Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
AM-GM inequality: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$
Let $a, b, c, d > 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.
I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
| By AM-GM $\frac{a^2}{b}+b\geq2a$.
Thus, $\sum\limits_{cyc}\frac{a^2}{b}\geq\sum\limits_{cyc}(2a-b)=\sum\limits_{cyc}a$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that:$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$ $\phi$ is the golden ratio
$$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$$
I try:
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\... | Note that $$\frac{1}{\dbinom{2n}{n}}=\left(2n+1\right)B\left(n+1,n+1\right)=\left(2n+1\right)\int_{0}^{1}x^{n}\left(1-x\right)^{n}dx
$$ hence $$\sum_{n\geq0}\frac{2^{n+3}}{\dbinom{2n}{n}\left(n+1\right)}=\sum_{n\geq0}\frac{2^{n+3}}{n+1}\left(2n+1\right)\int_{0}^{1}x^{n}\left(1-x\right)^{n}dx
$$ $$=8\int_{0}^{1}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
If $x = y$, $p$ = what? It is given that,
$$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$$
If $x=y$, $p$ =?
| We have: $\dfrac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p$
Let's replace $x$ with $y$:
$\Rightarrow p=\dfrac{\sqrt{2y+3y}+\sqrt{2y-3y}}{\sqrt{2y+3y}-\sqrt{2y-3y}}$
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{-y}}{\sqrt{5y}-\sqrt{-y}}$
$\hspace{9 mm} =\dfrac{\sqrt{5y}+\sqrt{y}\hspace{0.5 mm}i}{\sqrt{5y}-\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove or disprove the inequality $\tan x \geq x + x^3$ Given the inequality $\tan x \geq x + x^3$. Prove or disprove it.
$x \in (0, \pi/2)$.
Hints would be appreciated.
| The MacLaurin series of $\tan$ is given by
$$\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$$
Hence taking the difference $x^3 + x - \tan(x)$ becomes
$$\frac{2x^3}{3} - \frac{2x^5}{15} - \cdots$$
You could then suspect that for small $x$ (especially when $x^3 \gg x^5$), we will have that $x^3 +x > \tan(x)$.
For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$
\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A ... | Subtract the first line from the second:
$$\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}=\begin{vmatrix}
x-a & a-x & 0 &0 \\
a & x & b &c \\
m &n & x &p \\
m & n& p& x
\end{vmatrix}=(x-a)\begin{vmatrix}
1 &-1 & 0 &0 \\
a & x & b &c \\
m &n & x &p \\
m & n& p& x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int_0^{1/2} \log(1-x) \log(1-2x) \ dx$. While messing around with Wolfram Alpha, I discovered that
$$\int_0^{\frac{1}2} \log(1-x) \log(1-2x) \ dx = 1 - \frac{\pi^2}{24} - \frac{\log(2)}2.$$
I've tried all sorts of standard tricks, but I cannot seem to prove it. Can someone prove this beautiful identity?... | With the substitution $u = 1-2x$, $x = (1+u)/2$, $dx = -du/2$, we get $$\begin{align*} \int_{x=0}^{1/2} \log (1-x) \log (1-2x) \, dx &= -\frac{1}{2} \int_{u=1}^0 \log \left(\frac{1+u}{2}\right) \log u \, du \\ &= \frac{1}{2} \int_{u=0}^1 \log u \,(\log (1+u) - \log 2) \, du \\
&= \frac{1}{2} \left( -\log 2 \int_{u=0}^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$\lfloor \sqrt{n-100} \rfloor$ dividing $n$
For how many positive integers $100 < n \le 10000$ does $\lfloor \sqrt{n-100} \rfloor$ divide $n$?
Let $n = 100+k^2+c$, where $k$ is a nonnegative integer and $c$ an integer such that $0 \leq c < 2k+1$. Then $k = \lfloor \sqrt{n-100} \rfloor$ and $1 \leq k \leq 99$ and we h... | First, fix $k = \lfloor \sqrt{n-100} \rfloor \in [1,98]$. Then, $100+k^2 \le n \le 100+k^2+2k$.
Exactly two of the integers $\{100+k^2+c : c \in \mathbb{N}, 1 \le c \le 2k\}$ will be divisible by $k$ (since these are $2k$ consecutive integers).
Also, $100+k^2$ will be divisible by $k$ iff $100$ is divisible by $k$, whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$ Let $a,b \in \mathbb{R}^*$ Show that :
$$3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right)$$
Let $a,b\in\mathbb{R}^*$
let $t=\dfrac{a}{b}+\dfrac{b}{a}$ , we've :
$$
\begin{aligned}
3+\frac{a^... | your inequality is equivalent to $${\frac { \left( {a}^{2}-ab+{b}^{2} \right) ^{2}}{{b}^{2}{a}^{2}}}\geq 0$$ and this is true for all $$a,b>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle
$ABC$ is equilateral
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{... | Because
$$0=\sum\limits_{cyc}(2b^2a-a^2b-abc)=\sum\limits_{cyc}(a^3-abc-(b^3-2b^2a+a^2b))=$$
$$=(a+b+c)\sum\limits_{cyc}(a^2-ab)-\sum\limits_{cyc}b(a-b)^2=$$
$$=\sum\limits_{cyc}(a-b)^2\left(\frac{1}{2}(a+b+c)-b\right)=\frac{1}{2}\sum\limits_{cyc}(a-b)^2(a+c-b).$$
Id est, the given it's
$$(a-b)^2(a+c-b)+(b-c)^2(b+a-c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Square root positive/negative What is the square root of $9$? Is it always $\pm 3$ or just positive $3$?
Trying to find the solution set of this equation : $x-3 = \sqrt{x+3}$
I want to understand the concept of square root to solve the problem.
Thanks
| Well, let us start with something we all know and love, the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
And I have a question for you. Why is it that there is a $\pm$ in front of the square root? Isn't it already the case that we have things like $\sqrt9=\pm3$? And the answer is no. By definition, we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Recurrence relation with quadratic equation Situation: working on a homework problem for my discrete math class that I think is solved, but I am now wondering if my solution is right. This a is a recurrence relations problem with a quadratic equation to find the roots and then an extra term to deal with.
The question:
... | Yeah! Seems quite good. Btw here's another nice way to eliminate the extra term and work out the recurrence:
Multiplying the giving recurrence by a factor of 2, seems like a hint to to me.
$$2A_n = 2A_{n-1} + 2A_{n-2} + 2^{n+1}$$
And also observe that
$$A_{n+1} = A_{n} + A_{n-1} + 2^{n+1}$$
From where we can easily eli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would... | A faster approach. If we set $f(x)=\log\left(\frac{x+\sqrt{x^2+4}}{2}\right)$ and $x=2\sinh\theta$ we have
$$f(2\sinh\theta) = \log\left(\frac{2\sinh\theta+2\cosh\theta}{2}\right) = \log(e^\theta) = \theta $$
and since $\theta\mapsto 2\sinh\theta$ is a bijective odd function from $\mathbb{R}$ to $\mathbb{R}$,
$$ f(-x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$x-\frac1x+y-\frac1y=4\,$ has no rational solutions? Do there exist rational numbers $x,y$ such that $x-\frac1x+y-\frac1y=4?$
I think no, because the equation reduces to $\frac{x^2-1}{x}+\frac{y^2-1}{y}=4\implies x^2y+x(y^2-4y-1)-y=0$ in two variables is irreducible in the field $\mathbb{Q}$(Eisenstein) Is my reasoning... | WLOG, suppose that $x,y>0$ (otherwise $x'=-\dfrac{1}{x}$),then let $x= \tfrac ab, y= \tfrac cd$ with $a,b,c,d \in \mathbb{N}, \; \gcd (a,b)= \gcd (c,d)=1$. The equation is equivalent to $$\dfrac{a}{b}-\dfrac{b}{a}+\dfrac{c}{d}-\dfrac{d}{c}=4 \iff cd(a^2-b^2)+ab(c^2-d^2)=4abcd.$$
From here, since $ab|(a^2-b^2)cd+(c^2-d^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{... | $$
{3-\sqrt{5}x\over (1+\sqrt 5 \, x)^3} = \frac A {1+\sqrt 5\,x} + \frac B {(1+\sqrt 5\,x)^2} + \frac C {(1+\sqrt 5\,x)^3}
$$
Find $A$, $B$, and $C$ and then integrate each term separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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The sum of digits in a 2-digit number
The sum of digits in a two digit number formed by the two digits from $1$ to $9$ is $8$. If $9$ is added to the number then both the digits become equal. Find the number.
