Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &... | Since $x$ is an acute angle, $\cot x$ and $\tan x$ are positive.
Therefore, by A.M.-G.M. inequality,
$\frac {\tan x + \cot x}{2} \ge \sqrt{\tan x \times \cot x}$
Now, $\tan x \times \cot x = 1$ (by definition)
Therefore we obtain,
$\tan x + \cot x \ge 2 \sqrt{1}$
And so, $\tan x + \cot x \ge 2 $
Proof complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 3
} |
Law of Indices and Quadratic Expressions So I think I need some clarification about the rules for manipulating indices, in particular these two equivalences:
$(x^3)^2 = x^{(3)(2)} = x^6$
$a = a^1$
Take the expression:
$(5+5)^2$, which is equivalent to $(5+5)(5+5) = 100$.
Using the rules for indices above, I would have ... | Your error is here:
$$
(5^1+5^1)^2 \color{red}{\neq} 5^{(1)(2)}+5^{(1)(2)}.
$$
You can easily verify that these two numbers are not the same.
In general, we have $(a+b)^2=a^2+b^2+2ab$, so it is typically the case that $(a+b)^2\neq a^2+b^2$.
This is true:
$$
(5^1+5^1)^2 \color{red}{=} 5^{(1)(2)}+5^{(1)(2)}+2\cdot5^{1}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sequence with denominators of products of consecutive Fibonacci numbers I'm trying to figure out a way to solve the value of this:
$$\frac{1}{1\times 2}-\frac{1}{2\times 3}+\frac{1}{3\times 5}-\frac{1}{5\times 8}+\frac{1}{8\times 13}-\dots$$
The only thing I can come up with is a summation involving the $nth$ fibonacci... | You can do this using partial fractions.
Note that:
$$\begin{align}
\frac{1}{1 * 2} &= \frac{1}{1} + \frac{-1}{2} \\
\frac{1}{2 * 3} &= \frac{-1}{2} + \frac{2}{3} \\
\frac{1}{3 * 5} &= \frac{2}{3} + \frac{-3}{5} \\
\frac{1}{5 * 8} &= \frac{-3}{5} + \frac{5}{8} \\
\frac{1}{8 * 13} &= \frac{5}{8} + \frac{-8}{13}
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Surface integral problem: $\iint_S (x^2+y^2)dS$
*
*The problem statement, all variables and given/known data
$\iint_S (x^2+y^2)dS$, $S$is the surface with vector equation $r(u, v)$ = $(2uv, u^2-v^2, u^2+v^2)$, $u^2+v^2 \leq 1$
*Relevant equations
Surface Integral. $\iint_S f(x, y, z)dS = \iint f(r(u, v))\left |... | Given parametrization
$$\phi (u,v) = (2uv,{u^2} - {v^2},{u^2} + {v^2})$$
Given function:
$$f(u,v) = {u^2} + {v^2}$$
Using polar coordinates:
$$\begin{gathered}
u = r \cdot cos(\varphi ) \hfill \\
v = r \cdot \sin (\varphi ) \hfill \\
\end{gathered} $$
Transform function:
$$f(r \cdot cos(\varphi ),r \cdot \sin (\va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\lim_{x\rightarrow 1}x^{\frac{1}{n-x-x^2-x^3-\ldots -x^n}}=\frac{1}{\sqrt[k]{e}}$ Proving
$$\lim_{x\rightarrow 1}x^{\left(\frac{1}{n-x-x^2-x^3\cdots -x^n}\right)}=\frac{1}{\sqrt[k]{e}}$$
when $$k=1+2+3+\ldots+n$$
| Let $S$ be the denominator of the exponent so that $$S = n - x - x^{2} - \cdots - x^{n}$$ We have to calculate the limit $$\lim_{x \to 1}x^{1/S}$$ Let this desired limit be $L$ so that $$\begin{aligned}\log L &= \log\left(\lim_{x \to 1}x^{1/S}\right)\\
&= \lim_{x \to 1}\log x^{1/S}\text{ (by continuity of log)}\\
&= \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt... | mathisfun did a nice job of showing you how to reconcile your answers.
Another way of showing that $\sin(45^\circ) + \sin(15^\circ) = \sin(75^\circ)$ is to use the formula
$$\sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)$$
with $\alpha = 45^\circ$ and $\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How can I solve the system of equations? How can I solve the system of equations?
$$\begin{cases}
x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\
x^2 y^2-3 x y^3-3 x y+5 y^2=0.
\end{cases}$$
I have not any idea to solve.
| $$\begin{cases}
x^2 y^2+12 x y^3-18 x y-18y^4-4 y^2+27=0,&\\
x^2 y^2-3 x y^3-3 x y+5 y^2=0.
\end{cases}$$
We subtract the second equation from the first to get
$15xy^3-15xy-18y^4-9y^2+27=0$.
But $15xy^3-15xy=15xy(y^2-1)$, and $-18y^4-9y^2+27=-9(2y^4+y^2-3)=9(-2y^2-3)(y^2-1)$.
Hence $15xy^3-15xy-18y^4-9y^2+27$$=(y^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the least significant digit Explain why the number $(3^{27}\cdot7^{313}\cdot11^{121})^{1000}$ has 1 as its least significant digit. I know I am supposed to use $\pmod{10}$ but I am not sure how to combine the answers of the individual powers and then raise them to $1000$
| Notice that $3^3=27\equiv 7\pmod{10}$, $3^4\equiv 21\equiv 1\pmod{10}$. $7^2\equiv 9\pmod{10}$, $7^4\equiv 9^2\equiv 1\pmod{10}$. Also, $11^4\equiv 1^4\equiv 1\pmod{10}$. Now powers distribute over products, so
$$(3^{27}7^{313}11^{121})^{1000}=3^{27000}7^{313000}11^{121000}$$
and since the exponents are all multiples o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\displaystyle\lim_{x\to0^+}\left(\frac{\ln(4^x-3^x)-\ln(4^x-1)}{x}\right)(4^x-1)$? How to evaluate
$$L:=\lim_{x\to0^+}\left(\frac{\ln(4^x-3^x)-\ln(4^x-1)}{x}\right)(4^x-1)$$?
My solution:
$$\begin{align}
L&=\lim_{x \to 0^+} \frac{\ln\left(\frac{4^x-3^x}{4^x-1}\right)}{x}(4^x-1)=\\
&=\lim_{x \to 0^+} \f... | $$\begin{align}L&=\lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)\times\frac{4^x-1}{x}=\\
&=\lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)\times\lim_{x \to 0^+}\frac{4^x-1}{x}=\\
&=\ln 4 \cdot \lim_{x \to 0^+} \ln\left(1 - \frac{3^x-1}{4^x-1}\right)=\\
&=\ln 4 \cdot \ln \left( \lim_{x \to 0^+}\left[1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using ... | A slightly different perspective:
Use above drawing:
Label apex of isosceles triangle $A$, centre of circle $O$, extend $AO$ to intersect the circle in $C$. Length $AC = 2r$. Pick any point, say, on the circle's left part, call $ B$. Look at $\triangle ABC$.
$\triangle ABC$ is a right triangle (Thales circle)
Draw the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| Well,
$$\sqrt{155+1}=\sqrt{156}=\sqrt{4\times39}=\sqrt{4}\times\sqrt{39}=2\sqrt{39}$$
$$\sqrt{1555+1}=\sqrt{1556}=\sqrt{4\times389}=\sqrt{4}\times\sqrt{389}=2\sqrt{389}$$
And on..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
How to solve the congruence $ x^4 + x + 3 = 0 \pmod{3^3}$? $$x^4 + x + 3 = 0 \pmod{3^3}$$
I'm not sure how to this, I've tried many times but it never works for me :/ so, I hope someone will help me
| As the modulus is small, it isn't too hard to just check all the possible values of $x$. For $x = 0, 1, 2, 3$, the calculations are easy enough:
\begin{align*}
x = 0 &\implies x^4 + x + 3 = 3 \not\equiv 0 \bmod 9\\
x = 1 &\implies x^4 + x + 3 = 5 \not\equiv 0 \bmod 9\\
x = 2 &\implies x^4 + x + 3 = 21 \not\equiv 0 \bmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Confusion about a Linear Transformation question. Let $\beta := [M_1, M_2, M_3, M_4]$ be the ordered basis of $R^{2×2}$ defined by:
$$ M_1 := \begin{pmatrix}
1 & 0\\
0 & 0 \end{pmatrix},
M_2 := \begin{pmatrix}
0 & 1\\
0 & 0 \end{pmatrix},
M_3 := \begin{pmatrix}
0 & 0\\
1 & 0 \end{pmatrix},
M_4 := \begin{pmatrix}
0 & 0... | $$ \begin{pmatrix}
a & b\\
c & d \end{pmatrix}=a.\begin{pmatrix}
1 & 0\\
0 & 0 \end{pmatrix}+b.\begin{pmatrix}
0 & 1\\
0 & 0 \end{pmatrix}+c.\begin{pmatrix}
0 & 0\\
1 & 0 \end{pmatrix}+d.\begin{pmatrix}
0 & 0\\
0 & 1 \end{pmatrix}$$
Now $L$ is linear and then apply $L$ to each of these four matrices given $L(A)=A+A^T$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$ We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an ... | Use the Algebraic Limit Theorem for sequences:
*
*$\lim(ca_n)=c\cdot\lim a_n$
*$\lim(a_n+b_n)=\lim a_n + \lim b_n$
*$\lim(a_n b_n)=\lim a_n\cdot\lim b_n$
*$\lim\left(\frac{a_n}{b_n}\right)=\frac{\lim a_n}{\lim b_n}$ provided $b_n \ne 0$ and $\lim b_n \ne 0$
Rule (4) in the backwards direction, with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
deriving the sum of $x^n/(n+2)^2$ I am writing a research paper and I have stumbled upon an issue.
