Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$y'=\frac{y^2}{2x(y-x)}$ I'm trying to solve the following differential equation:
$$y'=\frac{y^2}{2x(y-x)}$$
It is supposed to have a relatively easy general solution, but I can't find it.
I've tried several things, the most promising of which is the change $z(x)=y(x)-x$, which yields (if I haven't made any mistake):
$... | $$y'=\frac { y^{ 2 } }{ 2x(y-x) } \\$$
Solve respect to the $x $
$$\\ \\ { y }^{ 2 }{ x }^{ \prime }-2x(y-x)=0\\ \\ { x }^{ \prime }-2\frac { x }{ y } +2\frac { { x }^{ 2 } }{ { y }^{ 2 } } =0\\ x=zy\\ { x }^{ \prime }=z+y{ z }^{ \prime }\\ z+y{ z }^{ \prime }-2z+2{ z }^{ 2 }=0\\ y{ z }^{ \prime }-z+2{ z }^{ 2 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1417278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmat... | For the general formula I get it to $e^A = I \cosh(\sqrt{\alpha\beta}) + A\sinh(\sqrt{\alpha\beta})/\sqrt{\alpha\beta}$ where the same square root is selected everywhere (the one under $\cosh$ doesn't matter due to symmetry).
You get this by checking the powers of $A$:
*
*$A^0 = I$
*$A^1 = A$
*$A^2 = (\alpha\beta)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,coul... | For brevity write $a=\cos A$, $b=\cos B$, $c=\cos C$. Then we have
$$C=\pi-A-B$$
and so
$$c=-\cos(A+B)=-ab+\sin A\sin B$$
and so
$$(c+ab)^2=(1-a^2)(1-b^2)\ .$$
Expanding, rearranging and using the identity you are given,
$$a^2+b^2+c^2=1-2abc=ab+bc+ca\ .$$
Now
$$(a-b)^2+(b-c)^2+(c-a)^2=2a^2+2b^2+2c^2-2ab-2bc-2ca$$
and ... | {
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"url": "https://math.stackexchange.com/questions/1419328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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simultaneous equation help Can someone help me with to solve this system of equations ?
$$
\left\{
\begin{array}{c}
y=x+1 \\
y^2+2x^2=2
\end{array}
\right.
$$
| We know from equation one that $y = x+1$. Plugging this value into equation two we get
\begin{align*}
y^2+2x^2 &= (x+1)^2+2x^2 \\
&=x^2+2x+1+2x^2\\
&=3x^2+2x+1 \\
\end{align*}
We know that $3x^2+2x+1 = 2$ from equation two. Simplifying we get $$3x^2+2x-1 = 0$$$$(x+1)(3x-1) = 0 \tag{factoring}$$
$$x = -1, \frac{1}{3}$$... | {
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"timestamp": "2023-03-29T00:00:00",
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Polar to Rectangular Coordinates Problem:
Transform the following equation from polar to rectangular coordinates.
\begin{eqnarray*}
\rho &=& \frac{2}{1 - \cos \theta} \\
\end{eqnarray*}
Answer:
Recall that:
\begin{eqnarray*}
\rho &=& {(x^2+y^2)} ^ {\frac{1}{2}} \\
\cos \theta &=& \frac{x}{\rho} =\frac{x}{(x^2+y^2)^{\f... | Let's check with $\theta=60^\circ$ so that $\rho=4$. Then: $x=2$, $y=2\sqrt3$, $y^2=12$. But $4(x+2)=16$ and $4x+4=12$. So you are right and the book is wrong.
| {
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"url": "https://math.stackexchange.com/questions/1423553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Prove by Induction: $8^n - 3^n$ is divisible by $5$ for all $n \geq 1$ Prove by Induction that for all $n \geq 1$ we have
$$8^n - 3^n \text{is divisible by 5} ...(*)$$
My proof so far
Step 1: For $n=1$ we have $8^1 - 3^1 = 8 - 3 = 5$ which is divisible by 5.
Step 2: Suppose (*) is true for some $n=k\geq 1$ that is... | By the binomial theorem, $8^n=(5+3)^n=5a+3^n$.
Induction is used to prove the binomial theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Simple Combination of balls I encounter this problem but I cannot figure out the solution. Please help.
Given that there are 4 white balls, 2 red balls and 1 black balls of the same size. 4 balls are drawn at random. Find the number of ways this can be done.
The given answer is 1 + 4C1 + 4C1 + 4C2 + 2 × 4! ÷ 2! = 39.
T... | For the answer where balls of the same color are indistinct and order does matter, look above to Dylan's answer.
Assuming all balls are distinct (even those of the same color) and assuming order does not matter
You can break this into cases:
$\begin{array}{|c|c|c||c|}
\hline
\text{White}&\text{Red}&\text{Black}&\text{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Get number of additions that lead to a specific sum with given summands Suppose you have $n$ variables. Each of these variables (e.g. $a$) have their own interval between $0$ and $a_i$ (whole numbers). The only rule is that all these variables have to add up to a given value $s$.
An example:
$n = 3$, so the three varia... | Suppose you have $k$ variables, and you want to add it up to $n$:
$$f_n(x) = \frac{1}{n!}\frac{d^n}{dx^n}\prod_{i=1}^k \frac{1-x^{a_i+1}}{1-x}$$
Then your solution is $f(0)$.
To come up with this, start with the generating functions for your constraints:
$$(1+x+...x^{a_i}) = \frac{1-x^{a_i+1}}{1-x}$$
Suppose we compute... | {
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Find all real functions $f$ such that $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x,y$ we have:
$f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$
I have solved this problem but the solution is "bruteforce", so I wanted to ask if there is a more elegant way... | I claim that the only possible solutions to the functional equation
$$f\big(xf(y)\big)=(1-y)f(xy)+x^2y^2f(y)\tag0\label0$$
are $f(x)=0$ and $f(x)=x-x^2$.
To show that, first let $y=1$ in \eqref{0} and you'll get $f\big(xf(1)\big)=x^2f(1)$. Now assuming $f(1)\neq0$ we'll have $f(x)=\frac{x^2}{f(1)}$ for every real numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be:
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-... | The question is not entirely clear. Your expression can be also writen as
$$
\frac{3 x}{\left(x^2+2 x+4\right)^2}+\frac{x+2}{x^2+2 x+4}+\frac{1}{(x-2)^2}
$$
hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability that the equation $x^2 + k_1 x + k_0 = 0$ has real solutions
$k_1$, $k_0$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $x^2 + k_1 x + k_0 = 0$ has real solutions?
This is a subproblem of another problem, an... | It has real solutions iff $\Delta=k_1^2-4k_0\ge 0$.
$$P=\frac{\sum_{i=1}^{20}\lfloor\frac{i^2}{4}\rfloor+80\cdot 100}{10000}$$
$(2k+1)^2\equiv 1\pmod{4},\, (2k)^2\equiv 0\pmod{4}$. Therefore (see here and here):
$$P=\frac{\frac{1^2+3^2+5^2+\cdots+19^2-10}{4}+\frac{2^2+4^2+6^2+\cdots+20^2}{4}+8000}{10000}$$
$$=\frac{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find solution to system of differential equations with initial conditions I have a system
$$
\begin{cases}
x_1'(t) = -x_2(t) \\
x_2'(t) = -x_1(t)
\end{cases}
$$
I have found the linearly independent solutions
$$
\begin{split}
\vec{x}(t) &= e^{\lambda_1 t} \vec{v}_1 = e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}, \\
\vec{x}(... | Apart from some confusing notation with indices $(1,2)$ - better replace that with e.g. $(a,b)$ - you've got everything OK. About the initial conditions:
$$
\vec{x}(t) = c_1 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}
\quad \Longrightarrow \quad
\vec{x}(0) = c_1 \begin{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you prove $37^{100} - 37^{20}$ is a multiple of $10$ using induction? I've tried making it $37^{5n} - 37^n$ is a multiple of $10.$ Then I made the base case be $n = 0,$ so $1 - 1 = 0$ which is a multiple of $10.$ I assumed $37^{5n} - 37^n$ is a multiple of $10$ for all $n \geqslant0.$ The inductive step is reall... | I know OP requested induction, but I would like to post this anyway on the off-chance that someone finds it interesting:
This is the same as proving that $10 {\large\mid} 37^{80}-1$, since $37^{100}-37^{20} = 37^{20}(37^{80}-1)$ and $37^{20}$ has only powers of $37$ as divisors since $37$ is prime.