My attempt:
Let the two digit number be $10x+y$ where, $x$ is a digit at tens place and $y$ is the digit at ... | Let the $2$-digit number be $\overline{ab} = 10a + b$.
If $9$ is added to the number then both the digits become equal.
This means that $a = b$, so the number plus $9$ is $10a + a = 11a$, which is always $0 \text{ mod } {11}$ (divisible by $11$).
Therefore $\overline{ab} + 9 \equiv 0 \text{ mod } {11} \Rightarrow \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
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Integration of inverse function We have to integrate the folowing
I tried and substitued x=atan$^2 \theta$
But stuck in that
| Hint: integrate by parts:
$$
\int \arcsin\sqrt{\frac{x}{x+a}}\ \mathrm dx=x\arcsin\sqrt{\frac{x}{x+a}}-\int \frac{1}{2} \sqrt{\frac{a x}{(a+x)^2}}\ \mathrm dx
$$
and let $\sqrt x=u$ to integrate the second integral:
$$
\frac{1}{2} \sqrt{\frac{a x}{(a+x)^2}}\ \mathrm dx=\frac{\sqrt{a} u^2}{a+u^2}\ \mathrm du
$$
The fina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Is $\int\frac{\mathrm d \, x}{x^2-1}= \operatorname{arctanh} ~x + C = \operatorname{arccoth} ~x + K$ correct? My professor gave us an integral but I think it's incorrect.
Is this correct?
$$
\DeclareMathOperator\arctanh{arctanh}
\DeclareMathOperator\arccoth{arccoth}
\int\frac{\mathrm d \, x}{x^2-1}= \arctanh ~x + C = \... | It is useful to rewrite $\DeclareMathOperator\arctanh{arctanh}\DeclareMathOperator\arccoth{arccoth} \arccoth x$ as an inverse function. In this case, let $f(x)=\coth x=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\frac{e^{2x}+1}{e^{2x}-1}$. To determine the inverse, just solve for $x$ in the equation $y=\frac{e^{2x}+1}{e^{2x}-1}$.
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\... | From :
$$
\cos(A)+\cos(B)+2\cos(C)=2
\\
\implies \cos(A)+\cos(B)=2-2\cos(C)
\\
\implies \cos(A)+\cos(B)=2[1-\cos(C)]
\\
\\\implies \cos(A)+\cos(B)=4\sin^2\left(\frac{C}{2}\right)
$$
Using Prosthaphaeresis Formulas :
$$
\cos A+\cos B=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2
$$
And substituting this formula in the first equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$... | Let's modify ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} $
to ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} =K$
And ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$
to $K = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Floor function and inequality I'm trying to work through a problem and I need to find a nice way to show that
$$\lfloor x^2\rfloor +\lfloor2x\rfloor\leq\lfloor x^4\rfloor$$
for $x\in[\sqrt[4]{3},\infty)$.
I know one possible way is to break down that interval into all the places where the three floor values increase b... | Hint: by definition of the greatest integer function:
$$
\begin{cases}
\begin{align}
\lfloor x^2\rfloor +\lfloor2x\rfloor & \leq x^2 + 2 x \\
\lfloor x^4\rfloor & \gt x^4 - 1
\end{align}
\end{cases}
$$
Therefore a sufficient condition for the inequality to hold is:
$$
x^2+2x \le x^4 - 1 \;\; \iff \;\; x^4-x^2-2x-1 = x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Zeros of a Fourth Degree Polynomial Given that $\left(\sqrt3+\sqrt5\right)$ is one zero of a fourth-degree polynomial with integer coefficients and leading coefficient 1, how can the constant term of this polynomial be found?
I know that $\left(\sqrt3-\sqrt5\right)$ must also be a root because it is the conjugate. How ... | We know that $\sqrt 3+\sqrt 5$ is a root of the polynomial $\sqrt 3+\sqrt 5=x$, so we begin from there and eliminate radicals to obtain a polynomial over $\Bbb Z$:
\begin{align}
\sqrt 3+\sqrt 5&=x\\
\sqrt 3&=x-\sqrt 5\\
3&=x^2-2\sqrt 5 x+5\\
2\sqrt 5x&=x^2+2\\
20x^2&=x^4+4x^2+4\\
0&=x^4-16x^2+4.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How is $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1$? I wondered, why this:
$$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$$
is equal to $\sqrt{2}-1$.
Can anyone explain me, why this is equal? :/
| Since I've expanded on my initial "joke comment", I might as well make it a full joke answer :) By that I mean, nobody in their right mind would take this approach to actually verify that the two quantities are equal: instead, what follows is a good, but limited, way to produce expressions with radicals that look diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Prove that if $n+1$ is divisible by $8$, then the sum of $n$'s factors is also divisible by $8$
Let $n$ be a positive integer. Prove that if $n+1$ is divisible by $8$, then the sum of $n$'s factors is also divisible by $8$.
If $n+1$ is divisible by $8$, then $n \equiv 7 \pmod{8}$, but I didn't see how to separate it ... | Let $n + 1 = 8m$.
We show first that if $d$ is a divisor of $n = 8m - 1$, then $d^2 - 1$ is divisible by $8$. Now $d$ must be odd (because $n$ is), so $d - 1$ and $d + 1$ are consecutive even numbers, so one must be a multiple of $4$, and there product $d^2 - 1$ must be a multiple of $8$.
Now $8m - 1$ cannot be a squa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \fra... | I encountered this limit when computing
$$
\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^n}\sum_{k=0}^n\frac{2^k}{k+1}
$$
in equation $(8)$ of this answer, and for which an asymptotic expansion is given in equation $(6)$ of this answer:
$$
\begin{align}
\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}k
&=\sum_{k=1}^n\frac1{\binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| You may prove that identity in a very geometric flavour, i.e. by bisecting twice a $30^\circ$ angle, through the Pythagorean theorem and the bisector theorem. It is best to keep the expressions of the involved lengths as simple as possible during the process. So, let we consider a triangle $BAE$ with $BA=\sqrt{3}, AE=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$ Can anybody help me prove this inequality ?
| It is not difficult to show that $x-\sqrt{\strut x^2+1}<0$ for all $x\in\mathbb{R}$. Then $$\left(x-\sqrt{\strut x^2+1}\right)\left(y-\sqrt{y^2+1}\right)>0.$$ We get $$xy+\sqrt{\strut x^2+1}\sqrt{y^2+1}>x\sqrt{y^2+1}+y\sqrt{\strut x^2+1}\tag{1}$$ From the other hand, since $(a-b)^2\geqslant 0$, then $$ab\leqslant\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to prove that $(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$ if $(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$ Let $x,y$ be real numbers such that :
$(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$.
Prove that :
$(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$.
I tried taking $x=y$. It simplifies everything a l... | HINT:
Write $$x = \frac{1}{2}(s-\frac{1}{s})\\
y = \frac{1}{2}(t-\frac{1}{t})$$
with $s$, $t>0$. Then
$$( x+ \sqrt{1+y^2})(y + \sqrt{1+x^2}) -1 = (s t -1) \frac { s t (s+t)^2 + (s-t)^2}{ 4 s^2 t^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $ \sin \alpha = \frac 45 $ and $ \cos \beta = \frac{5}{13} $, prove that $ \cos \frac{\alpha-\beta}{2} = \frac{8}{\sqrt {65}} $ I can solve it easily if I assume that $ 0 < \alpha, \beta < \frac{\pi}{2}$
But there is no mention of the quadrants in which $ \alpha $ and $ \beta $ lie in.
Is the question wrong ?
| Hint:
$\sin \alpha>0$ then $0<\alpha<\pi$ and $\cos \beta>0$ then $-\frac{\pi}2<\beta<\frac{\pi}2$ and $-\frac{\pi}2<-\beta<\frac{\pi}2$
Thus $$-\frac{\pi}2<\alpha-\beta<\frac{3\pi}2$$
and
$$-\frac{\pi}4<\frac{\alpha-\beta}2<\frac{3\pi}4$$
1) $\sin^2\alpha+\cos^2\alpha=1$, then $\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Need to prove divisibility of $7^p-5^p-2$ by $3$ by using Fermat's little Theorem Let $p > 2$ be a prime number. Prove that $$7^p-5^p-2$$ can be divided by $6p$.