I have to evaluate
$$\sum_{n=1}^{\infty} \frac{x^n}{(n+2)^2}$$
Here is what I did:
$$ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$
$$\sum_{n=1}^{\infty} x^{n+1} = \frac{x^2}{1-x}$$
Integrate once with respect to $x$.... | Not sure what you did, but all that work seems kinda silly to arrive at your result.
$$\sum_{n=1}^\infty\frac{x^n}{(n+2)^2}=\frac{1}{x^2}\sum_{n=1}^\infty\frac{x^{n+2}}{(n+2)^2}=\frac{1}{x^2}(-x-\frac{x^2}{4}+\sum_{n=1}^\infty\frac{x^n}{n^2})=\frac{1}{x^2}(-x-\frac{x^2}{4}+\text{Li}_2(x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that $ ax^2+bx+c=0 $ has at least one root in $(0,1)$ if $10a+12b+15c=0$
If $10a+12b+15c=0$, Prove that $$ ax^2+bx+c=0 $$ has at least one root in $(0,1)$.
Progress
I tried to solve this by Rolle`s theorem ($f'$ has a root between any two roots of $f$), but could not reach the result.
| Consider $x = \frac{10}{12}$:
$$\begin{align*}
f\left(\frac{10}{12}\right)
&= a\left(\frac{10}{12}\right)^2+b\left(\frac{10}{12}\right)+c\\
&= \frac{10}{144}\left(10a+12b+\frac{144}{10}c\right)\\
&= \frac{10}{144}\left(10a+12b+15c-\frac{6}{10}c\right)\\
&= -\frac{6}{144}c
\end{align*}$$
And consider $x=0$:
$$f(0) = 0^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Prove two identities relating to series Show that:
$$(1)\sum_{n=1}^\infty\ln\left(\cos \frac{x}{2^n}\right)=\ln\frac{\sin x}x$$
$$(2). \sum_{n=1}^\infty\frac1{2^n}\tan \frac{x}{2^n}=\frac1x-\cot x$$
Thank you in advance.
NOTE:
The origial problem (1) I stated is $\sum_{n=1}^\infty\lim_{x\to \infty}\left(\cos \frac{x}{2... | For the first one :
$$\sum\limits_{n=1}^{N} \ln \cos \frac{x}{2^n} = \ln \prod\limits_{n=1}^{N} \cos \frac{x}{2^n} = \ln \frac{\sin \frac{x}{2^N}\prod\limits_{n=1}^{N} \cos \frac{x}{2^n}}{\sin \frac{x}{2^N}} = \ln \frac{\sin x}{2^N\sin \frac{x}{2^N}}$$
where, we made use of the identity: $\sin y \cos y = \frac{1}{2}\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to compute $\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)$ I find this problem on facebook group.
$$\mbox{Is it possible to find exact value of}\quad
\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)\ {\large ?}.
$$
I think this is not telescope sum. And Wolfram A... | Let $$\tan x=\frac{1}{n^2} \text{and } \tan y=\frac{1}{2n^2}$$ so we have that $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac{1}{2n^2}\right)}=\frac{3n^2}{2n^4-1}$$ hence we have $$x+y=\arctan\frac{\frac{1}{n^2}+\frac{1}{2n^2}}{1-\left(\frac{1}{n^2}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$ converges with no trig functions Let
$$\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$$
How to show that it converges with no use of trigonometric functions?
(trivially, it is the anti-derivative of $\sin ^{-1}$ and therfore can be computed directly)
| Here is a proof in single-variable calculus language. On the interval $0\leq s \leq 1$, we have
$$
\sqrt{1-s^2} = \sqrt{(1+s)(1-s)} \leq \sqrt{2} \sqrt{1-s}
$$
So
$$
\frac{1}{\sqrt{1-s^2}} \geq \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-s}}
$$
We would like to use the Comparison Test, comparing $\int_0^1 \frac{ds}{\sqrt{1-s}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Find all the prime numbers that satisfy the following conditions There was a brainteaser in the Science Magazine from University of Hong Kong which is as follow:
Find all the prime numbers $p$ such that $\sqrt{\frac{p+7}{9p-1}}$ is rational.
I tried a few numbers and it seems to suggest that $11$ is a suitable candidat... | The fact that $p$ is prime is something of a red herring.
Suppose that $$\sqrt{\frac{x+7}{9x-1}}$$
is rational for some positive integer $x$.
Then for some positive integers $k$, $a$, and $b$, we have the equations
$$x+7=ka^2$$
and $$9x-1=kb^2$$
Multiplying the first equation by $9$ and subtracting the second equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1078489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
show that $n + \lfloor \sqrt{n} + \frac{1}{2}\rfloor$ is never a perfect square for all positive integers $n$ show that $n + \lfloor \sqrt{n} + \frac{1}{2}\rfloor$ is never a perfect square for all positive integers $n$
I am thinking of a proof by contradiction by assuming $n + \lfloor \sqrt{n} + \frac{1}{2}\rfloor = k... | Let $t=\lfloor\sqrt{n}+\frac{1}{2}\rfloor$, so $t-\frac{1}{2}\le\sqrt{n}<t+\frac{1}{2}$.
If $n+t=k^2$ for some integer $k$,
then $t^2-t+\frac{1}{4}\le n<t^2+t+\frac{1}{4}\implies t^2+\frac{1}{4}\le n+t<t^2+2t+\frac{1}{4}<(t+1)^2\implies$
$t^2<k^2<(t+1)^2$, which gives a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Equivalence of $\pi$ is the first positive zero of the taylor series for $\sin(x)$ and $\pi/4 = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$ For $x\in\mathbb{R}$, define $\sin (x) = x - x^3/3!+x^5/5!-\cdots$ and $\pi = 4(1-\frac{1}{3}+\frac{1}{5} -\frac{1}{7}+\cdots)$. Then show that $\sin(\pi/2) = 1$
In the p... | Here's a detailed outline for how to do this in a fairly standard line of development of trig functions from their power series definitions.
(A "plug in" combinatorial proof would be neat to see as well, of course.)
Step 1. Show the basic trig identities (double angle formulas and $\sin(x)^2 + \cos(x)^2 = 1$) for the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater... | $f(a) = 3a^2 +2a + (3b^2-2b+1) \to \triangle' = 1^2 - 3(3b^2-2b+1) = -9b^2 +6b - 2 < 0,
\forall b \text{ since } \triangle' = 3^2 -(-9)(-2) = 9 - 18 = -9 < 0 \to f(a) > 0, \forall a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ I was just playing around with a calculator, and came to the conclusion that:
$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$
Now I'm curious. Is it possible to evaluate the exact value of ... | $$y=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}\equiv\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$
$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$ But the Let the term in br... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 3,
"answer_id": 2
} |
How to solve $y=(xy'+2y)^2$? What kind of differential equation is this thing and how to solve it?
$$y=(xy'+2y)^2$$
$$y=x^2y'^2+4xyy'+4y^2$$
| the differential equation $y = (x\frac{dy}{dx} + 2y)^2$ is singular at $x = 0.$ one has to find the solution away from $x = 0$ and patch it up around
$x = 0$ and $y = 0$ or $y + \frac{1}{4}.$ there is a boundary layer at $x = 0.$
now we will find the outer solution that is valid away from $x = 0.$
$y = (x\frac{dy}{dx}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Can I get a closed-form of $\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$? Can I get a closed-form of
$$\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$$
| Let $A$ be the series. Then
\begin{align}A &= \sum_{k = 1}^\infty (-1)^k \frac{\zeta(2k)}{2^{2k-1}}\\
&= \sum_{k = 1}^\infty \frac{(-1)^k}{2^{2k-1}} \sum_{n = 1}^\infty \frac{1}{n^{2k}}\\
&= 2\sum_{n = 1}^\infty \sum_{k = 1}^\infty \left(-\frac{1}{4n^2}\right)^k\\
&= -2\sum_{n = 1}^\infty \frac{1}{4n^2+1}\\
& = -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find all positive integers $a,b,c,d$ with $a\le\ b\le c$ such that $a!+b!+c!=3^d$ ? How to find all positive integers $a,b,c,d$ with $a\le\ b\le c$ such that $a!+b!+c!=3^d$ ?
| Let $X = a!+b!+c!$.
