Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Asymptotic of integral I have this definite integral:
$$ \int_{0}^{\frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta , t\in (0,1)$$
where $n$ is an integer. I need to find the asymptotic as a function on $n$.
I suspect it should be $O(\frac{1}{\sqrt{n}})$ but wasn't able to complete the calcu... | One may write:
$$
\begin{align}
I(n)=\int_{0}^{\large \frac \pi 2} \left( 1
+ t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta&=\int_{0}^{\large \frac \pi 2} \left( 1 + t^2 \tan^2\theta
\right )^{-\frac{n-1}{2}} d \theta \\\\
&=\int_{0}^{\infty}
\frac{1}{1+x^2}\left( 1 + t^2 x^2 \right )^{-\frac{n-1}{2}} dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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What values of x does $\sum_{n=1}^{\infty} \frac{1}{n} (\frac{3}{(-1)^n+2})^n x^n$ converge for? In a past year paper there was a question asking to find the values of x in which $\sum_{n=1}^{\infty} \frac{1}{n} (\frac{3}{(-1)^n+2})^n x^n$ converges.
I tried using the root test, and so:
$\lim_{n \rightarrow \infty} (\... | First note that the coefficients of this power series is
$$a_n=\frac{1}{n}\left(\frac{3}{(-1)^n+2}\right)^n=\begin{cases}\frac{3^n}{n},&n\text{ odd}\\
\frac{1}{n},&n\text{ even}\end{cases}.$$
Then for sufficiently large $n$
$$\frac{3}{n^\frac{1}{n}}\leq \sup_{N\geq n}|a_N|^{\frac{1}{N}}\leq 3.$$
Thus the radius of conv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show $\frac{2x}{2+x}<\log(x+1)$
How to show $\frac{2x}{2+x}<\log(x+1)$ for $x>0$
Without differentiating, more elementary (it looks then more complicated but OK)
$\log(x+1)=\int\limits_1^{x+1}\frac 1udu$ and $\frac{2x}{2+x}=x\frac{1}{ 1+\frac x2}$ hence
$\log(x+1)-\frac{2x}{2+x}=\int\limits_1^{x+1}\frac 1u-\f... | Note that
$$ \frac{4}{(x+2)^2}-\frac{1}{x+1}=-\frac{x^2}{(x+1)(x+2)^2}<0$$
and hence
$$ \frac{4}{(x+2)^2}<\frac{1}{x+1}. $$
Integrating from $0$ to $x$ ($x>0$), one has
$$ \int_0^x\frac{4}{(t+2)^2}dt<\int_0^x\frac{1}{t+1}dt $$
which gives
$$ \frac{2x}{2+x}<\ln(x+1). $$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that ${4\over \pi}=\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}$ How do I prove this infinite product?
$${4\over \pi}=\prod_{k=1}^{\infty}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}$$
I try:
$$\prod_{k=1}^{\infty}\left(1+{1... | Hint. Let's set
$$
P_n:=\prod_{k=1}^{n}\left(1+{1\over 4k}\right)^2\left(2k+1\over 2k+1+(-1)^{k-1}\right)^{(-1)^{k-1}}.
$$ One may observe that
$$
P_{2n+1}=\left(1+{1\over 4(2n+1)}\right)^2\left(4n+3\over 4n+4\right)P_{2n}, \quad n\ge0,
$$ since
$$
\lim_{n \to \infty}\left(1+{1\over 4(2n+1)}\right)^2\left(4n+3\over 4... | {
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"url": "https://math.stackexchange.com/questions/2067363",
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Prob. 4, Chap. 3 in Baby Rudin: How to show that these are the limit superior and the limit inferior? Here's Prob. 4, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Find the upper and lower limit of the sequence $\left\{ s_n \right\}$ defined by $$s_1 = 0; \ s_{2m} = \frac{s_{2m-... | Hint. Show inductively that
$$
s_{2n}=\frac{1}{2}-\frac{1}{2^n}\quad\text{and}\quad s_{2n+1}=1-\frac{1}{2^n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$ without using L'Hospital We have to find the following limit.
$$\lim_{x\to0} \frac 8 {x^8} \left[ 1 - \cos\frac{x^2} 2 - \cos\frac{x^2}4 + \cos\frac{x^2}2\cos\frac{x^2}4 \right]$$
In this I thou... | Beside the good hint by lab bhattacharjee, as Henry W commented, Taylor series make the problem quite simple.
Starting with $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ and replacing $y$ successively by $\frac {x^2} 2$ and $\frac {x^2} 4$ $$\cos \left(\frac{x^2}{2}\right)=1-\frac{x^4}{8}+\frac{x^8}{384}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determining the number of ways a number can be written as sum of three squares I was going through Erich Friedman's "What's Special About This Number?" and there some numbers are classified based on the number of ways we can write them as sum of squares. I want to prove the following claim by Friedman:
129 is the smal... | For the system of equations.
$$x_1^2+x_2^2+x_3^2=x_4^2+x_5^2+x_6^2=x_7^2+x_8^2+x_9^2=x_{10}^2+x_{11}^2+x_{12}^2$$
Solutions can be parameterized.
$$x_1=a(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_2=b(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_3=c(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$
$$x_4=(ay^2-2byn+az^2-2czn-... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
| $10 \mid (a^2+ab+b^2)$ implies $10 \mid a^3-b^3 = (a-b)(a^2+ab+b^2)$, that is, $a^3 \equiv b^3 \bmod 10$.
Now, $x \mapsto x^3$ is a bijection mod $10$. Therefore, $a^3 \equiv b^3 \bmod 10$ implies $a \equiv b \bmod 10$.
Then $a^2+ab+b^2 \equiv 3a^2$ implies $10 \mid a$ and so $10 \mid b$.
Therefore, $1000 \mid a^3$ and... | {
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Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ I have got an olympiad problem which is as follow:
Compute $\frac{p(12)+p(-8)}{10}$ where $p(x)=x^4+ax^3+bx^2+cx+d$ and $p(1)=10$, $p(2)=20$, $p(3)=30$.
I have been told that answer is $1984$.
I thought applying the values and getting a relation between... | Let $q(x) = p(x)-10x$. Then, $q(x)$ is a monic polynomial, and $q(1) = q(2) = q(3) = 0$.
So the roots of $q(x)$ are $x = 1, 2, 3$ and $r$ for some real number $r$.
Hence, we can write $q(x) = (x-1)(x-2)(x-3)(x-r)$.
Thus, $p(x) = (x-1)(x-2)(x-3)(x-r)+10x$ for some real number $r$.
So, $p(12) = 11 \cdot 10 \cdot 9 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071259",
"timestamp": "2023-03-29T00:00:00",
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Show that:$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(\cdots\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{\phi\sqrt{5}}$ $\phi$ is the golden ratio
Show that
$$\prod_{n=1}^{\infty}\left({2n\over 2n-1}\right)^2\left(10n-6\over 10n-1\right)\left(10n-4\over 10n+1\right)={\pi\over2}\cdot{\phi\over5}\cdot\sqrt{... | Rewrite this as the product of $$\frac{(2n)^2}{(2n-1)(2n+1)} \cdot \frac{(10n)^2}{(10n-1)(10n+1)} \cdot \frac{(10n/4-1)(10n/4+1)}{(10n/4)^2} \cdot \frac{(2n+1)(10n-6)}{(2n-1)(10n+4)}.$$ Using the infinite product representation of sine, you can show that $$\frac{\pi}{m \sin (\pi / m)} = \prod_{n=1}^{\infty} \frac{(mn)^... | {
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"url": "https://math.stackexchange.com/questions/2072839",
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In how many $4$-digit numbers the sum of two right digits is equal to the sum of two left digits In how many $4$-digit numbers the sum of two right digits is equal to the sum of two left digits.
My attempt:We should find number of two pairs that can be digits of this number for choosing the place of digits we have $*8$... | This problem can be solved using Stars and Bars method.
Let the digits be $a,b,c,d$.
Then we must find integral solutions of $$a+b=c+d \Rightarrow a+b-c-d=0$$
with restrictions on $a,b,c,d\ $ i.e $1\le a\le 9$ and $0\le \{b,c,d\}\le 9$
The equation can be transformed by taking $a=x_1+1,\ b=x_2,\ c=9-x_3,\ d=9-x_4$ to:
... | {
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"url": "https://math.stackexchange.com/questions/2075664",
"timestamp": "2023-03-29T00:00:00",
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Finding the splitting field of the polynomial $x^5+2x^4+5x^2+x+4$ over $F_{11}$. I am working on the following problem:
Find the splitting field of the polynomial $f(x)=x^5+2x^4+5x^2+x+4$ over $F_{11}$.
What I have done:
So far I found that $-2$ is the only root of $f(x)$ in $F_{11}$. I have also factored $f(x)$ as $f(... | The polynomial $x^2+5x+1$ has no roots in $F_{11}$: indeed
$$
x^2+5x+1=x^2-6x+9-8=(x-3)^2-8
$$
and $8$ is not a square modulo $11$.
Similarly, $x^2-5x+2=x^2+6x+9-7=(x+3)^2-7$ and $7$ is not a square modulo $11$.
Adding a root of $x^2+5x+1$ is the same as adding a square root of $2$. The elements are of the form $a+b\sq... | {
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Finding necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$ My question is that:
Find the necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$.
