Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| The solutions of $8x^2+36x+36=0$ are $-1.5$ and $-3$, but the latter is not a solution of the original equation. This new, false solution comes when you multiply the equation by $(x+3)(2x+6)$, which is not the lcm of the denominators.
Note that $2x+6=2(x+3)$.
An example to show what's happening:
Take the equation
$$3x+... | {
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"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
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the sum of consecutive odd numbers If the sum of consecutive odd numbers starting with $-3$ until $2k+1$ equals $21$
What is the value of $k$ ?
I can solve this by trying the numbers $-3-1+1+3+5+7+9=21$ , so the last term is $7th$ so the $k$ value is $3$
But I could not solve this with formula, I know the odd numbers c... | Notice that
$$\begin{align}1^2&=1\\2^2&=1+3\\3^2&=1+3+5\\\vdots&\qquad\qquad\qquad\quad\ddots\\k^2&=1+3+5+\dots+(2k-1)\\(k+1)^2&=1+3+5+\dots+(2k-1)+(2k+1)\end{align}$$
Thus,
$$-3-1+1+3+\dots+(2k+1)=(k+1)^2-4$$
| {
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"url": "https://math.stackexchange.com/questions/2047073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Limit of a sequence Calculate the limit of the sequence $\{x_n\},$ defined as follows: $$x_n=\dfrac{a+aa+aaa+aaaa\cdots+aaaaaaa..aaa }{10^n},$$ where $a\in\{1,2\ldots,9\}.$
$aaaaaaa..aaa = a , n$ time
Can anyone help ?
| It looks like you mean the following:
$$x_n = \frac{a}{10^n}\sum_{k=1}^{n}\sum_{j=0}^{k-1}10^j$$
Assuming this is the case, first note that the inner sum can be computed in closed form. For any constant $c \neq 1$, we have
$$\sum_{j=0}^{k-1}c^j = \frac{c^k - 1}{c - 1}$$
so for $c = 10$ this is
$$\sum_{j=0}^{k-1}10^j = ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Which would you rather have? Q. Which would you rather have, a piece of an 8-inch pie that's been cut into sixths or a piece of a 10-inch pie that's been cut into eights?
A. This is a problem involving sectors. One-sixth of a pie is $\frac {1}{6}$ of $ 2\pi$ radians. The measure of the central angle is $\frac{1}{6} \... | You don't need to even use $\pi$ here. Circles behave like squares when it comes to area, so you can just calculate that
$$\frac{8^2}6 < \frac{10^2}8.$$
| {
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"url": "https://math.stackexchange.com/questions/2048416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_0^{\infty}\frac{e^x-1}{xe^x(e^x+1)}dx.$
Evaluate the integral
$$
\int_{0}^{\infty}{\mathrm{e}^{x} - 1 \over x\,\mathrm{e}^{x}\left(\mathrm{e}^x+1\right)}\,\mathrm{d}x\,.
$$
I have no idea how to approach integrals like this and I can't get any valuable result. Any hints will be appreciated.
| Notice that
\begin{align*}
\int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx
&= \int_{0}^{\infty} \left( \int_{0}^{1} e^{xt} \, dt \right) \frac{dx}{e^x(e^x+1)} \\
&= \int_{0}^{1} \int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dxdt.
\end{align*}
In order to compute the inner integral, we utilize the geometric seri... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Let $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$. Why does $a=b=c$? Let $a,b,c\in \mathbb{R}$ and $a,b,c \ne 0$ and $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$.
Why does $a=b=c$?
| $$2(a^2b^2+b^2c^2+c^2a^2)=2a^2bc+2b^2ca+2c^2ab\\\implies (a^2b^2+b^2c^2-2ab^2c)+(a^2b^2+c^2a^2-2a^2c)+(b^2c^2+c^2a^2-2abc^2)=0\\\implies (ab-bc)^2+(bc-ca)^2+(ca-ab)^2=0$$
From here you can only say $$ab=bc=ac$$
However since $a,b,c\neq0$ you get $a=b=c$
Bingo!
| {
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"question_score": "1",
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Given that $abc=1$, prove that $\frac{b}{c} + \frac{c}{a}+ \frac{a}{b} \ge \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$. I have tried to do a bunch of different versions of AM-GM on this and the equivalent versions that come from substituting in $abc=1$ and I always flip the inequality by doing it.
Help is greatly apprecia... | Put $a = \dfrac{x}{y}, b = \dfrac{y}{z}, c = \dfrac{z}{x}$, then $x,y,z > 0$, and $abc = 1$, the inequality to prove is: $\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2} \ge \dfrac{x}{z}+\dfrac{z}{y}+\dfrac{y}{x}$, or equivalently:$(xy)^3+(yz)^3+(xz)^3 \ge (xy)^2(xz)+(yz)^2(xy)+(xz)^2(yz)$. Put $m = xy, n = yz, p = xz$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving one of the Binomial Identities The binomial identity is
$$\sum^{\left \lfloor{n/3}\right \rfloor }_{k=-\left \lfloor{n/3}\right \rfloor} (-1)^k \binom{2n}{n+3k} = 2 \cdot 3^{n-1} $$
and this is valid for all positive integers $n$. What would be some proofs to show that this identity is true?
| Suppose we seek to prove that
$$\sum_{k=-\lfloor n/3\rfloor}^{\lfloor n/3\rfloor} (-1)^k {2n\choose n+3k}
= 2\times 3^{n-1}.$$
We start by introducing the integral
$${2n\choose n+3k} = {2n\choose n-3k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-3k+1}} (1+z)^{2n}
\; dz.$$
Observe that this vanishes for $3k... | {
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"url": "https://math.stackexchange.com/questions/2054777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the divergence of the field F: $\vec{F} = \frac{yj - zk}{\sqrt{y^2+ z^2}}$
Find the divergence of the field F:
$$\vec{F} = \frac{yj - zk}{\sqrt{y^2+ z^2}}$$
Normally I find dot the gradient with F like this:
$$\langle 0,y,-z\rangle\cdot\langle F_x, F_y, F_z\rangle$$
And this simplifies to:
$$\langle 0, 1, -1\r... | $\nabla \cdot (0\mathbf i + \frac {y}{\sqrt{y^2+z^2}}\mathbf j - \frac {z}{\sqrt{y^2+z^2}}\mathbf k) = \frac {\partial}{\partial x}0+\frac {\partial}{\partial y}\frac {y}{\sqrt{y^2+z^2}} + \frac {\partial}{\partial z} \frac {-z}{\sqrt{y^2+z^2}}\\
0+\frac {2y^2 + z^2}{(y^2+z^2)^\frac 32}-\frac {y^2 + 2z^2}{(y^2+z^2)^\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/(2^n) + ..... = (1/2)*2 = 1
1/4 + 1/8 + 1/16 + ..... + 1/(2^n) + ..... = [1/(2^2)]*2 = 1/2
1/8 + 1/16 + ..... + 1/(2^n) + ..... = [1/(2^3)]*2 = 1/(2^2)
So on and so forth.
Hence, S = 1 + 1/2 + 1/(2^2) + ..... + 1/(2^n) + ..... = 1 + 1 = 2
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find the quadratic equation from highest point on parabola? If I have a parabola, where the vertex is in $P=(2,0)$, and a point on the parabola is $Q=(-1,-6)$, how can I find the quadratic equation of the form:
$$f(x) = ax^2 + bx + c$$
| The vertex is the
extreme point,
at which $f'(x) = 0$.
Since
$f'(x) = 2ax+b$,
this is zero when
$x = -\frac{b}{2a}
$.
At this point
$\begin{array}\\
f(x)
&=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\
&=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c\\
&=\frac{b^2}{4a}-\frac{b^2}{2a}+c\\
&=-\frac{b^2}{4a}+c\\
\end{array}
$
Since this po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Solve the equation $5x^2+5y^2-8xy-2x-4y+5=0$ While I was solving the question given by Harsh Kumar I could not find the real roots just because I changed the sign of $5y^2$. I also proved that there don't exist any real solution of that equation.