*
*$a = 1$, otherwise $2 | a! \land 2| b! \land 2| c! \implies 2 | X \implies X \ne 3^d$
*$b \le 2$, otherwise $X \equiv 1 + 0 + 0 \equiv 1 \pmod 3 \implies X \ne 3^d$.
*If $b = 1$, then $1 \le c \le 2$, otherwise, $X \equiv 1 + 1 + 0 \equiv 2 \pmod 3
\implies X \ne 3^d$.
*If $b = 2$, then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
find x+y if $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $? Let $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $.
How find $x+y$? I tried a number of ways.
| Let's call $\alpha$ the value of $x+\sqrt{x^2+1}$ and $\beta$ the value of $y+\sqrt{y^2+1}$
Then we know that $x=\dfrac{\alpha^2-1}{2\alpha}$ and $y=\dfrac{\beta^2-1}{2\beta}$
We know also that $\alpha\beta=42$
Hence $x+y=\dfrac{\alpha^2-1}{2\alpha}+\dfrac{\beta^2-1}{2\beta}=\dfrac{(\alpha\beta-1)(\alpha+\beta)}{2\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 ... | Set $C = \displaystyle \sum_{i=0}^{999}5\cdot 10^i$.
Applying the summation formula for a geometric series,
$\quad \displaystyle C = 5 \, (1-10^{1000}) \,(1-10)^{-1}$
Continuing,
$\quad \displaystyle C \equiv 5 \, (1-3\times{3^3}^{333}) \, 5^{-1} \equiv 1 - 3\times(-1)^{333} \equiv 4 \pmod{7}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 9
} |
Let $C$ be the curve of intersection of the plane $x+y-z=0$ with ellipsoid $\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$. Let $C$ be the curve of intersection of the plane $x+y-z=0$ and the ellipsoid $$\frac{x^2}4+\frac{y^2}5+\frac{z^2}{25}=1$$ Find the points on $C$ which are farthest and nearest from the origin
When dea... | Since there's a computer result posted now, I guess I'll describe my approach further. From the set of "Lagrange equations" you describe, we can solve each one for $ \ \lambda \ $ to establish
$$ \lambda \ = \left ( \ 2 \ - \frac{\mu}{2} \right ) \ x \ = \left ( \ 2 \ - \frac{2\mu}{5} \right) \ y \ = \left( \ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Probability for a word to start with $\text{2,0,0,4}$ Let $A= \left\{ 2,2,4,4,0,0,0,0\right\}$. We arrange those $8$ numbers randomly. What is the probability to get a sequence starting with $2,0,0,4$?
The answer is:$$\frac{2\cdot 4 \cdot 3 \cdot 2 \cdot 4!}{8!}$$
The denominator is simple, there are $8!$ permutations... | The letters that are left to be distributed are {2, 4, 0, 0}. These can be arranged in $A=\binom{4}{2}\times\binom{2}{1}=\frac{2\times4!}{2!\times2!}$ different ways.
The total amount of ways those 8 numbers can be arranged is not 8!, but $B=\binom{8}{4}\times\binom{4}{2}=\frac{8!\times4!}{4!\times4!\times2!\times2!}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is sy... | The next step is to write, for example, $$\frac{x+1}{x^2+x+1} = \frac{1}{2} \cdot \frac{2x+1}{x^2+x+1} + \frac{1}{2} \cdot \frac{1}{x^2+x+1}$$ from which we then have $$\int \frac{x+1}{x^2+x+1} \, dx = \frac{1}{2} \log\left|x^2+x+1\right| + \frac{1}{2} \int \frac{dx}{(x+1/2)^2+(\sqrt{3}/2)^2},$$ and the second integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
testing for the convergence of a series I have to check if the following series converges. I thought that I have to use comparison test, but it did not lead me anywhere.
$$\sum_{n=2}^{\infty}\frac{ln^5(2n^7+13)+10sin(n)}{nln^6(n^{7/8}+2n^{1/2}-1)ln(ln(n+(-1)^n))}$$
| Note that for sufficiently large $n$
$$\frac{\ln^5{(2n^7+13)+10\sin(n)}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > \frac{\ln^5{(2n^7+13)-10}}{\ln^5{(n^\frac{7}{8}}+2\sqrt{n}-1)} > 1.$$
Also
$$ \lim_{n \to \infty}\frac{n^{\frac{7}{8}}+2\sqrt{n}-1}{n} = 0,$$
and for sufficiently large $n$ we have
$$ \ln(n^{\frac{7}{8}}+2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to proof that $\sum_{i=1}^{2^n} 1/i \ge 1+n/2$ I had troubles trying to prove that for every $n\ge1$
$$\sum_{i=1}^{2^n}\frac1i\ge 1+\frac n2$$
Can you give me a hint about the induction proof or show me in detail how can I prove it? I would appreaciate any help. Thanks!
| $$\sum_{i=1}^{2^{k+1}}\frac{1}{i} = \sum_{i=1}^{2^{k}}\frac{1}{i}+\sum_{i=2^k+1}^{2^{k+1}}\frac{1}{i} \\ = \sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^k+1}+\frac{1}{2^k+1} \ldots +\frac{1}{2^{k+1}-1}+ \frac{1}{2^{k+1}}\right) \\ \geq \sum_{i=1}^{2^{k}}\frac{1}{i}+\left(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}} \ldots +\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$ Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$
Thanks for any help
| To give another approach:
The sum can be calculated using the observation that $$\int_0^1\left(1-x^{\frac{1}{n}}\right)^kdx=\frac{1}{\binom{n+k}{n}}$$
which is related to the beta function. Using this we get:
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}=\frac{1}{4!}\sum_{n=0}^{\infty}\frac{(-1)^n}{\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find joint likelihood function of observations $x_1, x_2, \ldots, x_n$ and $y_1, y_2, \ldots, y_m$
Let $x_1,\ldots,x_n$ be observations from a normal distribution with mean $0$ and s.d $s_1$.
Similarly let $y_1,\ldots,y_m$ be observations from a normal distribution with mean $0$ and s.d $s_2$.
Find the combined likeli... | That is correct if $(x_1,\ldots,x_n)$ is independent of $(y_1,\ldots,y_n)$.
For the following to be valid, one must assume all $n+m$ observations are independent. One has
\begin{align}
L(s_1,s_2) & = \prod_{i=1}^n \left(\frac1{\sqrt{2\pi}\,s_1} \exp\left(\frac{-1}2 \left( \frac{x_i}{s_1} \right)^2 \right)\right) \prod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Investigate convergence of the series Investigate convergence of the series:
$$\left( \frac{n^2+3n+10}{n^2+5n+17} \right)^{n^2 (\sqrt{n+1}-\sqrt{n-1})}$$
It should be solvable with simple manipulations with the formula, i guess, but how to do that?
| Note that
$$
\sqrt{n+1}-\sqrt{n-1}=\frac2{\sqrt{n+1}+\sqrt{n-1}}\ge\frac1{\sqrt{n+1}}
$$
and by cross-multiplication, for $n\ge20$,
$$
\frac{n^2+3n+10}{n^2+5n+17}\le\frac{n}{n+2}
$$
Therefore, using Bernoulli's Inequality,
$$
\begin{align}
\left(\frac{n^2+3n+10}{n^2+5n+17}\right)^{n^2(\sqrt{n+1}-\sqrt{n-1})}
&\le\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Matrices, determinants, and applications to identities involving Fibonacci numbers Preamble
It is well known that since:
$$
\begin{pmatrix}
F_{n+1} \\
F_n \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
F_n & F_{n-1} \\
\end{pmatrix}
$$
it is valid that:
$$
\begin{pma... | The "strange matrix identity" at the bottom is just a linear recurrence among the cubed Fibonacci numbers:
$$F_{n+4}^3=3F_{n+3}^3+6F_{n+2}^3-3F_{n+1}^3-F_n^3$$
which looks somewhat less formidable than the matrix equation.
To derive this, we start with Binet's formula
$$ F_n = \frac{\phi^n-(-\phi^{-1})^n)}{\sqrt 5} $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find $ \binom {1}{k} + \binom {2}{k} + \binom{3}{k} + ... + \binom{n}{k} $
Find $$ \binom {1}{k} + \binom{2}{k} + \binom{3}{k} + ... + \binom {n}{k} $$ if $0 \le k \le n$
Any method for solving this problem? I've not achieved anything so far. Thanks in advance!
| For a polynom $P(x)=a_mx^m +\dots +a_1x+a_0$ denote $[x^k]P(x)=a_k$.