I know that answer is $k=-3$, as I know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
But I can't pr... | The condition necessary and sufficient for a polynomial $f(x)$ to be divisible by $(x-a)$ is that $f(a)=0$.
Let, $$f(x)=x^3+y^3+z^3+kxyz$$
For this polynomial to be divisible by $x+y+z$, it is necessary and sufficient that $f(-y-z)=0$.
However,
$$f(-y-z)=-(y+z)^3-kyz(y+z)+y^3+z^3=-(k+3)yz(y+z)=0$$
Simplifying it gives ... | {
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If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.
| The expression $ a^2 + b^2 - ab = c^2 $ is equivalent to
$$
a^2 + b^2 - ab = c^2 \longleftrightarrow a^2 - ab = c^2 - b^2 \longleftrightarrow a(a - b) = (c - b)(c + b)
$$
Now we multiply by $ b - c $ both sides (if $ b - c = 0 $ then there is nothing to prove):
$$
a(a - b)(b - c) = (c - b)(c + b)(b - c) = -(c - b)^2 (... | {
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$x + \frac1y = y + \frac1z = z + \frac1x$, then value of $xyz$ is? If $x,y,z$ are distinct positive numbers, such that $$x + \frac1y = y + \frac1z = z + \frac1x $$ then value of $xyz$ is?
$$A)\ 4\quad B)\ 3\quad C)\ 2\quad D)\ 1$$
My attempt:
1.I equaled the equation to '$k$'. Using the AM-GM inequality, I found tha... | Other answers have shown that the question is flawed. For fun, let's remove the condition that $x,y,z$ are positive, but keep the requirement that $x,y,z$ are distinct, and try to find all solutions $(x,y,z)$.
Trivially, $x,y,z \neq 0$. If we set $x+\dfrac{1}{y} = y+\dfrac{1}{z} = z + \dfrac{1}{x} = k$, for som real nu... | {
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"timestamp": "2023-03-29T00:00:00",
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finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$ finding value of $\displaystyle \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$
Substituting $x=\tan^2 \theta\;, dx = 2\tan \theta \sec^2 \theta$
integral is $=\displaystyle \int \frac{2\tan \theta \sec^2 \theta }{\tan^6 \theta \cdot \sec^3 \theta}d\theta= 2\int\frac{\cos^6 \the... | Here is another way...
First substitute $$u^2=1+x$$ so that $$I=2\int\frac{udu}{(u^2-1)^3u^3}$$
Then partial fraction decomposition gives $$I=2\times\int\left(-\frac{1}{u^2}-\frac{15}{16(u+1)}-\frac{7}{16(u+1)^2}-\frac{1}{8(u+1)^3}+\frac{15}{16(u-1)}-\frac{7}{16(u-1)^2}+\frac{1}{8(u-1)^3}\right) du$$
| {
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Prove that $A = \dfrac{f^2(1)+f^2(-1)}{2}$ is a composite number
Given $f(x) = a^{2016}x^2+bx+a^{2016}c-1$ where $a,b,c \in \mathbb{Z}$, suppose that the equation $f(x) = -2$ has two positive integer solutions. Prove that $A = \dfrac{f(1)^2+f(-1)^2}{2}$ is a composite number.
Let $g(x) = f(x)-2 = a^{2016}x^2+bx+a^{20... | Choosing $ a = 1$, $b = -3$, $c = 5$ gives two positive integer roots to $f(x) = 2$ ($x = 1$ and $x = 2$) and $[f^2(1)+f^2(-1)]/2 = 23$. This statement appears to be false.
If $f^2(x)$ means $f(x)^2$, then the statement fails for $a =1$, $b = -6$, $c = 11$, where the two roots are 2 and 4 and $[f(1)^2+f(-1)^2]/2 = 157$... | {
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Coefficient of $x^{50}$ in the expansion Find the coefficient of $x^{50}$ in the expansion of $$(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}+\cdots+1001x^{1000}$$
| Let $$S = (1 + x)^{1000} + 2x(1+x)^{999} +...+ 1000x^{999}(1+x)+ 1001 x^{1000}\tag1$$
This is an Arithmetic Geometric Series with $r = \frac{x}{1+x}$ and $d = 1$. Now $$\frac{x}{1+x}S = x(1 + x)^{999} + 2x^2(1 + x)^{998} +\cdots + 1000x^{1000} + \frac{1000x^{1001}}{1+x}\tag2$$
Subtracting we get,
$$(1 - \frac{x}{1+x}) ... | {
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For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?
1.$1$
2.$2$
3.$3$
4.$4$
5.$5$
My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both n... | $n = 1 \implies a^5+1 - a^3-a = a^3(a^2 -1) - (a-1) = (a-1)(a^4+a^3-1)$. Observe that $a \to 1^{-} \implies a^4+a^3 - 1 \to 1 $, thus for $\epsilon = 0.01$ we can find a $\delta > 0$ such that $ -\delta < a - 1 < 0$, and $|a^4+a^3-1 - 1| < 0.01$ or $a^4+a^3-1 > 0.99$, thus $(a-1)(a^4+a^3-1) < 0$ on $(1-\delta, 1)$.
You... | {
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Is probability that $x<2$ equal to the probability that $x^2<4$ given $1Assuming $x$ is a real number uniformly distributed over the interval $(1,3).$
so $x^2$ is also uniformly distributed over the interval $(1,9)${As for every $x=a\in (1,3) $ there exists $x^2=a^2\in (1,9)$}.
Probability that $x<2$ would be $\frac{1}... | You decided for any $x$ to match $x^2$, so you are claiming that $x<2$ is identical to $x^2<4$, but you can also match $4x-3$ in the interval $(1,9)$ for all $x$ in the interval $(1,3)$, which makes $x<2$ the same as $4x-3<5$.
That is the problem of defining probability on infinite group.
| {
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Different results with different methods to same limit I'm considering this limit:
$$\lim\limits_{x \to 0} \frac{x \cos(x) - \sin(x)}{x \sin^2(x)}$$
If I apply de l'Hôpital's rule I get $-\frac{1}{3}$, while if I immediately simplify the $x$ like this:
$$\frac{\cos(x) - \frac{\sin(x)}{x}}{\sin^2(x)}$$
I get $-\frac{1}{... | tl;dr: Your error is in the second step. You substitute $\frac{\sin x}{x}$ by its limit, $1$. But this is getting rid of the low-order terms, which matter.
One way to see it rigorously is to do a polynomial approximation near $0$, that is a Taylor expansion: when $x\to 0$,
$$
\frac{\sin x}{x} = 1-\frac{x^2}{6} + o(x^... | {
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Polynomial not equal to a square Can we prove that for any $a, b ,c$, there exists an integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer?
I think yes, because the range of polynomial is the whole of $\mathbb{R}$. Any ideas. By the way, is this related to elliptic curves by any chance? Thanks beforehand.
| This is a $1998$ putnam B6 problem.
An alternate approach: We write the assumed perfect square $n^3+an^2+bn+c $ in the form $(n^{3/2 }+ dn^{1/2}+f)^2$ giving us $$n^3+an^2+bn+c =n^3+2n^2d +2 (\sqrt {n})^3f+nd^2+2d\sqrt {n}f +f^2$$ Choosing $d=\frac {1}{2}a$ and $f=\pm 1$, we then, for $n $ sufficiently large, $$(n^{3/... | {
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Prove that there do not exist integers that satisfy the system
Prove that there do not exist integers $x,y,z,w,t,$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2\\(x+3)^2+w^2 &= (x+4)^2+t^2.\end{align*}
I thought about using a modular arithmetic argument. The given system is equival... | You are asking about four $x$ values. These are consecutive, so they assume the four values $0,1,2,3 \pmod 4.$ Let us give them new names,
$$ x_1 \equiv 1 \pmod 4, \; \; x_2 \equiv 2 \pmod 4, \; \; x_3 \equiv 3 \pmod 4, \; \; x_4 \equiv 0 \pmod 4. $$
$$ n = x_1^2 + y_1^2 = x_2^2 + y_2^2 = x_3^2 + y_3^2 = x_4^2 + y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the Volume integral $\iiint\left(x^2+y^2+z^2\right)\mathbb dv$. Suppose $a>0$ and $S = \{(x,y,z) \in \mathbb R^3 : x^2+y^2+z^2=a^2\}$ then
MY FIRST APPROACH: $$\int\int\int\left(x^2+y^2+z^2\right)\mathbb dv=a^2\int\int\int \mathbb dv=a^2.\frac{4}{3}\pi a^3=\frac 4 3\pi a^5$$.
MY SECOND APPROACH:If i use spher... | I think your mistake in first approach is that you take that as a line integral, whereas it is not: you can not put $\;x^2+y^2+z^2=a^2\;$ for the integrand, as this is a scalar function over the sphere $\;x^2+y^2+z^2=a^2\;$ , so you actually get (in spherical coordinates)
$$\iiint_S(x^2+y^2+z^2)dV=\int_0^a\int_0^{2\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the largest rectangle under the graph of $y = 1 - x^4$ using trigonometry I need to find the area of the largest rectangle possible under the curve of equation $1 - x^4$ with the base on the x-axis. The answers I've seen from other questions similar to this use calculus, but how would you solve it this using tr... | Let $(x,1-x^4)$ is a vertex of the rectangle, where $x>0$.
