Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Computing $\int \frac{1}{(1+x^3)^3}dx$ I tried various methods like trying to break it into partial fractions after factorizing, applying substitutions but couldn't think of any. How will we integrate this?
| HINT:
$$\int\dfrac{dx}{x^p(1+x^m)^n}=\int\dfrac1{mx^{p+m-1}}\cdot\dfrac{mx^{m-1}}{(1+x^m)^n}dx$$
$$=\dfrac1{mx^{p+m-1}}\int\dfrac{mx^{m-1}}{(1+x^m)^n}dx-\int\left(\dfrac{d\left(\dfrac1{mx^{p+m-1}}\right)}{dx}\cdot\int\dfrac{mx^{m-1}}{(1+x^m)^n}dx\right)dx$$
$$\implies\int\dfrac{dx}{x^p(1+x^m)^n}=-\dfrac1{m(n-1)x^{m+p-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Why is my solution of $e^{\sin x } - e^{- \sin x} - 4 = 0$ wrong? $$e^{\sin x } - e^{- \sin x} - 4 = 0$$
Substitute $e^{\sin x} = y$:
$$y - \frac{1}{y} - 4 = 0 \implies y^2 - 4y - 1 = 0$$
Solve for $y$:
$$y = 2 \pm \sqrt{5}$$
$e^{\sin x}$ can't be negative:
$$\therefore y = 2 + \sqrt{5} \implies e^{\sin x} = 2 + \sqr... | $e^{\sin x} = 2 + \sqrt{5}$ has no solutions!
In fact it is equivalent to $\sin(x)=\ln(2 + \sqrt{5})>\ln(4)>1$ which is impossible because $\sin(x)\in[-1,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove $\sum_{k=1}^n\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^n\frac{b_i^2}{a_i}$ How do I prove this?
$$\sum_{k=1}^n\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^n\frac{b_i^2}{a_i}$$
Here $a_i,b_i\in\Bbb R^+$. I guess the sum transform works, but I can't prove it.
| Here is An answer by toshihiro shimizu on AoPS:
I considered this problem long time and finally I found the following inductive solution. I'd like to know the direct solution.
We show the following stronger result by induction;
$$\begin{align*}
\sum_{k=1}^{n} & \frac{\left(b_{1}+b_{2}+\cdots+b_{k}\right)b_{k}}{a_{1}+a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\... | Hint:
Rationalize with $z=e^{ix}$ (and $dz=iz\,dx$),
$$\int \frac{dx}{\left (r+\cos x \right)^2}
=\int\frac{dz}{iz\left(r+\dfrac{z+z^{-1}}2\right)^2}
=\int\frac{4z\,dz}{i\left(z^2+2rz+1\right)^2}
=\int\frac{4(z+r-r)\,dz}{i\left((z+r)^2+1-r^2\right)^2}.
$$
This yields a term $\log({z^2+2rz+1})=\log(r+\cos x)$ and, depen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find Maclaurin series of $\frac{x^2+3}{x^2-x-6}$
Find the Maclaurin series of $\frac{x^2+3}{x^2-x-6}$.
So far I have:
$$\frac{x^2+3}{x^2-x-6}=1+\frac{x+9}{(x-3)(x+2)}=1+\frac{x+2}{(x-3)(x+2)}+\frac{7}{(x-3)(x+2)}=1-\frac{1}{(3-x))}+\frac{7}{(x-3)(x+2)}$$
How should I continue?
| As in this answer, suppose
$$
\frac{a+bx}{6+x-x^2}=\sum_{k=0}^\infty c_kx^k
$$
then
$$
\begin{align}
a+bx
&=\left(6+x-x^2\right)\sum_{k=0}^\infty c_kx^k\\
&=\sum_{k=0}^\infty c_k\left(6x^k+x^{k+1}-x^{k+2}\right)\\
&=\underbrace{\ \ \ \ 6c_0\ \ \ \ \vphantom{()}}_{a}+\underbrace{\left(6c_1+c_0\right)}_b\,x+\sum_{k=2}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this determinant equation in a simpler way
Question Statement:-
Solve the following equation
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 ... | Subtract the first row from the last and use Laplace's formula for the third row: $$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}= \begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
0 & 0 & 2 \\
\end{vmatrix} = 2(-1)^{3+3}\begin{vmatrix}
x & 2\\
4 & x
\end{vmatrix} = 2(x^2-8) = 2(x-2\sqrt{2})(x+2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
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Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $ Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $.
Attempt:
From FLT it can be concluded that either all of $ x , y , z$ and $w$ are multiples of 5 ( which is not possible since that would lead to $ x^4 + y^4 + z^4 -w^4 $ being a multiple of... | Note that fourth powers are always congruent to $0$ or $1$ mod $16$. This can be proved using case analysis. I'll list these cases:
*
*If $n$ is even, then $2 \mid n$ hence $16 \mid n^4$.
*If $n\equiv \pm1 \mod 16$, then $n^4 \equiv 1 \mod 16$
*If $n\equiv \pm3 \mod 16$, then $n^4 \equiv 3^4 \equiv 1 \mod 16$
*If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integration: $\int\frac{dx}{\sqrt{x(1-x)}}$ My teacher wrote this on the blackboard: $$\int_0^z \frac{1}{\pi \sqrt{x(1-x)}}dx= \frac{2}{\pi} \arcsin(\sqrt{z})$$
But when I try to calculate the integral:
\begin{align}
\int \frac{1}{\pi \sqrt{x(1-x)}}dx&=\int \frac{1}{\pi \sqrt{(\frac{1}{2})^2-(x-\frac{1}{2})^2}}dx\\
&= ... | Your answer should be $$\frac {1}{\pi}\arcsin(2z-1)\color{red}{+}\frac 12$$
In fact both answers are equivalent. This is because if $$\phi=\frac{2}{\pi}\arcsin(\sqrt{z}),$$ then$$\sin(\pi\phi)=2\sqrt{z}\cos(\arcsin(\sqrt{z}))=2\sqrt{z}\sqrt{1-z}\Rightarrow\phi=\frac{1}{\pi}\arcsin(2\sqrt{z-z^2})$$
Meanwhile, if $$\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let a, b, c be positive real numbers. Prove that Let a,b,c be positive real numbers. Prove that
$$\frac{a^3+b^3+c^3}{3}\geq\frac{a^2+bc}{b+c}\cdot\frac{b^2+ca}{c+a}\cdot\frac{c^2+ab}{a+b}\geq abc$$
I will post what I had solved originally, however it is unfortunately incorrect. Please help solve and/or aid in findi... | The left inequality.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $\prod\limits_{cyc}(a^2+bc)=2a^2b^2c^2+\sum\limits_{cyc}(a^3b^3+a^4bc)=8w^6+A(u,v^2)w^3+B(u,v^2)$,
$\prod\limits_{cyc}(a+b)=9uv^2-w^3$ and $a^3+b^3+c^3=27u^3-27uv^2+3w^3$, we see that
our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Show that $ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, [\arcsin{\frac{x}{a}} / \arccos{\frac{x}{a}}]= \frac{\pi^2}{8} $? I found this very interesting result:
$$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}} = \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arccos{\frac{x}{a}} = \frac{\pi^2}{8}, $$
for $a>0.$ I wond... | $$ \arcsin{\frac{x}{a}}=t $$
$$ \frac{dx}{\sqrt{a^2 - x^2}}=dt$$
$$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}}=\frac{(\arcsin{\frac{x}{a}})^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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By using a geometric series and a factorisation, compute the first three terms of this given Taylor expansion By using the geometric series
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
and the factorisation
$$\frac{1}{1-3x+2x^2}= \left(\frac{1}{1-2x}\right) \left(\frac{1}{1-x}\right)$$
compute the first three terms of the... | Alternative approach: for $x$ near zero we have that $|x(3-2x)|<1$ and
$$\frac{1}{1-3x+2x^2}=\frac{1}{1-x(3-2x)}=\sum_{k=0}^{\infty}(x(3-2x))^k\\=1+(x(3-2x))+(x(3-2x))^2+(x(3-2x))^3+o((x(3-2x))^3)\\
=1+(3x-2x^2)+(9x^2-12x^3+o(x^3))+(27x^3+o(x^3))+o(x^3)\\
=1+3x+7x^2+15x^3+o(x^3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to prove property of greatest integer function:
How can we prove that
$$\lfloor x + y \rfloor = \lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor$$
for all real $x$, where $ \lfloor x \rfloor$ denotes greatest integer less than or equal to $x$?