Clearly $2{\large\mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
... | Since :$$\left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ { x }^{ 2 } } +\sqrt [ 3 ]{ x } +1 \right) =x-1\\ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) =x-1$$
so we have
$$\lim _{ x\rightarrow 1 }{ \left( \frac { \sqrt [ 3 ]{ x } -1 }{ 2\sqrt { x } -2 } \right) = } \frac { 1 }{ 2 } \lim _{ x\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
| A little deus ex machina but this seems to be related to the golden ratio $\phi = \frac{1+\sqrt 5}{2}$, which is the solution of $x+\frac1x = 1$, and to the Fibonacci sequence $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657, \ldots$
In particular $x^3+\frac1{x^3} = 18$ has the so... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Solving quartic equations Given the following quartic equation:
$$x^4-2x^3-7x^2+8x+12=0$$
Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?
| Notice, $$x^4-2x^3-7x^2+8x+12=0$$ substituting $x=-1$ we get $$(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12=0\iff 0=0$$ Hence, $x=-1$ is a root of given quartic equation i.e. $(x+1)$ is a factor of $x^4-2x^3-7x^2+8x+12$, now we have
$$(x+1)(x^3-3x^2-4x+12)=0$$
further factorizing $x^3-3x^2-4x+12=(x-3)(x^2-4)$
$$(x+1)(x-3)(x^2-4)=0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Summation of simple sequence This is a fairly simple question, I'm sure, but I appear to be having trouble. What is the result of the following sequence:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+ .... + \frac{n}{2^{n}}.$$ ?
Thanks
| $$2S_n-S_n=\left(\frac11+\frac{2}{2}+\frac{3}{4}+\cdots+ \frac{n}{2^{n-1}}\right)-\left(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+ \cdots+ \frac{n}{2^{n}}\right)\\
=1+\frac{1}{2}+\frac{1}{4}+ \cdots+ \frac{1}{2^{n-1}}-\frac n{2^n}=2-\frac{n+2}{2^n}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore,
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any... | Suppose $7+5\sqrt{2}=(x+y\sqrt{2})^3 = (x^3+6 x y^2)+(3 x^2 y +2 y^3)\sqrt2 $.
If we're looking for integer solultions, then we must have $7=x^3+6 x y^2$ and $5=3 x^2 y +2 y^3$, because $\sqrt2$ is irrational.
Consider $5=3 x^2 y +2 y^3=(3x^2+2y^2)y$. Since $5$ is prime, $y$ must be $1$ or $5$ because $3x^2+2y^2 \ge 0$... | {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $3\mid n^3 − 4n$ Prove that $n^3 − 4n$ is divisible by $3$ for every positive integer $n$.
I am not sure how to start this problem. Any help would be appreciated
| Well, one can start off by factoring $n^3 - 4n$:
$n^3 - 4n = n(n^2 - 4)$
$= n(n + 2)(n - 2); \tag{1}$
next we note that every $n \in \Bbb Z$ is of one of the three forms
$n = 3q, \tag{2}$
$n = 3q + 1, \tag{3}$
or
$n = 3q + 2; \tag{4}$
this follows from the basic properties of Euclidean division applied to $n$ with $3$ ... | {
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integrate $\int\sqrt{x^2-1}dx$ unproven step In order to find $\int\sqrt{x^2-1}dx$ one makes substitutions $x=\sec(\theta)$ , $dx=\sec(\theta)\tan(\theta)d\theta$ and $\sqrt{x^2-1}$ = $\sqrt{\tan^2(\theta)}$. Then you find $\int{\tan^2(\theta) \sec(\theta)}d\theta$ directly as easy as $1+1=2$ it seems from many online ... | $\int\sqrt (x^2-1)dx$
(1)Use a substitution of $x=cosh \theta$, $dx=sinh\theta$
=$\int\sqrt(cosh^2-1).(sinh\theta)d\theta$
=$\int\sqrt sinh^2\theta.(sinh\theta)d\theta$
=$\int\sinh^2\theta d\theta$
Using double angle formulae: $cosh 2\theta=1+2sinh^2\theta$
Hence, $sinh^2\theta=\frac{1}{2}(cosh 2\theta-1)$
=$\frac{1}{2... | {
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Finding $\lim_{x\to 1 }\left( \frac{x-1-x\ln x }{x(x-1)^{2}}\right)$ without L'Hopital's rule I would like to calculate this limit but without using L'Hopital's rule
$$\lim_{x\to 1 }\left( \frac{x-1-x\ln(x) }{x(x-1)^{2}}\right)=-\dfrac{1}{2}$$
My thoughts:
Note that $\lim\limits_{x\to 1 }\dfrac{\ln(x)}{x-1}=1$ so I tri... | Let us first make $x=y+1$. So, $$\dfrac{x-1-x\ln(x) }{x(x-1)^{2}}=\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$$ Now, consider Taylor expansion around $y=0$ $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ So, the numerator write $$y-(y+1) \log (y+1)=-\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)$$
I am sure that ... | {
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how can I prove it is a convergent sequence and its limit is $\sqrt{a}$ Let we have the following sequence $$x_1=1$$
$$x_{n+1}=\frac{x_n^2+a}{2x_n}$$ such that $$a>0$$ how can I prove it is a convergent sequence and its limit is $\sqrt{a}$
| Calculate $ x_{n+1} - \sqrt{a} $,
$$ x_{n+1} - \sqrt{a} = \frac{(x_n - \sqrt{a})^2}{2x_n},$$
and
$$ x_2 = \frac{a + 1}{2} \geqslant \sqrt{a}.$$
Thus by induction we can prove that $\forall n \geqslant 2,\,\, x_n \geqslant \sqrt{a}$.
Now that
$$ x_{n+1} - \sqrt{a} = \frac{x_n - \sqrt{a}}{2x_n} \cdot (x_n - \sqrt{a}) ... | {
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Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating th... | Note that $\sqrt{c+1}\geq \sqrt{c-1}$ and so $\frac{1}{\sqrt{c}+\sqrt{c-1}}\geq \frac{1}{\sqrt{c}+\sqrt{c+1}}$. Write $\sqrt{c}-\sqrt{c-1}=\frac{1}{\sqrt{c}+\sqrt{c-1}}$ and $\sqrt{c+1}-\sqrt{c}=\frac{1}{\sqrt{c}+\sqrt{c+1}}.$ Thus you have the result.
| {
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solving ODE : $xy'=\sin(x+y)$ May I ask some hints or solution for solving $xy'=\sin(x+y)$ ??
My idea was substitution : $x+y=u$ , then it becomes $x(u'-1)=\sin u$
and still I can't approach further..
| Let $u=x+y$ ,
Then $y=u-x$
$y'=u'-1$
$\therefore x(u'-1)=\sin u$
$x\dfrac{du}{dx}-x=\sin u$
$x\dfrac{du}{dx}=\sin u+x$
$\dfrac{du}{dx}=\dfrac{\sin u}{x}+1$
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20Equ... | {
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Maths Challenge IMO I saw this problem on a maths challenge book.
Given any set $A=\{a_1 ,a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1+a_2+a_3+a_4$ by $S_A$. Let $n_A$ denote the number of pairs $(i,j)$ with $1\leq i\leq j\leq 4$ for which $a_i +a_j$ divides $S_A$. Find all sets of four ... | WLOG $a_1<a_2<a_3<a_4$. For every permutation $a,b,c,d$ of $a_1,a_2,a_3,a_4$ we have $$a+b\mid (a+b+c+d)-(a+b)=c+d\implies \boxed{a+b\mid c+d}$$
Since $a_3+a_4>a_1+a_2$ and $a_2+a_4>a_1+a_3$ we can not have $a_3+a_4\mid a_1+a_2$ and $a_2+a_4\mid a_1+a_3$, so $n_A\leq 4$.
Since $a_1+a_4\mid a_2+a_3\;\;(*)$ only iff $$k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $p^3+4$ is prime If $p$ and $p^2 +8$ are both prime number, prove that $p^3 +4$ is also prime.
I was thinking using two cases for even and odd. So $p$ is even thus there is no prime, the statement does not hold. So, $p$ is odd, $p$ has to be of the form of $2k+1$.
let $p=2k+1$ which is prime
$(2k+1)^2 + 8 =... | Hint: if you don't know the theory of quadratic residue, you can note that every prime $p>3$ can be written in the form $6n\pm1$. It has remainder $\pm1$ when divided by $3$. Note therefore that for $p>3$, $p^2+8$ is divisible by $3$.
Indeed $$p^2+8=(6n\pm1)^2+8\equiv (\pm1)^2\!+8\equiv 0\pmod{\!3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A given polynomial equation of 5 degree has three equal roots . Given equation is $x^5-10a^3x^2+b^4x+c^5=0$ which has 3 equal roots. What I know is that since its a 5th degree equation therefore it must have 5 roots out (of which 3 are equal).