Thus, by AM-GM $$S=2x(1-x^4)=2\left(x-x^5\right)=2\left(\frac{4}{5\sqrt[4]5}-\left(x^5+\frac{4}{5\sqrt[4]5}-x\right)\right)\leq$$
$$\leq\frac{8}{5\sqrt[4]5}-2\left(5\sqrt[5]{x^5\cdot\left(\frac{1}{5\sqrt[4]5}\right)^4}-x\right)=\frac{8}{5\sqrt[4]5}.$$
The equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Verify $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0Verify by Mean Value Theory or otherwise that $\cos{x}<\left(\frac{\sin{x}}{x}\right)^3$ for $0<x<\pi/2$.
I am unable to solve the problem. Please give me a solution of the problem.
| We need to prove that $f(x)>0$, where $f(x)=\frac{\sin{x}}{\sqrt[3]{\cos{x}}}-x$.
Indeed, by AM-GM $$f'(x)=\frac{\cos{x}\sqrt[3]{\cos{x}}+\frac{\sin^2x}{3\sqrt[3]{\cos^2x}}}{\sqrt[3]{\cos^2x}}-1=\frac{3\cos^2x+\sin^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\cos^4x}}=$$
$$=\frac{1+\cos^2x+\cos^2x-3\sqrt[3]{\cos^4x}}{3\sqrt[3]{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Function is small $o$ of $x^2$ I have to solve the following exercise:
Give $a$, $b$, $c \in \mathbb R$ such that
$$\frac{1}{1-\cos x} = \frac{a}{x^2} + b +cx^2 + o(x^2)$$
for $x\to 0$.
Here's my attempt:
I know that $$\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^... | hint
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+o(x^6)$$ and after factoring out by $x^2$ ,
$$\frac{1}{1-X}=1+X+X^2+o(X^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Approximating square roots using binomial expansion. Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.
Binomial expansion
$ (1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ... $
Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$
= 1 - x -\frac{1}{2}x^... |
We want to (manually) approximate $\sqrt{2}$ by using the first few terms of the binomial series expansion of
\begin{align*}
\sqrt{1-2x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-2x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{2}\\
&= 1-x-\frac{1}{2}x^2-\frac{1}{2}x^3+\cdots\tag{1}
\end{align*}
Here we look for a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Proving a sequence is convergent by convergence of odd and even subsequences? I want to use this method to prove convergence of: $a_{n}=\sqrt{2-a_{n-1}}$, $a_{0}=\frac{2}{3}$.
Here is my attempt at proof:
It can be proven inductively that $0< a_{n}<2$ for all $n$.
I want to show that each of the odd and even subsequenc... | $$
\begin{align}
&\, a_{n}=\sqrt{2-a_{n-1}}\,\Rightarrow\,a_{n}\ge0 \\[2mm]
&\, a_{n}^2=2-a_{n-1}\,\Rightarrow\,a_{n-1}=2-a_{n}^2=\left(\sqrt{2}-a_{n}\right)\left(\sqrt{2}+a_{n}\right)\,\ge0 \\[2mm]
&\, \qquad\Rightarrow\,\sqrt{2}-a_{n}\ge0\,\Rightarrow\,a_{n}\le\sqrt{2} \\[8mm]
&\, a_{n}-a_{n-2}=\left(2-a_{n+1}^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
... | \begin{equation}
\int \frac{dx}{x(x^2+1)^2}
\end{equation}
\begin{equation}
\frac{1}{x(x+i)^2(x-i)^2}=\dfrac{A}{x}+\dfrac{B}{x+i}+\dfrac{C}{x-i}+\dfrac{D}{(x+i)^2}+\dfrac{E}{(x-i)^2}
\end{equation}
*
*Let $x=0$. Then $A=\dfrac{1}{(0+i)^2(0-i)^2}=1$\
*Let $x=-i$. Then $B=\dfrac{1}{-i(-i-i)^2}=-\dfrac{1}{2}$\
*Let $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$ Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}... | Let's go back to here:
$$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)} =$$
$$\lim_{h\to0}\frac{x^2(\sin x - \sin(x+h)) + (2xh+h^2)\sin x}{h\sin x\sin(x+h)} =$$
$$\lim_{h\to0}\left(\frac{x^2(\sin x - \sin x\cos h - \cos x\sin h)}{h\sin x\sin(x+h)}+\frac{2x+h}{\sin(x+h)}\right) =$$
$$\lim_{h\to0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $a_{n+2}=3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$ is an integer Given the sequence
$$a_1=a_2=1;\ a_ {n+2} = 3a_n + 2\sqrt{2a_n^2 + 2a_{n+1}^2}$$
prove that $a_n$ is an integer for all $n\in\mathbb N$.
Attempt
It is enough to show that $2a_n^2 + 2a_{n+1}^2$ is a perfect square. That means it's an even perfect square and so... | $a_{n+2}\ =\ 3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$
$\implies\ \left(a_{n+2}-3a_n\right)^2\ =\ 8a_n^2+8a_{n+1}^2$
$\implies\ a_n^2-6a_{n+2}a_n-8a_{n+1}^2+a_{n+2}^2=0$
Treat this as a quadratic equation in $a_n$. The discriminant is
$\Delta_n\ =\ 36a_{n+2}^2+32a_{n+1}^2-4a_{n+2}^2 = 4\left(8a_{n+1}^2+8a_{n+2}^2\right)$
Thus
$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Coefficient in binomial expansion for negative terms Find the exponent for $x^2$ in the expansion of:
$(\frac{x}{3} - \frac{1}{x^3})^{10}$
What trips me up is that the second term is negative. Though I find using the binomial theorem that the expansion should be a sum of terms on the form:
${ 10 \choose k}(\frac{x}{3})... | $$\begin{align}
\left(\frac x3-\frac 1{x^3}\right)^{10}
&=\frac {x^{10}}{3^{10}}\left(1-\frac 3{x^4}\right)^{10}\\
&=\frac {x^{10}}{3^{10}}\sum_{r=0}^{10}\binom {10}r\left(- 3x^{-4}\right)^r
\end{align}$$
Putting $r=2$ gives $x^2$ term, hence coefficient of $x^2$ is
$$\frac 1{3^{10}} \binom{10}2(-3)^2=\frac {45}{3^8}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a conve... | Another way.
Let $a+b+c=3$.
Hence, we need to prove that
$$\sum\limits_{cyc}\frac{a}{9-2a}\geq\frac{3}{7}$$ or
$$\sum\limits_{cyc}\left(\frac{a}{9-2a}-\frac{1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{a-1}{9-2a}\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{a-1}{9-2a}-\frac{a-1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
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Sum the series: $1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$ We have the series$$1+\frac{1+3}{2!}+\frac{1+3+3^2}{3!}+\cdots$$How can we find the sum$?$
MY TRY: $n$th term of the series i.e $T_n=\frac{3^0+3^1+3^2+\cdots+3^n}{(n+1)!}$. I don't know how to proceed further. Thank you.
| The numerator is a geometric sum that evaluates to,
$$\frac{3^{n+1}-1}{3-1}$$
Hence what we have is,
$$\frac{1}{2} \sum_{n=0}^{\infty} \frac{(3^{n+1}-1)}{(n+1)!}$$
$$=\frac{1}{2} \sum_{n=1}^{\infty} \frac{3^n-1}{n!}$$
$$=\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{3^n}{n!}- \sum_{n=1}^{\infty} \frac{1^n}{n!} \right)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Differentiation first principles for cube Find the derivative of function $f(x) = \sqrt{x} + \dfrac{1}{x^3}$ from the first principles.
I tried to use the formula
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
$$
I try to expand for the $x^3$ first but it looks more and more complicated. The answer end up with $0$.... | Hint: (I'm assuming you mean $\require{cancel}f(x) = \sqrt{x} + \frac1{x^3}$.) Here's the "hard" computation for a few similar functions. You can see the pattern: in the end, there is a factor of $h$ in the numerator which will cancel with the $h$ that you are going to use as the denominator in the difference quotient,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that a sequence is increasing (check the idea)
4.5 A real sequence $\{x_n\}$ satisfies $7x_{n+1}=x_n^3+6$ for $n\ge1$. If $x_1=\frac{1}{2}$, prove that the sequence increases and find its limit. What happens if $x_1=\frac{3}{2}$ or if $x_1=\frac{5}{2}$?
In order to prove, that sequence is increasing, my idea is... | Given that $x_{n+1} = (x_{n}^3 + 6)/7$, when $x_1 = \frac{1}{2}$ we will find the limit in 3 steps:
*
*$\color{red}{0 < x_n < 1 \text{ for all $n$.}}$ Note that if $0 < x_n < 1$ then $0 < x_n^3 + 6 < 7$, which in turn implies that $0 < x_{n+1} < 1$. So, starting with $x_1 = \frac{1}{2}$, we will get the entire sequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all values of $ac-bd$ where $a, b, c, d$ are real numbers, $a^2c^2+a^2d^2+b^2c^2+b^2d^2=2017$, and $ad+bc=44$. Find all values of $ac-bd$ where $a, b, c, d$ are real numbers, $a^2c^2+a^2d^2+b^2c^2+b^2d^2=2017$, and $ad+bc=44$.
