Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt... | Here a substitution can help out with calculations. Let $ x = \sin t$, i.e. $t = \arcsin x$. First note that $x\in \langle -1,1 \rangle$, so $t\in \langle -\pi/2,\pi/2 \rangle$ and $\cos t>0$. This is important because we can then simplify $\sqrt{\cos^2t} = |\cos t| = \cos t$, which is used couple of times from now on.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $0 < a \le b$. Then $a^2 \le \left (\frac{2ab}{a+b} \right) ^2 \le ab \le \left( \frac{a+b}{2} \right) ^2 \le b^2$ In a introductory text about real analysis, in a chapter about the axioms of order (I hope this is the correct translation), I was faced with the following
Problem: Let $0 < a \le b$. Show that $a^2 \l... | It look quite correct for me, and the equality part is good since $a,b$ are both positive.
Just for showing a shorter (maybe) version for this by using two important but simple inequalies:
(1). $(a+b)^2 \geq 4ab$
And its variation in case of $a>0,b>0$ :
(2). $ 1 \geq \frac{4ab}{(a+b)^2}$
We have:
$$
b^2 \geq (\frac{a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4614921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how to solve $|z|^2 + z|z| + 2 = 2i Im(z)$? this is the process I follow:
*
*$z|z| = 2iImz -|z|^2- 2 $
*$z \sqrt{Re(z)^2 + Im(z)^2} = 2iy - (Re(z)^2 + Im(z)^2) - 2$
*$z\sqrt{x^2 + y^2} =2iy -(x^2+y^2)-2$
*$z=(2iy -(x^2+y^2)-2)/\sqrt{x^2 + y^2}$
then i get: $Im(z)=2y/\sqrt{x^2 + y^2}$ and $Re(z)=(-(x^2+y^2)-2)/\sq... | You may simplify the notation from the beginning. Call $z=a+bi$ and your equation becomes:
$$a^2+b^2+(a+bi)\sqrt{a^2+b^2}+2=2bi$$ or
$$[a^2+b^2+a\sqrt{a^2+b^2}+2]+[bi\sqrt{a^2+b^2}]=2bi$$
So, you can split it into the real part and the imaginary part.
$$a^2+b^2+a\sqrt{a^2+b^2}+2=0\quad (1)$$
$$b\sqrt{a^2+b^2}=2b \quad ... | {
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"url": "https://math.stackexchange.com/questions/4617200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$. Suppose $a$ and $b$ are elements of the group $G$. If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$.
I'm trying to prove it as follow.
Here's my solution:
If $a^2=e$ so $a=a^{-1}$. So we have $b^3=ab^2a$.
$$b^2=ab^3a$$
So I thought if I can show that $b^3b^2=b^2 b^3=e$ I ca... | If $a^2 = e$ and $a^{-1} b^2 a = b^3$, we can substitute $a^2 = e$ into the second equation to get $$a^{-1} b^2 a = b^3 = (a^2)^{-1} b^2 a = e^{-1} b^2 a = b^2 a$$
Then, multiplying both sides of this equation by $b^{-2}$, we get $$b^{-2} b^2 a = b^{-2} (b^2 a) = a$$
Now, we can substitute this result, $a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$ Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The given equation is equivalent to $$2^{2x}+\dfrac12\cdot2^{2x}=3^x\sqrt3+\dfrac{3^x}{\sqrt3}$$ which is $$\dfrac32\cdot2^{2x}=3^x\left(\sqrt3+\dfrac{1}{\sqrt3}\right)$$ The last equation can be written ... | Alternative approach:
*
*Let $r = \log_2(3).$
*Note that $2^a = 2^b \iff a = b.$
Solve the equation $$2^{2x}+2^{2x-1}=3^{x+0.5}+3^{x-0.5}$$
The LHS can be re-written as
$$2^{2x} \times \left[1 + \frac{1}{2}\right] = 2^{2x} \times \frac{3}{2}.$$
The RHS can be re-written as
$$3^{x + 0.5} \times \left[1 + \frac{1}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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polynomial approximation of a circle First of all, I am not a mathematician and my mathematics are fairly rusty. I would appreciate some help
I believe there is not polynomial equation of a circle (is that right?) but take a look at this picture
In it we can see two equations. One a circle $30x^2+30y^2=300000$ and the... | Expanding upon my comment about osculating circles ...
Suppose we have the second-degree polynomial function
$$y=ax^2+bx+c \tag1$$
that we want to be "parallel" to an origin-centered circle of radius $r$ near a point $R$ on that circle; let's say that the distance between the curves at $R$ should be $s$, and write $S$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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}... | By Cauchy and AM-GM inequality,
$$\sum\frac{a^3}{b+c+d}\ge\frac{\left(\sum a^2\right)^2}{2\sum ab}\ge\frac{\left(\sum a^2\right)^2}{3\sum a^2}=\frac{\sum a^2}3\ge\frac13.$$
The last "$\ge$" is equivalent to $\sum a^2\ge1$. By rearrangement inequality
$$a^2+b^2+c^2+d^2\ge ab+bc+cd+da=1.$$We're done.
| {
"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": 0
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Find the volume of the following region $E= (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) $
Find the volume of the following region $E= \{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 ≤1, \sqrt{2}(x^2 + y^2) ≤z≤ \sqrt{6}(x^2 + y^2) \} $
I figured that the region E is formed... | To expand on my comment, we first observe that the upper paraboloid $z=\sqrt6\,(x^2+y^2)$ meets the sphere $x^2+y^2+z^2=1$ in a cylinder with radius $\frac1{\sqrt3}$ :
$$\begin{cases}z = \sqrt6\,(x^2+y^2) \\ x^2 + y^2 + z^2 = 1 \end{cases} \implies \frac{z}{\sqrt6} + z^2 = 1 \implies z = \sqrt{\frac23} \\
\implies x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How fast does the sequence $a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx$ decays as $n \to \infty$? The question is as in the title.
The sequence $\{a_n \}$ define by
\begin{equation}
a_n := \int_{1/8}^{1/2} \Bigl(\frac{1}{2}+x^2\Bigr)^{1/2+n} dx
\end{equation}
decays to $0$ as $n \to \infty$, which i... | With the substitution $t = \log \left( {\frac{3}{4}} \right) - \log \left( {x^2 + \frac{1}{2}} \right)$, your integral becomes
$$
a_n = \left( {\frac{3}{4}} \right)^{n + 3/2} \int_0^{\log (16/11)} {{\rm e}^{ - nt} \frac{{{\rm e}^{ - t} }}{{\sqrt {3 - 2{\rm e}^t } }}{\rm d}t} .
$$
Now
$$
\frac{{{\rm e}^{ - t} }}{{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Taylor Series of $ f(z) = \frac{z^2-2z+5}{(z-2)(z^2+1)}$ in $1<|z|<2$. I have to find Taylor series of
$f(z)=\frac{z^2-2z+5}{(z-2)(z^2+1)}$, for $1<|z|<2$.
I have started with
\begin{eqnarray*}
f(z)&=&\textstyle\frac{z^2 -2z +5}{(z-2)(z^2+1)}\\[4pt]
&=&\textstyle\frac{z^2+1-2(z-2)}{(z-2)(z^2+1)}\\[4pt]
&=&\text... | First decompose your expression into partial fractions. $$\frac{z^2-2z+5}{(z-2)(z^2+1)}=\frac{A}{z-2}+\frac{Bz+C}{z^2+1}.$$ You will find that $A=1,B=0,C=-2.$ Then $$\frac{z^2-2z+5}{(z-2)(z^2+1)}=\frac{1}{(-2)(1-\frac{z}{2})}+\frac{-2}{1+z^2}$$ $$=(-1/2)(1-\frac{z}{2}+\frac{z^2}{4}-...)-2(1-z^2+z^4-...)$$ $$=\sum_{n=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+... | By the Buffalo way, WLOG $a=\min\{a,b,c\}$, then set $a=1+k$, $b=1+k+u$ and $c=1+k+v$. Along with the condition $a\ge1$, we get $k$, $u$, $v\ge0$. Plug in and we get
\begin{align}
&(a+b)(b+c)(c+a)(\text{left side}-\text{right side})\\[.6em]
={}&24k^5+40k^4u+40k^4v+120k^4+30k^3u^2+50k^3uv
\\{}+{}&160k^3u+30k^3v^2+160k^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
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Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (... | As this Art of Problem Solving thread suggests, write this using a form of partial fractions:
$$(Ar + B)(r^2 - r + 1) + (Cr + D)(r^2 + r + 1) = r^2 - 1$$
Now comparing the coefficients of $r^3$, $r^2$, $r$, and the constant term, we end up with $A + C = 0, B - A + D + C = 1, -B + A+D+C=0$, and $B + D = -1$.