The equation I made by mistake was: $$5x^2+5y^2-8xy-2x-4y+5=0.$$
Please... | We may rewrite this as
$$
4x^2 - 8xy + 4y^2 + x^2 -2x + 1 + y^2 - 4y + 4 = 0\\
(2x - 2y)^2 + (x-1)^2 + (y-2)^2 = 0
$$
and we see that this is a sum of three (real) squares that equals $0$. If the sum of three squares equals $0$, then that means that each of the squares are zero. But $(x-1)^2 = 0$ means $x = 1$ and $(y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Prove that $\frac{a}{2}x^2+bx+c=0$ has a root between $x_1$ and $x_2$. If $x_1$ and $x_2$ are non-zero roots of the equations $ax^2+bx+c=0$ and $-ax^2+bx+c=0$
respectively. Prove that $\frac{a}{2}x^2+bx+c=0$ has a root between $x_1$ and $x_2$.
Please help me ..
| Let $f(x)=\frac{a}{2}x^2+bx+c$, then $f(x_1)=-\frac{a}{2}x_1^2$ and $f(x_2)=\frac{3a}{2}x_2^2$. Consequently $f(x_1)f(x_2)<0$. By continuity it should have a root in $(x_1,x_2)$.
| {
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"url": "https://math.stackexchange.com/questions/2061976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Calculate the limit $\lim_{(x,y,z)->(0,0,0)}{\frac{xyz}{\sqrt{x} + \sqrt{y} + \sqrt{z}}}$ $$\lim_{(x,y,z)->(0,0,0)}{\frac{xyz}{\sqrt{x} + \sqrt{y} + \sqrt{z}}}$$
I tried thinking about a solution using the epsilon-delta definition: I assumed that the limit is 0 but I couldn't figure out a inequality that would lead me ... | Using the AM-GM inequality
$$0 \le \frac{xyz}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \frac{\left( \sqrt[3]{\sqrt{x}\sqrt{y}\sqrt{z}} \right)^{3/2}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \le \frac{\left( \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \right)^{3/2}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \frac{1}{3\sqrt{3}}\sqrt{\sqrt{x}+\sqrt{y}+\sqrt{z}}$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle i... | Consider the angle $960^\circ$. I can easily break this down into
$$360^\circ + 360^\circ + 180^\circ + 60^\circ$$
knowing that $360^\circ$, $180^\circ$, and $60^\circ$ are a full revolution, half revolution, and an acute angle, respectively. Instead of making a "note to self" regarding the number of revolutions, let's... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you find the derivative of $\sqrt{x^2+9}$ using the definition of a derivative? I had this question on a not-so-recent math test. It asked to find the derivative of $\sqrt{x^2+9}$ with the definition of a derivative. Using the chain rule, I can figure out that the derivative is $\frac x{\sqrt{x^2+9}}$, but how c... | By the definition of the derivative, what we want to evaluate is
$$\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}$$
We now multiply the top and bottom by the conjugate of the numerator (to simplify the square roots) and simplify to get
$$=\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}\frac{\sqrt{(x+h)^2... | {
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"timestamp": "2023-03-29T00:00:00",
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solve for $x$ in inverse trignometry $$
\operatorname{arccot} x + \operatorname{arccot} (n^2-x + 1) = \operatorname{arccot }(n - 1)
$$
In this we have to solve for value of $x$ .
I thought to convert arccot into arctan . Then add using the identity .
But its getting too long .
Is there any short method to solve it.... |
1) Take the cotangent from both side:
$cot(arccot(x)+arccot(n^2-x+1))=cot(arccot(n-1))$
2) Use $cot(A+B)={cot(A)cot(B)-1\over cotA+cotB}$
${cot(arccot(x))cot(arccot(n^2-x+1))-1\over cot(arccot(x))+cot(arccot(n^2-x+1))}=n-1 $
${x(n^2-x+1)-1\over x+n^2-x+1 }=n-1$
$ -x^2+(n^2+1)x=(n^2+1)(n-1)+1$
$A:=n^2+1$
$x^2-Ax+(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding out the number of solutions If $a,b,c$ are three distinct real numbers for which there exists $x,y,z$ such that $$x+y+z=1$$$$ax+by+cz=t$$$$a^2x+b^2y+c^2z=t^2$$then $a^3x+b^3y+c^3z-t^3=P(t)$,be a polynomial in $t$,then the question is to find out the number of solutions to P(t)=0 .
My attempt at the solution
We ... | you are given $a^{3}x+b^{3}y+c^{3}z−t^{3}=P(t)$.Equate $P(t)=0$.Then multiply $(a+b+c)$ on both sides and then simplify to get
$t^{3}-(a+b+c)t^{2}+(ab+bc+ca)t-abc=0 $ which shows that the roots are $a,b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be obser... | Differentiate to get $4 n x^{3} + 4$. This has the single root $x_{0} = - \dfrac{1}{\sqrt[3]{n}}$ which is a minimum for $f(x) = nx^4+4x+3$. We want $f(x_{0}) \le 0$, or
$$
0 \ge n x_{0}^{4} + 4 x_{0} + 3 = n (-\dfrac{1}{n}) x_{0} + 4 x_{0} + 3 = 3 (x_{0} + 1).
$$
Thus we must have
$$
- \dfrac{1}{\sqrt[3]{n}}= x_{0} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067147",
"timestamp": "2023-03-29T00:00:00",
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prove $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le\frac{1}{64abcd}$ Let $a,b,c,d$ be positive real no such that $a+b+c+d\le1$
prove that$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le \frac{1}{64abcd}$$
I did this way$$\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}\le\frac{1}{64abcd}$$
$\implies$ $a^2cd+b^2cd+c^... | $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}=\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}$
$\implies(ac+bd)(ad+bc)\le\frac{1}{64}$
Now $(ac+bd)(ad+bc)\le\frac{ac+bd+ad+bc}{4}$
$=\frac{(a+b)(c+d)}{4}\le\frac{(a+b+c+d)}{64}$
As $a+b+c+d\le1$
$\implies\frac{(a+b+c+d)^4}{64}\le\frac{1}{64}$
Hence proved
| {
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"url": "https://math.stackexchange.com/questions/2067312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two... | The Brahmagupta–Fibonacci identity says that every product of two sums of two squares is a sum of two squares in two different ways:
\begin{align}
(a^2+b^2)(c^2+d^2) & = (ac+bd)^2 + (ad-bc)^2 \\[6pt]
& = (ac-bd)^2 + (ad+bc)^2
\end{align}
For example:
\begin{align}
(2^2+3^2)(1^2+7^2) & = 23^2 + 11^2 \\[6pt]
& = 19^2 + 1... | {
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"url": "https://math.stackexchange.com/questions/2068122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 13,
"answer_id": 5
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Prove by induction that $n^3-n$ is divisible by $24$ for all odd positive integers Q: Prove by induction that $n^3-n$ is divisible by 24 for all odd positive integers
After proving the first part for n=1
Assume true for some positive integer $n=k$
ie $k^3-k=24x$ where x is an integer
Prove true for $n=k+2$
ie $(k+2)^3-... | We can see that when $n = 1 -----> n (n^2 − 1) = 1 (1 − 1) = 1 (0) = 0$
$n = 3 -----> n (n^2 − 1) = 3 (9 − 1) = 3 (8) = 24$
We can see that this is true for $n = 1$ and $3$.
Assume statement is true for $n = 2k−1$. Then we must show it is true for $n = 2k+1$.
Then for $n = 2k-1 -----> (2k−1) ((2k−1)^2 − 1) = 24m$, ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$ I found this inequality using unusual calculations in maths Olympics
and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.
Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Pro... | By applying the generalized mean inequality twice,
$$\left(\frac{\sqrt{x + y - x^2} + \sqrt{y + z - y^2} + \sqrt{z + w - z^2} + \sqrt{w + x - w^2}}{4}\right)^2\\
\le \frac{1}{4}(2x + 2y + 2z + 2w - x^2 - y^2 - z^2 - w^2)\\
= \frac{1}{2} - \frac{x^2 + y^2 + z^2 + w^2}{4}\\
\le \frac{1}{2} - \left(\frac{x + y + z + w}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring two complicated polynomials I have two expressions which I've derived by taking the derivative. Both of these expressions are from separate problems, and in both of these instances I'm having difficulty understanding the algebra that brings the expression to a factorized form. They are below:
Problem #1
Given... | Problem 1
$f(x)=5(2x+1)(2x-3)^4(x^2+x+1)^4+8(2x-3)^3(x^2+x+1)^5$
As in above equation we have $(2x-3)^3$ and $(x^2+x+1)^4$ common in both terms so take them out. And solve the remaining.