I noticed that $a^2c^2=(ac)^2$ and so on for all the terms in the polynomial. How does... | Well, $(ac-bd)^2+(ad+bc)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Infinite sum including logarithm I would like to calculate the following sum:
$$\sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right)$$
I do know that it converges but I have gone that far:
\begin{align}
& \sum_{n=1}^{\infty}\ln \left( \frac{n^2+2n+1}{n^2+2n} \right) \Longleftrightarrow \sum_{n=1}^\infty \ln ... | Rewrite the general term as $\ln\dfrac{(n+1)^2}{n(n+2)}$, use the functional property of logs and you'll obtain a telescoping product for partial sums:
\begin{align}\sum_{k=1}^{n}\ln \frac{(k+1)^2}{k(k+2)}&=\ln\frac{2^2}{1\cdot 3}+\ln\frac{3^2}{2\cdot 4}+\ln\frac{4^2}{3\cdot 5}+\dotsm\dotsm\dotsm\\&\phantom{=}+\ln\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Surface integral on one-sheeted hyperboloid I'm working on the following exercise:
"Let $\Sigma:=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=z^2+1,-1\leq z\leq3\}$ be a surface with orientation $\omega$.
Compute $\int\int_{\Sigma}(\nabla\times F)\cdot\omega\ dS$ knowing that $F=(-\frac{y}{3},-\frac{z}{3},-\frac{x}{3})$ and $\omeg... | By the divergence theorem, your integral equals
$$
\Phi = \iiint_E \underbrace{\nabla\cdot \nabla \times F}_{=0}\; dV-\iint_{S_1}\nabla \times F\; dS-\iint_{S_2}\nabla \times F\; dS,
$$
where $S_1$ and $S_2$ are the surfaces that close $\Sigma$ at $z=-1$ and $z=3$. $S_1$ can be parametrized as follows:
$$
\begin{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$ As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum.
Lagrange multiplier gives a dirty calculation so I can't handle it. Is ... | Let $x=u+v$ and $y=u-v$. Then $x^2+y^2=2(u^2+v^2)$, while the equation $x^3+3xy+y^3=1$ becomes $2u^3+6uv^2+3(u^2-v^2)=1$, or
$$2u^3+3u^2-1=-3v^2(2u-1)$$
Factoring the cubic on the left hand side, we obtain
$$(u+1)^2(2u-1)=-3v^2(2u-1)$$
So either $2u-1=0$, in which case $v$ can be anything, or else $(u+1)^2=-3v^2$, whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Infimum of two variable function on the closed unit disc I am trying to compute
$$\inf_{x,y} \frac{ax-by}{1+x^2+y^2}$$
subject to the constraint $x^2+y^2 \leq 1$. Here $a,b$ are any two fixed, real numbers. I am having trouble computing this using standard derivative techniques, and Wolfram alpha is unable to recognize... | By using the Cauchy-Schwarz inequality, one has
$$ |f(x,y)|\le\frac{\sqrt{a^2+b^2}\sqrt{x^2+y^2}}{1+x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}+\frac{1}{\sqrt{x^2+y^2}}} $$
and "=" holds if and only if $\frac{a}{b}=\frac{x}{-y}$.
Using
$$ a^2+b^2\ge2ab$$
("=" holds if and only if $a=b$) one has
$$ |f(x,y)|\le\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find all integer solutions for $xy+1=3(x+y)$ the original problem was to find all integer solutions for $xy+1=3(x+y)$.
$$xy+1=3(x+y)$$
$$xy+1=3x+3y$$
$$xy-3x=3y-1$$
$$x(y-3)=3y-1$$
$$x=\frac{3y-1}{y-3}$$
I was able to find all possible solutions using this method,
I was wondering if you can say that $3|xy+1$ or $x+y|x... | We want to find a factorization if possible, so consider $xy-3x-3y+k = k-1$ as a re-writing of the condition. Then we can take any value for $k$ to make a factorization possible, and $(x-3)(y-3) = xy-3k-3y+9$, so take $k=9$ to get
$$(x-3)(y-3) = 8$$
Then the integer ordered factor pairs of $8$ are
$$\{(-1,-8), (-2,-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is it possible to integrate $\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$ without involving complex numbers? The integral is $$\int \frac{\mathrm{d}x}{\sin^2 x+\sin x+1}$$
Consider
$$\int \frac{\mathrm{d}x}{ax^2+bx+c}$$
There are a total of three cases, depending on the discriminant of $ax^2+bx+c$. Two of which are sho... | Let $\displaystyle I = \int \frac {1}{\sin ^2x + \sin x + 1}dx$
Let $$\sin x = -\frac {(2-\sqrt {3})t + (2 +\sqrt {3})}{t + 1}\implies \cos x\ dx = \frac {2\sqrt {3}}{(t + 1)^2}dt$$
Then $$I = 2\sqrt 3 \int \frac {t + 1}{\left((6 - 3\sqrt 3)t^2 + (6 + 3\sqrt 3)\right)\sqrt {(4\sqrt {3} - 6)t^2 - (4\sqrt {3} + 6)}}\ dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}$ if $\measuredangle C = 90^{\circ}$
In$\triangle ABC$, $\angle C$ is a right angle. Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}.$ Legs are named in traditional way.
My Work
As $r = \frac{A}{s}$. So, here $r = \frac{ab}{a+b+c}... | The first part:
$$(a+b+c)(a+b-c)=(a+b)^2-c^2 = a^2+b^2 + 2ab -c^2=2ab$$
by Pythagora's theorem.
The second part: $r_c=\frac{A}{s-c}=\frac{ab}{a+b-c}$ and it works with absolutely the same calculation above.
| {
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"question_score": "2",
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How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$
Consider
$$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
I want to show that $I=3\pi$ and that $I=J$.
F... | $\begin{align}I&=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad\\
&=\int_{0}^{\infty} \dfrac{16x^4(1+x^2)^4}{(1+x^6)^4} dx\\
&=\int_{0}^{\infty} \dfrac{16\cdot {{x}^{12}}+64\cdot {{x}^{10}}+96\cdot {{x}^{8}}+64\cdot {{x}^{6}}+16\cdot {{x}^{4}}}{(1+x^6)^4} dx\\
\end{align}$
Perform the change of variable ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to i solve this Exponential equation How to solve this exponential equation?
$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
| $$\\ 7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}\\ 125\cdot { 5 }^{ x }-25\cdot { 5 }^{ x }=81\cdot { 3 }^{ x }-21\cdot { 3 }^{ x }\\ 100\cdot { 5 }^{ x }=60\cdot { 3 }^{ x }\\ { \left( \frac { 5 }{ 3 } \right) }^{ x }=\frac { 3 }{ 5 } \\ x=-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Variance of number of heads in three flips of a coin selected from a set of fair and unfair coins We have $10$ coins, $2$ are two-tailed, $2$ are two-headed, the other $6$ are fair ones. We (randomly) pick a coin and we flip it $3$ times. Find the variance of the number of gotten heads.
My attempt:
$X$ - number of head... | That's one approach. It is good. Here's another,
We know $X\mid C \sim \mathcal{Bin}(3, C)$ and $\mathsf P(C=c)=\tfrac 15\mathbf 1_{c=0}+\tfrac 35\mathbf 1_{c=1/2}+\tfrac 15\mathbf 1_{c=1}$
So $\mathsf E(C)= \tfrac 1{2}$ and $\mathsf {Var}(C)=\tfrac 1{10}$ and $\mathsf E(C^2)=\frac 7{20}$
By the Law of Total Proba... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of $(p,q)$ in $\mathbb R^2$ such that $x^4+px^2+q$ is divisible by $x^2+px+q$ I have found four through my attempt, but apparently the answer is that there are five pairs.
Mine were $(-1,0),(-1,1),(0,0),(0,1)$.
What am I missing?
| Fun problem here's another approach using Vieta's formulas:
The sum of the roots of $x^2+px+q$ is $-p$. The sum of the roots of $x^4+px^2 + q$ is $0$. Similarly the product of the roots of both polynomials is $q$. In particular this tells us that if $x^2+px+q$ divides $x^4+px^2 + q$ the quotient would have to be $x^2... | {
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In how many ways can we roll a red die, a yellow die, and a black die, and get a sum of $9$? I know I can use generating functions. Each of the die has a generating function $x+x^2+x^3+x^4+x^5+x^6$, and so I need to find the coefficient of $x^9$ in the generating function of their sum, $(x+x^2+x^3+x^4+x^5+x^6)^3$. I am... | Robert Frost's approach is probably the simplest, but here's a way to do it with generating functions.
We could just multiply it out. But we can save a bit of work by noticing that $(x + x^2 + x^3 + x^4 + x^5 + x^6)^3 = \left(\frac{x(1 - x^6)}{1 - x}\right)^3 = \frac{x^3 - 3x^9 + 3x^{15} - x^{21}}{(1 - x)^3}$.
We can ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Asymptotics for $k^c +k - 1 \choose k$ when $c\geq 1$ I am looking for the asymptotics of $n +k - 1 \choose k$ when $n = k^c$ for integer constant $c\geq 1$.