Adding th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which solution is correct: $\lim_{x \to \infty} {\frac{x^2+2x+3}{2x^2+x+5}}^\frac{3x-2}{3x+2}$? I was solving this question $\lim_{x \to \infty} {\frac{x^2+2x+3}{2x^2+x+5}}^\frac{3x-2}{3x+2}$.
Now, this is solved using $\lim_{x \to a} [1+f(x)]^\frac{1}{g(x)}=e^{\lim_{x \to a} \frac{f(x)}{g(x)}}$.
However, there seem to... | Here is how your second approach should work:
$$\frac{x^2+2x+3}{2x^2+x+5} = 1 + \left(\frac{x^2+2x+3}{2x^2+x+5}-1\right)$$$$ = 1 + \underbrace{\frac{-x^2+x-2}{2x^2+x+5}}_{\stackrel{x\to\infty}{\longrightarrow}-\frac 12}$$
So, you could not use "your rule with $e$", because it would require this term to tend to zero.
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | $$g(k)=\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{(4k+2)^4+4\cdot1^4}{(4k)^4+4\cdot1^4}$$
Using your formula only,
$\displaystyle x^4+4y^4=(x^2+2y^2)^2-(2xy)^2=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
$x=4k+2, y=1$
$\displaystyle\implies x^2+2xy+2y^2=(4k+2)^2+2(4k+2)+2=16k^2+24k+10\ \ \ \ (1)$
$x^2-2xy+2y^2=(4k+2)^2-2(4k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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)^... | It is possible to solve it without rational fractions. The denominator can be written as
$$\sin(x)+2\cos(x) = \sqrt 5\left(\frac{1}{\sqrt 5}\sin x + \frac{2}{\sqrt 5}\cos x\right) = \sqrt 5 \sin\left(x+\theta\right)$$
where $\sin\theta = \frac{2}{\sqrt 5}$. Now by the change of variable $x+\theta\rightarrow y$ we get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $f(\frac{2\pi k}{2^n \pm1})$ given $f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$. We are required to find $f(\frac{2\pi k}{2^n \pm1})$ given
$f(x)=\prod_{i=1}^{n-1}[2\cos(2^{i-1}x)-1],n\geq1$
Let's start by removing the product notation
$\Rightarrow f(x)=(2\cos x-1)(2\cos 2x-1)(2\cos 2^2x-1)...(2\cos 2^{n-1}x... | In fact you have solved roughly all of it .
You just need to know that:
*
*$\frac{2^n2\pi k}{2^n +1} +\frac{2\pi k}{2^n+1}=2\pi k$
*$\frac{2^n2\pi k}{2^n -1} -\frac{2\pi k}{2^n-1}=2\pi k$
So in the end you get the answer 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating Functions - Counting how many solutions for an exact $x^n$ Having trouble with this problem:
$$ x_1+x_2+x_3+x_4+x_5+x_6 = 13 $$
$$ x\ne 3 $$
Thanks in advance for any further help.
Being able to get to this point:
$$\begin{split}
f(x) &= (x^0+x^1+x^2+x^4+x^5+\cdots)^6\\
&= [(-x^3)+{x^0+x^1+x^2+(x^3)+x^4+x^5+... | The P.I.E approach needs to prepare some "ingredients".
There are solutions containing $4$ "bad" $x$'s, $3, 2, 1$ or none "bad" $x$'s.
The numbers involved are the following :
$$ {6\choose 4}{2\choose 1} = 30$$
$$ {6\choose 3}{6\choose 2} = 300$$
$$ {6\choose 2}{10\choose 3} = 1800$$
$$ {6\choose 1}{14\choose 4} = 6006... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solving $4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6$ for real $x$ I was given the following equation:
$$4^{x+\sqrt{x^2-2}}-5\cdot 2^{\sqrt{x^2-2}-1}=6; x\in \mathbb{R}$$
Here is how I tried to solve it:
$$2^{2x+2\sqrt{x^2-2}+1}-5\cdot 2^{\sqrt{x^2-2}}=12$$
$$2^{\sqrt{x^2-2}}(2^{2x+1+\sqrt{x^2-2}}-5)=2^2\cdot 3$$
Wh... | Comment: As you modified we have:
$2^{2x+1+2\sqrt{x^2-2}}-5\cdot 2^{\sqrt {x^2-2}}-12=0$
$2^{2x+1+2\sqrt{x^2-2}}=2^{2x+1}\cdot 2^{2\sqrt {x^2-2}}=2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2$
$2^{2x+1}\cdot (2^{\sqrt{x^2-2}})^2-5\cdot 2^{\sqrt{x^2-2}}-12=0$
Let $2^{\sqrt{x^2-2}}=y$ , we can write:
$(2^{2x+1})\cdot y^2-5y-12=0$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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
$$\int_0^{\infty} \frac{\ln ... | Note that $x^4+a^4= (x^2+\sqrt2 ax+a^2) (x^2-\sqrt2 ax+a^2)$
\begin{align}
&\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{x^2+b^2} d x \\
=&\int_{-\infty}^{\infty} \frac{\ln (x^2+\sqrt2 ax+a^2)}{x^2+b^2} \ d x\\
=& \int_{-\infty}^{\infty}\bigg( \int_0^{\pi/4}\frac{2ax\cos t}{x^2+2ax \sin t+a^2} dt+ {\ln (x^2+a^2)}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int \frac1{(x^2+a^2)^{m}}dx$ The integral $$\int\frac{1}{(x^2+a^2)^m}dx$$ can be expressed by a recursive formula of $$\frac{1}{2a^2(m-1)}\frac{x}{(x^2+a^2)^{m-1}} + \frac{2m-3}{2a^2(m-1)}\int\frac{dx}{(x^2+a^2)^{m-1}}$$ I do not understand how integration by part leads to this result. Specifically, s... | We have
$$J = \int\dfrac{dx}{(x^2+a^2)^m} = \dfrac1{a^2}\int\dfrac{(a^2+x^2)-x^2}{(x^2+a^2)^m}dx$$
and hence
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{m-1}} - \dfrac1{a^2}\int\dfrac{x^2}{(x^2+a^2)^m}dx$$
which gives
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{m-1}} + \dfrac{1}{2a^2(m-1)}\int x\,d\, \dfrac1{(x^2+a^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\sum_{cyc}\frac{\sqrt{y + z − x}}{\sqrt{y} + \sqrt{z} − \sqrt{x}} \leq 3$ for $x$, $y$, $z$ the sides of a triangle
Let $x,y$ and $z$ be the length of a triangle. Prove that
$$\frac{\sqrt{y + z − x}}{\sqrt{y} + \sqrt{z} − \sqrt{x}}
+ \frac{\sqrt{z + x − y}}{\sqrt{z} + \sqrt{x} − \sqrt{y}}
+ \frac{\sqrt{x + ... | For the second term, when we'll do the proving we'll shake hands with the expression $$z+x-y \leq (\sqrt{z}+\sqrt{x}-\sqrt{y})^2$$ This is not true for the triplet $(n-1,n,n+1)$
I'll leave it for the OP to answer the reason for this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4651299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,... | Here is one proof, using how ${ \log (1+x) }$ is "sandwiched" between its Taylor polynomials on ${ [0,1] }.$
Consider ${ \log(1+x) }$ on ${ (-1, \infty) }.$ Its Taylor polynomials about $0$ are ${ S _n (x) = x - \frac{x ^2}{2} + \ldots + (-1) ^{n+1} \frac{x ^n}{n}. }$
Notice errors ${ \varepsilon _n (x) := \log(1+x) - ... | {
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"timestamp": "2023-03-29T00:00:00",
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Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:
$$x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$
For cubic equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three ... | What has not been mentioned thus far is that one can in fact use any number of "auxiliary cubics" in the solution of the quartic equation. Don Herbison-Evans, in this page (Wayback link, as the original page has gone kaput), which was adapted from his technical report, mention five such possible auxiliary cubics.
