Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$(x^2+1)(y^2-1)dx+xydy=0$
Solve $(x^2+1)(y^2-1) \, dx+xy \, dy=0$
My solution :
$$\begin{align}
\frac{x^2+1}{-x} \, dx &= \frac{-y}{y^2-1} \, dy \\
\left(-x-\frac{1}{x}\right) \, dx &= -\frac{1}{2}\frac{2y}{y^2-1} \, dy \\
\int \left(-x-\frac{1}{x}\right) \, dx &= -\frac{1}{2} \int \frac{2y}{y^2-1} \, dy \\
\frac{x^2}{2}+\ln|x| &= \frac{1}{2}\ln|y^2-1| \\
y^2 &= e^{x^2+2\ln|x|}+1 \\
y &= \pm\sqrt{e^{x^2+2\ln|x|}+1}
\end{align}$$
I got these solutions using ode calculator, and I don't get why it is correct and where am I wrong?
$$y = \pm \frac{\sqrt{e^{-x^2+c_1}+x^2}}x$$
Help please.
Thanks !
| Here I propose a solution with which you can compare:
\begin{align*}
(x^{2} + 1)(y^{2} - 1) + xyy' = 0 & \Longleftrightarrow \frac{x^{2} + 1}{x} + \frac{yy'}{y^{2} - 1} = 0\\\\
& \Longleftrightarrow \frac{x^{2}}{2} + \ln|x| + \frac{1}{2}\ln\left|y^{2} - 1\right| = c_{0}\\\\
& \Longleftrightarrow \ln\left|y^{2} - 1\right| = -x^{2} - \ln(x^{2}) + c_{1}\\\\
& \Longleftrightarrow y^{2} - 1 = \frac{\exp\left(-x^{2} + c_{1}\right)}{x^{2}}\\\\
& \Longleftrightarrow y^{2} = \frac{\exp\left(-x^{2} + c_{1}\right) + x^{2}}{x^{2}}\\\\
& \Longleftrightarrow y(x) = \pm\sqrt{\frac{\exp\left(-x^{2} + c_{1}\right) + x^{2}}{x^{2}}}
\end{align*}
If you still have any questions, please let me know.
Hopefully this helps !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to evaluate the integral $\int_{0}^{\frac{\pi}{2}}\sqrt[n]{\tan\theta} d \theta$, where $n\geq 2$? In my post, I started to investigate the integral $\int_0^{\frac{\pi}{2}} \sqrt{\tan \theta} d \theta$ and then$\int \sqrt[3]{\tan \theta} d \theta$ in post.
After encountering the Beta Functions, I want to try to apply it to the integral.
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \sqrt{\tan \theta} d \theta &=\int_{0}^{\frac{\pi}{2}} \sin ^{\frac{1}{2}} \theta \cos ^{-\frac{1}{2}} \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{3}{4}\right)-1} \theta \cos ^{2\left(\frac{1}{4}\right)-1} \theta d \theta \\
&=\frac{1}{2} B\left(\frac{3}{4}, \frac{1}{4}\right) \\&=\frac{\pi}{\sqrt2}
\end{aligned}
$$
I then go further to $$
I_n=\int_{0}^{\frac{\pi}{2} }\sqrt[n]{\tan \theta} d\theta.
$$
Similarly
$$
\begin{aligned}
I_n&=\int_{0}^{\frac{\pi}{2}} \tan ^{\frac{1}{n} } \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \sin ^{\frac{1}{n} } \theta \cos ^{-\frac{1}{n} } \theta d \theta \\
&= \int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{n+1}{2 n}\right)-1} \theta \cos ^{2\left(\frac{n-1}{2 n}\right)-1} d \theta \\
&=\frac{1}{2} B\left(\frac{n+1}{2 n}, \frac{n-1}{2 n}\right)
\end{aligned}
$$
Applying the theorem $$
B(x, 1-x)=\pi \csc (\pi x), \textrm{ where } x\notin Z
$$
gives
$$
\boxed{\int_{0}^{\frac{\pi}{2}} \sqrt[n]{\tan \theta} d \theta =\frac{\pi}{2} \csc \left(\frac{n+1}{2 n} \pi\right)=\frac{\pi}{2} \sec \left(\frac{\pi}{2 n}\right)}
$$
which is unexpectedly beautiful and decent.
Furthermore
Replacing $\frac{1}{n}$ by $a$ yields
$$\boxed{
\int_{0}^{\frac{\pi}{2}} \tan ^{a} \theta d \theta =\frac{\pi}{2} \csc \left(\frac{a+1}{2} \pi\right)=\frac{\pi}{2} \sec \frac{a \pi}{2}}
$$
For example, $$
\int_{0}^{\frac{\pi}{2}} \sqrt[3]{\tan \theta} d \theta =I\left(\frac{1}{3}\right) =\frac{\pi}{2} \sec \left(\frac{ \pi}{6}\right)=\frac{\pi}{\sqrt{3}} $$
$$\int_{0}^{\frac{\pi}{2}} \sqrt[6]{\tan \theta} d \theta =I\left(\frac{1}{6} \right)=\frac{\pi}{2} \sec\left(\frac{\pi}{12}\right)=\pi \sqrt{2-\sqrt{3}}$$
$$
\int_{0}^{\frac{\pi}{2}} \tan ^{\frac{1}{e}}\theta d\theta =\frac{\pi}{2} \sec \frac{\pi}{2 e}
$$
$$
\int_{0}^{\frac{\pi}{2}} \tan ^{\frac{1}{\pi}} \theta d \theta=\frac{\pi}{2} \sec \frac{1}{2}
$$
checked by Wolframalpha.
My question: Is there a method without using Beta Functions?
| Do the change of variable $w = \tan \theta$
$$ I = \int_0^{\frac{\pi}{2}} \sqrt[n]{\tan \theta} d \theta = \int_{0}^{\infty} \frac{w^{\frac{1}{n}}}{1+w^2}dw $$
Recall the integral representation of the $\sec(x)$ function:
$$ \sec(x) = \frac{2}{\pi}\int_{0}^{\infty} \frac{t^{\frac{2x}{\pi}}}{t^2+1}dt \quad |x|<\frac{\pi}{2}$$
Of course this formula can be proven using the complete beta function but it can also be proven using contour integration around a branch point, the first answer here contains both proofs. There are other other methods prescinding the beta function, here is a good compilation.
If we put $\displaystyle x = \frac{\pi}{2n}$:
$$ \sec\left(\frac{\pi}{2n}\right) = \frac{2}{\pi}\int_{0}^{\infty} \frac{t^{\frac{1}{n}}}{t^2+1} dt$$
$$\Longrightarrow \int_{0}^{\infty} \frac{t^{\frac{1}{n}}}{t^2+1} dt = \frac{\pi}{2} \sec\left(\frac{\pi}{2n}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to find radius of convergence of given series? Consider the series
$$\frac{1}{3}+\frac{x}{5}+\frac{x^2}{3^2}+\frac{x^3}{5^2}+\frac{x^4}{3^3}+\frac{x^5}{5^3}+\cdots$$
I want to find the radius of convergence of this series.
$\textbf{Here is my work}$
I tried to figure out $a_n$ for the given power series but it did not work.
$$a_n=3^{-\frac{n+1}{2}|\sin(\frac{n\pi}{2})|} \times5^{-\frac{n}{2}|\sin(\frac{(n+1)\pi}{2})|}$$
$$L=\lim_{n\to \infty}\frac{a_{n+1}}{a_n
}
$$
Also I thought about finding radius of convergence separately for even and odd terms but I am not sure how is that going to help.
| The series:
$$
S=\frac{1}{3}+\frac{x}{5}+\frac{x^2}{3^2}+\frac{x^3}{5^2}+\frac{x^4}{3^3}+\frac{x^5}{5^3}+...
$$
Can be expressed as:
$$
S=\sum_{k=1}^{\infty} {a_k}
$$
Where:
$$
a_k=\frac{x^{k-1}}{3^k}+\frac{x^{k}}{5^k}
$$
Now, testing the convergency:
$$
\lim_{k \rightarrow \infty}\left|{\frac{a_{k+1}}{a_k}}\right|=
\lim_{k \rightarrow \infty}\left|{\frac{\frac{x^{k}}{3^{k+1}}+\frac{x^{k+1}}{5^{k+1}}}{\frac{x^{k-1}}{3^k}+\frac{x^{k}}{5^k}}}\right|=
\lim_{k \rightarrow \infty}\left|{\frac{\frac{1}{3^{k+1}}+\frac{x}{5^{k+1}}}{\frac{1}{3^k x}+\frac{1}{5^k}}}\right|=
\lim_{k \rightarrow \infty}\left|{\frac{1+x\left(\frac{3}{5}\right)^{k+1}}{\frac{3}{x}+3\left(\frac{3}{5}\right)^k}}\right|=...
$$
$$
...=\lim_{k \rightarrow \infty}\left|\frac{x}{3}\right|<1
$$
Hence, we get:
$$
\left|x\right|<3
$$
And finally radius of convergence is $r=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all the integers which are of form $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}, a,b,c\in \mathbb{N}$, any two of $a,b,c$ are relatively prime.
I have a question which askes to find all the integers which can be
expressed as
$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}$
where $a,b,c\in \mathbb{N} $ and any two of $a,b,c$ are relatively
prime.
My approach:
Since they told any two of $a,b,c$ are relatively prime, so:
$\displaystyle \tag*{} \begin{align} \text{gcd}(a,b) &= 1 \\ \text{gcd}(b,c) &= 1 \\ \text{gcd}(c,a) &= 1\end{align}$
We have:
$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c} = \dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}$
And we have to prove:
$\displaystyle \tag*{} \frac{a}{b} + \frac{b}{a}+ \frac{c}{a} + \frac{a}{c}+ \frac{b}{c} + \frac{c}{b} \in \mathbb{N} $
We have the following that if $\text{gcd}(m,n)=1$ and $m,n,k \in \mathbb{Z}$ then
$\displaystyle \tag*{} \frac{m}{n} + \frac{n}{m} = k$
has only $1$ positive solution, which is $m=n=1$
So, the only solution I found is $a=b=c=1$.
Is there anything I am missing? Any help would be appreciated, thank you. :)
| NB: While I was typing up my proof, @user provided a similar proof, leaving mine essentially a duplicate. However, I'll leave it since I may have written it out in slightly greater detail.
First, we may note that $c$ must divide $a+b$, etc. To see this, write the sum on common denominator
$$
\frac{b+c}a+\frac{a+c}b+\frac{a+b}c
=\frac{bc(b+c)+ac(a+c)+ab(a+b)}{abc}
$$
and note that, for $c$ to divide the numerator, after removing terms that are multiples of $c$, we are left with $c|ab(a+b)$, which forces $c|a+b$. Similarly, $a|b+c$ and $b|a+c$.
We may assume that $a\le b\le c$, where equality is only possible for the number 1. So let's first deal with the cases where equalities are an issue.
First, $a=b=c=1$ is a solution. Ie, $(a,b,c)=(1,1,1)$.
Next, if $a=b=1<c$, we need $c|a+b$, for which the only option is $c=2$. This is also a solution. Ie $(a,b,c)=(1,1,2)$.
So now we are left with the case $a<b<c$. Since $c|a+b$, but $a+b<2c$, we must have $c=a+b$. Next $b|a+c=2a+b$, so $b|2a$; however, $a<b$, so $2a=b$ since $2a$ is a multiple of $b$ but less than $2b$. This also forces $a=1$ since $a$ and $b=2a$ should be relatively prime. This makes $(a,b,c)=(1,2,3)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $\int_{0}^\infty \frac{1}{(x^2 +b^2)^4}\ dx$ I ran into this integral in the context of Quantum Mechanics, and I don't really know how to tackle it. Here, $b$ is simply a real constant, which we can assume is positive.
$$\int\limits_0 ^\infty\frac{1}{(x^2+b^2)^4}\ dx$$
It doesn't look like I can use "traditional" methods to solve it, so I was thinking to maybe try to transform it to complex integral somehow and apply Cauchy's theorem or something, but I'm unsure if that would even work.
Any nudge in the right direction would be appreciated!
| $$I=\int_0^\infty \frac{dx}{(x^2+b^2)^4}$$
First I’ll start by ridding the integrand of $b$
$$I=\int_0^\infty \frac{dx}{(x^2+b^2)^4}=\int_0^\infty\frac{dx}{b^8((\frac{x}{b})^2+1)}$$
Substitute $u=\frac{x}{b}$
$$I=\frac{1}{b^7}\int_0^\infty\frac{dx}{(x^2+1)^4}$$
Now we will start building a reduction formula for the indefinite integral $\int \frac{dx}{(x^2+1)^n}$
$$J_{n}=\int \frac{dx}{(x^2+1)^n}=\int \frac{x^2+1-x^2}{(x^2+1)^n}dx=\int\frac{dx}{(x^2+1)^{n-1}}-\int\frac{x\cdot x}{(x^2+1)^n}dx$$
We can recognize that the first integral is $J_{n-1}$, and we will use integration by parts on the second, where $u=x$ and $dv=\frac{x}{(x^2+1)^n}$
$$J_{n}=J_{n-1}-\left[\frac{-x}{(2n-2)(x^2+1)^{n-1}}+\frac{1}{2n-2}\int \frac{dx}{(x^2+1)^{n-1}} \right]$$
Simplify
$$J_{n}=J_{n-1}-\frac{1}{2n-2}J_{n-1}+\frac{x}{(2n-2)(x^2+1)^{n-1}}=\frac{2n-3}{2n-2}J_{n-1}+\frac{x}{(2n-2)(x^2+1)^{n-1}}$$
Now we’ll introduce a new sequence, $I_{n}$, and find a recursive formula using our information on $J_{n}$
$$I_{n}=\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{2n-3}{2n-2}I_{n-1}+\left. \frac{x}{(2n-2)(x^2+1)^{n-1}}\right |_0^\infty=\frac{2n-3}{2n-2}I_{n-1}$$
Therefore
$$\frac{I_{n}}{I_{n-1}}=\frac{2n-3}{2n-2}$$
$$\frac{I_{n}}{I_{n-1}}\frac{I_{n-1}}{I_{n-2}} \frac{I_{n-2}}{I_{n-3}}\cdots\frac{I_{3}}{I_{2}}\frac{I_{2}}{I_{1}}=\frac{2n-3}{2n-2}\frac{2n-5}{2n-4}\frac{2n-7}{2n-6}\cdots\frac{3}{4}\frac{1}{2}$$
On the left side we cancel almost everything. The right side can be rewritten using double factorials.
$$\frac{I_{n}}{I_1}=\frac{(2n-3)!!}{(2n-2)!!}$$
$$I_1=\int_0^\infty \frac{dx}{x^2+1}=\frac{\pi}{2}$$
Therefore
$$I_{n}=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}$$
We can calculate the original integral like this:$$
I=\frac{I_4}{b^7}=\frac{5!!}{6!!}\frac{\pi}{2}\frac{1}{b^7}=\frac{5\pi}{32b^7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Simplification of $\frac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$
Simplify
$$\dfrac{2\sqrt{21}-\sqrt{35}+5\sqrt{15}-16}{\sqrt7+2\sqrt5-\sqrt3}$$
Final solution should have rational denominators.
Suppose the solution is $X$, I have tried to make up an equation for $X^2$
$$X^2 = \frac{200}{-2\sqrt{15}+2\sqrt{35}-\sqrt{21}+10}$$
My idea is that solving for $X^2$, if can be simply done, can easily give us $X$.
Please help!
| Hint: we are generalising the method of rationalising a denominator of two terms. For instance, if our denominator was $\sqrt{a}-b$ then we would multiply (numerator and denominator) by $\sqrt{a}+b$ to get $a-b^2$. That is, we are using "difference of two squares".
Observe that $[(\sqrt{7}+2\sqrt{5})-\sqrt{3}][(\sqrt{7}+2\sqrt{5})+\sqrt{3}]=(\sqrt{7}+2\sqrt{5})^2-(\sqrt{3})^2=7+4\sqrt{35}+20-3$
Now what can we do?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Weird Maximization Problem For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$. Let $x_1, x_2, \dots, x_{216}$ be positive real numbers such that $\displaystyle\sum_{i=1}^{216} x_i=1$ and
$$\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}.$$Find the maximum possible value of $x_2.$
I simplified the condition to $\displaystyle\sum_{i=1}^{216}\dfrac{x_i^2}{1-a_i}=\dfrac{1}{215},$ but I'm not sure what to do next.
| Given $ \sum \frac{ x_i ^2 } { 1 - a_i} = \frac{1}{215}$, apply Cauchy Schwarz / Titu's lemma to get
$$ \sum \frac{ x_i ^2 } { 1 - a_i} \geq \frac{ ( \sum x_i) ^ 2 } { \sum 1 - a_i } = \frac{ 1^2 } { 215 } . $$
Since equality holds throughout, we conclude that $ \frac{ x_i } { 1-a_i}$ is a constant, say $k$.
Since $ 1 = \sum x_i = \sum (1-a_i)k = 215 k$, so $ k = \frac{1}{215}$.
Hence, $ x_ i = \frac{ 1 - a_i } { 215 }$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the minimum and maximum value of $F=|a-2b|+|b-2c|+|c-2a|$
Let $a, b, c \in \Bbb R$ satisfy $a^2+b^2+c^2=21$. Find the minimum and maximum value of $$F=|a-2b|+|b-2c|+|c-2a|$$
I found $7\le F \le \sqrt{399}$ but couldn't prove it. I was thinking of the following inequality:
$$|x_1+x_2+\cdots+x_n|\le|x_1|+|x_2|+\cdots+|x_n|\le \sqrt{n \left(x_1^2+x_2^2+\cdots+x_n^2\right)}$$
but they are not really efficient. Does anyone know how to solve this problem or know where it first appeared?
| The min and max bounds of $7$ and $\sqrt{399}$ are indeed correct. Let $$F_L = |a + b + c|$$ and $$F_U = \sqrt{15(a^2 + b^2 + c^2) - 12(ab + bc + ac)}$$
By the inequality you've proposed, we have $F_L \le F \le F_U$, and presumably you've performed the steps to show that $\min_{a^2 + b^2 + c^2 = 21} F_L = 7$ and $\max_{a^2 + b^2 + c^2 = 21}F_U = \sqrt{399}$; the values to obtain the min and max for these lower and upper bounds are $(a, b, c) = (4, 2, 1)$ and $(a, b, c) = (-3\sqrt{\frac{21}{19}}, \sqrt{\frac{21}{19}}, 3\sqrt{\frac{21}{19}})$. Substituting these in for $F$ also obtains the lower and upper bounds.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by induction that $(b^n-1)\cdots (b^n-b^{n-2})\ge b^{n(n-1)}-b^{n(n-1)-1}$ Let $b\ge 2$ and $n$ a natural number at least 1. Prove that
$$(b^n-1)\cdots (b^n-b^{n-2})\ge b^{n(n-1)}-b^{n(n-1)-1}$$
The base-case is obvious.
For the inductive case, assume this holds for $n$ and we show it for $n+1$. I started by writing $(b^{n+1}-1)\cdots (b^{n+1}-b^{n-1})$ and tried to write a chain of inequalities arriving at $b^{(n+1)n}-b^{(n+1)n-1}$. This goal is the same as $b^{n(n+1)-1}(b-1)$.
We can factor a $b$ out of every factor other than the first:
$$(b^{n+1}-1)\cdots (b^{n+1}-b^{n-1}) = b^{n-1}(b^{n+1}-1)(b^n-1)(b^n-b)\cdots (b^n-b^{n-2})$$
and apply the inductive hypothesis, so that the above is
$$\ge b^{n-1}(b^{n+1}-1)(b^{n(n-1)}-b^{n(n-1)-1}) $$
$$ = b^{n-1+n(n-1)-1}(b^{n+1}-1)(b-1) $$
$$= b^{n^2-2}(b^{n+1}-1)(b-1) $$
Therefore it suffices to show that
$$ b^{n^2-2} (b^{n+1}-1) \ge b^{n^2+n-1}$$
or
$$b^{n+1}-1\ge b^{n+1}$$
... which ... eh.
Since the only rounding I did was using the inductive hypothesis, it must be that I'm getting a bad result because I'm using the inductive hypothesis in a bad way--like because I'm using it in a product with a large number. But I don't see an alternate approach.
| I'm close,
but it's close to midnight here
so I'll stop with what I have
so far.
I have shown that the insquality
is equivalent to
$\prod_{k=2}^{n}(1-1/b^{k})
\ge 1-1/b
$.
I can't get a proof,
but I'm sure this is true.
Here's my derivation.
Want
$(b^n-1)\cdots (b^n-b^{n-2})\ge b^{n(n-1)}-b^{n(n-1)-1}
$
or
$\prod_{k=0}^{n-2}(b^n-b^k)\ge b^{n(n-1)}-b^{n(n-1)-1}
$
or
$\prod_{k=0}^{n-2}b^k(b^{n-k}-1)
\ge b^{n(n-1)-1}(b-1)
$
or
$b^{\sum_{k=0}^{n-2}k}\prod_{k=0}^{n-2}(b^{n-k}-1)
\ge b^{n(n-1)-1}(b-1)
$
or
$b^{(n-2)(n-1)/2}\prod_{k=0}^{n-2}(b^{n-k}-1)
\ge b^{n(n-1)-1}(b-1)
$
or
$\begin{array}\\
\prod_{k=2}^{n}(b^{k}-1)
&\ge b^{n^2-n-1-(n^2-3n+2)/2}(b-1)\\
&= b^{(n^2+n-4)/2}(b-1)\\
&= b^{n(n+1)/2-2}(b-1)\\
\end{array}
$
For $n=2$
this is
$b^2-1
\ge b(b-1)
$
which is true.
For $n=3$ this is
$(b^2-1)(b^3-1)
\ge b^{4}(b-1)
$
which is true
for $b \ge 2$.
For $n=4$ this is
$(b^2-1)(b^3-1)(b^4-1)
\ge b^{8}(b-1)
$
which is true
for $b \ge 2$.
Since
$\sum_{k=2}^n k
=n(n+1)/2-1
$,
dividing by
$b^{n(n+1)/2-1}
$,
this is
$\prod_{k=2}^{n}(1-1/b^{k})
\ge 1-1/b
$.
Letting $x=1/b$ this is
$\prod_{k=2}^{n}(1-x^k)
\ge 1-x
$
for $0 < x \le 1/2
$
or
$\dfrac1{1-x}
\ge \dfrac1{\prod_{k=2}^{n}(1-x^k)}
$
or
$\dfrac1{(1-x)^2}
\ge \dfrac1{\prod_{k=1}^{n}(1-x^k)}
$.
The right side looks like
the power series
for restricted partitions,
but I don't know where to go from here,
so I'll stop.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many instances of number $k$ in sums that add up to number $n$? Task in Combinatorics course:
$1 \leq k < n$. Prove that the number $k$ can be found $(n-k+3)2^{n-k-2}$ times in sums that add up to $n$.
Example: $n=4, k=2$.
$1+1+2$
$1+2+1$
$2+1+1$
$2+2$
These are the sums that add up to $4$ and the number $2$ can be found $5$ times.
Thus far, I have tried making sense of that formula. $n-k$ is the number or countables left if we have one $k$. $2^{n-1}$ is how many sums that add up to $n$ there are in total so in the second half of the formula it could be $2^{n-1}2^{k-1}$ that would add up to $2^{n-k-2}$. But none of that comes together. Why would there be a $+3$ in the first half.
| We have the marked generating function
$$Q(z,u) = \sum_{q=1}^n
\left(z+\cdots+z^{k-1} + u z^k + z^{k+1} + \cdots\right)^q.$$
The term $u^p z^m$ represents a composition that sums to $m$ and
contains $p$ instances of the number $k.$ Hence we differentiate with
respect to $u$ and set $u=1$ to get the count. Differentiation yields
$$\sum_{q=1}^n
q \left(z+\cdots+z^{k-1} + u z^k + z^{k+1} + \cdots\right)^{q-1}
z^k.$$
Set $u=1$ to get
$$[z^n] z^k \sum_{q=1}^n q \frac{z^{q-1}}{(1-z)^{q-1}}.$$
Note that with $k\ge 1$ we may extend $q$ beyond $n$ without getting any
additional contributions to the coefficient extractor. This yields
$$[z^n] z^k \sum_{q\ge 1} q \frac{z^{q-1}}{(1-z)^{q-1}}
= [z^{n-k}] \frac{1}{(1-z/(1-z))^2}
\\ = [z^{n-k}] (1-z)^2 \frac{1}{(1-2z)^2}
\\ = [z^{n-k}] \frac{1}{(1-2z)^2}
- 2 [z^{n-k-1}] \frac{1}{(1-2z)^2}
+ [z^{n-k-2}] \frac{1}{(1-2z)^2}
\\ = (n-k+1) 2^{n-k} - 2(n-k) 2^{n-k-1} + (n-k-1) 2^{n-k-2}
\\ = (4n-4k+4-4n+4k+n-k-1) 2^{n-k-2}
= (n - k + 3) 2^{n-k-2}.$$
This is with $n-k\ge 2.$ We get for $k=n-1$
$$(n-k+1) 2^{n-k} - 2(n-k) 2^{n-k-1}
= 2\times 2 - 2 = 2$$
which is true because the compositions are $(1,n-1)$ and $(n-1,1).$ We
get for $k=n$
$$(n-k+1) 2^{n-k} = 1$$
which is also true because there is just one composition $(n).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to tackle the integral $\int_{0}^{1} \sqrt{-1+\sqrt{\frac{4}{x}-3}} d x$? $ \text {Let } y=\sqrt{-1+\sqrt{\frac{4}{x}-3}}\textrm{ then ,}$
$ \displaystyle \begin{aligned}I&=16 \int_{0}^{\infty} \frac{y^{2}\left(y^{2}+1\right) d y}{\left(y^{4}+2 y^{2}+4\right)^{2}}\\&=4\left[3 \underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}+2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}} d y}_{J}+\underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}-2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}}}_{K} d y\right] \end{aligned}\tag*{} $
Now let’s play a little trick on the integral $ J$.
