Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How does $\cos x=\frac12(e^{ix}+e^{-ix})$? I have seen the following definition many times:
$$\cos x=\frac12(e^{ix}+e^{-ix})$$ However, it makes little sense to me as it appears far from obvious. Please help me understand this definition, either a derivation or explanation of how the values on the right equal $\cos x$.... | Taylor series of $\cos(x)$:
$$\begin{align}
\cos(x)&=\color{green}{1-\frac{x^2}{2!}+\frac{x4}{4!}-\cdots}
\end{align}$$
Taylor series of $e^x$:
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$
Taylor series of $e^{ix}$:
$$\begin{align}
e^{ix}&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\frac{(ix)^4}{4!}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Determine the elements of a triangle knowing relationships between lengths and angles A triangle's sides' lengths are three successive numbers (meaning b=a+1, c=a+2). The smallest angle is a half of the triangle's biggest angle.
Find the area of this triangle and its angles.
Solution:
Sides are 4, 5 and 6.
Area: $15\... | Hint:
Solve
$$\frac{\sin\theta}{a}=\frac{\sin3\theta}{a+1}=\frac{\sin2\theta}{a+2}.$$
$$\frac{\sin\theta}a=\frac{4\sin\theta\cos^2\theta-\sin\theta}{a+1}=\frac{2\sin\theta\cos\theta}{a+2}\implies\frac1a=\frac{4\cos^2\theta-1}{a+1}=\frac{2\cos\theta}{a+2}\\\implies a^2-3a-4=0\implies a=4$$ by eliminating $\cos\theta$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
| Put $u=x+\frac 1x$
and let
$u^\boxed{\tiny n}=x^n+\frac 1{x^n}$.
By expansion it is clear that
$$u^3=u^\boxed{\tiny{3}}+3u$$
and
$$u^5=u^\boxed{\tiny5}+5u^\boxed{\tiny3}+10u$$.
Eliminating $u^\boxed{\tiny3}$ gives
$$u^{\boxed{\tiny5}}=u^5-5u^3+5u=1-5+5=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.
Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.
| You've got the right idea about getting an inequality. Since exponential functions grow much faster than polynomials, we should expect $3^b$ to dominate $2b^2+1$ at some point. We can show that it happens at $b=2$ using induction.
Induction hypothesis: Suppose that $2b^2+1<3^b$ for some particular value $b>2$.
Inductio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
| I would emphasize that
$$ \sin (x + A) = \sin x \cos A + \cos x \sin A $$
You have
$$ \sin x \; \; \frac{1}{3} + \cos x \; \; \frac{\sqrt 8}{3} = 1 $$
We can take $A = \arctan \sqrt 8,$ so that $\cos A = 1/3$ and $\sin A = \sqrt 8 / 3. $ So, once again, you have
$$ \sin (x + \arctan \sqrt 8) = 1. $$
On e of the a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Find the value of the fraction
What common fraction is equivalent to $$\frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \text{ ?}$$
I didn't see how to relate the numerator to the denominator. We can't find the value of the n... | $$\sum_{n=1}^{\infty }\frac{1}{n^3}=\sum_{n=1}^{\infty }\frac{1}{(2n)^3}+\sum_{n=1}^{\infty }\frac{1}{(2n-1)^3}=\zeta(3)$$
so
$$\sum_{n=1}^{\infty }\frac{1}{(2n-1)^3}=\zeta(3)-\frac{1}{8}\zeta(3)=\frac{7}{8}\zeta(3)$$
$$\dfrac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Approximation of the Gamma function for small value It is well known that $\Gamma(u) \stackrel{u \to 0}{\sim} \frac{1}{u}$.
I am looking for more precise information on the behavior of $\Gamma(x)$ when $x$ is small, ie: $x\to 0$.
My question is then, are there accurate (for small value) inequalities for the Gamma funct... | $\Gamma(z)$ has a pole at zero, with Laurent series
$$
\frac{1}{z}-\gamma+ \left( {\frac {{\pi }^{2}}{12}}+{\frac {{\gamma}^{2}
}{2}} \right) z+ \left( -{\frac {\zeta \left( 3 \right) }{3}}-{\frac
{{\pi }^{2}\gamma}{12}}-{\frac {{\gamma}^{3}}{6}} \right) {z}^{2}+
\left( {\frac {{\pi }^{4}}{160}}+{\frac {\zeta \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sq... | $$\sqrt{n^2+n}-n = n(\sqrt{1+1/n}-1)$$
Which by the binomial theorem is equal to
$$n\left(-1 + \sum_{k=0}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right) = n\left(\sum_{k=1}^\infty\frac{\prod_{l=1}^k\left(\frac{3}{2} - l\right)}{k! n^k}\right)$$
(With the product equal to 1 for $k = 0$.) So now m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
equation of lines which intersect another line at given angle.
Find the equation of $2$ lines through the origin which intersect the line
$\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}$ at an angle of $\displaystyle \frac{\pi}{3}$
$\bf{My\; Try::}$ Let equation of line be $\displaystyle \frac{x-0}{a} = \f... | you must set
$$a=3+2t$$
$$b=3+t$$
$$c=t$$ with a real number $t$ since the second line intersect the line above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Need help with a trigonometry problem (w/ picture) I have this trigonometry problem I got when programming a code library for cameras in games. I made a picture in Paint to explain the problem as simple as possible. Here's a link:
The known values are random but it shouldn't be a problem in this case. However, I want ... |
The law of cosines:
$$
\begin{cases}
\text{a}^2=\text{b}^2+\text{c}^2-2\text{b}\text{c}\cos\left(\alpha\right)\\
\text{b}^2=\text{a}^2+\text{c}^2-2\text{a}\text{c}\cos\left(\beta\right)\\
\text{c}^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(\gamma\right)
\end{cases}
$$
And the law of sines:
$$\frac{\text{a}}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin... | Transform our integral by letting $\theta=\arcsin x$, then
$$
\begin{aligned}
\int_{0}^{1} \frac{\arcsin x}{x} d x &=\int_{0}^{\frac{\pi}{2}} \frac{\theta \cos \theta d \theta}{\sin \theta} \\
&=\int_{0}^{\frac{\pi}{2}} \theta d(\ln (\sin \theta)) \\
&=-\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta
\end{aligned}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
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Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.
$0^2 + 0 + 4 = 4 \\
1^2 + 1 + 4 = 6 \\
2^2 + 2 + 4 = 10 \\
3^2 + 3 + 4 = 5 \\
4^2 + 4 + 4 = ... | As we can use the formula for the roots of a quadratic over any field with characteristic $\;\neq2\;$, calculating this quadratic's discriminant we get
$$\Delta=1-4\cdot4=-15=-4=7\pmod{11}$$
and now check the squares modulo $\;11\;$ :
$$0^2=0\;,\;\;1^2=1\;,\;\;2^2=4\;,\;\;3^2=9\;,\;\;4^2=5\;,\;\;5^2=3\;,\;\;6^2=3$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Stuck on showing $\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$ converges I am trying to show if this series converges or diverges and I know it converges since for very large values of n,
$$\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$$ becomes $$\sum_{n=0}^\infty \frac{(-1)^n1}{n}$$
which is convergent from ... | I'll take the opportunity for showing some useful techniques for the manipulation of similar series.
Let $\omega=\frac{1+i\sqrt{3}}{2}$ and $\overline{\omega}=\omega^{-1}=\frac{1-i\sqrt{3}}{2}$. By the residue theorem
$$ \frac{x^2+3x-7}{x^3+1} = \frac{-3}{x+1}+\frac{2+\frac{2i}{\sqrt{3}}}{x-\omega}+\frac{2-\frac{2i}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$\sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right)=\frac{\log 2-1}{2}$ - looking for an elementary solution As stated in the title, I am looking for the most elementary proof of the following identity:
$$ \sum_{k\geq 1}\left(1-2k\,\text{arctanh}\frac{1}{2k}\right) =
\frac{\log 2-1}{2}\tag{1}$$
I have a pro... | $$
\begin{align}
& \zeta^{\ast}(2n)=\zeta(2n)-\frac{2}{2^{2n}}\zeta(2n) \Rightarrow \frac{\zeta(2n)}{2^{2n}}=\frac{\zeta(2n)-\zeta^{\ast}(2n)}{2} \Rightarrow \\[4mm]
& \sum_{n=1}^{\infty} \frac{\zeta(2n)}{2^{2n}(2n+1)} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{\zeta(2n)-\zeta^{\ast}(2n)}{2n+1} = \frac{1}{2}\sum_{n=1}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\s... | When I see expression where both $\alpha = a+b\sqrt{n}$ and $\beta =a-b\sqrt n$ occur, I immediately calculate $\alpha + \beta = 2a$ and $\alpha\beta = a^2-nb^2$ since they are guaranteed to be integers (more precisely, the minimal polynomial of both of them is $x^2 - (\alpha+\beta)x+\alpha\beta$ which might be helpful... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Convergence of the series: $\sum_{n=2}^{\infty}\frac{1}{n^2(\ln n)^2\left | \sin(n\pi\sqrt{2}) \right |}$
Does the series $\sum_{n=2}^{\infty}\frac{1}{n^2(\ln n)^2\left | \sin(n\pi\sqrt{2}) \right |}$ converge or diverge?