Given... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "165",
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How to build a linear equation system? How can one build a $3\times 3$ linear equation system ($3$ equations with $3$ variables) if the variables are known, for example, $a=1, b=2$ and $c=3$?
Thank you in advance!
| Another possibility is to start with a non-singular upper-left triangular matrix $T$ and the desired solution $\vec{x}=(x_1,x_2,x_3)^t$. Now one adds multiples of upper rows to lower ones to obtain a full matrix $A$, e.q.
\begin{gather*}
T=\begin{pmatrix}1& 2 & 3\cr 0& 4 & 5\cr 0 & 0 & 6\end{pmatrix},\;
\vec{x}=... | {
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For $n \in \mathbb{N}$ $\lfloor{\sqrt{n} + \sqrt{n+1}\rfloor} = \lfloor{\sqrt{4n+2}\rfloor}$ This is Exercise 3.20 from Apostol's book. Many of them seem tough and here is one of them which I am struggling with.
For $n \in \mathbb{N}$, prove that this identity is true: $$\Bigl\lfloor{\sqrt{n} + \sqrt{n+1}\Bigl\rfloor} ... | Note that $$\frac{1}{2} > \sqrt{n+\frac{1}{2}}-\sqrt{n} > \sqrt{n+1}- \sqrt{n+\frac{1}{2}} > 0,$$
since $f(x)=\sqrt{x+1/2}-\sqrt{x}$ is a decreasing function. The inequalities and the fact that $f(x)$ is decreasing follow from noting that $\sqrt{x+1/2} - \sqrt{x} = 1/2(\sqrt{x+1/2} + \sqrt{x}).$
So we can write
$$ 1> \... | {
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mystery regarding power series of $\frac{1}{\sqrt{1+x^{x}}}$ In the course of playing around with $\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+n^{n}}}$, I used w|α to obtain the power series for $f(x)=\frac{1}{\sqrt{1+x^{x}}}$
which is
\begin{align*}
\frac{1}{\sqrt{1+x^{x}}} =& \frac{1}{\sqrt{2}} - \frac{x\log(x)}{4\sqrt{2}}... | Well,
$$\frac1{\sqrt{1+x^x}}=\frac1{\sqrt{1+e^y}}$$
where $y=x\log x$ so it's unsurprising you get a series in terms of $y=x\log x$.
Then
$$\frac1{\sqrt{1+e^y}}=\frac1{\sqrt2}\frac1{\sqrt{1+u}}$$
where $u=\frac12(e^y-1)=o(y)$. This explains the fact that the coefficients
are rationals over $\sqrt2$.
Now consider $f(x)=... | {
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If $X$ and $Y$ are not divisible by 3, then $X^2 - Y^2$ is divisible by 3 Prove: If $X$ is not divisible by 3 and $Y$ is not divisible by 3, then $X^2 - Y^2$ is divisible by 3.
I cannot find a good way to prove this.. Tried contrapositive and contradiction but cannot see the answer. Anyone have any ideas?
| If $x$ and $y$ are not divisible by 3 then they must be of the form $x = 3k \pm 1$ and $y = 3t \pm 1$. Therefore $x^2 - y^2 = 9k^2 \pm 6k + 1 -9t^2 \mp 6t -1 = 3(3k^2 - 3t^2 \pm 2k \mp 2t).$ So $x^2 - y^2$ is divisible by 3.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2^{k} \mid {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \cdot 3^{k}$ I have to prove this question:
$$2^{k} \,\left|\, {2k \choose 0} \cdot 3^{0} + {2k \choose 2} \cdot 3^{1} + \cdots + {2k \choose 2i} \cdot 3^{i} + \cdots + {2k \choose 2k} \... | Let's talk about some general methods. This is a special case of the following problem: given that you understand a sequence $a_n$, how can you understand the sequence $b_n = \sum_{k=0}^n {n \choose k} a_k$? Generating function techniques will come to your rescue. If you let $A(x) = \sum_{n \ge 0} a_n x^n$, then
$$B... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have
$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$
$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 18
} |
Integral of $\frac1{\cos x}$ using t substitution Okay, so I'm trying to find $ \int \frac1{\cos x}\mathrm{d}x$ using the substitution $t = \tan\left(\frac{x}{2}\right)$.
I sub in the trig identity for $\sec$ as $\frac{1+t^2}{1-t^2}$ and then rearrange and substitute $\frac{\mathrm{d}t}{\mathrm{d}x} = \frac12 \left(1+ ... | We know that $$\cos x=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2} }=\frac{1-\tan^{2}\frac{x}{2}}{\sec^2 \frac{x}{2} }$$ Given that $$\int \frac{1}{\cos x}dx$$$$=\int \frac{1}{\frac{1-\tan^{2}\frac{x}{2}}{\sec^2 \frac{x}{2} }}dx$$ $$=\int \frac{\sec^2 \frac{x}{2}}{1- \tan^{2}\frac{x}{2}}dx$$ Let $$\tan \frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However... | $$\int \frac{1}{\sin x\cos x} dx = \int( \tan x+\cot x)\, dx= \int\frac{\sin x}{\cos x}\, dx+\int\frac{\cos x}{\sin x} \,dx$$ from where it should be fairly simple....(choose correctly the variables though)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 9
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What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| OK, first rewrite it as
$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$
Multiply both sides by $2 \cos(A/2)$
$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$
Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:
$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$
Using Simpson's rule on the second hand of the equation to s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\log _2(x-4) + \log _2(x+2) = 4$ Here is how I have worked it out so far:
$\log _2(x-4)+\log(x+2)=4$
$\log _2((x-4)(x+2)) = 4$
$(x-4)(x+2)=2^4$
$(x-4)(x+2)=16$
How do I proceed from here?
$x^2+2x-8 = 16$
$x^2+2x = 24$
$x(x+2) = 24$ Which I know is not the right answer
$x^2+2x-24 = 0$ Can't factor this
| The quadratic formula:
$$ x = \frac{ -b \pm \sqrt{b^2 - 4 a c} }{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-24) }}{2 \times 1} $$
4 + 96 in the square root.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Combinatorial proof of a binomial coefficient summation: $\sum_{k=1}^n \binom nk \binom n{k-1} = \frac12\binom{2n+2}{n+1} - \binom{2n}n$ Let $n$ and $k$ be integers with $1 \leq k \leq n$. Show that:
$$\sum_{k=1}^n {n \choose k}{n \choose k-1} = \frac12{2n+2 \choose n+1} - {2n \choose n}$$
I was told this is supposed t... | Using standard binomial identities, this can be proven as follows:
$$
\begin{align}
\sum_{k=1}^n \binom{n}{k}\binom{n}{k-1}&=\sum_{k=1}^n \binom{n}{n-k}\binom{n}{k-1}\\
&=\binom{2n}{n-1}\\
&=\binom{2n+1}{n}-\binom{2n}{n}\\
&=\frac{n+1}{2n+2}\binom{2n+2}{n+1}-\binom{2n}{n}\\
&=\frac{1}{2}\binom{2n+2}{n+1}-\binom{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Solving systems of linear equations over a finite ring I want to solve equations like this (mod $2^n$):
$$\begin{array}{rcrcrcr} 3x&+&4y&+&13z&=&3&\pmod{16} \\ x&+&5y&+&3z&=&5&\pmod{16} \\ 4x&+&7y&+&11z&=&12&\pmod{16}\end{array}$$
Since we are working over a ring and not a field, Gaussian elimination doesn't work. So ... | You can still use Gaussian elimination as long as you don't "divide" by things that are not relatively prime to the modulus. In this case, you can "divide" by any odd number, and perform all the usual computations. In this case, you can perform Gaussian pretty well:
\begin{align*}
\left(\begin{array}{ccc|c}
3 & 4 & 13 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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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... | You can do partial fractions in this way: $$\frac{x}{x^{3}-1} = \frac{Ax+B}{x^{2}+x+1} + \frac{C}{x-1}$$
Now by equating the coefficients you get:
*
*$A+C=0$
*$B+C-A=1$ and;
*$C-B=0$.