$\displaystyle \begin{aligned}J &=\int_{0}^{\infty} \frac{1+\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{\infty} \frac{d\left(y-\frac{2}{y}\right)}{\left[\left(y-\frac{2}{y}\right)^{2}+6\right]^{2}} \\&=\int_{-\infty}^{\infty} \frac{d u}{\left(u^{2}+6\right)^{2}}\\ &\stackrel{u=\sqrt6 \tan \theta}{=}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sqrt{6} \sec ^{2} \theta d \theta}{\left(6 \sec ^{2} \theta\right)^{2}}\\&=\frac{\pi}{12 \sqrt{6}} \end{aligned} \tag*{} $
For the integral $ K$ , we first split the interval into two.
$ \displaystyle \begin{aligned}K &=\int_{0}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{1-\frac{1}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y+\int_{1}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}}+\int_{3}^{\infty} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}} d y \\&=\int_{\infty}^{3} \frac{d u}{\left(u^{2}-2\right)^{2}}+\int_{3}^{\infty} \frac{d v}{\left(v^{2}-2\right)^{2}} \\&=0 \end{aligned} \tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=4\left(3 \cdot \frac{\pi}{12 \sqrt{6}}\right)=\frac{\pi}{\sqrt{6}}}\tag*{} $
My Question
Is there any other substitution or method to tackle the integral?
| Taking the inverse function $x=x(y)$ and integrating between $0$ and $\infty$ gives that:
$\int_0^\infty \frac{4}{y^{4}+2y^{2}+4}dy$,
Decomposing the rational function into the sum of several simple fractions gives:
$\frac{4}{(y^{2}+\sqrt{2}y+2)( y^{2}-\sqrt{2}y+2)}=$,
$\frac{A y+B}{ y^{2}+\sqrt{2}y+2}+\frac{C y+D}{ y^{2}-\sqrt{2}y+2}$,
$A=\frac{\sqrt{2}}{2},B=1, C=-\frac{\sqrt{2}}{2}, D=1$.
Integrating between $0$ and $\infty$, we get:
$\frac{\pi}{\sqrt{6}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Obtaining a tight bound for an Expectation w.r.t a uniform random variable Let $x\in [0,1]^{n+1}$, let $t > 0$, and let $u$ be a uniform random variable over {$1,\ldots, n, n+1$}, then I want to tightly bound
$$a_t(x) = E_u \left[ \mathrm{exp}\left\lbrace t\left(\frac{1}{n}\sum_{i\neq u}x_i - x_u \right) \right\rbrace \right]$$
with respect to a fixed value of $x$, or
$$\sup_{x\in[0,1]^{n+1}} a_t(x)$$
1st Attempt: (Hoeffding's Lemma)
This attempt bounds the supremum (and hence $a_t(x)$ for any x). If we let $Y = \frac{1}{n}\sum_{i\neq u}x_i - x_u$, then $E_u[Y] = 0$ and $ -1\leq Y \leq 1$ since $x_i\in [0,1]$ for all $i$. Then I can use Hoeffding's Lemma to get that:
$$a_t \leq e^{t^2\frac{(1 - (-1))^2}{8}} = e^{\frac{t^2}{2}}$$
I was wondering if one could obtain a tighter bound. Possibly a bound that depends on $n$. I tried to check whether $a_t$ is a convex or concave in $x$, so that maybe i could find the maximum analytically, but unfortunately it is neither.
Update: (Numerics)
I tried to find the supremum numerically and compare it to $e^{\frac{t^2}{2}}$. Since $a_t(x)$ is neither convex or concave, I maximized it with different initial points $x_0$ and averaged the maximum. I did this for each $n\in \left\lbrace 2,4\ldots, 40\right\rbrace$. I find it surprising that the difference initially increases rapidly but then somewhat plateaus for larger values of $n$. I tried this for several values of $t$ and it consistently followed this pattern.
Thank you!
Edit 1: Fixed a typo.
Edit 2: Added some numerics.
| I think we can get a fairly close bound by optimizing $a$ analytically. Note that
\begin{align*}
a_t(x) &= \frac{1}{n+1}\sum_{j = 1}^{n+1} \exp\left(t \left(\frac 1n \sum_{i \ne j} x_i - x_j \right) \right) \\
&= \frac{1}{n+1}\sum_{j = 1}^{n+1} \exp\left(t \left(\frac 1n \sum_{i} x_i - x_j - \frac 1n x_j\right) \right) \\
&= \frac{1}{n+1} \exp\left( \frac tn \sum_i x_i \right) \sum_i \exp\left(-t \left(1+\frac 1n \right)x_i\right) \\
&=: \frac{1}{n+1} f(x).
\end{align*}
For ease of notation, let $z := t(1+\frac 1n)$. Although $f$ may be neither concave nor convex, its maximum still occurs at a critical point. We compute
\begin{align*}
\frac{\partial}{\partial x_j} f &= te^{\frac tn \sum_i x_i} \left(\frac 1n \sum_i e^{-z x_i} - \left(1+\frac 1n\right) e^{-z x_j}\right).
\end{align*}
Therefore, at any critical point, we must have for all $j$ either $x_j = 0$, $x_j = 1$, or $x_j = -\frac 1z \ln(Y)$, where
\begin{align}
Y &= \frac 1{n+1} \sum_i e^{-z x_i} &(=e^{-z x_j})
\end{align}
We now attempt to find the critical point which is the maximizer. Let $J$ be the number of $x_j$ which are $0$, $K$ be the number of $x_j$ which are $1$, so $n+1-(J+K)$ are $-\frac 1z \ln(Y)$. We compute, from the definition of $Y$,
\begin{align*}
Y &= \frac 1{n+1} \sum_i e^{-z x_i} \\
&= \frac 1{n+1} (J + K e^{-z} + (n+1-J-K)Y),
\end{align*}
and solving gives
\begin{equation}
Y = \frac{J+Ke^{-z}}{J+K}.
\end{equation}
Then we compute
\begin{align*}
f(x) &= \exp\left( \frac tn \sum_i x_i \right) \sum_i e^{-zx_i} \\
&= \exp\left( \frac tn \left(K - \frac{1}{z} \ln(Y)(n+1-(J+K))\right) \right)(n+1)Y \\
&= (n+1) \exp \left( \frac tn K + \frac{J+K}{n+1} \ln(Y)\right),
\end{align*}
so we attempt to solve
\begin{align*}
\max_{J,K \ge 0}&\quad \frac tn K + \frac{J+K}{n+1} \ln(Y)\\
\text{s.t. } & \quad J+K \le n+1.
\end{align*}
Since $J$ and $K$ are integers, this could be done with a brute force search by checking at most $n^2$ values, but we can obtain an analytic upper bound by maximizing over $J, K \in \mathbb{R}_+$. Our Lagrangian is
\begin{align*}
\min_{\lambda \ge 0} \max_{J,K \ge 0} L(J,K,\lambda) &= \min_{\lambda \ge 0} \max_{J,K \ge 0} \left(\frac tn K + \frac{J+K}{n+1} \ln(Y) - \lambda (J+K-(n+1))\right)
\end{align*}
We first compute
\begin{align*}
\frac{\partial}{\partial J} Y &= \frac{K(1-e^{-z})}{(J+K)^2} \\
\frac{\partial}{\partial K} Y &= -\frac{J(1-e^{-z})}{(J+K)^2},
\end{align*}
so our first order conditions are
\begin{align*}
0 = \frac{\partial}{\partial J} L &= \frac{K(1-e^{-z})}{(n+1)(J+Ke^{-z})} + \frac{\ln(Y)}{n+1} - \lambda \\
0 = \frac{\partial}{\partial K} L &= \frac tn - \frac{J(1-e^{-z})}{(n+1)(J+Ke^{-z})} + \frac{\ln(Y)}{n+1} - \lambda.
\end{align*}
Setting them equal to one another yields
\begin{align*}
\frac tn - \frac{J(1-e^{-z})}{(n+1)(J+Ke^{-z})} &= \frac{K(1-e^{-z})}{(n+1)(J+Ke^{-z})}.
\end{align*}
Manipulating this expression shows
\begin{align*}
Y = \frac{J+Ke^{-z}}{J+K} &= \frac{1-e^{-z}}{z} \\
J &= \frac{1-e^{-z}-ze^{-z}}{z-1+e^{-z}}K \\
J+K &= \frac{z(1-e^{-z})}{z-1+e^{-z}}K.
\end{align*}
Substituting these expressions in, we see that we now need to find
\begin{align*}
\max_{K \ge 0} & \quad \frac tn K \left(1 + \frac{(1-e^{-z})}{z-1+e^{-z}}\ln \left(\frac{1-e^{-z}}{z}\right) \right) \\
\text{s.t.} & \quad K \le \frac{z-1+e^{-z}}{z(1-e^{-z})}(n+1).
\end{align*}
It is possible to show $1 + \frac{(1-e^{-z})}{z-1+e^{-z}}\ln \left(\frac{1-e^{-z}}{z}\right) > 0$ for all $z > 0$, so this expression is maximized by $K = \frac{z-1+e^{-z}}{z(1-e^{-z})}(n+1)$. Hence our maximizers are
\begin{align*}
K^* &= \frac{z-1+e^{-z}}{z(1-e^{-z})}(n+1) \\
J^* &= n+1-K \\
Y^* &= \frac{1-e^{-z}}{z}
\end{align*}
Plugging into our expression, we have
\begin{align*}
f(x) &\le (n+1) \exp \left( \frac tn K^* + \frac{J^*+K^*}{n+1} \ln(Y^*)\right) \\
&= (n+1) Y^*e^{\frac tn K^*} \\
&= (n+1) \left(\frac{1-e^{-z}}{z} \right)\exp \left( \frac{z-1+e^{-z}}{1-e^{-z}} \right),
\end{align*}
and therefore
\begin{align*}
a_t(x) &\le \left(\frac{1-e^{-z}}{z} \right)\exp \left( \frac{z-1+e^{-z}}{1-e^{-z}} \right),
\end{align*}
where $z = t(1+\frac 1n)$. I haven't checked precisely how this compares to the $e^{\frac{t^2}2}$ bound, but it should beat it quite handily for large $t$. As $n$ increases, $z$ initially decreases (relatively) rapidly but plateaus around $t$. Since the bound is an increasing function of $z$, this agrees with the behavior you saw numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Computation of maximum likelihood estimates Question
Solve for the maximum likelihood estimates of the distribution with parameters $\alpha$ and $\beta$ and the following pdf, where $x > 0$: $$f(x; \alpha, \beta) = \frac {\alpha} {\sqrt {2\pi \beta}} x^{-\frac 3 2} \exp\left[-\frac {(\alpha - \beta x)^2} {2\beta x}\right].$$
Hint: You should be able to write the parameter estimates in terms of $\overline{x}$ and $\overline{1/x}$, where $\overline{1/x} = \frac 1 n \sum^n_{i = 1} \frac 1 {x_i}$.
My working
$$\begin{aligned}
l(\alpha, \beta) & = n\ln\frac {\alpha} {\sqrt{2\pi \beta}} - \frac 3 2 \sum^n_{i = 1} \ln x_i - \sum^n_{i = 1} \frac {(\alpha - \beta x_i)^2} {2\beta x_i}\\
& = n\ln \alpha - \frac 1 2 n \ln(2\pi) - \frac 1 2 n \ln \beta - \frac 3 2 \sum^n_{i = 1} \ln x_i - \sum^n_{i = 1} \frac {(\alpha - \beta x_i)^2} {2\beta x_i}\\
\implies \frac {\mathrm{d}l} {\mathrm{d}\alpha} & = \frac n {\alpha} - \sum^n_{i = 1} \frac {2(\alpha - \beta x_i)} {2\beta x_i}\\
& = n \left(\frac 1 {\alpha} + 1\right) - \sum^n_{i = 1} \frac {\alpha} {\beta x_i}\\
& = n \left(\frac 1 {\alpha} + 1\right) - \frac {n\alpha} {\beta} \overline{1/x}\\
& = n\left(\frac 1 {\alpha} + 1 - \frac {\alpha} {\beta} \overline{1/x}\right)\\
\frac {\mathrm{d}l} {\mathrm{d}\beta} & = -\frac n {2\beta} - \sum^n_{i = 1} \frac {2(\alpha - \beta x_i)(-x_i)(2\beta x_i) - (\alpha - \beta x_i)^2(2x_i)} {4\beta^2 x^2_i}\\
& = -\frac n {2\beta} + \sum^n_{i = 1} \frac {(\alpha - \beta x_i)(2\beta x_i) + (\alpha - \beta x_i)^2} {2\beta^2 x_i}\\
& = -\frac n {2\beta} + \sum^n_{i = 1} \frac {(\alpha - \beta x_i)(\alpha + \beta x_i)} {2\beta^2 x_i}\\
& = -\frac n {2\beta} + \sum^n_{i = 1} \frac {(\alpha^2 - \beta^2 x^2_i)} {2\beta^2 x_i}\\
& = -\frac n {2\beta} + \frac {n\alpha^2} {2\beta^2} \overline{1/x} - \frac n 2 \overline{x}\\
& = \frac n 2 \left(\frac {\alpha^2} {\beta^2} \overline{1/x} - \frac 1 {\beta} - \overline{x}\right)
\end{aligned}$$
When
$$\begin{aligned}
\frac {\mathrm{d}l} {\mathrm{d}\alpha} & = 0,\\
\frac {\beta(\alpha + 1)} {\alpha^2} & = \overline{1/x}.
\end{aligned}$$
When
$$\begin{aligned}
\frac {\mathrm{d}l} {\mathrm{d}\beta} & = 0,\\
\alpha^2 \overline{1/x} - \beta & = \beta^2 \overline{x}.
\end{aligned}$$
Assuming my score equations are correct and I have not gone wrong anywhere, I am now stuck here. In particular, I am unable to see how I can further manipulate the two equations above to return just $\alpha$ and $\beta$.
Any intuitive explanations will be greatly appreciated :)
| Your calculations look right to me.
You can solve the first equation for $\beta$:
$$
\beta=\frac{\alpha^2}{\alpha+1}\overline{1/x}\;.
$$
Substituting that into the second equation yields
$$
\alpha^2\overline{1/x}-\frac{\alpha^2}{\alpha+1}\overline{1/x}=\left(\frac{\alpha^2}{\alpha+1}\overline{1/x}\right)^2\overline x\;.
$$
Now multiply by $\frac{\alpha+1}{\alpha^2\overline{1/x}}$ (i.e. divide by $\beta$) to obtain
$$
\alpha+1-1=\frac{\alpha^2}{\alpha+1}\overline{1/x}\cdot\overline x\;.
$$
Multiply by $\frac{\alpha+1}\alpha$:
$$
\alpha+1=\alpha\overline{1/x}\cdot\overline x\;.
$$
Now solving for $\alpha$ yields
$$
\alpha=\frac1{\overline{1/x}\cdot\overline x-1}\;.
$$
Finally, substituting into the equation for $\beta$ above yields
\begin{eqnarray}
\beta&=&\left(\overline{1/x}\cdot\overline x-1\right)^{-2}\frac{\overline{1/x}\cdot\overline x-1}{\overline{1/x}\cdot\overline x}\overline{1/x}
\\
&=&
\frac1{\left(\overline{1/x}\cdot\overline x-1\right)\overline x}\;.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove this $4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$ let $a,b,c,d\in R$,and such $a^2+b^2+c^2+d^2=4$, prove or disprove
$$4(a+b+c+d)+(a^3+b^3+c^3+d^3)\le 20$$
I try use Cauchy-Schwarz inequality have
$$4(a+b+c+d)\le 4\sqrt{4(a^2+b^2+c^2+d^2)}=16$$
but
$$(a^3+b^3+c^3+d^3)^2(1+1+1+1)\ge (a^2+b^2+c^2+d^2)^3$$
so we have
$$a^3+b^3+c^3+d^3\ge 4$$
| We can solve it using the Lagrangian mutliplier:
Given that: $a^2 + b^2 + c^2 + d^2 = 4$ and $a,b,c,d \in \mathbb{R}.$ Prove $$\displaystyle 4(a+b+c+d) + (a^3 + b^3 + c^3 + d^3 ) \le 20 $$
Let us define the constraint function: $\displaystyle g(x) = a^2 + b^2 + c^2 + d^2.$
And If we define a set $\displaystyle \bar{U} =\{ (a,b,c,d) : a^2 + b^2 + c^2 + d^2 \le 1000\}$
Then our constraint set will be $$\displaystyle \bar{S} =\{ x \in \bar{U} : g(x) = 4 \} $$
Also our objective function is : $\displaystyle f(x) = 4(a+b+c+d) + (a^3 + b^3 + c^3 + d^3 ) $.
What we are left to do is, maximize $f(x).$
Now we define a Lagrangian function as follows:
$$\displaystyle \mathcal{L}(x,\lambda) = f(x) - \lambda g(x)$$
Thus to find maxima of the Lagrangian function, we need to find $(x, \lambda) $ that it satify : $\displaystyle \nabla \mathcal{L}(x,\lambda) = 0 \ \text{and} \ \nabla g(x) \ne 0 $.
Futher we get :
$$
\displaystyle
\left(
\begin{array}[c]
*3a^2 + 4 \\
3b^2 + 4 \\
3c^2 + 4 \\
3d^2 + 4 \\
\end{array}
\right)
=
\lambda
\left(
\begin{array}[c]
*2a \\
2b \\
2c \\
2d \\
\end{array}
\right)
$$
Hence $\displaystyle (a,b,c,d) \in \{ \frac{\lambda - \sqrt{\lambda ^2 -12} }{3} , \frac{\lambda + \sqrt{\lambda ^2 -12} }{3} \} $
Therefore we have deal with four cases and the constraint, with the condition that $\displaystyle \lambda \in \mathbb{R} \geq \sqrt{12} $:
$\displaystyle 1. \ a=b=c=d \\ 2. \ a=b\ne c=d \\ 3. \ a=b=c \ne d$
On solving for $\lambda$ for each of the cases we get $ \lambda $ for case $2$ and $3$ dosen't satisfy the above metioned condition.
Hence the only solutions we get is for case $1$ : $ a = \pm 1 \ \text{and} \ \lambda = 3.5$ . Also at this point $\displaystyle \nabla g(x) \ne 0 $
Finally we have $\displaystyle \mathcal{L}(1,1,1,1,3.5) , \mathcal{L}(-1,-1,-1,-1,-3.5) $ as stationary points and fuction $f(x)$ will attain maxima at $(1,1,1,1)$.
Given that: $a^2 + b^2 + c^2 + d^2 = 4$ and $a,b,c,d \in \mathbb{R} \implies$ $$\displaystyle 4(a+b+c+d) + (a^3 + b^3 + c^3 + d^3 ) \le 20 $$
with equality at $a=b=c=d=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is one root extraneous when using complex number polynomial expansions of cos(4x) to find cos(π/8)? Using De Moivre's theorem, binomial expansion, and the Pythagorean identity, I have the following polynomial for $\cos 4\theta$:
$$\cos\left(4\theta\right) = 8\cos^4\theta -8\cos^2\theta +1 $$
I trying to find an exact value for $\cos\left(\frac{\pi}{8}\right)$.
Since $\cos\left(\frac{\pi}{2}\right)=0$, then $\cos\left(4\cdot\frac{\pi}{8}\right)=0$, so it must be true that
$$ 8\cos^4\left(\frac{\pi}{8}\right) -8\cos^2\left(\frac{\pi}{8}\right) +1 =0$$
and by the quadratic formula I have
$$ \cos \left(\frac{\pi}{8}\right) = \pm\frac{\sqrt{2\pm\sqrt{2}}}{2} $$
Since $\frac{\pi}{8}$ is in the first quadrant, we discard the negative root, so we have
$$ \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2\pm\sqrt{2}}}{2} $$
Now, $ \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2+\sqrt{2}}}{2} $ is true, but $ \cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2-\sqrt{2}}}{2} $ is false. (I know this from the approximate numerical value of $\cos\left(\frac{\pi}{8}\right)$.)
My question: How do we know $\frac{\sqrt{2-\sqrt{2}}}{2}$ is an extraneous root?
| Since $0<\frac\pi8<\frac\pi4$, and since $\cos|_{[0,\pi/2]}$ is strictly decreasing,$$1>\cos\left(\frac\pi8\right)>\cos\left(\frac\pi4\right)=\frac{\sqrt2}2.$$But$$\frac{\sqrt{2-\sqrt2}}2<\frac{\sqrt2}2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Ratio of radii of two circles inscribed in a right isosceles triangle. There is a right isosceles triangle $\triangle ABC$ with the vertex $B$ facing the hypotenuse.
A circle is inscribed into the triangle with radius $r_1$, then another circle with radius $r_2$ is inscribed in the leftover space close to either $A$ or $C$ but not $B$
What is the ratio $\large{\frac{r_1}{r_2}}$ equal to?
My Attempt:
Let's call the circle with radius $r_1$, $C_1$ and the other circle with radius $r_2$, $C_2$
The smaller circle shall be closer to vertex $A$.
The Length from $A$ to $C_2$s tangents will be called $h_1$, and from these tangents to $C_1$s tangents will be called $h_2$. The length of the legs of the triangle will be called $x$.
If we draw a line from $C_1$ to $A$ we will see $r_1$ and $r_2$ are bases of similar triangles.
This means $\large{\frac{h_1}{r_2}=\frac{h_1+h_2}{r_1}}$
If we ignore $C_2$ we can see the triangle is made up of four smaller triangles and a square, since the sum of the area of these shapes will be equal to the area of the triangle: $$\large{2r_1(x-r_1)+r_1^2=\frac{x^2}{2}\\2xr_1-r_1^2=\frac{x^2}{2}\\-r_1^2+2xr_1-\frac{x^2}{2}=0}$$ From the quadratic equation:$$\large{\frac{-2x\mp\sqrt{4x^2-2x^2}}{-2}\\x\mp\frac{x}{\sqrt{2}}}$$ Since we know $r_1$ must be less than $x$
$$r_1=x-\frac{x}{\sqrt{2}}$$
$h_1+h_2$ is exactly half of the hypotenuse, this means $h_1+h_2=\frac{x}{\sqrt{2}}$
From this it follows that $$\large{\frac{h_1+h_2}{r_1}=\frac{\frac{x}{\sqrt{2}}}{x-\frac{x}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}}$$
Since $\large{\frac{h_1}{r_2}=\frac{h_1+h_2}{r_1}}$, $\large{\frac{h_1}{r_2}=\frac{1}{\sqrt{2}-1}}$ and $\large{(\sqrt{2}-1)h_1=r_2}$
Because $h_1+h_2=\frac{r_1}{\sqrt{2}-1}$$$\large{h_2=\frac{r_1-r_2}{\sqrt{2}-1}}$$
If we extend a line like so...
We can see $\large{(h_2)^2+(r_1-r_2)^2=(r_1+r_2)^2}$
|
In $\triangle OPQ$, $ \displaystyle PQ = r_1 - r_2, OQ = r_1 + r_2, \angle POQ = \frac{\pi}{8}$
$\displaystyle \frac{r_1 - r_2}{r_1 + r_2} = \sin \frac{\pi}8$
$ \implies \displaystyle \frac{r_1}{r_2} = \frac{1 + \sin (\pi /8)}{1 - \sin (\pi /8)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Multiplication of Dirichlet Generating Functions Let $A(x)$ be a Dirichelet generating function given by:
$A(x) = \frac{a_1}{1^x}+\frac{a_2}{2^x}+\frac{a_3}{3^x}+...=\sum_{n=1}^{\infty}\frac{a_n}{n^x}$.
Given the Dirichelet generating functions $A(x)$, $B(x)$, and $C(x)$, where $C(x) = A(x) B(x)$, find a formula for $c_n$ in terms of the $a_i$ s and $b_j$ s.
--Attempt--
Find the first few terms of $c_n$ as:
$(\frac{a_1}{1^x})(\frac{b_1}{1^x})=\frac{a_1b_1}{1^x}$ and so the coefficient of $\frac{1}{1^x}$, that is, $c_1$, is $a_1b_1$
$(\frac{a_1}{1^x}+\frac{a_2}{2^x})(\frac{b_1}{1^x}+\frac{b_2}{2^x})=\frac{a_1b_1}{1^x}+\frac{a_1b_2+a_2b_1}{2^x}$ and so the coefficient of $\frac{1}{2^x}$ is $a_1b_2+a_2b_1$
$(\frac{a_1}{1^x}+\frac{a_2}{2^x}+\frac{a_3}{3^x})(\frac{b_1}{1^x}+\frac{b_2}{2^x}+\frac{b_3}{3^x})=\frac{a_1b_1}{1^x}+\frac{a_1b_2+a_2b_1}{2^x}+\frac{a_1b_3}{3^x}+\frac{a_2b_2}{4^x}+\frac{a_3b_1}{3^x}$ and so the coefficient of $\frac{1}{3^x}$ is $a_1b_3+\frac{4}{3}^{-x}a_2b_2+a_3b_1$
I'm confident in $c_1$ and $c_2$, but feel that I'm missing something for $c_3$ and on. Any push in the right direction would be much appreciated!
| We obtain
\begin{align*}
\sum_{n=1}^\infty \frac{\color{blue}{c_n}}{n^x}
&=\left(\sum_{k=1}^\infty\frac{a^k}{k^x}\right)\left(\sum_{l=1}^\infty\frac{b^l}{l^x}\right)\\
&=\sum_{n=1}^\infty\left(\sum_{{k\cdot l=n}\atop{k,l\geq 1}}\frac{a_k}{k^x}\,\frac{b_l}{l^x}\right)\\
&=\sum_{n=1}^\infty
\left(\sum_{{k=1}\atop{k|n}}^n\frac{a_k}{k^x}\,\frac{b_{n/k}}{\left(n/k\right)^x}\right)\\
&=\sum_{n=1}^\infty
\color{blue}{\left(\cdots\right)}\frac{1}{n^x}\\
\end{align*}
Do you see how to do the final step(s) and derive a formula for $\color{blue}{c_n}$? You might also want to calculate $c_3$ and $c_4$ from it to better see what's going on.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of infinite product $\lim_{x\to\infty} \prod_{n\in\mathbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2}$ What is the limit
$$\lim_{x\to\infty} \prod_{n\in\mathbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2} \:?$$
The infinite product is well-defined pointwise since $\frac{(x-1)^2+n^2}{(x+1)^2+n^2}-1 = O(1/n^2)$ for fixed $x$.