I have already proved that it can be divided by $2$ and $p$, but how can I prove that it can be divided by $3$ ?
| Fermat's little theorem asserts that if $p$ is prime, then for every integer $n$, we have :
$$n^p-n\equiv 0\quad (mod\,p)$$
Since $5$ and $7$ are primes :
$$7^p-5^p-2=(7^p-7)-(5^p-5)\equiv0\quad(mod\,p)$$
But $p$ is also odd, so that the relation : $7^p-5^p-2\equiv1-(-1)^p-2\quad (mod\,6)$ takes the form :
$$7^p-5^p-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\int \frac{dx}{x\sqrt{x^4-16}}$ While I know that the answer is $-\frac{1}{8}\arctan \frac{4}{\sqrt{x^4-16}},$ I couldn't reach it. When I solved it I reached a different answer, if someone can point at what point I made a mistake and what is the way to solve it I will be thankful.
I will start with taking $u=x^2$, so... | $\int \frac{1}{x\sqrt{x^4 - 16}} dx$
Multiply and dividide by $x^3$
$\int \frac{x^3}{x^4 \sqrt{x^4 - 16}} dx$
Put $x^4 - 16 = u^2$
$4x^3 dx = 2u du$
$x^3 dx = \frac{1}{2} u du$
On putting in integral,
= $\frac12 \int \frac{u}{(u^2 + 16) \sqrt{u^2}} du$
= $\frac12 \int \frac{u}{u(u^2 + 16)} du$
= $\frac12 \int \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all three-digit numbers $\overline{abc}$ such that 6003 digit number $\overline{abcabcabc.....abc}$ is divisible by 91? Find all three-digit numbers $\overline{abc}$ such that $6003$ digit number $\overline{abcabcabc......abc}$ is divisible by 91?Here $\overline{abc}$ occurs $2001$ times.I know the divisibility ru... | The given number can be written as follows,
$abc(1+10^3+10^6+\cdots+10^{6000})$
Now, $91|1001=1+10^3$ . The sum $S=1+10^3+10^6+\cdots+10^{6000}$ has $2001$ terms, therefore, $91$ and $(1+10^3)+10^6(1+10^3)+\cdots+10^{1999}(1+10^3)+10^{6000}$ are relatively prime $\implies$ $abc$ is a multiple of 91.
Therefore, the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is it true that $\,(1+\sin n)^{1/n}\to 1$? In order to show that $\lim_{n\to\infty}(1+\sin n)^{1/n}=1$, we need to show that $\,1+\sin n\,$ cannot not become too small. It suffices for example to show that
$$
1+\sin n\ge \frac{c}{n^k},
$$
for some $c,k>0$. This could be a consequence of showing that there exist $d,m>0$... | Consider $f(x)=(1+\sin x)^{\frac{1}{x}}$ . The binomial expansion is valid for $|\sin x|<1$.
Case I
$x\neq\frac{\pi}{2}+2\pi k$ and $x\neq\frac{3\pi}{2}+2\pi k$, where $k\in\mathbb{Z}$. Hence we may expand binomially
$f(x)=1+\frac{\sin x}{x}+\frac{\frac{1}{x}(\frac{1}{x}-1)}{2!}\sin^2x+\dots+\frac{\frac{1}{x}(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Evaluating an expression given values of symmetric polynomials Evaluate $\dfrac x{yz} + \dfrac y {xz} + \dfrac z y$
Given, $z+y+x=4, \qquad xyz=-60, \qquad xy+xz+yz=-17$
How do we do this? I found a common denominator, and substituted it for $-60$, but I am unaware of how to proceed.
Someone already asked the question... | \begin{align}
& \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} = \frac{x^2 + y^2 + z^2}{xyz}
\end{align}
$x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy + xz +yz)$
so: \begin{align} \frac{x^2 + y^2 + z^2}{xyz} = \frac{(x+y+z)^2 - 2(xy + xz +yz)}{xyz} = \frac{16 - 2(-17)}{-60} = \frac{50}{-60} = -\frac{5}{6}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I ha... | Actually the calculation for this sum is very simple and what we need is the sum of telescopic series. In fact
\begin{align}
\sum_{n=1}^\infty\frac{H_{n+2}}{n(n+2)}&=\frac12\sum_{n=1}^\infty H_{n+2}\left(\frac{1}{n}-\frac1{n+2}\right)\\\\
&=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{1}{n}\left(H_{n}+\frac{1}{n+1}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
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If $-1 < a < 1$ then $\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3$?
Prove that if $-1<a<1$, then :
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3 $
I do not know how to even approach this problem. Those fourth roots confuse me a lot.
Any help would be appreciated.
Also I would like to know how the left ... | Let $f:[-1,1] \to \mathbb{R}$, $f(x)=\sqrt[4]{1-x}+\sqrt[4]{1+x}+\sqrt[4]{1-x^2}$.
$f$ is obvously continous and differentable on $(-1,1)$
Then:
$$f'(x)=\frac{\sqrt[4]{(1-x)^3}-\sqrt[4]{(1+x)^3}-2x}{4\sqrt[4]{(1-x^2)^3}}$$
Of course $4\sqrt[4]{(1-x^2)^3} \geq 0$ for all $x\in[-1,1]$.
*
*for $x>0$ :
$\sqrt[4]{(1-x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\b... | it must be $$z_{1,2}=\frac{a^2+b^2}{2ab}\pm\sqrt{\left(\frac{a^2+b^2}{2ab}\right)^2-1}$$
and we can factorize to $$-\frac{(a-bz)(az-b)}{ab}$$ and $$ab\ne 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two sm... | Hint: Observe if you have a right triangle with legs $a, b$ and $h$ is as specified, then we see that
\begin{align}
\text{Area} = \frac{a\cdot b}{2} = \frac{h\cdot \sqrt{a^2+b^2}}{2} \ \ \Rightarrow \ \ h^2 = \frac{a^2b^2}{a^2+b^2}.
\end{align}
One can show that the integral
\begin{align}
\frac{2}{\pi}\int^\infty_0 \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int\frac{dx}{x+ \sqrt{x^2-x+1}}$ Evaluate :
$$\frac{dx}{x+ \sqrt{x^2-x+1}}$$
After dividing and multiply with $x-\sqrt{x^2-x+1}$, I get
$x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx$.
Is $\int \frac{\sqrt{x^2-x+1}}{x-1}dx$ is integrable in terms of elementary functions?
| We have $$I =\int \frac {\sqrt {x^2-x+1}}{x-1} dx =\frac {1}{2 }\int \frac {\sqrt {(2x-1)^2+3}}{x-1} dx $$ Substituting $u =2x-1$, we get, $$I =\frac {1}{2} \int \frac {\sqrt{u^2+3}}{u-1} du $$ Now perform hyperbolic substitution $u =\sqrt {3} \operatorname {sinh}(v)$ and simplifying gives us $$I =\frac {3}{2} \int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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A 9 dice are rolled at the same Time?what is the possible number of combinations, so that we have ( exactly )one 1 or at (least) a 4 or (both) A 9 dies are rolled at the same Time !
what is the possible number of combinations, so that we have ( exactly )one 1 or at (least) a 4 or (both)
My solution ;
Let C = exactl... | Rolling 9 identical dice, in general, corresponds to placing 9 identical balls in 6 boxes.