$f(x)=(2x-3)^3(x^2+x+1)^4\left[(5(2x+1)(2x-3)+8(x^2+x+1)\right]$
Problem 2
$h(t) = (t+1)^{2/3}(3)(2t-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to express a power series in closed form I am trying to express the power series $x + x^4/4! + x^7/7! + \cdots$ in closed form;
I have already worked out the power series $1 + x^3/3! + x^6/6! + \cdots$ to be $(e^x + e^{x(2\pi i/3)} + e^{x(2\pi i/3)^2})/3$ using the roots of unity and the series for $e^x$ to get the... | Apologies to the two earlier answers, I made a typo - as they noticed - in my original question; the question now reflects what the closed form for $1 + x^3/3! + x^6/6! + \cdots$ should be. From that we can either integrate or differentiate twice to get my desired closed form for $x + x^4/4! + x^7/7! + \cdots$ as follo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m?
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any he... | Indeed, we have
$$\lim_{x\to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-[-lx+10-m]\right)=0$$.
$$-l=\lim_{x\to\infty }\frac{\frac{x^3+4x^2+3x+5}{x^2+2x+3}}{x}=1$$
thus
$l=-1$
and
$$10-m=\lim_{x\to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-x\right)=2$$
therefore $m=8$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
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:
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\... | Too long for a comment.
$$F(a,b)=\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=\frac 1b \,\, _3F_2\left(1,1,\frac{b}{a};\frac{1}{2},\frac{b}{a}+1;\frac{1}{2}\right)$$ where appears the generalized hypergeometric function.
$$F(2,1)=\frac \pi 2$$ $$F(2,3)=\frac{3 \pi }{2}-4$$ $$F(1,1)=\pi -\frac{\pi ^2}{8}$$
| {
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"url": "https://math.stackexchange.com/questions/2072228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How does this separation of the variable from the fraction result in $3\cdot \frac{1}{x}$ & not $\frac{3}{x}$ I wish to write the following expression as the product of a whole number or a fraction and variable expression. The answer given in my textbook is as follows:
$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3... | The key is simply to notice that
$$\frac31=3$$
Thus,
$$\frac31\cdot\frac1x=3\cdot\frac1x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$ The result below has been disproven.
Let $n \in \mathbb{N}$. Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$.
We know that $p \neq 2,5$. We need to have \begin{align*}5^n &\equiv 2 \pmod{p}\\2^n &\equiv 5 \pmod{p}.\end{align*} Thi... | Proof that $1409 \mid \gcd(5^{65}-2,2^{65}-5$):
We have \begin{align*}5^5 &\equiv 307 \pmod{1409}\\5^{10} &\equiv 1255 \pmod{1409}\\5^{20} &\equiv 1172 \pmod{1409}\\5^{40} &\equiv 1218 \pmod{1409}\\5^{65} &\equiv 2 \pmod{1409}\end{align*} and \begin{align*}2^{12} &\equiv 1278 \pmod{1409}\\2^{24} &\equiv 253 \pmod{1409}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
If $x = y$, $p$ = what? It is given that,
$$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$$
If $x=y$, $p$ =?
| $
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$
Apply componendo and dividendo,
$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}+\sqrt{2x+3y}-\sqrt{2x-3y}}{\sqrt{2x+3y}+\sqrt{2x-3y}- \sqrt{2x+3y}+\sqrt{2x-3y}}=\frac{p+1}{p-1}
$
$
\frac{\sqrt{2x+3y}}{\sqrt{2x-3y}} = \frac{p+1}{p-1}
$
Put x=y
$
\frac{\sqrt{5y}}{\sqrt{... | {
"language": "en",
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"answer_count": 3,
"answer_id": 1
} |
If $x = 2\log_39 + \log_{27}5,$ then $3^x = ??$ If
$$x = 2\log_39 + \log_{27}5,$$
then $3^x = ??$
I know $2\log_39$ is $4$, however how do I find an exact answer to the whole equation??
I tried to change the base of the second one to $ 3 $, however it did not work out.
| Use the fact that $\log_{27}(5) = \frac{\log_3(5)}{\log_3(27)} = \frac{\log_3(5)}{3} = \log_3(5^{\frac{1}{3}})$.
Then $x = 2\log_3(9) + \log_{27}(5) = \log_3(81) + \log_3(5^\frac{1}{3}) = \log_3(81 \cdot 5^{\frac{1}{3}})$.
You obtain that $3^x = 81 \cdot 5^\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$
\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A ... | $$ \begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}=0 $$
Subtract the first row to the second row, subtract the third row from the fourth row:
$$ \begin{vmatrix}
x & a & b &c \\
a-x & x-a & 0 &0 \\
m &n & x &p \\
0& 0& p-x& x-p
\end{vmatrix}=0 $$
Factorize $(a-x)$ an... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle
$ABC$ is equilateral
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{... | $x+y+z\geq 3(xyz)^{\frac{1}{3}}$,when $x=y=z$ ,equal sign is established.So:$$2\sum\frac{c}{a}\geq\sum\frac{c}{a}+3$$ i.e.$$\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\leq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$$Without loss of generality, we assume:$c\geq b\geq a$,so $\frac1{a}\geq\frac1{b}\geq\frac1{c},$then we have:$$\frac{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": 2
} |
Maximum value of $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$ If $a,b,x$ are real and $$f(x)=2(a-x)(x+\sqrt{x^2+b^2}),$$ then find the maximum value of $f(x)$.
Is there any method to solve this question without differentiation because using differentiation I am getting an ugly expression.
| This is still differentiation, but less ugly:
$$f(x)=2(a-x)(x+\sqrt{x^2+b^2})\implies\log f = \log 2+\log(a-x)+\log(x+\sqrt{x^2+b^2})$$
$$\implies\frac{f'(x)}{f(x)}=\frac{-1}{a-x}+\frac{1}{\sqrt{x^2+b^2}}$$
$$f'(x_*)=0\implies a-x_*=\sqrt{x_*^2+b^2}\implies(a-x_*)^2=x_*^2+b^2$$
$$\implies x_*=\frac{a^2-b^2}{2a}$$
So, $... | {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 0
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Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots ... | The reason why we can dispose of the minus sign is that, firstly, $$(x+\sqrt{x^2-1})^{-1}=(x-\sqrt{x^2-1})$$
Therefore $$\ln(x\pm \sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})^{\pm1}=\pm\ln(x+\sqrt{x^2-1})$$
But $\operatorname{arcosh}(x)$ is defined as non-negative, so the minus sign is not required
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that:$\int_{0}^{1}{x+x^2+\cdots+x^{2n}-2nx\over (1+x)\ln{x}}dx=\ln{\left[\left({2\over \pi}\right)^n(2n)!!\right]}$ Prove that
$$\int_{0}^{1}{x+x^2+x^3+\cdots+x^{2n}-2nx\over (1+x)\ln{x}}dx=\ln{\left[\left({2\over \pi}\right)^n(2n)!!\right]}$$
$n\ge1$
My try:
$${x+x^2+\cdots+x^{2n}\over 1+x}={1\over 1+x}\cdot{x(1... | In THIS ANSWER, I showed that
$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \left(\frac{x^{2n}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)}$$
We can use this result to evaluate the integral
$$\begin{align}
\int_0^1 \left(\frac{x^{2n+1}-x}{1+x}\right)\,\frac{1}{\log(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079609",
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"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
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... | It simply means
\begin{align}
&\ln(-x+\sqrt{x^2+4})-\ln2=-\ln(x+\sqrt{x^2+4})+\ln2\\
\iff &\ln(x+\sqrt{x^2+4})+\ln(-x+\sqrt{x^2+4})=2\ln 2\\
\iff&\ln[(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)]=\ln4\\
\iff &(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)=4.
\end{align}
| {
"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": 4
} |
$x-\frac1x+y-\frac1y=4\,$ 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... | Here is a sketch of a naive proof that it has no solutions (no elliptic curves).