Could you give me some hints? Thanks for helping.
| Easier than it looks. Let $m$ be the closest integer to $n \sqrt 2.$ We have
$$ | 2 n^2 - m^2 | \geq 1. $$
they are integers and $\sqrt 2$ is irrational. $2 n^2 - m^2 \neq 0$ is an integer.
$$ | n \pi \sqrt 2 - m \pi | = \left| \frac{\pi (2 n^2 - m^2)}{n \sqrt 2 + m} \right| \geq \frac{\pi }{n \sqrt 2 + m} > \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
| Continuing from Momo's answer where was proposed the interesting change of variable $x=\tan(\theta)-1$, the integral reduces to $$I=\int \frac{d\theta}{\cos^2(\theta)\,(\sin(\theta)-\cos(\theta))}$$ Using the tangent half-angle substitution (just as Momo proposed), it reduces to $$I=\int\frac{2 \left(t^2+1\right)^2}{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What is the smallest n integer number such that 24^n does not divide 50! What is the smallest n integer number such that 24^n does not divide 50! (factorial) ?
This is what I've done so far, but it seems a bit complicated
| $24=\color\red{2^3}\cdot\color\green{3^1}$
The multiplicity of $\color\red{2}$ in $50!$ is $\sum\limits_{n=1}^{\log_{\color\red{2}}50}\Big\lfloor\frac{50}{\color\red{2}^n}\Big\rfloor=25+12+6+3+1=\color\red{47}$
The multiplicity of $\color\green{3}$ in $50!$ is $\sum\limits_{n=1}^{\log_{\color\green{3}}50}\Big\lfloor\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find and prove the limit $\lim_{n \to \infty}\frac{2^{n+2}+3^{n+3}}{2^n+3^n}$ I'm trying to find and prove the limit $$\lim_{n \to \infty}\frac{2^{n+2}+3^{n+3}}{2^n+3^n}$$
which according to WolframAlpha happens to be $27$.
The furthest I could get is
$$\frac{2^{n+2}+3^{n+3}}{2^n+3^n} = 4\left(1 + \frac{3^{n}}{2^n+3^n}... | $$2^{n+2}+3^{n+3}\sim_\infty 3^{n+3}, \qquad 2^n+3^n\sim_\infty 3^n,$$
hence
$$\frac{2^{n+2}+3^{n+3}}{2^n+3^n}\sim_\infty\frac{3^{n+3}}{3^n}=3^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Induction divisibility question Q. Prove by induction that $2^{3n-1} + 5(3^n)$ is divisible by $11$ for any even number $n$, where $n$ is an element of natural numbers.
What is have so far:
(base case): $p(2) = 77$, $77/11 = 7$. so base case holds
$p(k) = 2^{3k-1} +5(3^k) $
$p(k+2) = 2^{3k+5} + 5(3^{k+2}) $
$p(k+2) = 2... | Hint $\ {\rm mod}\ 11\!:\,\ 2x \equiv 8^n - 3^n\equiv (-3)^n-3^n\equiv 0\ $ by $\,n\,$ even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Use mathematical induction to prove an assertion The assertion: $n^3 + 5n$ is divisible by $6$
I have completed the basis step $(n=1)$ and the first part of the induction step $(n=k)$, but I am stuck on the second part $(n=k+1)$.
This is what I have so far:
For $n=k$: $k^3 + 5k = 6t$
For $n= k+1$:
$(k+1)^3 + 5(k+1)$
$=... | Proof by induction
First, show that this is true for $n=1$:
$1^3+5\cdot1=6$
Second, assume that this is true for $n$:
$n^3+5n=6k$
Third, prove that this is true for $n+1$:
$(n+1)^3+5(n+1)=$
$n^3+3n^2+3n+1+5n+5=$
$\color\red{n^3+5n}+3n^2+3n+1+5=$
$\color\red{6k}+3n^2+3n+1+5=$
$6k+3n^2+3n+6=$
$6k+6+3n^2+3n=$
$6(k+1)+3\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Show that $b = ln(\sqrt[3]{\frac{18}{27-k}})$ for $9 = \frac{k}{3-2e^{-b3}}$ I did:
$$9 = \frac{k}{3-2e^{-b3}} \Leftrightarrow \frac{k}{9} = 3-2e^{b3} \Leftrightarrow \frac{k}{9}-\frac{3}{1} = -2e^{-3b} \Leftrightarrow \frac{\frac{k-27}{9}}{\frac{-2}{1}} \Leftrightarrow -\frac{k-27}{18} = \frac{1}{e^{2b}} \Leftrightarr... | Please note that
$$ -\frac{18}{k-27} = -\frac{18}{-(27-k)} = \not-\frac{18}{\not-(27-k)} = \frac{18}{27-k}$$
Therefore
$$b = \ln\left(\sqrt[3]{-\frac{18}{k-27}}\right) = \ln\left(\sqrt[3]{\frac{18}{27-k}}\right)$$
And hence your solution is the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}$ We have
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{2^{7/3}}+\frac{\sqrt{3}}{2^{4/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag1$$
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-2}=\frac{3}{2}-\frac{\ln(\sqrt[... | What you need is $\enspace\displaystyle \arctan x+\arctan y=\arctan\frac{x+y}{1-xy}\enspace$ because the values in front of the two $\arctan$ are equal of the results of wolframalpha. Also the values in front of the two $\ln$ are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$ - my answer is wrong (why?) Need to find
$$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$$
One thing I use is
$$\lim_{x \to 1}\frac{x^{1/m}-1}{x^{1/n} - 1} = n/m$$
$$ \lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}} = 3\lim_{x \to 1}\frac{x^{1/... | Put $$t=x^{\frac{1}{6}}.$$
we compute
$$\lim_{t\to 1}\frac{3}{1-t}\left(\frac{1}{1+t+t^2}-\frac{1}{1+t}\right)$$
$$=\lim_{t\to 1}\left(\frac{1}{1-t}\right)\frac{-1}{2}$$
$=+\infty$ at $1^+$ and $-\infty$ at $1^-$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that the determinant of a special $4 \times 4$ matrix is nonzero? Show that if $a,b,c,d \in \mathbb R$ and at least one is different from $0$, then
$$\begin{vmatrix}
a & b & c & d\\
b & -a & d &-c \\
c & -d & -a &b \\
d & c & -b &-a
\end{vmatrix} \neq 0$$
| I like the Artem methode but this is a direct methode :
We can assume that $a\neq0$ :
\begin{eqnarray}
\begin{vmatrix}
a & b & c & d\\
b & -a & d &-c \\
c & -d & -a &b \\
d & c & -b &-a
\end{vmatrix} &=&\frac{1}{a}\begin{vmatrix}
a^2 & b & c & d\\
ab & -a & d &-c \\
ac & -d & -a &b \\
ad & c & -b &-a
\end{vmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$
I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\... | My approach is a little different. I would sub $x=u^{1/n}$ in the integral and get
$$I(n) = \int_0^1 \frac{dx}{1+x^n} = \frac1n \int_0^1 du \frac{u^{\frac1n-1}}{1+u} = \frac1n \int_0^1 du \, u^{1/n} \left (\frac1u - \frac1{1+u} \right ) \\= 1-\frac1n \int_0^1 du \frac{u^{1/n}}{1+u}$$
We then note that $u^{1/n} = e^{\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Fraction Decomposition I have the following problem:
Suppose $x + y + z = 0$. Show that $$\frac{x^5 + y^5 + z^5}{5}= \frac{x^3 + y^3 + z^3}{3} \times \frac{x^2 + y^2 + z^2}{2}$$ and $$\frac{x^7 + y^7 + z^7}{7}= \frac{x^2 + y^2 + z^2}{2} \times \frac{x^5 + y^5 + z^5}{5}$$
I thought this was a fun problem to tackle, bu... | You've surely been trying to express one of the unknowns using the others and plugging in. Try to preserve the symmetry between them instead:
$$(x+y+z)^3 = x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3y^2x + 3y^2z + 3z^2x + 3z^2y + 6xyz$$
But we know that the left hand side is a zero and, moreover, wherever something like $x+z$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given ${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$, find $\sin 4 x$. If
$${[1+ {(1+ x)}^{1/2}]×\tan(x) = \left[1+ {(1- x)}^{1/2}\right]}$$
then find the value of $\sin(4x)$.
