Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Recursive sequence depending on the parameter
For the given parameter $\mathbb R\ni t\geq 1$, the sequence is
defined recursively: $$a_1=t,\;\;a_{n+1}a_n=3a_n-2$$ $(a)$ Let $t=4$.
Prove the sequence $(a_n)$ converges and find its limit.
$(b)$ Which parameters $t\geq 1$ is the sequence $(a_n)$ increasing
for?
My... | If you have already proven $a_n > 2, \forall n$ , then
Let $b_n=a_n-2$, $b_n>0$. $a_{n+1} = 3 - \frac{2}{a_n} \Rightarrow b_{n+1} + 2=3-\frac{2}{b_n+2} \Rightarrow b_{n+1} = \frac{b_n}{2+b_n} < \frac{b_n}{2}$.
Therefore as $n\to \infty, b_n \to 0, a_n \to 2.\blacksquare$
The next is pure hindsight based on dan_fulea's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$ If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$
My attempt is as follows:-
$$a_0+a_1(x-2)+a_2(... | We have
\begin{eqnarray*}
\sum_{r=0}^{2n} (x-2)^r = \sum_{r=0}^{2n} b_r (x-3)^r.
\end{eqnarray*}
Expand the $(x-2)$
\begin{eqnarray*}
(x-2)^r=(x-3+1)^r= \sum_{s=0}^{r} \binom{r}{s} (x-3)^s.
\end{eqnarray*}
Invert the order of the plums
\begin{eqnarray*}
\sum_{r=0}^{2n} \sum_{s=0}^{r} \cdots = \sum_{s=0}^{2n} \sum_{r=s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ ... | Another method is $x^2-y^2=1995$ where $x,y \in I$ can be written as
$$(x+y)(x-y)=3×5×133$$
Hence $x+y=1995,665,399,133$ and $x-y=1,3,5,15$
Hence $(x,y)=(\pm 998,\pm 997),(\pm 334,\pm 331),(\pm 202,\pm 197),(\pm 74,\pm 59)$ where none is multiple of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Expectation and variance of the squared distance between $X$ and $Y$ Given that $X$ and $Y$ are two independent univariate random variables sampled uniformly from the unit interval [0,1]. I am trying to find the expected value and the variance of the random variable $Z = \mid X - Y \mid ^2 $ (the squared distance betw... | Variace is $E[(X-Y)^4] -[E[(X-Y)^2]]^2$
=$ E(X^4) + E(Y^4) + 6E(X^2)E(Y^2) -4E(X^3)E(Y) - 4E(X)E(Y^3) -\frac{1}{36}$
= $\frac{2}{5} + \frac{6}{9} - 1 - \frac{1}{36}$
=$\frac{7}{180}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to get expectation of a table $X$ and $Y$ are random variables with joint distribution table given as
$$\begin{array}{l|ccc}
& x=1 & x=2 & x=3 \\ \hline
y=1 & 3/12 & 1/12 & 3/12 \\
y=2 & 1/12 & 3/12 & 1/12
\end{array}$$
Calculate $\mathrm{Cov}(X, Y)$ and explain why $X$ and $Y$ are independent.
Attempt:
I k... | Your first intiution is correct, however the expected value of the random variable $X$ is given by the expected value of the marginal probability density function of X. It means that in order to get $E(X)$, you have to sum up on X, to get $E(Y)$ you have to do the same but on $Y$.
$$
\begin{array}{l|ccc}
& x=1 & x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integrate $\int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx.$ I am not able to solve the following integration, $$\int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx.$$ I have tried in different ways but most good one I think,
\begin{align*} \int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx &= \int_{0}^{\pi} \frac{\sin x}{\sin x (1+3^{\cos x})} \... | Just for the fun of it.
Dr Zafar Ahmed DSc provided the good solution.
For the fun of it, I did consider the more general problem of
$$\int_{a}^{\pi-a} \frac{dx}{1+k^{\cos x}} $$
Expanding the integrand as a Taylor series built around $x=\frac \pi 2$ we have
$$\frac{1}{1+k^{\cos x}}=\frac{1}{2}+\frac{\log (k)}{4} \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Find angle in triangle $ABC$ with cevian line $AD$, such that $AB=CD$.
As you can see in the picture, there is a triangle $ABC$ with $∠C=30°$ and $∠B=40°$. Now we assuming that $AB=CD$, try to find the exact value of $∠CAD$.
My attempt: Denote $∠CAD$ by $x$, we know that
$$\frac{\sin C}{AD}=\frac{\sin{x}}{CD},\quad\f... | An algebraic proof.
Consider the triangle $ACD$ and label $\varphi=\angle\, CAD$. Both $AC$ and $AD$ can be related to $AB$, i.e., $CD$, by the law of sines: $\frac{CD}{\sin \angle CAD}=\frac{AD}{\sin \angle ACD}=2AD$, thus $AD=\frac{AB}{2\sin \varphi}$. Also, $\frac{AC}{\sin \angle ABC}=\frac{AB}{\sin \angle ACB}=2AB$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Find the real value(s) solution of the equation $3^x = 3(x+6)$ I tried using Lambert W function but I got stuck while trying to set the question in the form $xe^x =y$.
| $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Consider
\begin{align}
3^x &= 3(x+6)
\tag{1}\label{1}
\end{align}
as a special case of
\begin{align}
a^x&=b\,x+c
,\quad a=3,b=3,c=18
\tag{2}\label{2}
\end{align}
and transform it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induct... | Let
$$a_n=\frac{x_n}{y_n}$$
Then
$$\frac{x_{n+1}}{y_{n+1}}=\frac{3x_n-y_n}{3y_n-x_n}$$
Assume
\begin{align}
&x_{n+1}=3x_n-y_n&
&y_{n+1}=3y_n-x_n
\end{align}
Then $y_n=3x_n-x_{n+1}$ from which $3x_{n+1}-x_{n+2}=3(3x_n-x_{n+1})-x_n$ which gives
$$x_{n+2}-6x_{n+1}+8x_n=0$$
This linear recurrence has general solution
$$x_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^... | $\frac{(x-8)(x-10)}{x-7}\ \frac{\ln(1+\frac{x}{7}-1)}{\frac{x}{7}-1}\times (\frac{x}{7}-1)$
as $\frac{x}{7}\to 1\Rightarrow\displaystyle\lim_{x\rightarrow7} \frac{\ln(1+\frac{x}{7}-1)}{\frac{x}{7}-1}=1$
So we are left with $\frac{(x-8)(x-10)}{7(x-7)}\times(x-7)$ can you do from here.
If you have to find limit of $1^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to compute an upper bound of $1 + \frac{1}{2^a} + \frac{1}{3^a} + \cdots$? Suppose $a>2$.
How to compute an upper bound of $1 + \frac{1}{2^a} + \frac{1}{3^a} + \cdots$ ?
Is $\frac{1}{a-2}$ an upper bound?
| When a = 2 we can say
$1 + \frac 14 (1 + \frac {4}{9}) + \frac {1}{16}( 1 + \frac {16}{25} + \frac {16}{36} + \frac {16}{49}) + \cdots < 1 + \frac 14 (2) + \frac 1{16} (4) + \cdots = 2$
when $a> 1,$
$\sum_\limits{n=0}^{\infty} \frac {1}{n^a} < \sum_\limits{n=0}^{\infty} (\frac {1}{2^{a-1}})^n = \frac {2^{a-1}}{2^{a-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Problem with computing $\int\frac{dx}{2x^2+5} $ by trigonometric substitution I so close to can solve this problem but I don't find the correct response:
$$\int\frac{dx}{2x^2+5}
$$
Always get the answer:
$$ \frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}}
$$
But the correct answer have more one square root multiplying the... | Setting $x\mapsto\ \sqrt{\frac{5}{2}}u$ yields:
$$\int_{ }^{ }\frac{dx}{2x^{2}+5}=\sqrt{\frac{1}{10}}\int_{ }^{ }\frac{1}{u^{2}+1}du=\sqrt{\frac{1}{10}}\arctan\left(u\right)+C=\sqrt{\frac{1}{10}}\arctan\left(\sqrt{\frac{2}{5}}x\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help... | Integrate as follows\begin{align}
& \int{\frac{x}{(2x^2-3) \sqrt{3x^2-2x+1}}}\ dx\\
=& \ \frac12\int \left(\frac{1}{2 x+\sqrt{6}}+\frac{1}{2 x-\sqrt{6}}\right)\frac1{\sqrt{3
x^2-2 x+1} }\ dx\>\>\>\>\>\>\>t_{\pm}=\frac1{2x\pm \sqrt6}\\
=& -\frac14\int \frac {1}{\sqrt{3-2(2+3\sqrt6)t_+ +2(11+2\sqrt6)t_+^2}}\ dt_+ \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Finding supremum & infimum of a given set
If exist, find, $\sup S,\inf S$\begin{equation}
S:=\left\{\left\lfloor\sqrt[|n|]{\frac{7}{3}}\right\rfloor\cdot
\frac{n^{2}-2 n-4}{n^{2}-n-6}: n \in
\mathbb{Z}\setminus\{-2,0,3\}\right\} \end{equation}
My attempt:
\begin{equation}f(n):=\left\lfloor\sqrt[|n|]{\frac{7}{3}}\r... | The points $n=0$, $n=-2$, and $n=3$ are excluded by definition. When $|n|\geq4$ the integer part of the square root is $1$ and the quotient is between $2/3 $ and $3/2$.