To take care of the conditions placed, use stars and bars for the following $2$ cases:
*
*Zero $1's$, at least one $4$:
Consider as if only boxes $2-6$ left, with one $4$ preplaced $\to\binom{8+5-1}{8}$
*Exactly one $1$:
Prep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$ What results?
$$\int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x+1}dx$$
my try : $u= \tan \frac{x}{2 } $
but : What is the short way?
| \begin{eqnarray}
\sin x+\cos x+1 &=& 1+\sin x+\cos x \\
&=& (\sin\frac{x}{2}+\cos\frac{x}{2})^2+(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}) \\
&=& (\sin\frac{x}{2}+\cos\frac{x}{2})2\cos\frac{x}{2}
\end{eqnarray}
We have by substitution $x=\dfrac{\pi}{2}-u$
\begin{eqnarray}
\int^{\frac{\pi}{2}}_0 \frac{\sin u}{\sin u+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How do we evaluate $\int\frac{dx}{\sin^{4}{x}+1}$? How do we evaluate $$\displaystyle\int{\dfrac{dx}{\sin^{4}{x}+1}}?$$
don't know how to solve this, please help.
| $\displaystyle \int \dfrac{dx}{\sin^4 x+1}$
Multiplying top and bottom with $\csc^4 x$ gives
$\displaystyle \int \dfrac{\csc^4 x}{1+\csc^4 x}\,dx$
$=\displaystyle \int \dfrac{\csc^2 x(1+\cot^2 x)}{1+(1+\cot^2 x)^2}\,dx$
Let $u=\cot x\implies du=-\csc^2 x\,dx$
$=\displaystyle \int \dfrac{-(1+u^2)}{1+(1+u^2)^2}\,du$
$=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$
We have the infinite series:$$\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$$
This is not my series: $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\fra... | Let
$$ f(x)=\sum_{n=1}^\infty\frac1{n(n+1)(n+2)}x^{n+2}. $$
Then
$$ f'(x)=\sum_{n=1}^\infty\frac1{n(n+1)}x^{n+1},f''(x)=\sum_{n=1}^\infty\frac1{n}x^{n},f'''(x)=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}. $$
So
$$ f''(x)=-\ln(1-x)$$
and
$$ f(1)=-\int_0^1\int_0^x\ln(1-t)dtdx=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Volume of a piece of ellipsoid I want to compute the volume of $A:=\{(x,y,z)\in\mathbb{R}^3:\frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{25}\leq 1; x\geq\frac{y^2}{9}+\frac{z^2}{25}\}$ (which I think is a piece of ellipsoid) so I set up the following integral: $\iiint_A 1 dxdydz=\int_{x=0}^{x=2}\int_{-3\sqrt{x}}^{3\sqrt{x}}\... | The intersection between the ellipsoid and the paraboloid is given by
$$
\frac{x^2}{4}+x=1\quad\Rightarrow\quad x=2(\sqrt{2}-1)>0
$$
So the projection of this intersection in the plane $x=0$ is the ellipse
$$
\frac{y^2}{9}+\frac{z^2}{25}=2(\sqrt{2}-1)
$$
Let $D$ be the region bounded by this ellipse:
$$
D=\{(y,z)|\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$ How do I solve for $A$? $$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$$
Since the denominator evaluates to $0$, I tried doing $$\lim_{x\to 0} [\sin(2x)-\sin(Ax)]=A^2 \cdot \lim_{x\to0}[x+x^3]$$ but it would go into $0=0$.
If I checked from the graph, t... | $$\begin{align}\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3}&=\lim_{x\to 0} \left[\frac{\sin(2x)-\sin(Ax)}{x+x^3}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\right]\\&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{\frac{x+x^3}{x}}\\
&=\lim_{x\to 0} \frac{2\frac{\sin(2x)}{2x}-A\frac{\sin(Ax)}{Ax}}{1+x^2}\\
&=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$... | The second way is correct but I would be more carefull and use like this:
$$1+2+...+k+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$
P.s: Not use the equality in the first place. The problem using the equality is that you need guarantee equivalence in every step what is not so easy in many problem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solve for x (quadratic equation) How do you solve the following equation:
\begin{align*}
\sqrt{2017 + \sqrt{2017 - x}} &= x
\end{align*}
I tried squaring it twice, but then I am left with quadratic equation that I can not solve.
| Let $2017-x=y^2$, where $y\geq0$.
Hence, $x+y>0$ and $2017+y=x^2$, which gives $y+x=x^2-y^2$ or $x-y=1$ and the rest is smooth.
We have $y=x-1$, $2017+x-1=x^2$ or $$x^2-x-2016=0$$ and since $x>0$,
we get the answer $\{\frac{1+\sqrt{8065}}{2}\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How many solutions does a trigonometric function have $0\le x \le 2\pi$? How do I algebraically determine how many solutions a trigonometric solution have $0\leq x\leq 2\pi$? So far I have been drawing graphs for each question and counted the solutions but I want a way to do this algebraically. Many people tell me to u... | If $0\le x \le 2\pi$ then $0 \le 3x \le 6 \pi$
$\sin (\theta) = -1/4$ will have two solutions in $0 \le \theta \le 2\pi$[$*$] so it will have two solutions in $2\pi \le \theta \le 4\pi$ and two solution is $4\pi \le \theta \le 6\pi$. So there are six solutions for $0 \le \theta = 3x \le 6\pi$ or in other words $0 \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using th... | Apart from a couple of threes that might be fours ? You are fine upto here
\begin{eqnarray*}
\tan(-\frac{5\pi}{16})=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/4)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/4)}{2}}}}.
\end{eqnarray*}
Now $\cos(5\pi/4)=\frac{ 1}{\sqrt{2}}$ & after a little bit algebra ...
\begin{eqnarray*}
\tan(-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that
$$
(a+b+c)^5 \leq 999^2 \implies a+b+c< \sqrt[5]{999^2}\leq 15
$$
and
$$
(100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871.
$$
Also
$$
(100 a+10b... | You should be able to quickly see that $(a+b+c)$ must be a square. Since $999<1024$, we know that $(a+b+c)<16$ so we have ${a+b+c}\in\{1,4,9\}$, and thus $\sqrt{(a+b+c)^5}\in\{1,32,243\}$, the last of which produces a valid solution. Since you form $\overline{abc}$ as a number I'll assume $a$ cannot be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Algebraic manipulation: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$ If you're up for a tedious algebraic manipulation, I'm stuck.
The following is true: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$.
However, I can't for the life of me figure out the how to sh... | The sum can rewrite in this way:
$$\frac{1}{(n-2)(n-1)n}\sum_{r=1}^n(n^2-nr-n-nr+r^2+r)=$$
$$=\frac{1}{(n-2)(n-1)n}\left(n^3-n(\frac{n(n+1)}{2})-n^2-n(\frac{n(n+1)}{2})+(\frac{n(n+1)(2n+1)}{6})+(\frac{n(n+1)}{2})\right)=$$
$$=\frac{1}{6(n-2)(n-1)n}(6n^3-3n^3-3n^2-6n^2-3n^3-3n^2+2n^3+3n^2+n+3n^2+3n)=$$
$$=\frac{1}{6(n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\s... | \begin{align}
\frac d {dx} \arcsin\left( \frac{2x}{1+x^2} \right) & = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac d {dx} \frac{2x}{1+x^2} & & \text{by the chain rule} \\[10pt]
& = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} & & \text{by the quotient rule} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
convergence of $\int_{\mathbb{R}^3}\frac{dx\,dy\,dz}{1+x^4+y^4+z^4}$ I need to determine whether the integral
$$\int_{\mathbb{R}^3}\frac{dx\,dy\,dz}{1+x^4+y^4+z^4}$$
converges or diverges.
I have no good idea for a substitution, but I tend to believe that the way to solve it is to find some upper bound of the function... | $$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\rho^4}\,d\rho = \pi^2\sqrt{2}$$
and in a similar way
$$\iiint_{\mathbb{R}^3}\frac{d\mu}{1+\frac{1}{3}(x^2+y^2+z^2)^2} = \int_{0}^{+\infty}\frac{4\pi\rho^2}{1+\frac{1}{3}\rho^4}\,d\rho = \pi^2 3^{3/4}\sqrt{2} $$
hence the give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the minimal-polynomial of $\begin{pmatrix} 0&1&0&1\\1&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$
$E:=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} A:= \begin{pmatrix} 0&1&0&1\\1&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix} A^2:= \begin{pmatrix} 1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix}$
My atte... | The quick way to deduce the minimal polynomial (or Jordan normal form or anything else) is to note that $A = B \otimes C$ where
$$
B = \pmatrix{1&1\\0&0}, \quad C = \pmatrix{0&1\\1&0}
$$
and $\otimes$ denotes the Kronecker product.