Aim is to establish the relationship between the constants $a,b$ and $c$.
Op... | Let $$f(x) = x^5 - 10a^3x^2 + b^4x + c^5 = (x - m)^3g(x)$$ where $m$ is the repeated root, and $g(x)$ is some second-order polynomial.
Then, differentiating and substituting $x = m$,
$$5x^4 - 20a^3x + b^4 = (x - m)^3g'(x) + 3(x - m)^2g(x)$$
$$5m^4 - 20a^3m + b^4 = 0$$
Differentiating another time, and similarly substi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta (2)$?
$$\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta(2)$$
By using numerical calculation, I found this relationship between the values of zeta function at even integers and $\zeta(2)$, but this needs proving, any help?
| Taking the derivative of
$$
\sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1}
$$
and multiplying by $x$ yields
$$
\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2}
$$
Therefore,
$$
\hspace{-1cm}\begin{align}
\frac83\sum_{n=1}^\infty\frac{n}{2^{2n}}\zeta(2n)
&=\frac83\sum_{n=1}^\infty\frac{n}{4^n}\sum_{k=1}^\infty\frac1{k^{2n}}\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
About Integration $\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx$ What i want to prove is following integral
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx=\frac{1}{2} \sqrt{\left(1+\sqrt{2}\right) \pi }
\end{align}
can you give some explicit method to obtain this result?
| So let us follow through on Eclipse Sun's suggestion. In doing so we will find that while the double integral can indeed be separated into two parts (two double integrals), neither of them turns out to be equal to zero.
Let
$$I = \int^\infty_{-\infty} e^{-x^2} \cos (x^2) \, dx = 2 \int^\infty_0 e^{-x^2} \cos (x^2) \, d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the following limit? $\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$ What is the following limit?
$$\lim\limits_{x\to 0}=\frac{(1+x)^5-1-5x}{x^2+x^5}$$ Should I calculate the exact value of $(1+x)^5$?
| Using $$\displaystyle (1+x)^5 = \binom{5}{0}+\binom{5}{1}x+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4+\binom{5}{5}x^5$$
and Using $$\displaystyle \binom{n}{r} = \frac{n!}{r!\times (n-r)!}\;,$$ Where $n!=n\times (n-1)\times (n-2)\times ...2\times 1$
So we get $$\displaystyle (1+x)^5 = 1+5x+10x^2+10x^3+5x^4+x^5$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Power series expansion of ODE So I have this equation:
$$(1-x^2)y''-2xy'+\lambda y=0$$
Where it is given that the solution can be written as the following convergent power series:
$$y(x) =\sum_{n=0}^\infty a_n x^n$$
And my goal is to arrive to this relationship of the coefficients:
$$a_{n+2} = \frac{n(n+1)-\lambda}{(n+... | $y′=\sum_{n=-1}^{\infty}a_{n+1}(n+1)x^n$
$y′'=\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n$
$\lambda y = 2xy'-(1-x^2)y''$
$\lambda y = 2x[\sum_{n=-1}^{\infty}a_{n+1}(n+1)x^n]-(1-x^2)[\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n]$
$\lambda y = [\sum_{n=-1}^{\infty}2a_{n+1}(n+1)x^{n+1}]-[\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n]... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| Let $a_n = x^n + \frac{1}{x^n}$. Then $x^2 = 3x - 1$ implies $a_{n+2} = 3a_{n+1} -a_n$ for all $n$ and also $a_1=3$. Since $a_0=2$, we get
$a_2 = 3a_1 -a_0 = 7$
$a_3 = 3a_2 -a_1 = 18$
$a_4 = 3a_3 -a_2 = 47$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Two apparently different evaluations of $\int \frac{x-1}{9x^2-18x+17}dx$
Evaluate the indefinite integral $$\int \frac{x-1}{9x^2-18x+17} \, dx .$$
This is an exercise from a book I'm studying. It gives the answer as:
$$\ln(9x^2 -18x+17)^\frac{1}{18} +C .$$
This is an easy integral. You just notice that the numerator... | Despite appearances to the contrary, there's no problem here: The two expressions actually differ by a constant, but this equivalence is buried in a few special identities involving the logarithm function, namely $$\log (ab) = \log a + \log b$$ and $\log (a^b) = b \log a$ (both for appropriate $a, b$).
More specificall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving a property of a prime number (Elementary number theory) Prove the following statement.
For all prime numbers $a$, $b$, and $c$, $a
^2 + b^2
\neq c^2$
.
What i tried
Proving by contradiction
Assume the negation of the statement
There exists prime numbers $a$, $b$, and $c$ such that $a
^2 + b^2
= c^2$
Since we kn... | A more elementary approach: since the only quadratic residues $\pmod{3}$ are $0$ and $1$, if $a^2+b^2=c^2$ at least one number between $a$ and $b$ has to be a multiple of three. With a similar argument $\pmod{4}$ and $\pmod{5}$, we have that $a^2+b^2=c^2$ implies that at least one number between $a$ and $b$ is even, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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What do I do next when trying to find the derivative of this fraction? I'm trying to find the derivative of this equation: $-\frac{3(x-6)}{2\sqrt{9-x}}$
The quotient rule: $\frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ where $g(x)$ and $f(x)$ are functions.
So I take out the constants and I'm le... | Until here: $$\frac{-3}{2}[\frac{(9-x)^{1/2}(1)-(x-6)[(\frac{1}{2})(9-x)^{-1/2}(-1)]}{9-x}]$$ you're good. This simplifies to: $$-\frac{3}{2}\left[\frac{\sqrt{9-x}+ \frac{1}{2}(x-6)\frac{1}{\sqrt{9-x}}}{9-x}\right] = -\frac{3}{2}\left[\frac{\color{red}{(9-x)}+ \frac{1}{2}(x-6)}{(9-x)^{3/2}}\right] = -\frac{3}{4} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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$x^{2000} + \frac{1}{x^{2000}}$ in terms of $x + \frac 1x$. If $x + \frac{1}{x} = 1$, then what is
$$ x^{2000} + \frac{1}{x^{2000}} = ?$$
| This is a not so elegant approach.
Let $a_n=x^n+\frac1{x^n}$ then
$$a_n\left(x+\frac1x\right)=a_{n+1}+a_{n-1}$$ or $$a_{n+1}=a_{n}-a_{n-1}$$
We have $a_1=1$ and $a_2=-1$, hence solving recursively - plugging - we obtain \begin{align}
a_{6n-4}=a_{6n-2}&=-1\\
a_{6n-3}&=-2\\
a_{6n}&=2\\
a_{6n-1}=a_{6n+1}&=1
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Solution to Complex Equation $\cos z = 2i$ I tried solving the equation $\cos{z}=2i$ where $z$ is a complex number. The solution $i$ have ended up with is $z$, $= (4k+1)\frac{\pi}{2} - i\ln{(2+\sqrt{5})}$. However the text book solution is $z=(2k+1)\frac{\pi}{2} - (-1)^k\ln{(2+\sqrt{5})}i$. Are the results geometrical... | $$\cos z = \frac{1}{2}\left(e^{iz}+e^{-iz}\right).$$ Solve first for $w=e^{iz}$.
$$w^2-4iw + 1=0$$
so:
$$w=\frac{4i\pm\sqrt{-16-4}}{2} = (2\pm\sqrt{5})i$$
So we have to deal with two cases with care. If $w=(2+\sqrt{5})i$, you get your solution:
$$z = (4k+1)\frac{\pi}{2} - i\ln(2+\sqrt{5})$$
If $w=(2-\sqrt{5})i$, then u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct
Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red... | Your belief that
$$\def\Q{\mathbb{Q}}
\Q(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+ \sqrt{3}):a,b\in \Q\}
$$
is wrong, as well as
$$
\Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt{2}+ c\sqrt{3}:a,b,c\in \Q\}
$$
The second claim can be immediately dismissed, because this would mean that $[\Q(\sqrt{2}+\sqrt{3}):\Q]=3$, but clearly $\Q(\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$ A straight line L intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$,then square of th... | Hint.
I give you a hint that you can apply to find the distance you're looking for.