Besides the values you calculated, you have $f(-3)=11/6$. So the supremum is actually a maximum, and it is equals $11/6$. The infimum is a minimum and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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2018 MathCounts: Let $D(k)$ be the number of diagonals for a polygon with $k$ sides. If $D(m) + D(n) = 125$, what is the value of $m+n$? The question was asked in the 2018 Raytheon MATHCOUNTS National Competition. It appears in the video around the 35:34 mark:
https://www.youtube.com/watch?v=dSnOLW_W6og&t=2134
Let $D(... | Let $m=x-n$. Then,
$$\frac{n(n-3)}{2}+\frac{m(m-3)}{2}=125$$
is equivalent to
$$x^2-(2n+3)x+2n^2- 250 = 0$$
which implies
$$x=\frac{2n+3\pm\sqrt{1018-(2n-3)^2}}{2}$$
So, there has to be an integer $k$ such that $$1018=(2n-3)^2+k^2$$
Here, both $2n-3$ and $k$ are odd.
Since the rightmost digit of $(\text{odd})^2$ is ei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$.
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
We have th... | First of all, as you observed, the condition can be written as
$$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$$
Let's call the expression $S$. Write $S$ as:
$$\sum_{cyc}(x-1)\left(\frac{1}{y^2}+\frac{1}{x^2}\right) - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
Since $x,y,z > 1$, we can a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three-variable generating series for strings I need to find the generating series of strings from $C = \{a, b, c\}$ without including the strings $aac$ and $baa$. Also, all $b$ substrings must have an odd length (so $bbb, cbc, aaab$ are all valid but $bbcb$ is not) and the $a$ substrings must have a minimum length of 3... | Here's a derivation of the one-variable generating function $f(z,z,z)$. Maybe it will help you.
Let $a_n$, $b_n$, and $c_n$ be the number of strings that start with $a$, $b$, and $c$, respectively. Then $a_0=b_0=c_0=1$, $a_1=a_2=0$, and, by conditioning on the length $k$ of the current run, we see that
\begin{align}
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Distinct Spanning Trees Say that you are asked to produce a labelled graph $G$ that has $E$ edges and exactly $k$ distinct spanning trees. In general, how much do we know about whether such a graph exists? For example, a labelled graph with $6$ edges and exactly $7$ distinct spanning trees seems to be impossible to cre... | Take for example the graph $G=(V,E)$ where $V = \{a,b,c,d\}$ and $E = \{ab, bc, bd, cd\}$. Then the adjacency matrix is
$$
A = \left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 \\
\end{array}
\right)
$$
and the degree matrix
$$
D = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Which of these answers is the correct indefinite integral? (Using trig-substitution or $u$-substitution give different answers) Answers obtained from two online integral calculators:
$$\begin{align}\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= -\sqrt{\dfrac{x + 1}{1 - x}} + \sqrt{\dfrac{x + 1}{1 - x}}x - 2\arc... | Now, both answers are correct. They merely look different. They differ by a constant.
Note 1...
$$
-\sqrt{\frac{x+1}{1-x}}+\sqrt{\frac{x+1}{1-x}}\;x
=
-\sqrt{\frac{x+1}{1-x}}\;(1-x)
=
-\frac{\sqrt{x+1}\;(1-x)}{\sqrt{1-x}}
\\=
-\sqrt{x+1}\sqrt{1-x}
=-\sqrt{(1+x)(1-x)}
=-\sqrt{1-x^2}
$$
Note 2...
$$
2\,\arcsin \left( \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2... | A Third (Magical) Solution by Cornel Ioan Valean
This time we assume we have in hand that $\displaystyle \int_0^1 \frac{\arctan^2(x)}{x}\log\left(\frac{x}{(1-x)^2}\right)=G^2$, which is calculated extremely easily at this link. Furthermore, we also assume we have at our disposal the value of the powerful harmonic seri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Prove equilateral triangle if vertices $a,b,c$ satisfy $|a|=|b|=|c|$ and $\sum_{cyc}\frac1{8-a/b-b/a-a/c-c/a}=0.3$
If $a,b,c\in \mathbb{C}$ are different complex numbers in pairs and $|a|=|b|=|c|=1$ and $P(a),Q(b),R(c)$ are the vertices of a triangle and $\displaystyle \sum_{\bf{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}... | Apply the Jensen's inequality
$$f(x_1) + f(x_2) + f(x_3) \ge 3f\left(\frac{x_1+x_2+x_3}3\right)$$
for the convex function $f(x)=\frac1x$ to get
$$0.3=\sum_{{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}
\ge \frac{1}{24-2\left(\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\right)}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \qua... | Note that, if $\cos x >0$, $x$ is in the 1st or 4th quadrant; if $\cos x< 0$, $x$ is in the 2nd or 3rd quadrant.
So, given $3\theta \in [-3\pi, 3\pi]$ and $\cos 3\theta = -\frac12< 0$, the angle $3\theta$ has two sets of values, with one set in the 2nd quadrant, i.e.
$$3\theta =\frac{2\pi}3 + 2\pi k = -\frac{4\pi}3 , ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$ How can one prove
$$
\sum_{k = 1}^{n}\frac{\left(-1\right)^{k - 1}\, 16^{k}\,
\left(k - 1\right)!\, k!\, \left(k + n - 1\right)!}
{\left[\left(2k\right)!\right]^{\, 2}\,\left(n - k\right)!} =
\f... | Remarking that
\begin{equation}
\sum_{k=1}^n\sin\left( \left( 2k-1 \right)x \right)=\frac{\sin^2nx}{\sin x}
\end{equation}
the proposed integral
\begin{align}
I_n&= \frac{2}{n}\int_0^{\frac{\pi }{2}}\frac{1-\cos (2 n x)}{\sin (x)} \, dx\\
&= \frac{4}{n}\int_0^{\frac{\pi }{2}}\frac{\sin^2nx}{\sin x},dx\\
&=\frac{4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving $\frac{5}{8} \cot36^\circ = \cos^3x$ without substituting the trig values for $36^\circ$
Find the value of $x$ such that
$$\frac{5}{8} \cot36^\circ = \cos^3x$$
The answer is $x=18^\circ$.
It's really messy to plug in the standard values of $\cos36^\circ$, $\sin36^\circ$ and miraculously guess a suitable ... | Yet an other way to proceed, using algebraic only manipulations after a quick reshape...
We want to show
$$
\frac 58\cdot \frac{\cos 36^\circ}{\sin 36^\circ}
\overset{!}=\sin^3 72^\circ \ .
$$
Let us denote by $c,s$ respectively the values for $\cos 36^\circ$ and $\sin 36^\circ$. Then we want:
$$
\frac 58\cdot\frac cs ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What does this series converge to? $\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$ What does the following expression converge to?
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$$
(It looks like this problem)
$\displaystyle\sum_{k=0}^n a_{2k+1}+a_{2k+2}-a_{k+1}$ ;
$\di... | $$\mathop {\lim }\limits_{n \to + \infty } \sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$
$$=\mathop {\lim }\limits_{n \to + \infty } \sum_{k=0}^n\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\mathop {\lim }\limits_{n \to + \infty } \sum_{k=0}^n\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.
Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i ... | Counterexample: For $t=8, n = 5$ and $(a_1,a_2,a_3,a_4,a_5)=(1,5,7,9,10)$ we have:
$$2^8=1^2+5^2+7^2+9^2+10^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}... | Since
$$
\frac{1}
{{k^2 }} < \int\limits_{k - 1}^k {\frac{1}
{{x^2 }}} dx\,\,\,\,\,\,\forall k \geqslant 2
$$
we have that
$$
\sum\limits_{k = n}^{3n - 1} {\frac{1}
{{k^2 }} < } \int\limits_{n - 1}^{3n - 1} {\frac{1}
{{x^2 }}} dx = \left[ { - \frac{1}
{x}} \right]_{n - 1}^{3n - 1} = \frac{1}
{{n - 1}} - \frac{1}
{{3n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$
I tried to prove it by ind... | It is trivial if you use $$\sum_{n=1}^\infty\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$$
Note that $x_1=1$ and $x_i\ge\dfrac{5}{4}\forall i\ge 2$. Hence, for $n\ge 4$,
$$LHS = \sum_{i=1}^n\dfrac{1}{i^2x_i^2}$$
$$\le1+\dfrac{16}{25}\sum_{i=2}^n\dfrac{1}{i^2}$$
$$<1+\dfrac{16}{25}\left(\dfrac{\pi^2}{6}-1\right)$$
$$<1.42<RHS$$
(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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May I know is there quick way to factorize multi-variables equations? I want to factorize such equation:
$$b^2c + bc^2 +a^2c+ac^2+a^2b+ab^2 +2abc$$ into product of linear factors.
May I know is there any quick way/trick to do so?I am very confused with such equation.
Hello users, my main point here is what to think w... | There is a nice article on this, Gary Brookfield (2016) Factoring Forms, The American Mathematical Monthly,
volume 123 number 4, pages 347-362. The main theorem was proved by Aronhold in 1849.
The general theorem, easy to state, is that a homogeneous cubic factors completely over the complexes if and only if the deter... | {
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"source": "stackexchange",
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Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get
$$t^2-\frac{3+i... | I guess the following helps you more:
$$\cos z=\cos(x+iy)=\cos x\cosh y-i\sin x \sinh y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve $y''+y=p(x)$ where $p$ is a polynomial function The equation is $$y'' + y=p(x)$$ Let $p(x)=\sum\limits_{k=0}^n a_k x^k$
The homogeneous solution to the equation is $y_h = a\cos(x)+b\sin(x)$. I can look for a particular solution of the form $y_p=\sum\limits_{k=0}^n b_k x^k$.
We get $$\sum\limits_{k=0}^{n-2} ((k+... | As you stated, you have $b_n=a_n$ and $b_{n-1}=a_{n-1}$. You also have that, for $0 \le k \le n - 2$ that
$$(k+2)(k+1)b_{k+2}+b_k=a_k \implies b_k = a_k - (k+2)(k+1)b_{k+2} \tag{1}\label{eq1A}$$
Thus, you can start from $k = n - 2$ and work down to $k = 0$ to get each value of $b_k$ in turn. For example with the next $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569599",
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"source": "stackexchange",
"question_score": "2",
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On various angular velocities of a straight line $y$ = $\frac{a}{100}\times x$, with $a$ varying from $-1000$ to $+1000$. Let a straight line be defined by $y$ = $\frac{a}{100}\times x$.
Let $a$ vary from $-1000$ to $+1000$
Desmos : https://www.desmos.com/calculator/xs2sowsaal
When $a$ is big ( in absolute value) imp... | Consider the circunference $\gamma:\:x^2+y^2=1$ and let $A(x_A,y_A)$ the intersection of $\gamma$ and $t$ where $t: \: y=\frac{a}{10^2}x$.
Now $(x_A,y_A)$ are the solution to: $$\left\{\begin{matrix}
x^2+y^2=1
\\ y=\frac{a}{10^2}x
\end{matrix}\right.$$
So, I have: $$\left\{\begin{matrix}
x=\sqrt{\frac{10^4}{10^4+a^2}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something ... | There's a remarkable identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
Applying this to $x=\sqrt[3]a,y=\sqrt[3]b,z=\sqrt[3]c$ gives that
$$a+b+c-3\sqrt[3]{abc}=0\implies a+b+c=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Egyptian fraction representation of $1$ where all denominators of the fractions are odd. Question: Is there an Egyptian fraction representation for $1$ where all the fractions have odd denominators?
I tried to generate one below:
$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{23}+\... | $$\frac 13+\frac 15+\frac 17+\frac 19+\frac{1}{15}+\frac{1}{19}+\frac{1}{21}+\frac{1}{25}+\frac{1}{173}+\frac{1}{1294257}+\frac{1}{2233466948475}$$ is a solution
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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prove that $\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)} \ge \frac{27}{(a+b+c)^2}$
Prove that $$\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)} \ge \frac{27}{(a+b+c)^2},$$ where $a,b,c$ are positive reals.
After applying AM-GM I got
$$
\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)}
\ge \frac{36}{(a+b+c)^... | By Holder $$\sum_{cyc}\frac{2}{b(a+b)}=\frac{1}{(a+b+c)^2}\sum_{cyc}b\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{b(a+b)}\geq\frac{27}{(a+b+c)^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a sequence is a Cauchy sequence using definition I am intending to show that the sequence $<f_n>$ where $f_n$ = $ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}........+\frac{(-1)^{n-1}}{n}$ is a Cauchy sequence using the definition of a Cauchy sequence.
My attempt: Let $\epsilon>0$ and $n>m$
$|f_n-f_m|=|\frac{(-1)^m}{... | If $(n-m)$ is odd, we may assume $m$ is even without loss of generality, since $\left\lvert k \right\rvert = \left\lvert -k \right\rvert$. Then \begin{align*}\left\lvert f_n - f_m\right\rvert &= \left\lvert \frac{1}{m+1} - \frac{1}{m+2} + \frac{1}{m+3} - \frac{1}{m+4} + \dotsb + \frac{1}{n-2} - \frac{1}{n-1} + \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double integral over square shaped region $$\iint_R(y-2x^2)dxdy$$ where $R$ is the region inside the square $|x|+|y|=1$.
So the area is:
\begin{align}
4×\left[\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}(y-2x^2)dydx\right]
&=4×\left[\int_0^1\int_{y=0}^{y=1-x}ydydx-2\int_0^1\int_{y=0}^{y=1-x}x^2dydx\right]\\
&=4×\left[\frac{1}{2}... | First, let's break it into two integrals:
$$
\underbrace{\iint_R y - 2x^2 \ \ dxdy}_{I} = \underbrace{\iint_R y \ \ dxdy}_{I_1} -2 \underbrace{\iint x^2\ \ dxdy}_{I_2}
$$
*
*As $y$ is an odd function, then the integral over the upper part is the negative of lower part:
\begin{align*}
I_{1} & = \iint_{R} y \ \ dxd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex number proof involving angles I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).
This is what I have done:
Expand (3 + i)^3
(3 + i)^3 = (3 + i)(3 + i)(3 + i)
= 9 + 3... | It is actually easier to use polar co-ordinates on this one.
Let's say that $B$ had the polar co-ordinates $(r_1, \theta_1)$ and $C$ had $(r_2, \theta_2)$. Then by the multiplication all you have to show is this:
$r_2=r_1^3 \implies r_2^2=r_1^6 \ldots (1)$
$\theta_2=3\theta_1 \ldots (2)$
Calculating $r_2$ would be the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$. Given $f=x^3+9x^2+24x-40n^3+40xn^2+94n^2-12x^2n-62nx-74n+20$ has real roots, show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$.