Since $B$ has minimal polynomial $x^2 - x$ and $C$ has minimal polynomial $x^2 - 1$, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proof convergence of series using limit of summand I have the following series: $\sum^\infty_{n=0}(\frac{n+(-1)^n}{2n+10})^{\frac{n}{3}}$ and I have to check wether it converges or not. Which convergence test should I take? I think the limit of the summand is zero so the series should converge but how do I prove that?
| $$
\begin{align}
\,& 0\le\left[\frac{n+\color{red}{(-1)^n}}{2n+10}\right]^{\frac{n}{3}}\le\left[\frac{n+\color{red}{1}}{2n+10}\right]^{\frac{n}{3}} \space\implies \\[4mm]
\,& 0\le\sum_{n=0}^{\infty}\left[\frac{n+(-1)^n}{2n+10}\right]^{\frac{n}{3}}\le\sum_{n=0}^{\infty}\left[\frac{n+1}{2n+10}\right]^{\frac{n}{3}} \sp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2... | So the area is $$\begin{align}A&=\int_{1}^2\big[(4x-x^2-1)-(x+1)\big]dx\\
&=\int_{1}^2(-x^2+3x-2)dx\\
&=\Big[\frac{-x^3}{3}+\frac{3x^2}{2}-2x\Big]_{1}^{2}\\
&=\left(-\frac{8}{3}+6-4\right)-\left(-\frac{1}{3}+\frac{3}{2}-2\right)\\
&=-\frac{8}{3}+2+\frac{1}{3}-\frac{3}{2}+2\\
&=-\frac{7}{3}+4-\frac{3}{2}\\
&=-\frac{7}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Calculating area of sphere with constraint on zenith Let $\mathbb S_R$ be the sphere of radius $R$ centered about the origin. Consider $$A_R= \left\{ (x,y,z)\in \mathbb S_R \mid x^2+y^2+(z-R)^2\leq 1 \right\}.$$
I want to calculate the area of this region of the sphere with radius $R$ about the origin. I already calcul... | I think @David K has addressed your point. Here is how I would proceed:
Parametrize the surface as follows:
\begin{cases}
x=x\\
y=y\\
z=\sqrt{R^2-x^2-y^2}
\end{cases}
with $(x,y)\in D$, where $D$ is projection of the intersection of both spheres in the $xy$-plane. When both spheres intersect, we have
\begin{cases}
x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit.
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding thi... | $\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}
$
Since
$a^3-b^3
=(a-b)(a^2+ab+b^2)
$,
$a-b
=\dfrac{a^3-b^3}{a^2+ab+b^2}
$.
Therefore
$a^{1/3}-b^{1/3}
=\dfrac{a-b}{a^{2/3}+a^{1/3}b^{1/3}+b^{2/3}}
$.
Since,
for $x > 0$,
$1 < (1+x)^{1/3}
< 1+x/3
$
(cube both sides),
we have,
if $u > 0$,
$n^{2/3}
< (n^2+u)^{1/3}
= n^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How do you project vectors? So, I am relatively new to the concept of vectors. I'm not to sure on how to "project a vector", as shown below.
For $\textbf{v} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$ and $\textbf{w} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, compute $\text{proj}_{\textbf{w}} \textbf{v}$.
And... For $\textbf... | Recall what is the application of the dot product between vectors in physics:
$$
\mathbf{v} \cdot \mathbf{w} = \text{scalar} = \left| {proj_\mathbf{w} \mathbf{v}} \right| \cdot \left| \mathbf{w} \right|
$$
then
$$
\left| {proj_\mathbf{w} \mathbf{v}} \right| = \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$ The following inequality is exercise 1.8 from this book.
For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$
I've managed to prove this via brut-force and Muirhead's inequality (Very unsatis... | It is a typo. In fact $27(S_1S_2-S_3)^2\geq 64S_2^3$ (which is correct) expands to
$$
\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^3
$$
with exponent $3$ instead of $2$ on the right-hand side.
Then the wanted inequality follows because
$$
a^2 + ab + b^2 = \frac 34 (a+b)^2 + \frac 14 (a-b)^2 \ge \frac 34 (a+b)^2
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
plugging in terms of a sequence I am having some trouble understanding how to plug terms into the following sequence. I was given the following,
$$a_1=1, a_{n+1}=\frac{a_n}{\sqrt{n}}$$ for all n greater or equal to 1.
Plugging in terms we get the following,
$$a_1=1$$
$$a_2=a_1=1$$
$$a_3=\frac{a_2}{\sqrt{2}}=\frac{1}{\s... | We have that for $n \in \mathbb Z^+$, $$a_{n+1} = \frac {a _n}{\sqrt {n}} \tag {1}$$
Thus, using the relation $(1)$ successively, we get, $$a_2 = a_{1+1} = \frac {a_1}{\sqrt {1}} = \frac {1}{1} =1 $$ $$a_3 = a_{2+1} = \frac {a_2}{\sqrt {2}} = \frac {1}{\sqrt {2}} $$ $$a_4 = a_{3+1 } = \frac {a_3}{\sqrt {3}} = \frac {1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions of $(b^2+1)(c^2+1)=a^2+1$ As the title says, I'm interested in integer solutions of the equation $(b^2+1)(c^2+1)=a^2+1$. Is it possible to parametrize the solutions as in the case of Pythagorean triples? If yes, then how would one proceed to find a parametrization in this case?
As suggested in the co... | Partial answer :
Let $n\in\mathbb{N}$ and :
$$(a,b,c)=(n,n+1,n^2+n+1)$$
We have :
$$(a^2+1)(b^2+1)=(n^2+1)(n^2+2n+2)=n^4+2n^3+3n^2+2n+2=(n^2+n+1)^2+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Compute the following without the calculator
$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$
Please solve the following equation without using calculator.
Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step... | Let $f(x)=4x^4-16x^3-17x^2+27x-3$. Then $g(x)=f(x)-4=(4x^2 - 20x + 7)(x^2 + x - 1)$.
Since $a=(5+3\sqrt{2})/2$ satisfies $4a^2-20a+7=0$ we know that $g(a)=0$ and hence
$f(a)=4$. So the answer is $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to use powers on matrices In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?
| Since first part is answered by Doug M , for the second bit we can approach by the method of induction.
We consider this matrix ,$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n}$
Lets check for n=2,
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Compute $5!25! \mod 31$ For an exercise, I was asked to compute $5!25! \mod 31$. I noticed that $5! = 120 \equiv -4 \equiv 27 \mod 31$. Therefore we have that
$$5!25! \equiv 27 \cdot 25! \mod 31.$$
Because of the congruence of Wilson, I also know that $30! \equiv -1 \mod 31$.
We have that $30! \equiv 30 \cdot 29 \cdo... | $$5!25!\equiv\left(15!\right)^2\equiv1^2=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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If $x,y,z>0$, prove that $\frac{x+y+z}{3\sqrt{3}} \geq \frac{yz+zx+xy}{\sum \limits_{cyc} \sqrt{x^2+xy+y^2}}$
If $x,y,z > 0$, prove that $\dfrac{x+y+z}{3\sqrt{3}} \geq \dfrac{yz+zx+xy}{ \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}}$
with equality if and only if $x=y=z$.
SOURCE :CRUX (Page Number 20 ; Question... | Let $x^2+y^2+z^2=k(xy+xz+yz)$.
Hence, $k\geq1$ and we need to prove that
$$\sum_{cyc}\sqrt{x^2+xy+y^2}\geq\frac{3\sqrt3(xy+xz+yz)}{x+y+z}$$ or
$$\sum_{cyc}(2x^2+xy)+2\sum_{cyc}\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\frac{27(xy+xz+yz)^2}{(x+y+z)^2}$$
Now, by C-S $$\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq x^2+\frac{x(y+z)}{2}+yz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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how to prove this question of eigen-values and eigen-vectors? If
$$
A=\begin{bmatrix}
\sin\theta & \csc\theta & 1 \\
\sec\theta & \cos\theta & 1 \\
\tan\theta & \cot\theta & 1 \\
\end{bmatrix}
$$
then prove that there does not exist a real value of $\theta$ for which characteristics roots of $A$ are $-1,1,3$
i tried to... | \begin{eqnarray}
a\sin\theta+b\cos\theta&=&\sqrt{a^2+b^2}\left[\frac{a}{\sqrt{a^2+b^2}}\sin\theta+\frac{b}{\sqrt{a^2+b^2}}\cos\theta\right]\\
&=&\sqrt{a^2+b^2}\left[\sin\phi\sin\theta+\cos\phi\cos\theta\right]\\
&=&\sqrt{a^2+b^2}\cos(\theta-\phi)
\end{eqnarray}
Where $$\phi=\arcsin\frac{a}{\sqrt{a^2+b^2}}$$
So the res... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is nth number on this series? I have series of
1+5+12+22+.....
where d=3n+1 where n=0,1,2...
but i can not find the nth number on this series ? can anyone explain how to get nth number on this series?
| We have $$T_1 =1$$ $$T_2 =5 =1+4$$ $$T_3 = 12 = 1+4+(4+3) $$ $$T_4 =22 =1+4+ (4+3) + (4+3+3) $$ $$T_5 =35 = 1+4+(4+3)+(4+3+3)+(4+3+3+3) $$ $$\vdots $$
We can thus write $$T_n = 1 + 4 +(4+3) +(4+2\times 3) + \cdots +(4+(n-2)\times 3) $$ $$= 1 + (n-1)\times 4 + (1+2+3+\cdots + (n-2))\times 3$$ $$=1+(n-1)\times 4 + (\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Calculating limit of recursive sequence I am preparing for a test and wanted to ask you
$a_0 = 1; a_{n+1} = \sqrt{a_n} + \frac{15}{4} $
I already showed its strictly monotonically increasing. Now im trying to calculate the limit.
$$\lim a_{n+1} = \lim a_n \Leftrightarrow a = \sqrt{a} + \frac{15}{4} \Leftrightarrow a =... | The missing part is $a_n$ has upper bound.