The options given are:
a) $x$
b) $4x$
c) $2x$
I tried applying many trigo identities but none of them is working and the radi... | For convenience, write
$$ s := \sin x \qquad c := \cos x \qquad m := 1 +\sqrt{1+x} \qquad n := 1 + \sqrt{1-x}$$
The initial equation can then be rewritten as
$$m s = n c \qquad\to\qquad m^2 s^2 = n^2 c^2 \qquad \to\qquad c^2 = \frac{m^2}{m^2+n^2} \qquad s^2 = \frac{n^2}{m^2 + n^2}$$
Note that $m$ and $n$ are both stric... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $TC = 7+2x+x^2$ and $TR = 10x$ Was wondering whether you might be able to take a look over my working and see if I'm doing this right! I was using the box method before, but have tried to do this one this way!
Find the break-even points in the case where total cost function $TC=7+2x+x^2$ and total revenue func... | Yes, your follow through is correct. Although I would like to point out one thing: You didn't have to use the quadratic formula to find the roots of the polynomial.
Whenever you get a polynomial, factoring usually is the quickest way to find the roots.
For your polynomial, $x^2-8x+7=0$, we see that we need two numbers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $ \int_{0}^{\infty} \frac{1}{x^3+x+1}dx$ I want to evaluate the following integral
$$ \int_{0}^{\infty} \frac{1}{x^3+x+1}\>dx$$
via Residue theorem. Or, any other methods are welcome!
Actually i just compute this via mathematica, but it seems
the command
Integrate[1/(x^3 + x + 1), {x, 0, Infinity}]
gives some ... | Note that $x^3+x+1$ has only one real root below per the Cardano’s formula
$$r= \sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{\sqrt{93}+9}{18}}= -0.6823
$$
Then, decompose the integrand accordingly to integrate as follows
\begin{align}
&\int_0^\infty \frac{1}{x^3+x+1}dx\\
=&\ \int_0^\infty \frac1{(x-r)(x^2+rx-1/r)}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot ... | Divide both sides by $2^{n+1}$.
$$9^{n+1}=27$$
$$n+1=\log_9(27)=\frac 32$$
$$n=\frac 1 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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If $x^2 + y^2 - 3 = 0$ has no rational solutions, then $x^2 + y^2 - 3^k = 0$ has no rational solutions.
*Explain why $x^2 + y^2 – 3 = 0$ not having any rational solutions (Exercise 20) implies $x^2 + y^2 – 3^k = 0$ has no rational solutions for $k$ an odd, positive integer. (Book of Proof by Hammack)
Please cr... | Essentially correct, but more complicated than required. You make some confusion about $k$ and $j$, though.
Suppose $x$ and $y$ are rational solutions of $x^2+y^2=3^k$, with $k$ an odd integer. Then $k=2j+1$ and so
$$
\left(\frac{x}{3^j}\right)^{\!2}+\left(\frac{y}{3^j}\right)^{\!2}=3
$$
where $x/3^j$ and $y/3^j$ are r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving a set of 2x2 matrices are linearly independent. I'm trying to prove the following matrices are linearly independent.
$ \beta = \left\{\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 3 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmat... | Hint
The basis of $\mathbb{M}_{2\times 2}(\mathbb{R})$ is as follow
$$B=\left\{e_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},e_2=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, e_3=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, e_4=\begin{bmatrix} 0 & 0 \\0 & 1 \end{bmatrix}\right\}$$
and
$$\begin{bmatrix} 1 & -1 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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maximum value of $xy+yz+zx.$ given $x+2y+z=4$ if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
| Since $z=4-x-2y$,
$$\begin{align}xy+yz+zx&=xy+y(4-x-2y)+(4-x-2y)x\\&=xy+4y-xy-2y^2+4x-x^2-2xy\\&=-x^2+(4-2y)x-2y^2+4y\\&=-(x^2+(2y-4)x)-2y^2+4y\\&=-\left((x+y-2)^2-(y-2)^2\right)-2y^2+4y\\&=-(x+y-2)^2+(y-2)^2-2y^2+4y\\&=-(x+y-2)^2-y^2+4\\&\le 4\end{align}$$
The equality is attained when $x+y-2=y=0$, i.e. $x=2,y=0,z=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ 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 ... | HINT
I would say - simply divide the number $1$ manually expression $(2+2x+x^2)$:
More attention and gradually comes out:
$1:(2+2x+x^2) = 1/2 - x/2 + x^2/4 - x^4/8 + x^5/8 - x^6/16 + \cdots$
$\Rightarrow f(x)=1/2 \cdot(1 - x + x^2/2 - x^4/4 + x^5/4 - x^6/8 + \cdots)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 6
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About the diophantine equation $y^3=8x^6+2x^3y-y^2$
How I can solve the equation $y^3=8x^6+2x^3y-y^2$ in integers?
I made the substition $x^3=z$ and got the equation $8z^2+2yz-y^3-y^2=0$. So I decided to apply the general formula for quadratic equations and thus I got $$z=\frac{-2y\pm \sqrt{32y^3+36y^2}}{16}=\frac{-y... | What you have so far means that $k=2m+3$, so $8y+9 = 4m^2+12m+9\Rightarrow y =\dfrac{m(m+3)}{2}\in\mathbb{Z}, \forall m$. Then, $x^3 = z = \dfrac{m(m+3)}{16}(-1\pm (2m+3) ) = \dfrac{m(m+1)(m+3)}{8}$ or $x^3 = -\dfrac{m(m+2)(m+3)}{8}$.
Now, the only options for the first case are $2x = m+1,m+2 $ and you can do the same... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all solutions to $xyz=1,000,000;$ $x,y,z \in \mathbb Z$ I had previously solved $xy=1,000,000;$ $x,y \in \mathbb Z$, I believe: 1,000,000 has 49 factors so there are 49 pairs since $x$ and $y$ could be both positive or both negative.
Please would you help me find as efficient and systematic a way as possible.
EDIT... | If $x$, $y$, and $z$ are positive integers with $xyz=2^6\cdot5^6$, then $x=2^a\cdot5^{a'}$, $y=2^b\cdot5^{b'}$, and $z=2^c\cdot5^{c'}$ with $0\le a,a',b,b',c,c'$ and $a+b+c=a'+b'+c'=6$. There are $1+2+3+4+5+6+7=28$ non-negative solutions to $a+b+c=6$, and likewise for $a'+b'+c'=6$, so there is a total of $28\cdot28$ p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does $\angle ACB$ change as the point $C$ changes? Let $A = (-4,0)$, $B = (4,0)$, $C = (x,y)$.
How would you determine how angle $\angle ACB$ changes in relation to $x$ and $y$? Must a further variable be introduced? If so, will knowing the distance between $C$ and the origin help? I've been puzzling at this for ab... | If $\theta = \angle ACB$, then $AC = (x + 4, y)$ and $BC = (x - 4, y)$, so
\begin{align*}
\cos\theta
&= \frac{AC \cdot BC}{\|AC\|\, \|BC\|} \\
&= \frac{x^{2} - 16 + y^{2}}{\sqrt{(x + 4)^{2} + y^{2}}\, \sqrt{(x - 4)^{2} + y^{2}}} \\[12pt]
&= \sqrt{\frac{(x^{2} - 16 + y^{2})^{2}}{[(x + 4)^{2} + y^{2}][(x - 4)^{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prime Factorization of $15^8 + 16$ I have created a problem and I am not able to find the answer. What is the prime factorization of $15^8 + 16$? The thing is I can find an answer with a calculator easily, but what is the step by step approach to the problem.
| \begin{align*}
n^8+n+1 &= (n^2+n+1)(n^6-n^5+n^3-n^2+1) \\
15^8+15+1 &= (15^2+15+1)(15^6-15^5+15^3-15^2+1) \\
&= 241 \times 10634401 \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Power Series representation of the function $f(x) = {x\over 2x^2 + 1}$?
Find Power Series representation of the function $f(x) = \dfrac x{2x^2 + 1}$?