Solving you get the values of $A= -\frac{1}{3}$, $B=\frac{1}{3}$ and $C=\frac{1}{3}$.
So, the required integral breaks down as $$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$ then $\tan x + \cot x=?$ Hello :)
I hit a problem.
If $\sin x + \cos x = \frac{\sqrt{3} + 1}{2}$, then how much is $\tan x + \cot x$?
| We have
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x}.$$
So you want to know how much is $\sin x\cos x$.
Squaring the equality you know, we have:
\begin{align*}
(\sin x + \cos x)^2 &= \left(\frac{\sqrt{3}+1}{2}\right)^2\\
\sin^2 x ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
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Why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$ I need some help understanding why $\frac{x}{x+1}=\frac{1}{x^{-1}+1}$.
I would be grateful if someone could explain. Be explicit.
Thank you!
| $$\frac{x}{x+1} = \frac{x \cdot 1}{x \cdot (1+\frac{1}{x})}= \frac{x}{x} \cdot \frac{1}{1+\frac{1}{x}} = 1 \cdot \frac{1}{1+\frac{1}{x}} = \frac{1}{1+x^{-1}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$n^3 \equiv n^5 \pmod{12} $? I am proving that
$$5n^3 + 7n^5 \equiv 0 \pmod{12}$$
It would suffice to show
$$n^3 \equiv n^5 \pmod{12}$$
How would I go about doing that?
I suppose I could just go through each $n \equiv r \pmod{12}$ with $r$ from $1$ to $11$ and show that $n^3 \equiv n^5 \pmod{12}$ for each, but that w... | $$\begin{align}n^5-n^3&=(n-2)(n-1)n(n+1)(n+2)-4n^3+4n\\\\
&=(n-2)(n-1)n(n+1)(n+2)-4(n-1)n(n+1)\end{align}$$
The first expression is a product of 5 consecutive numbers, hence is divisible by $5!$ i.e., by 120. Similarly, the second expression is divisible by $4\cdot (3!)$ i.e., by 24.
Also, $n^5-n^3 = n^2(n-1)n(n+1)$. C... | {
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Evaluating $\int_{0}^{2}\frac{1}{(x+2)(x-1)}dx$ $\int_{0}^{2}\frac{1}{(x+2)(x-1)}$
Will I be able to use partial fractions on the indefinite integral and then evaluate my answer as f(2)-f(0)?
Any help is appreciated as always. :)
EDIT: I looked this up and apparently the integral doesn't converge so now I need help ... | Hint 1:
$$\frac{1}{(x+2)(x-1)}=\frac{1}{3}\left(\frac{1}{x-1}-\frac{1}{x+2}\right)$$
Hint 2: Does this converge or diverge? Think about $$\int_{-1}^{1}\frac{1}{x}dx.$$ Does the value of this integral even make sense?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Question regarding solving polynomial of congruence? In my textbook, they said:
$$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... | Any integer is going to be congruent to one of $0,1,2,3,4$ modulo $5$. As the testing shows, only those integers which are congruent to $1$ modulo $5$ are solutions, and any integer congruent to $1$ mod $5$ is a solution.
Those tests are using an arbitrary $x$ each time, so you shouldn't think of them as a system of co... | {
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"source": "stackexchange",
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What is the first derivative of $y= \sin^2(\sqrt{x})$? I'm trying to find out what the graph of the function $y= \sin^2(\sqrt{x})$ looks like. Can someone please help me?
| The graph of $ y = \sin^2(\sqrt{ x }) $ on $x \in [0,15]$ from MATLAB is:
The first derivative of $\displaystyle y = \sin^2(\sqrt{ x })$ is $\displaystyle y^{\prime} = 2 \sin (\sqrt{ x }) \cos( \sqrt{x} ) \frac{1}{2}x^{ - \frac{1}{2} } $.
Simplifying gives $\displaystyle y^{\prime} = \frac{ \sin{ \sqrt{x} } \cos{ \sqr... | {
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Infinitely many integer solutions for the equations $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$ How do you show that the equation $x^3+y^3+z^3=1$ has infinitely many solutions in integers? How about $x^3+y^3+z^3=2$?
| For $x^3+y^3+z^3=1$ it is trivial - an infinite family of solutions is $(1,n,-n)$, and permutations of that.
For $x^3+y^3+z^3=2$ I'm not so sure there are infinitely many solutions. Are you just hypothesizing this, or do you know it to be true?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral of floor function: $\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$ How would you go about solving integral of a floor? The particular problem I have is:
$$\int \,\left\lfloor\frac{1}{x}\right\rfloor\, dx$$
| The function:
$$\left\lfloor\frac{1}{x}\right\rfloor$$
is equal to $n$ on the interval $\left(\frac{1}{n+1},\frac{1}{n}\right)$,so if we try to determine the integral from $t>0$ to $1$, we can let $n=\left\lfloor\frac{1}{t}\right\rfloor$ and we have constant value $1$ on range $(\frac{1}{2},1)$, constant value $2$ on r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How do I prove this inequality?: $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ where $a,b,c>0$ and $abc=1.$ I have been thinking over this problem for a couple of days, but I have no idea how to solve it in a simple way.
I am interested if there is a way only using elementary me... | Since $abc = 1$, any product term such as $ab$ or $\frac{1}{bc}$ can be rewritten as a singleton term (i.e., $ab = \frac{1}{c}$ and $\frac{1}{bc} =a$. Thus the inequality is equivalent to $$\left(a-1+\frac{1}{c}\right)\left(b-1+\frac{1}{a}\right)\left(c-1+\frac{1}{b}\right) \leq 1.$$
This inequality was Problem 2 on ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at nex... | Since $(n-1)^3 + n^3 + (n+1)^3 = 3n(n^2+2)$ we know that it is divisible by $3$. If we show that $n(n^2+2)$ is also divisible by $3$ then $(n-1)^3 + n^3 + (n+1)^3$ is divisible by $3^2 = 9$. There are three cases to check:
*
*$0(0^2+2) \equiv 0 \cdot 2 \equiv 0 \pmod 3$
*$1(1^2+2) \equiv 1 \cdot 0 \equiv 0 \pmod 3$... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 2
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Finding $\alpha$ such that $\tan\alpha = \frac{(1+\tan 1^{\circ})\cdot (1+\tan 2^{\circ})-2}{(1-\tan 1^{\circ})\cdot(1-\tan 2^{\circ})-2}$ I am preparing for my college entry test and I ran into this problem in my book:
$$\tan\alpha = \frac{(1+\tan 1^{\circ})\cdot (1+\tan 2^{\circ})-2}{(1-\tan 1^{\circ})\cdot(1-\tan 2^... | $$\begin{align*}
V&=\frac{\tan a\tan b+\tan a+\tan b-1}{\tan a\tan b-\tan a-\tan b-1}\\
&=\frac{\sin a \cos b + \cos a \sin b + \sin a\sin b - \cos a\cos b}{\sin a\sin b-\cos a\cos b-\sin a\cos b-\cos a\sin b}\\
&=\frac{\sin (a+b)-\cos(a+b)}{-\cos(a+b)-\sin(a+b)}
\end{align*}$$
Now I will use:
$$\sin x-\cos y = \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/37854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Power series of $\ln(x+\sqrt{1+x^2})$ without Taylor The answer is $$x-\frac{ 1}{2}\frac{ x^3}{3}+\frac{ 1\cdot 3}{2\cdot 4}\frac{ x^5}{5}-\frac{ 1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{ x^7}{7}+\cdots$$ But I can't see how. Unfortunately, "how" can't be using Taylor's formula, because that isn't introduced until the ne... | You can use the fact, as you write, that the derivative of such function is $$y' = \frac{1}{\sqrt{1+x^2}}$$
But from the binomial theorem you have that
$$\frac{1}{\sqrt{1+x^2}} = \sum_{k=0}^{\infty} {-\frac{1}{2} \choose k} x^{2k}$$
You can check by yourself that
$${-\frac{1}{2} \choose k} ={\left( { - 1} \right)^k}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/42295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
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... | André's approach is pretty simple. In case you really must follow the hint, here's how to get it: note, for example, that in the first fraction,
$$\frac{ab+ca}{a^2+ab+ca}$$
the only thing that is missing from having the numerator exactly equal to the denominator is an $a^2$; so adding it and taking it away again we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/44694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to prove that $a_n = \frac{2n+1}{n+1}$ is bounded? How do I prove that $\displaystyle a_n = \frac{2n+1}{n+1}$ is bounded?