I tried to do it by directly finding the roots in wolfram alpha for an... | Your polynomial equation is
$$\begin{equation}\begin{aligned}
f(x)&=x^3+9x^2+24x-40n^3+40xn^2\\
& \; \; \; +94n^2-12x^2n-62nx-74n+20
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note you have
$$\begin{equation}\begin{aligned}
f(5n) & = (5n)^3 + 9(5n)^2 + 24(5n) - 40n^3 \\
& \; \; \; \; \; + 40(5n)n^2 + 94n^2 - 12(5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6... | $$x = \frac{-3 - 2 \sqrt{6} \pm \sqrt{12 \sqrt{6} - 3}}{6}$$
There are $2$ more complex roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Proving that $\lim_{(x;y)\rightarrow (0;0)} \frac{xy(x-y)}{x^4+y^4} = 0$ using the squeeze theorem My attempt:
$$0 \le \frac{|xy(x-y)|}{|x^4+y^4|} = \frac{|x^2y-xy^2|}{|x^4+y^4|} \le \frac{|x^2y|+|xy^2|}{x^4+y^4} \le \frac{|x|^2|y|+|x||y|^2}{x^4+y^4} = \frac{|x|^2|y|+|x||y|^2}{(x^2+y^2)^2-(\sqrt{2}xy)^2} = ?$$
What do ... | The main reason you're having trouble is the limit doesn't exist. For example, if you approach $(0,0)$ along the line $x = 2y$, you have
$$\begin{equation}\begin{aligned}
\frac{xy(x-y)}{x^4+y^4} & = \frac{(2y)y(2y-y)}{(2y)^4+y^4} \\
& = \frac{2y^3}{17y^4} \\
& = \frac{2}{17y}
\end{aligned}\end{equation}\tag{1}\label{eq... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2... | In cases like $\,x^4+2x^3y+3x^2y^2+2xy^3+y^4\,$ where the difference between consecutive coefficients is "simpler" than the original sequence, it is worth trying to multiply by $\,x-y\,$:
$$
(x-y)(x^4+2x^3y+3x^2y^2+2xy^3+y^4) \,=\, x^5 + x^4 y + x^3 y^2 - x^2 y^3 - x y^4 - y^5
$$
The same pattern of "simpler" differenc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Finding the Laurent series without contour integration I would like to expand the following series around $x=0$. I know the answer already (one way is by simply plugging it into Mathematica) but I would like to understand how to get to the result.
$$\frac{1}{\sqrt{1+x} - 1} = \frac{2}{x} + \frac{1}{2} - \frac{x}{8} + \... | How about using the expansion (for $|x| < 1$)
$$
\sqrt{1+x} = 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6)
$$
Hence
$$
\frac{x}{\sqrt{1+x} - 1} = \frac{1}{1/2 - x/8 + x^2/16 - (5 x^3)/128 + (7 x^4)/256 + O(x^5)}
$$
which can be expanded
$$
\frac{x}{\sqrt{1+x} - 1} =2 + x/2 - x^2/8 + x^3/16 - (5 x^4)/1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that when $p$ varies, the midpoint of $PR$ lies on the curve $y^2=2x-4$. The line passing through the point $P$ ($p^2,2p$) on the curve $y^2=4x$ and the point $Q$(2,0) intersects the curve once again at point $R$, find the coordinates of point $R$ in terms of $p$.
I am able to solve this part of the question and ... | The midpoint of $PR$ in terms of $p$ is:
$$\left(\frac{p^2+4/p^2}{2}, \frac{2p-4/p}{2} \right)$$
and you can surely show that $\left(\frac{2p-4/p}{2} \right)^2 = 2 \left( \frac{p^2+4/p^2}{2} \right)-4$, right?
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c$ be side lengths of a triangle, $a+b+c=1$. Prove that $P=a^3+b^3+c^3+3abc<\frac{1}{4}$.
Let $a,b,c$ be side lengths of a triangle such that $a+b+c=1$. Prove that $$a^3+b^3+c^3+3abc<\frac{1}{4}\,.$$
I solved this question. However, I'd like to know if there's a neater solution that doesn't involve the subst... | Because it is a somewhat rare occurrence that the inequality could be derived easily/naively from the equality case, I'm posting this solution which is essentially the same as Michael's written backwards.
Observe that when $ a = b = 0.25, c = 0.5$, we get equality.
This suggests that we need to look at the triangle... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Choosing numbers without replacement What is the probability of choosing 4 numbers from a bag of numbers from 1-10 without replacement, so that the smallest number choosen is 4?
I was thinking $\left(\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}\cdot\frac{4}{7}\right)$.
Is it correct or am I wrong?
Thank you in advanc... | After $4$ is fixed, the remaining three numbers have to be from $5$ to $10$ inclusive. So the number of draws whose smallest number is $4$ is the number of draws from $5$ to $10$, or $\displaystyle\binom{6}{3}=20$. Then the desired probability follows as $\frac{20}{\displaystyle\binom{10}{4}}=\frac{20}{210}=\frac2{21}$... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$ Is the limit $0$ or $3$?
$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$?
Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?
| Use $$\frac{x^3-1}{x-1}=x^2+x+1$$ for $x\neq1$.
Also, by the definition of the limit if $x$ is closed to $1$, so $x\neq1$.
| {
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Verifying Stoke's Theorem using path integration along a parametrized ellipse
Verify the Stoke's Theorem by showing that the line integral
$$ \int\limits_{C} -y^3 dx + x^3 dy - zdz = \frac{3\pi}{2} \tag{1},$$
through direct computation, where $C$ is the intersection of the cylinder $x^2 + y^2 =1$ and the plane $x+y+z ... | Assuming that the $ z$ in $ \vec F$ is a typo, found in an exercise of integral transformation, I consider $ \vec F=-y^3\hat i + x^3\hat j - z^3\hat k$
Stoke's theorem tells us that $ \int_C \vec F.d\vec{r}=\iint_S (\vec\nabla\times\vec F) d\vec S$ where $ C$ is the curve bounded by the region $ S$.
So $ \vec\nabla\... | {
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Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem... | $$\begin{align}
&\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
=&\int^{\infty}_{0}\frac{2dx}{x^2+\frac{a^2b^2}{x^2}+(a^2+b^2)}\\
=& \int^{\infty}_{0}\frac{d(x+\frac{ab }x)}{(x+\frac{ab}{x})^2+(a-b)^2}
+ \int^{\infty}_{0}\frac{d(x-\frac{ab }x)}{(x-\frac{ab}{x})^2+(a+b)^2}\\
=& \int^{\infty}_{\infty}\frac{dt}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+... | If $x,y,z>0$, then the sides of any triangle are : Then $$a=y+z, b=x+z,c=x+y \implies b+c-a=2x, c+a-b=2y, a+b-c=2x$$
Then the LHS of the required inequality becomes:
$$\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)\ge \frac{1}... | {
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How to find/prove the maximum of a symmetric homogeneous function involving sides of a triangle? Let a, b and c be the side lengths of a triangle. Consider the following function
$$ f(a,b,c)=\frac{\sqrt{b\,c\,(a+c-b)(a+b-c)}}{a\,(a+b+c)} + \frac{\sqrt{a\,c\,(b+c-a)(a+b-c)}}{b\,(a+b+c)} + \frac{\sqrt{a\,b\,(b+c-a)(a+c-... | By AM-GM, we have $\sqrt{(a+c-b)(a+b-c)} \le \frac{a+c-b + a+b-c}{2} = a$
(or, $(a+c-b)(a+b-c) = a^2 - (b-c)^2 \le a^2$).
So, the first term is less than or equal to $\frac{\sqrt{bc}}{a+b+c}$.
Similarly, the second term is less than or equal to $\frac{\sqrt{ac}}{a+b+c}$,
and the third term is less than or equal to $\fr... | {
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How to determine the convergence of the series $\sum_{n=1}^\infty\frac{e^nn!}{n^n}$? $$\sum_{n=1}^\infty\frac{e^nn!}{n^n}$$
we cant use the ratio test nor the root test, because the limit will be 1.
the main question was to determine whether or not the series
$$\sum_{n=1}^\infty\frac{a^nn!}{n^n}$$
converge, for every ... | Without Stirling's formula
When the ratio test is inconclusive, we can try Gauss's test. (An extension of the ratio test.)
Let $u_n > 0$ and suppose
$$
\frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(n^{-r}), \qquad h \in \mathbb R,r>1
$$
If $h>1$, then $\sum u_n$ converges; if $h \le 1$ then $\sum u_n$ diverges.
In ou... | {
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How to find the solution to a multi-inifite summation? I'm familiar with techniques to solve a single infinite sum such as:
$$
\sum_{i=1}^{\infty} (1 + i) \cdot \frac{1}{2^i}
$$
Which ends up being equal to $3$, but I'm having trouble figuring out how to tackle a multi summation such as this:
$$
\sum_{i=0}^\infty \sum_... | A key to calculate this double sum lies in the fact that for absolutely convergent series $\sum_{i=0}^{\infty}a_i$ and $\sum_{j=0}^{\infty}b_j$ we have
$$\left(\sum_{i=0}^{\infty}a_i\right)\left(\sum_{j=0}^{\infty}b_j\right)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_j = \sum_{j=0}^{\infty}\sum_{i=0}^{\infty}a_ib_j$$
N... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$... | This problem is equivalent to proving that there exists some real number $r$ such that
\begin{equation}
\lim_{n \to \infty} f_n(r) = 0
\end{equation}
where $f_n = g_n^0(x)$ and $g^k_n$ is defined by the recurrence
\begin{equation}
g_n^k(x) =
\begin{cases} \sqrt {x+k} & \text{if }n=1 \\
\sqrt{x + k + g^{k+1}_{n-1}... | {
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"source": "stackexchange",
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If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$? Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.