So you have two lines $$\begin{cases} L_1 \equiv \frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10} \\
L_2 \equiv \frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}\end{cases}$$
For $L_1$, you can find a point $A_1 \in L_1$, for example $A_1=(-2,-6,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels l... | Using C-S inequality, we can obtain
$$\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)^2\leq3\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).$$
Since $(b+c)\left(\frac1{b}+\frac1{c}\right)\geq4$, we have $\frac{a}{b+c}\leq\frac1{4}\left(\frac{a}{b}+\frac{a}{c}\right)$, similarly we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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Limit Infinity Minus Infinity form without using infinite series Expansions Evaluate $$L=\lim_{x \to 0} \frac{1}{\sin^2x}-\frac{1}{\sinh^2x}$$ If we take L.C.M and use LHopital's Rule it becomes quite Tedious. I tried in this way.
$$
\sinh x=\frac{e^x-e^{-x}}{2}
$$
Now
$$\lim_{x \to 0}\frac{e^x-e^{-x}}{x}=2 \implies ... | Write the original expression as
\begin{eqnarray*}
\frac{1}{\sin ^{2}x}-\frac{1}{\sinh ^{2}x} &=&\frac{\sinh ^{2}x-\sin ^{2}x}{%
\sin ^{2}x\sinh ^{2}x} \\
&=&\left( \frac{\sinh x-\sin x}{x^{3}}\right) \times \left( \frac{\sinh
x+\sin x}{x}\right) \times \left( \frac{x}{\sin x}\right) ^{2}\times \left(
\frac{x}{\sinh x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not. verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not.
I tried to write the left term in a form $y^2 \equiv a \pmod{75}$, where $a \in \mathbb{Z}$ so then I can use quadratic repriocity laws to solve this problem. But i get a fraction:
$8x^2... | Hint:
$$
8x^2-2x-3=(4x-3)(2x+1)
$$
Is $2$ invertible mod $75$?
Since a full answer has been given in another post, I will add the following.
Indeed $2\cdot38\equiv1\pmod{75}$, so there are solutions.
Although not required by the question, we can find all the solutions as follows.
$$
\begin{align}
8x^2-2x-3
&=(4x-3)(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve combinatorics with variable set sizes? The question: How many ways to put $15$ students into groups, such that each group has $3$~$5$ students?
If the group was $3$ equal groups of $5$, then the answer would be $\cfrac{15!}{5! \cdot 5! \cdot 5! \cdot 3!}$, and if it was $5$ equal groups of $3$, then the an... | Since it is unclear whether the groups are labelled or unlabelled,
I'll treat them as labelled, and indicate the correction needed if they are unlabelled.
Groups can be$\;$ 5-5-5, $\;$ 5-4-3-3, $\;$ 4-4-4-3, $\;$ or$\;$ 3-3-3-3-3 , so # of ways for labelled groups is:
$$\frac{15!}{5!\cdot 5! \cdot5!}+\frac{15!}{5!\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximizing $3 x^2+2 \sqrt{2} x y$ with $x^4+y^4=1$ I want to have maxizing value of $3 x^2+2 \sqrt{2} x y$ when $x^4+y^4=1$, $x>0,y>0$.
How can I solve it.
| here is one way to do this: let us parametrize the constraint $$x^4 + y^4 = 1, x > 0, y > 0 \text{ by } x = \sqrt{\cos t}, y = \sqrt{\sin t}, 0 < t < \pi/2.$$ we need to maximise $$ f = 3x^2 + 2\sqrt 2xy=3\cos t + 2\sqrt 2\sqrt{\cos t \sin t} = 3\cos t + 2\sqrt{\sin 2t}, 0 < t < \pi/2$$
taking the derivative of $f$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $\lim_{n \longrightarrow \infty} \frac{3n^2+2n}{4+3n^2}=1$ I need show that $\forall \, \epsilon > 0$, exist $M \in \mathbb N$ such that, $\left|\frac{3n^2+2n}{4+3n^2}-1\right| < \epsilon$ if $n \geq M$.
Let's consider
$\left|\frac{3n^2+2n}{4+3n^2}-1\right|=\left|\frac{2n-4}{4+3n^2}\right|<\left|\frac{2n-4}{3... | If $n \geq 1$, then
$$
\bigg| \frac{3n^{2}+2n}{4+3n^{2}} - 1 \bigg|
=
\bigg| \frac{2n-4}{4+3n^{2}} \bigg|
\leq
\frac{2n}{4+3n^{2}} + \frac{4}{4+3n^{2}}
<
\frac{2}{n} + \frac{4}{n^{2}};
$$
taking any $\varepsilon > 0$, if $n \geq \max \{ \lceil \frac{4}{\varepsilon} \rceil + 1, \lceil \sqrt{\frac{8}{\varepsilon}} \rcei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if 2 divides $x^2-5$ then 4 divides $x^2-5$ so I have to prove this and I use two different types of proof and I came to a contradicting result. Can someone point out an error I made?
Using a direct proof:
If 2 divides $x^2-5$ than $x^2-5=2k$ from that $x^2=2(k+2)+1$. From that we know x is odd. since x is o... | It is clear that $x^2-5$ is not divisible by $2$ when it is of the form $4k-1$ or $4k-3$. So suppose $x^2-5 = 4k-2$. This quantity is certainly divisible by $2$ if there is such an integer $k$, which is why we need to do a bit more analysis to make sure something fishy isn't going on. From this we have $$x^2 = 4k+3$$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove this $\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$ Let $x,y>0$, and $x+y=2$, show that
$$\sqrt{x^2+3}+\sqrt{y^2+3}+\sqrt{xy+3}\ge 6$$
I tried using Minkowski inequality
$$\sqrt{x^2+3}+\sqrt{y^2+3}\ge\sqrt{(x+y)^2+(\sqrt{3}+\sqrt{3})^2}=\sqrt{4+12}=4$$
But
$$\sqrt{xy+3}\le\sqrt{\dfrac{(x+y)^2}{4}+3}=2$$
so i... | Let $\sqrt{x^2+3}+\sqrt{y^2+3}=t$.
We need to prove that
$$\sqrt{x^2+3}+\sqrt{y^2+3}-4\geq2-\sqrt{xy+3}$$ or
$$\frac{x^2+3+y^2+3+2\sqrt{(x^2+3)(y^2+3)}-16}{t+4}\geq\frac{1-xy}{2+\sqrt{xy+3}}$$ or
$$\frac{2(x^2+3)+2(y^2+3)-16-\left(\sqrt{x^2+3}-\sqrt{y^2+3}\right)^2}{t+4}\geq\frac{4-4xy}{4(2+\sqrt{xy+3})}$$ or
$$\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Proving binomial summation identity using generating functions An exercise for class requires me to prove the following identity using generating functions:
$$\sum_{k=0}^{m/2} (-1)^k {n \choose k} {n+m-2k-1 \choose n-1} = {n \choose m}$$ for all $m \leq n$ and $m$ is even.
I've tried a bunch of things but can't wrap my... | Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor m/2\rfloor} (-1)^k
{n\choose k} {n+m-2k-1\choose n-1}$$
where $n\ge m.$
Re-write this as
$$\sum_{k=0}^{\lfloor m/2\rfloor} (-1)^k
{n\choose k} {n+m-2k-1\choose m-2k}.$$
Now introduce
$${n+m-2k-1\choose m-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-2k+1}} (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum from the system of equations
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x +3y + 4z$.
This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x... | Hint:
If you add up the first equation and the second times $-1$, you get:
$$x-z=z^2-x^2=-(x-z)(x+z)$$
Edit:
Therefore, $x=y=z$, because otherwise you can divide by $x-y$ to get $x+y=-1$ and, using the first set of equations, $z^2+1=-1$ and $z^2=-2$, which is not possible. The same goes for y and z. Using this in the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically... | Since $ \sin\left(\frac{6\pi}7\right) = \sin\left(\frac{\pi}7\right) $, then the expression is positive because $ \sin\left(\frac{2\pi}7\right) - \sin\left(\frac{\pi}7\right) > 0 $.
Let $ I $ denote this expression, then $ I > 0 $. Consider
$ I^2 = \left[\underbrace{ \sin^2\left(\frac{2\pi}7\right) + \sin^2\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find all polynomial $f(x)$ such that $x^3-1\mid f(x)g(x)-1$ This is a problem that has haunted me for more than some days. Not all the time - but from time to time, and always on windy or rainy days, it suddenly reappears in my mind:
Find all polynomial $f(x)$,such $f(x)\in Z[x],\deg{(f(x))}\le 2$,and there exist $g(x... | Let $\rho = e^{2\pi i/3} = \frac{-1 + i\sqrt{3}}{2}$. In $\mathbb{C}[x]$, we have the factorisation $x^3-1 = (x-1)(x-\rho)(x-\rho^2)$, so if $x^3-1 \mid f(x)g(x) - 1$, we must have $f(1)g(1) = f(\rho)g(\rho) = f(\rho^2)g(\rho^2) = 1$. Since $f,g$ are assumed to be polynomials with integer coefficients, we know that we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1482417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\lim... | Hint: If you want another approach you could probably do a series of deconvolutions in the Fourier domain. The euclidean algorithm should work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Problem related to maximum and minimum value of a trigonometric function I have used the following technique to calculate the maximum value of the function...
but I couldn't proceed with next step..
can anyone guide me please?