I'm not sure how to tackle this...I'm supposed to find interval of convergence.
| $\sum x^n = \frac {1}{1-x}$ when $x$ is in the radius of convergence
$\sum (-x)^n = \frac {1}{1+x}\\
\sum (-x^2)^n =\sum (-1)^nx^{2n} = \frac {1}{1+x^2}\\
\sum (-1)^n(\sqrt 2 x)^{2n} =\sum (-1)^n(2^n) x^{2n} \frac {1}{1+2x^2}\\
x\sum (-1)^n(2^n)x^{2n} = \frac {x}{1+2x^2}\\
\sum (-1)^n(2^n)x^{2n+1} = \frac {x}{1+2x^2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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maximum value of the expression : $2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1$ I have to find out the maximum value of the expression :
$2x+3y+z$ as $(x,y,z)$ varies over the sphere $x^2+y^2+z^2=1.$
| By Cauchy Schwarz inequality $$(2x+3y+z)^2\le (4+9+1)(x^2+y^2+z^2)$$
$\therefore$ the maxium value that can be attained is when $2x=3y=z$ , and the value attained is $\sqrt{14}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral calculus sine functions: $\frac{1}{2\pi }\int_{-\pi }^{\pi }\frac{\sin\left((n+1/2)\,x\right)}{\sin\left(x/2\right)}\,dx = 1$ For an integer, $n$, how do I show the following?
$$
\frac{1}{2\pi }\int_{-\pi }^{\pi }\frac{\sin\left((n+1/2)\,x\right)}{\sin\left(x/2\right)}\,dx = 1.
$$
Can I use induction?
| To use induction, first establish a base case. If $n=0$, then we see trivially that
$$\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left((n+1/2)x\right)}{\sin(x/2)}\,dx=\frac{1}{2\pi}\int_{-\pi}^\pi \frac{\sin\left(x/2\right)}{\sin(x/2)}\,dx=1$$
Next, we assume that for some integer $N\ge 1$ we have
$$\frac{1}{2\pi}\int_{... | {
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"url": "https://math.stackexchange.com/questions/2037719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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$n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
| You should be able to do the third term in your decomposition.
For the second term, follow Mark Bennet's hint: write the fraction as $$\frac{x+2}{9(1+x+x^2)} = \frac{(x+2)(1-x)}{9(1-x^3)} = \frac{2-x-x^2}{9} \cdot \frac{1}{1-x^3},$$ then write the second factor as an infinite geometric series with common ratio $x^3$.
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Edit: Found another solution, removed old answer (it was incorrect anyway)
You have $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$. By simple algebraic manipulation you can get the
$$x^4-x^3+x^2-x+1 = 0$$
Now notice that $x^4 = x^3-x^2+x-1$ and multiplying both sides by $x$ you get $x^5 = x^4-x^3+x^2-x=-1$.
Therefore
$$x^{2000}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral evaluation $\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx$ - not able to evaluate residue Integral evaluation $\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1} dx$,
Here is my attempt:
Replace cos (ax) with $e^{iaz}$, then take the real part such that:
$\int_{-\infty}^{\infty} \frac{cos (ax)}{x^2+x+1... | In this case, you can take the integral to be
$$\operatorname{Re}{\int_{-\infty}^{\infty} dx \frac{e^{i a x}}{x^2+x+1}} $$
Assume $a \gt 0$ for now. The roots of the denominator are at $x_{\pm} = -\frac12 \pm i \frac{\sqrt{3}}{2} = e^{\pm i 2 \pi/3}$.
Because we assumed $a \gt 0$, we consider the contour integral
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $x$ : $(2-3\mathbb i)x^6 + 1 + 5\mathbb i = 0$ $x^6 = \frac{ 1 + 5\mathbb i}{-2+3\mathbb i}$
How do I convert the right hand side into polar form to find 6th roots?
For numerator we have argument $= \arctan(5) + \pi$
And for denominator argument $= \arctan(-3/2) + \pi$
But $\arctan(5)$ and $\arctan(-3/2)$ are... | I prefer to put z instead of x in the equation.
any way
$$Z^6=1-i\\z^6=\sqrt{((-1)^2+(1^2)}.exp(arctan(\frac{-1}{1}))=\\ z^6=\sqrt 2.e^{-\frac{\pi}{4}i}\\
z^6=\sqrt 2.e^{(-\large\frac{\pi}{4}i+2k\pi)}\\$$ now go to the power of $\frac{1}{6}$
$$\sqrt[6]{z^6}=(\sqrt 2.e^{\large(-\frac{\pi}{4}i+2k\pi)})^{\frac{1}{6}}\\... | {
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"url": "https://math.stackexchange.com/questions/2043936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lagrange multipliers: how could this be outside the restriction? I have to get the extrema of function $f(x,y)=\cos^2(x)+\cos^2(y)$ restricted to $x+y=\frac{\pi}{2}$. So this is what I did:
$$g(x,y)=x+y=\frac{\pi}{2}$$
$$\nabla f(x,y)=\lambda.\nabla g(x,y)$$
So I got the gradients of both functions:
$$\nabla f(x,y)=<-2... | Hint
From $\nabla f=\lambda \nabla g$ we get
\begin{cases}\sin 2x=-\lambda\\ \sin 2y=-\lambda\end{cases} Since $\sin t$ is $2\pi$-periodic we have that $\sin 2t$ is $\pi$-periodic. Thus we get that $2y=2x+k\pi,$ for some integer $k.$ That is, $y=x+\frac {k\pi}2.$ Using the constraint $x+y=\frac \pi2$ we get that $2x+\... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-0.0064$ (Motivation) As homework, we have been asked to prove that the following series converges: $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}$$
I did it in two ways:
*
*Using the alternating series test (Leibniz criterion), proving that $\frac{n^3}{4^n}$ is ... | As in other answers consider $$\sum_{n=1}^{+\infty}n^3 x^n$$ and rewrite $$n^3=n(n-1)(n-2)+a n(n-1)+bn$$ Expanding and identifying coefficients, you should get $a=3$ and $b=1$. So
$$\sum_{n=1}^{+\infty}n^3 x^n=\sum_{n=1}^{+\infty}n(n-1)(n-2) x^n+3\sum_{n=1}^{+\infty}n(n-1) x^n+\sum_{n=1}^{+\infty}n x^n$$
$$\sum_{n=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solution to $\mathbf{x}'=A\mathbf{x}$, with $A$ anti-symmetric implies $|\mathbf{x}|=|\mathbf{x}_0|$ Let $A$ be anti-symmetric. That is, $A^T=-A$. Show that the solution to the $\mathbf{x}'=A\mathbf{x}$ satisfies $|\mathbf{x}|=|\mathbf{x}_0|$. Hint: Compute $d/dt|\mathbf{x}|^2$.
The given solution:
$d/dt|\mathbf{x}|^... | The expression $2A|\mathbf{x}|^2$ makes no sense as $A$ is a matrix while $|\mathbf{x}|^2$ is a number so let's stop one step before it. The product rule gives us the equation $\frac{d}{dt} |\mathbf{x}|^2 = 2(A\mathbf{x}) \cdot \mathbf{x}$. Assuming that $\mathbf{x}$ is a column vector and calculating the dot product u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub? I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've work... | Hint: Let $u=36x^2+1$, then $du=72xdx$ and thus
$$\frac{x^5}{(36x^2+1)^{3/2}}dx=\frac{(u-1)^2}{36^2u^{3/2}}\frac{du}{72}.$$
Now we get something easier to handle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to solve system of equations involving square roots How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect.
$$
\left\{
\begin{array}{c}
\sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\
\sqrt{1 - x_1}... | HINT:-
Since the roots have to be positive, we see that each radical must reduce to the form $a\sqrt5$ for the first equation , and $b\sqrt3$ for the second equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
An interesting result in ratio and proportions If
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then
$$
\frac{a+c}{b+d}=k
$$
Also
$$
\frac{a^2}{b^2}=\frac{c^2}{d^2}=k^2
$$
And
$$
\left( \frac{a+c}{b+d} \right)^2=k^2
$$
Also
$$
\frac{a^2 + c ^2}{b^2+d^2}=k^2
$$
Hence
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
... | if we factorizing $$\frac{a^2+c^2}{b^2+d^2}-\left(\frac{a+c}{b+d}\right)^2=\frac{2 (a d-b c) (a b-c d)}{(b+d)^2 \left(b^2+d^2\right)}=0$$ and we get $$ad-bc=0$$ or $$ab-cd=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How should I evaluate the integral $\lim_{n\rightarrow\infty}n\int\frac{\cos(nx)}{1+x^2}\,dx$? Q1: How should I evaluate the integral $$\lim_{n\rightarrow\infty}n\int_{\mathbb{R}}\frac{\cos(nx)}{1+x^2}\,dx$$ (Hint: dominated convergent theorem and integration by parts) ?
Possible Solution:
\begin{align}
n\int\frac{\cos... | By $nx=t$ and using integration by parts $\int_{-\infty}^{\infty}n\frac{cos(nx)}{1+x^2}dx = \frac{sin(t)}{1+(t/n)^2}|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{sin(t)}{(1+(t/n)^2)}\frac{2t}{n^2}dt \\ $. Here the first term goes to zero and second term is bounded by $\int_{-\infty}^{\infty}\frac{1}{(1+(t/n)^2)}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving polynomial equation with calculus I saw a question stating that
Let $f(x)$ is a polynomial of degree $5 $.
