicate:
reducible polynomial modulo every prime
Ok so we have to prove the following: If $R=\mathbb{Z}_p$ for $p$ a prime, then, $x^4+1$ is reducible over $R[x]$. This is are my ideas (please let me know if there is an easier way to tackle this problem):
First if $p=... | If $p \equiv 1 \pmod{4}$, we have the factorization $$x^4 + 1 = (x^2 + i)(x^2 - i),$$ where $i \in \mathbb{F}_p$ is the element such that $i^2 = -1$.
When $p \equiv 3 \pmod{4}$, if $2$ is a square $\bmod{p}$ (say $a^2 = 2$), then
$$
x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1).
$$
If $2$ is not a square, then $-2$ is a squ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/106564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Area of a circle I've tried to find as a personnal exercise where the formula $A=\pi R^2$ comes from.
After drawing the problem, I've found that $A = 2\int\limits_{-R}^{R}\sqrt{R^2-t^2}dt$. How can I calculate this ?
| This is a classic case of trigonometric substitution. Set $t=R\sin\theta$, $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$.
Then $R^2 - t^2 = R^2(1-\sin^2\theta) = R^2\cos^2\theta$, hence $\sqrt{R^2-t^2} = \sqrt{R^2\cos^2\theta} = |R\cos\theta| = R\cos\theta$, because $R\geq 0$ and $\cos\theta\geq 0$ for $\theta\in[-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the value of a? Let $a$ and $b$ be two positive numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
If $4G = 5H$, what is the value of $a$?
| The Harmonic Mean of $a$ and $b$ is
$$\frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
The Geometric Mean of $a$ and $b$ is
$$\sqrt{ab}.$$
So, to state the problem you have in a way that would be actually intelligible would be:
Let $a$ and $b$ be positive numbers such that $a\gt b$; assume that
$$4\times\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3... | You can use a similar 'trick' to find the derivative of $ y=\sqrt[n] x $. The limit will involve multiplying the numerator $(x+h)^{1/n} - x^{1/n} $ by an appropriate expression to get $ (x + h) - x $ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Find the value of $A/B$? $$A = x^4 ( 1 + y^4 ) + y^4 ( 1 + z^4 ) + z^4 ( 1 + x^4 )$$
$$B = x^2 y^2 z^2$$
If $x$, $y$ and $z$ are real, which of the following can be the value of $A/B$?
Options
i) 0
ii) 2
iii) 4
iv) 8
| Hint:
Expand $A$ to be a sum of $6$ terms and use $\text{AM} \ge \text{GM}$.
For the sake of completeness, we add more detail.
Since we talk of the ratio $\frac{A}{B}$, we can assume that $B \neq 0$, and thus none of $x,y,z$ are non zero.
Now $A = x^4 + x^4y^4 + y^4 +y^4z^4 + z^4 + z^4x^4 \ge 6 x^2y^2z^2 = 6B$, using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $n^2+(n+k)^2+(n+2k)^2+ \ldots +(n+mk)^2$ have a general equation? Does $n^2+(n+k)^2+(n+2k)^2+\ldots+(n+mk)^2$ have a general formula?
e.g.
$$1^2+2^2+3^2+\ldots+n^2=n(n+1)(2n+1)/6$$
| The sum can we written as $\sum_{j=0}^m(n+jk)^2=\sum_{j=0}^mn^2+2jkn+j^2k^2$ and using the formulas $\sum_{j=0}^mj=\frac{m(m+1)}2$, $\sum_{j=0}^mj^2=\frac{m(m+1)(2m+1)}6$, we get
\begin{align*}\sum_{j=0}^m(n+jk)^2&=(m+1)n^2+knm(m+1)+k^2\frac{m(m+1)(2m+1)}6\\
&=(m+1)\left(n^2+knm+\frac{k^2m(2m+1)}6\right).
\end{align*}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find the sine and cosine from the tangent? I'm given the problem:
If $\cot(\theta) = 1.5$ and $\theta$ is in quadrant 3, what is the value of $\sin(\theta)$?
I looked at all the related answers I could find on here, but I haven't been able to piece together the answer I need from them.