I was able to prove that $\lfloor x + y \rfloor = \... | If $n$ is an integer and $t$ any real number, then it is straightforward to show that$\lfloor t + n \rfloor = \lfloor t \rfloor + n.$
Therefore, since $\lfloor x \rfloor$ is an integer,
\begin{align*}
\quad &\lfloor y + x - \lfloor x \rfloor \rfloor + \lfloor x \rfloor \\
= &\lfloor y + x \rfloor - \lfloor x \rfloor +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to calculate this limit with cube-roots? How to solve this limit?
$$\lim_{x\to \pm\infty}\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}$$
I figured this is the same as:
$$\lim_{x\to \pm\infty}(\sqrt[3]{x-1}+\sqrt[3]{x+1})(\sqrt[3]{x-1}-\sqrt[3]{x+1})$$
But that didn't help much I guess...
| $$A=\sqrt[3]{(x-1)^2}-\sqrt[3]{(x+1)^2}=(x-1)^{2/3}-(x+1)^{2/3}=x^{2/3}\left( \left(1-\frac 1x \right)^{2/3}-\left(1+\frac 1x \right)^{2/3}\right)$$ Use the generalized binomial theorem or Taylor series.
$$\left(1-\frac 1x \right)^{2/3}=1-\frac{2}{3 x}-\frac{1}{9 x^2}-\frac{4}{81
x^3}+\cdots$$ $$\left(1+\frac 1x \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might... | Let us show that $\sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k}$ is Cesaro summable. Once we establish this, then this is also Abel summable and the Cesaro sum is equal the Abel sum, which is
$$ \sum_{k=1}^{\infty}(-1)^{k-1}\sqrt{k} = -\operatorname{Li}_{-1/2}(-1) = (1 - 2^{3/2})\zeta(-1/2). $$
To this end, let $s_n = \sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
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$x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ real? Question: For $x,y \in \mathbb{C}$, suppose $x+y=xy=w \in \mathbb{R}^+$. Is $x^w+y^w$ necessarily real?
For instance, if $x+y=xy=3$, then one solution is $x = \frac{3 \pm i \sqrt{3}}{2}$, $y = \frac{3 \mp i \sqrt{3}}{2}$, but $x^3 + y^3 = 0$, which is real.
I've checked ... | Yes. Write $x=a+bi$, $y=c-di$, then clearly $b=d$ because $x+y$ is real.
So $x=a+bi$, $y=c-bi$. Then $xy=ac-abi+cbi+b^2$, so $a=c$ or $b=0$.
*
*If $a=c$, then $y = \overline{x}$, i.e. the complex conjugate of $x$.
So $x^w+y^w = x^w+(\overline{x})^w = x^w+\overline{x^w}$, which is
real.
*If $b=0$, $x$ and $y$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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In how many ways can I form a set of 6 elements made up of 0s, 1s, 2s, 3s, 4s and 5s such that the sum of its elements is a multiple of 6? I have a set $ S $ which is made up of 6 elements, each one being either 0, 1, 2, 3, 4 or 5 (note that repetition is allowed). I want to evaluate the following problems:
a) In how m... | As a set, $\{1,1,1,1,1,1\}=\{1\}$, hence the sum of its elements is $1$, not $6$.
If the actual problem is
In how may ways we may choose $x_1,x_2,x_3,x_4,x_5,x_6$ from
$\{0,1,2,3,4,5\}$ in such a way that $x_1+x_2+x_3+x_4+x_5+x_6$ is a
multiple of $6$?
Then the answer is just $\frac{6^6}{6}=\color{red}{6^5}$, si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exact Differential Equation - possible textbook mistake - $(3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0$ and $y(0)=2$ When solving the differential equation
$$
(3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0
$$
I verified that it is an Exact Equation. By saying that
$$
F_x = 3x^2+6xy^2\\
F_y = 6x^2y+4y^2
$$
and integrating both sides I f... | Your book solution is not correct $${ x }^{ 3 }+3{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 3 }=16\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+6{ x }^{ 2 }y{ y }^{ \prime }+3{ y }^{ 2 }{ y }^{ \prime }=0\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+\frac { dy }{ dx } \left( 6{ x }^{ 2 }y+3{ y }^{ 2 } \right) =0\\ \left( 3{ x }^{ 2 }+6{ x }{ y }^{ 2 } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $
$$\left(\frac{n}{2}\right)^n \gt n! \gt \left(\frac{n}{3}\right)^n \qquad n\ge 6 $$
I tried it prove it by mathematical induction but failed .
For $n=6$
$$(3)^6 \gt 6!$$
Now for $n=k$
$$\left(\frac{k}{2}\right)^k \gt k!$$
No... | $$\begin{align}
\left(\frac{k+1}2\right)^{k+1}&=\frac {k+1}2\left(\frac {k+1}2\right)^k\\
&>\frac{k+1}2\left[\left(\frac k2\right)^k+\binom k1\left(\frac k2\right)^{k-1}\frac12\right]\\
&>\frac{k+1}2[k!+k!]\\
&=(k+1)!
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove inequality $\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$ I was trying to figure out if the following function can serve as a mean (see mean value theorem):
$$\arccos \left( \frac{\sin y-\sin x}{y-x} \right)$$
And turns out that for $x,y \leq \pi$ it does serve as a mean admirabl... | This is not a full answer, just a possible way to prove the inequality.
We use the following form of the inequality:
$$\text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$
It makes sense to try infinite products, mainly because we get rid of the square root:
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 4,
"answer_id": 2
} |
Generating function for cubes of Harmonic numbers By generalizing the approach in Integral involving a dilogarithm versus an Euler sum. meaning by using the integral representation of the harmonic numbers and by computing a three dimensional integral over a unit cube analytically we have found the generating function ... | By using the functional equations for the trilogarithm we simplified the result as follows:
\begin{eqnarray}
&&S^{(3)}(x)= \\
&&\frac{ \text{Li}_3(x)}{(1-x)}+
3\frac{\text{Li}_3(1-x)-\zeta (3)}{(1-x)}+
\log(1-x)\frac{ \left(-2 \log ^2(1-x)+3 \log (x) \log(1-x)-\pi ^2 \right)}{2 (1-x)}
\end{eqnarray}
For a sanity check ... | {
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"url": "https://math.stackexchange.com/questions/1903520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equiv... |
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\begin{align}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&<2\tag{1}\label{1}.\end{align}
Let $\rho,\ r,\ R$
be the semiperimeter,
inradius and circumradius
of the triangle $ABC$, then
we can rewrite \eqref{1} as
\begin{align}
\frac a{2\rho-a}
+\frac b{2... | {
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"url": "https://math.stackexchange.com/questions/1903775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Max profit question, regarding negative exponents with factors $p= 10q^{-0,5}$
$C(q) = 5q$
In a monopoly the equation for maximum profits is: $p'(q)\cdot q + p(q) - C'(q) = 0$
First order condition:
$-0,5 \cdot 10q^{-1,5} \cdot q + 10q^{-0,5} - q = 0$
can't figure out onwards from here. How do I do with these negative... | As stated above in a comment I believe the correct equation for monopoly profit optimization should be $(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0$. We will solve this equation for $q$ below:
$(-.5)\cdot 10q^{-1.5}\cdot q + 10q^{-.5} - 5 = 0 \Rightarrow \frac{-5}{\sqrt{q}} + \frac{10}{\sqrt{q}} - 5 = 0$
Here we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full ... | From the condition by AM-GM we obtain:
$$\prod\limits_{cyc}\frac{a}{1+a}=\prod\limits_{cyc}\left(\frac{1}{1+b}+\frac{1}{1+c}\right)\geq\prod\limits_{cyc}\frac{2}{\sqrt{(1+b)(1+c)}}=8\prod\limits_{cyc}\frac{1}{1+a}$$
Id est, $abc\geq8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
show that $\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$ Let $a,c>0$ show that
$$\dfrac{c^2}{a}+\dfrac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$$
It seem AM-GM inequality.How?
| Write $a=b^2$ and $c=d^2$, and multiply through by $b^2d^2$. The inequality becomes
$$b^6+d^6+16b^3d^3\geq 9b^4d^2+9b^2d^4.$$
Dividing by $d^6$ and writing $x=\frac{b}{d}$:
$$x^6+1+16x^3\geq 9x^4+9x^2.$$
This follows from
$x^6+16x^3+1-9x^4-9x^2=(x-1)^4(x^2+4x+1)\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
how I can simplify this limit? So I got $f(x) = 1+\sin^2(x)-e^{x^{2}}$
and the limit :
$$\lim_{x\to 0^+} \cfrac{\ln(1+6x)f(x)}{x^a\cos(f(x))}$$
I have to simplify this limit :(in a form like $\lim_{x\to o} x^{n-a}$ , or something similar)
*
*I noticed that $\cos(f(x))$ for x->$0$ = $1$
*I can see => $\ln(1+6x)$... | By using Taylor expansions, we have that as $x\to 0^+$,
$$f(x) = 1+\sin^2(x)-e^{x^{2}}=1+(x-x^3/6+O(x^5))^2-(1+x^2+x^4/2+O(x^6))\\=
1+x^2-x^4/3-1-x^2-x^4/2+O(x^5)=-5x^4/6+O(x^5).$$
Hence
$$\frac{\ln(1+6x)\cos(f(x)}{x^af(x)}=\frac{(6x+O(x^2))\cos(-5x^4/6+O(x^5))}{x^a(-5x^4/6+O(x^5))}\\
=\frac{(6x+O(x^2))\cdot(1+O(x^8))}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Taylor/Maclaurin series type question I want to express $\cos(z)$ in a Taylor series centred on $z_0=\frac{\pi}{4}$.