$ x-1|f(x)+1 $ and $x+1|f(x)-1$ . Find $f(x)$ without using calculus.
And because I am not good at calculus , I was not able to solve the question with or without using calculus.
Please help me in solving... | Followup on previous comment:
The two conditions are equivalent to $f(1)=−1$ and $f(−1)=1$ but that's not enough to determine the polynomial univocally.
Let the quintic be $f(x) = ax^5+bx^4+cx^3+dx^2+ex+f$, then the two conditions give:
$$
\begin{align}
\begin{cases}
a+b+c+d+e+f & = -1 \\
-a+b-c+d-e+f &= 1
\end{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \fra... | Call $k$ the common value of
$$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$$
and then
$$\log(a^ab^bc^c)=a\log a+b\log b+c\log c=\left(a(b-c)+b(c-a)+c(a-b)\right)\cdot k=0$$
which implies that $a^ab^bc^c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Solve $\sqrt{3}^n=3^3$ I got answer but i could not understand, can anyone explain me ?
$\sqrt{3}^n=3^3$
$\implies{3}^{n/2}=3^3$
$\implies \frac{n}{2}=3\implies n=6$
How we get ${3}^{n/2}$ in the second line and how we get $\frac{n}{2}$ in third line.
| $$\sqrt{x} = x^{\frac{1}{2}}$$
Since if $\sqrt{x}\sqrt{x} = x$ by definition then it follows that $x^kx^k = x^1$ where $2k = 1$ so $k = \frac{1}{2}$. If you choose to agree with this, then
$$\sqrt{3^n} = \sqrt{3}^\frac{n}{2} = 3^3 \implies \frac{n}{2} = 3 \implies n = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but g... | If we transform the whole thing into a single fraction, then both numerator and denominator will be symmetric polynomials on three variables.
The good thing about them is that one can then use the Fundamental Theorem of Symmetric Polynomials to write $P(x,y,z)$ in a unique way as a polynomial $P'(e_1(x,y,z),e_2(x,y,z),... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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QR algorithm for "general" square matrices
*
*Can QR algorithm find repeat eigenvalues (https://en.wikipedia.org/wiki/QR_algorithm) ?
I.e. does it support the case when not all N eigenvalues for real matrix N x N are distinct?
*How to extend QR algorithm to support finding complex eigenvalues ?
*Is it possible to ... | Summary
The QR Decomposition resolves all of your matrix types.
a) Repeated eigenvalues
The three eigenvalues of $\mathbf{A}$ are
$$
\lambda \left( \mathbf{A} \right) = \left\{ 1, 1, 1 \right\}
$$
The decomposition is
$$
\begin{align}
\mathbf{A} &= \mathbf{Q\,R} \\
% A
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question:
Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}≥\frac{3}{2}$, for each $a,b,c>0$.
I'm not sure how to use Lagrange multipliers for it... any ideas?
| Let $f(x,y,z) = \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} $
$$\begin{eqnarray*}
\vec \nabla f \cdot \hat x &=& +\frac{1}{(y+z)} - \frac{y}{(x+z)^2}- \frac{z}{(x+y)^2}
\\ \vec \nabla f \cdot \hat y &=& -\frac{x}{(y+z)^2} + \frac{y}{(x+z)}- \frac{z}{(x+y)^2}
\\ \vec \nabla f \cdot \hat z &=& -\frac{x}{(y+z)^2} - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove that $\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$ I need to prove this equation
$$\sum_{k=0}^n (-1)^k\frac{{ {n}\choose{k}}{ {x}\choose{k}}}{{ {y}\choose{k}}} = \frac{{ {y-x}\choose{n}}}{{ {y}\choose{n}}}$$
where $x$, $y$ and $n$ are ... | Standing the conditions you gave, the $y \choose n$ is not null, and we can multiply by it both sides, giving
$$
\left( \begin{gathered}
y - x \\
n \\
\end{gathered} \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
y \\
n \\
\end{gathered} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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show that; strange sum yields triangular numbers $\sum_{k=1}^{n}\tan^2\left({k\pi\over 2n+1}\right)=T_{2n}$ $1,3,6,10,15,...$ for $n=1,2,3,...$ it is the n-th Triangular numbers.
I find it unsual that this sum yields even triangular numbers;
$$\sum_{k=1}^{n}\tan^2\left({k\pi\over 2n+1}\right)=T_{2n}$$
How can I show th... | As recommended in the comment section of OPs question we can follow the nice answer of Fallager.
We obtain
\begin{align*}
\left(\cos\frac{k\pi}{2n+1}+i\sin\frac{k\pi}{2n+1}\right)^{2n+1}&=(-1)^k\tag{1}\\
\\
\sum_{j=0}^{2n+1}\binom{2n+1}{j}\left(\cos\frac{k\pi}{2n+1}\right)^j\left(i\sin \frac{k\pi}{2n+1}\right)^{2n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint
Replace $d$ by a variable $x$; make use of the fact that the s... | Consider instead the polynomial in $x$
$P(x)=\det\begin{pmatrix}1&x&x^2&x^3&x^4\\1&a&a^2&a^3&a^4\\1&b&b^2&b^3&b^4\\1&c&c^2&c^3&c^4\\1&d&d^2&d^3&d^4\end{pmatrix}$.
If you use Laplace's expansion in the first row, you'll notice that $P(x)$ has degree $4$.
Also, $P(a)=P(b)=P(c)=P(d)=0$ because plugging $x=a,b,c,d$ create... | {
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"url": "https://math.stackexchange.com/questions/2068179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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Find the least possible integer for which $\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$ Question
Find the least possible no for which $$\sqrt[3]{n+1} - \sqrt[3] n< \frac 1{12}$$
How do I reach a stage where I can deduce definitely the smallest integer value of $n$. I keep getting stuck after I cube on both sides and I stil... | Let $n = x^3$
\begin{array}{c}
\sqrt[3]{n+1} - \sqrt[3] n < \frac 1{12} \\
\sqrt[3]{x^3+1} - x < \frac 1{12} \\
\sqrt[3]{x^3+1} < x + \frac 1{12} \\
x^3+1 < x^3 + \dfrac 14x^2 + \dfrac{1}{48}x + \dfrac{1}{1728} \\
\dfrac 14x^2 + \dfrac{1}{48}x - \dfrac{1727}{1728} > 0 \\
x^2 + \dfrac{1}{12}x - \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Do not use series expansion or L' Hospital's rule: $f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5}$ In the following function
$$f(x)=\frac{\sin 3x+A\sin 2x+B\sin x}{x^5},(x\neq 0)$$
is continuous at $x= 0$. Find $A$ and $B$. Also find $f(0)$.
I first thought of using L hospital. But my sir told me to
'Do not use series exp... | Hint. If one admits that
$$
\lim_{u \to 0}\frac{\sin u}u=1 \tag1
$$ then using
$$
\sin(2x)=2\sin x \cos x,\quad \sin(3x)=3\sin x-4\sin^3x,
$$ one may observe that, for $x \neq0$,
$$
\begin{align}
\frac{\sin 3x + A \sin 2x + B\sin x}{x^5}
&=\frac{\sin x}{x}\cdot\frac{4\left(\cos x+\frac{A}4 \right)^2+\left(B-1-\frac{A^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac{x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1$ for $a+b+c=m$
If $a+b+c=m$, prove that:
$$\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac {x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1.$$
My Attempt:
$a+b+c=m... | Note that
$$\frac{x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}} = \frac{x^a}{x^a} \frac{x^a}{x^a+x^b+x^c}.$$
Using a similar reasoning for the other two terms, we find that
$$\begin{align*} &\frac {x^{2a}}{x^{2a}+x^{a+c}+x^{a+b}}+\frac {x^{2b}}{x^{2b}+x^{a+b}+x^{b+c}}+\frac {x^{2c}}{x^{2c}+x^{b+c}+x^{a+c}}\\ &= \frac{x^a}{x^a+x^b+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Simple exercise on Galois theory Find the splitting field of $x^6-2x^4-8x^2+16$ over $\mathbb {F}_3$ and list the intermediate fields between the base camp and the splitting field.
| $x^6-2x^4-8x^2+16 = x^6+x^4+x^2+1$ in $\mathbb {F}_3[x]$.