From this we get:
$$\frac{(n+k-1)!}{k! (n-1)!} \approx \frac{\sqrt{2 \pi(n+k-1)}\left(\frac{n+k-1}{e}\right)^{n+k-1}}{\sqrt{2 \pi k}\left(\frac{k}{e}\right)^k \... | Note that $$\frac{k!}{(k^c+k-1)^k}\binom{k^c+k-1}{k}=\prod_{i=1}^k\frac{k^c+i-1}{k^c+k-1}=\prod_{i=1}^{k-1}\left(1-\frac{i}{k^c+k-1}\right)$$ Since $c>1$, the last product is asymptotically equivalent to $$\prod_{i=1}^{k-1}\left(1-\frac{i}{k^c}\right)\approx\prod_{i=1}^{k-1}\exp\left(-\frac{i}{k^c}\right)\approx\exp\le... | {
"language": "en",
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"source": "stackexchange",
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$2 \times 2$ Examples
Find examples of $2 \times 2$ matrices with $a_{12} = \frac{1}{2}$ for which (a) $A^2 = I$, (b) $A^{-1}=A^T$, and (c) $A^2 = A$.
What I thought was a trivial problem is turning out to be rather irritating. At the moment I am dealing with part (a). To solve the problem, I narrowed my search spac... | Let's start with what you have:
$$A = \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix}$$
So then
$$A^2 = \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix}\cdot \begin{bmatrix} a & \frac{1}{2} \\ -\frac{1}{2} & d \end{bmatrix} = \begin{bmatrix} a^2-\frac{1}{4} & \frac{1}{2}(a+d) \\ -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve differential equation $(x^2+xy)y'=x\sqrt{x^2-y^2}+xy+y^2$ Solve and find a particular solution that satisfies $y(1)=1$.
What is the type of this differential equation?
| $$(x^2+xy)y'=x\sqrt{x^2-y^2}+xy+y^2$$
$$y'=\dfrac{x\sqrt{x^2-y^2}+xy+y^2}{(x^2+xy)}$$
is an homogeneous differential equation. Let $u=\dfrac{y}{x}$ so
$$u'x+u=\dfrac{\sqrt{1-u^2}+u+u^2}{(u+1)}$$
or
$$\dfrac{u+1}{\sqrt{1-u^2}}du=\dfrac{dx}{x}$$
after integration (let $u=\sin t$) we have
$$t-\cos t=\ln Cx$$
with $y(1)=1$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $ \int \frac{1}{\sin{x}\cos^{3}x}dx $ $$ \int \frac{1}{\sin{x}\cos^{3}x}dx $$
$$\Rightarrow \int \frac{1}{\sin{x}\cos^{3}x}{\cos{x}\over \cos{x}}dx$$
$$\Rightarrow \int \frac{\sec^{4}{x}}{\tan{x}}dx$$
$$\Rightarrow \int \frac{\sec^{2}(1+\tan^{2}x)}{\tan{x}}dx$$
$$Substitution \tan{x}=t \Rightarrow \sec^{2}xdx=... | Another way, I am not sure it is an elegant way but is different than yours
$$\int \frac { 1 }{ \sin { x } \cos ^{ 3 } x } dx=\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } }{ \sin { x } \cos ^{ 3 } x } dx= } \int { \frac { \sin { x } }{ \cos ^{ 3 }{ x } } } dx+\int { \frac { 1 }{ \sin { x\cos { x } } } dx } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to integrate ${dx}/{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}}$
If
$\displaystyle\frac{dx}{dt}=-\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}$
and $x(0)=R$, find $t$ when $x=0$.
I have no idea how to integrate $\displaystyle\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\righ... | $$\displaystyle\frac{dx}{dt}=-\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{R}\right)}$$
$$ \int \frac{1}{\sqrt{\frac{2K}{m}\left( \frac{1}{x}-\frac{1}{R}\right)}}dx= \int - dt $$
$$ \frac{1}{\sqrt{\frac{2K}{m}}} \int \frac{1}{\sqrt{\frac{1}{x}-\frac{1}{R}}} dx=\int - dt$$
$$ \frac{1}{\sqrt{\frac{2K}{m}}} \int \frac{1}... | {
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"source": "stackexchange",
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Find the sum of the infinite series: $ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+...$
Find the sum of the infinite series
$$ 1 + \frac{1+2}{2!} + \frac{1+2+2^2}{3!} +\frac{1+2+2^2+2^3}{4!}+... ....$$
What I have done let
$$ S = \underbrace{\frac{1}{1!}}_{\text{1st Term}} + \underbrace{\frac{1... | The numerator is simply geometric progression,
$$1 + 2 + 2^2 + \cdots + 2^{n-1} = 2^n - 1$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{2^n - 1}{n!} = \sum_{n = 1}^{\infty}\frac{2^n}{n!} - \sum_{n = 1}^{\infty}\frac{1}{n!} = (e^2 - 1) - (e - 1) = e^2 -e$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of function with natural logarithm I need help solving this problem. I tried L'hospital and rearranging but nothing worked.
$$
\lim_\limits{x→∞} x^2\left(\ln\left(1 + \frac{1}{x}\right)- \frac{1}{x+1}\right)
$$
| We have
$$x(\frac{\ln (1+\frac{1}{x})}{\frac{1}{x}})-\frac{x^2}{1+x}$$
Let $x=\frac{1}{y}$. We have,
$$=\frac{1}{y}\frac{ \ln (1+y)}{y}-\frac{1}{y^2+y}$$
This is,
$$=\frac{1}{y} \frac{ \ln (1+y)}{y}-\frac{1}{y}+\frac{1}{y+1}$$
$$=\frac{1}{y}(\frac{\ln(1+y)}{y}-1)+\frac{1}{y+1}$$
$$=\frac{\ln(1+y)-y}{y^2}+\frac{1}{y+1}... | {
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Find the minimum real number $\lambda$ so that the following relation holds ($x>y$): $\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$
ّFind the minimum real number $\lambda$ so that the following relation holds for arbitrary real numbers $x,y$($x>y$): $$\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-... | For $\lambda$, any value above $0$ is possible.
$\lambda\dfrac{x^3-y^3}{(x-y)^3}\geq \dfrac{x^2-y^2}{(x-y)^2}$
$\lambda\dfrac{x^3-y^3}{(x-y)}\geq (x^2-y^2)$
$\lambda (x^2+xy+y^2) \geq x^2-y^2$
$\lambda \geq \dfrac{x^2-y^2}{x^2+xy+y^2} \cdots(1)$
$(x-y)^2 \gt 0 \implies x^2+y^2 \gt 4xy$
Also $(x^2-y^2) \gt 0$
Since n... | {
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Find the number of integer solutions of $x_1 + x_2 + x_3 = 16$, with $x_i \geq 0$, $x_1$ odd, $x_2$ even, $x_3$ prime. Find the number of integer solutions to $x_1+x_2+x_3=16$, with $x_i \geq 0$, $x_1$ odd, $x_2$ even , $x_3 $ prime.
My attempt:
Is the same as the number of ways to choose $16$ objects from $3$ distinc... | We need this:
$x_1+x_2+x_3=16$ and $x_1=2k_1+1,~x_2=2k_2,~k_1,k_2\ge0$ and $x_3\in\{2,3,5,7,11,13\}$.
Since $x_1+x_2$ is odd then $x_3$ should be odd so $x_3\in\{3,5,7,11,13\}$.
Let's find the answer for all $x_3$.
Note that: $x_1+x_2=n$, has $\left(\binom{n}{2}\right)$, since $n$ is odd then $x_1$ is odd and $x_2$... | {
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"source": "stackexchange",
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Proof in Arithmetic Progression My maths teacher at school asked a question which I am finding difficult to crack down.
We are given that $a^2 , b^2$ and $c^2$ are in AP. We need to prove that $\frac{a}{b+c} , \frac{b}{a+c}$ and $\frac{c}{a+b}$ are in AP.
This is what I tried.
Let the common difference of the AP be... | $$\frac{2b}{a+c}=\frac{a}{b+c}+\frac{c}{a+b}\Leftrightarrow \\
2b[b^2+b(a+c)+ac]=a[a^2+a(b+c)+bc]+c[c^2+c(a+b)+ab]\Leftrightarrow\\
2b^3+2b^2(a+c)=a^3+a^2(b+c)+c^3+c^2(a+b)\Leftrightarrow\\
2b^2(a+b+c)=a^2(a+b+c)+c^2(a+b+c)\Leftrightarrow \\
(a+b+c)(2b^2-a^2-c^2)=0$$
Can you finish?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know whe... | $$\frac{3^{11}-1}{2}=\frac{3-1}{2}(3^{10}+3^{9}+3^{8}+3^{7}+3^{6}+3^{5}+3^{4}+3^{3}+3^{2}+3+1)\equiv4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Sum with irrational powers and binomial coefficients What is the value of:
$$\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?$$
I am stuck because of the binomial coefficient there,... | $\sum_{k=1}^{n} {\left(\left(-1\right)^{n-k}{n+1 \choose k+1}\left(\frac{\sqrt5}{1+\sqrt5}\left(3+\sqrt5\right)^k-\frac{\sqrt5}{1-\sqrt5}\left(3-\sqrt5\right)^k-\frac{5}{2}\right)\right)}?
$
Here's a start.
I'm feeling too tired right now
to do more.
$\begin{array}\\
\sum_{k=1}^{n} {n+1 \choose k+1} x^k
&=\sum_{k=2}^{n... | {
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"timestamp": "2023-03-29T00:00:00",
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How many passwords of length 8 with digits non-decreasing sequence and start with 5? Passwords have length of 8 characters.
The usable characters are the 10 digits 0,1,2,3,4,5,6,7,8,9.
How many passwords with digits only have non-decreasing sequence of digits and start with 5?