Notice that for $k\le i \le n$
$$\dbinom{i}{k}=[x^k](1+x)^i$$
hence
\begin{align}
\sum_{i=1}^n\dbinom{i}{k}&=\sum_{i=k}^n\dbinom{i}{k}\\
&=\sum_{i=1}^n[x^k](1+x)^i\\
&=[x^k]\sum_{i=k}^n(1+x)^i\\
&=[x^k]\frac{(1+x)^{n+1}-(1+x)^k}{(1+x)-1}\\
&=[x^k]\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}$ for angles of a triangle Let the three internal angles of a triangle are $a,b,c$. Prove that
$$\cos(2a) + \cos(2b) + \cos(2c) \geq -\frac{3}{2}.$$
I'm looking for an elementary, geometric proof. So avoid derivatives and tools from optimalization theory if i... | If you are interested in an algebraic proof, mine goes as follows:
$\cos{2A} + \cos{2B} = 2 \cos{(A+B)} \cos{(A-B)} = %
2 \cos{(\pi - C)} \cos{(A-B)} = \\-2 \cos{C} \cos{(A-B)}$
Hence,
\begin{eqnarray*}
& \cos{2A} + \cos{2B} + \cos{2C} &\geq& - \dfrac{3}{2} \\
\iff & -2 \cos{C} \cos{(A-B)} + \cos{2C} &\geq& - \dfrac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is there a trick to prove that $(x^4+y^4)^{1/2} \leq x^2+y^2$? $(x^{4}+y^{4})^{1/2} \leq x^{2}+y^{2}$
I tried multiplying the original by 1 = $\displaystyle \frac{(x^{4}+y^{4})^{1/2}}{(x^{4}+y^{4})^{1/2}}$, but that just brings me back to the original.
Ive looked up sqrt and exponent laws and can't find anything helpf... | $$
\begin{align}
(x^2 + y^2)^2 &= x^4 + 2x^2y^2 + y^4\\
&= (x^4 + y^4) + 2x^2y^2
\end{align}
$$
Now compare $(x^4 + y^4)^{\frac{1}{2}}$ with $\left[(x^4 + y^4) + 2x^2y^2\right]^{\frac{1}{2}}$, knowing that for all $x, y$, $2x^2y^2\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Pythagorean triples So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.
I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.
Here is my attempt.... | Hint If $m,n$ are integers then $(m^2-n^2, 2mn, m^2+n^2)$ is a Pytagorean triple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Laurent series expansion, can one simplify this? I have to expand $f(z)=\frac{z-1}{(z^2+1)z}$ in an annulus $R(i,1,2)$.
$$f(z)=\frac{1}{z-i}\frac{1}{z+i}-\frac{1}{z-i}\Big(\frac{i}{z+i}-\frac{i}{z}\Big)$$
$$\frac{1}{z-i}\frac{1}{z+i}=\frac{1}{z-i}\cdot\frac{1}{2i}\cdot\frac{1}{1-(-(z-i)/2i)}=\sum_{n=0}^{\infty}(-1)^n... | From partial fractions, we have that
$$
f(z) = \frac{-1}{z} + \frac{1}{2}\frac{1+i}{z+i} + \frac{1}{2}\frac{1-i}{z-i}
$$
For the Laurent series in annulus, I obtained:
\begin{align}
\frac{-1}{z} + \frac{1}{2}\frac{1+i}{z+i} + \frac{1}{2}\frac{1-i}{z-i}
&= \frac{-1}{z-i}\frac{1}{1+\bigl(\frac{i}{z-i}\bigr)} + \frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I find the coefficient of $x^2$ in this polynomial, given its value at three points? We are given the following data about a polynomial $P(x)$ of unknown degree: $P(0) = 2$, $P(1) = −1$, $P(2) = 4$.
Determine the coefficient of $x^2$ in $P(x)$ if all the third-order differences are $1$.
| Since all third order differences are $1$ we know that this polynomial is of degree $3$ and leading coëfficiënt $\frac{1}{6}$.
$P(0) = 2$, so we know $P(x) = \frac{1}{6} x^3 + b x^2 + cx + 2$.
Because $P(1) = -1$ and $P(2) = 4$ we know that :
$ \frac{1}{6} + b + c + 2 = -1$
$ \frac{8}{6} + 4b + 2c + 2 = 4$.
The soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then ... | Hint: Let $x=\sinh a$ and $y=\sinh b$. Then, using the fact that $\cosh^2u-\sinh^2u=1$ and $\sinh u+\cosh u=e^u$, we arrive at $e^{a+b}=1\iff a+b=0$, assuming a and b are reals. Since $\sinh u$, just like $\sin u$, is an odd function, the proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Real solutions of $x^n + y^n = (x+y)^n$ I have to find all real solutions of the following equation:
$x^n + y^n = (x+y)^n$
Clearly for $n = 1$, the equation holds for every $x,y$ real numbers.
If $n$ is greater or equal to $2$, we do binomial expansion on RHS and from $x^n + y^n - (x+y)^n = 0 $ it follows that the roo... | First, if $x=0$ or $y=0$ then the equation is satisfied.
Second, if $x>0$ and $y>0$ then because by binomial expansion
$$ (x+y)^n = x^n + y^n + \text{strictly positive terms}$$
the equation can never be satisfied. On the other hand, if $x<0$ and $y<0$ then the equation is equivalent to $(-x)^n+(-y)^n=(-x-y)^n$ where al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A problem of inequality Let $a_1, a_2, a_3$; $b_1, b_2, b_3$; $c_1, c_2, c_3$; $d_1, d_2, d_3$ be all real numbers.
We need to show that
$$\begin{align}(a_1b_1c_1d_1 + a_2b_2c_2d_2 &+ a_3b_3c_3d_3)^4\\ &\leq (a_1^4+a_2^4+a_3^4)(b_1^4+b_2^4+b_3^4)(c_1^4+c_2^4+c_3^4)(d_1^4+d_2^4+d_3^4)\end{align}$$
I have used Cauchy-S... | Your inequality is an application of the Hölder's Inequality.
More specifically,
$$(a_1^4 + a_2^4 + a_3^4)^\frac{1}{4}\cdots(d_1^4 + d_2^4 + d_3^4)^\frac{1}{4} \ge \left[ a_1^{\left(4 \cdot \frac{1}{4}\right)}b_1^{\left(4 \cdot \frac{1}{4}\right)}c_1^{\left(4 \cdot \frac{1}{4}\right)}d_1^{\left(4 \cdot \frac{1}{4}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Eigenvalues of operator on $S_n$'s group algebra Take the group algebra of the symmetric group $S_n$ (or equivalently consider $S_n$'s regular representation) - I guess over $\mathbb{C}$.
If $e_{i,j} \in S_n$ denotes the element which swaps only the pair $i$ and $j$, consider the linear operator
$$
O = (e_{1,2} + e_{2,... | For $n=4$, the eigenvalues are $4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2$.
For $n=5$, the eigenvalues are the roots of
$(x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve the recurrence $T(n) = 2T(n-1) + n, T(1)=1, n\geq 2$ This question has been already solved here, I just want to figure out why I'm not being able to solve it using my method. Here's what I did -
$T(n)=2T(n-1)+n$
$T(n-1)=2T(n-2)+(n-1)$
$\therefore \,T(n)=2\{2T(n-2)+(n-1)\}+n$
$T(n)=4T(n-2)+3n-2$
$T(n-2)=2T(n-3)+(n... | look for a particular solution of the form $T(n) = an + b$ where $a,b$ need to be determined. we need $an + b = 2(an - a + b) + n$ equating the coefficient of $n$ gives you $a = -2$ and then $b = 2b -a$ so $b = a + -2$ and $T(n) = -n-2$ is a particular solutions. the solution to the homogeneous equation is $T(n) = C2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$. Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.... | In order that an integer $n$ can be represented as $c^2-b^2$, it suffices that $n\not\equiv{2}\pmod{4}$. In such a case $n$ can be written as the product of two divisors with the same parity, $n=pq$, and we can take $c=\frac{p+q}{2},d=\frac{p-q}{2}$. So it is sufficient to prove that for an infinite number of integers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1127860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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What values of $a$ and $b$ does this system have infinitely many solutions? As a disclosure, this question is more for me to confirm that I did my work correctly. More specifically, the "solution" provided to me claims there are two values of $a$ and $b$ that yield infinite solutions, but I found only one.
That said, ... | You can check easily if the solution is wrong. (It is not.) Put $a=b$. Then you have:
\begin{equation} A =
\begin{pmatrix}
a & 0 & a & |2 \\
a & 2 & a & |a \\
a & 2 & a & |a \\
\end{pmatrix}
\end{equation}
Reducing to (III - I) and (II - I)
\begin{equation} A =
\begin{pmatrix}
a & 0 & a & |2 \\
0 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find $\int_0^{1/4}\frac{1}{x\sqrt{1-4x}}\ln\left({\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}}\right)dx$ Let $H_n$ be the harmonic series. I want to find the value of $A=\displaystyle\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{1}{4}\right)^n\frac{H_n}{n} $.
From this paper : https://cs.uwaterloo.ca/journals/JIS/VOL15/Boya... | Set $x=1$ in
$$
\sum_{n=1}^\infty \frac{{2n\choose n}}{4^n}\frac{H_n}{n} x^n=2\operatorname{Li}_2\left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)$$
we get
$$\sum_{n=1}^\infty \frac{{2n\choose n}}{4^n}\frac{H_n}{n}=2\operatorname{Li}_2(1)=2\zeta(2)$$
For a different result, pick $x=-1$,
$$\sum_{n=1}^\infty (-1)^n\frac{{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Surface Area by Integration $$2\pi\int_{3}^6\left(\frac{1}{3}x^\frac{3}{2}-x^\frac{1}{2}\right)\left(1+\left(\frac{1}{2}x^\frac{1}{2}-\frac{1}{2}x^\frac{-1}{2}\right)^2\right)^\frac{1}{2}dx$$
I've managed to simplify this down to the equation below (not sure if it'll help), but I still can't integrate it.