Find the maximum and minimum value of the function $$f(x) = \sin^2(\cos x)+\cos^2(\sin x)$$
I... | Using Calculus
Let $$f(x) = \sin^2(\cos x)+\cos^2(\sin x) = 1-\cos^2(\cos x)+1-\sin^2(\sin x)\;$$
So we get $$\displaystyle f(x) = 2-\left[\cos^2(\cos x)+\sin^2(\sin x)\right] = 2-\frac{1}{2}\left[2\cos^2(\cos x)+2\sin^2(\sin x)\right]$$
So we get $$\displaystyle f(x) = 2-\frac{1}{2}\left[1+\cos(2\cos x)+1-\cos(2\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Fourier sine series for $x^3$ It is asked to find the Fourier Sine Series for $x^3$ given that
$$\frac{x^2}{2} = \frac{l^2}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \cos\left(\frac{n \pi x}{l} \right)$$
integrating term by term. (This result was found in another exercise). As suggested, I integrat... | Both terms have to agree at $x=0$, hence, simply, $C=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^... | $$
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
(x - 1)^2 + \left(\left(-\frac{3}{4}x - \frac{13}{2}\right) + 1\right)^2 = 25 \\
(x + 5)^2 + \left(\left(-\frac{3}{4}x - \frac{13}{2}\right) + 9\right)^2 = 25 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Why test of divisible by $12$ works with $3$ and $4$ but not with $2$ and $6 ?$ Test of divisible by $4 ,$ last two digit must be divisible by $4 ,$ since $100$ is always divisible by $4$ remaining two digit $,$ we need to check $.$
Test of divisible by $3 ,$ sum digits must be divisible by $3 .$ But I don't know why $... | Since we are working in a numerical system of basis $10$, an integer number $N>0$ can be written as $$N=a_n\times 10^n+a_{n-1}\times 10^{n-1}+\ldots +a_1\times 10 +a_0$$
The number $10=9+1$ leaves remainder $1$ under division by $3$, and so does the another powers of $10$, so $N$ has the form $$N=9m+a_{n}+a_{n-1}+\ldot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simple inequality with a,b,c I'm looking for proof of
$$\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)} \leq \sqrt{2}(a+b+c)$$
I tried using $m_g \leq m_a$, generating the permutations of $\sqrt{a(b+c)} \leq \frac{a+(b+c)}{2}$ and adding them together, but I get $\frac{3}{2}(a+b+c)$, and obviously $\frac{3}{2}>\sqrt{2}$, so ... | Using Cauchy-Schwarz inequality with the vectors:
$$\begin{cases}
u=(\sqrt{a},\sqrt{b},\sqrt{c})\\
v=(\sqrt{b+c},\sqrt{c+a},\sqrt{a+b})
\end{cases}$$ you get
\begin{align}[\sqrt{a(b+c)}+\sqrt{b(c+a)}+\sqrt{c(a+b)}]^2 &=(u.v)^2\\ &
\le \Vert u \Vert^2 \Vert v \Vert^2\\
&=(a+b+c)((b+c)+(c+a)+(a+b))\\
&=2(a+b+c)^2\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is $\angle AEB$? Let $E$ be a point inside the square $ABCD$. and $|EC|=3,|EA|=1,|EB|=2$
What is the angle $\widehat {AEB}$?
I can only find $|ED|=\sqrt{12}$
| There has to be an easier way, but this works...
Use the law of cosines to get:
$s^2 = 5 - 4\cos(AEB)$
$s^2 = 13 - 12\cos(BEC)$
$2s^2 = 10 - 6\cos(AEB+BEC)$
Eliminate $s^2$, and use $\cos(AEB + BEC) = \cos(AEB)\cos(BEC) - \sin(AEB)\sin(BEC)$. Exchange the sines for cosines, eliminate $\angle BEC$, and a whole bunch of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1489850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Suppose that n is an integer divisible by 24. Show that the sum of all the positive divisors of n−1 (including 1 and n−1) is also divisible by 24. Suppose that n is an integer divisible by 24. Show that the sum of all the positive divisors of n−1 (including 1 and n−1) is also divisible by 24.
| Note that $n-1\equiv -1\pmod{3}$ and $n-1\equiv -1\pmod{8}$.
It is clear that $n-1$ is not a perfect square. If $ab=n-1$, where $a$ and $b$ are positive, call $a$ and $b$ a couple, or, in the more business-oriented language of today, partners.
If $\{a,b\}$ are a couple, then since $ab\equiv -1\pmod{3}$, one of $a$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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an urn contains six ball of each of the three colors: red, blue, and green. An urn contains six balls of each of the three colors: red, blue, green. Find the expected number of different colors obtained when three balls are drawn:
a. with replacement;
b. without replacement.
The correct answers are:
a: 19/9
b: 2.1912
I... | We can also use recurrences, where A,B,C are the number of balls of three colors:
a)
\begin{align*}
f(a,b,c) = \left\{\begin{matrix}
3-\left(\lfloor\frac{a}{A}\rfloor + \lfloor\frac{b}{B}\rfloor+\lfloor\frac{c}{C}\rfloor\right)& \text{if }A+B+C-(a+b+c) = 3\\
\dfrac{a\cdot f(a-1,b,c)+b\cdot f(a,b-1,c)+c\cdot f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Given integer roots of $x^2+mx-n=0$ and $x^2-mx+n=0$. Show $6 \mid n$. Suppose that $m$ and $n$ are integers such that both the quadratic equations $x^2+mx-n=0$ and $x^2-mx+n=0$ have integer roots. Prove that $n$ is divisible by $6$.
I figured out the roots of the equations through quadratic formula-for the first one i... | We will be using the following special case of the Vieta Relations: the sum of the roots of $x^2+bx+c=0$ is $-b$, and their product is $c$.
Suppose to the contrary that $n$ is odd. The product of the roots of each equation is therefore odd, so each root is odd.
The product of the roots in the first equation is $-n$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Integration of $\frac{1}{(1+x^4)^\frac{1}{4}}$ This question has been puzzling me a lot. This is in the middle list of question(meaning its moderately tough). I have tried everything i could think of.
My try: $$\int \frac{1}{x(1+\frac{1}{x^4})^\frac{1}{4}}dx$$
$$-\frac{1}{4}\int \frac{-4x^4}{x^5(1+ \frac{1}{x^4})^\fra... | Let $$I = \int\frac{1}{(1+x^4)^{\frac{1}{4}}}dx\;,$$ Then Put
$$x=\sqrt{\tan z }\;\;\;, \;\; dx = \frac{\sec^2 z}{2\sqrt{\tan z}}dz$$
$$I=\int \frac{1}{\sqrt[4]{1+\tan^2 z }} \frac{ \sec^2 z } {2 \sqrt{\tan z}}dz=\frac{1}{2} \int \sec^{ \frac{3}{2}} z \tan^{-\frac{1}{2}} zdz$$
$$=\frac{1}{2} \int \frac{1}{\cos z \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$ Prove by induction: $\left(\frac{1}{n+1}+\frac{1}{n+2}+...\frac{1}{2n}\right)^2<\frac{1}{2},n\ge 1$
For $n=1$ inequality holds.
For $n=k$
$$\left(\frac{1}{k+1}+\frac{1}{k+2}+...\frac{1}{2k}\right)^2<\frac{1}{2}$$
For $n=... | $$ a_n = H_{2n}-H_{n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} $$
is an increasing sequence, since:
$$ a_{n+1}-a_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)}>0$$
hence:
$$ \forall n\geq 1,\qquad a_n \leq \lim_{n\to +\infty}a_n = \log(2) $$
and the problem boils down just to showing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Olympiad problem about finding minimum value with $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$ Let $x,y,z$ be positive real numbers such that $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$.