The motivation of the above limit comes from a complex integration. To show that the integration over a semicircle vanishes, I need to find the above limit.
| As an option you can also use the following approach to find the limit.
$$\ln\prod(x)=\ln\prod_{n\in \Bbb Z} \frac{(x-1)^2+n^2}{(x+1)^2+n^2}=\ln\prod_{n\in \Bbb Z}\bigg(1- \frac{4x}{(x+1)^2+n^2}\bigg)$$
$$=\sum_{n=-\infty}^\infty\ln\Big(1-\frac{4x}{(x+1)^2+n^2}\Big)$$
Decomposing the logarithm into the series $(\frac{4x}{(x+1)^2+n^2}<1$ at $x\to\infty$)
$$\ln\prod(x)=-\sum_{n=-\infty}^\infty\frac{4x}{(x+1)^2+n^2}-\frac{1}{2}\sum_{n=-\infty}^\infty\Big(\frac{4x}{(x+1)^2+n^2}\Big)^2-...$$
Given that $\sum_{n=-\infty}^\infty\frac{1}{n^2+a^2}=\frac{\pi}{a}\coth\pi a$
$$-\sum_{n=-\infty}^\infty\frac{4x}{(x+1)^2+n^2}=-\frac{4\pi x}{x+1}\coth\pi(x+1)\to-4\pi \,\,\text{at}\,\,x\to\infty$$
It is not difficult to show that all other terms of the logarithm decomposition have an additional power of $\frac{1}{x}$ and $\to 0$ at $x\to\infty$.
For example, the second term of the decomposition
$$\sum_{n=-\infty}^\infty\Big(\frac{x}{(x+1)^2+n^2}\Big)^2=-\frac{x^2}{2(x+1)}\frac{\partial}{\partial x}\sum_{n=-\infty}^\infty\frac{1}{(x+1)^2+n^2}$$
$$=-\frac{x^2}{2(x+1)}\frac{\partial}{\partial x}\frac{\pi}{x+1}\coth\pi(1+x)\to0\,\,\text{at}\,\,x\to \infty$$
Therefore,
$$\lim_{x\to\infty}\ln\prod(x)=-4\pi\,\, \Rightarrow\,\,\lim_{x\to\infty}\prod(x)=e^{-4\pi}$$
what, of course, also follows from the closed form of the product obtained by @jjagmath.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that if $m$ and $n$ are integers and $a$ and $b$ are elements from a ring, then $(m \cdot a) ( n \cdot b) = (mn) \cdot (ab)$
Show that if $m$ and $n$ are integers and $a$ and $b$ are elements from a ring, then $(m \cdot a) ( n \cdot b) = (mn) \cdot (ab)$
Source:Gallian 7th ed Pg. 243, Q.15
Consider $P(n)$ to be the statement that $\forall m \in M$ , $(m \cdot a) ( n \cdot b)=(mn) \cdot (ab)$ is true
Induction:
*
*Base case: $P(1) = (m \cdot a)( 1 \cdot b) = (m \cdot a)(b)$
Let $P(m)$ be the statement that $(m \cdot a)(b) = (m \cdot 1) (ab)$, then:
1.1. Base case : $(1 \cdot a)(b) =(ab)= (1 \cdot 1)(ab)$
1.2. Proving $P(m+1)$ from $P(m)$: $ (m+1)a \cdot b= (m)a \cdot b + (1) a\cdot b= (m \cdot 1) ab + 1 a \cdot b= m(a \cdot b) + a \cdot b = (m+1) (a \cdot b)$
*Proving $n+1$th case from $nth$:
$P(n+1) = ( m \cdot a) \left( (n+1) \cdot b \right) = (m \cdot a) \left( n \cdot b + 1 \cdot b\right)= (m\cdot a)( n \cdot b) + ( m \cdot a) (b)= (mn) \cdot (a b) + (m) \cdot (ab)= (mn+m) \cdot (ab)=\left( (m+1)n\right) \cdot (ab)$
Remark: I have used the result in the sub induction to simplify the last equality in the above.
I am not sure how to extend the above proof to $m,n \in \mathbb{Z}$.
Could I get some one check my work so far, and, explain how I could finish it?
| Perhaps show the simpler statements
$$\tag1(n\cdot a)\cdot b=n\cdot(a\cdot b)$$
and
$$\tag2a\cdot(n\cdot b)=n\cdot(a\cdot b)$$
and
$$\tag3m\cdot(n\cdot a)=(mn)\cdot a$$
(which you may already have done).
Then use these to see
$$(m\cdot a)\cdot(n\cdot b)
\stackrel{(1)}=m\cdot(a\cdot(n\cdot b))
\stackrel{(2)}=m\cdot(n\cdot(a\cdot b))
\stackrel{(3)}=(mn)\cdot(a\cdot b)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probabilistic interpretation. Prove that $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^n$
Prove that $\sum_{k=0}^{n} \binom{n+k}{k} \frac{1}{2^k} = 2^n$
Well, the solution given on the book is as follows:
We will solve this counting problem by a powerful and elegant interpretation of the result. First we divide the identity by $2^n$, getting
$$ \sum_{k=0}^{n} \binom{n+k}{k} \frac{1}{2^{n+k}} = \sum_{k=0}^{n} p_k = 1. $$
This is the sum of probabilities $p_k = \binom{n+k}{k} \frac{1}{2^{n+k}}$. Now,
$$ p_k = \frac{1}{2}\binom{n+k}{k} \frac{1}{2^{n+k}} + \frac{1}{2}\binom{n+k}{k} \frac{1}{2^{n+k}} = \mathbb{P}(A_k) + \mathbb{P}(B_k),$$
with the events
\begin{align*}
A_k &= \{\text{$(n+1)$ times head and $k$ times tail}\}, \\
B_k &= \{\text{$(n+1)$ times tail and $k$ times head}\}.
\end{align*}
However, I am not getting the way to calculate the probabilities of $A_k$ and $B_k$. I mean, how did we arrive at the conclusion that $\mathbb{P}(A_k) = \frac{1}{2}\binom{n+k}{k} \frac{1}{2^{n+k}}$ and $\mathbb{P}(B_k) = \frac{1}{2}\binom{n+k}{k} \frac{1}{2^{n+k}}$. I mean, how to calculate those probabilities?
| We toss a fair coin until we get a total of $n+1$ heads or $n+1$ tails, whichever comes first. Say we get $n+1$ heads first, but along the way we also get $k$ tails ($k\leq n$) so the total number of tosses is $n+k+1$. The last toss must be head since we stop tossing after the $n+1$-th head. So we only need to figure out the number of possibility to get $k$ tails from the first $n+k$ tosses, which is $\binom{n+k}{k}$. Divide by the number of all possible outcomes from $n+k+1$ tosses, which is $2^{n+k+1}$ to obtain the probability $\frac{1}{2^{n+k+1}}\binom{n+k}{k}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the partial fractions decomposition of $\frac{9}{(1+2x)(2-x)^2} $ So this is basically my textbook work for my class, where we are practicing algebra with partial fractions.
I understand the basics of decomposition, but I do not understand how to do it when then the denominator is a power of $x^2$?
e.g. this question - $$\frac{9}{(1+2x)(1-x)^2} $$
I understand that it will turn into- $$\frac{9}{(1+2x)(2-x)^2} = \frac {A}{1+2x} + \frac {B}{(1-x)}+ \frac {C}{(1-x)^2}$$ and then it will become
$$\frac{9}{(1+2x)(1-x)^2} =\frac{A(1-x)^2 +B(1+2x)(1-x)+C(1+2x)}{(1+2x)(1-x)^2}$$
but what do you do once you are at this step? The example on the textbook isn't very clear, so if anyone could tell me what I do after doing this, and why that is the case, I would be very thankful.
| You say
"I understand that it will turn into-
$\frac{9}{(1+2x)(2−x)^2}= \frac{A}{1+2x}+\frac{B}{(1−x)}+\frac{C}{1−x^2}$".
But this is just wrong! For one thing, the "2- x" has mysteriously turned into "1- x". For another you have "$1- x^2$" where you should have "$(2- x)^2$".
You need $\frac{9}{(1+ 2x)(2- x)^2}= \frac{A}{1+ 2x}+ \frac{B}{2- x}+ \frac{C}{(2- x)^2}$.
We have three unknown values so need three equations to solve for A, B, and C. There are many ways to get them. I prefer getting rid of the fractions by multiplying by $(1+ 2x)(1- x)^2$:
$9= A(2- x)^2+ B(1+ 2x)(2- x)+ C(1+ 2x)$
Now we can get three equations by taking x to be three different numbers. Choosing x= 2 and x= -1/2 make the equation very easy:
If x= 2, 2- x= 0 and we have 9= C.
If x= -1/2, 1+ 2x and we have 9= 9A/4 so A= 4.
0 is also easy- if x= 0 we have 9= 4A+ 2B+ C= 36+ 2B+ 9.
2B= -36 so B= -18.
$\frac{9}{(1+ 2x)(2- x)^2}= \frac{4}{1+ 2x}- \frac{18}{2- x}+ \frac{9}{(2- x)^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$
How to find the value of $3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{16} + ...}}}}$
I tried to solve it and found a relation that if I assume the given expression to be something say $x$, then following result holds true.
$$\boxed{x = 3 + \sqrt{3x}}\hspace{4cm}\bf ...(1.)$$
Can be proved as,
$$x = 3 + \sqrt{3(3 + \sqrt{3x})}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3x}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^3(3 + \sqrt{3x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7x})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^7(3 + \sqrt{3x})})}}$$
$$x = 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8 + \sqrt{3^{15}x}}}}$$
If I continue this process repeatedly,
$$x= 3 + \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^{8} + \sqrt{3^{16} +\sqrt{...}}}}}$$
But how can I make sure that $(1.)$ is absolutely correct? Also how can I make sure that "If I continue this process repeatedly" I would get the same expression?
| I think that OP's guess was okay, here's a different approach.
Let $a_1 = 3$ and let $a_{n+1} = 3 + \sqrt{3a_n}$ for $n\geq 1$. Then the first few elements are
\begin{align*}
a_1 &= 3,\\
a_2 &= 3+ \sqrt{3^2},\\
a_3 &= 3+\sqrt{3^2 + \sqrt{3^4}},\\
a_4 &= 3+ \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8}}},
\end{align*}
so it is reasonable to define
\begin{align*}
3+ \sqrt{3^2 + \sqrt{3^4 + \sqrt{3^8+\ldots}}} := \lim\limits_{n\to\infty} a_n.
\end{align*}
We first need to make sure that the limit exists. We will use the theorem saying that any monotone and bounded sequence is convergent, it is often useful for recursive sequences such as $a_n$.
Monotonicity: we will prove it by induction. First note that $a_2 \geq a_1$. Now assume that for some $n\geq 1$ we have $a_{n+1}\geq a_n$. Then
\begin{align*}
a_{n+2} - a_{n+1} = 3 + \sqrt{3a_{n+1}} - (3 + \sqrt{3a_{n}}) = \sqrt{3}(\sqrt{a_{n+1}} - \sqrt{a_n}) \geq 0,
\end{align*}
where the last inequality follows from the induction hypothesis and the fact that the square root is increasing. Thus, by induction we get that $a_{n+1}\geq a_n$ for all $n\geq 1$, so the sequence is increasing.
Boundedness: obviously the sequence is bounded from below by 0. To find a reasonable upper bound let us first compute what we think the limit should be (if a sequence is increasing, then its limit is also its upper bound). Note that
\begin{align*}
\lim\limits_{n\to \infty} a_{n} = \lim\limits_{n\to \infty} a_{n+1} = 3 +3\lim\limits_{n\to\infty}\sqrt{a_n},
\end{align*}
so if we let $x = \lim\limits_{n\to \infty} a_{n}$, we get that $x = \frac 32(3 + \sqrt{5}).$
To make the computations easier, we will show that $a_n \leq 27$ by induction ($27\geq x$ and it works nicely if we plug it into $\sqrt{3\cdot a_n}$). It is of course true for $a_1$. If $a_n\leq 27$ , then
\begin{align*}
a_{n+1} = 3 + \sqrt{3a_n} \leq 3 + \sqrt{3\cdot 27} = 3+9 = 12 \leq 27.
\end{align*}
Thus we get that $a_n$ is bounded and increases, so it has a finite limit. By the above computation, the limit is equal to $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the function for a sequence using differences I am trying to find a function generating/suitable for the following sequence:
${2,6,12,20,30,42,56...}$
The first order differences are:
$4, 6, 8, 10, 12, 14...$
and the second order differences are:
$2, 2, 2, 2, 2, 2...$
this means that:
$y''= 2 \implies y' = 2\cdot x + c$ and we can see that for $c = 2$ we can get the correct numbers i.e. the first order differences (replacing starting from $x = 1$). So $y' = 2\cdot x + 2$
This means that $y = x^2 + 2\cdot x + c$
But I can't find any value of $c$ that would give the original sequence.
Now I see that $y'$ could also be $y' = 2\cdot x + 4$ as it gives the correct numbers if we replace starting from $x = 0$ but then we have $y = x^2 + 4\cdot x + c$ and I again can't find any $c$ that matches the original sequence.
What am I doing wrong here?
| Let $\{a_n\}_{n\in\mathbb N}=\{2,6,12,20,30,42,56,\ldots\}$. Let $\{b_n\}_{n\in\mathbb N}$ be the sequence of the first-order forward differences of $\{a_n\}$, and $\{c_n\}_{n\in\mathbb N}$ the sequence of second-order differences. So for $n\ge1$,
$$b_n = a_{n+1} - a_n \\ c_n = b_{n+1} - b_n$$
As you observed, $c_n=2$ for all $n$. Then
$$\begin{align*}b_{n+1} &= b_n + 2 \\
&= b_{n-1}+2\times2 \\
&= b_{n-2}+3\times2 \\
&~~\vdots \\
&= b_1 + 2n
\end{align*}$$
which means $b_n = b_1+2(n-1)=2(n+1)$.
Now solve for $a_n$:
$$\begin{align*}
a_{n+1} &= a_n + 2(n+1) \\
&= a_{n-1} + 2((n+1) + n) \\
&= a_{n-2} + 2((n+1) + n + (n-1)) \\
&~~\vdots \\
&= a_1 + 2 \sum_{k=0}^{n-1} (n+1-k)\\
&= a_1 + n^2 + 3n
\end{align*}$$
so that $a_n = 2 + (n-1)^2 + 3(n-1) = n^2 + n$.
In case you are interested in the generating function, it takes the form
$$f(x) = \sum_{n=0}^\infty a_n x^n$$
Recall that if $|x|<1$, we have
$$\frac1{1-x} = \sum_{n=0}^\infty x^n$$
and finding $f(x)$ is just a matter of differentiating both sides twice.
$$\frac1{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1} = \sum_{n=0}^\infty (n+1)x^n$$
$$\frac2{1-x^3} = \sum_{n=0}^\infty (n^2+n) x^{n-1}$$
$$\implies f(x) = \frac{2x}{1-x^3} = \sum_{n=0}^\infty (n^2+n) x^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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$\int{\frac{x^3}{(x-1)(x-2)(x-3)}}dx$ Question:
$$Find\int{\frac{x^3}{(x-1)(x-2)(x-3)}}dx$$
What I did:
resolved into partial fraction;
$$A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) = x^3$$
$$A = \frac{1}{2},\hspace{0.2cm} B = -8\hspace{0.2cm}, C= \frac{27}{2}$$
$$\frac{1}{2}\int\frac{1}{x-1}dx \hspace{0.2cm}-8\int\frac{1}{x-2} + \hspace{0.2cm}\frac{27}{2}\int\frac{1}{x-3}$$
$$= \frac{1}{2}ln(x-1)\hspace{0.2cm} -8ln(x-2)\hspace{0.2cm}+ \frac{27}{2}ln(x-3) + c$$
But it was done differently in the textbook and other answers I see, they did long division and changed the form to $Quotient + \frac{Remainder}{Divisor}$:
$$\int{\frac{x^3}{(x-1)(x-2)(x-3)}}dx$$
$$= \int\left(1 + \frac{6x^2 - 11x + 6}{(x-1)(x-2)(x-3)}\right)dx$$
$$A = \frac{1}{2},\hspace{0.2cm} B = -8\hspace{0.2cm}, C= \frac{27}{2}$$
$$\int dx \hspace{0.2cm}+\frac{1}{2}\int\frac{1}{x-1}dx \hspace{0.2cm}-8\int\frac{1}{x-2} + \hspace{0.2cm}\frac{27}{2}\int\frac{1}{x-3}$$
$$=x + \hspace{0.2cm} \frac{1}{2}ln(x-1)\hspace{0.2cm} -8ln(x-2)\hspace{0.2cm}+ \frac{27}{2}ln(x-3)+ c$$
The answers are different, my question is, what did I do wrong in the first solution? To the best of my knowledge I think I did the partial fraction correctly.
| The equation$$A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)=x^3$$has no solutions, since the LHS is a quadratic polynomial, whereas the RHS is a cubic one. And actually$$\frac12(x-2)(x-3)-8(x-1)(x-3)+\frac{27}2(x-1)(x-2)=6x^2-11x+6.$$
| {
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"answer_id": 0
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Series expansion of $\text{Li}_3(1-x)$ at $x \sim 0$ My question is simple, but maybe hard to answer. I would like to have a series expansion for $\text{Li}_3 (1-x)$ at $x \sim 0$ in the following form:
$$\text{Li}_3 (1-x) = \sum_{n=0} c_n x^n + \log x \sum_{m=1} c_m x^m. \tag{1}$$
The first few terms are:
$$\text{Li}_3(1-x) = \zeta (3)-\frac{\pi ^2 x}{6}+\left(\frac{3}{4}-\frac{\pi ^2}{12}\right) x^2+\left(\frac{7}{12}-\frac{\pi
^2}{18}\right) x^3+\left(\frac{131}{288}-\frac{\pi ^2}{24}\right) x^4+\left(\frac{53}{144}-\frac{\pi
^2}{30}\right) x^5+ \left(-\frac{x^2}{2}-\frac{x^3}{2}-\frac{11 x^4}{24}-\frac{5 x^5}{12}\right) \log x+O\left(x^6\right), \tag{2}$$
however Mathematica fails to go beyond $\mathcal{O}(x^{15})$. The coefficients for the $\log$ term are easy to guess:
$$c_m = \frac{H_{m-1}}{m}. \tag{3}$$
I also managed to obtain the coefficient with the $\pi^2$:
$$- \frac{\pi^2}{6n}, \tag{4}$$
but I am struggling with the remaining coefficient, especially with so few coefficients. Any idea what this coefficient could be?
| I don't know if there is a closed-form expression for the coefficients, but the following is a way to get the first few terms of the series. The approach can be used to get as many terms as you want.
Starting with the fact that $$- \int_{0}^{x} \frac{\operatorname{Li}_{n}(1-t)}{1-t} \, \mathrm dt= \int_{1}^{1-x} \frac{\operatorname{Li}_{n}(u)}{u} \, \mathrm du = \operatorname{Li}_{n+1}(1-x) - \operatorname{Li}_{n+1}(1),$$ we have
$$\begin{align} \operatorname{Li}_{2}(1-x) - \frac{\pi^{2}}{6} &= - \int_{0}^{x}\frac{\operatorname{Li}_{1}(1-t)}{1-t} \, \mathrm dt\\ & = \int_{0}^{x}\frac{\ln(t)}{1-t} \, \mathrm dt \\ &= \int_{0}^{x} \ln(t) \left(1+\mathcal{O}(t) \right) \, \mathrm dt \\ &= x \left( \ln(x)-1 \right) + \mathcal{O}(x^{2}).\end{align}$$
Then $$\begin{align} \operatorname{Li}_{3}(1-x) -\zeta(3) &= - \int_{0}^{x}\frac{\operatorname{Li}_{2}(1-t)}{1-t} \, \mathrm dt \\ &= - \int_{0}^{x} \frac{\frac{\pi^{2}}{6}+t \left(\ln(t)-1 \right) + \mathcal{O}(t^{2})}{1-t} \, \mathrm dt \\ &= - \int_{0}^{x} \left(\frac{\pi^{2}}{6}+t \left(\ln(t)-1 \right) + \mathcal{O}(t^{2}) \right)\left(1+ t+\mathcal{O}(t^{2}) \right) \, \mathrm dt \\ &= -\frac{\pi^{2}}{6}x - \frac{\pi^{2}}{6} \frac{x^{2}}{2}+ \frac{x^{2}}{4} \left(1-2 \ln(x)\right) + \frac{x^{2}}{2} + \mathcal{O}(x^{3}). \end{align} $$
Therefore, we have $$\operatorname{Li}_{3}(1-x) = \zeta(3) - \frac{\pi^{2}}{6}x + \frac{x^{2}}{12} \left(-6 \ln(x) - \pi^{2}+9 \right) + \mathcal{O}(x^{3}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\iint_D \frac1{(x^2+y^2)^2 }dx dy$ over a region bounded by four circles
I'm thinking of doing the substitution $x = \frac{u}{u^2 + v^2}, y = \frac{v}{u^2 + v^2},$ but I'm not sure how to exactly compute the range of values $u$ and $v$ take. Clearly, $u=\frac{x}{x^2+y^2}$ and $v = \frac{y}{x^2 + y^2}.$ The Jacobian of the result is $-\frac{1}{u^2 + v^2}$, and so the resulting integral equals $\iint_{D'} dudv,$ where $D'$ is the region of possible values for the pairs $(u,v)$. In the circle $x^2 + y^2 - 2x= 0$, for instance, we have the point $(1,0)$, which corresponds to $u=1, v=0$. Also, we have the point $(2,0)$, corresponding to $u= \frac{1}2, v = 0$. I can't seem to generalize what values u and v can take on.
| Given that the first equation could be written
$$
\frac{x}{x^2+y^2}=\frac{1}{2} \implies u=\frac{1}{2}
$$
with similar reasoning $u\in\left[\frac{1}{4},\frac{1}{2}\right]$ and $v\in\left[\frac{1}{6},\frac{1}{2}\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4431482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sequence $f(m), f(f(m)),\ldots$ never contains a square I have the following problem, and I would like to know if my proof is correct, and if there is a faster way to prove the result.
Problem
Let $\{x\}$ denote the closest integer to the real number $x$. Define $f(n)=n+\{\sqrt n\}$. Prove that for every positive integer $m$, the sequence $$f(m), f(f(m)),\ldots$$ never contains the square of an integer.
Attempt
First suppose that $m=k^2$, then $f(m)=k^2+k$. If $f(m)=n^2$ was a square we would have $$4n^2+1=(2k+1)^2$$ But this is only possible when $n=0$. We may therefore assume that $m$ is not a square.
Let therefore $k^2<m<(k+1)^2$. Since $\{\sqrt{m}\}=k,k+1$ we have that $$f(m)<(k+1)^2+k+1<(k+2)^2$$
Therefore if $f(m)$ is a square we wust have $f(m)=(k+1)^2$. There are two possibilities:
*
*$m=(k+1)^2-k=k^2+k+1$ and $\{\sqrt{k^2+k+1}\}=k$
*$m=(k+1)^2-(k+1)=k^2+k$ and $\{\sqrt{k^2+k}\}=k+1$
Both of these are impossible.
$\{\sqrt{k^2+k+1}\}=k+1$
Proof: The above happens if $\sqrt{k^2+k+1}\geq k+\frac{1}{2}$. Squaring both sides yields $1\geq\frac{1}{4}$.
$\{\sqrt{k^2+k}\}=k$
Proof: The above happens if $\sqrt{k^2+k}\leq k+\frac{1}{2}$. Squaring both sides yields $0\leq\frac{1}{4}$.
| A stronger observation is that the given $f(n)$ is precisely the $n$-th non-square integer. I will try to sketch a proof-
Note that the number of non-square integers between $k$-th and $(k-1)$-th square integers is
$$k^2-(k-1)^2-1=2(k-1)$$
So, the number of non-square integers below the $k$-th square integer is
$$2\left (1+2+\dots +(k-1)\right)=k(k-1)$$
Now, we want $n$ such that
$$n=k(k-1)$$
i.e.,
$$k=\frac{1+\sqrt{4n+1}}2$$
So, the $n$-th non-square integer is
$$\left(\frac{1+\sqrt{4n+1}}2\right)^2-1\;\text{ if $\left(\frac{1+\sqrt{4n+1}}2\right)^2$ is an integer}\\
\text{or }\left\lfloor\left(\frac{1+\sqrt{4n+1}}2\right)^2\right\rfloor\;\text{ if $\left(\frac{1+\sqrt{4n+1}}2\right)^2$ is not an integer}$$
which is
$$n+\left(\frac{\sqrt{4n+1}-1}2\right)\text{ or }\left\lfloor n+\left(\frac{1+\sqrt{4n+1}}2\right)\right\rfloor$$
So, it suffices to show that
$$\langle n\rangle=\begin{cases}
\left(\frac{\sqrt{4n+1}-1}2\right)&\text{ if it is an integer}\\
\left\lfloor \frac{1+\sqrt{4n+1}}2\right\rfloor&\text{ if not}
\end{cases}$$
To do this, it is enough to see that the first expression is an integer iff $n=m(m-1)$ for some $m\in \mathbb N$. So, it reduces to proving
$$\langle m\sqrt{m-1}\rangle=m$$
and indeed
$$(m-1)-\frac 12<\sqrt{m(m-1)}<(m-1)+\frac 12$$
And, for the second exp0ression, we only need to observe that
$$\left\lfloor \sqrt{n+\frac 14}+\frac 12\right\rfloor=\left\lfloor \sqrt{n+\frac 12}\right\rfloor$$
Leaving aside a few details (which I leave for you to fill), this more or less completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving $x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$ for positive values with $xyz=1$
How can you prove that
$$x^2 + y^2 + z^2 + x + y + z \geq 2(xy+yz+zx)$$
given that $x,y,z > 0$ and $xyz = 1$.