According to the key it works as follows. However, I don't really understand the meaning.
Upper bound:
$\displaystyle a_n = \frac{2n+1}{n+1} < \frac{2n+2}{n+1} = \frac{2(n+1)}{n+1} = 2$
Lower bo... | For the upper bound part you could also reason as follows:
$$\frac{2n+1}{n+1}=\frac{n+n+1}{n+1}=\frac{n}{n+1}+\frac{n+1}{n+1}=\frac{n}{n+1}+1$$
Since $\dfrac n {n+1}$ comes as monotonically increasing on the natural numbers, and the limit of $\dfrac n {n+1}$ as $n$ increases without bound equals 1, it follows that the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/44950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$
How would you evaluate the following limit:
$$\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}$$
I tried to use this formula: $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, It didn't work.
Any hints?
| Edit: The main idea is to use the expansion
$$\sqrt[6]{1+x^{-1}}=1+\frac{1}{6}x^{-1}+O\left( x^{-2}\right) $$
which can be generated by changing variables $x=1/z$, expanding about $z=0$, $$\sqrt[6]{1+z}=1+\frac{1}{6}z+O\left( z^{2}\right) $$ and restoring $x$ via $%
z\rightarrow 1/x$.
Let $f(x)=\sqrt[6]{x^{6}+x^{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/45970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
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about $\operatorname{SO}(2)$ group When
$$
A =\begin{pmatrix} a & b \\ c & d \end{pmatrix},\quad a,b,c,d \in \mathbb{R}, \quad A^2 -2aA + I = 0
$$
Is $ A \in \operatorname{SO}(2)\; $ if $A^n \in SO(2)\;$ for some $n \in \Bbb N$?
$\operatorname{SO}(2)$ is special orthogonal group.
| The answer is yes I believe. Your requirement that $A^2-2aA+I=0$ forces $\det(A)=1$, so we know that if $A^n \in SO(2)$, then $\det(A^n)=\det(A)=1$, so we've eliminated the problem that Qiaochu mentioned. Now, it can be shown using a little bit of algebra that any matrix $B$ in $SO(2)$ is of the form $$\begin{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/56411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$ Prove :
$$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$
For $a,b,c,d>0$
| Since for $x>0$ we have $$\left(\frac{x^2+1}{x+1}\right)^3-\frac{x^3+1}{2}=\frac{(x-1)^4(x^2+x+1)}{2(x+1)^3}\geq0$$
and by P-M $$\left(\frac{x+1}{2}\right)^3\leq\frac{x^3+1}{2},$$
we obtain:
$$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\frac{a^3+d^3}{2}+\frac{b^3+c^3}{2}=$$
$$=\frac{a^3+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/59190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Supremum of a Set containing terms of a sequence I am trying to prove that the sequence defined by $x_n = a_n - b_n$ converges to zero where $a_n$ and $b_n$ are sequences such that
$a_n = \sqrt{a_{n-1}b_{n-1}}$ and $ b_n = \frac{a_{n-1} + b_{n-1}}{2}$.
By definition $a_1 = \sqrt{ab}$ and $b_1 = \frac{a + b}{2}$ where $... | To show that $x_n\le0$,
$$
\begin{align}
x_n&=a_n-b_n\\
&=\frac{a_n^2-b_n^2}{a_n+b_n}\\
&=\frac{a_{n-1}b_{n-1}-\frac{1}{4}(a_{n-1}+b_{n-1})^2}{a_n+b_n}\\
&=\frac{-\frac{1}{4}(a_{n-1}-b_{n-1})^2}{a_n+b_n}\\
&\le0
\end{align}
$$
But instead of showing that $a_n-b_n\to0$, it is simpler to show that $b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the minimum number of students
There are $p$ committees in a class (where $p \ge 5$), each consisting of $q$ members (where $q \ge 6$).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
It is easy to see that the maximum number of ... | For the case in which $p \le q+1$ an arrangement that yields the minimum number of students can be described as follows.
Let $P = \{\langle m,n \rangle:1 \le m \le p, 1 \le n \le q+1\}$, and let $S = \{\langle m,n \rangle \in P:m < n\}$. If $P$ is thought of as a $p \times (q+1)$ grid, $S$ is the part of it strictly ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
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
| $$\ S =\sum_{k=1}^n k^3 $$
\begin{align}
k^3 =& Ak(k+1)(k+2)+Bk(k+1)+Ck+D \\
k^3= & Ak^3+(3A+B)k^2+(2A+B+C)k+D
\end{align}
Therefore $A=1$, $B=-3$, $C=1$, $D=0$ and
\begin{align}
S =& \sum_{k=1}^n (k(k+1)(k+2)-3k(k+1)+k)\\
S =&\sum_{k=1}^n k(k+1)(k+2)-3\sum_{k=1}^nk(k+1)+\sum_{k=1}^nk \\
S =&\sum_{k=1}^n6\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "67",
"answer_count": 16,
"answer_id": 13
} |
Proof of the equality $\sum\limits_{k=1}^{\infty} \frac{k^2}{2^k} = 6$ Show that for $k$ running over positive integers
$$ \sum_{k=1}^\infty \frac{k^2}{2^k}=6 .$$ We can use finite calculus.
| Solution
For convenience, we denote $$a_k=\frac{k^2+2k+3}{2^{k-1}},~~~k=1,2,\cdots$$
Then $$\frac{k^2}{2^k}=a_k-a_{k+1}.$$
Thus
\begin{align*}\sum_{k=1}^n \frac{k^2}{2^k}&=(a_1-a_2)+(a_2-a_3)+\cdots+(a_n-a_{n+1})=a_1-a_{n+1}\\&=6-\frac{(n+1)^2+2(n+1)+3}{2^n}\\&=6-\frac{n^2+4n+6}{2^n}.\end{align*}
It follows that \begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/62422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 6
} |
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... | Let me try a slightly different tack... Let us use the terminology of modular arithmetic, which is the same thing you wrote, but denoted differently. Let $n \equiv 0\ \mathrm{mod}\ 3$. Then $n^2 \equiv 0\ \mathrm{mod}\ 3$. Next, for $n \equiv 1\ \mathrm{mod}\ 3$, we have $n^2 \equiv 1\ \mathrm{mod}\ 3$. Lastly, for $n ... | {
"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": 3
} |
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... | Obviously $x^1+1\equiv 0 \pmod p$ has exactly one solution.
For each $n$ it must hold that $x^{2^{n+1}}+1\equiv 0 \pmod p$ has at most twice as many solutions as $x^{2^n}+1\equiv 0\pmod p$ -- otherwise one of the $2^n$th roots must have more than two square roots in $\mathbb F_p$, which is is impossible.
Fermat's littl... | {
"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": 0
} |
Is there a more efficient method of trig mastery than rote memorization? I would like to get alot better at trig than I am. What is the best/most efficient method?
Thanks much in advance
Joe
| I would emphasize how to derive trigonometric identities from a few ones. Learn:
*
*how to derive the relations between the direct functions of the same angle from the definition of the trigonometric functions and the Pythagorean formula;
*maxima, minima, zeroes and period of each function;
*if a function is odd ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/65548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
How can I differentiate this? I was trying to differentiate this: $\frac{1}{2} \sin^{-1} \frac{2x}{1+x^2}$ but I am really stuck.I obtained an answer that does not match with the one given in the book.Your help is appreciated.[Edit: I used the formula $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$ to get $\frac{1+x^2... | $y=\sin(x)\Rightarrow\frac{\mathrm{d}y}{\mathrm{d}x}=\cos(x)=\sqrt{1-y^2}\therefore\frac{1}{\sqrt{1-y^2}}=\frac{\mathrm{d}x}{\mathrm{d}y}=\frac{\mathrm{d}}{\mathrm{d}y}\sin^{-1}(y)$
The rest is the chain-rule:
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{2}\sin^{-1}\left(\frac{2x}{1+x^2}\right)
&=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/67702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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... | For problem 2, you can do better than manually checking by using the Chinese Remainder Theorem.