Please help.... | You did a good bit of the hard work yourself, but these other computations may help:
\begin{align}
\frac{\pi}{3}\int_3^6\sqrt{x^4-4x^3-2x^2+12x+9}\;dx &= \frac{\pi}{3}\int_3^6 (x^2-2x-3)\;dx\tag{simplify}\\[1em]
&= \frac{\pi}{3}\int_3^6 x^2\;dx - \frac{2\pi}{3}\int_3^6x\;dx -\pi\int_3^61\;dx\\[1em]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving $\cos x + 3^{1/2} \sin x = 1$ for $0\leq x \leq 360^{\circ}$ $\cos x + \sqrt3 \sin x = 1$
Not sure what the step would be to get an answer.
| If we square both sides of the equation:
$$
\Big(\cos(x) + \sqrt{3}\sin(x)\Big)^2 = 1^2 \\
\cos^2(x) + 3\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1 \\
\cos^2(x) + \sin^2(x) + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1 \\
1 + 2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 1 \\
2\sin^2(x) + 2\sqrt{3}\sin(x)\cos(x) = 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
A level Integration: $\int\frac{x^3}{\sqrt{x^2-1}}dx$ Using the substitution $x=\cosh (t)$ or otherwise, find
$$\int\frac{x^3}{\sqrt{x^2-1}}dx$$
The correct answer is apparently
$$\frac{1}{3}\sqrt{x^2-1}(x^2+2)$$
I seem to have gone very wrong somewhere; my answer is way off, can someone explain how to get this answe... | Let $\displaystyle u=x^2, dv=\frac{x}{\sqrt{x^2-1}}dx$ $\;\;$and $\;\;du=2xdx, v=\sqrt{x^2-1}$ to get
$\displaystyle\int\frac{x^3}{\sqrt{x^2-1}}\;dx=x^2\sqrt{x^2-1}-\int2x\sqrt{x^2-1}\;dx=x^2\sqrt{x^2-1}-\frac{2}{3}(x^2-1)^{3/2}+C$
Alternatively, let $u=\sqrt{x^2-1}, \;\;du\displaystyle=\frac{x}{\sqrt{x^2-1}}dx, \;\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Proof that the harmonic series diverges (without improper integrals)
Show that $$\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)$$ Use this to deduce that the series $$\sum\limits_{k=1}^\infty \frac{1}{k}$$ diverges.
Hint: Use the estimate $$\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x$$
Note that all instan... | we can use the fact that $$\dfrac{1}{k-1} + \dfrac{1}{k}+\dfrac{1}{k+1} = \dfrac{1}{k}+\dfrac{2k}{k^2 - 1} > \dfrac{1}{k}+\dfrac{2}{k} = \dfrac{3}{k}\text{ for } k \ge 2$$ to show that $s=1 + \frac12 + \frac13 + \cdots$ cannot be finite, therefore the harmonic series must diverge.
$\begin{align}
s &= \dfrac{1}{1} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Number of paths in 3D coordinates A cute problem which is an extension of a well-known counting problem:
Find the number of paths of length $12$ from $(0,0,0)$ to $(4,4,4)$ passing through adjacent lattice points (for two ajacent points in the path, the distance between them is $1$), which avoid $(1,1,1), (2,2,2), (3,3... | We can use inclusion exclusion. let $f(S)$ be the number of paths that pass through all the elements of $S$. Then we want:
$\color{blue}{f(\emptyset)}-\color{red}{f((1,1,1))}-\color{green}{f((2,2,2))}-\color{purple}{f((3,3,3))}+\color{blue}{f((1,1,1),(2,2,2))}+\color{red}{f((1,1,1),(2,2,2))}+\color{green}{f((1,1,1),(3,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factorizing $(x+a)(x+b)(x+c)$ I was solving questions related to polynomial factorization. I have learnt the remainder and factor theorems, and some basic identities.
There was a question like this one:
$$p(x)=x^3+8x^2+19x+12$$
The method we have been taught is to first find the factors of the last number, here it is $... | Since $$P(-1)=(-1)^3+8(-1)^2+19(-1)+12=0$$
Therefore $x+1$ is one of the factors. After dividing $p(x)$ by $x+1$ we have
$$x^3+8x^2+19x+12=(x+1)(x^2+7x+12)$$
Since $$x^2+7x+12=(x+3)(x+4)$$ We are going to have
$$p(x)=x^3+8x^2+19x+12=(x+1)(x+3)(x+4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding Smallest x and y to Satisfy Equation Find the smallest natural numbers $x$ and $y$ such that $$7^2x=5^3y$$
I'm unsure how to proceed with this question. Could someone explain the process for determining the answer?
Added from the comments:
$x=5^3$ would satisfy $5^3$ divides $7^2x$
| First consider left side. $5^3$ divides right side, so $5^3$ must also divides left side. Now $7^2$ and $5^3$ are relatively prime, so $5^3$ occurs in $x$, in other word:
$$x=5^3 \cdot x'$$
For some natural $x'$. $x'$ is natural, so $x' \geq 1$, so:
$$x=5^3 \cdot x' \geq 5^3$$
The same way we get:
$$y \geq 7^2$$
But if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\frac{abc}{1+bc}+\frac{bcd}{1+cd}+\frac{cda}{1+ad}+\frac{dab}{1+ab}\le \frac{1}{17}$ Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\dfrac{abc}{1+bc}+\dfrac{bcd}{1+cd}+\dfrac{cda}{1+ad}+\dfrac{dab}{1+ab}\le \dfrac{1}{17}$
My attempt:
I figured out that if each of the eleme... | Here is my approach. I am stuck at a place but I hope this is a good method and I don't reach a dead end. The LHS is equivalent to $\frac{a}{1/bc+1}+\frac{b}{1/cd+1}+\frac{c}{1/da+1}+\frac{d}{1/ab+1}$. From AM-GM, we have $1+1/bc>2/\sqrt{bc}$. Inverse the inequality with inverting the inequality sign and then multiply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove the limit of a sequence using definition Find the limit of the following sequence and justify it using the definition:
$$x_n = \frac{1}{n}sin\frac{nπ}{4}+\frac{n}{2n^2+cosn} $$
The definition:
X is the limit of the sequence $x_n$ if the following condition holds:
For each real number Ɛ > 0, there exists a natura... | You are correct, the limit is zero. To prove it by definition, let $\varepsilon > 0$ and set $N > \frac{2}{\epsilon}$. If $n > N$, then
\begin{align}|x_n - 0| &= \left|\frac{1}{n}\sin \frac{n \pi}{4} + \frac{n}{2n^2 + \cos n}\right|\\
& \le \frac{1}{n}\left|\sin \frac{n\pi}{4}\right| + \left|\frac{n}{2n^2 + \cos n}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Calculate integral using beta and gamma functions I have to calculate the following integral using beta and gamma functions:
$$
\int\limits_0^1 \frac{x\,dx}{(2-x)\cdot \sqrt[3]{x^2(1-x)}}
$$
I came up with this terrible solution. Firstly, let's break it into two parts:
$$
\int\limits_0^1 \frac{(x-2)\,dx}{(2-x)\cdot \sq... | For the second integral, make the substitution $ \displaystyle x = \frac{u}{1+u}$.
Then
$$ \begin{align} 2 \int_{0}^{1} \frac{dx}{(2-x) \sqrt[3]{x^2(1-x)}} &= 2 \int_{0}^{\infty} \frac{du}{(2+u)u^{2/3}} \\ &= \int_{0}^{\infty} \frac{du}{\left(1+ \frac{u}{2}\right)u^{2/3}} \\ &= \frac{2}{2^{2/3}} \int_{0}^{\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solution of equation $[1+\frac{x}{b}]e^{-x/b}=z$ Can we solve this equation
$$\left(1+\frac{x}{b}\right)e^{-x/b}=z$$
We have to determine value of $x$ in term of $z$.
Problem occur while calculating the following integral.
$$\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}... | Your original problem is
$f(a, b, r)
=\frac{a}{b^{2}}\int_{0}^{\infty}(1-(1+\frac{x}{b})e^{-\frac{x}{b}})^{a-1}x^{r+1}e^{-\frac{x}{b}}dx
$.
I'll play with it and see what I get.
Letting
$x/b = y$,
so
$dx = b dy$,
$f(a, b, r)
=\frac{a}{b}\int_{0}^{\infty}(1-(1+y)e^{-y})^{a-1}(by)^{r+1}e^{-y}dy
=ab^r\int_{0}^{\infty}(1-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find matrix from Eigenvectors and Eigenvalues A matrix $A$ has eigenvectors
$v_1 = \left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)$
$v_2 = \left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)$
with corresponding eigenvalues $\lambda_1$= 2 and $\lambda_2$= -3, respectively.