We can easily prove that the equality holds when $x = y = z = 1$
I could able to prove the result when one of $x,y,z$ is $1$ considering $x,y,z$ as $x,1/x, 1$ using the inequality $x + 1/x \geq 2$ for any positive number of $x$.
But couldn't able to find a full proof.
| Let $$f(x,y,z):=x^2 + y^2 + z^2 + x + y + z - 2(xy+yz+zx) \,.$$
Suppose WLOG that $x \ge y \ge z$, and write $g=\sqrt{xy}\,. \;$
Then
\begin{eqnarray} f(x,y,z)-f(g,g,z)&=& x^2+y^2-2g^2+(x+y-2g)(1-2z) \\ &=&
(\sqrt{x}-\sqrt{y})^2 \cdot \Bigl( (\sqrt{x}+\sqrt{y})^2+1-2z\Bigr) \ge 0 \,,
\end{eqnarray}
since $\sqrt{x}+\sqrt{y} \ge 2\sqrt{z}. \;$ Moreover,
$$f(g,g,z)=z^2+2g+z-4gz \,,$$
so, recalling that $zg^2=zxy=1$, we get
$$g^4 f(g,g,z)=1+2g^5+g^2-4g^3=(g-1)^2(2g^3+4g^2+2g+1) \ge 0 \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the nth root of i I saw a video some time ago showing a nice method for showing that $\sqrt{i} = \pm\left(\frac1{\sqrt2} + i \frac1{\sqrt2}\right)$. The teacher assumes that $\sqrt{i}$ can be equated to some complex number $a +bi$, and solves accordingly:
$$\sqrt{i} = a+bi$$
$$\iff i = a^2-b^2+ i 2ab$$
$$\implies
\begin{cases}
a^2-b^2=0 \\
2ab=1
\end{cases}
$$
Solving this latter system of equations yields $a = \pm\frac1{\sqrt2}$ and $b = \pm\frac1{\sqrt2}$.
Seeing this made me curious about the general case, $\sqrt[n]i = a + bi$. I first tried isolating $a$ and $b$ by using the teacher’s method:
$$\sqrt[n]{i} = a+bi \\
\iff i = (a+bi)^n \\
= a^n + i{n\choose 1}a^{n-1}b + i^2 {n\choose 2}a^{n-2}b^2 + … + i^{n-2} {n\choose n-2}a^2b^{n-2} + i^{n-1} {n\choose n-1}ab^{n-1} + i^nb^n
$$
But this quickly becomes complicated.
The next step would be to group terms with respect to whether they have a factor of $i$ or not, but that would require knowing the divisibility of $n$. Otherwise, we wouldn’t be able to know what $i^n$ is, or any subsequent terms for that matter.
Moreover, even if were to isolate certain divisibility cases for $n$, this would yield an at-least-nth degree system of equations that would be really hard to solve.
How could I proceed from here?
| Let me begin by saying that this problem is pretty much always attacked using deMoivre's Theorem as in the other answers, but I thought it was still worth pointing out that your approach to the problem using the binomial expansion can be pushed further due to the cyclic behavior of powers of $i$.
$$
\begin{align}
i^1 &=i \\
i^2 & -1\\
i^3 &=-i\\
i^4 &=1\\
i^5 &=i\\
\end{align}
$$
after which the pattern repeats. So if we were to write your sum as $\sum_{k=0}^n i^k a^{n-k}b^k$ we can collect terms based on the residue of $k$ modulo $4$
$$
\begin{align}
\sum_{k=0}^n i^k a^{n-k}b^k&=\sum_{0\le k \le n \\k \equiv 0 \bmod 4}^n a^{n-k}b^k+i\sum_{0\le k \le n \\k \equiv 1 \bmod 4}^n a^{n-k}b^k-\sum_{0\le k \le n \\k \equiv 2 \bmod 4}^n a^{n-k}b^k-i\sum_{0\le k \le n \\k \equiv 3 \bmod 4}^n a^{n-k}b^k \\
&= \sum_{0\le k \le n \\k \equiv 0 \bmod 2}^n (-1)^{k/2}a^{n-k}b^k+i\sum_{0\le k \le n \\k \equiv 1 \bmod 2}^n (-1)^{(k-1)/2}a^{n-k}b^k \\
&=\sum_{j=0}^{\lfloor{n/2}\rfloor} (-1)^{j}a^{n-2j}b^{2j}+i\sum_{j=1}^{\lfloor{(n+1)/2}\rfloor} (-1)^{(j-1)}a^{n-2j+1}b^{2j-1}
\end{align}
$$
so for a given $n$ we have to solve
$$
\begin{align}
\sum_{j=0}^{\lfloor{n/2}\rfloor} &(-1)^{j}a^{n-2j}b^{2j}=0 \\
\sum_{j=1}^{\lfloor{(n+1)/2}\rfloor} &(-1)^{(j-1)}a^{n-2j+1}b^{2j-1}=1
\end{align}
$$
Taking $n=3$ we get
$$
\begin{align}
a^3-3ab^2&=0 \\
3a^2b-b^3&=1
\end{align}
$$
from which it's not difficult to obtain the solutions $-i, \frac{\sqrt 3 +i}{2}, \frac{-\sqrt 3 +i}{2}$
But for $n=4$ and
$$
\begin{align}
a^4-6a^2b^2+b^4&=0 \\
4a^3b-4ab^3&=1
\end{align}
$$
things start getting harder. And by the time we get to $n=5$ and
$$
\begin{align}
a^5-10a^3b^2+5ab^4&=0 \\
5a^4b-10a^2b^3+b^5&=1
\end{align}
$$
the solutions are getting difficult without the use of symbolic algebra software.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
I got different answer of $~\int \frac{1}{(x^2+1)^2}\mathrm{dx}$ $$\begin{align}
&\int {1 \over (x^2+1)^2 } \mathrm{dx} ~~ \leftarrow~~ \text{I assume}~~x\ne0 ~~ \text{since it is trivial as it is held} \\
&=\int {x^2+1-x^2 \over (x^2+1)^2 } \mathrm{dx}\\
&=\int \left\{ {(x^2+1) \over (x^2+1)^2 }- {x^2 \over (x^2+1)^2 } \right\} \mathrm{dx} \\&=
\int {1 \over (x^2+1) } \mathrm{dx}- \int {x^2 \over (x^2+1)^2 } \mathrm{dx}\\&=
\arctan(x)+\mathrm{const_1}-\int x^2 (x^2+1)^{-2} \mathrm{dx} \\&=
\arctan(x)+ \mathrm{const_1}-\left\{ x^2\cdot {(-1)(x^2+1)^{-1} \over 2x } - \int (2x) \cdot {(-1)(x^2+1)^{-1} \over 2x } \mathrm{dx} \right\}\\&=\arctan(x)+ \mathrm{const_1} - \left\{ -{1 \over 2 }x {1 \over (x^2+1) }+\int {1 \over (x^2+1) } \mathrm{dx} \right\} \\&=\arctan(x)+ \mathrm{const_1}+ {x \over 2 (x^2+1) }-\int {1 \over x^2+1 } \mathrm{dx}\\&=
\arctan(x)+ \mathrm{const_1}+ {x \over 2(x^2+1) }- \left(\arctan(x)+ \mathrm{const_2} \right) \\&={x \over 2(x^2+1) }+ \underbrace{\mathrm{const_3} }_{ \mathrm{const_1}-\mathrm{const2} } \end{align}$$
But the answer in the book(A First Course in Calculus by Serge Lang) says the correct form is
$$ \int {1 \over (x^2+1)^2 } \mathrm{dx}= \underbrace{\color{fuchsia}{{x \over 2(x^2+1) } + {1 \over 2 }\arctan(x)}}_{\text{I assume arbitrary const ommited} } $$
Where I've made mistake(s)?
ADD
I am currently in outside so I will to be late to respond.
| The fifth line on the integration by parts is incorrect. I wish you can understand how to carry out integration by parts by showing the solution. Hope that it helps.
$$
\begin{aligned}
\int \frac{1}{\left(x^{2}+1\right)^{2}} d x&=\int-\frac{1}{2 x} d\left(\frac{1}{x^{2}+1}\right) \\
&=-\frac{1}{2 x\left(x^{2}+1\right)}-\frac{1}{2} \int\left(\frac{1}{x^{2}}-\frac{1}{x^{2}+1}\right) d x \\
&=-\frac{1}{2 x\left(x^{2}+1\right)}+\frac{1}{2 x}+\frac{1}{2} \arctan x+C \\
&=\frac{x}{2\left(x^{2}+1\right)}+\frac{1}{2} \arctan x+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A gamma summation: $\sum_{n=0}^{\infty} \frac{2}{\Gamma ( a + n) \Gamma ( a - n )} = \frac{2^{2a-2}}{\Gamma ( 2a - 1 )} + \frac{1}{\Gamma^2 (a)}$ Let $a \notin \mathbb{Z}$ and $a \neq \frac{1}{2}$. Prove that
$$\sum_{n=0}^{\infty} \frac{2}{\Gamma \left ( a + n \right ) \Gamma \left ( a - n \right )} = \frac{2^{2a-2}}{\Gamma \left ( 2a - 1 \right )} + \frac{1}{\Gamma^2 (a)}$$
Attempt
Using the fact that
\begin{align*}
\frac{1}{\Gamma\left ( a+x \right ) \Gamma \left ( \beta - x \right )} &= \frac{1}{\left ( a+x-1 \right )! \left ( \beta-x-1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \frac{\left ( a + \beta-2 \right )!}{\left ( a + x -1 \right )! \left ( \beta - x -1 \right )!} \\ &=\frac{1}{\Gamma \left ( a + \beta - 1 \right )} \binom{a + \beta - 2}{a + x -1}
\end{align*}
the question really boils down to the sum
$$\mathcal{S} = \sum_{n=0}^{\infty} \binom{2a-2}{a+n-1}$$
To this end,
\begin{align*}
\sum_{n=0}^{\infty} \binom{2a-1}{a+n-1} &=\frac{1}{2\pi i} \sum_{n=0}^{\infty} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a+n}}\, \mathrm{d}z \\
&= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1 + z \right )^{2a-1}}{z^a} \sum_{n=0}^{\infty} \frac{1}{z^n} \, \mathrm{d}z \\
&= \frac{1}{2\pi i} \oint \limits_{|z|=1} \frac{\left ( 1+z \right )^{2a-1}}{z^{a-1} \left ( z-1 \right )} \, \mathrm{d}z
\end{align*}
using the handy identity $\displaystyle \binom{n}{k} = \frac{1}{2\pi i } \oint \limits_{\gamma} \frac{\left ( 1+z \right )^n}{z^{k+1}} \, \mathrm{d}z$. I think I'm on the right track, but I'm having a difficult time evaluating the last contour integral. Any help?
| To supplement Claude Leibovici's answer, I will use Euler's integral representation of the Gaussian hypergeometric function to show that $$_2F_1(1,1-a;a;-1)=\frac{1}{2} \left(\frac{2^{2a-2} \, \Gamma^{2}(a)}{\Gamma(2a-1)} +1 \right) $$ for at least $a >1$. This is where the series $\sum_{n=0}^{\infty} \frac{2}{\Gamma ( a + n) \Gamma ( a - n )} $ converges absolutely.
$ \begin{align} _2F_1(1,1-a;a;-1) &= \, _2F_1(1-a,1;a;-1) \\ &= \frac{1}{B(1,a-1)} \int_{0}^{1} (1-x)^{a-2} (1+x)^{a-1} \, \mathrm dx \\ &= (a-1) \int_{0}^{1} (1-x)^{a-2}(1+x)^{a-2}(1+x) \, \mathrm dx\\ &= (a-1) \left( \int_{0}^{1} (1-x^{2})^{a-2} \, \mathrm dx +\int_{0}^{1} x (1-x^{2})^{a-2} \, \mathrm dx \right) \\ &= (a-1) \left( \frac{1}{2} \int_{0}^{1} (1-u)^{a-2} u^{-1/2} \, \mathrm du + \frac{1}{2}\int_{0}^{1} v^{a-2} \, \mathrm dv \right) \\ &= (a-1) \left( \frac{1}{2} \, B \left(\frac{1}{2},a-1\right) + \frac{1}{2(a-1)} \right) \\ &= \frac{a-1}{2} \left(\frac{ \sqrt{\pi} \, \Gamma(a-1)}{\Gamma \left(a- \frac{1}{2}\right)}+\frac{1}{a-1} \right) \\ &\overset{(1)}= \frac{a-1}{2} \left(\frac{\sqrt{\pi} \, \Gamma(a) 2^{2(a-1/2)-1} \Gamma \left(a-\frac{1}{2}+\frac{1}{2} \right)}{(a-1) \Gamma \left( 2\left(a-\frac{1}{2}\right)\right) \sqrt{\pi}}+\frac{1}{a-1} \right) \\ &= \frac{1}{2} \left(\frac{2^{2a-2} \Gamma^{2}(a)}{\Gamma(2a-1)} +1 \right). \end{align}$
$(1)$ Legendre duplication formula
| {
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"url": "https://math.stackexchange.com/questions/4446515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Find $\sqrt{\frac{1}{2}-f(1)}+\dots+ \sqrt{\frac{1}{2}-f(99)}$ Define $f(n)=\sqrt[2]{n^4+\frac{1}{4}}-n^2.$
Find $$\sqrt{\frac{1}{2}-f(1)}+\dots+ \sqrt{\frac{1}{2}-f(99)}$$
I tried to simply $f(n).$ So rationalising, we get $$\sqrt{\frac{1}{2}-f(n)}\sqrt{\frac{1}{2}+f(n)}=\sqrt{\frac{1}{2}+n^2-\sqrt{n^4+\frac{1}{4}}}\sqrt{\frac{1}{2}+n^2+\sqrt[2]{n^4+\frac{1}{4}}}=\sqrt{(\frac{1}{2}+n^2)^2-n^4-\frac{1}{4}}=\sqrt{n^2}=n$$
So $$\sqrt{\frac{1}{2}-f(n)}=\frac{n}{\sqrt{\frac{1}{2}+f(n)}}.$$
EDIT: Using the hints given below, we have to find We have to find $$\sum_{n=1}^{99} \frac{n}{\sqrt{n^2+1/4}+1/2}=\sum_{n=1}^{99} \frac{n({\sqrt{n^2+1/4}-1/2})}{(\sqrt{n^2+1/4}+1/2)(\sqrt{n^2+1/4}-1/2)}=\sum_{n=1}^{99}\frac{n({\sqrt{n^2+1/4}-1/2})}{n^2}= \frac{\sqrt{n^2+1/4}-1/2}{n}$$
Any solutions?
|
Claim:
$$\sqrt{\frac12-f(n)}=\frac12\left(\sqrt{(n+1)^2+n^2}-\sqrt{n^2+(n-1)^2}\right).$$
Proof:
Let
$$x=\sqrt{\frac12-f(n)}=\sqrt{\frac12+n^2-\sqrt{n^4+\frac14}},\quad y=\sqrt{\frac12+n^2\color{red}{+}\sqrt{n^4+\frac14}},$$
then by simple calculations,
\begin{align*}
\begin{cases}
x^2+y^2&=2n^2+1\\
xy&=n
\end{cases}
\end{align*}
so
\begin{align*}
\begin{cases}
y+x=\sqrt{x^2+y^2+2xy}&=\sqrt{2n^2+2n+1}\\
y-x=\sqrt{x^2+y^2-2xy}&=\sqrt{2n^2-2n+1}
\end{cases}
\end{align*}
The claim is proved by solving $x$.
Now the problem is reduced to a telescoping sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$ I am asked to prove that $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$
However, I am asked to prove it using the fact that
$$\frac{\pi}{2}\tan\left(\frac{\pi}{2}z\right)=\sum_{m \text{ odd}}\left(\frac{1}{m-z}-\frac{1}{m+z}\right),$$
where $z\in \mathbb{C}$, which is something I proved in a previous exercise.
My first thought was using the fact that
$$\frac{1}{m-z}-\frac{1}{m-z}=\frac{2z}{m^2-z^2}$$
and therefore
$$\sum_{m \text{ odd}}\left(\frac{1}{m-z}-\frac{1}{m+z}\right)=\sum_{m \text{ odd}}\frac{2z}{m^2-z^2}=\sum_{n=0}^\infty \frac{2z}{(2n+1)^2-z^2}.$$
This last series is similar to the one I am aiming at, but I don't know how to transform it into the one that I want. Can someone help me?
| Choose $z=\frac{a}{2}$ and take the limit as $a\rightarrow 0$.
We know that
$$
\lim_{x\rightarrow 0}\frac{\tan(x)}{x}=1,
$$
so
$$
\lim_{a\rightarrow 0}\frac{\frac{\pi}{2}\tan(\frac{\pi a}{4})}{a}=\frac{\pi^2}{8},
$$
which could be verified by L'Hospital's rule. We can replace $\tan(\frac{\pi a}{4})$ with the above summation, i.e.,
$$
\frac{\pi}{2}\tan\left(\frac{\pi a}{4}\right)=\sum_{n=0}\frac{a}{(2n+1)^2-\frac{a^2}{4}}.$$
Now, we can substitute and clean up.
$$
\begin{align}
\lim_{a\rightarrow 0}\frac{\sum_{n=0}\frac{a}{(2n+1)^2-\frac{a^2}{4}}}{a}&=\frac{\pi^2}{8}\\
\lim_{a\rightarrow 0}\sum_{n=0}\frac{1}{(2n+1)^2-\frac{a^2}{4}}&=\frac{\pi^2}{8}\\
\sum_{n=0}\lim_{a\rightarrow 0}\frac{1}{(2n+1)^2-\frac{a^2}{4}}&=\frac{\pi^2}{8}\\
\sum_{n=0}\frac{1}{(2n+1)^2}&=\frac{\pi^2}{8}
\end{align}
$$
Disclaimer: Thanks to Dan Velleman for his/her insight.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$x^2 - (m - 2)x + 6 = 0$ : Possible values of $m$.
One of the roots of the equation $x^2 - (m - 2)x + 6 = 0$ lies between
$(0, 2)$ and the other root lies between $(3, 4)$. It is known that
$m$ is an integer. What is the sum of all the possible values of $m$?
Let $a$ be the root that lies between $(0, 2)$ and let $b$ be the root that lies between $(3, 4)$. I can write it mathematically as
$0 \lt a \lt2$ and $3 \lt b \lt4$ and also we can say that $3 \lt a+b \lt6$.
$a+b=m-2$ and this means that $3 \lt m-2 \lt6 \Rightarrow 5 \lt m \lt 8$. So the possible integer values of $m$ should be $6$ and $7$ and the possible sum should be $13$.
But as per the solution this is not the correct answer. What have I done wrong? Please help !!!
Thanks in advance !!!
| The problem with using Viete's relations the way you did for the inequality is that they apply even when the roots are complex. So $ \ x^2 - (6-2)x + 6 \ = \ 0 \ $ has the roots $ \ 2 \ \pm \ i\sqrt2 \ \ , $ which sum to $ \ 4 \ \ \ , $ satisfying $ \ 3 \ < \ a+b \ < \ 6 \ \ . $
One way we can look at the quadratic polynomial is to consider its "vertex form" describing an "upward-opening" parabola with $ \ y-$intercept $ \ (0 \ , \ 6 ) \ \ . $ "Completing the square" produces
$$ x^2 \ - \ (m-2)x \ + \ 6 \ \ = \ \ \left(x \ - \ \frac{m-2}{2} \right)^2 \ + \ \left(6 \ - \ \frac{(m-2)^2}{4} \right) \ \ , $$
which tells us that the vertex is the $ \ x-$intercept of the parabola for
$$ 6 \ - \ \frac{(m-2)^2}{4} \ \ = \ \ 0 \ \ \Rightarrow \ \ m \ \ = \ \ 2 \ \pm \ 2·\sqrt6 \ \ . $$
Our quadratic equation then has real roots for $ \ m \ < \ 2 \ - \ 2·\sqrt6 \ $ and $ \ m \ > \ 2 \ + \ 2·\sqrt6 \ \ . $ But the $ \ y-$intercept at $ \ (0 \ , \ 6 ) $ means that we cannot has positive real roots for $ \ m \ < \ 2 \ - \ 2·\sqrt6 \ \ , $ since the vertex is "to the left" of the $ \ y-$axis, leaving us with $ \ m \ > \ \approx \ 6.899 \ \ . $
With $ \ m \ $ required to be an integer, we can assess cases directly. For $ \ m \ = \ 7 \ \ , $ the equation is $ \ x^2 - (7-2)x + 6 \ = \ (x-2)·(x-3) \ = \ 0 \ \ ; $ unfortunately, neither of the roots are in the specified open intervals. The next integer case leads to $ \ x^2 - (8-2)x + 6 \ = \ 0 \ \ , $ with roots $ \ 3 - \sqrt3 \ \approx \ 1.27 \ $ and $ \ 3 + \sqrt3 \ \approx \ 4.73 \ \ , $ the larger of these being "to the right" of the interval $ \ (3 \ , \ 4 ) \ \ . $ Using larger integers simply shifts the vertex "lower and to the right", making the larger of the roots even larger. Thus, we confirm Z Ahmed's conclusion that there are no possible integer values of $ \ m \ $ to sum. (Trick question...)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .
| By the fundamental theorem of algebra, we expect there to be 3 solutions.
If $z$ is a solution to $z^3 = -8$, then $|z^3| = |-8| = 8$, but since $|ab| = |a| * |b|$ and $|z|$ is a real number $\geq 0$, this implies $|z|^3 =8$ so $|z| = 2$.
Therefore we have the magnitude of all solutions to $z^3 = -8$. Now, we just need to match the angle. We have $(r e^{i \theta})^3 = r^3 e^{3 i \theta}$, and we already know that $r = 2$, so we just need $3 \theta = \pi (\mod 2 \pi)$.
$\theta = \pi/3$ is obviously a solution, for the other two solutions, add $2 \pi/3$ and $4 \pi/3$.
Therefore, the solutions are $2 e^{i \pi/3}, 2 e^{3 \pi i / 3} = -2$ and $2 e^{5 \pi i/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Investigate whether the series $\sum_{n=1}^\infty \frac{1+x^2 \sin (5x)}{\sqrt{n^3}}$ converges uniformly on $[-4,2]$ or not. Investigate whether or not the series $\sum_{n=1}^\infty \frac{1+x^2 \sin (5x)}{\sqrt{n^3}}$ converges uniformly on $[-4,2]$.
Weierstrass M-Test.
Let $(f_n)$ be a sequence of functions on $D \in \Bbb R$ to $\Bbb R$.
Let $(M_n)$ be a sequence of positive real numbers such that $|f_n(x)|\le M_n$ for all $x \in D, n \in \Bbb N$. If the series $\sum_{n=1}^\infty M_n$ convergent, then $\sum_{n=1}^\infty f_n$ uniformly convergent on $D$.
Attempt:
Notice that for all $x \in [-4,2]$, we have:
*
*$x^2 \le 16$, which is achieved when $x=-4$.
*$|\sin(5x)| \le 1$.
Therefore, for all $x \in [-4,2]$ and any $n \in \Bbb N$, we have
\begin{align*}
\left|\frac{1+x^2 \sin(5x)}{\sqrt{n^3}}\right| \le \frac{1+|x^2 \sin(5x)|}{\sqrt{n^3}}\le \frac{1+x^2}{\sqrt{n^3}} \le \frac{1+16}{\sqrt{n^3}} = \frac{17}{\sqrt{n^3}}.
\end{align*}
Hence, we get a sequence of positive real numbers
$\left(\frac{17}{\sqrt{n^3}} \right)$ such that $\left|\frac{1+x^2 \sin(5x)}{\sqrt{n^3}}\right| \le \frac{17}{\sqrt{n^3}}$ for all $x \in [-4,2]$ and $n \in \Bbb N$.
Now, the series $\sum_{n=1}^\infty \frac{17}{\sqrt{n^3}}$ is convergent by the $p$-Series Test.
Thus, by the Weierstrass M-Test, we conclude that the series $\sum_{n=1}^\infty \frac{1+x^2 \sin (5x)}{\sqrt{n^3}}$ converges uniformly on $[-4,2]$. Q.E.D.
Does this approach correct? If not, how to get it correctly?
Thanks in advanced for any comments and helps.
| Your approach is right. I see no point to be deducted.
| {
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"url": "https://math.stackexchange.com/questions/4463047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Understanding vieta jumping. Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $$\frac{a^2 + b^2}{ab+1}$$
is the square of an integer.
I have a few questions about the proof. First here it is ,
Assume $$\frac{a^2 + b^2}{ab+1}=k$$ is not a perfect square. Rearranging, we get a quadratic in $a$
$$a^2 −kb·a+(b^2 −k)=0. \quad (*)$$
Clearly $(*)$ has two solution the first one is $a$ so let the second be $s$. Then by Vieta we have $$s=kb-a=\frac{b^2-k}{a}$$ The first equation shows $s$ is an integer, and from $*$ $s$ is positive.
Now since $*$ symmetric we can assume $a>b$ Ignore $a=b$ because it gives a perfect square. Hence $s<a$ so we have a descent. Right?
In my book the author said this So, do we have a descent? If we repeat the process of (s, b), we would get a quadratic in s, and we pick the second root. Do you see an issue? But I don’t really see any issue.
| Problem: Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$. Show that $$\frac{a^2+b^2}{ab+1}$$ is the square of an integer (perfect square).
Expression $$\frac{a^2+b^2}{ab+1}$$ is symmetric with respect to $a$ and
$b$, then without loss of generality we can suppose $a\geq b$.