The sequence of eight digits $D_1, D_2, D_... | One way to choose a password is to just choose where the strict increases in the password are, and how big they are. For example, the password
(5, 5, 5, 7, 7, 7, 8, 8)
has strict increases after positions 3 and 6, the first being a jump of size 2, and the second of size 1. Also, any sequence of positions to increase a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum\limits_{cyc}\frac{a^2-bd}{b+2c+d}\geq0$
Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a^2-bd}{b+2c+d}+\frac{b^2-ca}{c+2d+a}+\frac{c^2-db}{d+2a+b}+\frac{d^2-ac}{a+2b+c}\geq0$$
This inequality is a similar to the following inequality of three variables.
Let $a$, $b$ and $c$ be po... | Update
By chance, I saw a nice proof as follows.
\begin{align}
\sum_{\mathrm{cyc}} \frac{a^2-bd}{b+2c+d}
&\ge \sum_{\mathrm{cyc}} \frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d}\\
&= \sum_{\mathrm{cyc}} \Big(\frac{a^2-\frac{(b+d)^2}{4}}{b+2c+d} + \frac{b+d-2c}{4}\Big)\\
&= \sum_{\mathrm{cyc}} \frac{a^2-c^2}{b+2c+d}\\
&= \Big(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How prove $ax(a+x)+by(b+y)+cz(c+z)\ge 3(abc+xyz)$ let $a,b,c,x,y,z\ge 0$ and such
$$a+b+c=x+y+z$$
show that
$$ax(a+x)+by(b+y)+cz(c+z)\ge 3(abc+xyz)$$but it does not help for a proof of the starting inequality (at least I don't see, how it helps).
I tried also BW, but we get there something, which impossible to kill dur... | It's enough to prove that $x^2a+y^2b+z^2c+xyz\geq4abc$.
Let $x^2a+y^2b+z^2c+xyz<4abc$, $x=kp$, $y=kq$ and $z=kr$, where $k>0$ and
$p^2a+r^2b+q^2c+pqr=4abc$.
Hence, $p^2a+r^2b+q^2c+pqr=4abc>x^2a+y^2b+z^2c+xyz=k^2(p^2a+r^2b+q^2c+kpqr)$,
which says that $0<k<1$ and $a+b+c=x+y+z=k(p+q+r)<p+q+r$,
which is contradiction be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use Newtons iterative process to show the following For the function
$$ f: = \cot(\sqrt x) + \frac{1}{\sqrt{x}}$$
with the initial approximate
$$x_0 = \pi^2\left(n-\frac{1}{2}\right)^2$$
Show that after one iteration
$$ x_{n+1} = \pi^2\left(n-\frac{1}{2}\right)^2\left(1+ \frac{1}{1+ \pi^2\left(n-\f... | From your resulto of $x_1$, I followed the calculations and get the same result and a usefull thing should be this.
If we use that, calling $\Delta := \pi(n-1/2)$
$$\left(1 + \frac{1}{1+\Delta^2}\right)^2 = 1 + \frac{2}{1 + \Delta^2} + \frac{1}{(1 + \Delta^2)^2}$$
From your result we have that
$$x_1 = \Delta^2\left(1 ... | {
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How to solve $x^2 +\left(\frac{x}{x-1}\right)^2 =8$? I tried to solve this question but it turns into a 4th degree equation and I could only get one solution for this equation,i.e., 2.
It is to be evaluated for solutions. Thanks.
| $$x^2 +\frac{x^2}{(x-1)^2} = 8$$
$$x^2 +\frac{x^2}{(x^2-2x + 1)} = 8$$
$$x^2 = (8 - x^2)(x^2-2x + 1)$$
$$x^2 = (8 - x^2)x^2-(8 - x^2)2x + (8 - x^2)$$
$$x^2 = 8x^2 - x^4 -16x + 2x^3 - x^2 + 8$$
$$0 = - x^4 -16x + 2x^3 + 6x^2 + 8 $$
$$(x-2)^2(x^2+2x-2)= 0$$
$$x= 2$$
$$x= \sqrt{3}-1 $$
$$x= -1 - \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$a\sqrt{\sin x}+b\sqrt{\cos x}≤(a^{4/3}+b^{4/3})^{3/4}$ prove that :
$$a,b>0\\,0<x<\pi/2$$
$$a\sqrt{\sin x}+b\sqrt{\cos x}≤(a^{4/3}+b^{4/3})^{3/4}$$
my try :
$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\frac{a}{b}\sqrt{\cos x})$$
$$\frac{a}{b}=\tan y$$
$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\tan y\sqrt{\c... | Using AM-GM inequality,
$a^{4/3}+b^{4/3} \ge 2\sqrt{a^{4/3}b^{4/3}}$
$a^{4/3}+b^{4/3} \ge 2ab^{2/3}$
$\left(a^{4/3}+b^{4/3}\right)^{3/4} \ge 2^{3/4}{ab^{1/2}} \cdots(1)$
Using AM-GM inequality,
$a\sqrt{\sin x}+b\sqrt{\cos x} \ge 2\sqrt{ab\sqrt{\sin x \cos x}}$
$a\sqrt{\sin x}+b\sqrt{\cos x} \ge 2(ab)^{1/2}(\sin x\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\time... | Systematically work order by order
\begin{eqnarray*}
(x+y+z)^2= x^2+y^2+z^2+2(xy+yz+zx)
\end{eqnarray*}
right firstly we do not need to write everything out; We shall use the following short hand
\begin{eqnarray*}
(\sum x)^2= \sum x^2+2\sum xy
\end{eqnarray*}
so $\sum xy=2$
\begin{eqnarray*}
(\sum x)(\sum x^2)= \sum x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$ Evaluate:
$$S_n=1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$$
a_n are the individua... | By setting $H_n = \sum_{k=1}^{n}\frac{1}{k}$ we have to compute $\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2$. We may notice that
$$ \sum_{n=1}^{N}\frac{H_n}{n}=\sum_{1\leq m\leq n\leq N}\frac{1}{mn}=\frac{H_N^2+H_N^{(2)}}{2}\tag{1}$$
and for the same reason
$$ \sum_{n=1}^{N}\frac{H_n^{(2)}}{n^2} = \frac{1}{2}\left[\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
All possible ways scores $(1,2,3)$ add up to $n$ Could someone please explain mathematical explanation behind this?
You can win three kinds of basketball points, 1 point, 2 points, and 3 points. Given a total score $n$, print out all the combination to compose $n$.
Examples:
For n = 1, the program should print followin... | It is a Fibonacci like sequence.
Let $\,A_{\small n}\,$ be the total number of $\,(1,\,2,\,3)\,$ combinations that compose $\,n\,$, Then:
$$ A_{\small 1}=1,\,A_{\small 2}=2,\,A_{\small 3}=4,\quad\color{red}{A_{\small n}=A_{\small n-1}+A_{\small n-2}+A_{\small n-3}} \\[4mm] \Rightarrow\quad \left\{A_{\small n}\right\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to simplify $\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$ How would you go about simplifying the expression $\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$?
| $$\frac{75}{8}\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}=\frac{75}{8}\sqrt{\frac{25a^3-9a^3}{225}}=\frac{75}{8}\frac{\sqrt{16}\sqrt{a^3}}{\sqrt{225}}=\frac{75}{8}\frac{4}{15}\sqrt{a^3}=2.5\sqrt{a^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of a infinite series $$\frac{2}{2} + \frac{2\cdot 5}{2\cdot 9} + \frac{2\cdot 5\cdot 10}{2\cdot 9\cdot 28} + \cdots + \frac{2\cdot 5\cdot 10 \cdots (n^2+1)}{2\cdot 9\cdot 28\cdots (n^3+1)}\tag1$$
For this series $(1)$, how would one go about applying the comparison test to check for convergence or divergence?
| The general term of the series, $a_n$ is given by
$$a_n=\prod_{k=1}^n\frac{(k^2+1)}{(k^3+1)}$$
Then, we see that
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2+1}{(n+1)^3+1}\to 0\implies \text{the series converges}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding a basis for a set of matrices in a vector space. I am trying to find a basis for the following vector space:
V = {2x2 matrices A | $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$A = A$\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$}
So far, I have augmented $\bigl( \beg... | No, it is not. For example, also $\;A\;$ and all its powers commute with A, and also for example
$$\begin{pmatrix}-1&1\\1&0\end{pmatrix}\;\ldots$$
To solve this, write
$$B:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;\text{so that}\;\;AB=BA\iff \begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Number of ways in which 3 people can throw a normal die to have a total score of 11
Number of ways in which 3 people can throw a normal die to have a total score of 11
My approach:
The answer can be obtained by finding the coefficient of $x^{11}$ in the expansion of $(x+x^2+x^3+x^4+x^5+x^6)^3$.
General term $... | Here's a pen-and-paper approach.
First, let's compute the number of ways to throw $2$ through $12$ with two dice.
\begin{eqnarray}
N(2)&= 1 &: \{1+1\}\\
N(3)&= 2 &: \{1+2,2+1\}\\
N(4)&= 3 &: \{1+3,2+2,3+1\}\\
N(5)&= 4 &: \{1+4,2+3,3+2,4+1\}\\
N(6)&= 5 &: \{1+5,2+4,3+3,4+2,5+1\}\\
N(7)&= 6 &: \{1+6,2+5,3+4,4+3,5+2,6+1\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$
Prove that $$\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)}{2c^2+(a+b)^2} \le 8$$.