Determine Ab ... | By definition of eigenvalue and eigenvector, we have
$$\tag{1}A\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)=2\left(
\begin{array}{c}
2 \\
1 \\
\end{array}
\right)\mbox{ and }A\left(
\begin{array}{c}
1 \\
-1 \\
\end{array}
\right)=-3\left(
\begin{array}{c}
1 \\
-1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What are all the twin primes $p$ and $q = p + 2$ for which $pq - 2$ is also prime? It seems that $p = 3$ and $q = 5$ ($pq - 2 = 13$) are the only solutions. However I'm having a difficult time proving this.
I have that all primes can be represented as $3k + 1$ or $3k + 2$ so if $p = 3k + 2$ then $q = 3k + 4$ and
\beg... | Hint $\,\ {\rm mod\ 3}\!:\ p\not\equiv \color{#c00}0\,\Rightarrow\, \color{#0a0}q\,(pq\!-\!2) = \overbrace{(p\,+\,2)}^{\large 0\ {\rm if}\ p\,\equiv\,\color{#c00}{\bf 1}\ }\,\overbrace{(p\,(p+2)-2)}^{\large\ 0\ {\rm if}\ p\,\equiv\,\color{#c00} 2}\equiv 0 $
hence we conclude $\ 3\nmid \color{#0a0}q\,\Rightarrow\,3\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Why does $\int\frac{1}{2x+1}dx=\frac{1}{2}\ln|2x+1|+C$? The way I am thinking is as follows:
$$\int\frac{1}{2x+1}\,dx = \int\frac{1}{2}\frac{1}{x+\frac{1}{2}}\,dx = \frac{1}{2}\int\frac{1}{x+\frac{1}{2}}\,dx = \frac{1}{2}\ln\left|x+\frac{1}{2}\right|+C$$
However, the textbook answer is $\frac{1}{2}\ln|2x+1|+C$. Where i... | Let us given
$$ \int{1\over2x+1}dx$$
Let us do the following substitution
$$ u = 2x+1 $$
So,
$$ du = 2dx $$
Finally,
$$ \int{1\over2x+1}dx = \frac12\int{1\over{u}}du = \frac12\ln\lvert{u}\rvert+C=\frac12\ln\lvert{2x+1}\rvert+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the value of the infinite sum $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ Can anyone help me to find what is the value of $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ when it tends to infinity
The first i ... | Hint. Observe that the general term of your series is
$$\frac{(-1)^n}{3n+1}. $$
Then you may write
$$
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^n}{3n+1}&=\sum_{n=0}^{\infty}(-1)^n\int_0^1 x^{3n} dx\\\\
&=\int_0^1 \sum_{n=0}^{\infty}(-1)^nx^{3n} dx\\\\
&=\int_0^1 \frac{1}{1+x^3} dx\\\\
&=\int_0^1 \left(\frac{1}{3 (1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Area of triangle bounded by line and degenerate "crossed lines" conic The question is
Show that the two lines given by
$$(A^2 - 3B^2)x^2 + 8ABxy +(B^2 - 3A^2)y^2=0$$
and the line given by $$Ax+By+C=0$$ determine an equilateral triangle of area $$\frac{C^2}{\sqrt{3}\;(A^2+B^2)}$$
I tried factorizing the pair of... | $$( A^2 - 3 B^2 ) x^2 + 8 A B x y + ( B^2 - 3 A^2 ) y^2 = 0 $$
are a pair of straight lines through the the origin making angle $ \pi/3 $ or $2 \pi/3 $ with each other at the origin, just like two sides along diameter of circum-circle of a regular hexagon.
They are factorisable into two straight lines as it is easy to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}... | By using some telescopic sums, we can be even more accurate. We may notice that:
$$ \frac{1}{n^2}-\frac{1}{n(n-1)} = -\frac{1}{n^2(n-1)} $$
and if $n>1$:
$$ \frac{1}{n^2(n-1)}-\frac{1}{(n-1)n(n+1)}=\frac{1}{(n-1)n^2(n+1)} $$
so:
$$ \sum_{n=1}^{N}\frac{1}{n^2} = 1+\sum_{n=2}^{N}\frac{1}{n(n-1)}-\sum_{n=2}^{N}\frac{1}{(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
The Cauchy product of $\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$ with itself. In Apostol's Mathematical Analysis there is the following exercise. Show that the Cauchy product of
$$\sum_{n=0}^\infty{(-1)^{n+1}\over n+1}$$
with itself is the series
$$2\sum_{n=1}^\infty{(-1)^{n+1}\over n+1}\left(1+\frac12+\cdots+\frac1n\r... | The Cauchy product will be
$$\sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^{k+1}}{k+1} \frac{(-1)^{n-k+1}}{n-k+1} = \sum_{n = 0}^\infty \sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)}.$$
Now
\begin{align}\sum_{k = 0}^n \frac{(-1)^n}{(k+1)(n-k+1)} &= (-1)^n \sum_{k = 0}^n \frac{1}{n+2}\left(\frac{1}{n-k+1} + \frac{1}{k+1}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
showing certain vertices form an equilateral triangle Suppose $a,b,c$ lie in the unit circle of the complex plane and satisfy $a+b+c = 0$. Then, $a,b,c$ form the vertices of an equilateral triangle.
Try
So, I want to show that $|b-a|=|c-a|=|b-c|$. We are given $|a| = |b| = |c| = 1$ and $a+b+c=0$. Since $|z| = z \overli... | Without loss of generality you can assume that $a=1$, because the centre of mass of the three points, the origin, is invariant under their rotation about it. This implies that $c = \overline b$, and that the real component of $b$ and $c$ is $-1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How do I calculate: $\int \frac{dx}{3\sin^2 x+5\cos^2x}?$ How can I calculate this integral ?
$$\int \frac{dx}{3\sin^2 x+5\cos^2x}=\text{?}$$
Thank you! I've tried using universal substitution but the result was too complicated to be somehow integrated. Can you please give me a useful hint ?
| At first substitute: $$\int \frac{dx}{3\sin^2\left(x\right)+5\cos^2\left(x\right)}dx = \int \frac{1}{3\sin^2\left(\arctan\left(u\right)\right) + 5\cos^2\left(\arctan\left(u\right)\right)}\frac{1}{1+u^2}du$$ where $x=\arctan\left(u\right)$ and $dx=\frac{1}{1+u^2}du$.
Then we can write this as $$\int \frac{1}{\left(u^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know... | Using the Euclidean division we get
$$\begin{array}\\x^4-3x^3+2&\Bigg|&x-1\\
-(x^4-x^3)&\Bigg|&x^3\\
=-2x^3+2&\Bigg|&-2x^2\\
-(-2x^3+2x^2)\\
=-2x^2+2&\Bigg|&-2x\\-(-2x^2+2x)\\=-2x+2&\Bigg|&-2
\end{array}$$
so we find that
$$x^4-3x^3+2=(x-1)(x^3-2x^2-2x-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$y''+y'^{2}+y=0$ equation solution How would you solve this differential equation $y''+y'^{2}+y=0$? I can't apply the ansatz method (or more formally apply the characteristic polynomial method). Thanks
| Consider $y=ax^2+bx+c,$ then we have $y'=2ax+b, (y')^2=4a^2x^2+4abx+b^2, y''=2a$ which leads to
$$c=-2a-b^2\\
-ax^2-bx=4a^2x^2+4abx\\
-ax-b=4a^2x+4ab\\
-ax-b=4a(ax+b)\\
a=-\frac 14\text{ or }ax+b=0$$
$b$ is a free variable when $a=-\frac 14$, leading to a set of solutions $y=-\frac 14x^2+bx+\frac 12-b^2$. When $ax+b=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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When is the number of $N$'s factors $1 + \sqrt{N}$? (Answer: Only $N = 4$ and $N = 16$.)
The following question arose in a course for pre-service and in-service elementary school teachers:
For what $N \in \mathbb{N}$ is it the case that the number of $N$'s factors is $1 + \sqrt{N}$?
The initial observations were tha... | Just to rephrase Robert Israel's answer into more palatable language for the level at which I am presently teaching:
There are two functions to think about here.
The first is the divisor function, $N \rightarrow d(N) := \#\{N$'s distinct divisors$\}$.
The second is the function $N \rightarrow 1 + \sqrt{N}$.
The issue w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Proof $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant I would appreciate if somebody could help me with the following problem:
Question: Defined by $a_{1} =1,a_{2}=2$ and $a_n a_{n+2}=a_{n+1}^2+1(n\geq 1)$
Proof. $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant for $n\geq 1$
I tried (Mathematical In... | The formula $$\frac{a_n^2+a^2_{n+1}+1}{a_na_{n+2}}=3$$ holds for $n=1,2$. Assume it is true for $k$ and show it is true for $k+1$. Here is another approach:
$$
\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}a_{k+2}}\cdot\frac{a_{k}}{a_{k}}
=\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k}a_{k+2}}\cdot\frac{a_{k}}{a_{k+1}}
=\frac{a_{k+1}^2+a^2_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$ I have a few students that are having trouble understanding that
$$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$
specifically that
$$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-... | I would suggest including more terms at the end of the products. That makes it easier to see the pattern and to see why it's just $(2n+1)$ that's left at the end.
$$
\require{cancel}
\begin{align}
\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} &=
\frac{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove... | You can rearrange it into two possible ways.