Let consider first case $a=b$. Then
$$\frac{a^2+b^2}{ab+1}=\frac{2a^2}{a^2+1}=2-\frac{2}{a^2+1}$$ This
expression can be integer at integer positive $a$ only if $a=1$, then
$$\frac{a^2+b^2}{ab+1}=1$$ is perfect square.
Now consider case $a>b$. Let suppose that $a$ is minimum positive
integer such that exists positive integer $b<a$ such that
$\frac{a^2+b^2}{ab+1}$ is integer but is not perfect square. Let
consider expression $$s=\frac{b^3-a}{ab+1}$$
$$s=\frac{b^3-a}{ab+1}=b\frac{a^2+b^2}{ab+1}-a\Rightarrow s\in\mathbb{Z}$$
$$s=\frac{b^3-a}{ab+1}>-\frac{a}{ab+1}\geq-\frac{a}{a+1}>-1$$ $$s>-1
\land s\in\mathbb{Z}\Rightarrow s\geq 0$$ $$s=0\Rightarrow
a=b^3\Rightarrow \frac{a^2+b^2}{ab+1}=b^2$$ which contradicts to the fact
that $\frac{a^2+b^2}{ab+1}$ is not perfect square. Then $s>0$.
$$s=\frac{b^3-a}{ab+1}<\frac{b^3}{b^2}=b$$ Let consider expression
$$\frac{b^2+s^2}{bs+1}=\frac{a^2+b^2}{ab+1}$$ Then $b$ is positive
integer such that exists positive integer $s<b$ such that
$\frac{b^2+s^2}{bs+1}$ is integer but is not perfect square. Then $b<a$ contradicts with the fact that $a$ is minimum positive integer with such
property. Then there is no minimum positive integer with such property,
then there is no positive integers $a,b$ with such property.
| {
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"url": "https://math.stackexchange.com/questions/4463376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Denesting radicals $\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}$ and $\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}}$ I am trying to do denesting radicals:$$\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}$$ and $$\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}}$$
I tried to find Ramanujan polynomial like this link denesting radicals
But it doesn't work. I also tried to solve the system of equation $\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$ but it led to a scary-looking one.
| (Too long for a comment.)
this is the system of equations after raise power of 3
$$
\begin{align}
p(a,b,c) = a^3+3b^3+9c^3+18abc+22 &= 0
\\ q(a,b,c) = a^2b+3ac^2+3b^2c-5 &= 0
\\ r(a,b,c) = a^2c+ab^2+3bc^2-3 &= 0
\end{align}
$$
Not something you'd do by hand, but that's an algebraic system which can always be reduced to one single polynomial equation using resultants, and then checked for rational solutions using the rational root theorem.
In this case for example, with WA's assistance you get 1 & 2
$\to$ 3 with the only rational root $c=1\,$:
$$
\begin{align}
u(b,c) = \text{res}_a (p,q) &= -3 b^9 + 27 b^6 c^3 - 22 b^6 - 81 b^5 c + 648 b^3 c^6 - 66 b^3 c^3
\\ &\quad - 486 b^2 c^4 - 198 b^2 c - 243 b c^2 + 81 c^9 + 396 c^6 + 484 c^3 - 27
\\v(b,c) = \text{res}_a(q,r) &= 3 b^7 c - 5 b^5 - 18 b^4 c^4 + 9 b^3 c^2 + 15 b^2 c^3 + 9 b^2 + 27 b c^7
\\ &\quad - 30 b c - 27 c^5 + 25 c^2
\\ w(c) = \text{res}_b(u,v) &= -9 (c - 1) (c^2 + c + 1) (3 c^3 + 1) (9 c^3 - 8) (\small{\dots\text{<degree 63 factor>} \dots})
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The closed form of $\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+1\right)^{n}} d x$ We are going to deal with the integral
$$
I_{n}=\int_{0}^{\infty} \frac{\cos x}{\left(x^{2}+1\right)^{n}} d x
$$
by differentiation on its related integral
$$
J(a)=\int_{0}^{\infty} \frac{\cos x}{x^{2}+a} d x,
$$
where $a>0$
Using contour integration along anti-clockwise direction of the path
$$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) $$
$$
\begin{aligned}
J(a) &=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos x}{x^{2}+a} d x \\
&=\frac{1}{2} \operatorname{Re} \int_{\gamma} \frac{e^{i z}}{z^{2}+a} d z \\
&=\frac{1}{2} \operatorname{Re}\left[2 \pi i \operatorname{Res}\left(\frac{e^{i z}}{z^{2}+a}, \sqrt{a} i\right)\right] \\
&=\frac{1}{2} \operatorname{Re}\left[2 \pi i \frac{e^{i(\sqrt{a} i)}}{2 \cdot \sqrt{a} i}\right] \\
&=\frac{1}{2} \operatorname{Re}\left[\frac{\pi}{\sqrt{a}} e^{-\sqrt{a}}\right] \\
&=\frac{\pi}{2 \sqrt{a}} e^{-\sqrt{a}}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ by $(n-1)$ times yields
$$
\left.I_{n}=\frac{\pi}{(n-1)!}J^{(n-1)}(1)=\frac{(-1)^{n-1} \pi}{2(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{e^{-\sqrt{a}}}{\sqrt{a}}\right)\right|_{a=1}
$$
Noting that
$$
\frac{d}{d a}\left(e^{-\sqrt{a}}\right)=-\frac{1}{2 \sqrt{a}} e^{-\sqrt{a}},
$$
we have $$
I_{n}=\left.\frac{(-1)^{n} \pi}{(n-1) !} \cdot \frac{d^{n}}{d a^{n}}\left(e^{-\sqrt{a}}\right)\right|_{a=1}
$$
By the formula, we can find
$$
I_{1}=\frac{\pi}{2e};I_{2}=\frac{\pi}{2 e};
I_{3}=\frac{7 \pi}{16 e} ;
I_{4}=\frac{37 \pi}{96e} ;
I_{5}=\frac{133 \pi}{384 e};\cdots\
I_{10}=\frac{12994393 \pi}{53084160 e}
$$
Suggestions on how to deal with the last derivative is highly appreciated.
| From this answer of mine: $$I_{n+1}=\frac{\pi/e}{2n!}\sum_{k=0}^n\frac{2^{-n-k}(n+k)!}{k!(n-k)!}=\frac{\pi/e}{2^{2n+1}n!}\sum_{k=0}^n\frac{2^k(2n-k)!}{k!(n-k)!}.$$
| {
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When $a^2 + b^2 + c^2 + d^2 = 1$ and the expression $(a - b)(b - c)(c - d)(d - a)$ reaches its minimum value, determine the value of product $abcd$.
Consider real number $a, b, c, d$ such that $a^2 + b^2 + c^2 + d^2 = 1$. When the expression $(a - b)(b - c)(c - d)(d - a)$ reaches its minimum value, determine the value of product $abcd$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
So I took the mock exam which was organised by our school today... It didn't go well. ｡ﾟ( ﾟஇ‸இﾟ)ﾟ｡ That's why this question is here.
Here's what I'd attempted in the official amount of time that was given, (so no external thoughts, I still have more subjects that need to be taken care of, sorry~)
(I lied, there are some afterthoughts sprinkled in here.)
First of all, $$\begin{aligned} (a - b)(b - c)(c - d)(d - a) &= [(ac + bd) - (ab + cd)][(ac + bd) - (bc + da)]\\ &= (ac + bd)^2 - (ab + bc + cd + da)(ac + bd)\\ &+ (ab + cd)(bc + da) \end{aligned}$$
(From this point on, it's all what I've thought of after the official time period has ended.)
Let $ac + bd = x$. The minimum value of the quadratic function $$x^2 - (ab + bc + cd + da)x + (ab + cd)(bc + da)$$ is $\dfrac{-\Delta}{4} = \dfrac{4(ab + cd)(bc + da) - (ab + bc + cd + da)^2}{4} = -\left[\dfrac{(a - c)(b - d)}{2}\right]^2$.
The equal sign occurs when $x = \dfrac{ab + bc + cd + da}{2} \iff 2(ac + bd) = ab + bc + cd + da \ (1)$. Now, we just need to find the maximum value of $\left[\dfrac{(a - c)(b - d)}{2}\right]^2$.
(∩ᗒ.ᗕ)⊃━☆ﾟ.❉*, $\left[\dfrac{(a - c)(b - d)}{2}\right]^2 \le \dfrac{(a - c)^2 + (b - d)^2}{8} \le \dfrac{a^2 + b^2 + c^2 + d^2}{4} = \dfrac{1}{4}$.
The equal sign occurs when $a = b = -c = -d \ (2)$. Combining $(1)$ and $(2)$, we have that $$\left\{ \begin{aligned} 2(ac + bd) &= ab + bc + cd + da\\ a = b &= -c = -d \end{aligned} \right. \implies a = b = c = d = 0$$
This feels wrong, and of course, it is. Let's elaborate on $(1)$ further, in that, $$\begin{aligned} 2(ac + bd) = ab + bc + cd + da &\iff 2[(ac + bd) - (bc + da)] = ab - bc + cd - da\\ &\iff 2(a - b)(c - d) = (a - c)(b - d) \end{aligned}$$
Now, does this mean anything? It is with deep regret that I have to inform you that, I don't freaking know.
Hmmm~ this doesn't seem to work. Let's try something different, for example, assuming that $a \ge b \ge c \ge d$, we have that $$\begin{aligned} (a - b)(b - c)(c - d)(d - a) &\ge \dfrac{[(a - b) + (b - c) + (c - d)]^3(d - a)}{27} = \dfrac{-(a - d)^4}{27} \end{aligned}$$
The equal sign occurs when $a - b = b - c = c - d \iff \left\{ \begin{aligned} b = \dfrac{d + 2a}{3}\\ c = \dfrac{2d + a}{3} \end{aligned} \right. \ (3)$.
(From this point on, it's all what I've thought of after the official time period has ended, the sequel.)
Plugging the above results into $a^2 + b^2 + c^2 + d^2 = 1$, we have that $7d^2 + 4da + 7a^2 = \dfrac{9}{2}$, and since $(7d^2 + 4da + 7a^2) - \dfrac{5(d - a)^2}{2} = \dfrac{9(d + a)^2}{2} \ge 0, \forall d, a \in \mathbb R$, it means that $$\dfrac{5(d - a)^2}{2} \le \dfrac{9}{2} \iff (d - a)^2 \le \dfrac{9}{5} \iff\dfrac{-(a - d)^4}{27} \ge -\dfrac{3}{25}$$
The equal sign occurs when $d = -a \ (4)$.
Combining $(3)$ and $(4)$, we have that $$\left[ \begin{aligned} (a; b; c; d) &= \left(\dfrac{3\sqrt{5}}{10}; \dfrac{\sqrt{5}}{10}; -\dfrac{\sqrt{5}}{10}; -\dfrac{3\sqrt{5}}{10}\right)\\ (a; b; c; d) &= \left(-\dfrac{3\sqrt{5}}{10}; -\dfrac{\sqrt{5}}{10}; \dfrac{\sqrt{5}}{10}; \dfrac{3\sqrt{5}}{10}\right) \end{aligned} \right. \implies abcd = \dfrac{9}{400}$$
According to WolframAlpha, the minimum value of the above expression is $-\dfrac{1}{8}$, happening when $$(a; b; c; d) = \left(-\dfrac{\sqrt 3 + 1}{4}; \dfrac{\sqrt 3 + 1}{4}; \dfrac{\sqrt 3 - 1}{4}; -\dfrac{\sqrt 3 - 1}{4}\right) \implies abcd = \dfrac{1}{64}$$
So my friend was right after all...（；¬＿¬) How am I supposed to do this in 90 minutes?
All anger aside, that's all for now. This took way more time to write down than it needed to. Anyhow, have a wonderful tomorrow, everyone~
By the way, the choices were $-\dfrac{1}{64}$; $-\dfrac{1}{8}$; $\dfrac{1}{8}$ and $\dfrac{1}{64}$.
| Let $f(a,b,c,d) := (a-b)(b-c)(c-d)(d-a)$.
We have $f_{\min} < 0$, by noting that $f < 0$ when $a > b > c > d$.
Since the constraint is symmetric and $f$ is cyclic, WLOG, assume that $a = \max(a, b, c, d)$. At minimum, we have $a - b > 0$ and $d - a < 0$, so $(b - c)(c - d) > 0$
which results in $a > b > c > d$
or $a > d > c > b$. Note also that $f(a, b, c, d) = f(a, d, c, b)$. Thus, we only need to consider the case when $a > b > c > d$.
Let $x = a - b, \, y = b - c, \, z = c - d$.
Then $x, y, z > 0$ and
$c = d + z, \, b = d + y + z, \, a = d + x + y + z$.
We have
$$f = - xyz (x + y + z).$$
From $a^2+b^2+c^2+d^2 = 1$, we have
$$4d^2 + (2x + 4y + 6z)d + x^2 + 2xy + 2xz + 2y^2 + 4yz + 3z^2 - 1 = 0.$$
This is a quadratic equation in $d$ which has a non-negative discriminant i.e.
$$-12x^2 - 16xy - 8xz - 16y^2 - 16yz - 12z^2 + 16 \ge 0 $$
which is written as (noting $f = -xz \cdot y(x + y + z)$)
$$-12(z + x)^2 + 16xz + 16 \ge 16y (x + y + z). \tag{1}$$
Using $zx \le (z+x)^2/4$ and (1), we have
$$16 - 8(x + z)^2 \ge 16y (x + y + z). \tag{2}$$
From (2), we have $16 > 8(x + z)^2$.
Using (2) and $zx \le (z+x)^2/4$ and AM-GM, we have
\begin{align*}
f &= -xyz(x + y + z)\\
&\ge - xz\cdot (1 - (x+z)^2/2)\\
&\ge - \frac{(x + z)^2}{4} \cdot (1 - (x + z)^2/2) \\
&= - \frac{1}{8}(x + z)^2 (2 - (x + z)^2)\\
&\ge - \frac18 \cdot \frac{[(x + z)^2 + 2 - (x + z)^2]^2}{4}\\
&= - \frac18
\end{align*}
with equality if and only if $x = z$ and $(x + z)^2 = 1$ and $y(x + y + z) = 1 - (x + z)^2/2$ i.e. $x = z = 1/2$ and $y = \frac{\sqrt 3 - 1}{2}$.
It is not difficult to get
$a = \frac{\sqrt 3 + 1}{4}, b = \frac{\sqrt 3 - 1}{4}$ and $c = -b, d = -a$.
Thus, $abcd = 1/64$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Creating my own change of variables to evaluate an integral The question asks me to evaluate the integral $$\iint_{R} e^{\frac{x+y}{x-y}} dA$$ where $R$ is trapezoid region with the vertices $(1,0), (2,0), (0,-2), (0,-1)$. I'm supposed to suggest a possible transformation and integrate and sketch the two regions.
My work :
Let the transformation be
$u=x-y$, $v=x+y$
Then with some algebra, I get $x=\frac{u+v}{2}$, and $y=-\frac{1}{2} (u-v)$
$J(u,v)=\begin{vmatrix}
\frac{1}{2} & \frac{1}{2}\\
-\frac{1}{2} & \frac{1}{2}
\end{vmatrix}=\frac{1}{2}$
When I sketch the region I have something like this on the xy plane
On the uv plane the transformation looks like:
So the integral becomes
$$\int_{1}^{2}\int_{-u}^{u} e^{\frac{v}{u}}*\frac{1}{2} dv du$$
$$\frac{1}{2}\int_{1}^{2}u\Big(e-\frac{1}{e}\Big)du$$
$$=\frac{1}{2}\Big(e-\frac{1}{e}\Big)*\frac{3}{2}=\frac{3}{4}\Big(e-\frac{1}{e}\Big)$$
Does this look correct?
| Your approach is nice and the calculation is well done. We can check it, by calculating it slightly different and look if the results coincide. We apply the identity
\begin{align*}
\frac{x+y}{x-y}=1+\frac{2y}{x-y}
\end{align*}
and consider
\begin{align*}
\iint_{R} e^{\frac{x+y}{x-y}} dA=e\iint_{R} e^{\frac{2y}{x-y}} dA\tag{1}
\end{align*}
We use the variable transformation
\begin{align*}
u&=y\qquad\qquad\qquad x=u+v\\
v&=x-y\qquad\qquad\ y=u
\end{align*}
The trapezoid regions have vertices
\begin{align*}
\mathrm{Tr}_{(x,y)}&=\{(0,-2),(0,-1),(1,0),(2,0)\}\\
\mathrm{Tr}_{(u,v)}&=\{(-2,2),(-1,1),(0,1),(0,2)\}\\
\end{align*}
and the transformed region is given by the graphic below.
The Jacobian determinant is
\begin{align*}
J(u,v)=
\begin{vmatrix}
x_u&x_v\\
y_y&y_v\\
\end{vmatrix}
=
\begin{vmatrix}
1&1\\
1&0
\end{vmatrix}=-1
\end{align*}
We obtain with the right-hand side of (1)
\begin{align*}
\color{blue}{e\iint_{R} e^{\frac{2y}{x-y}} dA}
&=e\int_1^2\int_{-v}^0e^{\frac{2u}{v}}|-1|\,du\,dv\\
&=e\int_{1}^2\left.\frac{v}{2}e^{\frac{2u}{v}}\right|_{-v}^0\,dv\\
&=\frac{e}{2}\int_{1}^2v\left(1-e^{-2}\right)\,dv\\
&=\frac{1}{2}\left(e-\frac{1}{e}\right)\left.\frac{1}{2}v^2\right|_{1}^2\\
&\,\,\color{blue}{=\frac{3}{4}\left(e-\frac{1}{e}\right)}
\end{align*}
in accordance with OPs result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find solution of a Polynomial Equation If $4b^2+1/b^2=16$ then how do I find the solution of $b^4+4/b^4-63/b^2$?
From $4b^2+1/b^2=16$, I got $$(2b+1/b)^2 = 12 \tag{1}$$ and $$(2b-1/b)^2 = 20 \tag{2}.$$
By solving equation (1), $$2b+1/b = 2\sqrt{3}$$ and by solving equation (2), $$2b-1/b = 2\sqrt{5},$$
but I couldn't find a way to proceed further.
4 options are given below. From these 4, I have to choose one as answer.
a) -1/4
b) -2
c) 3
d) 1/4
[Edited]
Thank you all for the suggestions you made. Now I have found the answer. The following is how I came across the solution.
Solution:
| $$4b^2 + \frac{1}{b^2} -16 =0$$
$$4b^4 - 16b^2 + 1 = 0$$
$$let \;u = b^2$$
$$4u^2 - 16u + 1 = 0$$
$$u = \frac{4 \pm \sqrt{15}}{2}$$
$$b^2 = \frac{4 \pm \sqrt{15}}{2}$$
$$b = \sqrt{\frac{4 + \sqrt{15}}{2}}, \sqrt{\frac{4 - \sqrt{15}}{2}}, -\sqrt{\frac{4 + \sqrt{15}}{2}}, -\sqrt{\frac{4 - \sqrt{15}}{2}}$$
Then substitute into $b^4 + \frac{4}{b^4} - \frac{63}{b^2}$ as required.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\frac{z^4+z^3+z^2}{(z-2)^3} = \sum_{n=-\infty}^{n=\infty} a_n(z-2)^n$, how can I find $a_{-2}$? If $\sum_{n=-\infty}^{n=\infty} a_n(z-2)^n$ is the laurent series of the function $f(z)=\frac{z^4+z^3+z^2}{(z-2)^3}$ for $z \in \mathbb{C} \setminus\{2\}$ ,then find $a_{-2} ?$
My attempt :By Taylor theorem, we have $f(z)=\sum_{-\infty}^{\infty}a_n(z-z_0)^n$ where $$a_n= \frac{1}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}} dz$$
$$\implies a_{-2}= \frac{1}{2\pi i}\int_C \frac{f(z)}{(z-2)^{{-2}+1}} dz=\frac{1}{2\pi i}\int_C \frac{f(z)}{(z-2)^{-1}} dz$$
$$=\frac{1}{2\pi i}\int_C \frac{\frac{z^4+z^3+z^2}{(z-2)^3}}{(z-2)^{-1}} dz$$
$$=\frac{1}{2\pi i}\int_C \frac{z^4+z^3+z^2}{(z-2)^2} dz$$
$$=\frac{1}{2\pi i} 2\pi i({4z^3+3z^2+2z})|_{z=2} =48$$
Therefore $a_{-2}=48$
| Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain by applying the binomial theorem
\begin{align*}
\color{blue}{[(z-2)^{-2}]}&\color{blue}{\frac{z^4+z^3+z^2}{(z-2)^3}}\\
&=[(z-2)^{-2}]\frac{((z-2)+2)^4+((z-2)+2)^3+((z-2)+2)^2}{(z-2)^3}\\
&=\binom{4}{1}2^3+\binom{3}{1}2^2+\binom{2}{1}2^1\\
&=32+12+4\\
&\,\,\color{blue}{=48}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$ given $x = \frac{1}{2-\sqrt{3}}$? It is given that $x = \frac{1}{2-\sqrt{3}}$. Find the value of $x^6 - 2\sqrt{3}x^5 - x^4 + x^3 - 4x^2 + 2x - \sqrt{3}$.
Well I tried rationalising and I came to know that $x = 2 + \sqrt{3}$. And I know that directly putting the values wont help either but I am not able to factorize the polynomial or manipulate it to help me.
I would be grateful if anybody could help me.
| Let $f(x) = x^6 -2\sqrt3 x^5 - x^4 + x^3 -4x^2 + 2x -\sqrt3$. Then, $f(2 + \sqrt3)$ is the remainder when $f(x)$ is divided by $x - (2+\sqrt3)$.(Remainder theorem )
Since, $$ f(x) = (x - (2+\sqrt3))(x^5 + (2-\sqrt3)x^4 + x^2 - (2-\sqrt3)x + 1) + 2$$
The remainder is $2$, thus $f(2+\sqrt3) = 2$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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How should I find $a_n$ knowing that $a_n = a_{n-1} + a_{n-3}$ I tried using a quadratic formula by using the constants of the recursive formula. Then when I get the solutions of the quadratic function, I would insert the $x$ values gotten to $a_n = a_1 \cdot (x_1)^n+a_2 \cdot (x_2)^n$. After, I would get some initial values such as $a_0$ and $a_3$ and make a system to solve. Unfortunately the $a_n$ formula gotten didn’t work.
| Here's an elementary way to find the roots of your characteristic equation.
$$
\begin{align}
r^3 - r^2 -1 &= 0 \\
\implies (y + \frac{1}{3})^3 - (y + \frac{1}{3})^2 - 1 &= 0 \quad \text{ where $y := r - \frac{1}{3}$} \\
\implies y^3 - \frac{y}{3} - \frac{29}{27} & = 0 \\
\implies 27y^3 - 9y - 29 &= 0 \\
\implies t^3 - 3t - 29 &= 0 \quad \text{ where t := 3y}
\end{align}
$$
Now let us find two real numbers $u$ and $v$ that satisfy $u + v = t$. Notice that $t^3 - 3uv\cdot t - (u^3 + v^3) = 0$. So we have
$$
\begin{align}
u^3 + v^3 & = 29 \quad \text{and,} \\
3uv & = 3 \\~\\
\implies u^3(29 - u^3) &= 1 \\
\implies u^6 - 29u^3 + 1 &= 0 \\~\\
\implies u^3 = \frac{29 \pm \sqrt{837}}{2} &= \frac{-29 \pm 3\sqrt{93}}{2}
\end{align}
$$
Without loss of generality, let $u = \sqrt[3]{\frac{29 + 3\sqrt{93}}{2}}$ and $v = \sqrt[3]{\frac{29 - 3\sqrt{93}}{2}}$. So,
$$
\begin{align}
t = t_1 := \sqrt[3]{\frac{29 + 3\sqrt{93}}{2}} + \sqrt[3]{\frac{29 - 3\sqrt{93}}{2}}
\end{align}
$$
Now, we can generate the other two roots of $t^3 - 3t - 29 = 0$ from $u$ and $v$.
Let $\omega = \frac{-1 + i\sqrt{3}}{2}$. Notice that $\omega$ and $\omega^2$ are the primitive cube roots of unity. Also notice that if we let $t = (u\omega + v\omega^2)$ or $t = (u\omega^2 + v\omega)$, we still get the same relation $t^3 - 3uv\cdot t - (u^3 + v^3) = 0$.
Hence the other two roots of $t^3 - 3t - 29 = 0$ are $t_2 := u\omega + v\omega^2$ and $t_3 := u\omega^2 + v\omega$.
We can find the corresponding value of $r$ for each of the roots since $r = \frac{t}{3} + \frac{1}{3}$.