Using the formula $\sum_{k=0}^{\infty} \dfrac{f^{(k)}(z_0)z^k}{k!}
$ and the fact that $f^{(k)}(z)=\cos(z+\frac{k\pi}{2})$ in this case I found that the Taylor series is $$\sum_{k=0}^{\infty} \frac{\cos(... | It is convenient to continue your calculation one line before the last line.
Note, that
\begin{align*}
\cos(z)=\frac{1}{\sqrt{2}}
\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(z-\frac{\pi}{4}\right)^{2k}
- \frac{1}{\sqrt{2}}
\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\left(z-\frac{\pi}{4}\right)^{2k+1}\tag{1}
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a sq... | $$\left( \frac { \sqrt { x } }{ \sqrt { y } } -\frac { \sqrt { y } }{ \sqrt { x } } \right) ^{ 2 }\ge 0$$
$$\Rightarrow\frac{ { x } }{ { y } }+\frac { { y } }{ { x } } \ge 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Shank's Baby-Step Giant-Step for $3^x \equiv 2 \pmod{29}$ Problem: Solve $3^x \equiv 2 \pmod{29}$ using Shank's Baby-Step Giant-Step method.
I choose $k=6$ and calculated $3^i \pmod {29}$ for $i=1,2,...,6$.
$$3^1 \equiv 3 \pmod {29}$$
$$3^2 \equiv 9 \pmod {29}$$
$$3^3 \equiv 27 \pmod {29}$$
$$3^4 \equiv 23 \pmod {29}$$... | I found flaw in my calculus for the second expression in second list, it should be
$$ 2 \cdot 3^{-12} = 2 \cdot (3^{-6})^2 \equiv 2 \cdot 22^2 = 2 \cdot 484 \equiv 11 \pmod {29}$$
Hence, $x = 5 + 12 = 17 $ which is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Woodbury formula problem Hello guys i have this linear algebra problem and i want some help on how i should proceed with it. i think the aim is to use the Woodbury formula
Suppose $u,v \in \mathbb{R}^n, \quad A \in \mathbb{R}^{n*n}$ is invertible and consider for scalars $\alpha , \beta \in \mathbb{R} : \\$ $B(\alpha,\... | We start with recalling two standard results:
*
*If $A$ is invertible then $\begin{pmatrix} A & B \\ C & D \end{pmatrix}$ is invertible iff $D-CA^{-1}B$ is invertible. This follows directly from $ \begin{pmatrix} I & 0 \\ -CA^{-1} & I \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} I & -A^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Unable to justify solution for a problem with exponential constant $e$ involved. For the question,
$e^{2x} + e^x - 2 = 0.$
I was asked to solve for $x$.
What I performed,
$e^{2x} + e^x = 2.$
$e^x(e^x + 1) = 2.$
For the 2 solutions involved,
\begin{align}
e^x &= 2\\
\ln2 &= x\\
.69 &= x
\end{align}
OR
\begin{align}
e^x... | $e^x (e^x+1) = 2$ does not imply that $e^x = 2$ or $e^x+1 = 2$. This type of reasoning only works if the right-hand side is zero. (This is why it's called the zero-product property.)
In general, if $ab = 2$ then we could have $a=2$ and $b=1$, or $a=1$ and $b=2$, or $a=4$ and $b=1/2$, or $a=2\pi$ and $b=\frac1\pi$, et... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove inequality $ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$ for $a,b,c\ge 0$ Prove that:
$$ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$$
I already know that this can be proven using Cauchy Schwarz, but I don't really see how to apply it here. I'm looking for hints.
| Expanding the two sides and cancelling like terms gives
$$a^7b+ab^7+b^7c+bc^7+c^7a+ca^7\geq a^5b^3+a^3b^5+b^5c^3+b^3c^5+c^5a^3+c^3a^5.$$
This follows from Muirhead's inequality, where we set $n=2, a_1=5,a_2=3,b_1=7$, and $b_2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right... | You can try this approach: Calculate sum of first $k$ odd numbers, namely from 1 to $2k-1$.
Denoting it by $S_{k}$. By induction prove this is $k^2$.
Now the desired sum is $S_{2n}-S_n= 4n^2-n^2=3n^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet... | First, let's make a common denominator of $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}$.
Camera? Action!
$$\require{cancel}\begin{aligned}\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}&=\frac{\sin^4{\theta}+\cos^4{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{\left(\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Prove that $\binom{k}{p} \equiv \left\lfloor \frac{k}{p} \right\rfloor \pmod p$ for all odd prime number $p$ and $k\ge 3$ I don't really know how to start this exercise. Do I have to use p-adic valuation ?
If it's the case it will give $\nu_p(\binom{k}{p})= \sum \limits_{r=1}^{\infty}\left(\left\lfloor \frac{k}{p^r} \r... | I try to explain better my previous comment on induction's argument:
let's suppose $\binom{k}{p}=n$, so that $np\leq k <np+p$;
it's easy to prove $\binom{k+1}{p}=\frac{k+1}{k+1-p}\binom{k}{p}\equiv 1\left\lfloor \frac{k}{p} \right\rfloor\pmod{p}$,
if $k+1\not\equiv 0\pmod p$, and in this case we are done since $k+1<pn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Is there a general rule for how to write high order polynomials in matrix form?
Is there a general rule for how to write high order polynomials in matrix form?
For example a linear combination of parameters:
$$a_1 x_1+a_2 x_2+a_3 x_3 + \cdots+ a_n x_n$$
Can be written as
$$\sum^n_{i=1} a_i x_i = \vec{a}^T\vec{x} $$
... | Linear Form
$$
\begin{pmatrix}
a & b \\
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}=
ax+by$$
Quadratic Form
$$
\begin{pmatrix} x & y \end{pmatrix}
\begin{pmatrix}
a & b \\
b & c
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}=
ax^2+2bxy+cy^2$$
Cubic Form
$$
\begin{pmatrix} x & y \end{pmatrix}
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Use De Moivre’s formula to show that $\cos^2 20^{o}+\cos^2 40^{o}+\cos^2 60^{o}+\cos^2 80^{o}=\frac{7}{4}$ I would appreciate if somebody could help me with the following problem.
Q: Use De Moivre’s formula to show that
$$
\cos^2 20^{o}+\cos^2 40^{o}+\cos^2 60^{o}+\cos^2 80^{o}=\frac{7}{4}
$$
I tried to let $z=\cos 20... | Set $z=\cos20^{\circ}+i\sin20^{\circ}$. Then,
$$
z^{2} = \cos40^{\circ}+i\sin40^{\circ}, \quad z^{3} = \cos60^{\circ}+i\sin60^{\circ}, \quad z^{4} = \cos80^{\circ}+i\sin80^{\circ}
$$
and
$$
z^{-1} = \cos20^{\circ}-i\sin20^{\circ}, \qquad z^{-2} = \cos40^{\circ}-i\sin40^{\circ}, \quad z^{-3} = \cos60^{\circ}-i\sin60^{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| What are the roots of $x^2-2x-3$? They should also be roots in the 3rd degree polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 1
} |
The value of \zeta(4) with fourier series Given a function as follows :
$f(x) = \pi^2 - x ^2$ on $|x|<\pi$ and $f(x+2\pi)=f(x)$
I did find its Fourier expansion
$f(x)=\frac{2}{3}\pi^2$+$\sum_{n \geq 1}\frac{4}{n^2}(-1)^{n+1}cosx$
And by putting $x=\pi$, I got the zeta of 2 , $\zeta(2) = \sum_{n \geq 1}\frac{1}{n^2} = \... | $f(x) = \pi^4-x^4$, $f(x+2 \pi)=f(x)$
$$f(x) = \sum_{n=0}^{\infty} a_n \cos{n x} $$
$$a_0 = \frac1{2 \pi} \int_{-\pi}^{\pi} dx \, (\pi^4-x^4) = \frac{4 \pi^4}{5}$$
$$a_n = \frac1{\pi} \int_{-\pi}^{\pi} dx \, (\pi^4-x^4) \cos{n x} = (-1)^{n+1} \frac{8 (n^2 \pi^2-6)}{n^4}$$
Thus,
$$f(x) = \frac{4 \pi^4}{5} - 8 \sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Where am I going wrong when attempting to evaluate the integral $\int { 3tan^{ 2 }(x)sec(x)dx } $ Evaluate the integral:
$$\int 3\tan^2(x)\sec(x) \,dx $$
We begin by taking the constant outside of the integral to get:
$$3\int \tan^2(x)\sec(x)\,dx $$
Next, we use the identity $\tan^2 (x)=\sec^2 (x)-1$ in order to get:
$... | It seems that you forgot to multiply $\int\sec^3(x)dx$ by $3$ :
$$\begin{align}&\int 3\tan^2(x)\sec(x) dx\\&=3\int \sec^3(x)dx-3\int \sec (x)dx\\&=\color{red}{3}\cdot \frac {\sec(x)\tan(x)+\ln|\sec(x)+\tan(x)|}{2}-3\ln|\sec(x)+\tan(x)|+C\\&=\frac 32\sec(x)\tan(x)-\frac 32\ln|\sec(x)+\tan(x)|+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof of the inequality $2^{n} < \binom{2n}{n} < 2^{2n}$? As review for a midterm I am asked to prove the inequality:
$2^{n} < \binom{2n}{n} < 2^{2n}, n > 1.$
What I have is a two-part inductive proof. It is not hard to show for $2^{n} < \binom{2n}{n}$:
Base step:
Let $n=2$:
$2^{2} < \frac{(2n)!}{2!2!} < 2^{2*2}$
$4 < ... | Another approach. Using the Stirling's approximation we have $$\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}\leq n!\leq n^{n+\frac{1}{2}}e^{1-n}
$$ hence $$\dbinom{2n}{n}=\frac{\left(2n\right)!}{n!^{2}}\leq\frac{e}{\sqrt{2}\pi n^{1/2}}4^{n}<\color{red}{4^{n}}
$$ and $$\frac{\left(2n\right)!}{n!^{2}}\geq\frac{\sqrt{\pi}2^{2n+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find the average rate of change between two points on a contour map Disclaimer: This a homework question for a multivariable calculus course.