Find the minimum value of
$$\frac{x^2y^2} {z^3(x^2+y^2)}+\frac {y^2z^2} {x^3(y^2+z^2)}+\frac {z^2x^2} {y^3(z^2+x^2)}$$
I'm pretty sure that the answer... | Let
$$a=\dfrac{1}{x},b=\dfrac{1}{y},c=\dfrac{1}{z},a^2+b^2+c^2\ge 1$$
Use Cauchy-Schwarz
$$\sum_{cyc}\dfrac{x^2y^2}{z^3(x^2+y^2)}=\sum_{cyc}\dfrac{c^3}{(a^2+b^2)}\ge \dfrac{(a^2+b^2+c^2)^2}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}$$
since
$$c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)\le\dfrac{2}{3}(a+b+c)(a^2+b^2+c^2)\le\dfrac{2}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Simple integral: $\int \frac {1+r}{-r^2+r-1} dr$. I was solving a much longer exercise, and while solving an ODE, I got this integral $$\int \frac {1+r}{-r^2+r-1} dr$$
I think this must be pretty simple, but I couldn't solve it, my substitutions didn't work and the polynomial in the denumerator does not have real root... | $r^2-r+1 = \frac{r^3+1}{r+1}$, hence $r^2-r+1=(r-\omega)(r-\omega^2)$, with $\omega=\exp\left(\frac{2\pi i}{6}\right)$.
By the residue theorem:
$$ \text{Res}\left(\frac{r+1}{r^2-r+1},r=\omega\right) = \frac{\omega+1}{2\omega},$$
$$ \text{Res}\left(\frac{r+1}{r^2-r+1},r=-\omega\right) = \frac{-\omega+1}{-2\omega},$$
hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Parametrization of $a^2+b^2+c^2=2d^2$ Is a complete parametrization of primitive solutions to the equation
$$a^2+b^2+c^2=2d^2\qquad a,b,c,d \in \mathbb{Z}$$
known? A reference would be great.
Solutions to $a^2+b^2=c^2$ give solutions to the equation above, but I know that there are other solutions.
| Look for a single rational solution to $x^2+y^2+z^2=2$. Say, $(x_0,y_0,z_0)=(-1,0,1)$.
Then take any three integers $(u,v,w)$ and seek solutions of form $(-1+tu,tv,1+tw)$. You get:
$$1-2tu+t^2u^2 + t^2v^2 + 1+2tw + t^2w^2=2$$
solving, excluding $t=0$, you get:
$$t=\frac{2(u-w)}{u^2+v^2+w^2}$$
Substituting that bach in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Summing reciprocal logs of different bases I recently took a math test that had the following problem:
$$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)
$$
\frac{1}... | $$
\frac{1}{\log_{2}50!} + \frac{1}{\log_{3}50!} + \frac{1}{\log_{4}50!} + \dots + \frac{1}{\log_{50}50!}
$$
$$=\log_{50!}2+\log_{50!}3+\log_{50!}4+...+\log_{50!}50$$
$$=\log_{50!}(2\cdot3\cdot4\cdot...50)$$
$$=\log_{50!}(50!)$$
$$=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Writing an integer as a sum of two square in many ways, with consecutive arguments Let $n\in{\mathbb N}$. I call $n=x_1^2+y_1^2=x_2^2+y_2^2=\ldots x_r^2+y_r^2$
(where $(x_1,y_1),(x_2,y_2),\ldots,(x_r,y_r)$ are distinct uples
in ${\mathbb N}^2$) a multi-decomposition of $n$, of length $r$. If the
$x_1,x_2,\ldots,x_r$ fo... | Although the answer is not quite on the topic.
$$x_1^2+y_1^2=x_2^2+y_2^2=x_3^2+y_3^2=x_4^2+y_4^2$$
For the equation is possible to write General standard formula - moreover it is symmetric. Interestingly, so monjo to record for any number of summands.
$$x_1=(a^2+b^2)(t^2+k^2)(n^2+q^2)p$$
$$y_1=(a^2+b^2)(t^2+k^2)(n^2+q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
MyApproach
Since I cannot form pattern above
I simplified this as $21^3$ = $3^3$ $. $ $7^3$
Similarly I did $27^3$ = $9^3$ $.$ $3^3$
Taking both I get $3^3$($7^3$+$9^3$)... | This is essentially a simpler version of lab's answer:
For any $a,b$, $(a-b)^3+(a+b)^3 = 2a^3 + 6ab^2 \equiv 0 \mod a$ for all $a,b$. In you case, you have two such pairs with $a=24$ (and $b=1,3$ respectively). So the answer is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the least squares solution of the linear system...... Hi I'm looking for the least squares solutions of....
$$ \begin{pmatrix} 1&-1 \\ -1&2 \\ -1&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix} $$
So Assuming this goes by $ A\vec{x}=\vec{b}$
Then using.... $A^T A \vec{... | The normal equations are correct:
$$
\mathbf{A}^{*}\mathbf{A} x = \mathbf{A}^{*}b.
$$
The error is in the solution which should be
$$
x = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}b.
$$
The inverse is
$$
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1} {6}
\left(
\begin{array}{cc}
5 & 3 \\
3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\ri... | $$
n\log\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)=1+\frac xn-\frac1{2n}+O\left(\frac1{n^2}\right)
$$
Therefore,
$$
\left(1+\frac1n+\frac{x}{n^2}+O\left(\frac1{n^3}\right)\right)^n=e\left(1+\frac xn-\frac1{2n}\right)+O\left(\frac1{n^2}\right)
$$
Thus
$$
\begin{align}
&\frac{\overbrace{\left(1+\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Solve the system of conguruences: $x = 2 \bmod7, x = 5 \bmod12, x = 8 \bmod 25$ Question:
Solve the system of conguruences: $x = 2 \bmod7, x = 5 \bmod12, x = 8 \bmod 25$
So far I have
The solution: $X= B_1X_1C_1 + B_2X_2C_2 + B_3X_3C_3$ and that the answer is -something- $\mod2100$
where
$B_1 = 300, B_2 = 175, B_3 = 8... | Let us first begin by trying to reduce this to a system of two congruencies.
Focus our attention first to the following two:
$\begin{cases} x\equiv 2\pmod{7}\\ x\equiv 5\pmod{12}\end{cases}$
We know that $\gcd(7,12)=1$ and so there should be some $a,b$ such that $7a+12b=1$. We can find the $a$ and $b$ necessary as a b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
| Using analysis: define $f(x)=x^3+x$. Since $f'(x)=3x^2+1>0$, the function is strictly increasing, hence, injective, and the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Sum of the rows of Pascal's Triangle. I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us
$0C0$ in first row.
$1C0$ and $1C1$ in the second, and so on and so forth. However, I s... | This can also be proven by induction. In order to do so, it is very helpful to prove the following identity:
$$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.$$
This can be proven combinatorially, but for the sake of expediency,
\begin{align*}
\binom{n}{k} + \binom{n}{k+1} &= \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Is every integer a mixed sum of three squares? Lagrange's four-square theorem states that every natural number can be represented as the sum of four integer squares $n = a^2 + b^2 + c^2 + d^2$.
Question: Is every integer a mixed sum of three integer squares $n = \pm a^2\pm b^2 \pm c^2$ ?
Note that the signs are indep... | You can write every number $n$ in the form $a^2+b^2-c^2$. Just pick $a$ so that $n-a^2$ is odd and then solve
$$\begin{align}
b+c&=n-a^2\\
b-c&=1
\end{align}$$
for $b$ and $c$:
$$\begin{align}
b&={n-a^2+1\over2}\\
c&={n-a^2-1\over2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bounds for $S=\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}$ Let $x,y,z\in[0,2]$ and $x+y+z=3$. What are the maximum and minimum of $$S=\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}?$$
When $x=y=z=1$, we have $S=3\sqrt{2}$.
For the upper bound, we can use the AM-GM inequality to get $$S\leq\frac{1}{2}(x+y+1+y+z+1+z+x+1)=\frac{... | A little numerical investigation gives that the maximum is attained for $x=y=z=1$ and the minimum for $x=1,y=2$ and $z=0$. An intuitive proof of this fact can be achieved by noticing that $F(x,y,z) = \sum_{cyc} \sqrt{x(y+1)}$ is concave in every variable and thus, the minimum should be attained at extremal points and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $f(x)=x^4$ is continuous on $[0,\infty)$ I understand that we want $\varepsilon$ so that $|x^2+y^2||x+y||x-y|<\varepsilon$, and for $|x+y||x-y|<\varepsilon$ I can choose $\delta$ so that $\delta(2|y|+\delta)<\varepsilon$, but I don't know what to do with the $|x^2+y^2|$ part.
| If $y=0$, then $|x^4-y^4|=x^4< \epsilon$ whenever $|x-y|=|x|<\epsilon^{1/4}$. So, take $\delta=\epsilon^{1/4}$.
If $y\ne 0$, then take $|x-y|<y/2$ so that $y/2<x<3y/2$.