We can assume x is positive, otherwise we would change x to -1/x. The same with y.
$x^2y+x(y^2-4y-1) -y =0$
$D=(y^2-4y-1)^2+4y^2$ should be a square of a rational number.
If $y=\frac{m}{n}$ (assuming m and n are co-prime), then multiplying... | {
"language": "en",
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"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Integral $\int \frac{dx}{\sin^2 (x) + \sin(2x)}$ We have to do the following integration
$$\int \frac{dx}{\sin^2 (x) + \sin(2x)}$$
I tried as
In the numerator, $1 = \sin^2(x)+\sin(2x)+\cos^2(x)-\sin(2x)$
But now how to proceed?
| First use the fact that $\sin(2x) = 2\sin(x)\cos(x)$ to obtain
$$\int \frac{1}{\sin^2(x) + 2\cos(x)\sin(x)}dx$$
Divide numerator and denominator by $\sin^2(x)$:
$$\int \frac{\frac{1}{\sin^2(x)}}{1 + 2\frac{\cos(x)}{\sin(x)}} = \int \frac{\csc^2(x)}{1 + 2\cot(x)}dx$$
Then you can use the substitution $u = 2\cot(x)+1$; $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of the reciprocals
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\df... | There are $3$ single digit inverses, for a total of $\frac{11}{6}$.
There are $3\cdot4=12$ two digits inverses, which don't exceed a total of $\frac{11}{6}\cdot\frac4{10}$, (as $\frac1{10}+\frac1{11}+\frac1{12}+\frac1{13}\le\frac4{10}, \frac1{20}+\frac1{21}+\frac1{22}+\frac1{23}\le\frac4{20}, \cdots$).
There are $3\cdo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How to prove this sum $\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$ Let $n\ge 2$ postive integer, show that
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$$
I have done this works
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k}\bin... | Note the Catalan numbers are defined as
\begin{align*}
C_{n}&=\frac{1}{n+1}\binom{2n}{n}\qquad\quad n\geq 0\\
\end{align*}
with generating function
\begin{align*}
C(z)=\sum_{n=0}^\infty C_nz^n=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)
\end{align*}
The LHS of OPs binomial identity is a Cauchy-product of Catalan numbers
\b... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Divisibility of $(4^{2^{2n+1}}-3)$ by 13. The question is to simply proof that $13\space|(\space4^{2^{2n+1}}-3)$ for all $n\in\mathbb{N}$. Since the chapter in my book is about the Euler-phi function and Euler's theorem, my guess would be to look at $4^{2^{2n+1}}\equiv3\space \pmod{13}$ and do something with Euler's th... | As you correctly note, to compute
$$
4^{2^{2n+1}} \pmod{13}
$$
you can compute
$$
2^{2n+1} \pmod{12}.
$$
It is advisable to use CRT, and compute separately modulo $3$ and modulo $4$.
Modulo $3$ you have
$$
2^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \pmod{3},
$$
whereas modulo $4$ you have
$$
2^{2n+1} \equiv 0 \pmod{4},
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\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\... | Inspired by Dr. Sonnhard's answer.
$$cos A + cos B + 2cos C = 2$$
=> $$cos A + cos B = 2 (1 - cos C)$$
=> $$(b²+c²-a²)/2bc + (c²+a²-b²)/2ca = 2[1 -
(a²+b²-c²)/2ab]$$
=> $$a(b²+c²-a²) + b(c²+a²-b²) = 2c[2ab - (a²
+b²-c²)]$$
=>$$ ab² + ac² -a³ + bc² + ba² - b³ = 2c[2ab -
(a²+b²-c²)]$$
=> $$ab² + ba² + ac² + bc² - a³ - b³... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-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}$$... | Let $\frac{p}{C}=\frac{\cos\alpha}{A}=\frac{\sin\alpha}{B} = k$
Then we have,
$\cos\alpha = kA$
On squaring we have,
$\cos^2\alpha = k^2A^2$
Also $\sin\alpha = kB$
On squaring $\sin^2\alpha = k^2B^2$
Adding these two squared equations,
$\sin^2\alpha + \cos^2\alpha = k^2A^2 + k^2B^2$
$\sin^2\alpha + \cos^2\alpha= k^2(A^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Alternative approaches to showing that $\gamma=\int_0^\infty \left(\frac{1}{1+x^a}-\frac{1}{e^x}\right)\,\frac1x\,dx$, $a>0$ Starting from the limit definition of the Euler-Mascheroni constant $\gamma$ as given by
$$\gamma=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)\tag 1$$
we can show that $\gamma$ h... | The same idea as in my previous answer applies. Notice that
$$ f(x) = \frac{1}{1+x} \quad \Rightarrow \quad \int_{\epsilon}^{\infty} \frac{f(x)}{x} \, dx = -\log\epsilon + \log(1+\epsilon) = -\log\epsilon + o(1) $$
as $\epsilon \to 0^+$. Now from the linked answer above, we find that
\begin{align*}
f(x) = e^{-x}
&\quad... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
solving Integration of trignometry In the following integral
$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$
My try: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.
| A big partial fraction decomposition!!:
We use Weierstrass substitution and use $u =\tan \frac {x}{2} $ to get $$I =\int \frac {1}{\sec x+\mathrm {cosec}x } dx =\int \frac {\sin x\cos x}{\sin x+\cos x} dx= 4\int \frac {u (u^2-1)}{(u^2+1)^2 (u^2-2u-1)} du =4I_1 $$ Now we have $$I_1 =\int \frac {u (u^2-1)}{(u^2+1)^2 (u-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $\frac{a_1}{a_2}+ \frac{a_2}{a_3}+\frac{a_3}{a_4}+...+\frac{a_{n-1}}{a_n} \le \frac{n}{2}$ Let $n \ge 2$ be a positive integer and let $a_1, a_2, ... a_n$ be positive numbers such that $$ a_1\le a_2, a_1+a_2\le a_3, a_1+a_2+a_3\le a_4, ... ,a_1+a_2+...+a_{n-1}\le a_n$$
prove that
$$\dfrac{a_1}{a_2}+ \dfrac... | The idea is "local adjustments". If $k$ is the smallest integer such that $a_1+\cdots+a_{k-1} < a_k$, we adjust up $a_1$, $\cdots$, $a_{k-1}$ with a scale factor of $s=\frac{\sum_{i=1}^{k}a_i}{2\sum_{i=1}^{k-1}a_i}$ for all, and adjust down $a_k$ by a scale factor of $t=\frac{\sum_{i=1}^{k}a_i}{2a_k}$, and achieve a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Absolute maximum of $f(x) = \sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$ My maths tutor solved this by using $f(x) = \sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$ and treating $A(2,3)$ and $B(1,0)$ as fixed points and $P(x^2,x)$ as a moving point and using the difference between the sides PA and PB of the triangle PAB... | No, it's not so hard.