MY ATTEMPT:I want to make a relation between $a,b,c$. By trial I found that if we put $... | For non-negatives $a$, $b$ and $c$ let $a+b+c=3$.
Hence,
$$8-\sum_{cyc}\frac{(2a+b+c)^2}{2a^2+(b+c)^2}=\sum_{cyc}\left(\frac{8}{3}-\frac{(a+3)^2}{2a^2+(3-a)^2}\right)=$$
$$=\frac{1}{3}\sum_{cyc}\frac{(a-1)(7a-15)}{a^2-2a+3}=\frac{1}{3}\left(\sum_{cyc}\frac{(a-1)(7a-15)}{a^2-2a+3}+4(a-1)\right)=$$
$$=\sum_{cyc}\frac{(a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\... | $$ \begin{align}\int \frac{1}{(1+x^2)\sqrt{x^2 - 1}} \, dx &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sqrt{\cosh^2 u - 1}} \, du && x = \cosh u \\ &= \int \dfrac{\sinh u}{(\cosh^2 u + 1)\sinh u} \, du \\ &= \int \dfrac{1}{\cosh^2 u + 1}\, du \\ && \end{align}$$
After which things become simple in terms of hyperbolic tsng... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum... | $$S=\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
For the converge use Ratio Test.
$$2S=\sum_{n=1}^{+\infty}\frac{2n+1+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}x^{2n+1}+\sum_{n=1}^{+\infty}\frac{x^{2n+1}}{2n+1}$$
The first sum is clearly an Infinite Geometric Series.
Now for the second part $$\ln(1+x)-\ln(1-x)=2\sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Integration on $ \int \sec^3 (2x) dx$ The answer is this but the coefficient is $\frac{1}{4}$. Why?
My steps:
$$\int \sec^3(2x)dx$$
Let $u = 2x$, then $\frac{1}{2}du = dx$
$$\frac{1}{2} \int \sec^3(u) du $$
Using int my parts: Let $a = \sec(u)$, then $da = \sec(u)\tan(u)du$. let $dv = \sec^2 (u) du$, then $v = \tan(... | What you derived is that $\int \sec^3 (u)~du=\text{blah}$. Then note that $\int \sec^3(2x)~dx=\frac12\int \sec^3(u)~du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$ Problem : Prove (or disprove) that, $\gcd((a^b-1)/(a-1), (a^c-1)/(a-1))=1$ (greatest common divisor),
when $a, b, c $ are prime numbers and $a, b, c \geq 3, b \neq c$. $(a^b-1), (a^c-1)$ are factors of $(a^{bc}-1)$.
Click here to see the related question .
| assume $c > b$
Also gcd $(a, 1 + a + ... + a^b) = 1$
Let $x = (a^c - 1)/(a-1) = 1 + a + a^2 + ... + a^{c-1}$
Let $y = (a^b - 1)/(a-1) = 1 + a + a^2 + ... + a^{b-1}$
$gcd(x,y) = gcd(x-y, y) = gcd((a^b + ... + a^{c-1}), y) = gcd ((a^b(1 + a + ... + a^{c-b-1}), y) = gcd((1 + a + ... + a^{c-b-1}), y)$
as we keep doing, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find limit of $\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$ if $f(x)=-\sqrt{25-x^2}$ I have a question , if then find $$\lim_{x\to 1} \frac{f(x) - f(1)}{x-1}$$
I got $f(1)=-\sqrt {24}$. Should I get limit = $\frac{\sqrt{24} - \sqrt{25-x^2}}{x-1}$ but answer is $\frac{1}{\sqrt{24}}$ . Should I use $f'(x)$? If I use ... | First, let us evaluate $f(1)=f(x)=-\sqrt{25-1^2}=-2\sqrt{6}$
So we wish to evaluate,
$$\lim _{x \to 3}\frac{-\sqrt{25-x^2}+2\sqrt6}{x-1}=\frac{-\sqrt{25-3^2}+2\sqrt6}{3-1}=\frac{-4+2\sqrt6}{2}= -2 + \sqrt{6}$$
$EDIT$
$$\lim _{x \to 1}\frac{-\sqrt{25-x^2}+2\sqrt6}{x-1}$$
Rationalizing
$$\lim _{x \to 1}\frac{(-\sqrt{25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Differential Equation involving System of Equations Find a solution to the boundary value problem \begin{align}y''+ 4y &= 0 \\ y\left(\frac{\pi}{8}\right) &=0\\
y\left(\frac{\pi}{6}\right) &= 1\end{align}
if the general solution to the differential equation is $y(x) = C_1 \sin(2x) + C_2 \cos (2x)$.
I was able to comput... | \begin{align*}
\frac{\sqrt 2 C_1}{2} + \frac{\sqrt 2 C_1}{2} &= 0 \\
C_1 + C_2 &= 0 \\
C_2 &= -C_1 \\
\frac{\sqrt{3}C_1}{2} + \frac{C_2}{2} &= 1 \\
\sqrt{3}C_1 + C_2 &= 2 \\
\sqrt{3}C_1 - C_1 &= 2 \\
(\sqrt{3} - 1)C_1 &= 2 \\
C_1 &= \frac{2}{(\sqrt{3} - 1)} \\
C_2 &= -\frac{2}{(\sqrt{3} - 1)} \\
\end{align*}
Which is w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation of divergent series $1-2^p+3^p-4^p+\ldots$ I am trying to obtain that, for $p=1,2,3\ldots$ ,
$$
s_p\equiv\sum_{n=1}^\infty (-1)^{n-1} n^p = \frac{2^{p+1}-1}{p+1}B_{p+1},
$$
where $s_0=1/2$ and $B_m$ are the Bernoulli numbers. The (divergent) series is regularized by means of the Euler summation
$$
\sum_{n=1}^... | We need to reshuffle the generating function as follows:
$$\begin{aligned}
\frac{e^z}{e^z+1}=&\ \frac{1}{1+e^{-z}}=\frac{(e^{-z}-1)^2}{(e^{-z}+1)(e^{-z}-1)^2}=\frac{e^{-2z}-2e^{-z}+1}{(e^{-z}-1)(e^{-2z}-1)}\\
=&\ \frac{1}{e^{-z}-1}-\frac{2}{e^{-2z}-1}=\frac{1}{z}\left(\frac{-2z}{e^{-2z}-1}-\frac{-z}{e^{-z}-1}\right)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$ Consider the integrals $(1)$ and $(2)$, how does on show that
(1): $I=J$
(2): and $I=J={\ln(7+4\sqrt{3})\over 4\sqrt3}?$
$$\int_{0}^{\pi/2}{\tan x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx=I\tag1$$
$$\int_{0}^{\pi/2}{\tan^3 x\over \sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\... | To prove that $I = J$ , note that
$$I-J = \int^{\pi/2}_0 \frac{\tan x-\tan^3 x}{\sqrt{(1+\tan^2 x)(1+\tan^6 x)}}\mathrm dx = \int^{\pi/4}_{-\pi/4} \frac{\tan (\pi/4-x)-\tan^3 (\pi/4-x)}{\sqrt{(1+\tan^2 (\pi/4-x))(1+\tan^6 (\pi/4-x))}}\mathrm dx$$
We need to prove the follwing function is odd
$$f(x) = \frac{\tan (\pi/4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Finding a basis for a column space Let A be the matrix: $$\begin{pmatrix} 1&2&3&2&1&0\\2&4&5&3&3&1\\1&2&2&1&2&1 \end{pmatrix}$$.
Show that {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} is a basis for the column space of A. Find a "nice ba... | Building on the insights of @puhsu, you know you have 2 vectors that span the image, columns 1 and 3. Can you construct the target vectors in this basis?
$$
\left[ \begin{array}{r}
1 \\ 4 \\3
\end{array} \right]
=
\alpha
\left[ \begin{array}{r}
1 \\ 2 \\ 1
\end{array} \right]
+
\beta
\left[ \begin{array}{r}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximum value of expression: $\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$ What is the maximum value of
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$$
where $a,b,c$, and $d$ are real numbers?
| The main goal of my approach to the solution was to bring out an equality of the form $ab+bc+cd \le k(a^2+b^2+c^2+d^2)$, and thus, having our maximum value of this expression to be $k$
An initial attempt at the AM-GM inequality yields:
$\frac{1}{2}a^2+b^2+c^2+\frac{1}{2}d^2 \ge ab+bc+cd$
This result, as it turns out, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factoring a quartic polynomial over the integers with roots that are not integers The quartic polynomial
$$
1728(x - 3) - x^2(12^2 - x^2)
$$
factors "nicely" as
$$
(x^2 - 12x + 72) (x^2 + 12x - 72) = (x^2 - 12x + 72)(x - 6\sqrt{3} + 6)(x + 6\sqrt{3} + 6) \, .
$$
(Note that $1728 = 3(24^2)$.) How is this factorization ... | $$
\begin{align}
1728(x - 3) - x^2(12^2 - x^2) & = x^4 - 144 x^2 + 1728 x - 5184 \\
& = x^4 - 12^2 x^2 + 12 \cdot 12^2 x - 36 \cdot 12^2 \\
& = x^4 - 12^2(x^2 - 2 \cdot 6 \,x + 6^2) = \\
& = x^4 - 12^2(x-6)^2 = \\
& = \big(x^2 - 12(x-6) \big)\big(x^2 + 12(x-6)\big) = \cdots
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \c... | Well, let's the start the manipulation at the left hand side. Using the identity
$$\sin^2\theta=\frac{1-\cos 2\theta}{2}$$
we get
$$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2a+2b)}{2}+\frac{1-\cos(2a-2b)}{2}\\
&=1-\frac{1}{2}\bigg[\cos(2a+2b)+\cos(2a-2b)\bigg]\\
&=1-\frac{1}{2}\bigg[(\cos 2a\cos 2b-\sin 2a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
} |
Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$
Let $x,y$ be positive integers satisfying $2x^2-y^2 = 1$. Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$.