The problem:
Find the average rate of change between $A$ and $C$ using the given contour map.
The average rate of change for a contour map is given by $\frac{\Delta altitude}{\D... | The textbook answer is incorrect. The correct solution is:
$\Delta altitude = -9 - (-3) = -6$.
$\Delta horizontal = \sqrt{(6-2)^2 + (5-4)^2} = \sqrt{17}$
$\frac{\Delta altitude}{\Delta horizontal}=\frac{-6}{\sqrt{17}}=\frac{-6\sqrt{17}}{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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integral points on a simple elliptic curve I'm trying to find all integer points on an elliptic curve $y^2=x(x^2-2x-3)$ or $y^2=x(x+1)(x-3) $. I guess there are only 3 integer points: $(-1,0), (0,0), (3,0) $. It looks like quite an elementary number theory problem, but I couldn't think up any proof... How could I prove... | Here's an elementary argument. Suppose none of $x$, $x+1$, or $x-3$ is $0$, and that their product is a perfect square. Note that for any prime $p\ge5$, $p$ can divide at most one of $x$, $x+1$, and $x-3$. Thus the only primes that could have an odd power in the prime factorization of any of these three numbers are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$... | Define $a := \cos^2 A$, and $b := \cos^2 B$, and $c := \cos^2 C$. Then
$$\cos A = \tan B \quad\to\quad \cos^2 A = \tan^2 B = \sec^2 B - 1 \quad\to\quad a = \frac{1}{b} - 1 \quad\to\quad b = \frac{1}{1+a}$$
Likewise,
$$c = \frac{1}{1+b} \qquad\text{and}\qquad a = \frac{1}{1+c}$$
so that, adding a slightly-gratuitous $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
| If $(4\cos2A+1)^2=9-8\cos A,$
$9-8\cos A=16\cos^22A+8\cos2A+1=8(1+\cos4A)+8\cos2A+1$
$\iff0=\cos A+\cos2A+\cos4A$
$=\cos A+2\cos A\cos3A=\cos A(1+2\cos3A)$
If $\cos A=0,A=(2n+1)90^\circ$ where $n$ is any integer
Otherwise, $$\cos3A=-\dfrac12=\cos120^\circ$$
$\implies3A=360^\circ m\pm120^\circ$ where $m$ is any integer
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$
If $A=\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}$ and $B = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ and $a,b,c>0.$ Then prove that $A\geq 1+B$
$\bf{My\; Try::}... | we need to prove that
$$\frac{3(1+a)(1+b)(1+c)}{\sum\limits_{cyc}(ab+2a+1)}\geq1+\frac{3abc}{ab+ac+bc}$$ or
$$\sum\limits_{cyc}(2a^2b^2-2a^2bc+a^2b+a^2c-2abc)\geq0$$ or
$$\sum\limits_{cyc}(a-b)^2(c^2+c)\geq0$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/1939188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Why is cross product defined in the way that it is?
$\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector? Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?
So why is cross product defined in the way that it... | As far as $\sin \theta$ and $\cos \theta$ are concerned,
Using the law of cosines,
\begin{align}
\|\overrightarrow{v_2}\|^2 + \|\overrightarrow{v_1}\|^2
-2\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \cos \theta
&= \|\overrightarrow{v_2} - \overrightarrow{v_1}\|^2
\\
2x_1 x_2 + 2y_1 y_2 + 2z_1 z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 13,
"answer_id": 6
} |
Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopita... | This answer does not use L'Hopital (personal taste), only a standard identity restated below, the binomial theorem, and a straightforward Taylor expansion to first order at $0$.
Using the identity $1-x^{2n+1} = (1-x)\sum_{k=0}^{2n} x^k$, we can rewrite
$$\begin{align*}
\frac{23}{1-x^{23}} - \frac{11}{1-x^{11}}
&= \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 2
} |
Trigonometry find area of triangle Triangle ABC is right angeled with right angle at corner C and angle a at corner A. Calculate triangle area if we know that c=|AB|=10 and tan a = 3/2
I get side |AC| = 2
side |CB| = 3
and side |AB| = 10 but this can not be case because it's not pythagoras definition then
Greetings fr... | From the tangent of $a$, you know that $\frac{BC}{AC} = \frac{3}{2}$.
So let $BC = 3x$ and $AC = 2x$
By Pythagoras, $AC^2 + BC^2 = AB^2 \implies 13x^2 = 100 \implies x^2 = \frac{100}{13}$
Area of triangle = $\frac{1}{2} AC \cdot BC = 3x^2 = \frac{300}{13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fourier series of $f(x)=\pi-x$ $$f(x)=\pi-x \qquad x \in [0,2 \pi[$$
$$a_0=\frac{1}{\pi} \ \int_0^{2 \pi}f(x) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi} (\pi-x) \ dx=0$$
$$a_n=\frac{1}{\pi} \ \int_0^{2 \pi} \cos(nx) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi}(\pi \ \cos(nx)-x \ \cos(nx)) \ dx=$$
$$\frac{1}{\pi} \ \Big( \ \Big[\frac... | Slight errors:
$$
\begin{align}
%
a_{0} &= \frac{1}{\pi} \int_{-\pi }^{\pi } f(x) \, dx = 2\pi \\[5pt]
%
b_{k} &= \frac{1}{\pi } \int_{-\pi }^{\pi } f(x) \sin (k x) \, dx = (-1)^k \frac{2}{k \pi}
%
\end{align}
$$
The decay of the amplitudes is linear:
The series expansion looks like
$$
\pi - x = \pi - 2 \sin (x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplify this expression fully How would i simplify fully the following expression?
$\dfrac{{\sqrt 2}({x^3})}{\sqrt{\frac {32}{x^2}}}$
So far i have got this
$\dfrac{{\sqrt 2}{x^3}}{{\frac{\sqrt 32}{\sqrt x^2}}}$ = $\dfrac{{\sqrt 2}{x^3}}{{\frac{4\sqrt 2}{x}}}$
Am not quite sure if this is correct however, could someon... | $\frac{\sqrt{2}x^3}{\sqrt{\frac{32}{x^2}}} = \frac{\sqrt{2}x^3}{\frac{\sqrt{32}}{\sqrt{x^2}}} = \frac{\sqrt{2}x^3}{\frac{4\sqrt{2}}{x}} = \frac{x^3}{\frac{4}{x}} = \frac{x^4}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Limit $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$ $ \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1}$
How would I solve this limit question? Thanks in advance.
| Another possible way to do it.