For the root $t_1$, the corresponding value of $r$ is:
$$
\begin{align}
r = r_1 := \frac{1}{3} \left( 1 + \sqrt[3]{\frac{29 + 3\sqrt{93}}{2}} + \sqrt[3]{\frac{29 - 3\sqrt{93}}{2}} \right ) \approx 1.46557
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $\int_{0}^{\pi} \ln (b \cos x+c)$ without using Feynman’s integration technique? I shall find the integral by Feynman’s Technique Integration on a particular integral
$\displaystyle I(a)=\int_{0}^{\pi} \ln (a \cos x+1) d x,\tag*{} $
where $-1\leq a \leq 1.$
$\displaystyle \begin{aligned}I^{\prime}(a) &=\int_{0}^{\pi} \frac{\cos x}{a \cos x+1} d x, \\&=\frac{1}{a} \int_{0}^{\pi} \frac{(a \cos x+1)-1}{a \cos x+1} d x \\&=\frac{\pi}{a}-\frac{1}{a} \int_{0}^{\pi} \frac{d x}{a \cos x+1} \\&\stackrel{t=\tan \frac{x}{2}}{=} \frac{\pi}{a}-\frac{1}{a} \int_{0}^{\infty} \frac{1}{1+\frac{a\left(1-t^{2}\right)}{1+t^{2}}} \cdot \frac{2 d t}{1+t^{2}} \\&=\frac{\pi}{a}-\frac{2}{a} \int_{0}^{\infty} \frac{d t}{(1-a) t^{2}+(1+a)} \\&=\frac{\pi}{a}-\frac{2}{a \sqrt{1-a^{2}}} \tan^{-1}\left[\frac{\sqrt{1-a} t}{\sqrt{1+a}}\right]_{0}^{\infty} \\&=\frac{\pi}{a}-\frac{\pi}{a \sqrt{1-a^{2}}}\end{aligned}\tag*{} $
Integrating both sides w.r.t. $a$ yields
\begin{aligned}\int I^{\prime}(a) d a &=\pi\int\left(\frac{1}{a}-\frac{1}{a \sqrt{1-a^{2}}}\right) da \\& \stackrel{a=\sin \theta}{=} \pi\int\left(\frac{1}{\sin \theta}-\frac{1}{\sin \theta \cos \theta}\right) \cos \theta d \theta \\&=\pi\int \frac{\cos \theta-1}{\sin \theta} d \theta\\&I(a) =\pi \int \frac{-\sin ^{2} \theta}{\sin \theta(\cos \theta+1)} d \theta\\&=\pi \ln (1+\cos \theta) +C\end{aligned}
Putting $a=0$ gives $C=-\pi\ln 2$ and hence
$$
\boxed{\int_{0}^{\pi} \ln (a \cos x+1) d x =\pi \ln \left[1+\cos \left(\sin ^{-1} a\right)\right]= \pi \ln \left(\frac{1+\sqrt{1-a^{2}}}{2} \right)}
$$
I now want to generalize it to
$$I(b,c)=\displaystyle \int_{0}^{\pi} \ln (b \cos x+c),\tag*{} $$
where $c\neq 0$ and $-1\leq \frac{b}{c} \leq 1.$
$$
\begin{aligned}
I(b,c)&=\int_{0}^{\pi} \ln (b \cos x+c) \\
&=\int_{0}^{\pi} \ln \left[c\left(\frac{b \cos x}{c}+1\right)\right] \\
&=\pi \ln c+\int_{0}^{\pi} \ln \left(\frac{b}{c} \cos x+1\right) d x \\
&=\pi \ln c+I\left(\frac{b}{c}\right)
\end{aligned}
$$
Putting $a=\frac{b}{c}$ yields
$$\boxed{\int_{0}^{\pi} \ln (b \cos x+c) = \pi\left[\ln c+\ln \left(1+\sqrt{1-\frac{b^{2}}{c^{2}}}\right)\right] = \pi \ln \left(\frac{c+\sqrt{c^{2}-b^{2}}}{2}\right)}
$$
For example,
$$
\int_{0}^{\pi} \ln (\cos x+1)=\pi \ln \left(\frac{1}{2}\right)=-\pi \ln 2;
$$
$$
\int_{0}^{\pi} \ln (\sqrt{3} \cos x+2) d x=\pi\ln \frac{3}{2}
$$
Is there any method other than Feynman’s integration technique?
| Note that, with $r= \frac cb -\sqrt{\frac{c^2}{b^2}-1}$
$$\ln (b \cos x+c)=\ln\frac{c+\sqrt{c^2-b^2}}2+\ln\left(1+2r\cos x+r^2\right)
$$
Thus, per $\int_0^\pi \ln\left(1+2r\cos x+r^2\right)dx=0$
$$\int_{0}^{\pi} \ln (b \cos x+c)dx
=\pi \ln\frac{c+\sqrt{c^2-b^2}}2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding Laurent series for $f(z)=\frac{4z^2+2z-4}{z^3-4z}$ around $z=2$ Having
$$f(z)=\frac{4z^2+2z-4}{z^3-4z}$$
find the Laurent series in $z=2$
the scope of $z$ is $0<|z-2|<2$
here is my approach:
$f(z)=\frac{4z^2+2z-4}{z^3-4z}=\frac{1}{z}+\frac{2}{z-2}+\frac{1}{z+2}=\frac{1}{z-2+2}+\frac{2}{z-2}+\frac{1}{z-2+4}$
but now I can't figure out a way to transform the denumenators to something to make use of maclaurin series like $\frac{1}{1-(z-2)}$
for the first and last fraction I wrote something like $\frac{1}{2(1-(-\frac{z-2}{2}))}$ and $\frac{1}{4(1-(-\frac{z-2}{4}))}$ I'm not sure about these and for the middle one I don't know what to write I appreciate any help
| Just another way to do it $$f(z)=\frac{4z^2+2z-4}{z^3-4z}$$ Let $z=x+2$ and consider now
$$g(x)=\frac{2 \left(2 x^2+9 x+8\right)}{x \left(x^2+6 x+8\right)}=\frac{1}{x+2}+\frac{1}{x+4}+\frac{2}{x}$$
Now use the McLaurin series around $x=0$
$$\frac 1{x+a}=\sum_{n=0}^\infty (-1)^n a^{-(n+1)}\,x^n$$ Which, by the end, makes
$$g(x)=\frac 2x+\sum_{n=0}^\infty (-1)^n \,2^{-2 (n+1)} \left(2^{n+1}+1\right) \,x^n$$ Replace $x$ by $z-2$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
I need $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
$((2207^2 -2)^2 - 2)^2 -2)^2 ... - 2)$
quite big
But I realized I didn't have to simplify it, I just have to check if
$((((((2207^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2$ is divisible by 3
| Say we have checked that $x^n + y^n \equiv 2 \mod d$ for some $n$, $d$. Then $x'=x^n$, $y'=y^n$ are roots of an equation
$$T^2 - (2 + s \cdot d)\,T+1$$
Now, for all such equations, the sums $x'^k + y'^k$ will be the same $\mod d$. For the particular equation $T^2 - 2 T + 1$ they equal $2$. We conclude that $x^{k n} + y^{k n} \equiv 2 \mod d$ for all $k$.
$\bf{Added:}$ In our case, $x^4 + y^4 = 47 \equiv 2 \mod 45$. We conclude that $x^{4k} + y^{4k} \equiv 2 \mod 45$ for all $k\ge 1$.
Or: $x^3 + y^3 = -18 \equiv 2 \mod 20$, so $x^{3k} + y^{3k} \equiv 2 \mod 20$ for all $k\ge 1$.
$\bf{Added:}$ (This should come first) For the equation $T^2+1=0$ the sums $x^{4k} + y^{4k}=2$. So for any equation $T^2 + a T + b=0$, congruent to $T^2+1$ $\mod d$ the sums $x^{4k} + y^{4k} \equiv 2 \mod d$.
Take home idea: For congruent equations $P_1(T) \equiv P(T) \mod d$ the corresponding symmetric functions will be congruent. Use congruent equations for which those symmetric functions are easy to calculate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to evaluate $\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}} d x$ for positive constants $a$ and $c$? First of all, we deal with the simple case.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(ax^{2}+1\right)}{1+x^{2}} d x \stackrel{x=\tan \theta}{=} & \int_{0}^{\frac{\pi}{2}} \ln \left(1+a \tan ^{2} \theta\right) d \theta \\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(\frac{\cos ^{2} \theta+a \sin ^{2} \theta}{\cos ^{2} \theta}\right) d \theta \\
=& \int_{0}^{\frac{\pi}{2}} \ln \left(\cos ^{2} \theta+a \sin ^{2} \theta\right) d \theta-2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta \\
\end{aligned}
$$
Using the result in my post and $2 \int_{0}^{\frac{\pi}{2}} \ln (\cos \theta) d \theta =-\pi \ln 2$ yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\ln \left(1+ax^{2}\right)}{1+x^{2}} d x &=\pi \ln \left(\frac{1+\sqrt{a}}{2}\right)+\pi \ln 2 =\pi \ln (1+\sqrt{a})
\end{aligned} \tag*{(*)}
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}}d x \\
=& \int_{0}^{\infty} \frac{\left.\ln \left[c(\frac{a}{c} x^{2}+1\right)\right]}{1+x^{2}} d x \\
=& \int_{0}^{\infty} \frac{\ln c}{1+x^{2}} d x+\int_{0}^{\infty} \frac{\ln \left(\frac{a}{c} x^{2}+1\right)}{1+x^{2}}
\end{aligned}
Replacing $a$ by $\frac{a}{c}$ in $(*)$ yields
$$ \boxed{\int_{0}^{\infty} \frac{\ln \left(a x^{2}+c\right)}{1+x^{2}}d x =\pi \ln (\sqrt{a}+\sqrt{c})}$$
My question: Can we go further to evaluate a more general integral
$$\int_{0}^{\infty} \frac{\ln \left(ax^{2}+bx+1\right)}{1+x^{2}} d x?$$
| As suggested by @Ali Shadhar, we can use Feynman's technique to evaluate the integral.
$$I(b)=\int_0^\infty\frac{\ln(ax^2+bx+1)}{x^2+1}\,dx$$
For now, fix $a\gt0$ and $b\ge0$. We will make further assumptions, if required, throughout the solution.
Differentiating w.r.t. $b$,
$$\begin{align}I'(b)&=\int_0^\infty\frac x{(x^2+1)(ax^2+bx+1)}\,\mathrm dx\\&=\frac1{(a-1)^2+b^2}\int_0^\infty\frac{(1-a)x+b}{x^2+1}-\frac{a(1-a)x+b}{ax^2+bx+1}\,\mathrm dx\\&=\frac{(1-a)}{(a-1)^2+b^2}\underbrace{\int_0^\infty\frac x{x^2+1}-\frac{2ax+b}{2(ax^2+bx+1)}\,\mathrm dx}_{=-\ln(a)/2}+\frac b{(a-1)^2+b^2}\int_0^\infty\frac1{x^2+1}-\frac{1+a}{2(ax^2+bx+1)}\,\mathrm dx\end{align}$$
First integral follows directly using the antiderivative.
For absolute convergence of the second integral, we make the assumption that the denominators have no real roots. Thus $b^2\lt4a$.
Now taking the antiderivative, it follows that
$$I'(b)=\frac{(a-1)\ln a}{2((a-1)^2+b^2)}+\frac{\pi b}{2((a-1)^2+b^2)}-\frac{b(1+a)}{\sqrt{4a-b^2}((a-1)^2+b^2)}\arccos\frac b{2\sqrt a}$$
Integrating from $0$ to $b$,
$$I(b)=I(0)+\frac{\ln a}2\arctan\frac b{a-1}+\frac\pi4\ln\Big(\frac{b^2+(a-1)^2}{(a-1)^2}\Big)-\int_0^b\frac{x(1+a)}{\sqrt{4a-x^2}(x^2+(a-1)^2)}\arccos\frac x{2\sqrt a}\,\mathrm dx$$
$I(0)$ follows from the OP. I'm not quite sure how to evaluate the last integral. Some help regarding that would be appreciated. Note that the integrals are problematic if we directly put $a=1$. In that case, we will need to first take limits and then do the final integrals.
Comment if any errors and I'll fix them.
| {
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Proof that $\exp(-x^2) +\exp(-(x-y)^2)\leq 1 + 2\exp(-y^2)$ As stated in main text $$\exp(-x^2) +\exp(-(x-y)^2)\leq 1 + 2\exp(-y^2) ,\quad \forall x,y$$ with $0<x<y$, although it most likely also holds for all $x,y\in\mathbb{R}$.
The question came up in comparing effects of gaussian kernels. I've checked numerically for $x,y \in [0,10]$ and it always held. Local maxima of the left hand side is around $x = y\approx 1.9$.
The inequality is in a sense not tight, i.e. $\exists c<2$ with $\exp(-x^2) +\exp(-(x-y)^2)\leq 1 + c\exp(-y^2)$, however I've no idea what this $c$ would be (and I also don't need the tight version). But because it's not tight I gave up on most inequalities I know of such as all the AM-GM or Cauchy-Schwarz, because they're all tight in that sense.
| Let $w = x^2$ and $z = 2x (y - x)$.
We have $w, z > 0$ and $y = x + \frac{z}{2x}$.
The desired inequality is written as
$$\mathrm{e}^{-w} + \mathrm{e}^{-\frac{z^2}{4w}} \le 1 + 2\mathrm{e}^{-w - z - \frac{z^2}{4w}}. \tag{1}$$
We split into three cases.
Case 1: $z \ge 2$
Using $\mathrm{e}^{u} \ge 1 + u$ for all $u \ge 0$, we have
$$\mathrm{e}^{-w} + \mathrm{e}^{-\frac{z^2}{4w}}
\le \frac{1}{1 + w} + \frac{1}{1 + \frac{z^2}{4w} } \le \frac{1}{1 + w} + \frac{1}{1 + \frac{2^2}{4w} } = 1.$$
The desired result follows.
Case 2: $z \le \ln 2$
We have
$$1 + 2\mathrm{e}^{-w - z - \frac{z^2}{4w}}
\ge 1 + \mathrm{e}^{-w - \frac{z^2}{4w}}.$$
Also, we have
$$1 + \mathrm{e}^{-w - \frac{z^2}{4w}} - \mathrm{e}^{-w} - \mathrm{e}^{-\frac{z^2}{4w}}
= (1 - \mathrm{e}^{-w})(1 - \mathrm{e}^{-\frac{z^2}{4w}}) \ge 0.$$
The desired result follows.
Case 3: $\ln 2 < z < 2$
The desired inequality is written as
$$\mathrm{e}^{-w} + \mathrm{e}^{-\frac{z^2}{4w}}(1 - 2\mathrm{e}^{-w - z}) \le 1.$$
Using $\mathrm{e}^{-z} \ge (1 - z/4)^4 > 0$ (equivalently $\mathrm{e}^{-z/4} \ge 1 - z/4 > 0$), it suffices to prove that
$$\mathrm{e}^{-w} + \mathrm{e}^{-\frac{z^2}{4w}}\Big(1 - 2\mathrm{e}^{-w}(1 - z/4)^4\Big) \le 1.$$
Using $\mathrm{e}^{u} \ge 1 + u + \frac12 u^2$ for all $u \ge 0$, we have
$$\mathrm{e}^{-\frac{z^2}{4w}}
\le \frac{1}{1 + \frac{z^2}{4w} + \frac12(\frac{z^2}{4w})^2}.$$
It suffices to prove that
$$\mathrm{e}^{-w} + \frac{1}{1 + \frac{z^2}{4w} + \frac12(\frac{z^2}{4w})^2}\Big(1 - 2\mathrm{e}^{-w}(1 - z/4)^4\Big) \le 1$$
or
$$\left(1 - \frac{2(1 - z/4)^4}{1 + \frac{z^2}{4w} + \frac12(\frac{z^2}{4w})^2}\right) \mathrm{e}^{-w} + \frac{1}{1 + \frac{z^2}{4w} + \frac12(\frac{z^2}{4w})^2} \le 1.$$
Using $\mathrm{e}^{u} \ge 1 + u + \frac12 u^2$ for all $u \ge 0$, we have
$$\mathrm{e}^{-w} \le \frac{1}{1 + w + \frac12 w^2}.$$
It suffices to prove that
$$\left(1 - \frac{2(1 - z/4)^4}{1 + \frac{z^2}{4w} + \frac12(\frac{z^2}{4w})^2}\right) \frac{1}{1 + w + \frac12 w^2} + \frac{1}{1 + \frac{z^2}{4w} + \frac12(\frac{z^2}{4w})^2} \le 1$$
or (after clearing the denominators)
$$3wz^4 + 4z^2(2w - z)^2 + 128w(z - 1)^2 \ge 0$$
which is clearly true.
We are done.
| {
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Evaluate the indefinite integral $\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}$ without trigonometric substitution. In order to find
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}
$$
we set $t=\arctan x$. Then $x=\tan t$ and $dt=\frac{dx}{x^2+1}$, so
$$
\int\frac{dx}{(x^2+1)\sqrt{x^2+1}}=\int\frac{dt}{\sqrt{\frac{1}{\cos^2t}}}=\int\cos tdt\\
=\sin t+C=\sin(\arctan x)+C
$$
Now, since
$$
\sin(\arctan x)=\sqrt{\frac{\tan^2(\arctan x)}{\tan^2(\arctan x)+1}}
$$
the answer is $\frac{x}{\sqrt{x^2+1}}+C$.
My Question: Is there another way to find this integral without using trigonometry?
| $$\int\frac{1}{(x^2+1)\sqrt{x^2+1}}dx\overset{x=\frac1y}=
-\int \frac y{(1+y^2)^{3/2}}dy=\frac1{\sqrt{1+y^2}}+C
$$
| {
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Graphing Functions Algebra Suppose $f(x),g(x),h(x)$ are all linear functions, and $j(x)$ and $k(x)$ are defined by $$j(x) = \max\{f(x),g(x),h(x)\},$$ $$k(x) = \min\{f(x),g(x),h(x)\}.$$ This means that, for each $x$, we define $j(x)$ to be equal to either $f(x),$ $g(x),$ or $h(x),$ whichever is greatest; similarly, $k(x)$ is the least of these three values.
Shown below is the graph of $y=j(x)$ for $-3.5\le x\le 3.5$.
Let $\ell$ be the length of the graph of $y=k(x)$ for $-3.5\le x\le 3.5$. What is the value of $\ell^2$?
Since the function $j(x)$ is made up of 3 parts. I first tried to find the equations of each of the lines. The horizontal line is just $j(x)=2 [-2\le{x}\le{2}]$. The other two lines have almost the same equation - the only thing that differs is their slope. I used point-slope form to find that the equation of the left-most line is $j(x)=-2x-2 [-3.5\le{x}\le{-2}]$, and the equation of the right-most line is $j(x)=2x-2 [2\le{x}\le{3.5}]$. Therefore, the slopes of the two lines, respectively, are $-2$ and $2$.
Then, I tried finding the total length of the function $j(x)$. I drew a perpendicular line from the top point of the right-most function until it met the line $y=2$. The distance between the point where it meets and the red dot is $1.5$. Since the slope is $2$, the change in $y$ goes up by $(1.5)(2)=3$. The same is true with the left-most function. Then, I used the Pythagorean theorem, $\sqrt{1.5^2+3^2}$, to find the lengths of the two slanted sides. I got $\sqrt{11.25}$. So, the total length is $4+2\sqrt{11.25}$.
However, I'm don't know how to find the relationship between $j(x)$ and $k(x)$ (aka how the maximum and minimum are related).
|
Since the functions $f(x),g(x),h(x)$ are linear, every interval that $j(x)$ maps to a line is in the domain of one of the linear functions.
If we extend the lines corresponding to the $3$ functions, (and choose the function with the smallest value for each $x$) we can see that $k(x)$, from left to right, is the rightmost function for $j(x)$ upto its intersection point with the leftmost line of $j(x)$ and then the leftmost function of $j(x)$.
The left and right functions are:
\begin{eqnarray}
L_1&=&-2x-2\\
L_2&=&2x-2\\
L_1&\overset{\tiny \text{set}}{=}&L_2\\
x&=&0\\
L_1(0)&=&L_2(0)=-2\\
L_1(3.5)&=&-9\\
L_2(3.5)&=&5\\
\end{eqnarray}
Now we can calculate $\ell$ using Pythagoras' theorem:
\begin{eqnarray}
\ell&=&\sqrt{3.5^2+(-2-(-9))^2}+\sqrt{3.5^2+(-2-5)^2}\\
&=&2\sqrt{3.5^2+7^2}\\
\ell^2&=&4(3.5^2+7^2)\\
&=&(2\cdot 3.5)^2+4\cdot 7^2\\
&=&5\cdot 7^2\\
&=&5\cdot 49\\
&=&245\\
\end{eqnarray}
| {
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Prove that $\prod\limits_{\mathrm{cyc}}\left(1+\frac{1}{\sqrt{ab}}\right)^a\geq 2^{a+b+c+d}$ for $a+b+c+d \le 4$
Let $a,b,c,d$ be positive real numbers satisfying $a+b+c+d\leq 4$. Show
$$\left(1+\frac{1}{\sqrt{ab}}\right)^a\left(1+\frac{1}{\sqrt{bc}}\right)^b
\left(1+\frac{1}{\sqrt{cd}}\right)^c\left(1+\frac{1}{\sqrt{da}}\right)^d\geq
2^{a+b+c+d}.$$
My attempt: I tried using the generalized GM-HM inequality,
$$RHS\geq \left(\frac{a+b+c+d}{\sum_{cyc}\frac{a\sqrt{ab}}{1+\sqrt{ab}}}\right)^{a+b+c+d}$$
and it remains to show that
$$a+b+c+d\geq 2\sum_{cyc}\frac{a\sqrt{ab}}{1+\sqrt{ab}}.$$
Any advice from here? Any idea for a different approach is also very welcome!
| Remarks: Here is a proof. The proof of (5) is not nice. Hope to see a nice proof.
Taking logarithm, the desired inequality is written as
$$\sum_{\mathrm{cyc}} a\ln \frac{1 + \frac{1}{\sqrt{ab}}}{2}
\ge 0. \tag{1}$$
Using $\ln x \ge \frac{2(x-1)}{1+x}$ for all $x > 0$ (easy to prove), we have
$$\ln \frac{1 + \frac{1}{\sqrt{ab}}}{2}
\ge \frac{8}{3} \cdot \frac{1}{3\sqrt{ab} + 1} - \frac23. \tag{2}$$
From (1) and (2), it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab} + 1} - \frac23(a + b + c + d) \ge 0$$
or
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab}(a+b+c+d) + (a+b+c+d)} \ge \frac23. \tag{3}$$
Since $a+b+c+d \le 4$, from (3), it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab}\cdot 4 + (a+b+c+d)} \ge \frac23. \tag{4}$$
Since the inequality (4) is homogeneous, WLOG, assume that $a + b + c + d = 4$. It suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{8}{3} \cdot \frac{a}{3\sqrt{ab}\cdot 4 + 4} \ge \frac23$$
or
$$\sum_{\mathrm{cyc}} \frac{a}{3\sqrt{ab} + 1} \ge 1. \tag{5}$$
A proof of (5) is given at the end.
We are done.
Proof of (5):
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sum_{\mathrm{cyc}} \frac{a}{3\sqrt{ab} + 1}
\ge \frac{\left[\sum_{\mathrm{cyc}} \sqrt a \, (\sqrt a + \sqrt d)\right]^2}{\sum_{\mathrm{cyc}} (\sqrt a + \sqrt d)^2 (3\sqrt{ab} + 1)}. \tag{6}$$
From (5) and (6), it suffices to prove that
$$\left[\sum_{\mathrm{cyc}} \sqrt a \, (\sqrt a + \sqrt d)\right]^2 \ge \sum_{\mathrm{cyc}} (\sqrt a + \sqrt d)^2 (3\sqrt{ab} + 1).$$
After homogenization, it suffices to prove that
$$\left[\sum_{\mathrm{cyc}} \sqrt a \, (\sqrt a + \sqrt d)\right]^2 \ge \sum_{\mathrm{cyc}} (\sqrt a + \sqrt d)^2 \left(3\sqrt{ab} + \frac{a + b + c + d}{4}\right).$$
Letting $a = x^2, b = y^2, c = z^2, d = w^2$, it suffices to prove that, for all $x, y, z, w \ge 0$,
$$\left[\sum_{\mathrm{cyc}} x (x + w)\right]^2 \ge \sum_{\mathrm{cyc}} (x + w)^2 \left(3xy + \frac{x^2 + y^2 + z^2 + w^2}{4}\right). \tag{7}$$
WLOG, assume that $w = \min(x, y, z, w)$.
Let $s = z - w, t = y - w, r = x - w$. Then $s, t, r \ge 0$. The inequality (7) is written as
$$A w^2 + Bw + C \ge 0 \tag{8}$$
where
\begin{align*}
A &= 4\, \left( r-t \right) ^{2}+4\, \left( s-t \right) ^{2}+4\,{r}^{2}+4\,
{s}^{2}, \\[6pt]
B &= {\frac { \left( 3\,{s}^{2}-2\,st+3\,{t}^{2} \right) {r}^{3}}{{s}^{2}-2
\,st+3\,{t}^{2}}}+{\frac {r \left( rs-3\,rt+2\,{s}^{2}-4\,st+6\,{t}^{2
} \right) ^{2}}{{s}^{2}-2\,st+3\,{t}^{2}}}\\[8pt]
&\qquad +4\,{s}^{3}+3\,{s}^{2}t+t
\left( 3\,s-2\,t \right) ^{2},\\[6pt]
C &= \frac14\,{r}^{2} \left( 2\,r-3\,t \right) ^{2}+\frac18\, \left( 4\,s-3\,t
\right) ^{2}{r}^{2}+\frac58\,{t}^{2}{r}^{2}+ \left( 3\,{s}^{2}t-8\,s{t}^{
2}+3\,{t}^{3} \right) r\\
&\qquad +{s}^{4}+3\,{s}^{3}t+4\,{s}^{2}{t}^{2}-3\,s{t}^
{3}+{t}^{4}.
\end{align*}
It is easy to prove that $A, B, C \ge 0$.
Thus, the inequality (8) is true.
We are done.
| {
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Let $a_n$ be a sequence defined by: $a_n = \min(m)$ $s.t.$ $\sum_{i = 1}^m \sqrt i > n^2$. Does $\sum \frac{1}{a_n}$ converges? I am trying to show convergence (Or divergence) of the following sum:
Let $a_n$ be a sequence defined by:
$a_n = \min(m)$ $s.t.$ $\sum_{i = 1}^m \sqrt i > n^2$.
Does $\sum \frac{1}{a_n}$ converges?
If we define $b_n = \min(m)$ $s.t.$ $\sum_{i = 1}^m \ i > n^2$, We get $b_n > n$, since $\sum_{i = 1}^ni = \frac{n(n+1)}{2} < n^2$, So $a_n > n$ for sure.
But im not sure how evaluate $a_n$ any further.