$x^6+x^4+x^2+1 = \dfrac{x^8-1}{x^2-1} = (x^2 + 1) (x^4 + 1)$
Therefore, the splitting field of $x^6-2x^4-8x^2+16$ is the same as the splitting field of $x^4 + 1=(x^2 + x + 2) (x^2 + 2 x + 2)$, which is $\mathbb {F}_9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Solving a system of linear differential equations with repeated eigen values I have this problem where to solve the system,
$$x'=4x+y-z$$
$$y'=2x+5y-2z$$
$$z'=x+y+2z$$
using a linear algebraic solution. I have found the eigen values of the
$$\begin{bmatrix}
4 & 1 & -1 \\
2 & 5 & -2 \\
1 & 1 & 2... | You have correctly calculated your eigenvalues. To avoid ambiguous notation, denote your eigenvalues as follows:
$\lambda_1=5$
$\lambda_2=3$
$\lambda_3=3$.
Using your eigenvalues, use the fact that:
$A\mathbf{x}=\lambda_1 \mathbf{x}$, $A\mathbf{x}=\lambda_2 \mathbf{x}$ and $A\mathbf{x}=\lambda_3 \mathbf{x}$, where $\ma... | {
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"url": "https://math.stackexchange.com/questions/2071916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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The sum of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. What can $p$ be?
The summation of all integers from $1$ to $p$ is divisible by $p$ and all prime numbers before $p$. Find all possible solutions for $p$, with proof. Here $p$ is a prime number.
This is a preparation probl... | Let $p_n$ denote the $n$-th prime. Note that $1+2+\cdots+p_n = p_n\cdot\dfrac{p_n+1}{2}$.
By Bertrand's postulate, $p_n < 2p_{n-1}$. Hence, $0 < \dfrac{p_n+1}{2} < \dfrac{2p_{n-1}+1}{2} < 2p_{n-1}$.
Therefore, $\dfrac{p_n+1}{2}$ is not divisible by $2p_{n-1}$.
If $n \ge 3$, then $2$ and $p_{n-1}$ are distinct primes... | {
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"url": "https://math.stackexchange.com/questions/2072199",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Why $x^5y\equiv y^5x\pmod {240}$ if $x,y$ have similar parity? Let $x,y$ be any 2 integers of similar parity. Why do we have :$$x^5y\equiv y^5x\pmod {240}$$
Any hint anyone?
| $$x^5y-xy^5=(x^5-x)y-(y^5-y)x$$
Now $p|(a^p-a)$ by Fermat's little theorem for prime $p$
$a^5-a$ is divisible by $a^3-a$
If $x,y$ are even, we are done.
For odd $a,a=2c+1$(say)
$(2c+1)^2=8\dfrac{c(c+1)}2+1=8d+1$ for some integer $d$
$(2c+1)^4=(8d+1)^2=16(d+4d^2)+1$
$\implies(2c+1)^{4m+1}\equiv2c+1\pmod{16}$
$\implies a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Multiplicative inverse using irreducible polynomial I am trying to compute the multiplicative inverse of:
$x^4 + x^2 + x$
Using the irreducible polynomial:
$x^8 + x^4 + x^3 + x + 1$
Over $GF(2)=\mathbb{F}_2$.
I tried applying the extended euclidean algorithm but to no avail. At some point during the algorithm I end up ... | $\mathbb{F}_2[x]/(x^8+x^4+x^3+x+1) \cong \mathbb{F}_{2^8}$.
If $\alpha \in \mathbb{F}_{2^8}$ and $\alpha\ne0$, then $\alpha^{2^8-1}=1$, and so $\alpha^{-1}=\alpha^{2^8-2}$.
Therefore, the inverse of $x^4 + x^2 + x$ in $\mathbb{F}_{2^8}$ is $(x^4 + x^2 + x)^{2^8-2}$.
This is an answer, even if not a very practical one.... | {
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"timestamp": "2023-03-29T00:00:00",
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How to quickly solve $y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx$? I'm currently trying to solve $$y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx,$$ which is a GRE math subject test problem. I was able to get the answer ($\sqrt{2}$) by breaking up the inte... | Hint. The function
$$
x \mapsto \sqrt{1+x^2}\sin^3x\cos^3x, \qquad x \in \left[-\frac{\pi}4, \frac{\pi}4\right]
$$ is odd as is $x \mapsto \sin (x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $f(x) = x^3$ is injective? This is my attempt. Is it right? Is there a simpler way? Is there a way which relies on less background knowledge?
A function is injective iff $f(x) = f(y) \implies x = y$. This is equivalent to $x \not = y \implies f(x) \not = f(y)$. We will prove this latter statement by contr... | Suppose $$x^3=y^3$$ then $$(x-y)(y^2+xy+x^2)=0$$ so either $x=y$ or $$x^2+xy+y^2=\left(x+\frac y2\right)^2+\frac {3y^2}4=0$$
But this is strictly non-negative and is zero only if $x=y=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proving the inequality I need to prove this:
$\binom{n}{2}a^2 + \binom{m}{2}b^2
\leq
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \right) ^2$
I tried this (WLOG $n\leq m$):
$\binom{n}{2}a^2 + \binom{m}{2}b^2 = \frac{n(n-1)}{2}a^2 + \frac{m(m-1)}{2}b^2
\\
\binom{n+m}{2} \left( \frac{n\cdot a + m\cdot b}{n+m} \... | I have to rush off now; if I remember, I'll come back and finish this answer.
Combinatorially: I'll interpret everything as areas.
Say we have $n$ sticks of length $a$, and $m$ sticks of length $b$. Then $a^2$ is the area of the square we get by placing two $a$-length sticks orthogonally; there are $\binom{n}{2}$ ways ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.
| We have $ a^2+b^2-ab=c^2$, then $ (a-b)^2=c^2-ab\ge 0$. So $ c^2\ge ab $. This implies that at least one of $ c\ge a $ or $ c\ge b $ is true.
First case: $ c\ge a $. Then $ b^2-ab=b (b-a)=c^2-a^2\ge 0$, so $ b\ge a $. Thus we have $b^2-c^2=ab-a^2=a (b-a)\ge 0$, so $ b\ge c $. Therefore $(a-b)(b-c)\le 0$.
Second case: ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Another way of proving :$\int_{0}^{1}{x-x^3+x^5-x^7\over (1+x^4)\ln{x}}dx=-\ln{2}$ Prove that
$$\int_{0}^{1}{x-x^3+x^5-x^7\over (1+x^4)\ln{x}}dx=-\ln{2}$$
My try
$x-x^2+x^3+x^5-x^7=x(1-x^2)+x^5(1-x^2)=x(1+x^4)(1-x^2)$
$$\int_{0}^{1}{x(1+x^4)(1-x^2)\over (1+x^4)\ln{x}}dx$$
Applying Frullani theorem
$$\int_{0}^{1}{x-x^3... | Maybe some one will be interested in that
We can show
$$ \int^1_0 \frac{x^a-1 }{\log x} dx =\log \left ( a+1\right)$$
Using differentiation under the integral sign.
Generally
$$ \int^1_0 \frac{x^a-x^b }{\log x} dx =\int^1_0 \frac{x^a-1-(x^b-1)}{\log x} dx =\log \left ( \frac {a+1}{b+1}\right)$$
Hence
$$ \int^1_0 \frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly
Say a rational number is $kingly$ if it is the square of another rational number. Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly.
If $\dfrac{n+7}{25n+7}$ is kingly, then we can write $\dfrac{n+7}{25n+7... | Option 1: $n+7$ and $25n+7$ are both perfect squares
Not possible:
All perfect squares are equivalent to $0,1$ or $4$ modulo $5$
$25n+7\equiv 2\pmod 5$
Option 2: $(n+7)$ divides $(n+25)$ and the ratio is a perfect square.
$\lim_\limits {n\to \infty} \frac {n+7}{25n+7}= \frac 1{25}$
There are 3 possible values that $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082283",
"timestamp": "2023-03-29T00:00:00",
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Show that $x^3+2y^3+4z^3=0 $ has no non trivial solutions without using infinite descent The question is to show that the equation $$x^3+2y^3+4z^3=0 $$ has no (non trivial) integer solutions. I know it can be done using infinite descent, but how do you do it using only the modular arithmetic argument?
| Mod $9$ also solves it.
By Binomial Theorem $(3k\pm 1)^3\equiv \pm 1\pmod{9}$,
so $x^3\equiv\{0,\pm 1\}\pmod{9}$
(also $x^3\equiv \{0,\pm 1\}\pmod{8}$, so mod $8$ solves it similarly, as a comment has suggested).