Define $x=1+y$ which makes $$A=\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt[3]{1+y}-1}{\sqrt{1+y}-1}$$ and use the generalized binomial theorem or Taylor series which write $$(1+y)^a=1+a y+\frac{1}{2} a(a-1) y^2+O\left(y^3\right)$$ which then gives $$A=\frac{1+\frac{y}{3}-\frac{y^2}{9}+O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$ Solve in positive integers the system of equations:
$$\begin{cases}
2x^2=4y^2+3z^2+2;
\\
13x=4y+3z+29
\end{cases}$$
My work so far:
I used wolframalpha: $x=3,y=1,z=2$.
| $$2x^2=4y^2+3z^2+2\tag1$$
$$13x=4y+3z+29\tag2$$
(1)
$$2x^2=4y^2+3z^2+2 \Rightarrow x>\max\{y,z\}$$
(2)
$$13x=4y+3z+29<4x+3x+29 \Rightarrow 6x<29 \Rightarrow x<5$$
(1) $z -$ even. (2) $x -$ odd. Hence $x \in \{1,3\}$. But $x>y,z\ge1$. Hence $$x=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and $f\left(x+f(x+2y)\right)=f(2x)+f(2y)$
Find all functions $f:\mathbb Z \rightarrow \mathbb Z$ such that $f(0)=2$ and
$$f\left(x+f(x+2y)\right)=f(2x)+f(2y)$$
for all $x \in \mathbb Z$ and $y \in \mathbb Z$
My work so far:
1) $x=0$ $$f\lef... | Using your results, we find that
$$\tag1f(x+f(x+2y))=2x+2y+4\qquad\text{for }x,y\ge0 $$
In particular,
$$\tag2 f(x+f(x))=2x+4\qquad\text{for }x\ge0$$
Let $S=\{\,k\in\Bbb Z\mid f(2k)=2k+2\,\}$. You essentially showed that $k\in S\implies k+1\in S$ and hence from the given $0\in S$, we have $\Bbb N_0\subseteq S$.
With $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The function $f(n)$ is defined for all integers $n$, such that $f(x) + f(y) = f(x + y) - 2xy - 1$ for all integers $x$ and $y$ and $f(1) = 1$ The function $f(n)$ is defined for all integers $n$, such that $f(x) + f(y) = f(x + y) - 2xy - 1$ for all integers $x$ and $y$ and $f(1) = 1$. Find $f(n)$.
I started plugging sm... | We find the first few values of $f(n)$.
Setting $x = y = 0$, we get $2f(0) = f(0) - 1$, so $f(0)= -1$.
Setting $y = 1$, we get
[f(x) + f(1) = f(x + 1) - 2x - 1,]
so $f(x + 1) = f(x) + 2x + f(1) + 1 = f(x) + 2x + 2$ for all integers $x$.
Then
\begin{align*}
f(2) &= f(1) + 2 \cdot 1 + 2, \\
f(3) &= f(2) + 2 \cdot 2 + 2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the matrix equation $\sin(X)=\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$ Solve the equation $\sin(X)=\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$, where $X\in M_2(\Bbb C)$ and $a\in \Bbb C$.
I discussed the case whether $X$ is diagonalisable or not.
If $X$ is diagonalisable, we have $X\sim \begin{pmatrix}x & 0\\0 & y\... | As you correctly infer, there is no diagonalizable solution $X$.
Let $A = \left(\begin{smallmatrix}1&a\\0&1 \end{smallmatrix} \right)$.
Notably, we compute
$$
\sin\pmatrix{\lambda&1\\0&\lambda} = \pmatrix{\sin\lambda &\cos\lambda \\0&\sin \lambda}=:M(\lambda)
$$
and so, $\sin(X) = A$ has a solution if and only if there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Tricky Limits Problem () Problem:
If $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=0$ then find the value of $3a+b$.
My attempt: $\lim_{x \to 0}{\sin2x\over x^3}+a+{b\over x^2}=\lim_{x \to 0}{\sin2x\over 2x}({2\over x^2})+a+{b\over x^2}={2+b+ax^2\over x^2}$.From this we can conclude that $a=0$ and $b=-2$, hence $3a+b... | We have that
\begin{align}
\lim_{x \rightarrow 0} \frac{\sin 2x}{x^3} + a + \frac{b}{x^2}
& = \lim_{x \rightarrow 0} \ a + \frac{\sin 2x + bx}{x^3} \\
& = \lim_{x \rightarrow 0} \ a + \frac{2 \cos 2x + b}{3x^2},
\end{align}
which gives that $b=-2$, otherwise the second term blows up. Continuing,
\begin{align}
\lim_{x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do you find the area of a parallelogram with the vertices? How do you find the area of a parallelogram with the following vertices; $A(4,2)$, $B(8,4)$, $C(9,6)$ and $D(13,8)$.
| For this, we plan to use the Shoelace formula.
Shoelace Formula: Given the coordinates of vertices of a polygon, its area is found by $$A=\frac 12\left|\sum_{i=1}^{n-1}x_iy_{i+1}+x_ny_1-\sum_{i=1}^{n-1}x_{i+1}y_i-x_1y_n\right|$$
Or, in other words, we have $$A=\frac 12|x_1y_2+x_2y_3+\ldots x_{n-1}y_n+x_ny_1-x_2y_1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Find all functions $f(x)$ such that $f\left(x^2+f(y)\right)$=$y+(f(x))^2$ Let $\mathbb R$ denote the set of all real numbers. Find all function $f: R\to \ R$ such that $$f\left(x^2+f(y)\right)=y+(f(x))^2$$ It is the problem. I tried to it by putting many at the place of $x$ and $y$ but I can't proceed. Please someb... | Observe
*
*$f(x^2+f(0)) = [f(x)]^2$
*$f(f(x)) = x+[f(0)]^2$.
Plugging $0$ into expression 1 yields
\begin{align}
f_2(0):=f(f(0)) = [f(0)]^2
\end{align}
and plugging $f(0)$ into expression 1 yields
\begin{align}
f(f(0)^2+f(0)) =[f_2(0)]^2 = [f(0)]^4.
\end{align}
Moreover, by expression 2, we have
\begin{align}
f(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Verify that $\int_0^{\pi/2} \frac{{\rm d}\theta}{(1-m^2\cos^2{\theta})^2}= \frac{(2-m^2)\pi}{4(1-m^2)^{3/2}}$ for $0I am told that the integral
\begin{align}
I&=\int_0^{\pi/2} \frac{1}{(1-m^2\cos^2{\theta})^2}\,{\rm d}\theta\\
&= \frac{(2-m^2)\pi}{4(1-m^2)^{3/2}}
\end{align}
Where $0<m<1$. I want to verify this however... | Let $t = \tan \theta = \frac {\sin \theta} {\cos \theta}$. Combining this with the identity $\cos^2 \theta + \sin^2 \theta = 1$ and keeping in mind that $\theta \in [0, \frac \pi 2]$ (where $\cos \theta \ge 0$), we obtain $\cos^2 \theta = \frac 1 {1+t^2}$. With this substitution the integral becomes
$$\int \limits _0 ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\int^{\pi}_0 \ln \left(\frac{b-\cos x}{a- \cos x}\right)dx=\pi \ln (\frac{b+\sqrt{b^2-1}}{a+\sqrt{a^2-1}})$ Given for $\alpha>1$
$$\int^{\pi}_0 \frac{dx}{\alpha-\cos x}=\frac{\pi}{\sqrt{\alpha^2-1}}$$
Prove that $$\int^{\pi}_0 \ln \left(\frac{b-\cos x}{a- \cos x}\right)dx=\pi \left(\frac{b+\sqrt{b^2-1}}{a+\... | First, notice the integrand can be written as
$$
\begin{align}
\ln (\frac{b-\cos x}{a- \cos x}) &=\ln(b-\cos x)-\ln(a-\cos x)\\
& =\ln(y-\cos x)|^{y=b}_{y=a}\\
& =\int^b_a\frac{1}{y-\cos x}dy
\end{align}
$$
So the integration equals
$$
\begin{align}
\int^{\pi}_0 \ln (\frac{b-\cos x}{a- \cos x})dx &=\int^{\pi}_0 [ \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Formula for $\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}$ Supposedly, the infinitely nested radical $$\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}\tag1\label{1}$$
converges to $$\frac {A-1}{6}+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt 3}\right)\tag2$$
where $A=\sqrt{4a-7}$. In fact, that's how Ramanujan arrived at the... | Here's how I duplicated
Ramanujan's result,
with a lot of help from Wolfy.
Start with
$x=\sqrt{a-\sqrt{a+\sqrt{a+x}}}
$.
Squaring,
$x^2-a
=-\sqrt{a+\sqrt{a+x}}
$.
Squaring again,
$(x^2-a)^2
=a+\sqrt{a+x}
$
or
$(x^2-a)^2-a
=\sqrt{a+x}
$.
A final squaring gives
$((x^2-a)^2-a)^2-a
=x
$.