Using induction, we'll show $a_n \le \frac {25} 4, \forall n \ge 1$.
For $n=1$ it's obvious. Suppose it's true for $n$. Then $a_{n+1}=\sqrt{a_n} + \frac{15}{4} \le \frac 5 2 + \frac {15} 4 = \frac {25} 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to establish the identity of the infinite sum How to prove the following identity?
$$\sum_{n=-\infty}^{\infty}\frac{1}{(z+n)^2 +a^2} = \frac{\pi}{a}\cdot\frac{\sinh 2\pi a}{\cosh 2\pi a - \cos 2\pi z}$$
| The trick here is to use $$\frac{1}{(z+w)^2+a^2} \times \pi\cot(\pi
w)$$ as shown at the following MSE
link. The sum term
is also quadratric in $n$ so the estimates of the integrals presented
there apply to the present case as well.
We get for the residues at $w = -z \pm ia$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the Taylor series for $f(x) =\ln(1+x)$ Find the Taylor series for $f(x) = \ln (1+x)$ centered at $x = 0$ using the formula for Taylor Series.
|
$$\ln(1+x)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac {x^4}4+\&\text{c}=\sum\limits_{r=0}^{\infty}\dfrac {(-1)^{r}x^{r+1}}{r+1}$$
The Taylor Series for $f(x)$ at the point $x=a$ is
$$f(x)=f(a)+f^1(a)(x-a)+\dfrac {f^2(a)}{2!}(x-a)^2+\dfrac {f^3(a)}{3!}(x-a)^3+\cdots+\dfrac {f^r(a)}{r!}(x-a)^r$$
Since we want it at $x=0$, $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$
The question is to evaluate $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$$
I could rewrite the denominator as $$\t... | The product can be rewritten as
$$\prod_{r=1}^{\infty} \frac{\cos{\left ( 2^{-r} a \right )}}{1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )}} $$
Use the tangent half-angle formula to simplify the denominator:
$$\begin{align}1+\tan{\left ( 2^{-r} a \right )} \tan{\left ( 2^{-r+1} a \right )} &= 1+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)... | Why not to use Taylor series $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\sin(3x)=3 x-\frac{9 x^3}{2}+\frac{81 x^5}{40}+O\left(x^6\right)$$
$$\frac{\sin 3x}{x^3}+\frac{a}{x^2}+b =\frac{a+3}{x^2}+\left(b-\frac{9}{2}\right)+\frac{81 x^2}{40}+O\left(x^3\right)$$ Then $???$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Proof that expression is positive if conditions is met Is there a way to show that this expression is always positive as long as $b>0$ and $r>x$ ? Assume $r>0$ and $x>0$
\begin{equation}
b-1+\sqrt{(1+b)^2-\frac{4rb}{x}}
\end{equation}
It's simple with $b>1$ but I can't figure it out if it is true for lower values.
| The statement is obviously not true if $(1+b)^2 - \frac {4br}x < 0$.
Case 1: $(1+b)^2 - \frac {4br}x < 0$ or in other words
$(1+b)^2 < \frac {4br}x$ or
$\frac {(1+b)^2*|x|}{4|b|} < r$. (Assuming $b \ne 0$. If $b = 0$ then $(1+b)^2 - \frac {4br}x = (1+b)^2 = 1$.)
The statement is false and yields non-real number.
Cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does removing all numbers in the harmonic series with a units digit of 9 affect the series? The harmonic sequence is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\ldots$ diverges. There is a simple reason why:
$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\ldots$ is ... | The series $\sum\limits_{n=1}^\infty\frac{1}{2n}$ diverges because $\sum\limits_{n=1}^N\frac{1}{2n}=\frac12\sum\limits_{n=1}^N\frac{1}{n}$. You get rid of all odd units digits this way, still diverging.
The series $\sum\limits_{n=1}^\infty\dfrac{1}{10^k n}$ for any fixed $k\in\mathbb N$ gives you $0$s in the first $k$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Tangent at a given point I have a function $\frac{x^2+2}{x-1}$. I want to find a tangent at a given point of $x=1+\sqrt{3}$.
At first, I found a value of the function at given point. This is what I got: $2+\frac{6}{\sqrt{3}}$
Then, I found a derivative of this point and I think it's $0$. So, for me the tangent is $y=(2... | The derivative of a function $f$ at a point $x_0$ is the slope of the tangent line to $f$ at $x_0.$
So, let $f(x)=\frac{x^2+2}{x-1}$ and $x_0=1+\sqrt{3}.$
$$\Rightarrow f'(x)=\frac{(x-1)\cdot 2x - (x^2+2)\cdot 1}{(x-1)^2}=\frac{x^2-2x-2}{(x-1)^2}.$$
and $$f'(x_0)=f'(1+\sqrt{3})=\frac{(1+\sqrt{3})^2-2(1+\sqrt{3})-2}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solving: $3^m-2=n^2$
Solve $3^m-2=n^2$ for positive pairs of integers $(n,m)$.
My try:
$3^m-6=n^2-4\implies 3(3^{m-1}-2)=(n+2)(n-2)\implies 3=n+2\ or\ 3=n-2$ so we got $n=\pm 5 , \pm 1$. Used $\pm$ as we have $n^2$.
Putting $n=\pm 5 , \pm 1$ gives $m=3,1$ respectively. We kick out negative ones as problem ask for pos... | This is too long for a comment. My approach is not "elementary" since I use factorization in quadratic rings, more exactly, I'll work in $\Bbb{Z}[\sqrt{-2}]$ which is known to be a Euclidean domain and hence a UFD (unique factorization domain). Moreover, in this ring every element $x+y\sqrt{-2}$ has norm $N(x+y\sqrt{-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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The number of 3 digit numbers of the form xyz such that $x
The number of 3 digit numbers of the form $xyz$ such that $x<y$ and $z\leq y$ is N such that n is a 3 digit number of the form $abc$ then find $a+c-b$
My approach:
Case 1:
$x<y$
$z<y$
$x,y,z\neq 0$
Choosing 3 digits out of 9 numbers: $\binom 93$, greatest ... | A trial and error method looks better here.
Note that, here $$1\le x \le 8$$ $$\max(x,z) \le y\le 9$$ $$0\le z\le 9$$
Now, for $x=1$, and for all $z$, $y$ can take $8+8+8+7+6+5+4+3+2+1=52$ values.
Again, for $x=2$, and for all $z$, $y$ can take $7+7+7+7+6+5+4+3+2+1=49$ values.
So, it can be checked that, in general for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\min\int_0^\pi {f^2(x)dx}$, assume $\int_0^\pi{f(x)\sin xdx} = \int_0^\pi{f(x)\cos xdx} = 1.$ $f(x)$ is continuous on $[0,\pi]$ and $\int_0^\pi{f(x)\sin xdx} = \int_0^\pi{f(x)\cos xdx} = 1.$
Find $\min\int_0^\pi {f^2(x)dx}.$
I try to solve this problem by this:
$$\begin{array}{l}
{\left( {\int\limits_0^\pi {f(x)... | Let us consider the Fourier series of $f$
\begin{align}
f(x) = \frac{1}{2}a_0+\sum^\infty_{n=1} a_n \cos nx+ \sum^\infty_{n=1} b_n \sin nx
\end{align}
then that means
\begin{align}
\int^\pi_0 f^2(x)\ dx= \frac{1}{4}a_0^2+\frac{\pi}{2}\sum^\infty_{n=1}(a_n^2+b_n^2).