$x^3+2y^3+4z^3\equiv 0\pmod{9}$
is equivalent to $x^3+2y^3\equiv -4z^3\pmod{9}$
$x^3+2y^2\equiv \{\pm 3,\pm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$
Prove that $$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$
To do this I unified three terms on the left side with a common denominator and then factored the numerator (w... | Claim:
$$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$
Proof:
Note that the Lagrange interpolating polynomial through $(a,p),(b,q),(c,r)$ is given by:
$$f(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}p+\frac{(x-a)(x-c)}{(b-a)(b-c)}q+\frac{(x-a)(x-b)}{(c-a)(c-b)}r$$
We seek the value o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A Viéte-like infinite product for $\sin (\pi/7)$ In the article "The unruly $\sin (\pi/7)$ of Samuel Moreno, the following infinite product is given:
Also, the same article shows that $\sin (\pi/7)$ is equal to the infinite nested radical
$ \sin (\pi/7) = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{2 - \sqrt{2 + \sqrt{2 - \s... | With the help of Half angle cosine formula it is possible to express many sine and cosine angles as finite or infinite nested square roots of 2 here.
In this perspective $2\sin\frac{\pi}{7}$ can be solved as follows
$2\sin\frac{\pi}{7}$ = $\sqrt{2-2\cos\frac{2\pi}{7}}$ = $\sqrt{2-\sqrt{2+2\cos\frac{4\pi}{7}}}$ = $\sqrt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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finding the geometry of complex numbers when you have terms in cube When I solve for the following $$\left|{\frac{1+z}{1-z}}\right|=2$$
I get $$ 3x^2 +3y^2-10x +3=0 $$ and this isnt the equation of a circle nor a line so what geometry is that?
the correct possibilities point to either a circle of radius 4/3 or 5/4 or ... | For this case:
$$3x^2 +3y^2-10x +3=0 \rightarrow 3\left(x^2 -\frac{10}{3}x\right)+3y^2+3=0$$
$$3\left[\left(x -\frac{5}{3}\right)^2-\frac{25}{9}\right]+3y^2+3=0$$
$$3\left(x -\frac{5}{3}\right)^2-\frac{25}{3}+3y^2+3=0$$
$$\left(x -\frac{5}{3}\right)^2+(y-0)^2=\left(\frac{4}{3}\right)^2$$
It is a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that given fraction is power of two I do not really know where to start with the following:
Prove that
$$\frac{\displaystyle{ \prod_{k = 1}^{2n - 1} k^{\min(k, 2n - k)}}}{\displaystyle{\prod_{k = 1}^{n - 1}(2k + 1)^{2n - 2k - 1}}}$$ is a power of two.
Could you give me a hint which helps to solve this pr... | Let $f(n)$ be the given expression. Then $f(1) = 1$ and $f(n+1)/f(n) = 2^{2n}$:
$$\frac{f(n+1)}{f(n)} = \frac{(n+1)^{n+1}(n+2)^n \dotsb (2n-1)^3(2n)^2(2n+1)}{3^{2n-1} \dotsb (2n-1)^3(2n+1)} \cdot \frac{3^{2n-3} \dotsb (2n-1)}{(n+1)^{n-1}(n+2)^{n-2} \dotsb (2n-1)}$$
which simplifies to
$$\frac{f(n+1)}{f(n)} = \frac{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A line intersects a hyperbola at the points $(-2,-6)$ and $(4,2)$ and one of the asymptotes of the hyperbola at $(1,-2)$. Find the centre.
A line intersects a hyperbola at the points $(-2,-6)$ and $(4,2)$. It also intersects one of the asymptotes of the hyperbola at the point $(1,-2)$.
Find the centre of the hyperbo... | Your intuition can be stated as the proposition below:
If $A$ and $B$ are two different points on a hyperbola and the midpoint of $AB$ lies on the asymptotes, then this midpoint is the center of the hyperbola.
Proof: Note that this proposition is irrelevant to coordinate systems, thus without loss of generality, the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Incorrect solution in limits $$
\lim_{n\to \infty}\left(\frac{1}{n^4}{+3^{\frac{2}{2+n}}}\right)^{n}
$$
So i re-write it like:
$\lim_{n\to \infty}e^{n\ln{\frac{1}{n^4}\ln3^{\frac{2}{2+n}}}}$ $=$ $e^{\frac{2n}{2+n}\ln{\frac{1}{n^4}\ln3}}=e^{{2}\ln{\frac{1}{n^4}\ln3}}$
So here, $\ln{\frac{1}{n^4}}$ give us minus infini... | Note that
$$
\mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{a}
{n}} \right)^{\,n} = e^{\,a} \quad \quad \mathop {\lim }\limits_{n\; \to \;\infty } \left( {1 + \frac{{f(n)}}
{n}} \right)^{\,n} \ne \mathop {\lim }\limits_{n\; \to \;\infty } e^{\,f(n)}
$$
that's the fault in your derivation. Consider inst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index? Let $\sigma(x)$ denote the sum of the divisors of $x$, and denote the abundancy index of $x$ as
$$I(x) = \dfrac{\sigma(x)}{x},$$
and the deficiency of $x$ as
$$D(x) = 2x - \sigma(x).$$
If the equation $I(a)=b/c$ has no solution $a \in \mathbb{N}$, then $b/c$... | Too long to comment.
One can prove that
$$\frac{2n}{n+D(n)}<\frac{(2a+2)n−aD(n)}{(a+1)n+D(n)}<I(n)\tag1$$
$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}\tag2$$
hold for any $a>0,b\gt -1$. Note here that $a,b$ are not necessarily integers.
However, it seems that we cannot prove the conjecture usi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
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Determinant composed from polynomials $p_1(x) = x + a$ and $p_2(x) = x^2 + bx + c$ Let $p_1(x) = x + a$ and $p_2(x) = x^2 + bx + c$ be two polynomials with real
coefficients, and $x_1$ and $x_2$ be two arbitrary real numbers. Consider the
following determinant $$D(x) = \begin{vmatrix}
1 & p_1(x_1) & p_2(x_1)\\
... | Note that since $\det$ is multi linear and alternating, we have
\begin{eqnarray}
D(x) &=& \det \begin{bmatrix}
1 & x_1+a & x_1^2+bx_1+c \\
1 & x_2+a & x_2^2+bx_2+c \\
1 & x+a & x^2+bx+c \
\end{bmatrix} \\
&=& \det \begin{bmatrix}
1 & x_1 & x_1^2+bx_1+c \\
1 & x_2 & x_2^2+bx_2+c \\
1 & x & x^2+bx+c \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of the exponential functions: $\lim_{x\to 0} \frac{e^x-e^{x \cos x}} {x +\sin x}$ I want to find the limit of this function by simply using algebraic manipulation. Though I have computed the limit through L' Hospital's method but still I want to compute the limit purely by function's manipulation to yield a form ... | $$\begin{align*}
& \lim_{x \rightarrow 0} \frac{e^x - e^{x\cos x}}{x + \sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x - 1 + 1 - e^{x\cos x}}{x + \sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x+\sin x}
+ \lim_{x \rightarrow 0} \frac{1-e^{x\cos x}}{x+\sin x} \\
& = \lim_{x \rightarrow 0} \frac{e^x-1}{x} \cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Compute the integral $\int_0^{\pi/2}\sin(1+\cos^2x)\mathrm{d}x $ My attempt.
$$\int\limits_0^{\pi/2}\sin(1+\cos^2x)\mathrm{d}x $$
Let $t=1+\cos^2x \Rightarrow \mathrm{d}t -2\sin x\cos x~ \mathrm{d}x. $
We have $\cos^2x=t-1 \Rightarrow \cos x=\sqrt{t-1}$ and $\sin x = \sqrt{2-t}.$ So $$\mathrm{d}x = \frac{-\mathrm{d}t}{... | Use the integral representations of the Bessel function of the first kind at $0$,
$$\int_{0}^{\pi/2}\cos\left(z\cos \theta\right)\,\mathrm{d}\theta=\frac{\pi}{2}J_0\left(z\right).$$
So,
$$\begin{align*}
\int_0^{\frac{\pi}{2}}\sin\left(1+\cos^2x\right)\mathrm{d}x&=\int_0^{\frac{\pi}{2}}\sin\left(\frac{1}{2}\cos\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions? Is there any easy way or shortcut to decompose $$\frac{1}{X^{3}\cdot (X-2)^{3}}$$ into partial fractions ? because dealing with the usual way of replacing and giving values to $X$ is too clumsy in this case , so I was wondering if there ... | Rewriting it,
$$\left(\frac{1}{x(x-2)}\right)^3$$
Using Partial fractions,
$$\left(\frac{1}{x(x-2)}\right)^3 = \left(-\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x-2}\right)\right)^3$$
Applying $(a-b)^3 = a^3-b^3+3a^2b-3ab^2$
$$\frac{1}{x^3(x-2)^3} = -\frac{1}{8}\left(\frac {1}{x^3}-\frac{1}{(x-2)^3}+\frac{3}{x^2(x-2)}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the bes... | $$(x+y)^4=1 \\
x^4+4x^3y+6x^2y^2+4x^3+y^4 =1$$
Now, by AM-GM we have
$$x^3y \leq \frac{x^4+x^4+x^4+y^4}{4}\\
x^2y^2 \leq \frac{x^4+y^4}{2}\\
xy^3 \leq \frac{x^4+y^4+y^4+y^4}{4}\\
$$
Thus
$$1=x^4+4x^3y+6x^2y^2+4x^3+y^4 \leq x^3+(3x^4+y^4)+3(x^4+y^4)+(x^4+3y^4)+y^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 10,
"answer_id": 6
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Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$. How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.