Any hints will be appericiated.
| For $a>b>0$ we have
$$a^3-b^3=(a-b)(a^2+ab+b^2)\ge a^2(a-b)$$ $$a-b={a^2-b^2\over a+b}\ge {a^2-b^2\over 2a}$$ Hence for $a=i^{1/2}$ and $b=(i-1)^{1/2}$ we get
$$i^{3/2}-(i-1)^{3/2}\ge i [i^{1/2}-(i-1)^{1/2}]\ge {1\over 2} i^{1/2}$$ Adding the terms gives
$$\displaystyle 2m^{3/2}\ge \sum_{i=1}^m i^{1/2}$$ If the sum is greater than $n^2,$ then $2m^{3/2}\ge n^2, $ which implies $m\ge 2^{-1}n^{4/3}. $ Hence $$a_n\ge {1\over 2}n^{4/3}$$
Remark The estimate is optimal concerning the exponent $4/3,$ that means $a_n\le cn^{4/3}.$ To this end it suffices to apply the inequalities (for $a>b>0$)$$a^3-b^3=(a-b)(a^2+ab+b^2)\le 3a^2(a-b)$$ $$a-b={a^2-b^2\over a+b}\le {a^2-b^2\over a}$$
| {
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Simpler approach when calculating $\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x}$? I know this is a really simple problem, but I can't figure out what logical mistake I am making on my approach :
$$\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x}$$
Dividing numerator and denominator with $4^x$,
$$ \lim_{x\to0} \frac{({\frac{e^2}{4}) }^x-{(\frac{e}{4}})^x}{1 - ({\frac{1}{2}})^x} $$
$\left(\frac{1}{2}\right)^x$ and $\left(\frac{e}{4}\right)^x$ converge to $0,$ so I thought this limit converges to $1$.
(+ additional approach)
$$
\lim_{x\to0} \frac{e^{2x} - e^x}{4^x - 2^x} \\
= \lim_{x\to0} \frac{e^{2x} - e^x}{x} \cdot \frac{x}{4^x - 2^x} \\
= \lim_{x\to0} \left(\frac{e^{2x} - 1}{x} - \frac{e^x - 1}{x} \right) \cdot \frac{1}{\left(\frac{4^x - 1}{x} - \frac{2^x - 1}{x} \right)}
$$
And this becomes $(2-1) \times \frac{1}{\ln4 - \ln2 } = \frac{1}{\ln2}$, using the definition of derivative.
I wonder if there's a simpler solution without using L'Hopital's rule.
| If you know $\exp(x)=1+x+o(x)$ (which is consequence of $\exp'(x)|_{x=0}=1$), then using $a^x=e^{x\ln a}$:
$$\lim_{x\to 0}\frac{e^{2x}-e^{x}}{4^x-2^x}=\lim_{x\to 0}\frac{1+2x+o(2x)-1-x-o(x)}{1+x\ln 4 +o(x\ln 4)-1-x\ln 2-o(x\ln 2)}=\\
\lim_{x\to 0}\frac{2x-x+o(x)}{x(\ln 4-\ln 2)+o(x)}=\lim_{x\to 0}\frac{1+o(1)}{\ln 2+o(1)}=\frac{1}{\ln 2}$$
| {
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How many positive integers $n$ are there such that $2n$ and $2n^2+1$ are both perfect squares? How many positive integers $n$ are there such that $2n$ and $2n^2+1$ are both perfect squares?
$n=2$ is the only solution I can find. Are there others?
| Not a perfect solution... but maybe can be a hint.
$2n$ is perfect square, so let $n=2k^2$.
Therefore, $8k^4+1$ is also a perfect square.
let $8k^4+1=l^2$.
Then, $l^2-8k^4=1.$
let $k^2=m$, which leads to:
$l^2-8m^2=1.$
This is a Pell equation. solving this, we get:
$(l, m)=(3, 1), (17, 6), ..., \left( \dfrac{(3+\sqrt8)^{\alpha}+(3-\sqrt{8})^{\alpha}}{2}, \dfrac {(3+\sqrt{8})^{\alpha}-(3-\sqrt{8})^{\alpha}} {2\sqrt{8}} \right), ...$ .
So, the number of $n$ will be the number of the perfect square of $\dfrac{(3+\sqrt8)^{\alpha}-(3-\sqrt8)^{\alpha}}{2\sqrt8}$. This can be also written:
$ (3+\sqrt{8})^{\alpha-1} + (3+\sqrt{8})^{\alpha-2}(3-\sqrt{8})+\cdots+(3+\sqrt{8})(3-\sqrt{8})^{\alpha-2}+(3-\sqrt{8})^{\alpha-1} $.
Use that $(3+\sqrt{8})(3-\sqrt{8})=1.$
| {
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Continuous mapping from a nowhere dense set to dense set Let $X$ be a complete metric space, $B$ be a (closed) nowhere dense set in $X$, $f:B\rightarrow X$ be a continuous mapping. Is it possible that $f(B)$ is dense in $X$?
I think that is possible. But I can't give a counter-example. Any help will be appreciated.
| Since $B$ is nowhere dense, it inherits the discrete topology from $X$, which means that any function from $B$ will be continuous. Thus any surjection from $\mathbb{Z}$ onto $\mathbb{Q}$ serves as an example.
But we can do better than this. It is actually possible for $f$ to be continuous on all of $X$, with $f(B)$ still dense! Let $X$ be the positive reals, with $f(x)=x\sin(x)$. Define $$B = \bigcup_{n \in \mathbb{N}} \{2\pi n^3 + \sin^{-1}(\frac{k}{n^4}): k \in \mathbb{N}, k < n^2\}. $$ One can see that $B$ is nowhere dense: every interval of length $2\pi$ contains only finitely many points of $B$. To show $f(B)$ is dense, we need to construct a $b \in B$ with $|b\sin(b) - x| < \epsilon$ for any arbitrary $x, \epsilon > 0$. For large $n$, we have that $\frac{1}{n^2} < \frac{\epsilon}{2}$. For larger $n$, there is some integer $k$ for which $|\frac{2\pi k}{n} - x| < \frac{\epsilon}{2}$. For still larger $n$, we also can guarantee that $k < n^2$.
Therefore $B$ contains the point $b = 2\pi n^3 + \sin^{-1}(\frac{k}{n^4})$. Now, $f(b) = b \sin(b) = b \cdot \frac{k}{n^4} = \frac{2\pi k}{n} + \frac{k}{n^4}\sin^{-1}(\frac{k}{n^4})$. Thus,
$$ |b - x|
= \bigg| \frac{2\pi k}{n} + \frac{k}{n^4}\sin^{-1}(\frac{k}{n^4}) - x \bigg|
\leq \bigg| \frac{2\pi k}{n} - x\bigg| + \bigg|\frac{k}{n^4}\sin^{-1}(\frac{k}{n^4}) \bigg|. $$
The first term is less than $\frac{\epsilon}{2}$ by choice of $k$. The second term is also less than $\frac{\epsilon}{2}$, because $\sin^{-1}(\frac{k}{n^4})$ is bounded by 1, and $\frac{k}{n^4} < \frac{n^2}{n^4} = \frac{1}{n^2} < \frac{\epsilon}{2}$. Therefore $|b - x| < \epsilon$.
There are probably better examples than this. This related post offers a hypothesis which, if true, would give a far more elegant example.
| {
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Centered hexagonal numbers This is one of Brilliant's daily challenges. What I see is four arithmetic progressions. I did my calculations according to the formulas:
$$S_{n1} = \frac n2(2a + (n − 1) × d_1),$$ $d_1=1$.
$$ S_{n2} = \frac n2(2a + (n − 1) × d_2),$$ $d_2=2$.
$X = S_{n1} + S_{n2} - 3$(so that I don't count the central dot four times) = $78 \cdot 2 + 144 \cdot 2 - 3 = 441$. The problem is Brilliant doesn't have such an answer. Did I make a mistake somewhere?
The first several centered hexagonal numbers are
$$1,7,19,37,61…$$
What is the 12th centered hexagonal number?
| For now, we will not add the single dot in the center.
Let $f(n)$ be the formula for the number of dots of the $n$th hexagon. Keep in mind that this is not the centered hexagonal number yet.
Notice that there are always six dots for each hexagon. Also, notice that for the $n$th hexagonal number, in a particular side excluding the vertices, there are $n - 1$ dots. Since hexagons have six sides, we have $6n - 6$ dots.
Hence, we have
\begin{align*}
f(n) &= 6n - 6 + 6 \\
f(n) &= 6n
\end{align*}
Now, let $P(n)$ be the centered hexagonal number without the single dot in the center. We can see that $$P(n) = f(0) + f(1) + f(2) + \cdots + f(n).$$ Simplifying the right-hand side in terms of $n$,
\begin{align*}
P(n) &= f(0) + f(1) + f(2) + \cdots + f(n) \\
P(n) &= 6(0) + 6(1) + 6(2) + \cdots + 6n \\
P(n) &= 6(0 + 1 + 2 + \cdots + n) \\
P(n) &= 6\left(\frac{n(n + 1)}{2}\right) \\
P(n) &= 3n(n + 1)
\end{align*}
Adding the dot in the center, we now have $P(n) = 3n(n + 1) + 1$. In this formula, the dot is also a hexagon, although its "size" makes it essentially a point. Since we are looking for the 12th hexagonal number, by replacing $n$ by $11$ (since the $+1$ is the $12$th hexagonal number), we get $397$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solve $ \frac{1}{xy} = \frac{x}{yz} + 1, \:\: \frac{1}{yz} = \frac{y}{zx} + 1, \:\: \frac{1}{zx} = \frac{z}{xy} + 1. $ Solve the following system for positive integers:
$$ \frac{1}{xy} = \frac{x}{yz} + 1, $$
$$ \frac{1}{yz} = \frac{y}{zx} + 1, $$
$$ \frac{1}{zx} = \frac{z}{xy} + 1. $$
Attempt:
Equivalent to
$$ z = x^{2} + xyz, $$
$$ x = y^{2} + xyz, $$
$$ y = z^{2} + xyz. $$
Multiplying all:
$$ xyz = (x^{2}+xyz)(y^{2}+xyz)(z^{2}+xyz) = (xyz)^{2} + (xy)^{3}z + (xz)^{3}y + (yz)^{3}x + x^{4}(yz)^{2} + y^{4}(xz)^{2} + z^{4}(xy)^{2} $$
| We see that the first equation is equivalent to:
$$\frac{1}{xy} = \frac{x}{yz} + 1\ \Rightarrow z = x^{2}+xyz \ \Rightarrow z(1-xy) = x^{2}$$
Since $x^{2}$ and $z$ are positive, we must have $1-xy > 0$. However, since $x$ and $y$ are positive integers, this is impossible, so the system has no solution. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243$ , find the value of $x$. $\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243=3^5$
$\Rightarrow \frac{16^{x-3}+14}{4^{x-3}+2}= {3^3}^5=3^{243}$
I am not able to proceed from here. Please help !!!
The options given for this problem are :-
*
*$x$ is a rational number.
*$x$ is a natural number.
*$x$ is an even natural number.
*$x$ is a rational number less than $0$.
Thanks in advance !!!
| Note that, for $x = 5$, the LHS of $\frac{16^{x-3}+14}{4^{x-3}+2}$ becomes $\frac{256+14}{16+2}=15$, with the LHS being less than $15$ for $x \lt 5$ (with $\lim_{x\to -\infty}\frac{16^{x-3}+14}{4^{x-3}+2}=\frac{14}{2} = 7$). Thus, $x \gt 5$. Also, we have that
$$\begin{equation}\begin{aligned}
\frac{16^{x-3}+14}{4^{x-3}+2} & = \frac{16(16^{x-4})+14}{4(4^{x-4})+2} \\
& = \frac{8(16^{x-4})+7}{2(4^{x-4})+1}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
If $x$ is an integer, then since $8(16^{x-4})+7 \equiv 7 \pmod{8}$ and $2(4^{x-4})+1 \equiv 1 \pmod{8}$, the fraction in \eqref{eq1A} also being an integer (i.e., $3^{243}$) means it's congruent to $7$ modulo $8$. However, since $3^2 \equiv 1 \pmod{8}$, we instead have $3^{243} \equiv 3(3^{242}) \equiv 3(3^2)^{121} \equiv 3 \pmod{8}$.
This contradiction shows $x$ can't be an integer. Thus, it must either be a positive irrational or a non-integer rational. Since none of the provided second to fourth options are valid possibilities, this means the first option must be the correct one, i.e., $x$ is a non-integral rational number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $ $$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $$
is there more efficient and elegant way? Other than the solution below.
a solution:
Let $ I = \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $,
$$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = \int \frac{\sin^{8}(x) + \cos^{8}(x) - \cos^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = x - \int \frac{\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$
so we have $2I$ is:
$$ 2I = x + \int \frac{\sin^{8}(x)-\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$
Now,
$$ \sin^{8}(x) - \cos^{8}(x) = \left[ \sin^{4}(x) - \cos^{4}(x) \right] \left[ \sin^{4}(x) + \cos^{4}(x) \right] $$
$$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right]\left[ \sin^{2}(x) + \cos^{2}(x) \right] \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 \sin^{2}(x)\cos^{2}(x) \right] $$
$$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] $$
For the denominator:
$$ \sin^{8}(x) + \cos^{8}(x) = \left[ \sin^{4}(x) + \cos^{4}(x) \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$
$$ = \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$
$$ = \left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4} $$
Now
$$ 2I =x+ \int \frac{ \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] }{\left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4}} dx $$
Let $U = \sin(x) \cos(x)$, then $U'(x) = \cos^{2}(x) - \sin^{2}(x)$, so
$$ 2I = x- \int \frac{\left[1 - 2 U^{2} \right] }{\left[1 - 2 U^{2} \right]^{2} - 2 U^{4}} dU $$
$$ = x- \int \frac{\left[ 2 U^{2} - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU $$
And from here I can continue using Partial Fraction:
$$- \int \frac{\left[ 2 U^{2} - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU = -\frac{1}{2}\int \frac{1}{ (1 - (2 + \sqrt{2})U^{2})} + \frac{1}{(1 - (2 - \sqrt{2})U^{2})} dU $$
and then for each of the 2 terms we use Partial Fraction again. First, let $\alpha = \sqrt{2+\sqrt{2}}U$, $\beta = \sqrt{2-\sqrt{2}}U$, so we have
$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}}\int \frac{1}{ 1 - \alpha^{2}} d \alpha - \frac{1}{2\sqrt{2-\sqrt{2}}} \int \frac{1}{(1 - \beta^{2})} d \beta $$
then we have
$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}} \left( \int \frac{1/2}{1-\alpha} d \alpha + \int \frac{1/2}{1+\alpha} d\alpha \right) - \frac{1}{2\sqrt{2-\sqrt{2}}} \left( \int \frac{1/2}{1-\beta } d \beta + \int \frac{1/2}{1 + \beta } d \beta \right) $$
$$ = -\frac{1}{4\sqrt{2+\sqrt{2}}} \left( -\ln(1-\alpha) + \ln(1+\alpha) \right) - \frac{1}{4\sqrt{2-\sqrt{2}}} \left( -\ln(1-\beta) + \ln(1+\beta) \right) $$
| Instead, we are going to evaluate
$$
\begin{aligned}
A=\int \frac{\cos ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x=\int \frac{1}{\tan ^{8} x+1} d x\stackrel{t=\tan x}{=} \int \frac{d t}{\left(1+t^{2}\right)\left(1+t^{8}\right)}
\end{aligned}
$$
Noting that
$$
\frac{1}{(1+z )(1+z^{4})}
= \frac{1}{2} \cdot \frac{1}{z+1}+\frac{1}{4} \cdot \frac{1-z}{z^{2}+\sqrt{2} z+1}+\frac{1}{4} \cdot\frac{1-z}{z^{2}-\sqrt{2} z+1}
$$
Replacing $z$ by $t^2$ gives
$$
\int\frac{d t}{\left(1+t^{2}\right)\left(1+t^{8}\right)}=\frac{1}{2} \underbrace{\int \frac{d t}{t^{2}+1}}_{x+C_1}+\frac{1}{4} \underbrace{\int \frac{1-t^{2}}{t^{4}+\sqrt{2} t^{2}+1}}_{J} d t+\frac{1}{4} \underbrace{\int \frac{1-t^{2}}{t^{4}-\sqrt{2} t^{2}+1}}_{k} d t
$$
$$
\begin{aligned}
J&=-\int \frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+\sqrt{2}} d t\\&=-\int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-(2-\sqrt{2})} \\&=-\frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{t+\frac{1}{t}-\sqrt{2-\sqrt{2}}}{t+\frac{1}{t}+\sqrt{2-\sqrt{2}}}\right|+C_{2}
\end{aligned}
$$
Similarly,
$$
K=-\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{t+\frac{1}{t}-\sqrt{2+\sqrt{2}}}{t+\frac{1}{t}+\sqrt{2+\sqrt{2}}}\right|+C_{3}
$$
$$
\begin{aligned}
A=& \frac{1}{2} x-\frac{1}{8 \sqrt{2-\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2-\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2-\sqrt{2}} \tan x+1}\right| \\&-\frac{1}{8 \sqrt{2+\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2+\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2+\sqrt{2}} \tan x+1}\right|+C_4
\end{aligned}
$$
$$\begin{aligned}\int \frac{\sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} dx=&x-A\\=& \frac{1}{2} x+\frac{1}{8 \sqrt{2-\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2-\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2-\sqrt{2}} \tan x+1}\right|\\& +\frac{1}{8 \sqrt{2+\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2+\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2+\sqrt{2}} \tan x+1}\right|+C \end{aligned}$$
| {
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"url": "https://math.stackexchange.com/questions/4487862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$a+b+c=1$, prove $(1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $, without AM-GM $a,b,c \in \mathbb{R}^{+}$, if $a+b+c=1$, prove that
$$ (1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $$
but without AM-GM as the only tool.
Collecting data:
Since $a+b+c=1$, we cannot have one of the variable to be greater than or equal to 1. So
$$ \boxed{0 < a,b,c < 1} $$
Also by AM-GM: $1=a+b+c \ge 3(abc)^{1/3}$, or
$$ \boxed{abc \le \frac{1}{27}} $$
Next,
$$ (1-a)(1-b)(1-c) = [1 - (a+b) + ab](1-c) = \boxed{ 1 - (a+b+c) + (ab + ac + bc) - abc }$$
$$ =\boxed{ (ab+ac+bc)-abc} $$
Next,
$$ (1+a)(1+b)(1+c) = [1 + (a+b) + ab](1+c) = 1 + (a+b+c) + (ac+ab+bc) + abc$$
$$ = \boxed{2 + (ac+bc+bc) + abc} $$
so the inequality is equivalent with:
$$ 2 + ac+bc+ab + abc \ge 8(ab+ac+bc) - 8abc $$
or
$$ \boxed{ 2 + 9abc \ge 7 (ab+ac+bc) } \:\: \leftarrow \:\: \text{to be proven} $$
Next, $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc) = 1$,
$$ \boxed{ab+ac+bc = \frac{1- (a^{2} + b^{2}+ c^{2})}{2}} $$
which means
$$ 7(ab+ac+bc) = \frac{7}{2} -\frac{7}{2}(a^{2}+b^{2}+c^{2}) = 2 + \frac{3 - 7(a^{2}+b^{2}+c^{2})}{2}$$
and it is left to prove
$$ \frac{3-7(a^{2}+b^{2}+c^{2})}{2} \le 9abc $$
$$ \boxed{3 \le 18 abc + 7(a^{2}+b^{2}+c^{2})} \:\: \leftarrow \:\: \text{to be proven}$$
| Proceeding along the OP:
It suffices to prove that
$$2 + 9abc \ge 7(ab + bc + ca).$$
Degree three Schur inequality yields
$$(a + b + c)^3 - 4(a + b + c)(ab + bc + ca) + 9abc \ge 0$$
that is
$$1 - 4 (ab + bc + ca) + 9abc \ge 0.$$
It suffices to prove that
$$2 + 4(ab + bc + ca) - 1 \ge 7(ab + bc + ca)$$
or
$$1 \ge 3(ab + bc + ca)$$
which is true since $(a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca$ $= \frac12[(a-b)^2+(b-c)^2 + (c-a)^2]\ge 0$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Given graph of the function $f(x)=x^3+ax^2+bx+c$ What is the $x$ value at the local minimum point? This is a problem from a timed exam, so I prefer approaches that lead to answer quickly.
Graph of the function $f(x)=x^3+ax^2+bx+c$ is as follow. What is the
$x$ value at the local minimum point?
$1)\frac12\qquad\qquad2)2\qquad\qquad3)\frac32\qquad\qquad4)3$
Here is my approach,
Suppose we have minima at $x_0$. We have,$$f(0)=4\Rightarrow c=4$$
$$f'(x)=3x^2+2ax+b\qquad\text{Since $f'(0)=0\rightarrow b=0$}$$Hence $f'(x)=x(3x+2a)$, So $x_0=-\frac{2a}3$. We have $f(x)=x^3+ax^2+4$. $$f(-\frac{2a}3)=0\Rightarrow -\frac{8a^3}{27}+\frac{4a^3}9+4=0\Rightarrow a=-3$$
Finally $x_0=\frac{-2a}3=2$.
Although one can get the answer with this approach in one or two minutes, I'm looking for quicker ways to solve the problem. Actually, first time I tried to solve the problem, I noticed that we can write $f(x)=(x+m)(x-n)^2$ where $m,n\in \mathbb{R}^+$. According to graph of function and noting that $f(0)=mn^2=4$, I can guess it the function is $f(x)=(x+1)(x-2)^2$. But assuming I'm in the exam, at this point should I quickly mark the answer and go to the next question or this approach is unreliable and I got the correct answer with a bit of luck?!
| Not a great deal faster, but you could also use the fact that the local extrema in the cubic function are symmetric about its inflection point, given by $ \ f''(x) \ = \ 6x + 2a \ = \ 0 $ $ \Rightarrow \ x_{infl} \ = \ -\frac{a}{3} \ \ . $ Since the local maximum is on the $ \ y-$axis (if the diagram is to be at all trusted), then the local minimum is at $ \ x_{min} \ = \ -\frac{2a}{3} \ \ . $ The function value at the inflection point is $ \ f\left(-\frac{a}{3} \right) \ = \ -\frac{a^3}{27} + \frac{a^3}{9} - \frac{ab}{3} + 4 \ = \ 2 \ \ . $ You will want to use your observation that $ \ f'(0) \ = \ 0 \ \Rightarrow \ b \ = \ 0 \ \ , $ so we have $ \ \frac{2·a^3}{27} \ = \ -2 \ \Rightarrow \ a^3 \ = \ -27 $ $ \Rightarrow \ x_{min} \ = \ -\frac{2·[-3]}{3} \ = \ +2 \ \ . $
Your educated guess that $ \ x^3 - 3x^2 + 4 \ = \ (x + 1)·(x - 2)^2 \ $ is correct; acting on that during the exam depends on how much of a gambler you are: Viete tells you that $ \ c \ = \ 4 \ = \ -r·s^2 \ \ , $ so $ \ r \ = \ -1 \ \ , \ \ s \ = \ +2 \ $ would be worth the risk if you're desperate for time...
ADDENDUM: It might also be noted that once you see that $ \ c \ = \ 4 \ \ , $ the Rational Zeroes Theorem lets you "reject" choices (3) and (4) (now... does $ \ \frac{1}{2} \ $ seems plausible...?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Ten natural numbers such that the sums of each choice of nine of them are $82, 83, 84, 85, 87, 89, 90, 91, 92$ In a Whats App group the following puzzle was asked
Ten (not necessarily different) natural number are such that if all but one of them is added, the possible results (depending on which one is omitted) are 82, 83, 84, 85, 87, 89, 90, 91, 92.
Non-uniqueness of the answer was also mentioned there. Naively, I set up 9 linear equations on 10 variables. By fixing, one of them I could get
the set of 10 numbers as 1,2,3,4,6,8,9,10,11,39. If we leave out first 9 numbers one by one to get the sum of remaining 9 as required. But leaving out 39 does not do so.
What could be the proper approach and the answer?
| According to the statement we may assume that the 10 numbers are $$x-5,x-4,x-3,x-2,x,x+2,x+3,x+4,x+5$$
plus a repeated one, say $x+k$, where $k\in\{0,\pm 2,\pm 3,\pm 4,\pm 5\}$. Let $S$ be the sum of those $10$ numbers, then
$$S=(x-5)+(x-4)+\dots+(x+4)+(x+5)+(x+k)=10x+k.$$
Summing all the nine relations together, we obtain
$$S-(x-5)+S-(x-4)+\dots +S-(x+4)+S-(x+5)\\=82+83+84+ 85+87+ 89+90+91+92$$
that is $S-x=(82+83+84+85+87+89+90+91+92)/9=87$, or $10x+k-x=87$, $9x=87-k$ which implies that $87-k$ should be divisible by $9$. Hence $k=-3$ and therefore $x=10$. It follows that a solution to the puzzle is
$$5,6,7,7,8,10,12,13,14,15.$$
P.S. In Arthur's answer, if we keep subtracting $8$ from $39$ and adding $1$ to the other numbers we get:
$$1,2,3,4,6,8,9,10,11,39\\
2,3,4,5,7,9,10,11,12,31\\
3,4,5,6,9,10,11,12,13,23\\
4,5,6,7,10,11,12,13,14,15\\
5,6,7,8,10,12,13,14,7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational. Note that my attempted method below is distinct from the solutions in this question.
I also know this is generally true for $\sqrt p + \sqrt q + \sqrt r$ where $p,q,r$ are prime, but I am asking if my particular method works.
Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.