If $a+b+c=0$, then $a^3+b^3+c^3 = \ldots $
A. $\;0\quad$ B. $\;1\quad$ C. $\;a^3b^3c^3\quad$ D. $\;3abc$
Source: 4/12/2020, Competitiv... | Let $a=1$, $b=1$, $c=-2$. So $a+b+c=0$ and $a^3+b^3+c^3=-6$.
Therefore, choices (1), (2), and (3) (whose values are $0$, $1$, and $-8$, respectively) are wrong. But choice (4) has $3abc=-6$. Therefore choice (4) is the correct answer.
| {
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Inequality $a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$ I found a problem where now I essentially have to show the following inequality
$$a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$$ where $a,b,c$ are the sides of a triangle.
I have tried a lot of approaches but am not making headway.
Any hint in the right direction w... | The inequality to be proven is equivalent to
$$\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}}\leq 1\,.$$
By the Weighted AM-GM Inequality,
$$\begin{align}&\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{b+c-a}{b}\right... | {
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Integral with negative radicand $ \int \frac{dx}{\sqrt{-a^2-x^2}}$ I saw this integral:
$$
\int \frac{dx}{\sqrt{-a^2-x^2}}.
$$
I tried to solve it by taking $-1$ as common factor which leaded me to:
$\frac{1}{i\sqrt{x^2+a^2}}$ which leads to:
$-i\ln(\sqrt{x^2+a^2}+x)$
However I placed it on Wolfram Alpha and it says th... | Note $\arctan (i z) = i \text{arctanh(}z) =i \ln \sqrt{\frac{1+z}{1-z}}$.
Thus
$$
\arctan\left(\frac{x}{\sqrt{-a^2-x^2}}\right)
=\arctan\left(-i \frac{x}{\sqrt{a^2+x^2}}\right)\\
= -i \ln \sqrt{\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}-x}}
= -i \ln {\frac{\sqrt{a^2+x^2}+x}{a^2}}\\
\hspace{-4cm}=-i\ln(\sqrt{x^2+a^2}+x)+C$$... | {
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"source": "stackexchange",
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A series of multiplication leads to $\frac{1}{2} = 2$ I'm presented with the equation $\frac{a+b}{a} = \frac{b}{a+b}$
Performing cross multiplication yields $a^2+2ab+b^2 = ab$
Subtracting $ab$ from both sides, we get $a^2+ab+b^2 = 0$
Multiplying both sides by (a-b) and simplifying:
$(a-b)(a^2+ab+b^2) = 0 * (a-b)$
$a^3... | You multiplied with $0=a-b$. That is wrong. You are just recovering this solution. Just to give you a similar example, say you have $$x-1=0$$
If I multiply both sides by $x+1$, I get $$x^2-1=0$$ or $$x=\pm 1$$Obviously $x=-1$ is not a solution. You introduced it when you multiplied with $x+1$.
| {
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Generating function expansion I am working on a generating function for a sequence problem and I am stuck with expanding.
I have
$$\frac{1}{4(1-x)^2} - \frac{1}{4(x+1)^2} $$
From my notes, the first term will be $\sum_{n=0}^{\infty}\frac{1}{4}(n+1)x^n.$
I am stuck on the second term where the denominator is $(x+1)^2$ ... | You can observe
$\frac{1}{4(1-x)^2} - \frac{1}{4(x+1)^2}=\frac{x}{(1-x^2)^2}=\frac{x}{1-x^2}\sum_{n=0}^\infty x^{2n}$
$=\sum_{n=0}^\infty \frac{x^{2n+1}}{1-x^2}$
| {
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Solve $x^4y^{\prime\prime} = (y-xy^\prime)^3, y(1) = y^\prime(1) = 1$ $\lambda^4x^4\lambda^{n-2}y^{\prime\prime} = (\lambda^ny-\lambda x\lambda^{n-1}y^\prime)^3 \Rightarrow \lambda^{n+2}x^4y^{\prime\prime} = (\lambda^n(y-xy^\prime))^3$.
$\lambda^{n+2} = \lambda^{3n} \Rightarrow n+2 = 3n \Rightarrow n = 1$.
Let $x = e^... | Since $x=e^t$, $x=1$ exactly when $t=0$. So, since $y=ue^{t}$, the boundary condition $y(1)=1$ translates into $u(0)e^0=1$ which means $u(0)=1$.
Then use the fact that $y'=u'+u$. The boundary condition $y'(1)=1$ means $u'(0)+u(0)=1$. Since we already have that $u(0)=1$, this means $u'(0)=0$.
You have already partially ... | {
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Stuck when transforming and solving this Given abc=1 ( all positive real numbers). Prove that:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$
My attempt:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b ... | The equality $a^2b+b^2c+c^2a\geq a+b+c$ we can prove by AM-GM:
$$\sum_{cyc}a^2b=\frac{1}{3}\sum_{cyc}(2a^2b+c^2a)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{(a^2b)^2c^2a}=$$
$$=\sum_{cyc}\sqrt[3]{a^5b^2c^2}=\sum_{cyc}a.$$
Another way.
Let $a=\frac{x}{y}$ and $b=\frac {y}{z},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac... | {
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Prove $\log_{a} c + \log_{b} c = \log_{a+b} c$ if and only if $1 + \log_{b} a = \log_{a+b} a$ If a, b and c are positive numbers, than equality $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ is true if and only if $$1 + \log_{b} a = \log_{a+b} a$$
Prove it!
I have looked at the solution but it is not clear for me.
We will ... | WE have:
$$1+ log_b (a)= log_{a+b} (a) $$
Let $a=c$ then:
$$ log _a(c)= log _c (c)=1$$
therefore:
$$log _a (c) + log _b (c)= log _{a+b} (c)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.
| by using the long division
$$\frac{6x^3-x^2-3x+2010}{1}=\frac{6x^3-x^2-3x+2010}{3x^2+x}=(3x^2+x)(2x-1)-\frac{2x-2010}{3x^2+x}$$
$$=(1)(2x-1)-\frac{2x-2010}{1}=2009$$
| {
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"source": "stackexchange",
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Calculating matrix determinant This is a very easy determinant to calculate, but I get two different results when I calculate it in two different ways.
\begin{equation}
A = \begin{bmatrix} 1 & 2 & 3 \\
0 & 1 & 2 \\
2 & 2 & 4
\end{bmatrix}
\end{equation}
When I used Laplace expansion right away I got:
\begin{equation}
\... | The operation $r_1 \leftarrow 2r_1 - r_3$ does not preserve the determinant. Indeed, if you check the rules, the determinant is preserved if you subtract a multiple of a row from the actual other row (and not a multiple of it).
Suppose you did $r_1 \leftarrow r_1 - r_3$, then did Laplace expansion, things would work. S... | {
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Prove and give geometric meaning of $|\sqrt{a^2+b^2} - \sqrt{a^2+c^2} | < |b-c| $
Carefully, prove that
$$ | \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
and give a geometric interpretation.
pf
We observe that $| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | = \dfrac{ |b^2 - c^2 | }{\sqrt{a^2+b^2} + \sqrt{b^2+c^2} } \leq \dfr... | As pointed out in the comments, the points $(0,0), (a,b), (a,c)$ form a triangle. Using the triangle inequality,
$$\sqrt{a^2 + b^2} + |b-c| \gt \sqrt{a^2 + c^2}$$
and
$$\sqrt{a^2 +c^2} + |b-c| \gt \sqrt{a^2 +b^2}$$
$$\implies \left|\sqrt{a^2+b^2}-\sqrt{a^2+c^2} \right| \lt |b-c|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt fo... | Hint:
$$5\cdot\dfrac{5^{\tan^2x}-1}{\tan^2x}\cdot\dfrac{\tan^2x}{1-\cos^2x}=5\cdot\dfrac{5^{\tan^2x}-1}{\tan^2x}\cdot\dfrac1{\cos^2x}$$
Now as $a=e^{\ln a}$ for $a>0$
$$\lim_{h\to0}\dfrac{a^h-1}h=\cdots=\ln a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Primes dividing an integer polynomial $x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8$ are congruent to 1 mod $4$ I need to prove that if $p\neq 2$ divides the integer polynomial
$$x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8\in\mathbb{Z}[x,y],$$ $gcd(x,y)=1,$ then $p\equiv 1\pmod{4}.$ Any ideas?