Indeed, you first want to compute $x^9$ modulo $\phi(100)=\phi(2^2\times 5^2) = 2\times 4\times 5$. But working modulo $40$ is equivalent, by the Chinese Remainder Theorem, to working modulo $8$ and modulo $5$. The advantag... | {
"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": 1
} |
Compute $1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.
Sol:
$P(n... | Everything is OK except for the very last line. You somehow lost a factor of two. The penultimate line is already the result you want, since $2k^2+5k+3=(k+1)(2k+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/72636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find critical points of a function: $y = x \sqrt{4 - x^2}$ I am having trouble finding the critical points of this function, I was wondering if someone could help me out.
So I have a function $y = x\sqrt{4 - x^2}$. I know that I have to take the derivative then set that equal to 0 to find the critical points.
$$\begi... | First, notice that the domain of your function is $[-2,2]$. We will only be working there.
Now, when you take the derivative, there are other critical points besides the stationary points: you must also exclude any point in the domain where the derivative is not defined. In this case, notice that if $x=-2$ or $x=2$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/75500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Left and Right Cosets Let $H$ be a subgroup of a group $G$ and let $a, b \in G$. I need to give a counterexample or proof of the following statement:
If $aH = bH$, then $Ha = Hb$
Proof:
For every $h \in H, ah = bh$
$
\begin{align}
ah &= bh \newline
ahh^{-1} &= bhh^{-1} \newline
a &= b \newline
ha &= hb
\end{alig... | Counterexample: We want to use a non-Abelian group such as $S_{3}$
$\mu_{1} = \pmatrix{1&2&3\\1 &3 &2}$, $\mu_{2} = \pmatrix{1&2&3\\3 &2 &1}$,
$\mu_{3} = \pmatrix{1&2&3\\2 &1 &3}$
$\rho_{0} = \pmatrix{1&2&3\\1 &2 &3}$, $\rho_{1} = \pmatrix{1&2&3\\2 &3 &1}$,
$\rho_{2} = \pmatrix{1&2&3\\3 &1 &2}$
Let $H = \{\rho_{0}, \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
quotient rule difficulties I'm trying to use the quotient rule to differentiate $\frac{r}{\sqrt{r^2+1}}$ but I'm getting the wrong answer. So far I have
$$\begin{align*}
\frac {d}{dr} \frac{r}{\sqrt{r^2+1}} &= \frac {\sqrt{r^2+1} \frac {d}{dr} r - r \frac {d}{dr} \sqrt{r^2+1}} {(\sqrt{r^2+1})^2} \\\\\\\\
&= \frac {\... | You are all right, just only one step away from the finish line.
$$ (r^2+1)^{-1/2} - r^2 (r^2+1)^{-3/2} $$
$$ = \dfrac{1}{\sqrt {r^2+1}} - \dfrac {r^2}{ (r^2+1)^{3/2}}= \dfrac{(r^2+1)}{ (r^2+1)^{3/2}} - \dfrac {r^2}{ (r^2+1)^{3/2}}=\dfrac {1}{ (r^2+1)^{3/2}}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/77724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving that $f(n)$ is an integer using mathematical induction I want to prove that
$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$
is an integer for every integer $n \geq 1$.
I define P(n) to be:
$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$ is an integer.
For my basis step, P(1) is true because
$$\frac{1^3}{3}+\frac... | Here is a proof not by induction, but which may be instructive.
Using repeated differences on the first few values of $P(n)$ we get
$$
\begin{array}{llll}
0 & 1 & 10 & 59 & 228 & 669 & 1630 & \\
1 & 9 & 49 & 169 & 441 & 961 & \\
8 & 40 & 120 & 272 & 520 & \\
32 & 80 & 152 & 248 & \\
48 & 72 & 96 & \\
24 & 24 & \\
0 & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Laurent series around certain points I want to compute the laurent series of
i) $(z-3)\sin(\frac{1}{z+1}); (z\ne -1)$ at $z_{0}=-1 $.
ii) $\displaystyle{\frac{z}{(z-1)(z-2)}}; ( z\ne 1,2) $ at $z_{0}=1$
and classify their singularities.
A Laurent series has the form: $\displaystyle{f(z)= \sum_{n=1}^{\infty}c_{n}\fra... | You're series expansion of $\sin(z)$ isn't quite right. It should be $\sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}$. Keep in mind that the expansion of $z-3$ about $z=-1$ is $z-3=-4+(z+1)$ Therefore,
$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \left(-4+(z+1)\right)\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Could someone please help me write $z= \frac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form? Could someone please help me write $z = \dfrac{(2+2i)^3}{(1+i\sqrt{3})^4}$ on polar form?
$|z|=\sqrt{2}$
But how do I proceed to get the argument?
$\mathrm{arg}(z)=\mathrm{arg}(2+2i)^3 - \mathrm{arg}(1+i\sqrt{3})^4$
Thanks /Dav... | Hints: Use $\arctan 1=\frac{1}{4}\pi $ and $\arctan \sqrt{3}=\frac{1}{3}\pi $ in
$$\arg \left( \left( 2+2i\right) ^{3}\right) =3\arg \left( 2+2i\right)
=3 \arctan \frac{2}{2}=3\times \frac{1}{4}\pi $$
and
$$\arg \left( (1+i\sqrt{3})^{4}\right) =4\arg (1+i\sqrt{3})=4\arctan \frac{\sqrt{3}}{1}=4\times \frac{1}{3}\pi .$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/78971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Hard algebra problem Given
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$
Now, it is necessary to find
$$\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}=?$$
Is this possible and how? a,b,c are given constants.
I think, ? is probably a complicated function $F=F(a,b,c)$
| Without loss of generality we can assume $a\ge b\ge c>0$.
Then
$$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}\le\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}\le\frac{x^2}{a^2c^2}+\frac{y^2}{b^2c^2}+\frac{z^2}{c^4}.$$
Noting that
$$\frac{x^2}{a^4}+\frac{y^2}{a^2b^2}+\frac{z^2}{a^2c^2}
=\frac1{a^2}\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How many ways a composite number can be resolved into two factors which are prime to each other? Let N denote the number, and suppose $N=a^p \times b^q \times c^r \cdots$, where $a,b,c,\cdots$, are different prime numbers and $p,q,r,\cdots$ are positive integers.Then it is clear that each term of the product
$$(1+a+a^... | We obtain all ordered pairs $(x,y)$ such that $xy=N$ and $\gcd(x,y)=1$ by choosing a subset of the set of prime divisors of $N$, and giving all these primes to $x$, and giving the rest of the primes to $y$. A set of $n$ elements has $2^n$ subsets, so there are $2^n$ ways to produce a suitable $x$, and hence $2^n$ ord... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.
| It is false.
Take $(c,d)=(1,1)$. Then $a^2 + b^2 = a^3 + b^3 = 2$ has real solutions with $(a+b) \neq 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/84855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Equivalence class of an ideal Let $K$ denote the number field $\mathbb{Q}(\sqrt{15}).$ According to standard lore, we have that $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{15}]$. Moreover,
$2\mathcal{O}_{K} = \langle 2, 1+\sqrt{15}\rangle^{2}$
and
$3\mathcal{O}_{K} = \langle 3, \sqrt{15}\rangle^{2}.$
Resorting to the law of qu... | Starting from $(2, 1+\sqrt{15})(\alpha) = (3, \sqrt{15})(\beta)$, we can multiply both sides by $(3,\sqrt{15})$ to get:
$$(2, 1+\sqrt{15})(3,\sqrt{15}) = (3\beta/\alpha)$$
Multiplying the ideals on the left together
$$(6, 2\sqrt{15}, 3 + 3\sqrt{15}, 15 + \sqrt{15}) = (3\beta/\alpha)$$
So the ideal on the left side is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/92694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion? Is there a way to evaluate this limit:
$$\lim_{x \to 0} \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$
without using de l'Hôpital rule and series expansion?
Thank you,
| You should know the following limits:
$$\begin{split}\lim_{y\to 0} \frac{\sin y}{y} &= 1\\ \lim_{y\to 0} \frac{e^y-1}{y} &= 1\\ \lim_{y\to 0} \frac{\tan y}{ y} &= 1\\ \lim_{y\to 0} \frac{(1+y)^\theta -1}{y} &= \theta \qquad \text{(}\forall \theta \in \mathbb{R}\text{)}\\
\lim_{y\to 0} \frac{1-\cos y}{y^2} &= \frac{1}{... | {
"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": 0
} |
$\lfloor (2+\sqrt{3})^n \rfloor $ is odd
Let $n$ be a nonnegative integer. Show that $\lfloor (2+\sqrt{3})^n \rfloor $ is odd and that $2^{n+1}$ divides $\lfloor (1+\sqrt{3})^{2n} \rfloor+1 $.