I know that $\sin^2... | Indeed, we know that
$$1.5 = \cot\theta = \frac{\cos\theta}{\sin\theta}$$
hence
$$1.5\sin\theta = \cos\theta.$$
Squaring both sides we have
$$2.25\sin^2\theta = \cos^2\theta$$
and since $\cos^2\theta = 1-\sin^2\theta$ we have
$$\begin{align*}
2.25\sin^2\theta &= 1-\sin^2\theta\\
2.25\sin^2\theta + \sin^2\theta &= 1\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$ How can I evaluate the integral
$$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$
I tried manipulating the known integral
$$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$
but couldn't do anything with it.
| You can use double integration:
$$\int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{{1 + x}}dx} = \int\limits_0^1 {\int\limits_0^{ - x} {\frac{{du \cdot dx}}{{\left( {1 + u} \right)\left( {1 + x} \right)}}} } $$
$$\int\limits_0^1 {\int\limits_0^x {\frac{{dm \cdot dx}}{{\left( {m - 1} \right)\left( {1 + x} \right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?
Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$
Side work:
$\sin{\theta} = \frac{3}{x}$
$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$
$dx = ... | The derivative of $\csc \theta$ is not $\csc \theta \tan \theta$ it is $-\csc \theta \cot \theta$.
After your edit:
Try writing $\sin^{-1} (3/x)$ in the form of $\tan^{-1}$.
Also you have the integral of $\cos 2\theta$ wrong.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove: $\int_0^{\frac{\pi}{2}}t(\frac{\sin nt}{\sin t})^4dt<\frac{\pi^2n^2}{4}$ I have a question about integral. Prove:
$$\int_0^{\frac{\pi}{2}}t\left(\dfrac{\sin(nt)}{\sin(t)}\right)^4dt<\dfrac{\pi^2n^2}{4}$$
I have tried several methods including $\sin(t)\geq\frac{2t}{\pi}$, but I can't work it out.
| First note that
$$I:=\int_0^{\frac{\pi}{2}}x\left(\frac{\sin nx}{\sin x}\right)^4\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{x}{\sin x}\frac{|\sin nx|}{\sin x}|\sin nx|\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x.$$
We can obtain the following inequalities with little effort :
$$1<\frac{x}{\sin x}<\frac{\pi}{2}(0<x<\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Is there a way to create a summation function for this? I have an algorithm that generates terms as follows:
$$\begin{align*}
\text{Term 1}: &2\times 3\times\left(\frac{1}{2}+\frac{1}{3}\right) + 4\\
\text{Term 2}: &2\times 3\times 4\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + 3\times 5\\
\text{Term 3}: ... | Term $k$ is $$(k+2)!(H_{k+2}-1)+\frac12(k+1)!(k+3)\;,$$ where $H_k$ is as usual the $k$-th harmonic number, $\sum_{i=1}^k\frac1k$, so the sum of the first $n$ terms is $$\sum_{k=1}^n\Big((k+2)!(H_{k+2}-1)+\frac12(k+1)!(k+3)\Big)\;.$$ You may of course replace $\frac12(k+1)!(k+3)$ by $\dfrac{(k+3)!}{2(k+2)}$ if you wish... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$? I have tried something to solve the series
$$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$
My approach is :
$$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^... | This kind of looks like you want to appeal to Vandermonde's convolution, or at least the method you'd use to prove it. It can be applied directly as follows:
Let $S = \sum\limits_{k=0}^n k\binom{n}{k}^2$ be the sum we want to compute. Note that $S = \sum\limits_{k=0}^n (n-k) \binom{n}{n-k}^2 = \sum\limits_{k=0}^n (n-k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/122147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
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If $f(x) = \frac{4^x}{4^x+2},$ find the value of $\sum \limits_{i=1}^{1999} f\left(\frac{i}{1999}\right) $
If
$$f(x) = \frac{4^x}{4^x+2} $$
then find the value of
$$f\left(\frac{1}{1999}\right) + f\left(\frac{2}{1999}\right) + f\left(\frac{3}{1999}\right) +\cdots+f\left(\frac{1999}{1999}\right).$$
I tried it b... | Hint:
Observe
$$f(x)+f(1-x)=1$$