Setting
$a=2$
gives
$x
=((x^2-2)^2-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Closed form of a sum $ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$ Consider a sum:
$$ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$$
with $x$ and $y$ being (non-zero) constants. Is it possible to obtain a nice closed form of this expression?
| ------ Complementing Jack's answer -----
Actually, since we have the formula for the infinite summation
$$
S\left( {x,y} \right) = \sum\limits_{0\, \leqslant \,k} {\;\frac{1}
{{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \frac{1}
{{2y^{\,2} }} + \frac{{\pi \coth \left( {\pi y/x} \right)}}
{{2xy}}
$$
then, for a par... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How should I calculate the determinant? $\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|=
\left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+
\left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$
I tried to calculate the determina... | $${\begin{vmatrix}1&a & b &c+d\\1 &b &c &d+a \\1 &c &d &a+b\\1&d &a &b+c &\end{vmatrix}} \space c_2+c_3+c_4 \to c_4 \\
{\begin{vmatrix}1&a & b &a+b+c+d\\1 &b &c &a+b+c+d \\1 &c &d &a+b+c+d\\1&d &a &a+b+c+d &\end{vmatrix}} \space factor \space (a+b+c+d)=\\(a+b+c+d)
{\begin{vmatrix}1&a & b &1\\1 &b &c &1 \\1 &c &d &1\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
y'' + y = -sin(x) $y'' + y = -\sin(x)$
$y(0) = 0 $
$y'(0) = 0$
I first solved for the homogeneous solution to get:
$y(x) = c_1 \sin(x) + c_2 \cos(x)$
then took the derivative of that:
$y'(x) = c_1 \cos (x) - c_2 \sin(x)$
This is where I am not sure where to go...
$y(0) = c_1 \sin(0) + c_2 \cos(0)
= 0 + c_2 $
so ... | $y'' + y = -\sin x$
solve the homogeneous equation
$y = c_1 \sin t + c_2 \cos x + y_p$
Undetermined coefficients.
$y = A x\sin x + Bx\cos x$
Why $A x \sin x + B x\cos x,$ and not the more simple-minded $A \sin x + B \cos x$?
$A \sin x + B \cos x$ is already part of the homogeneous solution. So, when we plug it into... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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find partial fraction $(2x^2-1)/((x^2-1)(2 x^2+3))$
Find partial fraction of $\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}$?
My attempt:
I did google and I tried to solved it as :
Let’s first get the general form of the partial fraction decomposition.
$\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}=\cfrac{A}{(x+1)}+\cfrac{B}{(x-1)}+\cfra... | Multiply out and compare coefficients. You'll find for example that the coefficient of $x^3$ on the right hand side is $2A+2B$. As in the left hand side there are zero $x^3$ you conclude $2A+2B=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the Derivative: This problem for my Calculus 1 class has got me stumped. I am not sure on where to start for this problem. Any help would be much appreciated.
$y=x\tanh^{-1}(x) + \ln(\sqrt{1-x^2})$
| $y=x\tanh^{−1}(x)+\ln(\sqrt{1−x^2})$
differentiation is linear
$\frac {dy}{dx} = \frac d{dx} (x\tanh^{−1}(x)) +\frac d{dx}(\ln(\sqrt{1−x^2})$
next we need to know the product rule and the chain rule
$\frac {dy}{dx} = \tanh^{−1}(x) + x\frac d{dx} \tanh^{−1}(x) +\frac {1}{\sqrt{1−x^2}} \frac d{dx}\sqrt{1−x^2}\\
\frac {dy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
General solution to the polynomial of form $x^a(1-x)^b=c$ Given $x\in(0,1)$, I am interested in finding the solution of polynomials in the form of $x^a(1-x)^b=c,$ where $a,b$ are both positive integers, and $c$ is carefully chosen so that a root in $(0,1)$ exists.
For $a+b<5$ the roots can be found painstakingly with ... | The equation $x^a(1-x)^b=c$ can be written as a generalized trinomial equation
$$
Ax^p+x=1\qquad(p=-\frac{a}{b},~A=c^{{1}/{b}}).
$$
Ramanujan showed that the root of the equation $Ax^p+x=1$ is given by the power series
$$
x=1-A+\sum_{n=2}^\infty\frac{(-A)^n}{n!}\prod_{k=1}^{n-1}(1+pn-k).
$$
So, according to these formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How do I prove that $[\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]....+[\frac{x+n-1}{n}]=[x]$ How do I prove that $[\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]....+[\frac{x+n-1}{n}]=[x]$
How to start?
thanks
| HINT
Use the fact that $[t]$ is the only integer such that $t - 1 \lt [t] \le t $
$\frac{x}{n} -1 +\frac{x+1}{n} - 1+\frac{x+2}{n} - 1…+\frac{x+n-1}{n} -1 \lt [\frac{x}{n}]+[\frac{x+1}{n}]+[\frac{x+2}{n}]…+[\frac{x+n-1}{n}] \le \frac{x}{n} + \frac{x+1}{n}+\frac{x+2}{n}…+\frac{x+n-1}{n}$
UPDATE
The OP question can be pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find the minimum value of this Expression (Without Calculus!) Problem:
Suppose that $x,y$ and $z$ are positive real numbers verifying $xy+yz+zx=1$ and $k,l$ are two positive real constants. The minimum value of the expression:
$$kx^2+ly^2+z^2$$ is $2t_0$, where $t_0$ is the unique root of the equation $2t^3+(k+l+1)t-k... | We need to find a maximal value of $m$, for which the inequality
$$kx^2+ly^2+z^2\geq m(xy+xz+yz)$$
is true for all reals $x$, $y$ and $z$ or
$$z^2-m(x+y)z+kx^2+ly^2-mxy\geq0$$
for which we need $m^2(x+y)^2-4(kx^2+ly^2-mxy)\leq0$ or
$(4k-m^2)x^2-2(2m+m^2)xy+(4l-m^2)y^2\geq0$ for all reals $x$ and $y$, for which we need... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration Calculus Problem How to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{\sin^2{x}}{1+\sin^2x}dx$$
My Try:
I substituted $\displaystyle \sin^2x=\frac{(1-\cos2x)}{2}$ but it gave a complicated expression.
Using the identity$\sin^2x+\cos^2x=1$, the integral becomes:
$$ \int_{0}^{\frac{\pi}{2}}\frac{\sin^2{x}}{1+\sin... | Note
$$ \int \frac{ \sin^2 x + 1 - 1}{1 + \sin^2 x } = \int dx - \int \frac{ dx }{1 + \sin^2 x} = x - \int \frac{ ( \sin^2 x + \cos^2 x ) dx }{1 + \sin^2 x} = x - \int \frac{ \sin^2 x }{1 + \sin^2 x} - \int \frac{ \cos^2 x dx}{1 + \sin^2 x } $$
Thus, we have (now putting the limits of integration)
$$ 2 \int_0^{\pi/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$?
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$
$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left si... | If you use the formula for the triple angle:
$-8\sin^3(50)=2\sin(150)-6\sin(50)=1-6\sin(50)$
so your last square root becomes
$\sqrt{9\sin^2(50)-6\sin(50)+1}=\sqrt{(3\sin(50)-1)^2}=3\sin(50)-1$
as the last expression is positive. So:
$3\sin(50)-1=a\sin(50)+b$
Now if $a$ and $b$ are rational/integer, we have $a=3$, $b=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Unique Factorization in Number Ring I am having a difficulty understanding ideals in a number ring; if I have (2), does it mean that it is a ring generated by $2$ (e.g. {$2k, k \in \mathbb{Z}$}).
If so, why is $(2)=(2,1+\sqrt{-5})^2$? It seems like the ideal $(2,1+\sqrt{-5})^2$ generated by $4$, $2+2\sqrt{-5}$, and $(1... | Given two ideals $A=(a_1,\ldots)$ and $B=(b_1,\ldots)$, $A=B$ if and only if each $a_i$ is in $B$ and each $b_i$ is in $A$.
Here:
$A=(2)$ and
$B=\left(2,1+\sqrt{-5}\right)^2=\left(2,1+\sqrt{-5}\right)\left(2,1+\sqrt{-5}\right)=\left(4, 2+2\sqrt{-5}, -4+2\sqrt{-5}\right)$.
We have:
$4=2\cdot2\in A$,
$2+2\sqrt{-5}=\left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$n$-derivative of function $f(x)=e^x \sin x$ at $x=0$ I have function $f(x) = e^x \sin{x}$ and must found $f^{(n)}(0)$
$f'(x) = e^x(\sin{x} + \cos{x}) $
$f''(x) = 2 e^x \cos{x}$
$f'''(x) = 2 e^x (\cos{x} - \sin{x})$
$f''''(x) = -4 e^x \sin{x}$
$f'''''(x) = -4 e^x (\sin{x} + \cos{x})$
I think $f^{(n)}(0) = \alpha (-1)^n... | Let us try to discover a general formula. I see two possible approaches.
The first one is to use induction: assume that $f^{(n)} (x) = \Bbb e ^x (a_n \cos x + b_n \sin x)$. We have $a_0 = 0$ and $b_0 = 1$. Then
$$f^{(n+1)} = \Bbb e ^x (a_n \cos x + b_n \sin x) + \Bbb e ^x (- a_n \sin x + b_n \cos x) = \Bbb e ^x [(a_n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
length of the shortest path connecting a point and another point on circle.