Edit (the OPs reaction to the hint t... | A proof can be based on the following
Fact. If an odd prime $p$ is a factor of the sum $x^2+y^2$ such that $x$ and $y$ are not multiples of $p$, then $p\equiv1\pmod4$.
Proof. Without loss of generality $p\nmid y$. Therefore we can divide the congruence
$$
x^2\equiv -y^2\pmod p
$$
by $y^2$, and conclude that there exist... | {
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Square root of prime is irrational. Is this a valid proof? I know similar questions exist, but I want to know, if this is a valid proof.
A prime number has 1 and it self as divisors. So the subset of a prime number is $ D(p) = \{1, p\}$
Now I want to prove $\sqrt{p} = \frac{a}{b}$ , to get the opposite.
Also $gcd(a,b)... | We have that $\sqrt{p}=\dfrac{a}{b}$ with $a$ and $b$ being relatively prime integers. Thus, $p=\dfrac{a^2}{b^2}$ and so $a^2=p\cdot b^2$.
$a^2$ must be divisible by $p$ because $p, a^2,$ and $b^2$ are all integers and so $a$ must be divisible by $p$ as well. If we define $a=pc$ where $c$ is also an integer, then $p=\d... | {
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"timestamp": "2023-03-29T00:00:00",
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Integral $\lim _{n\to \infty}\int _0^1\sqrt{\frac{1}{x}+n^2x^{2n}}\,dx$ Evaluate $\displaystyle \lim \limits _{n\to \infty}\int \limits _0^1\sqrt{\frac{1}{x}+n^2x^{2n}}\,dx$.
The integral is non asymptotic. The convergence is non uniform at both $x=0$ and $x=1$. I'm not sure how to proceed.
| We have
\begin{align}
I_n &= \int_0^1 \sqrt{\frac{1}{x} + n^2 x^{2n}} \mathrm{d} x \\
&\le \int_0^1 \left(\sqrt{\frac{1}{x}} + nx^n\right) \mathrm{d} x \\
&\le \int_0^1 \left(\sqrt{\frac{1}{x}} + (n+1)x^n\right) \mathrm{d} x \\
&= \left(2\sqrt{x} + x^{n+1}\right)\Big\vert_0^1 \\
&= 3.
\end{align}
Also, we have
\begin{a... | {
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"answer_id": 0
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Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.
I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate... | Strigthforward approach:
First we combine $\frac{1}{(\omega^{k}-1)^3}$ with $\frac{1}{(\omega^{15-k}-1)^3}$ to get rid of $i$ in the denominator.
$$\sum\limits_{k=1}^{14}\frac{1}{(\omega^k-1)^3}=
\sum\limits_{k=1}^{7}\frac{(\omega^{15-k}-1)^3+(\omega^{k}-1)^3}
{\left((\omega^k-1)(\omega^{15-k}-1)\right)^3}=$$
$$\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How do I show that $\sum_{\text{cyc}}\frac{3a^2-2ab+3b^2}{(a+b)^2} < \frac{9}{4}(\frac{a}{c} + \frac{c}{a}) - \frac{3}{2}$? Let the real numbers $a,b,c \in \mathbb(0, \infty\ )$, $a\leq b\leq c$. Prove that $$\frac{3a^2-2ab+3b^2}{(a+b)^2}+\frac{3b^2-2bc+3c^2}{(b+c)^2}+ \frac{3c^2-2ac+3a^2}{(c+a)^2} \leq\frac{9}{4}(\fra... | By your work $$\sum_{cyc}\frac{3a^2-2ab+3b^2}{(a+b)^2}\leq\sum_{cyc}\left(\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2}\right)\leq\frac{9}{4}\left(\frac{a}{c}+\frac{c}{a}\right)-\frac{3}{2}$$
because $$\frac{a}{b}+\frac{b}{a}\leq\frac{a}{c}+\frac{c}{a}$$ (it's just $bc-a^2\geq0$) and
$$\frac{b}{c}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate logarithmic integral $\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ I need help to evaluate the integral with the residue theorem:
$$\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$$
where a,b>0 real numbers.
I think I could consider the contour integral where C is the half circle in the... | Let $ a,b $ be reals such that $ 0<b<a $, we have the following :
\begin{aligned}\int_{0}^{+\infty}{\frac{\ln{\left(1+a^{2}x^{2}\right)}}{1+b^{2}x^{2}}\,\mathrm{d}x}&=\int_{0}^{+\infty}{\int_{0}^{1}{\frac{a^{2}x^{2}}{\left(1+b^{2}x^{2}\right)\left(1+a^{2}x^{2}y\right)}\mathrm{d}y}\,\mathrm{d}x}\\&=\int_{0}^{+\infty}{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $2z+p^2+qy+2y^2=0$ using Charpit's method Show by solving $2z+p^2+qy+2y^2=0$ using Charpit's method that $y^2[(x-a)^2+y^2+2z]=b$.
My efforts :
The given equation is
\begin{equation}
F=2z+p^2+qy+2y^2=0.
\end{equation}
Charpit's equation:
Charpit's auxiliary equations are
$ \dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+... | $$2z+p^2+qy+2y^2=0 \tag 1$$
You found :
$$z=-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2+d \tag 2$$
In order to check your result, put your result (2) into equation (1) :
$p=a-x$
$q=\frac{b}{y^3}$
$$2\left(-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2+d\right)+(a-x)^2+\frac{b}{y^3}y+2y^2=2d$$
This is not $=0$. So your resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factorizing symmetric expressions Factorize $$\sum_{\text{cyc}}a^4(b-c)$$ I used cyclic/symmetric method to find that $(a-b), (b-c), (c-a)$ are factors of the above equation but I am not able to completely factorize this. I used long division method but I found it quite lengthy so is there any shorter method than to ap... | $$\sum_{cyc}a^4(b-c)=a^4(b-c)+bc(b^3-c^3)-a(b^4-c^4)=$$
$$=(b-c)(a^4+b^3c+b^2c^2+bc^3-a(b^3+b^2c+bc^2+c^3))=$$
$$=(b-c)(a(a^3-b^3)+b^2c(b-a)+bc^2(b-a)+c^3(b-a))=$$
$$=(b-c)(a-b)(a^3+a^2b+ab^2-bc^2-b^2c-c^3)=$$
$$=(b-c)(a-b)((a-c)(a^2+ac+c^2)+b(a-c)(a+c)+(a-c)b^2)=$$
$$=(a-b)(a-c)(b-c)(a^2+b^2+c^2+ab+ac+bc).$$
Another w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving for an unknown $c$ in relation to a ratio of gamma functions I have been working with ratios of gamma functions and I am surprised how difficult it is to make even elementary conclusions. I am hoping it is just the learning curve.
Consider the following problem. Find a positive real $c$ for an integer $x > 1$... | One may use Newton's method to solve $\ln \Gamma(2x+3-c) = 2\ln x + \ln \Gamma(2x+1)$ to find $c$ (see [2]):
Choose the initial $c_0$,
$$c_{k+1} = c_k - \frac{\ln \Gamma(2x+3-c_k) - 2\ln x - \ln \Gamma(2x+1)}{-\Psi(2x+3-c_k)}, \ k=0, 1, 2, \cdots$$
where $\Psi(x) = (\ln \Gamma(x))'$ is the digamma function.
(Remark: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
On the nested square roots $\sqrt{1^2+\sqrt{2^2+\sqrt{3^2 ...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}$ I saw this on quora:
Use induction to show that
$\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}
\le 2
$.
My question is simpler.
If
$x_n
=\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}
$,
how can ... | [Not intended as a complete answer. It doesn't relate $x_n$ and $x_{n+1}$ as requested by OP, in part because that's not the "right" thing to look at.]
It is often harder to deal with inserting the variable just at the very extreme end of the nested roots. This makes intuitive sense in part because the function is s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the limit of $ \frac{\tan^2x-x^2}{x^2\tan^2x} $ How to find the limit of $ \frac{\tan^2x - x^2}{x^2\tan^2x}$ as $x$ approaches $0$.