I wasn't sure how to use the fact that $x,y$ are positive integers satisfying $2x^2-y^2 = 1$. We could use the theory of Pell's e... | The solutions of this Pell equation are
$$ \pmatrix{x_n\cr y_n\cr} = \pmatrix{ 3 & 2\cr 4 & 3\cr}^n \pmatrix{1\cr 1\cr} $$
for nonnegative integers $n$. If $M = \pmatrix{3 & 2\cr 4 & 3\cr}$, we have
$M^6 \equiv I \mod 70$. Thus $(x_n, y_n) \mod 70$ (and therefore mod $10$ and mod $7$) is periodic with period $6$. We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Apostol Calculus Vol 2 Exercise 8.17 Q No. 3 Evaluate the directional derivative of $f$ for the points and directions specified
$f(x,y,z)=x^2+y^2-z^2$ at $(3,4,5)$ along the curve of intersection of the two surfaces $2x^2+2y^2-z^2=25$ and $x^2+y^2=z^2.$
The answer is supposed to be $0.$ I am not getting the right answe... | I think your unit direction vector should be : $\left(
\begin{array}{c}
-\frac{4}{5}\\
\frac{3}{5}\\
0\\
\end{array}
\right)$
(maybe I'll include picture later).
And that the answer should be : $0$ accordingly : $\left(0=\left(
\begin{array}{c}
-\frac{4}{5}\\
\frac{3}{5}\\
0\\
\end{array}
\right) \cdot \left(
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the roots of $f(x)=x^2+x+1$ modulo 7, and modulo 13, and modulo 91 Find the roots of $f(x)=x^2+x+1$ modulo $7$, and modulo $13$, and modulo $91$
I think for mod $7$ and $13$, it can be done by trial and error. But how about mod $91$?
| For any odd modulus,
$$x^2+x+1\equiv0\iff4x^2+4x+4\equiv0\iff(2x+1)^2\equiv-3$$
For the prime modulus $7$,
$$(2x+1)^2\equiv-3\equiv4\implies2x+1\equiv\pm2\implies4(2x+1)\equiv\pm8\implies x+4\equiv\pm1$$
which implies $x\equiv2$ or $x\equiv4$ mod $7$.
For the prime modulus $13$,
$$(2x+1)^2\equiv-3\equiv36\implies2x+1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\... | Your work is correct, but for the last step write:
$$
\left[ 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1
$$
because you have just done the integration as antiderivative: $F(x)=2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}$ and you have only to evaluate the primitive at the two limits of integration, so that your d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Polynomial division. If an f(x) is divided by x²+4, has 2x+3 of remainder.
If an f(x) is divided by x²+6, has 6x-1 of remainder.
If an f(x) is divided by (x²+4)(x²+6), has S(x) of remainder, then find S(4)!
I have this.
I wrote it in polynomial formula.
$f(x)=(x²+4).R(x)+(2x+3)$
And yes, i have x²=-4.
But, i had to ex... | Approach using complex numbers.
Notice that $x^2+4=0$ when $x=\pm 2i$ and $x^2+6=0$ when $x=\pm i\sqrt{6}$ since we have that $$f(x)=(x^2+4)R(x)+2x+3$$
From this we have that $$f(2i)=4i+3\\f(-2i)=-4i+3$$
Also we have that
$$f(x)=(x^2+4)Q(x)+6x-1\\f(i\sqrt{6})=6i\sqrt{6}-1\\f(-i\sqrt{6})=-6i\sqrt{6}-1$$
From this we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration of double integral Integration of double integral of.
\begin{equation}
\int_{-\frac{1}{4}}^{0}{\int_{\frac{1}{2}-\sqrt{x+\frac{1}{4}}}^{\frac{1}{2}+\sqrt{x+\frac{1}{4}}}} {e^{y^2}dy dx} + \int_{0}^{2}{\int_{-1+\sqrt{x+1}}^{1/2+\sqrt{x+\frac{1}{4}}}} {e^{y^2}dy dx} + \int_{2}^{8}{\int_{-1+\sqrt{x+1}}^{2}... | If we reverse the order of integration the integrands can be transformed into $ye^{y^2}\,dy$ and $y^2e^{y^2}\,dy$.
Here is the region of integration from desmos.com/calculator
First divide the region into five sections having unique left and right bounding functions of $y$
Giving the following region and subregions
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to write the form of partial fraction decomposition of the function. How to write the form of partial fraction decomposition of the function.
$$I= \dfrac{2\,x^3+24\,x^2+20\,x+10}{x^2+10\,x+25}$$
I used long division I get$$Quotient= 2\,x+4\\Remainder =-70\,x - 90\\$$
But don't know how to solve further steps. Thank... | Dividing the function you get$\text{Quotient}= 2\,x+4\\\text{Remainder} =-70\,x - 90\\\text{Divisor}= x^2+10\,x+25\tag*{}$So,$2\,x^3+24\,x^2 +20\,x+10=(2x+4)(x^2+10x+25)+(-70x-90)\tag*{}$Rewrite the function,$I = \dfrac{(x^2+10\,x+25)(2\,x+4)+(-70\,x-90)}{x^2+10\,x+25}\\=(2\,x+4) -\dfrac{70\,x+90}{(x+5)^2}\tag*{}$Now d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Polar form of a complex with square root
Find the polar form of the roots of the polinomial
$p(z)=iz^2-2z+1+2i$
I found the roots $z_1=-i-\sqrt{i-3}$ and $z_2=-i+\sqrt{i-3}$ but I don't know how to deal with the complex numbers inside the square root in order to isolate both real and imaginary parts.
| To compute the square root of a complex number: $(x+iy)^2=-3+i$, expand the l.h.s. and identify the real and imaginary parts:
$$x^2-y^2=-3, \quad xy=\frac12.$$
You simplify the computation observing the square of the modulus of $x+iy$ is the modulus of $-3+i$:
$$x^2+y^2=\sqrt{10}.$$
So we have a linear system in $x^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Probability Interview Question - Brain teaser
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?
So I have
$$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2... | I've reasoned like this, as usual starting from the simplest case:
A has a 3-sided dice, B a 2-sided one.
$P(\mathbf{B} wins)$ = $P(\mathbf{B}_{dice}=1)$ * $P(\mathbf{A}_{dice}=1)$ +
$P(\mathbf{B}_{dice}=2)$ * $P(\mathbf{A}_{dice}<=2)$ =
= $\frac{1}{2}$ * $\frac{1}{3}$ + $\frac{1}{2}$ * $\frac{2}{3}$ =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 5
} |
Who came up with the identity $a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$ Though we can prove this it is not something that comes up intutively.
Our ancestors must have been interested in factorising $a^3+b^3+c^3$ but why find it for $a^3+b^3+c^3-3abc$ ?
| I don't know the history of this identity, but here is a derivation based on the properties of determinants.
$$\begin{vmatrix}
a &c &b\\
b &a &c\\
c &b &a
\end{vmatrix} =
\begin{vmatrix}
a+b+c &a+b+c &a+b+c\\
b &a &c\\
c &b &a
\end{vmatrix} = (a+b+c) \begin{vmatrix}
1 &1 & 1\\
b &a &c\\
c &b &a
\end{vmatrix} $$
Expandi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Prove: If $a$ and $b$ are odd perfect squares, then $a + b$ is not a perfect square.
Proof by Contradiction:
If $a$ and $b$ are odd perfect squares then $a = (2k+1)^2$ and $b = (2r + 1)^2$.
Assume $a + b$ is a perfect square.
\begin{align... | In you second line you have shown that a+b mod 4 = 2. Is this possible if a+b is a perfect square ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that sum is an integer. Prove that: $\sqrt{2-\frac{1}{1^2+\sqrt{1^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{2^2+\sqrt{2^4+\frac{1}{4}}}}+\sqrt{2-\frac{1}{3^2+\sqrt{3^4+\frac{1}{4}}}}+\cdots+\sqrt{2-\frac{1}{119^2+\sqrt{119^4+\frac{1}{4}}}}$ is an integer.
I noticed that $n^4+\frac{1}{4}$ is rather interesting, because i... | Since
$$\frac{1}{k^2+\sqrt{k^4+\frac 14}}=\frac{k^2-\sqrt{k^4+\frac 14}}{(k^2)^2-(k^4+\frac 14)}=-4k^2+2\sqrt{4k^4+1}$$
we have
$$\sum_{k=1}^{n}\sqrt{2-\frac{1}{k^2+\sqrt{k^4+\frac 14}}}=\sum_{k=1}^{n}\sqrt{2+4k^2-2\sqrt{4k^4+1}}\tag1$$
Here, since we have that
$$4k^4+1=4k^4+4k^2+1-4k^2=(2k^2+1)^2-(2k)^2=(2k^2+2k+1)(2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"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.