We need to prove that
$$\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}\leq\sqrt{10}$$ or
$$\sqrt{x^4-3x^2-6x+13}\leq\sqrt{10}+\sqrt{x^4-x^2+1}$$ or
$$1-3x-x^2\leq\sqrt{10(x^4-x^2+1)},$$
which is obvious for $x^2+3x-1\geq0$, but for $x^2+3x-1\leq0$ we need to prove that
$$(x^2+3x-1)^2\leq10(x^4-x^2+1)$$ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How do we show that $\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2$ I was going through the the top votes questions and I saw this quite of interesting post posted by @Sangchul Lee. I was just messing around with it and found a slightest Variation of Ahmed's inte... | Substitute : $u=\arctan\left(x\sqrt{x^2+2}\right)$
You will get : $$\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{\sqrt{x^2+2}+\frac{x^2}{\sqrt{x^2+2}}}{x^2\left(x^2+2\right)+1}$$
Thus :
$$\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\int_{0}^{\arctan(\sqrt{3})}{u\over 2}\cdot{du}=\left[\frac{u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Solve $\cos (2x+3y)=\frac{1}{2}$ and $\cos (3x+2y)=\frac{\sqrt{3}}{2}$ We have to solve for $x$ and $y$:
$$\left \{
\begin{align*}
\cos (2x+3y) &= \frac{1}{2} \\
\cos (3x+2y) &= \frac{\sqrt{3}}{2}
\end{align*}
\right.$$
I got
$$\left \{
\begin{align*}
2x+3y &= 2n\pi \pm\frac{\pi}{3} \\
3x+2y &= 2m\pi \... | $1)$ You have:
$$2x+3y = 2n\pi \pm\frac{\pi}{3}$$
Multiply by $-2$ and get:
$$-4x-6y = -4n\pi \mp\frac{2\pi}{3} \quad (I)$$
$2)$ You have:
$$ 3x+2y= 2m\pi \pm\frac{\pi}{6}$$
Multiply by $3$ and get:
$$9x+6y= 6m\pi \pm \frac{\pi}{2}\quad (II)$$
$(I)+(II)$:
$$5x=(6m-4n)\pi \mp\frac{2\pi}{3}\pm \frac{\pi}{2}$$
PS: $\mp\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \fra... | Write the expression as
$$\frac{\sum_{k=1}^{n}2^k/k}{2^n/n}.$$
Note the denominator $\to \infty.$ That rings the Stolz-Cesaro bell, so consider
$$\frac{2^{n+1}/(n+1)}{2^{n+1}/(n+1) -2^n/n} = \frac{1}{1 -(n+1)/(2n)} \to \frac{1}{1/2} = 2.$$
By Stolz-Cesaro, the desired limit is $2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\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})}$$
But, I am not being able to use it. Please help me.
| Hint
$$\tan{(15^{\circ})}=2-\sqrt{3}$$
and let$x=\tan{7.5^{\circ}}$.we have
$$\dfrac{2x}{1-x^2}=2-\sqrt{3}\Longrightarrow (2-\sqrt{3})x^2+2x-(2-\sqrt{3})=0$$
so we have
$$x^2+2(2+\sqrt{3})x-1=0$$
then we have
$$(x+2+\sqrt{3})^2=1+(2+\sqrt{3})^2=8+4\sqrt{3}=2(\sqrt{3}+1)^2$$
then we have
$$x=\sqrt{2}(\sqrt{3}+1)-2-\sqrt... | {
"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": 3
} |
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 ?
| Hint
For $x \ge 0$ and $y \ge 0$ use
AM-GM
$$x^2+(y^2+1) \ge 2\sqrt{x^2(y^2+1)}=2x\sqrt{y^2+1}$$
$$(x^2+1)+y^2 \ge 2\sqrt{y^2(x^2+1)}=2y\sqrt{x^2+1}$$
Sum both and prove that the equality is not possible.
Also think about $x<0$ or $y<0$. That should be easy.
Can you finish?
| {
"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": 1
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Find all $n \in \mathbb{Z}$ such that $\sqrt{\frac{4n-2}{n+5}}$ is a rational number How does one approach this problem, where
all $\sqrt{\frac{4n-2}{n+5}}$ is a rational number when $n \in \mathbb{Z}$.
| Let $p,q$ be relatively prime positive integers such that $$\frac{4n-2}{n+5} = \frac{p^2}{q^2}$$
(note that $p\neq 0$ because $4n-2=0$ doesn't have integer solution.)
Then, we have $4n-2 = q^2k$, $n+5=p^2k$, $k\in\Bbb Z$. Now, $$4n-2 = 4(n+5) - 22\implies (4p^2-q^2)k = 22 \implies (2p-q)(2p+q)k = 22.$$
Notice that $2p-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Very interesting integral limit I found this interesting problem on AoPS forum but no one has posted an answer. I have no idea how to solve it.
$$
\int_0^\infty \sin(x^n)\,dx
$$
For all positive rationals $n>1$, $I_n$ denotes the integral as above.
If $P_n$ denotes the product
$$
P_n=\prod_{r=1}^{n-1}I_{\bigl(\!\f... | since
$$T_{\frac{r}{n}}=\int_{0}^{+\infty}\sin{(x^{\frac{r}{n}})}dx=\Gamma\left(\dfrac{r}{n}+1\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}},r=1,2,,\cdots,n-1$$
proof see:3.1
Lemma $1$:
$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}=\dfrac{\sqrt{n}}{2^{n-1}}$$
Use this well... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Positive rational numbers with sum of their squares equal to $n$
Prove that for all positive integers $n$ there exist $n$ distinct, positive rational numbers with sum of their squares equal to $n$.
For $n = 1$ we can just take $1$. For $n = 2$ we can take $\left(\dfrac{1}{5}\right)^2+\left(\dfrac{7}{5}\right)^2 = 2$.... | You can get $x^2+y^2=1$ infinitely many ways using Pythagorean triples, just make sure all the triples are distinct, then add the expressions. If $n$ is odd, use an initial $1$ not from a triple.
Edit 2: (previous idea didn't work) Suppose that
$$x=\frac{k^2-2k-1}{k^2+1},\ y=\frac{k^2+2k-1}{k^2+1}. \tag{1}$$
Then $x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How can the limit of this $F(x,y)$ is $0$ when it is taken along $y=x^2$ or $ x=y^2$?
Find $$\lim_{(x,y) \to (0,0)}\frac{3x^{2}y}{x^{2}+y^{2}}$$ if it
exists.
From my textbook, it said that the limits along the parabolas $$y=x^{2}\text{ and } x=y^{2}$$ also turn out to be 0.
I couldn't figure out why, as if we set... | On $y=x^2$ we have
$$
\frac{3x^2y}{x^2+y^2}=\frac{3x^4}{x^2+x^4}=
\frac{3x^4}{x^2(1+x^2)}=\frac{3x^2}{1+x^2}
$$
which has limit $0$ when $x\to0$.
On $x=y^2$ we have
$$
\frac{3x^2y}{x^2+y^2}=\frac{3y^5}{y^4+y^2}=\frac{3y^5}{y^2(y^2+1)}
=\frac{3y^3}{y^2+1}
$$
which again has limit $0$ for $y\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I ha... | recall: $\displaystyle H_a = \int_0^1 \frac{1-x^a}{1-x}\,\mathrm{d}x$, and integration by parts once we have
$$\int_0^1 x^{a-1} \ln (1-x)\,\mathrm{d}x = -\frac{H_a}{a}$$
Thus,
\begin{align*}&\sum\limits_{n=1}^{\infty} \frac{H_{n+a}}{n(n+a)} = -\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^{1} x^{n+a-1} \ln (1-x)\,\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two sm... | Let
$$ I=\int^\infty_0 \frac{dx}{(a^2+b^2x^2)(b^2+a^2x^2)}.\tag{1}$$
Changing variable from $x\to\frac1{x}$ gives
$$ I=\int^\infty_0 \frac{x^2dx}{(a^2+b^2x^2)(b^2+a^2x^2)}.\tag{2}$$
So $a^2$(1)+$b^2$(2) gives
$$ a^2I+b^2I=\int^\infty_0 \frac{dx}{b^2+a^2x^2}=\frac{\pi}{2ab}$$
and hence
$$ I=\frac{\pi}{2ab(a^2+b^2)}.$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluation of $\int\frac{dx}{x+ \sqrt{x^2-x+1}}$ Evaluate :
$$\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?
| There is a substitution, called the Euler substitution, which solves your integral rather neatly. In your case we set $t = \sqrt{x^2 - x + 1} + x$; after subtracting $x$ from both sides and squaring, we get $x^2 -x + 1 = t^2 - 2xt + x^2$, which can be solved (after subtracting $x^2$ from both sides) to get $x = \frac{t... | {
"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": 1
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Partitions of $n$ with exactly 3 parts I'm learning about generating functions, and came across this question: find the generating function for the number of partitions of a number $n$, into exactly 3 parts. I just solved a problem where the condition was that each part is no greater than 3, but I'm kind of stuck on th... | Hint: The following can be found somewhat more detailed in this answer.
A generating function for the number of partitions with exactly three parts is
\begin{align*}
&\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\
&\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\
&\qquad=\frac{x^3}{(1-x)(1-x^2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Showing a vector belongs in the span Without using dimension, I have to show
From: question
$\overrightarrow{x} \in \mathbb{R}^3$ is also in span $\{v_1, v_2, n\}$ where $n = v_1 \times v_2$. Where $v_1, v_ 2\in \mathbb{R}^3$ as well.