I don't know how to approach and starting differentiating the numerator and denominator many times which after the 4th time gave me the result $\frac{2}{3}$. I want to know if ther... | Note that as $x \to 0$, we have $\tan(x) = x + \dfrac{x^3}{3} + O(x^5)$ (this is really a 2 line calculation). So,
\begin{align}
\tan^2(x) &= x^2 + \dfrac{2x^4}{3} + O(x^6)
\end{align}
Hence,
\begin{align}
\dfrac{\tan^2x - x^2}{x^2 \tan^2x} &= \dfrac{\frac{2x^4}{3} + O(x^6)}{x^4 + O(x^6)} \\
&= \dfrac{\frac{2}{3} + O(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the value of $\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}$? Let $x,y,z$ are real numbers such that $x,y,z \geq -1$, $xyz\not= 0$ and
\begin{matrix} \sqrt{x+1} + \sqrt{y+2} +\sqrt{z+3}= \sqrt{y+1} +\sqrt{z+2} + \sqrt{x+3} \\ \sqrt{y+1} + \sqrt{z+2} +\sqrt{x+3} = \sqrt{z+1}+\sqrt{x+2} + \sqrt{y+3} \end{matrix}
What... | The only answer that will satisfy the two equations involving square roots is that $$x=y=z$$. $$x$$,$$y$$,and$$z$$ can be any real numbers between -1 and 0 and above 0. If $$x=y=z$$, $$x-y=0$$, $$y-z=0$$, and $$z-x=0$$. So the result will be $$0/z+0/x+0/y$$ and since $$x,y, and z$$ do not equal zero. The resulting equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $|R_{n,0}| \leq \frac{|x|^{n+1}}{(n+1)(x+1)}$
The problem is: Prove that if $-1 < x \leq 0$, the the Taylor's remainder $R_{n,0}$ corresponding to $f(x) = \log(x+1)$ satisfies $$\displaystyle |R_{n,0}| \leq \frac{|x|^{n+1}}{(n+1)(x+1)} $$
First, I know that Taylor's remainder formula is $$\displaystyle R_{... | Here is an easy way
to get the series
for
$\ln(1-t)$.
$\sum_{k=0}^{n-1} t^k
=\dfrac{1-t^n}{1-t}
$
so
$\dfrac1{1-t}
=\sum_{k=0}^{n-1}t^k+\dfrac{t^n}{1+t}
$.
Integrating from
$0$ to $x$,
$\begin{array}\\
\int_0^x\dfrac1{1-t}dt
&=\sum_{k=0}^{n-1}\int_0^x t^kdx+\int_0^x \dfrac{t^n}{1-t}dt\\
\text{or}\\
-\ln(1-x)
&=\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$ I have to find the sum of the given series $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdot... | Let $s_n$ be the sum of the first $n$ terms of the series. It is clear that if $s_{3n}$ converges, then the series converges to the same sum. $$\begin{align}
s_{3n}&=\sum_{k=1}^{2n}\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}\\
&=\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^{n}\frac1{2k}\\
&=\sum_{k=1}^{4n}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$ I need to calculate the integral
$$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$$
being $\Gamma=S_1\cap S_2$, given:
*
*$S_1=\{(x,y,z)\in\mathbb{R}:2x+2y+z=3\}$
*$S_2=\{(x,y,z)\in\mathbb{R}:z=9-x^2-y^2\}$
In my problem i'm asked to solve it us... | Direct approach:
Let's study the intersection by equating the $z$:
$$3-2x-2y=9-x^2-y^2$$
$$x^2+y^2-2x-2y=9-3=6$$
$$(x-1)^2+(y-1)^2=(2\sqrt2)^2$$
Hence I would use the parametrization $x=1+2\sqrt2\cos \theta, y = 1+2\sqrt2\sin \theta, z=-1-4\sqrt2\sin \theta - 4\sqrt2 \cos \theta.$
$$\frac{dx}{d\theta} = -2\sqrt2 \sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by T... | The ratio test works fine using your way
$$a_n=\cos ^{n^3}\left(\frac{1}{n}\right)\implies \log(a_n)=n^3\log\left(\cos \left(\frac{1}{n}\right) \right)$$
$$\cos \left(\frac{1}{n}\right)=1-\frac{1}{2 n^2}+\frac{1}{24 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(\cos \left(\frac{1}{n}\right) \right)=-\frac{1}{2 n^2}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find $n$ and $m$, if $n,m$ are natural numbers, such that $m^6 + 279 = 2^n$. Suppose $n, m \in \mathbb{N}$. If $m^{6}+279=2^{n}$, find $n, m$.
What are some ways to approach this question? Instinct told me to take logarithm but doesn't work to well.
| Suppose that $n$ is odd. Then
$$2=2*1^k\equiv 2*4^k=2^{2k+1}=2^n=m^6+279\equiv m^6=m^2*m^2*m^2(\text{mod }3)$$
If $m$ is divisible by $3$, then this is $0$ modulo $3$. If not, then
$$\equiv 1*1*1=1\ (\text{mod }3)$$
Since this is a contradiction either way, we conclude $n$ is even. Then we can simplify the equation to
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
calculate $\sum_{n=0}^\infty \frac{3^n}{n!(n+3)}$ using power series let $f(x)=\frac{e^x-1-x-\frac{x^2}{2}}{x}$, because $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, $f$ can be expressed as $$f(x) = \frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x-\frac{x^2}{2}}{x}=\frac{\sum_{n=3}^\infty \frac{x^n}{n!}}{x}=\sum_{n=0}^\infty \f... | Try using the following instead so you avoid that mess:
$$x^2e^x=\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$$
$$\int x^2e^x \; dx=x^2e^x-2xe^x+2e^x+C=\sum_{n=0}^{\infty} \frac{x^{n+3}}{n!(n+3)}$$
At $x=3$:
$$9e^3-6e^3+2e^3-2=5e^3-2=\sum_{n=0}^{\infty} \frac{3^{n+3}}{n!(n+3)}$$
Notice that the series on the right is what y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..]
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse
Let the point $(h,k)$ lie ... | According to $\frac{hx}{4}+\frac{ky}{9}=1$, the horizontal and vertical lines corresponding to circular points $(h,k) = (0,10),\> (10,0)$ are $ y= \frac9{10}$, $ x = \frac4{10}$. which also corresponds to the elliptical axes $a= \frac4{10}$ and $b= \frac9{10}$. Thus, the equation of the standard ellipse is
$$\frac{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\mathrm{tr}(AX^{-1}A'X)$ Given $A=\begin{bmatrix}
a_1 & & \\
& a_2 & \\
&&a_2
\end{bmatrix}$ and $B\in\mathbb{R}^{3\times2}$, where $a_1,a_2>1$, and $X=AX(I+BB'X)^{-1}A$.
1) Find $\mathrm{tr}(AX^{-1}AX)$ in terms of $a_i$.
2) Show that $\frac{x_1\cdot\begin{vmatrix}
x_2 & x_{23}\\
x_{23}&x_3
\end{vmatrix}}{|X|... | Assume that $\det X \ne 0$.
From $X = AX(I + BB^\mathsf{T}X)^{-1}A$,
we have $I + BB^\mathsf{T}X = AX^{-1}AX$
and $BB^\mathsf{T} = AX^{-1}A - X^{-1}$.
Let $Y = X^{-1}$. We have $BB^\mathsf{T} = AYA - Y$.
Let
$$Y=\begin{bmatrix}
y_1 & y_{12} & y_{13}\\
y_{12} & y_2 & y_{23}\\
y_{13}&y_{23}&y_3
\end{bmatrix}, \quad
B =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$ Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$$
Well from their s... | Credits to Calvin Lin for the hint to apply Titu's Lemma, but I don't find the remaining part of the proof obvious.
Applying Titu's lemma yields:
\begin{align*}
\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} &= \frac{a^4}{ab+ac}+\frac{b^4}{bc+bd}+\frac{c^4}{cd+ca}+\frac{d^4}{da+db} \\
&\geq \frac{(a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compositeness test using $S_k=2S_{k-1}-3S_{k-2}+2S_{k-3}$ recurrence relation Can you prove or disprove the following claim:
Let $S_k=2S_{k-1}-3S_{k-2}+2S_{k-3}$ with $S_0=0$ , $S_1=1$ , $S_2=1$ . Let $n$ be an odd natural number greater than $2$. Let $\left(\frac{D}{n}\right)$ be the Jacobi symbol where $D$ represent... | The roots of $x^3-2x^2+3x-2=0$ are ${1, \frac{1+\sqrt{-7}}{2}, \frac{1-\sqrt{-7}}{2}}$
Let $\alpha = \frac{1+\sqrt{-7}}{2},\quad \beta = \frac{1-\sqrt{-7}}{2}$
Thus, the formula can be written as $S_n = A\cdot1^n + B\cdot\alpha^n + C\cdot\beta^n$ with $A,B,C.$
We can determine (A,B,C) using $S_0,S_1,S_2$ below
$A = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I solve $f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$? $$f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$$
I have been up on this problem for an hour, but without any clues.