So, $$\begin{align}&f\left(\dfrac 1 {1999}\right)+f\left(\dfrac 2 {1999}\right)+ \cdots+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac{1999}{1999}\right)\\&=f\left(\dfrac 1 {1999}\right)+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac 2 {1999}\right)+f\left(\dfrac {1997} {1999}\right)+\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Let $a,b \in {\mathbb{Z_+}}$ such that $a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$, Prove $a=b$ Let $a,b \in {\mathbb{Z_+}}$ such that $$a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$$
Prove $a=b$
| consider the prime factorisations of $a = \prod_p p^{\nu_p(a)}$ and $b = \prod_p p^{\nu_p(b)}$. Your assumptions yield
*
*$\nu_p(a) \le \nu_p(b^2) = 2\nu_p(b)$ for each $p$
*$3\nu_p(b) \le 4\nu_p(a)$
*$\ldots$
*$(4n+1)\nu_p(a) \le (4n+2)\nu_p(b)$ for each $p$, $n$
*$(4n+3)\nu_p(b) \le (4n + 4)\nu_p(a)$, each $p$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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How does $(x_n - \sqrt{2})^2/(2x_n ) \geq 0$ mean $x_n \geq \sqrt{2}$ when $x_1=1$? I don't understand one step of proving of the Babylonian Method for $a=2,x_1=1$, and
$$x_{n+1} =\frac{1}{2}( x_n + \frac{a}{x_n}). \quad (n=1,2,\ldots)$$
$$x_{n+1} - \sqrt{2} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - \sqrt{2} = ... | There seems to be confusion on what you think they claim, and what they actually claim. The step you are talking about is as follows.
Conclude that $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.
What is claimed here is that if $x_n \geq \sqrt{2}$, then $x_{n+1} \geq \sqrt{2}$. We know this holds, since, as sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the value of : $\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$
Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
$$\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$$
So, I have an intermediate form of $\infty - \infty$ and I tried multipl... | Multiplying and dividing by $\sqrt{4x^2 + 4} + (2x + 2)$ you obtain:
$$\begin{split}
\sqrt{4x^2 + 4} - (2x + 2) &= \frac{(\sqrt{4x^2 + 4} - (2x + 2))(\sqrt{4x^2 + 4} + (2x + 2))}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{(4x^2 + 4) - (2x+2)^2}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{-8x}{\sqrt{4x^2 + 4} + (2x + 2)} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Different approaches to evaluate this determinant How to evaluate this determinant $$\det\begin{bmatrix}
a& b&b &\cdots&b\\ c &d &0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots& 0\\c&0&\cdots&0&d
\end{bmatrix}?$$
I am looking for the different approaches.
| Your (upper) arrowhead matrix can be decomposed as follows:
$$\begin{pmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{pmatrix}=\color{red}{\begin{pmatrix}a-b-c&&&&\\&d&&&\\&&d&&\\&&&\ddots&\\&&&&d\end{pmatrix}}+\color{blue}{\begin{pmatrix}1&c\\&c\\&c\\&\vdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How to derive a differential equation of an ellipse I am quite new to differential equations and derivatives. I want to derive an differential form for equation of an ellipse. If i start with an ordinary ellipse equation
\begin{equation}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\end{equation}
How do i derive it then to get th... | The equation $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag{1}$$
has two variables: $\{ x, y \}.$ By "derive," it seems that you mean
$ \frac{dx}{dy}.$
Well, differentiating equation $(1)$ w.r.t $y,$ we get:
$$
\frac{d}{dy} \frac{x^2}{a^2} + \frac{d}{dy} \frac{y^2}{b^2} = \frac{d}{dy} 1
\tag{2}$$
First, note that $\dfrac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Showing $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$ How do I show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$?
The answer is to:
Let $A = x+y+z$, then $z=A-x-y$. Maximize $f(x,y)=xy(A-x-y)$ on region enclosed by $x=y=0$ and $x+y=A$
What I don't understand is the idea behind this method.