My attempt:
$$ u_{n}=(2+\sqrt{3})^n+(2-\sqrt{3})^n=\sum_{k=0}^n{n \choose k}2^{n-k}(3^{k/2}+(-1)^k3^{k/2})\in\mathbb{2N} $$
$... | For the first claim:
Here is a more general discussion.
As is often the case,
none of this is original,
and I did work it out
on my own for fun.
To make one root of
$f(x)
=x^2-2ax+b
=0$
with
$a, b > 0$
in $(0, 1)$.
The roots are
$x_{\pm}
=\dfrac{2a\pm\sqrt{4a^2-4b}}{2}
=a\pm\sqrt{a^2-b}
$.
$\begin{array}\\
x_{-}
&=a-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
approximation formula for the integral Get an approximation formula for the following integral:
$$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
| $$
\sum_{k=1}^n \left( \frac{1}{35} \right)^{k-1}\int_0^{\frac{\pi}{2}}\cos^{2(n-k)+1}(y) \cdot \sin^{2(k-1)}(y) \, dy
$$
We have that
$$B(x,y) = 2 \int_0^{\pi/2} \sin^{2x-1} \theta \cos^{2y-1} \theta d\theta$$
So we can write your expression as
$$\sum\limits_{k = 1}^n {{{\left( {\frac{1}{{35}}} \right)}^{k - 1}}} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/94337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? How do I prove that $1$,$\sqrt{2},\sqrt{3}$ and $\sqrt{6}$ are linearly independent over $\mathbb{Q}$? $\mathbb{Q}$ is the rational field.
I want to know the detail about the proof. Thanks in advance.
Actually I know any two... | I have no doubt that many slick solutions will be given here. I'll try to post an elementary one; which uses perhaps the most straightforward approach I can imagine. (The only things that are needed are some algebraic manipulation and basic properties of rational numbers; such as that the only solution of $x^2=6y^2$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/96946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
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A version of Hardy's inequality involving reciprocals. How can one prove for any sequence of positive numbers $a_n, n\ge1,$ we have
$$\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}$$
Added later:
Apparently, this is a version of Hardy's inequality. The above is the case $p=-1$. ... | Here is a proof.
We try using induction, but as usual, a direct approach seems to fail and we have to try and prove a stronger statement.
So we try and pick a positive function $f(n)$ such that
$$ \frac{f(n)}{a_1 + a_2 + \dots + a_n} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/98483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
integral resulting in Bessel function Prove that
$$\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right)$$ where $J_{1}$ is the Bessel function of the first kind of order 1.
Some calculations I have done
$$\int_{0}^{\infty} \sin \left(x\right) ... | Let $f(a) = \int_0^\infty \sin(x) \sin\left(\frac{a}{x}\right) \mathrm{d} x$, where $a\in \mathbb{R}$. Without loss of generality we can assume $a > 0$.
Observe that
$$ \begin{eqnarray}
a f^{\prime\prime}(a) = \int_0^\infty \sin(x) \left(-\sin\left(\frac{a}{x} \right) \right) \frac{a}{x} \frac{\mathrm{d} x}{x} \st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?
I have reduced this problem to $$ 2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$
but now, evaluating this integral is giving me some p... | Let $u=\sqrt{\tan(x)}$. Then $u^2 = \tan(x)$ and $2 u \mathrm{d} u = (1+ \tan^2(x)) \mathrm{d} x$. Thus
$$
\int_0^{\frac{\pi}{2}} \sqrt{\tan(x)} \mathrm{d} x = \int_0^\infty \frac{2u^2}{1+u^4} \mathrm{d} u
$$
Since $1+u^4 = (1 + \sqrt{2} u + u^2)( 1- \sqrt{2} u + u^2)$, partial fraction decomposition applies:
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/100253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 8,
"answer_id": 5
} |
Help calculating the probability of dealing a custom deck of cards Imagine a deck that consists of two types of cards. There are 2 type A cards and 9 Type B cards, totaling 11 cards. Assuming that 1 card will be dealt to 5 players. What are the probabilities that Type A will be dealt zero, one, and two times?
I hope I'... | Note that three events $A_0$, $A_1$, and $A_2$ of players receiving 0, 1 or 2 cards of type A are exclusive and partition the entire event space (there are only 2 cards of type A). This means that
$$
\mathbb{P}(A_0) + \mathbb{P}(A_1) + \mathbb{P}(A_2) = 1
$$
There total $\binom{11}{5}$ ways to deal 1 card to eac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/102369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ Could someone please explain how to solve this:
$$x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$$
not the answer only, but a step-by-step solution.
I tried to solve it, with the help of Khan Academy, but still I have no idea how to correctly solve it.
Thank you so much in advance!
| The idea is to represent the polynomial $p(x) = x^4 - 10x^3 + 21x^2 + 40x - 100$ as a product of simple factors. By Vietta formula the free terms equals to the product of roots. Thus, if we confine ourselves to integer roots, they should divide $100$. Start with $x=2$, and verify that $p(2) = 16 - 10 \times 8 + 21 \tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Diagonalize a matrix We have a matrix $A =
\left(
\begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{array}
\right)$.
How do you find a diagonal matrix $D$ and an orthogonal matrix $P$ so that:
$D=P^tAP$?
| $A =
\left(
\begin{array}{ccc}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{array}
\right)$.
You know
$D =
\left(
\begin{array}{ccc}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{array}
\right)$.
Now,
$Ax = \lambda x$
x = 2 (Case 1)
$$Ax = 2x$$
$$Ax - 2x = 0$$
$$(A - 2I)x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/105643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
value of $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$? Is there anything known about the value of the series $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\frac{1}{1+2+3+4+5+6}+\cdots$ ?
| Yes, in fact, everything is known about it. Step 1: get a formula for the denominators (I assume the 6th one is supposed to have a 6 in it, not stop at 5 like the 5th one). Step 2: apply partial fractions to get a telescoping series. Enjoy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solving the exponential equation: $3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ I have this exponential equation that I don't know how to solve:
$3 \cdot 2^{2x+2} - 35 \cdot 6^x + 2 \cdot 9^{x+1} = 0$ with $x \in \mathbb{R}$
I tried to factor out a term, but it does not help. Also, I noticed that:
$2 \cdot 9^... | from user21436 you have
$$t=\dfrac{35s\pm\sqrt{(-35s)^2-4(12)(18s^2)}}{24}
=\dfrac{35s\pm\sqrt{s^2[(-35)^2-4(12)(18)]}}{24}
=\frac{35\pm 54}{24}s = \frac{2}{3}s\ or \frac{9}{4}s
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/108447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Number of 5 letter words over a 3 letter alphabet using each letter at least once This is similar to my previous question, Number of 5 letter words over a 4 letter group using each letter at least once.
The only difference is that there are 3 letters to choose from instead of 4.
However, I've run into a problem.
Using... | You forgot that you can have strings with two copies of one letter, two of another, and one of the third. There are $3$ ways to choose which of the letters is used only once, and $5$ ways to place it in the string. There are then $\binom42=6$ ways to choose which two of the remaining four places get the two copies of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/108994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the name for the polynomials of the form : $ P_n(x)=2^{-n} \cdot ((x+\sqrt {x^2-4})^n+ (x-\sqrt {x^2-4})^n)$? Polynomials of the form :
$ T_n(x) =2^{-1} \cdot ((x+\sqrt {x^2-1})^n+ (x-\sqrt {x^2-1})^n)$
are known as Chebyshev polynomials of the first kind .