Show, by left side, that
$$\frac{x^3-y^3}{x-y} = x^2+xy+y^2,$$
or
$$\frac{x^3-y^3}{x^2+xy+y^2} = x-y.$$
You may read about "Long Division of Polynomials". See also LINK for knowing the process.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 9
} |
Solve $16x^{-3}=-2$
Solve $16x^{-3}=-2$.
My working:
\begin{align}
16x^{-3}&=-2\\
\frac{1}{16x^{3}}&=-2\\
\frac{16x^3}{16x^3}&=-32x^3\\
1&=-32x^{3}\\
-32x^{3}&=1\\
-32x&=\sqrt[3]{1}\\
-32x&=1\\
x&=\frac{-1}{32}
\end{align}
Is this right? What have I done wrong?
| Obviously you could check that $x=-\frac1{32}$ doesn't solve $16x^{-3} = -2$.
Instead you can solve as follows
$$\begin{align*}
16 x^{-3} & = -2 & \text{reciprocal } a=b \Leftrightarrow \frac1a = \frac1b \\
\Leftrightarrow \frac1{16} x^3 & = -\frac12 & \text{multiply by } 16\\
\Leftrightarrow x^3 & = -8 & \text{take c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Critical numbers of the function: $x\sqrt{5-x}$ Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$
On the interval: [-6,4]
Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.
Derivative of the function:
$$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$
$$ \frac {\fr... | Your derivative is wrong. Use the product rule to separate the $x$ from the $\sqrt{5-x}$.
$$\frac{d}{dx}\left(x\sqrt{5-x} \right)
=x\frac{d}{dx}\left(\sqrt{5-x} \right)+\sqrt{5-x}\frac{d}{dx}\left(x \right)$$
$$=x\frac{1}{2\sqrt{5-x}}(-1)+\sqrt{5-x}\cdot 1$$
$$=\frac{-x}{2\sqrt{5-x}}+\frac{2(5-x)}{2\sqrt{5-x}}$$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
chain rule derivative This is an derivatives Problem.
find the tangent to $y=\sqrt{x^2-x+5\;}\;$ at $x=5$.
$y = \boxed{\;?\;}$
What I did was first find y by plugin x into the equation. The answer is $y = 5$. Then I found the derivative of both $x^2-x+5$ and $\sqrt{x^2-x+5\;}\;$ the answers for those are $2x-1$ and $... | You're almost there. You have applied the chain rule to get the derivative. You just have to substitute at the point: $x:=5$.
$\begin{align}
y &= \sqrt{x^2-x+5}
\\[2ex]
{y}_x' & = {(x^2-x+5)}_x' \cdot {(u^{1/2})}_u'\mid_{u:=(x^2-x+5)}
\\[1ex] &= (2x-1) \cdot \tfrac 1 2 (x^2-x+5)^{-1/2}
\\[2ex]
{y}_x'|_{x:=5} & = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the interval after substitution Given this problem
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+9^{\sqrt[4]{x}+1}\geq 9^{\sqrt{x}}$$
$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+3^{2\sqrt[4]{x}+2}\geq 3^{2\sqrt{x}}\\8\cdot 3^{\sqrt{x}+\sqrt[4]{x}-2\sqrt{x}}+3^{2\sqrt[4]{x}+2-2\sqrt{x}}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+3^{2\s... | you already have
$t=3^{\sqrt[4]{x}-\sqrt{x}} \ge \dfrac{1}{9} \iff \sqrt[4]{x}-\sqrt{x} \ge -2 $
$p= \sqrt[4]{x} \ge 0, \implies p-p^2\ge -2 \iff -1\le p \le2 \cap p\ge0 \implies 0 \le p \le 2 \implies 0\le x \le 16 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Equation solving; three expressions as squares If positive integer $x,y$ satisfy,
$$2x^2+x=3y^2+y$$ then $x-y,2(x+y)+1,3(x+y)+1$ are perfect squares.
I somehow managed to prove $x-y$ is a perfect square but couldn't prove the others.
| $2x^2+x=3y^2+y\Longrightarrow (x-y)(3x+3y+1)=x^2$. Now observe if a prime $p$ divides both the factors, $x-y$ and $3x+3y+1$ on the LHS, then $p^2$ divides $x^2$ and so $p$ divides $x$.
Now, $p$ divides $x$ and $p$ divides $x-y$ implies $p$ divides $y$. But this is a contradiction as $p$ divides $3x+3y+1$. So we can co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Given the $ x+y+z =3$ Prove that $ x^2+y^2+z^2 \geq 3xyz$ Given the $ x+y+z =3$ and x, y and z are positive numbers. How to prove that $ x^2+y^2+z^2 \geq 3xyz$.
I tried many methods but I failed.
I did the AM-HM in-equality, but failed.
$\frac{\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}}{3} \geq \frac{3}{\frac{xy}{z}+... | Using AM-GM and Cauchy-Schwarz inequalities, For $x,y,z > 0 \Rightarrow 3 = x+y+z \geq 3\sqrt[3]{xyz} \Rightarrow 1 = 1^3 \geq xyz \Rightarrow 3^2 = (x+y+z)^2 \leq 3(x^2+y^2+z^2) \Rightarrow 3xyz \leq 3 \leq x^2+y^2+z^2 \Rightarrow 3xyz \leq x^2+y^2+z^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculating the limit of a sequence using Stirling's approximation I have the following limit:
$$
\lim_{n \to \infty} \frac{(2n)!\sqrt{n}}{n!^24^n}
$$
Now, in order to get somewhere further in calculating this limit, I used Stirling's approximation that led me to the following:
$$
\lim_{n \to \infty} \left[\frac{(2n)!}... | Take:
$$
lim_{n \to \infty} \frac{(2n)!\sqrt{n}}{n!^24^n} = lim_{n \to \infty} \frac{(2n)!}{n!^2} \cdot \frac{\sqrt{n}}{4^n}
$$
Recall Stirling's Approximation:
$$n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$$
Whereby:
$$(2n)! \sim \sqrt{2\pi 2n}\left(\frac {2n} e\right)^{2n} = \sqrt{4\pi n}\left(\frac {2n} e\right)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The sum of the squares is less than or equal to the square of the sums for all n. I am trying to understand this proof. Rather an important part of the proof. I have already shown this is true for $n=2$ and am assuming the $a_n$ case is true.
$$(a_1^2+a_2^2+...+a_n^2) \le (a_1+a_2+...+a_n)^2$$
Want to show that
$$(a_... | inductive step:
the claim being correct for
$$(a_1^2+a_2^2+...+a_n^2) \le (a_1+a_2+...+a_n)^2$$
implies
$$(a_1^2+a_2^2+...+a_n^2+a_{n+1}^2) \le (a_1+a_2+...+a_n+a_{n+1})^2$$
Proof
\begin{align}
a_1^2+a_2^2+...+a_n^2+a_{n+1}^2 &=(a_1^2+a_2^2+...+a_n^2)+a_{n+1}^2\\
& \leq (a_1+a_2+...+a_n)^2+a_{n+1}^2 \mbox{ (using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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How to solve $\int\frac{dx}{\sqrt{4-x^2}}$ with trigonometric substitution? The integral
$$\int\frac{dx}{\sqrt{4-x^2}}$$
I've found the variable
$$x=2\sin\theta$$
$$x^2=4\sin^2\theta$$
$$dx=2\cos\theta\,d\theta$$
Which gave me by substitution
$$\int\frac{2\cos\theta}{4-\sqrt{4\sin^2\theta}}\,d\theta$$
$$\int\frac{\cos\... | Do you need to use trigonometric substitution? Because you can solve it easier using regular substitution. First:
$$\int{\frac{dx}{\sqrt{4(1-\frac{x^2}{4})}}}=\frac{1}{2}\int{\frac{dx}{\sqrt{1-(\frac{x}{2})^2}}}$$
We use substitution $\frac{x}{2}=t ; \frac{dx}{2}=dt$.