| One starts with
$$\begin{align}
\log(1+x)=&x-{x^2\over 2}+{x^3\over 3}+o(x^3)\\
\sin{X}=&X-{X^3\over 6}+o(X^3)
\end{align}$$
Now make $X=\log(1+x)$ in the development of the sine function to get
$$\begin{align}\sin(\log(1+x))=&x-{x^2\over 2}+{x^3\over 3}-{\left(x-{x^2\over 2}+{x^3\over 3}\right)^3\over 6}+o(x^4)\\
=&x-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiability: What if Left-hand and Right-hand Limit are Equal in $x$ but differ from $f(x)$? Earlier today I asked this question: Derivative of Function with Cases: $f(x)=x^2\sin x^{-1}$ for $x\ne0$
Following the answers to my question I worked on my problem. Among other things I showed that the left hand limit o... | 1.There is no real number $\frac {1}{0}.$ Any computations using it are false or meaningless.
*If $f(x)=x^2 \sin 1/x$ for $x>0$ and $f(x)=10$ for $x\leq 0$ then for $x>0$ we have $$\frac {f(x)-f(0)}{x-0}=\frac {x^2\sin 1/x -10}{x}=x(\sin 1/x) -10/x $$which has no limit as $x\to 0.$ So $f$ has no "upper" derivative a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Alternative form of Eisenstein integers I just recently got into number theory and algebra so my knowledge is very limited. I stumbled upon the Eisenstein integers $\mathbb{Z}[\omega]$. They are of the form $\varepsilon = a + b \omega$, where $\omega = e^{2 \pi i/3} = -1/2 + i \sqrt{3}/2$.
Now my problem:
I want to sh... | Your end number is $$\frac{(a+2k)+bi\sqrt{3}}{2}$$ which is of the right form.
Another way to argue is that if $x+y\sqrt{-3}$ is an integer where $x,y\in \mathbb{Q}$ then $x-y\sqrt{-3}$ is also an integer and thus
$2x$ and $x^2+3y^2$ are integers. This implies that $3(2y)^2$ is an integer and thus $2y$ is an integer s... | {
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"url": "https://math.stackexchange.com/questions/2097658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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Prove $a_{n+2}=3a_n+2\sqrt{2a_n^2+2a_{n+1}^2}$ is an integer Given the sequence
$$a_1=a_2=1;\ a_ {n+2} = 3a_n + 2\sqrt{2a_n^2 + 2a_{n+1}^2}$$
prove that $a_n$ is an integer for all $n\in\mathbb N$.
Attempt
It is enough to show that $2a_n^2 + 2a_{n+1}^2$ is a perfect square. That means it's an even perfect square and so... | We have that:
$$a_{n+2}-3a_n=2\sqrt{2a_n^2+2{a_{n+1}}^2}\Rightarrow (a_{n+2}-3a_n)^2=(2\sqrt{2a_n^2+2{a_{n+1}}^2})^2\Rightarrow\\ {a_{n+2}}^2-6a_na_{n+2}+9a_n^2=8a_n^2+8{a_{n+1}}^2\Rightarrow {a_{n+2}}^2+{a_{n+1}}^2=-a_n^2+6a_na_{n+2}+9{a_{n+1}}^2\Rightarrow \\{a_{n+2}}^2+{a_{n+1}}^2=-a_n^2+6a_n(3a_n+2\sqrt{2a_n^2+2{a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?
How many subsets of set $\{1,2,\ldots,10\}$ contain at least one odd integer?
My Working:
What I can think of is subtracting the no. of subsets that don't contain a single odd number from the total number of subsets as if we calculate ... | Others have already explained the easy solution, here is an alternative more similar to what you tried.
We want to know how many subsets contain exactly $k$ odd integers, from $k = 1$ to $5$, and sum.
*
*$k = 1$: $\binom{5}{1} = 5$ possible subsets of $\{1,3,5,7,9\}$
*$k = 2$: $\binom{5}{2} = 10$ possible subsets o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
Find the remainder when $140^{67}+153^{51}$ is divided by $17$ Find the remainder when $$140^{67}+153^{51}$$ is divided by $17$.
$$140\equiv 4 \pmod {17}$$
$$67\equiv 16 \pmod{17}$$
$$153 \equiv 0 \pmod{17}$$
$$51\equiv 0 \pmod{17}$$
$$\Rightarrow 140^{67}+153^{51}\equiv 4^{16}+0 \equiv 1\pmod{17}$$
Solution should be ... | One has $140\equiv 4\pmod{17}$. Now $4^4=256\equiv 1\pmod{17}$ and $67=16\cdot 4+3$ so
$$140^{67}\equiv (4^4)^{16}\cdot 4^3\equiv 4^3\equiv13 \pmod{17}$$
Similarly one has $153\equiv 0\pmod{17}$ and so $153^{51}\equiv 0\pmod{17}$.
So the answer is indeed $13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.
*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$
I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, bu... | Polynomial long division:
\begin{array}{c|cc}
& &\color{red}{x^3} & \color{green}{-2x^2}& \color{blue}{-9x}& \color{orange}{+1}& \\ \hline
10x+2 & 10x^4 & -18x^3 & -94x^2& -8x& +2 \\
& \color{red}{10x^4} &\color{red}{+2x^3}\\
& & -20x^3 & -94x^2& \vdots & \vdots\\
& & \color{green}{-20x^3} & \color{green}{-4x^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Triangle inequality in C Show that $\left|Im(2+ z^{c} -4z^2) \right| \leq 9.5$ When $ \left| z \right| \leq \frac {3}{2}$
$z^c$= compliment of z
$\left|Im(2+ z^{c} -4z^2) \right| \leq \left| 2+z^{c} - 4z^2 \right|$
I have tried to split it up directly i have tried to force complete the square i always get a weird valu... | You're trying to prove the unprovable. Let $z=re^{it}$, such that
$$Im(2+ z^{c} -4z^2) =Im(z^c)-4Im(z^2)=-r\sin (t)-4r^2sin(2t).$$
For $r=3/2$, this expression is $\approx-10.076$ at $t\approx 0.814$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int\frac{dy}{(y^2 + ay + b)^{3/2}}$ Need the indefinite integral with $a,b$ constants. This is not a homework question. Need it for a physics problem.
| $∫\frac{dy}{(y2+ay+b)^{3/2}} = ∫\frac{dy}{((y + \frac{a}{2})^2+\frac{1}{4}(4b-a^2))^{3/2}} = |\frac{a}{2} + y = u , dy = du| = ∫\frac{du}{(u^2+\frac{1}{4}(4b-a^2))^{3/2}} = | t = \frac{2u}{\sqrt{4b - a^2}} , dt = \frac{2du}{\sqrt{4b-a^2}} = \frac{4}{4b-a^2}∫\frac{dt}{(t^2+1)^{3/2}}= | t = tan(z) , dt = \frac{dz}{cos^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate the following indefinite integral Calculate: $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx.$$
I have no idea how to start it.
| Let $u = x^2-x+1$, we have $u' = 2x-1$. Notice
$$\begin{align}
2x^2+x+\frac12 &= 2(x^2-x+1) + \frac32 (2x-1) = 2u + \frac32 u'\\
6x^2-7x+\frac{13}{2} &= 6(x^2-x+1) - \frac12(2x-1) = 6u - \frac12 u'
\end{align}
$$
The integrand of the integral can be rewritten as
$$\begin{align}
& \frac{(2u + \frac32 u')\cos(2x) + (6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Find the limit of $\frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$
Let $\{x_n\}_{n\geq 1}$ be defined as $x_n = \frac{n}{n^3+1}+\frac{2n}{n^3+2}+\dots+\frac{n\cdot n}{n^3+n}$. Then $\lim\limits_{n\to\infty} x_n$ is?
I want to find limit of this problem by a very specific method . I am uploading the pic... | Starting where you left: for $n\geq 1$,
$$
x_n = \frac{1}{n}\sum_{k=1}^n \frac{\frac{k}{n}}{1+\frac{k}{n^3}} \tag{1}
$$
While this furiously looks like a Riemann sum, it is not one due to the $\frac{k}{n^3}$ in the denominator. But then, we can use the squeeze theorem: as $1\leq k\leq n$,
$$
\frac{1}{1+\frac{1}{n^2}}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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