\end{align}
Since
\begin{align}
\int^\pi_0 f(x) \cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Reducibility of $x^4+1$ in prime fields In Dummit and Foote's "Abstract Algebra" there is a proof for the reducibility of $x^4+1$ in which it's stated:
For odd primes p, $p^2\equiv1\pmod 8$. Hence $x^8-1\mid x^{p^2-1}-1 $.
Why does $x^8-1\mid x^{p^2-1}-1$ follow?
| As a side answer, not to the original question, you can get this result fairly explicitly.
If $-1=a^2$, for some $a$, then $x^4+1=(x^2+a)(x^2-a)$.
If $2=a^2$ for some $a$, then $x^4+1=(x^2+ax+1)(x^2-ax+1)$.
If $-2=a^2$ for some $a$, then $x^4+1=(x^2+ax-1)(x^2-ax-1)$.
But if $-1$ and $2$ are not squares, modulo $p$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Intersection of two lines - Have I got it right? I have been asked to solve where these two lines intersect, it has been a very long time since I have done this, my numbers seem too big. Can someone check these for me?
2x + y = 6
x + 2y = 12
(find x) (plug into first equation)
x + 2y =... | By elimination:
$$\begin{cases}2x + y = 6,\\
x + 2y = 12.\end{cases}$$
Subtract twice the first equation from the second and get
$$-3x=0.$$
By substitution:
From the first equation, $y=6-2x$, and plugging in the second, $x+12-4x=12$, giving $$-3x=0.$$
By Cramer:
$$x=\frac{\begin{vmatrix}6&1\\12&2\end{vmatrix}}{\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2173508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$. Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.
I have shown it is not possible for $a=4$, but not for any a greater than $4$.
| Rewrite the equation into
$$7^a\cdot(x+y)=2^a\cdot xy$$
From this, we conclude that there is some positive integer $k$ with
\begin{align}
xy&=k\cdot 7^a\tag{1}\label{eq1}\\
x+y&=k\cdot 2^a\tag{2}\label{eq2}
\end{align}
Now, from equation $\eqref{eq1}$, we conclude that if $p\neq 7$ is prime and $p|k$, then $p|x$ or $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes... | Based on this (On the arctangent inequality.) answer, we have:
$$\frac{\arctan x}{x} \geq 1/2$$
for $x \in (0,1]$.
So letting $x = \frac{1}{n}$, we have $\arctan \frac{1}{n} \geq \frac{1}{2n}$ for each $n\geq 1$, so by the comparison test, your series diverges
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Factor $(x+y)^4+x^4+y^4$ Title says it all, I just want to know how to factor $(x+y)^4+x^4+y^4$. I only know that it's possible to factor, but got no idea how to do it. If it were a single-variable polynomial I could try to find rational roots or something, but I'm lost with this one.
| $$(x+y)^4+x^4+y^4=(x^2+y^2+2xy)^2+(x^2+y^2)^2-2(xy)^2=\\
[(x^2+y^2+2xy)^2-(xy)^2]+[(x^2+y^2)^2-(xy)^2]=\\
(x^2+y^2+xy)(x^2+y^2+3xy)+(x^2+y^2+xy)(x^2+y^2-xy)=\\
(x^2+y^2+xy)(2x^2+2y^2+2xy)=2(x^2+y^2+xy)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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Prove that $F(xy) =F(x) + F(y)$ when $F(x)$ is not $\ln(x)$ So we have this function for $x > 0$.
$$\int_{1}^{x} \frac{1}{t}\text{d}t$$
Show that $F(xy) = F(x) + F(y)$ without assuming $F(x) = \ln(x)$.
I came so far as this point, but I can't crack the last step.
$$\int_{1}^{x} \frac{1}{t}\text{d}t + \int_{x}^{xy} \fr... | Spoiler below... Don't forget to change both upper and bottom integral bounds when using an integral sustitution
This is the full answer...
$$ F(xy):=\int_{1}^{xy} \frac{1}{t}dt $$
$$ let \,\,\,\begin{matrix} t = x \cdot u & x \cdot u = xy \\ dt = x \cdot du & x \cdot u = 1\end{matrix}
\,\,\,OR\,\,\,
\begin{m... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Decomposition into partial fractions of an inverse of a generic polynomial with three distinct roots. Let $d \ge 2$ be an integer. Let $\left\{ m_j \right\}_{j=1}^d$ be strictly positive integers and $\left\{ b_j \right\}_{j=1}^d$ be parameters. Define the following quantity:
\begin{equation}
{\mathfrak F}_d(x) := \fra... | The result is given below:
\begin{eqnarray}
&&{\mathfrak F}_d(x) =
\sum\limits_{k=1}^d
\sum\limits_{\begin{array}{rrr} 1&\le l_{d-2}& \le m_k \\l_{d-2}& \le l_{d-3} &\le m_k\\&\vdots&\\l_1&\le l_0&\le m_k\end{array}}
\prod\limits_{j=-1}^{d-3} \binom{l_j + m_{f_{k,j}}-1-l_{j+1}}{m_{f_{k,j}}-1} \frac{1}{\left(b_k-b_{f_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly ap... | We can try to maximize $f(x,y)=x^3+y^3$ subject to
\begin{align}
&x>0\\
&y>0\\
&x^2+y^3-x^3-y^4\geq 0
\end{align}
Since $\nabla f(x,y)=(3x^2,3y^2)$, $f$ always increases away from the origin in the first quadrant. Hence, a maximum must occur on $x^2+y^3-x^3-y^4=0$.
At this point we can use Lagrange multipliers. A maxim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find all pairs of prime numbers $p$ and $q$ such that $\,p^2-p-1=q^3.$ I'm preparing a mathematical olympiad and our group is stuck in this problem. Here is all we've done:
First, let's rewrite the equation
$$p^2-p-1=q^3\Rightarrow p(p-1)=(q+1)(q^2-q+1)$$
It's obvious that $q$ must be less than $p$. Then, $p|(q^2-q+1)$... | Since we want $$p^2-(2+3k+k^3)p+(3k^2+3k+1)=0$$ to have integer solutions its Discriminant has to be a perfect square
i.e. $k^2(k^4+6k^2+4k-3)$ has to be a perfect square thus
$k^4+6k^2+4k-3$ has to be a perfect square
But $(k^2+3)^2 \le k^4+6k^2+4k-3<(k^2+4)^2$ thus we have
$4k-3=9 \Rightarrow k=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Use extended Euclidean algorithm to find $j,k$ such that $52j+15k=3$ Is this questions suggesting that $\gcd(52,15)=3$ i.e. $52j+15k=3$?
if it is then why am I getting $1$ when I am computing the $\gcd$.
$$
\gcd(52,15)
= \gcd(15,7)
= \gcd(7,1)
= \gcd(1,0)
= 1
$$
Am I going in the right direction or not?
| First, we find $\gcd(\color{#c00}{52}, \color{#0a0}{15})$ using the Euclidean Algorithm.
$$
\begin{align*}
\color{#c00}{52}&=3 * \color{#0a0}{15} + \color{#00c}{7}\\
\color{#0a0}{15}&=2 * \color{#00c}{7} + 1\\
\color{#00c}{7}&=1*\color{#00c}{7}+0
\end{align*}
$$
So, $\gcd(\color{#c00}{52}, \color{#0a0}{15})=1$. You cor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Proving the given identity For every continuous periodic function $F(\theta)$ with period $2 \pi$ and for $0 < \rho < 1,$ prove that
$$\lim_{\rho \rightarrow 1}\frac{1}{4\pi \rho} \int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right... | We can prove the identity by elementary methods.
Since $F(\theta )$ and $\cos n\theta $ are periodic
\begin{align}
&\int_{-\pi}^{\pi} F(\theta ) \left( \frac{1-\rho^2}{1-2 \rho \cos n\theta + \rho^2} - \frac{1-\rho^2}{1+2 \rho \cos n\theta + \rho^2} \right)~d\theta\\
=&\int_{0}^{2\pi} F(\theta ) \left( \frac{1-\rho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find the integral surface of given differential equation. Find the integral surface of the differential equation $(x-y)p+(y-x-z)q=z$ passing through the circle C: $z=1, x^2+y^2=1$
Clearly the Lagrange's auxillary equations are
$\frac{dx}{P} = \frac{dy}{Q}. =\frac{dz}{R}$
Where P=$(x-y)$ ,Q=$y-x-z$ & R=$z$
on com... | Your calculus is correct. The characteristic equations are :
$$\begin{cases}
x+y+z=c_1\\
\frac{x-y+z}{z^2}=c_2
\end{cases}$$
The general solution of the PDE, expressed on the form of implicit equation, is :
$$F(X,Y)=0 \quad \begin{cases}
X=x+y+z\\
Y=\frac{x-y+z}{z^2}
\end{cases}$$
where $F(X,Y)$ is any differentiable e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Counting the Number of Real Roots of $y^{3}-3y+1$ Here's my question:
How many real roots does the cubic equation $y^3-3y +1$ have?