Consider the polynomials of the form :
$P_n(x)=2^{-n... | They are called Pedja polynomials of the first kind. They satisfy $P_{n+1}(x) = x P_n(x) - P_{n-1}(x).$ The Pedja polynomials of the second kind are $$R_{-1}= 0, \; R_0(x) = 1, \; R_1(x)=x, \; R_2(x) = x^2 - 1$$ and solve $R_{n+1}(x) = x R_n(x) - R_{n-1}(x).$ Together they have
$$ P_n^2(x) - (x^2 - 4) R_{n-1}^2(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does $1+\frac{1}{1^k+2^k}+\frac{1}{1^k+2^k+3^k}+\frac{1}{1^k+2^k+3^k+4^k}+\frac{1}{1^k+2^k+3^k+4^k+5^k}\cdots$ telescope? after the post $1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+\cdots$, I had to ask
Can we telescope the series $$1+\frac{1}{1^k+2^k}+\frac{1}{1^k+2^k+3^k}+\frac{1}{1^k+2^k+... | This is not a perfect answer, but a hint. In sum of powers I've a (a bit old) treatize on the Eulerian numbers, which allow, in connection with the binomial-coefficients, to express the sums-of-like-powers as polynomials. Inserting these in the denominators of your sum should allow to do partial sums like in your linke... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/111611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve $\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$ I need to solve the limit
$$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$$
when $a,b,c \gt 0$.
I'm looking for ways to avoid $\frac{1}{x}$ power.
| HINT:
$$
\left( a^x + b^x - c^x \right)^\frac{1}{x} = a \left( 1 + \left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x \right)^\frac{1}{x} = a \left( 1 + x \cdot \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x} \right)^\frac{1}{x}
$$
Compute $\lim_{x \to 0} \frac{\left(\frac{b}{a}\right)^x - \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/111744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\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... | I understand that the point of this exercise is to apply the limit definition of the derivative to a function where the limit calculation is "tricky". But it's worth noting that if $F(x,y)=0$ identically (as in $y-\sqrt[3]{x}=0$ in this problem) then $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$.
So given that $x=y^3$, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
how can I write T in matrix form Let V be $ C[x]/(x^3+5x^2+6x+2) $ and let $ T:V→V $ be defined by
$T(p(x))=(x+1)p(x) $. I need to find the bases for the kernel (null space) and image (range) of T.
I have to write T in matrix form right? After that we can easily find the bases. How can we write it in matrix form? Al... | Here's a more naive approach, for when you don't see how the kernel factors: Using $\{1,x,x^2\}$ as an ordered basis, $1$ maps to $1+x$, $x$ maps to $x+x^2$, and $x^2$ maps to $x^2+x^3=x^2+x^3-x^3-5x^2-6x-2=-4x^2-6x-2$, so the matrix is: $$\begin{bmatrix} 1 & 0 & -2 \\ 1 & 1 & -6 \\ 0 & 1 & -4 \end{bmatrix}$$
The image... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$ Show
$$4 \cos^2{\frac{\pi}{5}} - 2 \cos{\frac{\pi}{5}} -1 = 0$$
The hint says "note $\sin{\frac{3\pi}{5}} = \sin{\frac{2\pi}{5}}$" and "use double/triple angle or otherwise"
So I have
$$4 \cos^2{\frac{\pi}{5}} - 2 (2 \cos^2{\frac{\pi}{10}} - 1) - 1$$
$$4 \... | You don't have to use the hint. Let $x = {\pi \over 5}$. Then $\cos(3x) = -\cos(2x)$ since $3x + 2x = \pi$. Since $\cos(3x) = 4\cos^3(x) - 3\cos(x)$ and $\cos(2x) = 2\cos^2(x) - 1$, you get
$$4\cos^3(x) - 3\cos(x) = - ( 2\cos^2(x) - 1)$$
This simplifies to
$$4\cos^3(x) + 2\cos^2(x) - 3\cos(x) - 1 = 0$$
This factors in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/113466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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A problem about parametric integral How to solve the following integral.
$I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx$ where $|\theta|<1$
| I shall find the integral by Feynman’s Technique Integration on a particular integral
$\displaystyle I(a)=\int_{0}^{\pi} \ln (a \cos x+1) d x,\tag*{} $
where $-1\leq a \leq 1.$
$\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\pi} \frac{\cos x}{a \cos x+1} d x, \\&=\frac{1}{a} \int_{0}^{\pi} \frac{(a \cos x+1)-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Writing $3\sin(x)+4\cos(x)$ in the form of $r \sin(x-a)$ I need to write $3\sin(x) + 4\cos(x)$ in the form of $r\sin(x-a)$.
Expanding $r\sin(x-a): r\sin(x)\cos(a)-r\sin(a)\cos(x)$
Comparing the two forms (The original equation and the expanded form): $3=r\cos(a)$ and $4=-r\sin(a)$.
Getting $r$:
$$\begin{align*}
3^2 + ... | $$\begin{align*}
f(x) &= 3 \sin(x) + 4 \cos(x) \\
f(x) &= 5\left(\frac{3}{5} \sin(x) + \frac{4}{5} \cos(x) \right)
\end{align*}$$
Let $\sin(y) = 4/5$ and $\cos(y) =3/5$, for some real number $y$.
$$\begin{align*}
f(x) &= 5\left(\cos(y)\sin(x) + \sin(y)\cos(x)\right)\\
f(x) &= 5\sin(x+y)&\text{(where }y = \arcsin(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
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 = ... | Added: It has been pointed out that there may be some ambiguity about the integral the OP is trying to find. The introduction has an $x^2$ at the bottom, but in the solution the OP seems to be using $x^3$. That is much easier. The substitution $x=3\sec\theta$ transforms the integrand into a constant times $\frac{\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Absolute maximum value of $\sin^2(x)-\sin(x)$ in $[0,\frac{3\pi}{2}]$ I thought it does not have absolute maximum, but wanted to just check and see why
| We are interested in $w^2-w$, with the proviso that $-1\le w\le 1$.
Completing the square, we find that
$4w^2-4w=(2w-1)^2+3$. So the minimum value of $4w^2-4w$ is $3$, reached when $w=1/2$. The maximum value in our interval is reached when $w$ is as negative as possible, namely at $w=-1$.
So the minimum value of $w^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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} = ... | you have concluded that for $x_n > 0$,$$x_{n+1} - \sqrt{2} = {{((x_n−\sqrt{2})^2)}\over{(2x_n)}}≥0. $$ This means $$x_{n+1} - \sqrt{2} ≥ 0$$ Now since n is arbitrary, any value of n will satisfy this equation , put $n = n -1$ in this equation, you will get
$$x_{n} - \sqrt{2} ≥0$$
| {
"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": 0
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How to change degree of elements in a field I'm working through an assignment and need to write several elements as polyomials of degree <= 2.
($x^2+1$)($x+1$)
within the field
$\mathbb Z$$_3$[x] / ($x^3 + 2x + 1$)
And am unsure how exactly to go about doing this - could anyone give me some pointers as to what the... | Divide the polynomial $(x^2+1)(x+1)$ (multiplied out) by the polynomial $x^3+2x+1$, remembering to work with the right field of coefficients. The remainder (or more properly, its equivalence class) is what you want.
How to do the division? Basically like ordinary polynomial division. $x^3+2x+1$ goes into $x^3+x^2+x+1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rearranging a formula, transpose for A2 - I'm lost Given the formula:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show ... | \begin{align}
q &= A_1\sqrt\frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} \\
\frac{q^2}{A_1^2} &= \frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} \\
\left(\frac{A_1}{A_2}\right)^2-1 &=\frac{2ghA_1^2}{q^2} \\
\frac{A_1}{A_2} &= \sqrt\frac{2ghA_1^2 + q^2}{q^2} \\
A_2 &= A_1\sqrt\frac{q^2}{2ghA_1^2 + q^2} \\
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Show $f(x) = x^3 - \sin^2{x} \tan{x} < 0$ on $(0, \frac{\pi}{2})$ This is the last of a homework problem set from Principles of Mathematical Analysis (Ch. 8 #18(a)) that I've been working/stuck on for a few days:
Define $f(x) = x^3 - \sin^2{x}\tan{x}.$
Find out whether it is positive or negative for all $x \in (0, \... | Power series approach works, though it requires some numerical computation. Since you seemed to be interested in asymptotics and approximations, you might find this interesting.
It is enough to show that
$$ \sin ^3 x \gt x^3 \cos x,\quad 0 \lt x \lt \frac{\pi}{2} $$
Now the power series for $\displaystyle \sin^3 x$ is
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/127021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
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