Then, we have
$$\begin{align}
|x^4-y^4|&=|x-y|(x+y)(x^2+y^2)\\\\
&<|x-y|\left(\frac52 y\right)\left(\frac{13}{4}y^2\right)\\\\
&=|x-y|\frac{65}{8}y^3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Value of the limit without (or with, but giving rigorous arguments) using the Taylor expansion of sin I'm trying to evaluate the limit as $N\to \infty.$
$$\frac{ \left(\dfrac{\sin \frac{1}{N}} {\frac{1}{N}}\right)^{N} -1 }{\frac{1}{N}}.$$
Note first that, using L'Hôpital, one can easily show that the numerator goes ... | Just develop carefully the Taylor expansion.
$$\sin \frac{1}{N}=\frac{1}{N}-\frac{1}{6N^3}+O(\frac{1}{N^5})$$
Then
$$\left(\frac{\sin \frac{1}{N}}{\frac{1}{N}}\right)^N=\exp \left[N \log\left( 1-\frac{1}{6N^2}+O(\frac{1}{N^4})\right)\right]=\exp\left[N\left(-\frac{1}{6N^2}+O(\frac{1}{N^4})\right)\right]=\exp\left(-\fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$? My integral is
$$I=\int\sqrt[3] {\frac{1}{(x+1)^2(x-1)^4}} dx$$
and hence
$$I=\int\frac{1}{(x-1)(x+1)}\sqrt[3] {\frac{x+1}{x-1}}dx $$
$\cos2\theta$ substitution wont be helpful here because of the cube root. Should I apply by parts ?
| First notice that
$$\eqalign{
I &= \int {\root 3 \of {{1 \over {{{(x + 1)}^2}{{(x - 1)}^4}}}} } dx \cr
&= \int {{1 \over {\left( {x - 1} \right)\left( {x + 1} \right)}}\root 3 \of {{{x + 1} \over {x - 1}}} } dx \cr
&= \int {{1 \over {{{\left( {x - 1} \right)}^2}}}{{x - 1} \over {x + 1}}\root 3 \of {{{x + 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Proving a trigonometric inequality in a triangle In $\Delta ABC$, prove that $$\sin\left(\frac{\pi-A}{4}\right) \sin\left(\frac{\pi-B}{4}\right)\sin\left(\frac{\pi-C}{4}\right) \geq \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ Some ideas on how to start manipulating. I tried breaking it in sum, but it led to nothing... | Hint: In $\triangle{ABC}$ we have,
\begin{align*}\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}} &=1+4\sin{\left(\frac{B+C}{4}\right)}\sin{\left(\frac{A+C}{4}\right)}\sin{\left(\frac{A+B}{4}\right)}\\
\cos{A}+\cos{B}+\cos{C}&=1+4\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Basic proofs for epsilon delta limits: $\lim_{x \to a} \frac{x}{1+x}=\frac{a}{1+a}$ Hello I am wondering if my approach and understanding of the epsilon delta definition is correct.
For example, if I wanted to use it to show that $$\lim_{x \to a} \frac{x}{1+x}=\frac{a}{1+a}$$ for $a \in \mathbb{R}, a \neq -1$
Then here... | You write
we want that for all $\epsilon \gt 0$ such that $0 \lt |x-a| \lt \delta$ , then $|f(x)-L| \lt \epsilon$ for a suitable $\delta$.
What you actually want is: For every $\epsilon > 0$, there corresponds a $\delta > 0$ such that for all $x$, $0 < |x - a| < \delta$ implies $|f(x) - L| < \epsilon$. Also, since $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\... | Using $x=e^{\ln x}$ we take the log. So the limit is the exponential of the limit of the log, which is the limit of $\frac{n-1}{n+1}\ln(1-\frac{1}{n^2+1}$. That fraction goes to zero, so the log is asymptotic to it, hence the limit is the limit of $\frac{n-1}{n+1}\frac{1}{n^2+1}$, which is easily seen to be 0. Hence yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Limits of finite sums I know that: $$\int_0^1 1 - x^2 dx = \frac{2}{3}$$
And that represents the area below the curve, delimited by the lines $x= 0$ and $x = 1$
But $$\int_0^1 1 - x^2 dx = \lim_{n \to \infty} \left( 1 - \frac{2n^3 + 3n^2 +n}{6n^3} \right) $$
And this limit is obtained by calculating an approximation of... | You're not quite summing it right. If, for example, we use $a=2, b=3$ in:
$$\sum_{k = 1}^n f \left( \dfrac{k(b - a)}{n} \right) \left( \dfrac{b - a}{n}\right)$$
the function in the sum runs from $\dfrac1n$ to $1$.
Instead use:
$$\sum_{k = 1}^n f \left( a+\dfrac{k(b - a)}{n} \right) \left( \dfrac{b - a}{n}\right)$$
Your... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
trigonometry: why this is not a triangle?
As the photo shown, angle x doesn't nesscarry to be 90 degree.But when y=0, x=90 degree.
So here is my problem, what happen if I increase y, does x becomes larger (greater than 90) or less than 90 degree?
My attempt:
My reasoning is that when you increase y (say set y=1). Th... | Suppose that $y$ is a little more than $0$, so that the base is a little less than $12$. What height would be needed in order to have a right triangle with base $12-y$ and hypotenuse $13$? The Pythagorean theorem says that is would have to be
$$\sqrt{13^2-(12-y)^2}=\sqrt{25+24y-y^2}\;.$$
Is this more or less than $5+y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Diagonal Scaling of $ A \in {\mathbb{R}}^{2 \times 2} $ Positive Definite and Its Conditional Number Given a Positive Definite Matrix $ A \in {\mathbb{R}}^{2 \times 2} $ given by:
$$ \begin{bmatrix}
{A}_{11} & {A}_{12} \\
{A}_{12} & {A}_{22}
\end{bmatrix} $$
And a Matrix $ B $ Given by:
\begin{bmatrix}
\frac{1}{\sqrt{... | Since multiplying by a constant doesn't change the Condition Number one could make the choice of:
$$ A = \begin{bmatrix}
a & 1 \\
1 & b
\end{bmatrix} $$
Hence the matrix $ B $ becomes:
$$ B = \begin{bmatrix}
\frac{1}{\sqrt{a}} & 0 \\
0 & \frac{1}{\sqrt{b}}
\end{bmatrix} $$
And the Matrix $ C $ becomes:
$$ C = \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$ $\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$
I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?
| Hint:
\begin{equation*}
\frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}=\left( \frac{\sin x-x}{x^{3}}%
\right) \left( \frac{\sin x+x}{x}\right) \left( \frac{x}{\sin x}\right) ^{2}
\end{equation*}
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\left( \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $ \int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x$. $$
\int \frac{\sin x+\cos x}{\sin^4x+\cos^4x}\,\text{d}x.