The length of the shortest path that begins at $(2,5)$ touches the x-axis and then
end at point on the circle $x^2+y^2+12x-20y+120=0$
$\bf{My\; Try::}$ Equation of circle in standard form :: $(x+6)^2+(y-10)^2=4^2$
So any point on the circle i... | Let $C(2, -5)$ be the reflection of point $A$ across the $x$-axis. Notice that $AB = CB$, so we now only need to minimize $CB + BP$. But there are no restrictions on the path from $C$ to $P$ (since any such path must eventually cross the $x$-axis), so the shortest path between the two fixed points is a straight line. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothi... | $\log_23$ vs $\log_36$
Multiply each term by $5$:
$5\log_23$ vs $5\log_36$
Apply logarithm rules:
$\log_23^5$ vs $\log_36^5$
Simplify:
$\log_2243$ vs $\log_37776$
Conclude:
$\log_2243<\log_2256=8=\log_36561<\log_37776$
Hence $\log_23<\log_36$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Mathematical Induction for Recurrence Relation I have solved the following recurrence relationship:
$T(1) = 1$
$T(n) = T(n-1) + n + 2$
so
$T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2$
I am now trying to perform mathematical induction to prove this.
$Basis:$
$T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 $
$Induction:$
$T(k+1) = T... | If induction is not a prerequisite, the following is an alternative that simply telescopes the sequence.
$$T(n) - T(n-1) = n + 2$$
$$\sum_{k=2}^{n} \big(T(k) - T(k-1)\big) = \sum_{k=2}^{n}(k + 2)$$
$$T(n) - T(1) = \frac{(n-1)(n+6)}{2}$$
$$T(n) \;=\; 1 + \frac{n^2 + 5 n - 6}{2} \;=\; \frac{1}{2}n^2 + \frac{5}{2}n - 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove :(P → Q) ∨ (Q → P) using natural deduction Allowed inference rules: ∨-I, ∨-E, ∧-I, ∧-E, →-I, →-E, ¬-I, ¬-E
I tried to prove a contradiction by assuming $¬ ((P → Q) ∨ (Q → P))$ but got stuck, or am I doing it in the wrong way?
Edit:
My proof attempt
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qqu... | Your start is good:
$1.\qquad¬ ((P → Q) ∨ (Q → P))\qquad \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad Assum$
$2.\qquad \qquad P → Q\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ Assum$
This second assumption is good too, since you basically want to do a DeMorgan on line 1, and to do thos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Absolute and conditional convergence of a series with $\tan$ $$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big)$$
I tried to do something like this but in the end it don't help:
$$\sum_{n=1}^{\infty}\tan{\Big(\pi \sqrt[4]{n^4 + (-1)^n n^2 + 4}} \Big) =
\sum_{n=1}^{\infty}\tan{\Big(\pi n \sqrt[4... | First, don't try to sum the series when you are transforming its terms. Work on the terms, find equivalents or developments, and then, apply a theorem stating the status of the series.
Here, you took the right start :
$$\sqrt[4]{n^4+(-1)^nn^2+4} = n\left(1+\frac{(-1)^n}{n^2}+\frac{4}{n^4}\right)^{\frac{1}{4}} = n\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all $\alpha$ such that the series converges Find all values of $\alpha$ such that series $$\sum^\infty_{n=1} \left( \frac{1}{n \cdot \sin(1/n)} - \cos\left(\frac{1}{n}\right) \right)^\alpha$$ converges.
I used Maclaurin for $\sin$ and $\cos$ and got:
$$a_n = \left( \frac{1}{1 - \dfrac{1}{3!n^2} + \ldots} - 1 + \f... | Hint
$$n\sin(\frac{1}{n})=1-\frac{1}{6n^2}(1+\epsilon_1(n))$$
$$\frac{1}{n\sin(\frac{1}{n})}=1+\frac{1}{6n^2}(1+\epsilon_2(n))$$
$$\cos(\frac{1}{n})=1-\frac{1}{2n^2}(1+\epsilon_3(n))$$
thus, when $n\to +\infty$, the general term of your series $u_n$, satisfies
$$u_n \sim (\frac{2}{3n^2})^\alpha $$
and by the limit comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Pythagoras always divisible by 3 Lets look at the pythagoras $a^2+b^2=c^2$ for the integers $a,b,c$. Proof that one of components $a,b,c$ is always divisible by 3. How do i prove that?
| Assume none of $a,b,c$ is a multiple of $3$.
then
$a \equiv \pm 1$ mod$ (3)$
$b \equiv \pm 1$ mod $(3)$ and
$c\equiv \pm 1$ mod $(3)$
thus
$a^2\equiv 1$ mod $(3),$
$b^2\equiv 1$ mod $(3)$
$a^2+b^2 \equiv 2$ mod$ (3)$
but
$c^2=a^2+b^2 \equiv 1$ mod$ (3)$
and this is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that a polynomial family is a base to a subspace QUESTION
Taking $P4 = ax^4+bx^3+cx^2+dx+e$ the vector space of polynomials of degree $4$ or lower and $H$ the subspace of $P_4$ such that its elements satisfy the conditions $p(1)= 0$, $p(-1)=0$
Show that $B = \{q_1,q_2,q_3\}$ is a base of $H$ if the polynomials are... | Note that given the conditions $p(1)=p(-1)=0$, it follows that $H$ is the subspace of $P_4$ consisting of all fourth degree polynomials divisible by both $x-1$ and $x+1$ and therefore by $x^2-1$. Using the division algorithm we know that for all $p(x)\in P_4$
\begin{eqnarray}
p(x)&=&ax^4+bx^3+cx^2+dx+e\\
&=&[ax^2+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you evaluate the integral $\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$? i'm required to evaluate this integral. I've tried factorizing but it doesn't lead me to anywhere.
$$\int\frac{x^2-1}{(x^4+3 x^2+1) \tan^{-1}\left(\frac{x^2+1}{x}\right)}\,dx$$
I've also tried letting $u = \fr... | An useful identity to remember is
$$\frac{dx}{x} = \frac{d(x+x^{-1})}{x-x^{-1}} = \frac{d(x-x^{-1})}{x + x^{-1}}$$
Using the first part of this identity, you can rewrite the integral as
$$\begin{align}
& \int \frac{x(x^2-1)}{(x^4+3x^2+1)\tan^{-1}\left(x + x^{-1}\right)}\frac{d(x+x^{-1})}{x-x^{-1}}\\
= & \int \frac{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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The polynomial has a root in the interval $(a_1,a_1+1).$ Let $a_1,a_2,\ldots,a_k$ and $p$ are positive real numbers where $a_1$ is minimum. If we solve $$\frac{1}{x-a_1}+\frac{1}{x-a_2}+\ldots \frac{1}{x-a_k}-1=0 \quad(1)$$ we have a $k$-degree polynomial.
Consider another polynomial, given by: $$\frac{1}{x-a_1}+\frac{... | Here is part of (a):
If $x < a_1 \le a_i$, then $x-a_i <0$ and so $\displaystyle\frac{1}{x-a_1}+\frac{1}{x-a_2}+\cdots \frac{1}{x-a_k} < 0 < 1$.
Therefore, there is no root smaller than $a_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Manipulation of conditions of roots of a quadratic equation. How do I write $a^5+b^5$ in terms of $a+b$ and $ab$. Also is there any general way of writing $a^n+b^n$ in terms of $a+b$ and $ab$?
| $$(a-b)^2=(a+b)^2-4ab$$ so that
$$a,b=\frac{a+b\pm\sqrt{(a+b)^2-4ab}}2.$$
Then
$$a^5+b^5=\left(\frac{a+b+\sqrt{(a+b)^2-4ab}}2\right)^5+\left(\frac{a+b-\sqrt{(a+b)^2-4ab}}2\right)^5$$ that you can evaluate by the binomial theorem (every other term will cancel out).
Let $m:=(a+b)/2,p=ab$, and after simplification,
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Distinct roots of $ax^2-bx+c=0$ in $(0,1)$, where $a,b,c\in \mathbb{Z}^+$
Question Statement:-
Let $a,b,c$ be positive integers and consider all the quadratic equations of the form $ax^2-bx+c=0$ which have two distinct real roots in $(0,1)$. Find the least positive integers $a$ and $b$ for which such a quadratic equat... | The brute force approach works out fairly quickly in this case.
*
*$f(0) \gt 0$ $\implies$ $c \gt 0$ and $f(1) \gt 0$ $\implies$ $b \lt a + c$
*the product of the roots must be in $(0,1)$ $\implies$ $c < a$
*the discriminant $\Delta \gt 0$ $\implies$ $b^2 \gt 4 ac$
It follows that $c \ge 1$, $a \ge c+1$ and $b \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Solving the differential equation $dy/dx=\frac{6}{x+y}$
Let $ \text{ }\dfrac{dy}{dx}=\frac{6}{x+y}$ where $y(0)=0$. Find the value of $y$ when $x+y=6$.