We can let $x = \langle x_1, x_2, x_3 \rangle$, but that isn't very helpful.
I know t... | I'll use the notation from the previous answer, since I don't like subscripts. We are given two vectors $\mathbf{u}$ and $\mathbf{v}$, and we want to show that any given vector $\mathbf{x}$ is in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$, where $\mathbf{n} = \mathbf{u} \times \mathbf{v}$.
First let's focus o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How do we evaluate $\int\frac{dx}{\sin^{4}{x}+1}$? How do we evaluate $$\displaystyle\int{\dfrac{dx}{\sin^{4}{x}+1}}?$$
don't know how to solve this, please help.
| By substituting $x=\arctan t$ the problem boils down to
$$ \int \frac{1+t^2}{(1+t^2)^2+t^4}\,dt = \int\left(\frac{1+i}{4}\cdot\frac{1}{t^2+\frac{1+i}{2}}+\frac{1-i}{4}\cdot\frac{1}{t^2+\frac{1-i}{2}}\right)\,dt $$
leading to:
$$ \frac{1}{2\sqrt{2}}\left(\sqrt{1-i}\,\arctan\frac{t}{\sqrt{\frac{1-i}{2}}}+\sqrt{1+i}\,\arc... | {
"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": 1
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Finding Limit with Taylor series Find the limit using Taylor-series , $f(x) =\frac{\ln(1+x^2)-x^2}{\sqrt{1+x^4}-1}$.
I calculated the limit of of the $\ln(1+x^2)$ which is equal to
${x^2} - \frac{2x^3}{3} + ...O(x^n)$
and $\ln(1+x^2) -x^2 $ = $\frac{-2x^3}{3} + ...$
but when i calculate $\sqrt{1+x^4}$ i get $0$ for ... | Hint
Use the generalized binomial theorem to show that $$\sqrt{1+x^4}=(1+x^4)^{\frac 12}=1+\frac{x^4}{2}-\frac{x^8}{8}+\cdots$$
Another way could be to consider Taylor series $$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+O\left(t^3\right)$$ and replace $t$ by $x^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$... | $\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$
$\sin\frac{y}{2} = 0$
So far so good
contining... suppose $\sin\frac{y}{2} \ne 0$
$(1-\frac{8}{3}\cos\frac{y}{2})=0\\
\frac{8}{3}\cos\frac{y}{2})=1\\
\cos\frac{y}{2}=\frac 38\\
\frac y2=\arccos\frac 38\\
x = \sin(\arccos\frac 38)\\
x = \sqrt {1 - \frac {9}{64}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"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:
$$... | Proving for sum =k+1
RhS= $\dfrac {(k)(k+1)}{2} + k + 1 $
Which gives
=$\dfrac {(k+1)(k+2)}{2}$
= $\dfrac {([k+1])([k+1] +1)}{2}$
Which means it hold for n and also n+1 !
| {
"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": 1
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Find all $n\in \mathbb{Z}^+ : 7|5^{6n}+5^n+2$ Question: We want to find all $n\in \mathbb{Z}^+$ such that $7|5^{6n}+5^n+2$.
Answer: We have that $7|5^{6n}+5^n+2 \iff 5^{6n}+5^n+2 \equiv 0 \bmod 7$.
But $\phi(7)=6\wedge\text{gcd}(5,7)=1$ and we can apply Fermat's theorem, so $5^6 \equiv1 \bmod 7\implies 5^{6n}\equiv1 \b... | From $5^n \equiv 4 (\bmod{7})$ write $(-2)^n \equiv 4 (\bmod{7})$ so that
$(-2)^{n-2} \equiv 1 (\bmod{7}).$ The order of $-2 (\bmod{7})$ is $6$, so $6\mid n-2$. Therefore $n =6k+2$ for any integer $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Equation involving fields Let $ K $ be a field and $f:K-\left \{ 0 \right \}\rightarrow K-\left \{ 0 \right \}$ a function with $f(f(x))=x^{-1},\forall x\in K-\left \{ 0 \right \}$ and $ f(1)\neq 1.$
Knowing that $ f^{2}(x)-f(x)+1=0 $ has an unique solution in $K-\left \{ 0 \right \} $, determine $ f(2)$.
I haven't fou... | Suppose that $a$ is a root of the polynomial $x^2-x+1$.
Then $(x-a)$ divides the polynomial $x^2-x+1$.
This means that $(x-a)(x-a)=x^2-x+1$ (because the root is unique).
it follows that $x^2-2a+1=x^2-x+1$ and so $2a=1$ and $a^2=1$.
we conclude that $1=1^2=(2a)^2=4a^2=4$.
So the field must have characteristic $3$, and s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Problem of Complex Numbers in Geometry using Roots of Unity Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
$x^2 + px + q = 0$
for some integers $... | $$w^7-1=(w-1)(w^6+w^5+w^4+\dots+w+1) $$
So $$w^6+w^5+w^4+\dots+w+1 =0 \tag{1}$$
Now use Vieta's Formula.
First note that the sum of roots is $-p$, so we have that
$$-p=\alpha+\beta=w^6+w^5+w^4+w^3+w^2+w=-1$$
From $\text{(1)}$ Also,
$$q=(\omega + \omega^2 + \omega^4) \times (\omega^3 + \omega^5 + \omega^6)$$
Expanding... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3$
Prove that $a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3$
SOURCE : "A Brief Introduction to Olympiad Inequalities" by Evan Chen
It was one of the practice problems. Equality case is easy. I tried AM-GM and Muirhead, but could not seem to find a suitable proof for... | Using $\text{AM-GM}$ this can be solved. Note that by $\text{AM-GM}$ $$4a^7+3b^7=a^7+a^7+a^7+a^7+b^7+b^7+b^7 \ge 7a^4b^3$$$$4b^7+3c^7=b^7+b^7+b^7+b^7+c^7+c^7+c^7 \ge 7b^4c^3$$$$4c^7+3a^7=c^7+c^7+c^7+c^7+a^7+a^7+a^7 \ge 7c^4a^3$$
Adding the three and dividing by seven, we have $$a^7+b^7+c^7 \ge a^4b^3+b^4c^3+c^4a^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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... | The problem is straightforward, it's just a form which is more trouble to rationalize that it's worth. And there are several equivalent expressions depending upon which identities are used.
$$ \tan\left(-\frac{5\pi}{16}\right)=-\frac{\sin\left(\frac{5\pi}{8}\right)}{1+\cos\left(\frac{5\pi}{8}\right)}$$
and
$$ \sin\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that
$$
(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... | Let
$$x = \sqrt{a+b+c} = \sqrt[5]{\overline{abc}},\quad a\in \{1, 2\dots9\}, \quad b,c\in\{0,1\dots9\},$$
then
$$\begin{cases}
\sqrt{1} <= x < \sqrt{27}\\
\sqrt[5]{100} < x < \sqrt[5]{999}\\
x\in\mathbb N,
\end{cases}$$
$$\begin{cases}2.5 < x < 3.99\\
x\in\mathbb N,
\end{cases}$$
$$x=3,$$
$$\begin{cases}
\overline{abc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to calculate a number of $x$ values such that $f(x)$ is a quadratic residue modulo $m$? Let $a, b$ be any fixed positive integers, $f(x) = (3x + 3a)^2 - b$
How to calculate a number of $x$ values such that $f(x)$ is a quadratic residue modulo $m$ without actually finding them?
Edit: I updated the question to make i... | Step 1: You can ignore the $-b$ since for each residue of $(3x+3a)^2-b$ there is a corresponding residue of $(3x+3a)^2$ whichi s just $b$ more than the former.
Step 2: Similarly you can ignore the $a$ since for each value of $(3x+3a)$ there is a value $(3y)$ with $y=x+a \pmod m$.
So we want to find the number of quadra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all possible values of $x$ if $\ln(x) = \sin(x)$. My Mathematics Teacher gave me the following problem :
Find all the possible values of $x$ if $\ln(x) = \sin(x)$.
I tried graphing both $\ln(x)$ and $\sin(x)$. I found that they intersect at a single point $\approx (2.22, 0.8)$. However, I am unable to find a "ma... | As said in answers and comments, the solution requires numerical methods.