... | For your first question, the answer is that it doesn't matter. We could just as well let $A=\frac{x+y+z}{3}$. Then $z=3A-x-y$, and so on. The calculations end up very similar. Indeed things are a little prettier, I think, if we let $\frac{x+y+z}{3}=A$. But it makes no real difference.
The point is that we fix $x+y+z$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/128983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Solving $\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}}$ How to get from
$$\lim_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = \lim_{n\to\infty} \frac{\frac{\frac{2}{n}}{\tan{\frac{2}{n}}}\cdot\fra... | $$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = 2}$$
There should be a typo in the book. ( the second term in the bottom should have $n^3$ )
$$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\c... | {
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"url": "https://math.stackexchange.com/questions/133814",
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"source": "stackexchange",
"question_score": "1",
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How to find the rational form of a $2\times 2$ rotation matrix Given the matrix
$$A = \begin{pmatrix}
\cos x & \sin x\\ -\sin x & \ \cos x \end{pmatrix}$$
I have to express it in rational form
The characteristic polynomial is $y^2 - 2y\cos x+1 $
Accordingly, the rational form $R$ if $x$ is neither $0$ or $2 \p... | The characteristic polynomial of
$$\left(\begin{array}{rr}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{array}\right)$$
is, as you note, $t^2 -2(\cos\theta)t + 1$.
The discriminant of this polynomial is
$$4(\cos^2\theta) - 4 = 4(\cos^2\theta - 1).$$
Therefore, the discriminant is always nonpositive, and is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$
Evaluate the integral: $$ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$$
The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.
| Observing the inverse of the function $f(x) = \sqrt[3]{1 - x^7}$ on the interval $[0,1]$ is $f^{-1} (x) = \sqrt[7]{1 - x^3}$, using the result
$$\int^b_a f(x) \, dx + \int^{f(b)}_{f(a)} f^{-1} (x) \, dx = b f(b) - a f(a),$$
since $f(a) = f(0) = 1$ and $f(b) = f(1) = 0$ it immediately follows that
$$\int^1_0 \sqrt[3]{1 ... | {
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"url": "https://math.stackexchange.com/questions/139393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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Finding derivative of $\sqrt{9-x}$ I am trying to find the derivative of $\sqrt{9-x}$ using the definition of a derivative
$$\lim_{h\to 0} \frac {f(a+h)-f(a)}{h} $$
$$\lim_{h\to 0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h} $$
So to simplify I multiply by the conjugate
$$\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdo... | Everything you have done is right except for the last step.
$$\begin{align}
&\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}=\\
&\lim_{h\to0} \frac{9-(a+h)-(9-a)}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{9-a-h-9+a}{h(\sqrt{9-(a+h)}+\sqrt{9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Derivative of $\ln(x\sqrt{x^2-1})$ I am trying to find the derivative of $\ln(x\sqrt{x^2-1})$ but I can not get what the book gets.
I get $$\frac{1}{x \sqrt{x^2-1}} \cdot \sqrt{x^2-1} + x\cdot\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x$$
which I reduce to
$$\begin{align}
&\frac{1}{x\sqrt{x^2-1}}\sqrt{x^2-1} + x^2(x^2-1)^\fr... | You obtained the correct derivative, but you need parentheses as such:
$$\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x \frac{1}{2}(x^2-1)^\frac{-1}{2}2x\Bigr)$$
Clean this up a bit to get
$$\tag{1}
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x^2 (x^2-1)^\frac{-1}{2} \Bigr).
$$
You were ok up to this point. The r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/143550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$ Let $0 <a < 1$ and let $b,c \in \mathbb{N}$, evaluate $$\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$$
How to evaluate in terms of $a$,$b$ and $c$?
| Since the question only cares about natural numbers of $b$ and $c$ , it should be no problem about any binomial series expansions.