Let $x+y=v$. Thus $$ \text{ }\dfrac{dy}{dx}=\dfrac{dv}{dx}-1$$
Therefore $$ \dfrac{dv}{dx}=\dfrac{6+v}{v}$$
On separating the variables and integrating, I get
$$y=6... | $u = x + y \implies u' = 1 + y' $. Thus,
$$ u' - 1 = \frac{6}{u} \implies u' = \frac{6+u}{u} \implies \int \frac{ u du }{6 + u} = x +C \implies u - 6 \ln|u+6| = x + C $$
Since $u = x+y$, then
$$ x + y - 6 \ln | x + y + 6 | = x + C $$
Since $y(0) = 0$, then
$$ 0 + 0 - 6 \ln 6 = C \implies C = - 6 \ln 6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prime factor of $A=14^7+14^2+1$ Find a prime factor of $A=14^7+14^2+1$. Obviously without just computing it.
| Hint: I've seen the 3rd cyclotomic polynomial too many times.
$$
\begin{aligned}
x^7+x^2+1&=(x^7-x^4)+(x^4+x^2+1)\\
&=x^4(x^3-1)+\frac{x^6-1}{x^2-1}\\
&=x^4(x+1)(x^2+x+1)+\frac{(x^3-1)(x^3+1)}{(x-1)(x+1)}\\
&=x^4(x+1)(x^2+x+1)+(x^2+x+1)(x^2-x+1)
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is h... | You can simply match the terms of second order, third order and fourth order to the corresponding terms in the answer:
For second order terms: by multiplication $(A+B)^2=A^2+AB+BA+B^2$ and using $AB=BA$ we see that
$$
\frac{(A+B)^2}{2!} = \frac{A^2+B^2}{2} + AB
$$
For third order terms we have by multiplication $(A+B)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Number of Integer Solutions to $k^2-2016=3^n$ How many pairs of integers (k,n) satisfy the equation
$k^2-2016=3^n$
The only thing I could think to do was factor out 3^2 from the entire equation to get
$(k/3)^2-224=3^{n-2}$
but that doesn't lead to anything useful. I know that (45, 2) is a solution, but that was from t... | Note that $2016=224\times 9$ and $x^2\equiv 0,1 \pmod 4$. If $x^2\equiv 0 \pmod 4$, then $3^n\equiv 0 \pmod 4$, but this is impossible. So it must be $x^2\equiv 1 \pmod 4$. Since $3^{2r+1}\equiv 3 \pmod 4$ and $3^{2r}\equiv 1 \pmod 4$ we deduce that $n$ is even. So, let's say $n=2m$ with $m\ge 0$. On the other hand, si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $e^{-ax} \le (1-x)^a + \frac{1}{2}ax^2$ for $a > 1$ and $0 \le x \le 1$: Prove that for all $a > 1$ and $0 \le x \le 1$:
$$
e^{-ax} \le (1-x)^a + \frac{1}{2}ax^2
$$
My very limited start:
$f(x) = (1-x)^a + \frac{1}{2}ax^2 - e^{-ax} \ge 0$
$f'(x) = ax + a e^{-ax} - a(1-x)^{a-1}$
| First, note from the well known $e^{t} \geqslant 1+t$ that
$$e^{-ax} \geqslant (1-x)^a = (1-x)^{a-1}- x(1-x)^{a-1} \geqslant (1-x)^{a-1}-x$$
Multiply by $-a$ to get
$$-a e^{-ax} \leqslant -a(1-x)^{a-1} + ax$$
Integrating this in $[0, x]$ gives
$$e^{-ax}\color{red}{-1} \leqslant (1-x)^a\color{red}{-1} + \frac12ax^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
$\bf{M... | HINT:
As $\arcsin\dfrac13<\arcsin\dfrac12=\dfrac\pi6$
$3\arcsin\dfrac13=\arcsin\dfrac{23}{27}=\arctan\dfrac{23}{10\sqrt2}$
$\arcsin\dfrac35=\arctan\dfrac34$
As $\dfrac{23}{10\sqrt2}\cdot\dfrac34>1,$
using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.
| Let $f(z)=w=x+iy$ and show that $(x,y)$ is on a circle, provided $z \in \mathbb{R}$. First, solve for $z$ so we can get a look at its real and imaginary parts: $$w=\frac{(iz+2)}{(4z+i)} \\
(4z+i)w= iz+2 \\
z(4w-i)=2-iw \\
z=\frac{2-iw}{4w-i} = \frac{2-iw}{4w-i}\frac{4\overline{w}+i}{4\overline{w}+i}=\frac{8\overline{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up ... | Note that each term tends to $0$ but the number of terms tends to $\infty$.
Hint. On may write
$$
\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1}=\frac1n \cdot \sum_{k=1}^n\frac1{1+\frac{k}{n}}
$$ and recognize a Riemann sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
| Remember Eulers theorem applied to the prime $5$ : $$ a^4 = 1 \mod 5 \iff a \mod 5 \neq 0$$ so we have $$(a^4+b^4+c^4+d^4) \mod 5 = 0 \iff a=b=c=d=0 \mod 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving inequations so that $x \rightarrow a^x$
The exponential function $x \rightarrow a^x$, with $0 < a < 1$ is
decrescent in $\mathbb{R}$.
Taking this property in consideration, solve the following
inequalities:
a. $(\frac{1}{3})^{2x} \le (\frac{1}{3})^{x+1}$
b. $0.1^{2-x}>0.1^{3x}$
I did:
a.$$2x \le n+1 \Left... |
When the sides of an inequality are multiplied (or divided) by a negative number, the inequality turns to the opposite side.
It appears you took the $\log$ of the sides to write your inequalities. in the first case, for instance,
$$\log\left(\left(\frac{1}{3}\right)^{2x}\right) \le \log\left(\left(\frac{1}{3}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})\exp(-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2)\,dv$
Compute $$\int_{-\infty}^{+\infty}(1+\frac{1}{v^2})e^{-\frac{u^2}{2\sigma_1^2\sigma_2^2}(\sigma_2v+\frac{\sigma_1}{v})^2}\,dv$$
Does this integral has a close form solution? What if $\sigma_... | Let $I$ denote the integral. Then by the substitution $v \to \frac{1}{v}$ we find that
\begin{align*}
I
&= \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2\sigma_2^2}\left(\sigma_2 v+\frac{\sigma_1}{v} \right)^2 \bigg\} \,dv \\
&\qquad+ \int_{-\infty}^{+\infty} \exp \bigg\{ -\frac{u^2}{2\sigma_1^2\sigma_2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e... | You're correct that Leibniz's Rule is a sound way forward. Note that we have
$$\begin{align}
e^{2x}\sin^2(x)=\frac12 e^{2x}-\frac12 e^{2x}\cos(2x)
\end{align}$$
Then, taking the $n$'th order derivative, we have
$$\begin{align}
\frac{d^n}{dx^n}\left(e^{2x}\sin^2(x)\right)&=\frac{d^n}{dx^n}\left(\frac12 e^{2x}-\frac12 e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 ... | Herein, we present an approach using partial fraction expansion. Proceeding, we write
$$\begin{align}
\frac{1}{x^2+2x+2}&=\frac{1}{i2}\left(\frac{1}{x+1-i}-\frac{1}{x+1+i}\right)\\\\
&=\text{Im}\left(\frac{1}{x+1-i}\right)\\\\
&=\text{Im}\left(\frac{1}{1-i}\frac{1}{1+\frac{x}{1-i}}\right)\\\\
&=\sum_{n=0}^\infty \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{nt\pi}{3}) $ Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{n\pi t}{3}) $
I am unsure how to simplify this in the best form, anyone have any ideas?
| $$ (\cos\frac{n\pi}{2}-1) \cos(\frac{nt\pi}{3}) = \\ \frac{1}{2} \cos\left( n \left(\frac{\pi t}{3} - \frac{\pi}{2}\right)\right)+\frac{1}{2} \cos\left( n \left(\frac{\pi t}{3} + \frac{\pi}{2}\right)\right) -\cos \left( \frac{n \pi t}{3} \right) $$
So all terms are cosines.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Use algebra to calculate the perimeter of a right angle triangle Use algebra to calculate the perimeter of a right angle triangle where the hypotenuse is $(x+4)$, the base is $(x+3)$ and the remaining side is $(x-4)$.
Thanks to the earlier comments I know that I need to use Pythagoras.
So far I know that the Theorem sa... | You're almost done.
$$\begin{align*}(x - 4)^2 + (x + 3)^2 &= (x + 4)^2 \\
\implies x^2 - 8x + 16 + x^2 + 6x + 9 &= x^2 + 8x + 16 \\
\implies x^2 - 10x + 9&= 0 \\
\implies (x - 9)(x - 1) &= 0 \end{align*}$$
So either $x = 9$ or $x = 1$. We know $x \neq 1$ because that would make a side of the triangle $x - 4 = -3$ which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"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.