Since you graphed the functions, you noticed that they intersect close to $x=2$. Then, in order to approximate the solution, you could have used Taylor expansions (as you though about) but building them around $x=2$.
You have $$\log(x)=\log (2)+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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 easiest way to find a minimal polynomial is to put the matrix in Jordan normal form. The minimal polynomial is $p(x) = \prod (x-\lambda)^{d_\lambda}$, where the product runs over the eigenvalues $\lambda$ and $d_\lambda$ is the size of the largest Jordan block for that eigenvalue (not the multiplicity of the eigenv... | {
"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": 1
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Simple yet confusing probability problem. A bag contains x green candies and y red candies. A candy is selected at random from the bag and it's color is noted. It is then replaced into the bag with an additional ten candies of the same color. A second candy is then randomly chosen. Find the probability that the second ... | We have x green candies and y red.
Case 1-
Green candy is picked in first attempt.
$\frac x{x+y}$
Now green candies x + 10.
Red candy is picked in second attempt.
$\frac {y}{x+y+10}$
Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} $
Case 2-
Red candy is picked in first attempt.
Probability = $\frac y{x+y}$
Now re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to prove $\ln(1+x)\ge x/(1+x/2)$ when $x > 0$? I think the equation establishes only when $x = 0$, so the condition should be x>=0. The first derivative of $\ln(1+x)-x/(1+x/2)$ is $0$ at $x=0$. So the inequality looks good to establish. Is there a better way to prove it?
| To show
$\ln(1+x)
\ge x/(1+x/2)
$.
Since
$\ln(1+x)
=\int_0^x \dfrac{dt}{1+t}
$,
this is equivalent to
$\dfrac1{x}\int_0^x \dfrac{dt}{1+t}
\ge \dfrac1{1+x/2}
$.
Since
$\dfrac1{1+x/2}$
is the value of
$\dfrac1{1+t}
$
at the midpoint,
I'll combine the two halves.
$\begin{array}\\
\int_0^x \frac{dt}{1+t}
&=\int_0^{x/2} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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How many numbers smaller than $10^6$ contain exactly three '$9$'s and have an odd sum of digits? i came up with an idea that i choose
Even Even Even 9 9 9 - and i sort it in ways such that
all even are the same - $\frac{6!}{3!*3!}$
two are the same - $\frac{6!}{3!*2!}$
all are different - $\frac{6!}{3!}$
other one is... | With your idea:
*
*Three even digits and the 9:
*
*The same even digit every time (5 choices: $0, 2, 4, 6, 8$ and permutation): $5 \cdot \frac{6!}{3! \cdot 3!}$
*Two even digits are the same: $5 \cdot 4 \cdot \frac{6!}{3! \cdot 2!}$
*The three even digits are different: $5 \cdot 4 \cdot 3 \cdot \frac{6!}{3!}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
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... | Hint: You need to be doing top curve minus bottom curve i.e;
$$(4x-x^2-1) - (x+1) = 3x-x^2-2$$
| {
"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": 2
} |
Mathematical Induction for an alternating series It's been a while since I have done a problem like this. I have this problem
$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$
So I have gotten this far:
Base Case:
$$n=1$$
$$(-1)^2+1^2 = \frac{ (-1)^2 1(2) }{2}$$
$$ 1 = 1$$
Assume:
$$\sum_{i=0}^n (-1)^{i+1}... | If $n$ is even,
$$\begin{align}
\sum_{i=0}^n (-1)^{i+1}i^2&=-0^2+1^2-2^2+3^2-4^2+\cdots+(n-1)^2-n^2\\
&=(1-2)(1+2)+(3-4)(3+4)\cdots+((n-1)-n)((n-1)+n)\\\\
&=-(1+2+3+4+\cdots+(n-1)+n)\\
&=-\frac{n(n+1)}2\end{align}$$
If $n$ is odd,
$$\begin{align}
\sum_{i=0}^n (-1)^{i+1}i^2&=-0^2+1^2-2^2+3^2-4^2-\cdots-(n-1)^2+n^2\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How do they got the solution for this $2\times 2$ matrix?
For $y'=\frac{1}{x}\begin{pmatrix}0 &1 \\ 2 & -1\end{pmatrix}y$ the solutions are given by $y_1=\begin{pmatrix}x \\ x\end{pmatrix},y_2=\begin{pmatrix}x^{-2} \\ -2x^{-2}\end{pmatrix}$ with $x\in \Bbb R,y\in \Bbb R^2$
I would like to know how they got this solut... | Let $x = e^u$. Then we have
$$ \def\d#1#2{\frac{d#1}{d#2}}
\d yu = \d yx \cdot \d xu = y' \cdot x $$
Hence, your equation reads
$$ \d yu = \begin{pmatrix} 0 & 1 \\ 2 & -1 \end{pmatrix}y $$
Now solve this linear system as usual and resubstitue $e^u$ by $x$, giving the stated solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is $x^2 \equiv -1 \pmod{365}$ solvable? I know that a similar question already exists, but I have a different question to ask.
We want to examine if $x^2 \equiv -1 \pmod{365}$ has a solution.
My thought is: $365=5\cdot 73$. So,The congruence $x^2 \equiv -1 \pmod{365}$ has solution, if and only if, the congrueces $x^2 \... | Yes. Let a solution to $x^2+1 \equiv 0 \pmod {5}$ be $r_{1}$, and let a solution to $x^2+1 \equiv 0 \pmod {73}$ be $r_{2}$
Then note that by CRT, we have that there exists such $x$ that $$x \equiv r_{1} \pmod {5}$$ $$x \equiv r_{2} \pmod {73}$$
Exists. Then note that for such $x$, $$x^2+1 \equiv 0 \pmod {5}$$ $$x^2+1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the limit of the expression
Find
$$ \lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{\log_2( 1+x^{-2} + x^{-3})}$$
Firstly I replace $x$ with $1/t$. Then $t$ tends to $0$. And then I fix numerator with formulas for limits considering $t \to 0$.
But I don't know what to do with denominator.... | Since it is the denominator term that is the source of concern, note that after the substitution $x=1/t$, we have
$$\begin{align}
\log_2(1+x^{-2}+x^{-3})&=\log_2(1+t^2+t^3)\\\\
&=\left(\frac{\log_2(1+t^2+t^3)}{t^2+t^3}\right)\,(t^2+t^3)\\\\
&=\left(\frac{\log(1+t^2+t^3)}{\log(2)\,(t^2+t^3)}\right)\,(t^2+t^3)
\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$ without induction
Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$
I know this can be easily proved by induction. But I am looking for another approach. How do I prove this withou... | If we consider
$$a_n = \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$
we have:
$$ a_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right) = \frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2} $$
hence:
$$ 4n a_n^2 \leq \exp\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to prove in this trig problem i have to prove this
$$\frac{\cos 3x}{\sin 2x \sin 4x}+\frac{\cos 5x}{\sin 4x \sin 6x}+\frac{\cos 7x}{\sin 6x \sin 8x}+\frac{\cos 9x}{\sin 8x \sin 10x} = \frac{1}{2}\csc x(\csc 2x - \csc 10x)$$
i tried taking lcm but does not leads to anything. i believe i have to write numerator as so... | The right side looks like a result of a telescopic sum, so we'll try to prove
$$ \frac{\cos(nx)}{\sin((n-1)x)\sin((n+1)x)}=\frac{1}{2}\csc(x)(\csc((n-1)x)-\csc((n+1)x)) $$
Why this? It may not be true, but if it is, we have
$$\begin{align*}
& \frac{\cos 3x}{\sin 2x\sin 4x} + \frac{\cos 5x}{\sin 4x\sin 6x} +\frac{\cos 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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$3$ numbers $x,y,z$ pairwise relatively prime, sum of any two is multiple of third, find $abc$
Three distinct positive integers $x,y,z$ are pairwise relatively prime, and the sum of any two is a multiple of the third one. Find $xyz$.
I don't know how to start.
| Suppose WLOG $x < y < z$. Then $z \mid x + y < 2 z$, and there are no other multiples of $z$ between $z$ and $2 z$. It follows that $x + y = z$.