$\int_0^a\dfrac{x^b+x^c}{(1+x)^{b+c+2}}dx$
$=\int_0^a\dfrac{x^b}{(1+x)^{b+c+2}}dx+\int_0^a\dfrac{x^c}{(1+x)^{b+c+2}}dx$
$=\int_0^a\dfrac{(x+1-1)^b}{(x+1)^{b+c+2}}dx+\int_0^a\dfrac{(x+1-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/144982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Splitting polynomials I have a polynomial ${\frac{{{{({z^2})}^p} \pm {p^p}}}{{{z^2} \pm p}}}$ where $p$ is an odd prime number, and I know it splits into two factors $$ \sum_{i = 0}^{p - 1} a_i z^i \text{ and } \sum_{i = 0}^{p - 1} ( - 1)^i a_i z^i $$
For example, when $p=5$
$$
\begin{eqnarray*}
\frac{x^{10}-5^5}{x^2-... | Here is the solution I found, which to me is unsatisfactory.
This solution uses Table 24 from Riesel’s “Prime Numbers and Computer Methods for Factorization”. Look up the number you are interested in in the n column, and note down the two sets of coefficients given there: they are $U_n(x)$ and $V_n(x)$. Multiply each c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How did they get this result? Please, explain these computations:
1) $-\left(\frac{1}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{3}}{2} = \cos x$
How did we get $\frac{\sqrt{3}}{2}$ from $-\left(\frac{1}{2}\right)^2 +1$?
2) $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{2}}{2} = \cos x$
How did we g... | Here's another way:
$$ \cos^2(x) = 1 - (\frac{1}{2})^2 $$ and
$$ \cos^2(x) = 1 - \sin^2(x) $$ so immediately we have
$$ \sin(x) = \pm\frac{1}{2} $$ then, since $\sin(x) = \frac{opp}{hyp}$ we have from the reference triangle, $$ \cos(x) = \frac{adj}{hyp} = \pm\frac{\sqrt{3}}{2} $$ and also
$$ \tan(x) = \frac{opp}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to solve an equation involving the floor of a number? I'm looking for a solving procedure for this type of exercises.
If $[x]$ represents the floor of $x$, solve the equation:
$$\left[\frac{6x+5}8\right]=\frac{15x-7}5$$
Choose the correct answer:
a) {$\frac{4}5$}
b) {$\frac{3}4$}
c) {$\frac{7}{15}$,$\frac{4}5$}
d... | Let $\mathbb{Z} \ni k = \lfloor \frac{6 x+ 5}{8} \rfloor$. Then the equation says $x = \frac{k}{3} + \frac{7}{15}$. Now we write the equation for $k$:
$$
k = \left\lfloor \frac{k}{4} + \frac{39}{40} \right\rfloor
$$
It now remains to test few small values of $k$. Values $k=0$ and $k=1$ work out.
These correspond ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/149237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Taylor expansion around infinity of a fraction I want to Taylor expand the following function:
$f(x)=\frac{1}{a-x^2}$ when $x \rightarrow \infty$.
I know the result (from Wolfram Alpha) to be $-\frac{1}{x^2} - \frac{a}{x^4} + O(\frac{1}{x^6})$
but I have no idea how to calculate it. Any help is greatly appreciated!
| One way is as follows. Set $x= \dfrac1y$ and expand it around $y=0$. $$g(y) = \dfrac{1}{a - \dfrac1{y^2}} = \dfrac{y^2}{ay^2-1} = - y^2 \dfrac1{1-ay^2} = -y^2 \left( 1 + ay^2 +a^2y^4 + a^3 y^6 + \cdots\right)$$
Now replace $y$ by $1/x$ to get that
$$f(x) = -\dfrac1{x^2} \left( 1 + \dfrac{a}{x^2} + \dfrac{a^2}{x^4} + \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/152570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Multiple roots of a polynomial over a field of characteristic $p$ I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$
have a root of multiplicity $>1$ over the field of characteristic $p$.
$ p=2, 3, 5, 7 $
Please help.
For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ ... | The discriminant of $f(x)$ is $55269 = (3^3)(23)(89)$. So the characteristics in which $f(x)$ has a multiple root are $3$, $23$ and $89$. In fact, $f(x) = x(x+1)^3$ in characteristic $3$, $(x^2+7x+8)(x+8)^2$ in characteristic $23$ and $(x+8)^2(x+28)(x+45)$ in characteristic $89$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.