Can someone please help me solving this?
| It is quite straight forward after rewriting the integrand:
$$\frac{\cos x (1+4\cos 2x )}{\sin x (1+4\cos^2 x)}= \frac{\cos x}{\sin x}\left(1 - \frac{4\sin^2 x}{1+4\cos^2 x}\right)$$
$$= \frac{\cos x}{\sin x} - \frac{2\sin 2x}{3+2\cos 2x}$$
Note, that $(3+2\cos 2x)' = -4\sin 2x$ and integrate:
$$\int \frac{\cos{x}(1+4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Sum of Squares for $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$ $\color{red}{\textrm{Update}}$
To clarify: I hope to see a (simple) SOS solution similar to the one of the following 3-variable problem, no other assumptions e.g. $a\ge b\ge c\ge d$.
Let us see an example. In the link below, by letting $y = \mathrm{mid}... | A very nice SOS solution by KaiRain@AoPS:
We have
\begin{align*}
&a^2+b^2+c^2+d^2+abcd+ 1 - ( ab+bc+cd+da + ac+bd)\\
=\,& \frac{\left(2a^2-2ab+2b^2-ac-ad-bc-bd+2 \right)^2+3 \left(a-b \right)^2 \left(c-d \right)^2 }{4 \left(a^2-ab+b^2+1 \right)}\\
&\quad + \frac{abcd \left(a-b \right)^2+cd \left(ab-1 \right)^2 +\left(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$? Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $... | Let $R$ be the ring generated by the symmetric polynomials and $f=a^2b+b^2c+c^2a$.
Certainly $f$ also contains $g=ab^2+bc^2+ca^2$ since $f+g$ is symmetric.
Let
$$u_{m,n}=a^mb^n+b^mc^n+c^ma^n.$$
To prove that $R$ contains all cyclic polynomials,
it suffices to prove that $u_{m,n}\in R$ for all $m$ and $n$. Indeed it
suf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\int_{0}^{\pi/2} (\frac{1}{\sin^3{\theta}} - \frac{1}{\sin^2{\theta}})^{1/4} \cos{\theta} d\theta = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$ Well, I have shown that $B(n, n+1) = \frac{(\Gamma(n))^2}{2\sqrt{2n}}$
From there I could deduce that $B(1/4, 5/4) = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$, then $n=1/4$.
I a... | Since you already received a good answer, another possible solution using the substitution proposed by @gt6989b in comments.
$$\int \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2 \sqrt[4]{(1-u) u}+2 \sqrt[4]{u} \,\,\, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};u\right)$$
$$\int_0^1 \left(\frac1{u^3} - \frac1{u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.
| Primitive solutions to $a^2+ab+b^2=c^2.$
Let $x=a/c, y=b/c.$
$x^2+xy+y^2=1\tag{1}$
Substitute $x = t , y=1+kt$ to equation $(1)$, then we get $$t = -\frac{(1+2k)}{(1+k+k^2)}$$
Thus, we get a parametric solution.
$(x,y,z) = (-\frac{(1+2k)}{(1+k+k^2)}, -\frac{(k-1)(k+1)}{(1+k+k^2)})$
Hence $(a,b,c)=(1+2k, k^2-1, 1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$ how should I procced from here having the equation $$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$$
I am getting to the following $$\frac{1}{2^{\frac{3}{2}}}\ln \left(y+\sqrt{2}\right)-\frac{1}{2^{\frac{3}{2}}}\ln \left(y-\sqrt{2}\right)=\frac{x^3}{3}+4x+c$$ from here ... | $$\frac{y+\sqrt{2}}{y-\sqrt{2}}=\exp\left[2^{3/2}\left(\frac{x^3}{3}+4x+c\right)\right]$$
to make a few steps easier I will call the RHS $f(x)$:
$$y+\sqrt{2}=(y-\sqrt{2})f(x)$$
$$y(1-f(x))=-\sqrt{2}(1+f(x))$$
and so:
$$y=-\sqrt{2}\frac{1+f(x)}{1-f(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Question about area of circle word problem Picture of problem
Please refer to the picture of the problem
I am trying to figure out: What percent of the total area is worth ten points?
So this is my attempt: the area of the 10 point region is pi * (4)^2 = 50.24
the total radius is 4+ 3+3+3 = 13
so the total area is 13^... | there's three areas worth $10$ points.
The bullseye has area $\pi r_1^2 = \pi 4^2 = 16\pi$.
The area of the entire dart board has area $\pi r_2^2 = \pi (4+3*3)^2= 13^2\pi =169\pi$
The area of the three inner rings is $\pi r_3^2 = \pi (4+2*3)=10^2\pi =100\pi$.
So the area the $4$th ring only is the total area of the dar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Base conversion problems Let $b$ be an integer greater than 2, and let $N_b = 1_b + 2_b + \cdots + 100_b$ (the sum contains all valid base $b$ numbers up to $100_b$). Compute the number of values of $b$ for which the sum of the squares of the base $b$ digits of $N_b$ is at most 512.
Since we are adding up to $100$ in b... | Note that from your formula, we can deduce that $N_b = \frac{10100_b}{ 2}$. Let us carry out the division in base $b$, distinguishing two cases:
*
*If $b$ is even, then we can note that since
$$N_b = \frac{10100_b}{ 2} = \frac{10000_b}{2} + \frac{100_b}{2} = \frac{b^4}{2} + \frac{b^2}{2} = \frac{b}{2}(b^3) + \frac{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Probability question on Properties of Expectation Problem
There are $x$ red and $y$ blue candies in a jar. A child randomly eats one candy everyday. If a child eats a red candy, the child will put a blue candy into the jar.
If a child eats a blue candy, the child doesn't put another candy into the jar.
Let $X$ denote ... | The expected value of $B$, the number of remaining blue candies just after all red candies are replaced, if we start with $x$ red candies and $y$ blue candies appears to be
$$\mathbb{E}_{x,y}[B] = \frac{y}{x+1} + H_x \approx \frac{y}{x + 1} + \log(x)$$
where $H_x$, the $x$th harmonic number, is equal to $\sum_{i = 1}^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can I solve this definite integral: $\int_{0}^{a}\frac{x^4dx}{\sqrt{a^2-x^2}}$
Evaluate $$\int_{0}^{a}\dfrac{x^4dx}{\sqrt{a^2-x^2}}$$
I tried taking $t$ as
$$t = \sqrt{a^2-x^2}$$
Thus my final integral became
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt$$
but I couldn't go any further in solving this integral.
I also tried by... | There are already 4 solutions. instead, I am going to evaluate the general integral $$I(k)=\int_{0}^{a} \frac{x^{k}}{\sqrt{a^{2}-x^{2}}} d x, \quad \textrm{ where }k=0,1,2, \cdots$$
by establishing a reduction formula on $k$.
$$
\displaystyle \begin{aligned}
I(k) &=\int_{0}^{a} \frac{x^{k}}{\sqrt{a^{2}-x^{2}}} d x \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2... | We have$:$ \begin{align*} \text{LHS}-\text{RHS}\\&=\frac{1}{3} \sum\limits_{cyc} \left( 2\,{a}^{4}+10\,{a}^{2}{b}^{2}+2\,{a}^{2}bc+5\,{a}^{2}{c}^
{2}+2\,a{b}^{3}+4\,ab{c}^{2}+{b}^{4}+4\,{b}^{3}c+24\,{c}^{4} \right)
\left( a-b \right) ^{2}\end{align*}
Another solution.
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ We have$:$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx$ A physics problem says that the following integral is said to evaluate to: $$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx = \frac{2x+3}{8}\sqrt{x^2-x+1} + \frac{1}{16}\ln(2x+1+2\sqrt{x^2-x+1})$$
I can not seem to find a way that evaluates to this. I know that a integral of this for... | Equate numerator as $x^2=A(x^2-x+1)+B(2x-1)+C$
which gives $A=1, B=\frac12$ & $C=-\frac12$. Now proceed as follows
$$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx$$
$$=\int \frac{(x^2-x+1)+\frac12(2x-1)-\frac12}{\sqrt{x^2-x+1}} dx$$
$$=\int\left( \sqrt{x^2-x+1} +\frac12\frac{(2x-1)}{\sqrt{x^2-x+1}}-\frac12\frac{1}{\sqrt{x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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.