Now $x \mid y + z = x + 2 y$, so $x \mid 2 y$, and since $\gcd(x, y) = 1$, we have $x \mid 2$, and thus $x = 1$ or $x = 2$.
Similarly $y \mid x + z = 2 x + y$, so as above $y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"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... | We use the so called uvw method for proving inequality involving symmetric polynomials of $3$ variables.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, our inequality it's $kw^6+A(u,v^2)w^3+B(u,v^2)\geq0$,
where $A$ and $B$ polynomials such that $A(0,0)=B(0,0)=0$.
Let $\theta^3=1$,... | {
"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": 1
} |
Prove that this is an abelian group
$G$ is a group. $f:\:G\to G,\:a\mapsto a^3$ is a group monomorphism. Prove that $G$ is abelian.
My idea is that it's sufficient to prove that $(aba^{-1}b^{-1})^3=e$. So
$$
(aba^{-1}b^{-1})^3=a^3b^3a^{-3}b^{-3}
$$
and I am stuck. Or
$$
(aba^{-1}b^{-1})^3=a^3(ba^{-1}b^{-1})^3=a^3ba^{... | Here's my version (revised) . . .
Since the map $\:x\mapsto x^3$ is a monomorphism,
\begin{align*}
&\hspace{1pt}(ab)^3 = a^3b^3,\text{ for all }a,b \in G.\tag{1}\\[6pt]
&\hspace{1pt}\text{If }a,b \in G,\text{ and }a^3 = b^3,\text{ then }a = b.\tag{2}\\
\end{align*}
Let $a,b \in G.$
\begin{align*}
\text{Then}&&a^3b^3a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving a summation result using strong induction I was recently taught strong induction in class and overall I'm still learning the topic. I am currently stuck on a problem and I don't even know where to begin with it. A starting place or hint would be great.
| Fix $k$ and apply induction on $n$. So, let us assume that $$\sum_{j=1}^\color{blue}nj(j+1)(j+2)\dots(j+k-1)=\frac{\color{blue}n(\color{blue}n+1)(\color{blue}n+2)\dots(\color{blue}n+k)}{k+1}.$$ Then
$$\begin{align}
&\sum_{j=1}^{\color{red}{n+1}}j(j+1)(j+2)\dots(j+k-1)\\
&=[(n+1)(n+2)(n+3)\dots(n+1+k-1)]+\sum_{j=1}^nj(j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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 $\alpha=\displaystyle\frac{5+\sqrt{3}}2, \beta=\displaystyle\frac{5-\sqrt{3}}2$.
$$\alpha+\beta=5; \alpha\beta=\frac 74$$
$\alpha,\beta$ are roots of the quadratic
$$x^2-(\alpha+\beta)+\alpha\beta=0\\
x^2-5x+\frac 74=0\\
4x^2-20x+7=0$$
i.e.
$$\overbrace{4\alpha^2-20\alpha+7}^{f(\alpha)}=0$$
Now consider the follo... | {
"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": 1
} |
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?
| $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}= 2 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6} \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}$
$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6= 2^6 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6} \\ \sin\fra... | {
"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": 1
} |
Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
... | Note that three sides $a,b,c$ in a triangle satisfy
$$ a<b+c, b<a+c, c<a+b. $$
So one has
$$ a^2<a(b+c), b^2<b(a+c), c^2<c(a+b). $$
Adding these three inequalities gives
$$ a^2+b^2+c^2<2ab+2bc+2ca. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
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 $$\t... | Lets start with:
$\tan \frac {a}{2^{r-1}} = \frac {2\tan \frac {a}{2^{r}}}{1-\tan^2 \frac {a}{2^{r}}}$
and that gets everything in terms of $\frac {a}{2^{r}}$
$u = \frac {a}{2^{r}}$
$\frac {\sin u}{\tan^2 u\frac {2 \tan u}{1-\tan^2 u}+\tan u}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(1 +\tan^2 u)}\\
\frac {\sin u (1-\tan^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How can I find all the matrices that commute with this matrix? I would like to find all the matrices that commute with the following matrix
$$A = \begin{pmatrix}2&0&0\\ \:0&2&0\\ \:0&0&3\end{pmatrix}$$
I set $AX = XA$, but still can't find the solutions from the equations.
| >>> from sympy import *
>>> A = diag(2,2,3)
>>> X = MatrixSymbol('X',3,3)
>>> Matrix(A*X - X*A)
Matrix([
[ 0, 0, -X[0, 2]],
[ 0, 0, -X[1, 2]],
[X[2, 0], X[2, 1], 0]])
If $\rm A X = X A$, then $x_{13} = x_{23} = x_{31} = x_{32} = 0$. The other five entries are unconstrained.
We can also vec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
How to find all solutions for $x^3=6x+6$ Could anyone help me to find how to find all solution for $x^3=6x+6$?
| Use Cardano's method:
Set $y=u+v$. As you have two unknowns instead of one, you can add a constraint on $u,v$, in order to simplify the equation. This equation becomes
$$(u+v)^3-6(u+v)-6=u^3+v^3+(u+v)(3uv-6)-6=0.$$
Adding the condition $3uv=6$, i.e. $uv=2$, you obtain the equation $u^3+v^3-6=0$, whence the system
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ 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 :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)... | We need to use the limit $$\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}\tag{1}$$ which can be easily established either via L'Hospital's Rule or Taylor's theorem. Now using $(1)$ we have $$\frac{\sin 3x - 3x}{x^{3}}\to - \frac{9}{2}\tag{2}$$ and hence $$\frac{\sin 3x}{x^{3}}+\frac{a}{x^{2}}+b\to 0\tag{3}$$ is equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proof that expression is positive if conditions is met Is there a way to show that this expression is always positive as long as $b>0$ and $r>x$ ? Assume $r>0$ and $x>0$
\begin{equation}
b-1+\sqrt{(1+b)^2-\frac{4rb}{x}}
\end{equation}
It's simple with $b>1$ but I can't figure it out if it is true for lower values.
| This is false. Consider $b = \frac{1}{10}, x = 1, r = \frac32$. Clearly, we have $b,x,r>0$ and $r>x$.
$$\sqrt{(1+b)^2-\frac{4rb}{x}} = \sqrt{\left(1+\frac{1}{10}\right)^2-\frac{4(3/2)(1/10)}{1}} = \sqrt\frac{121-60}{100} = \frac{\sqrt{61}}{10}$$
$$b-1+\sqrt{(1+b)^2-\frac{4rb}{x}} = -\frac9{10} + \frac{\sqrt{61}}{10} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that the sequence of Fibonacci quotients is Cauchy Consider the sequence defined by $a_1 = 1$ and $a_n =\frac{1}{1 + a_{n−1}}$
for all $n ≥ 2$. In general, this sequence can also be described as the sequence
of quotients of succesive terms in the Fibonacci sequence.
Prove that $(a_n)_{n\in \Bbb N}$ is a Cauchy se... | Given the Fibonacci sequence where $a_{n+1} = a_n + a_{n-1}$, we have with $b_n = a_n/a_{n-1},$
$$\begin{align} |b_{n+1} - b_{n}| &= \left|\frac{a_{n+1}}{a_n}- \frac{a_{n}}{a_{n-1}} \right| \\ &= \left|\frac{a_{n+1}a_{n-1} - a_n^2}{a_n a_{n-1}} \right| \\ &= \left|\frac{a_na_{n-1} + a_{n-1.}^2 - a_n a_{n-1} - a_n a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rule for multiplying rational numbers (follow-up) I was thinking about the rule where
$\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ and why it worked, and I found this question: Rule for multiplying rational numbers.
The best answer was:
Did:
One knows that $a/b$ may be defined as the number $x$ such that
$bx=a$,... | A completely formal proof depends on the particular axioms and definitions you work with, but as far as the intuition of it goes, consider the following points.
*
*The fraction $\frac{1}{b}$ denotes the multiplicative inverse of $b \ne 0\,$, which is to say that $x=\frac{1}{b}$ is the unique solution to $b \cdot x =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence m... | $c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.
$2ab = (a+b)^2 - c^2$
$2ab = (a+b+c)(a+b-c)$
Let $n = a+b+c$, and the above becomes:
$2ab = n(n-2c)$
So the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 6,
"answer_id": 0
} |
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.