I graphed the function and it crossed the x-axis $3$ times. But my professor doesn't want a graphical explanation. So in that case, I was looking at the Fundamental Theorem of Algebra an... | Set $x=2\cos\varphi$; then the equation becomes
$$
4\cos^3\varphi-3\cos\varphi=-\frac{1}{2}
$$
that is,
$$
\cos3\varphi=\cos\frac{2\pi}{3}
$$
Thus we get
$$
3\varphi=\frac{2\pi}{3}
\qquad\text{or}\qquad
3\varphi=\frac{2\pi}{3}+2\pi
\qquad\text{or}\qquad
3\varphi=\frac{2\pi}{3}+4\pi
$$
and therefore
$$
x=2\cos\frac{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 8
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Calculating Euler Number limit Please, so far I did
$$\lim_{x\to +\infty}\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}},$$ but I can write
$$\frac{x^2-x+1}{x+2}=1+\frac{x^2-2x-1}{x+2}=1+\frac{1}{\frac{x+2}{x^2-2x-1}}.$$
But
$$\lim_{x\to +\infty}\frac{x+2}{x^2-2x-1}=0,$$ so I can not use
$$e =\lim_{N\to \infty}(1+\fr... | Let $\displaystyle y=\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}}$ then
$$\ln y= \frac{1}{x-1}\ln(x^2-x+1)-\frac{1}{x-1}\ln(x+2)$$
$$\lim_{y\rightarrow \infty} \ln y=\lim_{x\rightarrow \infty}\frac{2x-1}{x^2-x+1}-\lim_{x\rightarrow \infty}\frac{1}{x+2}=0,\ L'Hopital$$
Then $\lim_{y\rightarrow \infty}y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric inequality with weird angle I have this inequality $\log_{\tan x}\sqrt{\sin^2x-\frac{5}{12}}<1$ that I would like to solve.
So $\tan x>0$ and $ \tan x\ne1$. First I try when $\tan x>1$.
Then I have $\sqrt{\sin^2x-\frac{5}{12}}<\tan x$
I have to solve the system
*
*$\sin^2x-\frac{5}{12}\ge0$ (I can't... | $1)$ If $\tan x>1\to \frac{\pi}{4}<x<\frac{\pi}{2}$ or $\frac{5\pi}{4}<x<\frac{3\pi}{2}$ $(1)$ then:
$$\sqrt{\sin^2x-\frac{5}{12}}<\tan x\to \sin^2x-\frac{5}{12}<\tan^2x\\
\sin^2x-\tan^2x<\frac{5}{12}\to -\frac{\sin^4x}{\cos^2 x}<\frac{5}{12}\\
-12\sin^4x<5(1-\sin^2x)\to12\sin^4x-5\sin^2x+5>0$$
which is true for any $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$ For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$
I think if we let $x=\tan A;y=\tan B;z=\tan C$, then
$$P\Leftrightarrow \... | Like How to show that the triangle is equilateral triangle?,
$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$ with $A+B+C=\pi$
Now use this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
integrate $\int\frac{x\cdot dx}{(x^3+1)^2}$ What methods are there to integrate: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$
I know about partial fractions:
$$\int\frac{x\cdot dx}{(x^3+1)^2} $$
$$= \int\frac{x\cdot dx}{((x+1)(x^2-x+1))^2} $$
$$= \int \left(\frac{A}{x+1}+\frac{Bx+C}{(x+1)^2} + \frac{Dx+E}{x^2-x+1} + \frac{Fx^3... | $$
\begin{align}
\int\frac{x\,\mathrm{d}x}{\left(x^3+1\right)^2}
&=-\frac13\int\frac1x\,\mathrm{d}\frac1{x^3+1}\tag{1}\\
&=-\frac1{3x\left(x^3+1\right)}-\frac13\int\frac{\mathrm{d}x}{x^2\left(x^3+1\right)}\tag{2}\\
&=-\frac1{3x\left(x^3+1\right)}-\frac13\int\left(\frac1{x^2}-\frac{x}{x^3+1}\right)\,\mathrm{d}x\tag{3}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find all solutions in positive integers to $a^b -b^a = 3$
Find all solutions in positive integers to $a^b -b^a = 3$
It appears that the only solution is $a=4, b=1$.
Modular arithmetic is not of much help (apart from deriving that $a,b$ must have opposite parity).
We know that $a^x$ dominates the polynomial $x^a$ bu... | Consider $$f_a(x)=x\ln(a)-a\ln(x)$$ with $a,x\ge 3$
We have $$f_a'(x)=\ln(a)-\frac{a}{x}>0$$ for $x>\frac{a}{\ln(a)}$
In particular, $f_a(x)$ is strictly increasing for $x\ge a$ and we have $$f_a(a+1)=(a+1)\ln(a)-a\ln(a+1)$$
The function $g(x)=(x+1)\ln(x)-x\ln(x+1)$ has derivate $g'(x)=\ln(x)-\ln(x+1)+\frac{x+1}{x}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$. This was the original excercise:
Prove that $$\frac{\sin{x}+\tan{x}}{2} \geq x$$ where $x \in (0,\frac{\pi}{2})$
This is how I did it:
Knowing that
\begin{align}\sin{0}&=0\\
\cos{0}&=1\\
\frac{\mathrm{d} \sin{x}}{\mathrm{d} x}&=\cos{x}\\
\frac{\mathr... | An alternative proof for the desired inequality without using
Taylor series would be to notice that
$$
f(x) = \frac 12 (\sin x + \tan x)
$$
satisfies $f(0) = 0$ and
$$
f'(x) = \frac 12 \left(\cos x + \frac{1}{\cos^2 x} \right) \ge
\sqrt{\cos x \cdot \frac{1}{\cos^2 x}} = \frac{1}{\sqrt{\cos x}}
\ge 1
$$
for $0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factorisation of quartic equation. I'm given a logarithm problem which is $$\log_{2} (x^3+1)-2\log_{2}x=\log_{2}(x^2-x+1)-2$$
I'm stuck in the step of $x^4-5x^3+x^2-4=0$
By many times of trial and error, I got $(x^2-4x-4)(x^2-x+1)=0$
My question, is there any standard way to factorise the equation without trials and er... | Hint: the quartic can be avoided altogether by noting that $x^3+1=(x+1)(x^2-x+1)\,$. Given that $x^2-x+1 \gt 0$ on $\mathbb{R}$ it follows that $x^3+1$ and $x+1$ have the same sign, which must be positive for the $\log_2$ to be defined. Then the equation simplifies to:
$$
\require{cancel}
\log_{2} (x+1)+\cancel{\log_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is
$$\begin{align}
x\quad&=\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{... | Cubic polynomials are ubiquitous in computer-aided design and computer graphics.
They also are the basis for computer fonts.
Finite element analysis is based on polynomial functions. Isogeometric analysis uses NURBS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 13,
"answer_id": 10
} |
Inequality involving the tangent function How can one show that
$$x \tan x > \frac{4x - \pi}{\pi - 2x},$$
for $x \in \left(0,\frac{\pi}{2}\right)$.
Clearly, for $x \in \left(0,\frac{\pi}{4}\right)$ as $4x - \pi < 0$ and $\pi - 2x > 0$ the term $\frac{4x - \pi}{\pi - 2x}$ is negative while $x \tan x$ is always positive... | For $0<x\leq\frac{\pi}{4}$ our inequality is obviously true.
Let $\frac{\pi}{4}<x<\frac{\pi}{2}$ and $t=\tan{x}$.
Hence, $t>1$ and we need to prove that
$$(\pi-2x)x\tan{x}>4x-\pi$$ or
$$2x^2\tan{x}+(4-\pi\tan{x})x-\pi>0$$ or
$$x<\frac{-4+\pi\tan{x}+\sqrt{16+\pi^2\tan^2x}}{4\tan{x}}$$
or $f(x)>0$, where
$$f(x)=\frac{-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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.