$$
What I have tried:
I tried writing denominator as
$ \sin^4x+\cos^4x = 1-2\sin^2x\cos^2x $ and $ 2\sin^2x\cos^2x = \frac{1}{2}\sin^2(2x) $
so the integral becomes,
$$
\int\frac{\sin x+\cos ... | Let $$\displaystyle I = \int\frac{\sin x+\cos x}{\sin^4x +\cos^4 x}dx = \int\frac{\sin x+\cos x}{1-2\sin^2 x\cos^2 x}dx$$
So we get $$I = 2\int\frac{\sin x+\cos x}{2-(\sin 2x)^2}dx = 2\int\frac{\sin x+\cos x}{\left(\sqrt{2}-\sin 2x\right)\cdot \left(\sqrt{2}+\sin 2x\right)}dx$$
So we get $$I = \frac{1}{\sqrt{2}}\int \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find image of $D=\{z: |z|<1\}$ under $f(z)={z\over z-1}$. Find image of $D=\{z: |z|<1\}$ under $f(z)={z\over z-1}$. I know it is supposed to be $\Re z <{1\over 2}$, but unfortunately I've been trying to show this for too long and my sanity along with my concentration drops monotonically. I would really appreciate any h... | Let $z=x+iy$ with $x,y\in\mathbb{R}$. We have
$$\frac{z}{z-1}=\frac{x+\mathrm{i}y}{(x-1)+\mathrm{i}y}=\frac{x(x-1)+y^2}{(x-1)^2+y^2}-\mathrm{i}\frac{y}{(x-1)^2+y^2}$$
so
\begin{align}
\Re\frac{z}{z-1} & =\frac{x(x-1)+y^2}{(x-1)^2+y^2}\\
& =\frac{x^2+y^2-x}{x^2+y^2-2x+1}\\
& < \frac{1-x}{2-2x}\\
& =\frac{1}{2}\frac{1-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the splitting field of $(X^3-2)(X^3-3)(X^2-2)$ What is the splitting field of $$(X^3-2)(X^3-3)(X^2-2)\ \ ?$$
The roots are $$\{\sqrt[3]2,\ \ j\sqrt[3]2,\ \ j^2\sqrt[3]2,\ \ \sqrt[3]3,\ \ j\sqrt[3]3,\ \ j^2\sqrt[3]3,\ \ \sqrt 2,\ \ -\sqrt 2\}$$
where $j=e^{\frac{2i\pi}{3}}$. So it splitting field is included in... | Yes. It is clear that $(X^3-2)(X^3-3)(X^2-2)$ splits over $\mathbb Q(j,\sqrt[3]2,\sqrt[3]3,\sqrt 2)$. On the other hand, any splitting field of $(X^3-2)(X^3-3)(X^2-2)$ over $\Bbb{Q}$ must contain $\sqrt[3]2,\sqrt[3]3,\sqrt 2$ and $\frac{j\sqrt[3]2}{\sqrt[3]2}=j$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$ $\bf{My\; Try:}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx = \underbrace{\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{1}}+\underbrace{\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{2}}.$$
Now Here $$\displaystyle I_{2} = \int_... | $$\int\frac{x\ln(x)}{(x^2+1)^2}\space\text{d}x=$$
Integrate by parts $\int f\text{d}g=fg-\int g\text{d}f$: $f=\ln(x),\text{d}g=\frac{x}{(x^2+1)^2}\space\text{d}x, \text{d}f=\frac{1}{x},g=\frac{1}{2(x^2+1)}$:
$$-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\frac{1}{x(x^2+1)}\space\text{d}x=$$
Substitute $u=x^2$ and $\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to calculate this infinite sum? $$ \sum_ {n=0}^\infty \frac {1}{(4n+1)^2} $$
I am not sure how to calculate the value of this summation. My working so far is as follows:
Let $S=\sum_ {n=0}^\infty \frac {1}{(4n+1)^2}$.
$\Longrightarrow S=\frac{1}{1^2}+\frac{1}{5^2}+\frac{1}{9^2}+...$
$\Longrightarrow S=(\frac{1}{1^2... | Step 1: The sum of all the reciprocal squares is $\zeta(2)=\frac{\pi^2}{6}$.
Step 2: The sum of all the reciprocal even squares is $\frac{1}{4}\zeta(2)=\frac{\pi^2}{48}$.
Step 3: Their difference, the sum of all the reciprocal odd squares, is $\frac{\pi^2}{8}$.
Step 4: Catalan's constant is $G=\frac{1}{1^2}-\frac{1}{3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Differentiate an exponential integral Would you guide me differentiating this integral for $m$:
$$\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x$$
noting that $a ,... | $$I_1=\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x=$$
$$I_1=\frac{e^{-\frac{\left(4a^2-b^2c^2\right)(a-m)^2}{4a^2b^2}}\cdot\sqrt{\pi}\left(1+c\sqrt{\frac{1}{c^2}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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estimating absolute value of complex function Let $f(z) = \frac{1}{z^2 - 2z + 4 }$ where $|z| = R > 1 $. How can I estimate $|f(z)|$ to get an upper bound of the form $\frac{ M}{|z|^p }$ where $M > 0$ and $p>1$ ? I know
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{|z^2 - 2z|} = \frac{1}{|z||z-2|}$$
and here is... | Clearly $|z^2 -2z+4|=|z^2-2(z-2)|=|2(z-2)-z^2|\geq ||2z-4|-|z^2||$.
$\Rightarrow$
$$ |f(z)| = \frac{1}{|z^2 - 2z + 4| } \leq \frac{1}{||2z-4|-|z^2| |}$$
So by triangle inequality we have that $|2z-4|\geq |2z|-|4|$
Since $|z|=R>1$ , there exists $a \in \mathbb R^+$ such that $4=a|z|$ and $a>1$ .
Therefore $$ |f(z)| = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of a fraction with a square root
Find $$\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}$$ (without L'Hopital)
Where is the following wrong? (The limit is 6.)
\begin{align}\lim_{x \to 2} \frac{4-x^2}{3-\sqrt{x^2+5}}& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{9-x^2-5}}= \\
& =\lim_{x \to 2} \frac{(2-x)(2+x)}{\sqrt{-x^2... | You messed up in the first step when you said $3-\sqrt{x^2+5}=\sqrt{9-x^2-5}$. You seem to be trying to use $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$ even though this is almost always false (as an example, take $a=b=1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a recurrence relation without the characteristic equations I am new to solving recurrence relations and I am presented with the following two problems
(1) $$a_n = (3n-1)a_{n-1}, a_0 = 3$$
and
(2) $$a_{n+3}-6a_{n+2} + 11a_{n+1}-6a_{n} = 0, a_0 =1, a_1 =1, a_2 = -2$$
For (1), I try to find a pattern so $a_n = (... | $$a_n = (3n-1)a_{n-1}=(3n-1)(3n-4)a_{n-2}=...=a_{n-m}\prod_{k=1}^m(3n-3k+2)\\
=a_{n-n}\prod_{k=1}^n(3n-3k+2)=a_0\prod_{k=1}^n(3n-3k+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the integral part of $S$. It is given that $S=\sqrt{2012\sqrt{2013\sqrt{2014\sqrt{\cdots \sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}}}}}$. Find the integral part of $S$.
Thanks in advance!
| This is a partial answer!
Define
$$
S(n) := \sqrt{n \sqrt{(n+1) \dotsm \sqrt{n^2}}} = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k-1}}.
$$
So far I can only prove that
$$
n^2 \leq S(n)^2
$$
for every $n \geq 1$. Indeed, first observe that this is trivial for $n = 1$, so let's assume $n \geq 2$. Then note that
$$
S(n)^2 = \prod_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Why is $\left( {\begin{array}{*{20}{c}} B & B \\ B & B \\ \end{array}} \right)$ positive semidefinite? Let $B\in M_n$ be positive semidefinite .
Why is $\left( {\begin{array}{*{20}{c}}
B & B \\
B & B \\
\end{array}} \right)$ positive semidefinite?
| Let $x = \left( \begin{matrix} x_1 \\ x_2 \end{matrix} \right) \in \mathbb{C}^{2n}$ with $x_1 ,x_2 \in \mathbb{C}^n$. Then we have
\begin{align}
\langle x, \left( \begin{matrix} B & B \\ B & B \end{matrix} \right)x \rangle &= \langle \left( \begin{matrix} x_1 \\ x_2 \end{matrix} \right) , \left( \begin{matrix} Bx_1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.
| $$I=\int_{0}^{+\infty}\frac{dx}{(x+1)^3+1}=\int_{1}^{+\infty}\frac{dx}{x^3+1}=\int_{0}^{1}\frac{x}{1+x^3}\,dx $$
The roots of $1+x^3$ lie at $-1,\xi=\frac{1+i\sqrt{3}}{2},\bar{\xi}=\frac{1-i\sqrt{3}}{2}$ and they are simple. Since:
$$ \text{Res}\left(\frac{x}{1+x^3},x=-1\right) = -\frac{1}{3}$$
the partial fraction dec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\int \frac{dx}{3^x-8}$ I have a problem with solving this Integrate
$$\int \frac{dx}{3^x-8}$$
At first I use substitute $$3^x =t $$
$$\log_3{t} = x$$
$$\frac{1}{t\ln{3}}dt = dx$$
Next:
$$\int \frac{dt}{t^2\ln3} - \int\frac18dx$$
$$=\frac{1}{\ln3}\int \frac{dt}{t^2} - \int\frac18dx$$
$$=\frac{1}{\ln3} \left(\... | Notice $\frac{1}{3^x - 8} = -\frac{1}{8 - 3^x} = - \frac{1}{8} \frac{8 - \mathbf{3^x} + \mathbf{3^x} }{(8-3^x)} = -\frac{1}{8} \left( 1 + \frac{3^x}{8-3^x} \right)$ and so
$$ \int \frac{1}{3^x - 8} = -\frac{1}{8} \left( x + \int \frac{3^x dx}{8 - 3^x} \right) = -\frac{x}{8} +\frac{1}{8 \ln 3} \int \frac{d(8-3^x)}{8 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | The error is in the very first implication. If $0/0$ is not equal to zero, there is no reason why $0/0$ must equal some $x$. There is no reason to believe that we can do this division and get a number.
Therefore you have a proof that $0/0$ cannot equal any nonzero $x$. Combine this with a proof that $0/0$ cannot eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
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"answer_id": 7
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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.