Now we have:
$$\int{\frac{dt}{\sqrt{1-t^2}}}=arcsin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Compute the infinite product $\prod\limits_{n=2}^\infty \left(1+\frac{1}{2^n-2}\right)$ I am trying to compute the infinite product
$$
\prod\limits_{n=2}^\infty \left(1+\frac{1}{2^n-2}\right) .
$$
Wolfram Alpha says the result is $2$, but I can't seem to figure out why.
| Write
$$1 + \frac{1}{2^n - 2} = \frac{2^n - 1}{2^n - 2} = \frac{1}{2} \frac{2^n - 1}{2^{n-1} - 1}$$
so by telescoping,
$$\prod_{k = 2}^N \left(1 + \frac{1}{2^n - 2}\right) = \frac{1}{2^{N-1}}(2^N - 1) = 2 - \frac{1}{2^{N-1}}.$$
Since $\frac{1}{2^{N-1}} \to 0$ as $N \to \infty$, we have
$$\prod_{n = 2}^\infty \left(1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Solve $ab = 2(a+b)$ using modular arithmetic
I have the following equation which I want to solve where $a,b \in \mathbb{N}$:
\begin{align*}ab &= 2(a+b) \\ \iff ab &= 2a + 2b \\ \iff ab - 2a &= 2b \\ \iff a(b-2) &= 2b \\ \iff a &= \frac{2b}{b-2}\end{align*}
Therefore I must solve for $a,b$ such that all pairs $\displa... | HINT
$$a =\dfrac{2b}{b-2} = 2 + \dfrac{4}{b-2}$$
That means, for $a$ to be an integer we must have $(b-2) | 4$
So $b-2$ has to be a factor of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1184247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$ I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx. $$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this close... | It can be observed that $x^{2} + x + 1 = (x-a)(x-b)$ where $a = e^{2\pi i/3}$ and $b = e^{-2\pi i/3}$. Now
\begin{align}
I &= \int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ (x-a)(x-b) } \, dx
= \frac{1}{a-b} \, \int_{1}^{\infty} \left( \frac{1}{x-a} - \frac{1}{x-b} \right) \, (\ln(x))^{2} \, dx.
\end{align}
From Wolfram Alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
Multivariable taylor polynomial $$f(x, y) = e^{2x+xy+y^2}$$
Find the 2nd order taylor polynomial to the above function about (0,0)
The formula is:
$$P(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)+\frac 12[f_{xx}(x-a)^2+2f_{xy}(x-a)(y-b)+f_{yy}(y-b)^2]$$
$$f_x=e^{2x+xy+y^2}(2+y)$$
$$f_y=e^{2x+xy+y^2}(x+2y)$$
$$f_{xx}=e^{2x+x... | We have:
$$f(x, y) = e^{2x+xy+y^2}$$
Finding partials and evaluating them at the point $(a, b) = (0,0)$, yields:
*
*$f_x(x, y) = (y+2) e^{x y+2 x+y^2} \implies f_x(0,0) = 2$
*$f_y(x, y) = e^{x y+2 x+y^2} (x+2 y) \implies f_y(0,0) = 0$
*$f_{xx}(x, y) = (y+2)^2 e^{x y+2 x+y^2} \implies f_{xx}(0,0) = 4$
*$f_{xy}(x, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Differentiate $\,y = 9x^2 \sin x \tan x:$ Did I Solve This Correctly? I'm posting my initial work up to this point.
Criticism welcomed!
Using the formula $(fgh)' = f'gh+fg'h + fgh'$, differentiate$$y = 9x^2\sin x \tan x$$
$$\begin{align} y' &= 9\frac d{dx}(x^2)\sin x \tan x + 9x^2 \frac d{dx} (\sin x) \tan x + 9x^2... | \begin{aligned}
\frac{d}{dx}(9x^2 \sin(x) \tan(x)) & = \frac{d}{dx}[9x^2]\sin(x) \tan(x) + 9x^2 \frac{d}{dx}[\sin(x)]\tan(x)+9x^2 \sin(x) \frac{d}{dx}[\tan(x)] & \\
& = 9(2x) \sin(x) \tan(x)+9x^2 \cos(x) \tan(x) + 9x^2 \sin(x) \sec^2(x)&
\end{aligned}
and $\cos(x)\tan(x) = \sin(x)$, so:
\begin{aligned}
\frac{d}{dx}(9x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188870",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Prove that $\displaystyle\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $-1Prove that $\displaystyle\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $-1<x<1$.
So far I have got $y=\tanh^{-1}(x)\Longleftrightarrow\tanh(y)=x$ .
Differentiating, $\displaystyle\tanh^{-1}(y)=x \Rightarrow \text{... | The hyperbolic arctangent of $x$, $y=\operatorname{arctanh} x$, is the solution of $\tanh y=x$, i.e.
$$ \tanh y=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}=x,\tag{1}$$
leading to:
$$ e^{2y} = \frac{x+1}{x-1}\tag{2}$$
so that:
$$\operatorname{arctanh} x = y = \frac{1}{2}\log e^{2y} = \frac{1}{2}\,\log\left(\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
Prove: If $a$, $b$, and $c$ are consecutive integers such that $a< b < c $ then $a^3 + b^3 \neq c^3$.
My Attempt: I start with direct proof.
Let $a,b,c$ be consecutive integers and $a< b < c $, there exists a ... | $(k-1)^3+k^3=(k+1)^3\\\implies k^3=(k+1)^3-(k-1)^3\\\implies k^3=(k+1-k+1)(k^2+2k+1+k^2-1+k^2-2k+1)\\\implies k^3=2(3k^2+2)$
Let $p$ be a prime that divides $k$.
If $p$ be odd then $k^3=2(3k^2+2)\implies p\mid3k^2+2\implies p\mid 2$
Hence $k$ can't have an odd prime divisor. Therefore $k=2^t$. $(?)$
Now assume that $t>... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Infinite sequence series. Limit If $0<x<1$ and
$$A_n=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\ldots+\frac{x^{2^n}}{1-x^{2^{n+1}}},$$
then $\lim_{n\to\infty} A_n$ is
$$\text{a) }\ \dfrac{x}{1-x} \qquad\qquad \text{b) }\ \frac{1}{1-x} \qquad\qquad \text{c) }\ \frac{1}{1+x} \qquad\qquad \text{d) }\ \frac{x}{1+x}$$
How to do thi... | Using the factorization $1 - x^{2^{k+1}} = (1 - x^{2^k})(1 + x^{2^k})$ we decompose
\begin{align}&\frac{x}{1 - x^2} + \frac{x^2}{1 - x^4} + \cdots + \frac{x^{2^n}}{1 - x^{2^n}}\\
&= \left(\frac{1}{1 - x} - \frac{1}{1 - x^2}\right) + \left(\frac{1}{1 - x^2} - \frac{1}{1 - x^4}\right) + \cdots + \left(\frac{1}{1 - x^{2^n... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $ Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $
Can someone please explain to me how to go about doing this?
| $% Predefined Typography
\newcommand{\paren} [1]{\left({#1}\right)}
\newcommand{\polar}[2] {\left(#1 ~\angle~ #2\right)}
$
$$\sqrt{3}\sin(\theta) - \cos(\theta)$$
First write as $\cos() + \cos()$:
$$\sqrt{3}\sin(\theta) - \cos(\theta) = \sqrt{3}\cos\paren{\theta - \frac{\pi}2} + \cos(\theta + \pi)$$
Then add (polar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1... | $\big[$Another (standard) method$\big]$
When we have integrals of the form: $\displaystyle \int \frac{dx}{a+b\sin (x) + c\cos (x)} $ we can perform the substitution: $t=\tan (x/2)$. Then, we have: $\displaystyle \sin(x) = \frac{2t}{1+t^2} $ , $\displaystyle \cos (x) = \frac{1-t^2}{1+t^2} $ , $\displaystyle dx= \frac{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 9,
"answer_id": 6
} |
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac... | We begin with an integral representation:$$S:=\sum_{n\ge0}\frac{(-1)^n}{(2n+1)^3}=\frac12\int_0^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$The integrand is even, so$$S=\frac14\int_{-\infty}^\infty\frac{x^2e^{-x}}{1+e^{-2x}}dx.$$Or with $e^{-x}=\tan t$,$$S=\frac14\int_0^{\pi/2}\ln^2(\tan t)dt.$$Since $\int_0^{\pi/2}\tan^{2k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Proving $\tan^210^\circ+\tan^250^\circ+\tan^270^\circ=9$ I need help proving the following identity.
$$\tan^210^\circ+\tan^250^\circ+\tan^270^\circ=9$$
I am not sure if it is even true.
| If $\tan3y=\tan30^\circ$
$3y=n180^\circ+30^\circ$ where $n$ is any integer
$y=60^\circ n+10^\circ$ where $n=0,1,2$
For $n=2,y=130^\circ,\tan130^\circ=\tan(180^\circ-50^\circ)=-\tan50^\circ$
Now $\tan3y=\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}$
and consequently, $\dfrac{3\tan y-\tan^3y}{1-3\tan^2y}=\dfrac1{\sqrt3}$ as $\tan3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous Prove that $f(x) = \sqrt{x^2 + x}$ where $x \in [0, +\infty)$ is uniformly continuous.
So lets take:
${\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}^2 \, \leqslant \,\, {\mid \sqrt{x^2+x} - \sqrt{y^2+y} \mid}{\mid \sqrt{x^2+x} + \sqrt{y^2+y... | Let's work with $f$ on an interval $[a,\infty[$, with $a>0$.
$$f'(x) = \frac{2x + 1}{2\sqrt{x^2 + x}} \le \frac{2x + 1}{2x} = 1 + \frac{1}{2x} \le 1 + \frac{1}{2a} $$
Now using MVT, $f$ becomes $(1 + \frac{1}{2a})$-Lipschitz on $[a,\infty[$. But Lipschitz $\implies$ Uniformly continuous.
For $[0,a]$, use the fact that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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