Assume that $\sqrt 2 + \sqrt 3 + \sqrt 5 = \frac pq$ for some $p,q \in \mathbb Z$, in lowest terms, with $q \neq 0$. Then at most one of $p$ and $q$ can be even. Then
\begin{align*}
\sqrt 2 + \sqrt 3 + \sqrt 5 &= \frac pq\\
\\
(\sqrt 2 + \sqrt 3 + \sqrt 5)^2 &= \left(\frac pq \right)^2\\
\\
2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=\frac{p^2}{q^2}\\
\\
2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=p^2
\end{align*}
which shows that $p^2$ is even, so $p$ is even and let $p=2k$ for some $k\in \mathbb Z$. Substituting this we get
\begin{align*}
2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=(2k)^2\\
\\
q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=2k^2\\
\\
q^2 &= 2 \left(\frac{k^2}{5+\sqrt 6 + \sqrt{10} + \sqrt{15}}\right),
\end{align*}
which shows as well that $q$ is even, a contradiction. Hence $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.
| There are infinite primes of the form $p=120k+61$ by Dirichlet's theorem. For any of them we have that $3$ and $5$ are quadratic residues, since $p\equiv 1\pmod{15}$, while $2$ is not a quadratic residue since $p\equiv 5\pmod{8}$. Assume that $\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{a}{b}$ and take a prime $p$ of the previous form with $p> \max(5,b)$. Since $\left(\frac{2}{p}\right)=-1$ while $\left(\frac{3}{p}\right)=\left(\frac{5}{p}\right)=1$, the minimal polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ over $\mathbb{F}_p$ has degree $2$. In particular $\sqrt{2}+\sqrt{3}+\sqrt{5}$ cannot be an algebraic number with degree $1$ over $\mathbb{Q}$, i.e. a rational number.
Alternative approach. Always assuming $\sqrt{2}+\sqrt{3}+\sqrt{5}=q\in\mathbb{Q}$ we have
$$ 8+2\sqrt{15}=(\sqrt{3}+\sqrt{5})^2 = (q-\sqrt{2})^2 = (q^2+2)-2q\sqrt{2},$$
$$ 2\sqrt{15}+2q\sqrt{2} = (q^2-6),$$
$$ (60+8q^2) + 8q \sqrt{30} = (q^2-6)^2, $$
$$ \sqrt{30} = \frac{(q^2-6)^2-(60+8q^2)}{8q} $$
but the last line implies $\sqrt{30}\in\mathbb{Q}$ which we know not to be the case.
Both approaches can be easily extended for proving that $\sum_{k=1}^{n}\sqrt{p_k}\not\in\mathbb{Q}$ for any collection of primes $p_k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ Value of p such that $\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$ is some finite | non-zero number.
My approach is as follow
$\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {{x^{\frac{{3p}}{3}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$
$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p}}}}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right) \Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt[3]{{{x^{3p + 1}} + {x^{3p}}}} + \sqrt[3]{{{x^{3p + 1}} - {x^{3p}}}} - 2\sqrt[3]{{{x^{3p + 1}}}}} \right)$
How do we proceed
| $$L=\mathop {\lim }\limits_{x \to \infty } \left( {{x^p}\left( {\sqrt[3]{{x + 1}} + \sqrt[3]{{x - 1}} - 2\sqrt[3]{x}} \right)} \right)$$
$$L=\mathop {\lim }\limits_{x \to \infty } \left( {{x^{p+1/3}}\left( {\sqrt[3]{{1 + 1/x}} + \sqrt[3]{{1 - 1/x}} - 2} \right)} \right)$$
Use $(1+z)^k=1+kz+\frac{k(k-1)}{2}z^2+O(z^3)$ when $z$ is very small, then let $z=1/x$
$$L=\lim_{z\rightarrow 0}~ z^{-p-1/3}\left(1+\frac{z}{3}-\frac{z^2}{9}+O(z^3)+1-\frac{z}{3}-\frac{z^2}{9}+O(z^3)-2\right)$$
$$L=z^{-p-1/3}\left(-\frac{2z^2}{9}+O(z^3)\right)$$
Let $-p-1/3+2=0$, then
$$L=-\frac{2}{9}$$
Hence $p=\frac{5}{3}$ makes the limit finite equal to $-\frac{2}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Binomial related problem $\left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{50}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{40}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{30}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right) = $
Where ${}^n{C_r} = \left( {\begin{array}{*{20}{c}}
n\\
r
\end{array}} \right)$
My approach is as follow $\left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{50}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{40}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{30}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right) = \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.{}^5{C_r}.{}^{50 - 10r}{C_5}} $
Using the technique of some previous problem I tried to solve by absorption
$\sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.\frac{{5!}}{{r!\left( {5 - r} \right)!}}.\frac{{\left( {50 - 10r} \right)!}}{{5!\left( {45 - 10r} \right)!}}} = \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.\frac{{\left( {45 - 9r} \right)!}}{{r!\left( {45 - 10r} \right)!}}.\frac{{\left( {50 - 10r} \right)!}}{{\left( {5 - r} \right)!\left( {45 - 9r} \right)!}}} \Rightarrow \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.{}^{45 - 9r}{C_r}.{}^{50 - 10r}{C_{5 - r}}} $
Not able to proceed further
| We can generalize the problem in this form:
\begin{align*}
a_{m, n} = \sum_{k \geq 0} (-1)^{n + k} \binom{n}{k}\binom{mk}{n} =\ ?
\end{align*}
For $n = 5$ and $m = 10$, the LHS is the same as the given question. Let
\begin{align*}
f(x) = \sum_{n \geq 0}a_{m, n}x^n
\end{align*}
be the generating function of the sequence $\{a_{m, n}\}_{n \geq 0}$ for a fixed $m$. Then we must find the coefficient of $x^n$ in $f(x)$:
\begin{align*}
f(x) = \sum_{n \geq 0}a_{m, n}x^n &= \sum_{n \geq 0}\left(\sum_{k \geq 0} (-1)^{n+k} \binom{n}{k}\binom{mk}{n}\right)x^n\\
&= \sum_{n \geq 0}\sum_{k \geq 0} (-1)^{n+k} \binom{n}{k}\binom{mk}{n}x^n\\
&= \sum_{k \geq 0}\sum_{n \geq 0} (-1)^{n+k} \binom{n}{k}\binom{mk}{n}x^n\\
&= \sum_{k \geq 0}(-1)^k\binom{n}{k}\sum_{n \geq 0}\binom{mk}{n}(-x)^n\\
&= \sum_{k \geq 0}(-1)^k\binom{n}{k}(1 - x)^{mk}\\
&= \sum_{k \geq 0}\binom{n}{k}\big(-(1 - x)^{m}\big)^k\\
&= \big(1 - (1 - x)^m\big)^n\\
&= \big(1 - (1-x)\big)^n\big(1 + (1 - x) + \cdots + (1-x)^{m-1}\big)^n\\
&= x^n\big(1 + (1 - x) + \cdots + (1-x)^{m-1}\big)^n
\end{align*}
Therefore, for finding the coefficient of $x^n$ in $f(x)$, we must find the constant coefficient in $\big(1 + (1 - x) + \cdots + (1-x)^{m-1}\big)^n$. But the constant coefficient achieves when $x = 0$. So it equals $\big(1 + (1 - 0) + \cdots + (1-0)^{m-1}\big)^n = m^n$. Thus we have
\begin{align*}
\bbox[5px, border: 2px solid blue]{a_{m, n} = \sum_{k \geq 0} (-1)^{n + k} \binom{n}{k}\binom{mk}{n} = m^n}
\end{align*}
So for $n = 5$ and $m = 10$, the result is $10^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Decomposition of a $2 \times 2$ matrix into the difference of 2 positive semi-definite matrices. For the matrix $$A = \begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix}$$ with norm $||A|| = \sum_{i,j = 1,2} |a_{ij}|$
Show any decomposition of $A = C - B$ with $B, C$ being positive semi-definite matrices, satisfies
$||B|| + ||C|| \geq 4$.
I've started with setting the conditions $c_{11} = b_{11}, c_{22} = b_{22}, c_{ij} = 1 + b_{ij} $ off the diagonal.
I then worked through $x^T A x = x^T(C-B)x$, with each term $\geq 0$, and worked through that while applying the above conditions, however I just arrive at a trivial answer $2 x_1 x_2 = 2x_1 x_2$, so I'm unsure how to proceed with this problem.
| We are given A=C-B with :
$$ A = \begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix} $$
Let us take B & evaluate C :
$$ B = \begin{bmatrix}
a & b \\
b & c \\
\end{bmatrix} $$
$$ C = A+B = \begin{bmatrix}
a & b+1 \\
b+1 & c \\
\end{bmatrix} $$
With $ B $ & $ C $ Positive Semi-Definite, we get (XXX) :
$ ac-b^2 \ge 0 $
$ ac-(b+1)^2 \ge 0 $
$ ac \ge b^2 $
$ ac \ge (b+1)^2 $
Addition gives $ 2ac \ge b^2 + (b+1)^2 $ &
$ 2ac \ge b^2 + (b+1)^2 $
Minimum value of RHS is at $ b=-1/2 $ , where $ b^2 + (b+1)^2 = 1/4 + 1/4 = 1/2 $ , which gives :
$ 2ac \ge 1/2 $
$ |2a| |2c| \ge 1 $
Standard InEquality gives :
$ |2a| + |2c| \ge $
$ |a| + |c| \ge 1 $ (XXX)
It is easy to see that (YYY) :
$ ||B|| = |a|+|c|+|b|+|b| $ &
$ ||C|| = |a|+|c|+|b+1|+|b+1| $
$ ||B||+||C|| = ( |a|+|c|+|b|+|b| ) + ( |a|+|c|+|b+1|+|b+1| ) $
$ ||B||+||C|| = 2( |a|+|c| ) + 2( |b|+|b+1| ) $ (YYY)
Use (XXX) in (YYY) to get the minimum value 4.
$ ||B||+||C|| \ge 2( 1 ) + 2( 1 ) $
$ ||B||+||C|| \ge 4 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\lim\limits_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$? As we know, the result of some expressions and series is equal to $\frac{\pi^2}{6} $ that the most important of them is $\zeta(2)$
Now, I have founded an equation whose limit at point $x=\infty$ is equal to $\frac{\pi^2}{6} : $
$$\lim_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$$
I want to know how can we prove it?
| We have that
$$\sin \left(\frac{\pi}{\sqrt{x}}\right)=\frac{\pi}{\sqrt{x}}-\frac16\frac{\pi^3}{x\sqrt{x}}+O\left(\frac{1}{x^2\sqrt{x}} \right)$$
and then
$$\frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi\sqrt{x}}{\frac{\pi}{\sqrt{x}}}\left(1+\frac16\frac{\pi^2}{x}+O\left(\frac{1}{x^2} \right)\right)-x=\frac{\pi^2}6+O\left(\frac{1}{x} \right)\to \frac{\pi^2}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Showing that $\sum_{k=0}^{n}{m+k\choose m}^{-1}=\frac{m}{m-1}\left[1-{m+n\choose m-1}^{-1}\right],m>1.$ Let $$S_{m,n}=\sum_{k=0}^{n}{m+k\choose m}^{-1}$$
Then $S_{0,n}=n+1, S_{1,n}=H_{n+1}.$
For $m>1$, let us consider the integral $$I_{m,k}=\int_{0}^{1} (1-x^{1/m})^k dx$$
which by using $x=\sin^{2m} t$ and Beta function, can be expressed as
$$I_{m,k}=2m\int_{0}^{\frac{\pi}{2}}\sin^{2m-1} t ~ \cos^{2k+1} t~ dt= \frac{\Gamma(m+1) \Gamma(k+1)}{\Gamma(m+k+1)}={m+k\choose m}^{-1}.$$
Next, we can write $$S_{m,n}=\sum_{k=0}^{n} I_{m,k}=\int_{0}^{1}dx \sum_{k=0}^n (1-x^{1/m})^k=\int_{0}^{1} x^{-1/m}[1-(1-x^{1/m})^{n+1}]~ dx.$$
$$\implies S_{m,n}=\frac{m}{m-1}-\int_0^1 x^{-1/m}(1-x^{1/m})^{n+1} dx$$
Again by using $x=\sin^{2m} t$ and Beta-integral, we get
$$S_{m.n}=\frac{m}{m-1}-2m\int_{0}^{\frac{\pi}{2}} \sin^{2m-3}~\cos^{2n+3} ~dt=\frac{m}{m-1}-\frac{\Gamma(m)\Gamma(n+2)}{\Gamma(m+m+1)}. $$
Upon simplification we have $$S_{m,n}=\frac{m}{m-1}\left[1-{m+n\choose m-1}^{-1}\right],m>1.$$
The question us how else this result can be obtained?
| We seek to show that for $m\gt 1$
$$S_{m,n} = \sum_{k=0}^n {m+k\choose m}^{-1}
= \frac{m}{m-1}
\left[1-{m+n\choose m-1}^{-1}\right].$$
We have for the LHS using an Iverson bracket:
$$[w^n] \frac{1}{1-w}
\sum_{k\ge 0} {m+k\choose m}^{-1} w^k.$$
Recall from MSE
4316307
that with $1\le k\le n$
$${n\choose k}^{-1}
= k [z^n] \log\frac{1}{1-z}
(z-1)^{n-k}.$$
We get with $m\ge 1$ as per requirement on $k$
$$m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m+1}} \log\frac{1}{1-z}
[w^n] \frac{1}{1-w}
\sum_{k\ge 0} w^k z^{-k} (z-1)^k
\\ = m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m+1}} \log\frac{1}{1-z}
[w^n] \frac{1}{1-w} \frac{1}{1-w(z-1)/z}
\\ = m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m}} \log\frac{1}{1-z}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{z-w(z-1)}.$$
Now residues sum to zero and the residue at infinity in $w$ is zero by
inspection, so we may evaluate by taking minus the residue at $w=1$ and
minus the residue at $w=z/(z-1).$ For $w=1$ start by writing
$$- m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m}} \log\frac{1}{1-z}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} \frac{1}{w-1}
\frac{1}{z-w(z-1)}.$$
The residue then leaves
$$- m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m}} \log\frac{1}{1-z}
= -m \frac{1}{m-1}.$$
On flipping the sign we get $m/(m-1)$ which is the first term so we are
on the right track. Note that when $m=1$ this term will produce zero. For the residue at $w=z/(1-z)$ we write
$$- m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m}} \frac{1}{z-1} \log\frac{1}{1-z}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{w-z/(z-1)}.$$
Doing the evaluation of the residue yields
$$- m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m}} \frac{1}{z-1} \log\frac{1}{1-z}
\frac{(z-1)^{n+1}}{z^{n+1}} \frac{1}{1-z/(z-1)}
\\ = m \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{m+n+1}} \log\frac{1}{1-z}
(z-1)^{n+1}
\\ = m [z^{m+n}] \log\frac{1}{1-z}
(z-1)^{n+1}.$$
Using the cited formula a second time we put $n := m+n$ and $k := m-1$
to get
$$m \frac{1}{m-1} {m+n\choose m-1}^{-1}.$$
On flipping the sign we get the second term as required and we have the
claim.
Remark. In the above we have $m\gt 1.$ We get for $m=1$
$$[z^{n+1}] \log\frac{1}{1-z}
(z-1)^{n+1}
= \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n+2}} \log\frac{1}{1-z}
(z-1)^{n+1}.$$
Now we put $z/(z-1) = v$ so that $z = v/(v-1)$ and $dz = -1/(v-1)^2 \;
dv$ to get
$$- \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+2}} \log\frac{1}{1-v/(v-1)}
(v-1) \frac{1}{(1-v)^2}
\\ = \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+2}} \frac{1}{1-v} \log(1-v).$$
On flipping the sign we obtain
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+2}} \frac{1}{1-v} \log\frac{1}{1-v} = H_{n+1},$$
again as claimed. This particular value follows by inspection, of
course.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
how to find maximum value without differentiating let x be positive real number, find max possible value of the expression
$$y = \frac{x^2 + 2 - \sqrt{x^4 + 4}}{x}$$
it can be found by differentiating, but is there no other way of finding it, like using AM $\geq$ GM.
or any other method.
i tried
$$y = x + \frac{2}{x} - \sqrt{x^2 + \frac{4}{x^2}}$$
but it gives nothing
| $$y(x) = x + \frac{2}{x} - \sqrt{x^2 + \frac{4}{x^2}} = x + \frac{2}{x} -\sqrt{(x + \frac{2}{x})^2-4}$$
$$t=x+\frac{2}{x}$$
$$y(t)=t-\sqrt{t^2-4} =\frac{4}{t+\sqrt{t^2-4}}$$
which is clearly monotonicly decreasing.
so we need to find the minimal value of $x+\frac{2}{x}$
by AM-GM:
$$\frac{x+\frac{2}{x}}{2}\geq \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inscribing an ellipse in an irregular convex pentagon Using the methods of projective geometry, identify the unique ellipse that is inscribed in a given convex pentagon.
Suppose the vertices of the pentagon are: $(1, 0), (4, 2), (3, 6), (-1, 5), (-1, 1)$. Find the equation of the unique inscribed ellipse that is tangent to all five sides of this convex pentagon.
| It's easier to find the dual conic first.
Rewrite every tangents in the form:
$$X_i x+Y_i y+1=0$$
Then $(X_i,Y_i)$ are the five points defining the dual conic:
$$
\begin{align}
0 &= \det
\begin{pmatrix}
X^2 & XY & Y^2 & X & Y & 1 \\
X_1^2 & X_1 Y_1 & Y_1^2 & X_1 & Y_1 & 1 \\
\vdots & & & & & \vdots \\
X_5^2 & X_5 Y_5 & Y_5^2 & X_5 & Y_5 & 1 \\
\end{pmatrix} \\ \\
&= AX^2+2HXY+BY^2+2GX+2FY+C \\
\end{align}$$
The required ellipse is
$$
\begin{align}
0 &= -\det
\begin{pmatrix}
0 & x & y & 1 \\
x & A & H & G \\
y & H & B & F \\
1 & G & F & C \\
\end{pmatrix} \\ \\
&= ax^2+2hxy+by^2+2gx+2fy+c
\end{align}$$
Note that $a$ is the co-factor of entry $A$ of the $3\times 3$ block matrix, etc.
For the vertical tangent, try to let $Y_i=N$ and only the terms with highest order in $N$ survive.
See also the case of quadrilateral here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that the function $f(x)=x+\sqrt{x}$ is one-to-one Show that the function $f(x)=x+\sqrt{x}$ is one-to-one.
I know that for showing that a function is one-to-one I have to prove that if $f(a)=f(b)$ then $a=b$.
Then I'm trying that in here but I get stuck.
$$f(a)=f(b)$$
$$a+\sqrt{a}=b+\sqrt{b}$$
$$a-b=\sqrt{b}-\sqrt{a}$$
How to do I show from here that $a=b$?
I've tried square both sides, completing the square and haven't worked. :(
I will appreciate a detail to understand, thanks in advance.
| Picking up from your last step, assuming $ a, b \neq 0$, we have
\begin{align*}
a - b = \sqrt{b} - \sqrt{a} & \iff (a - b)(\sqrt{b} + \sqrt{a}) = b - a \\
& \iff (a - b)(\sqrt{a} + \sqrt{b} + 1) = 0 \\
& \iff a - b = 0 \quad \text{or} \quad \sqrt{a} + \sqrt{b} + 1 = 0 \\
& \iff a = b \quad \text{(Since $ \sqrt{a} + \sqrt{b} + 1 \neq 0 $ for any $ a $ or $ b $)}
\end{align*}
So all in all we have $ f(a) = f(b) $ if and only if $a = b $ or $ a = b = 0$, or we can absorb the latter condition into the former one and says $ f(a) = f(b) $ if and only if $ a = b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Number of subgroups of order $4$ in the direct product $S_2\times S_4$. I have the following problem:
Find the number of subgroups of 4 elements in the direct product of
permutation groups $S_2 \times S_4$
I started with writing down all the elements in $S_4$ group. Here we have $4!=24$ elements:
$$S_4=\{ (1)(2)(3)(4),\\ (1)(2)(34),(1)(3)(24), (12)(34), (13)(24), (1)(4)(23), (14)(23), (12)(3)(4), (13)(2)(4), (14)(2)(3), \\ (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(132),\\
(1243), (1234), (1342), (1324), (1423), (1432) \}$$
The group $S_2$ is simple: $\{(1)(2), (12)\}$
In our subgroup we must have the neutral element $\{ (1)(2), (1)(2)(3)(4)\}$. But how can I find out how another $3$ elements can look like?
I understand, that we have to fulfill here 2 conditions:
*
*The product of any two elements must be in the subgroup
*The inverse of each element must be in the subgroup as well
| Within the $S_4$ factor of $G$ there are three $C_4$ subgroups and four $C_2\times C_2$ subgroups.
The remaining subgroups of order $4$ all intersect the $S_4$ factor in a group of order $2$.
There are just three $C_4$ groups outside of $S_4$, generated by $(1,2)(3,4,5,6)$, $(1,2)(3,4,6,5)$ and $(1,2)(3,5,4,6)$.
There are three $C_2 \times C_2$ subgroup that intersect $S_4$ in $\langle(3,4)(5,6)\rangle$, namely $\langle (1,2),(3,4)(5,6) \rangle$, $\langle (1,2)(3,4),(3,4)(5,6) \rangle$, and $\langle (1,2)(3,5)(4,6),(3,4)(5,6) \rangle$, and similarly for the other two intersections of that type.
Finally, there are two $C_2 \times C_2$ subgroup that intersect $S_4$ in $\langle(3,4)\rangle$,namely $\langle (1,2),(3,4) \rangle$ and $\langle (1,2)(5,6),(3,4) \rangle$, and simialrly for the other $5$ intersections of that type.
So the total number of subgroups of order $4$ is
$$3+4+3 + (3\times 3) + (2 \times 6) = 31.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $2/\sin(x)+3/\cos(x)$ using Cauchy-Schwarz inequality I want to find the minimum of $f(x)=\dfrac{2}{\sin(x)}+\dfrac{3}{\cos(x)}$ when $0<x<\pi/2$ using the C-S inequality.
By using derivatives it is easy to show that the minimum is
$$(2^{2/3}+3^{2/3})^{3/2}$$
but i have no idea how to prove this using C-S ineq,my attempts ends with an inequality of opposite direction.
| Theorem Let $a,\,b>0$. For $x\in(0,\,\pi/2)$, $(a^3\csc x+b^3\sec x)^2\ge(a^2+b^2)^3$, with equality iff $\tan x=a/b$.
Comment The present problem is $a=2^{1/3},\,b=3^{1/3}$.
Proof by Cauchy-Schwarz Use $u^2\ge\frac{(u\cdot v)^2}{v^2}$ with$$u=\left(\begin{array}{c}
a^{3/2}\csc^{1/2}x\\
b^{3/2}\sec^{1/2}x
\end{array}\right),\,v=\left(\begin{array}{c}
a^{1/2}\sin^{1/2}x\\
b^{1/2}\cos^{1/2}x
\end{array}\right),$$which are parallel iff $\tan x=a/b$. Then$$v^2=a\sin x+b\cos x=\sqrt{a^2+b^2}\cos(x-\arctan(a/b))\le\sqrt{a^2+b^2},$$which again is saturated at $\tan x=a/b$. I'll leave you to finish the calculation counting powers of $a^2+b^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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evaluate beta function I know:
$$\int_{0}^{\frac{\pi}{2}} \sin^{10}(x)\mathrm dx=\frac{1}{2}\cdot \mathrm B\left(11/2, 1/2\right)$$
and
$$\Gamma(1/2)=\sqrt{\pi}$$
But what is the following calculation based on?
$$\int_{0}^{\frac{\pi}{2}} \sin^{10}(x)\mathrm dx=\frac{1\cdot3\cdot5\cdot7\cdot9\cdot\pi}{2\cdot4\cdot6\cdot8\cdot10\cdot2}=\frac{63\pi}{512}$$
https://www.goseeko.com/blog/what-are-beta-and-gamma-functions/
| Using the power reduction formula for sine (7), we have
$$\int_0^{\frac\pi2} \sin^{10}(x) \, dx = \frac1{2^{10}}\binom{10}5 \int_0^{\frac\pi2}\,dx - \frac1{2^9} \sum_{k=0}^4 \binom{10}k \int_0^{\frac\pi2} \cos((10-2k)x) \, dx$$
Observe that for integer $k$,
$$\int_0^{\frac\pi2} \cos((10-2k)x) \, dx = \frac1{10-2k} \sin\left(\frac{(10-2k)\pi}2\right) = \frac1{10-2k} \sin((5-2k)\pi) = 0$$
So we have
$$\int_0^{\frac\pi2} \sin^{10}(x) \, dx = \frac\pi{2^{11}} \binom{10}5 = \frac{63\pi}{512}$$
Note that
$$\begin{align*}
\frac\pi{2^{11}} \binom{10}5 &= \frac{\pi \cdot 10!}{2^{11} \cdot 5! \cdot 5!} \\[1ex]
&= \Gamma\left(\frac12\right)^2 \cdot \frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{2^{11}\cdot5\cdot4\cdot3\cdot2\cdot1\cdot5\cdot4\cdot3\cdot2\cdot1} \\[1ex]
&= \Gamma\left(\frac12\right) \cdot \Gamma\left(\frac{11}2\right) \cdot \frac{10\cdot8\cdot6\cdot4\cdot2}{2^6\cdot5\cdot4\cdot3\cdot2\cdot5\cdot4\cdot3\cdot2} \\[1ex]
&= \Gamma\left(\frac12\right) \cdot \Gamma\left(\frac{11}2\right) \cdot \frac{1}{2\cdot5!} \\[1ex]
&= \frac{\Gamma\left(\frac12\right) \Gamma\left(\frac{11}2\right)}{2\Gamma(6)} \\[1ex]
&= \frac12 \operatorname{B}\left(\frac12,\frac{11}2\right)
\end{align*}$$
| {
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Why $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ results in pythagorean triples? As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2*0)^2+(0^2-1)^2=(0^2+1)^2$
$(2*1)^2+(1^2-1)^2=(1^2+1)^2$
$(2*2)^2+(2^2-1)^2=(2^2+1)^2$
$(2*0)^2+(0-1)^2=(0+1)^2$
$(2*1)^2+(1-1)^2=(1+1)^2$
$(2*2)^2+(4-1)^2=(4+1)^2$
$0^2+1^2=1^2$
$2^2+0^2=2^2$
$4^2+3^2=5^2$
$0+1=1$
$4+0=4$
$16+9=25$
$1=1$
$4=4$
$25=25$
| All Pythagorean triple have a "leg" that is a multiple of
$\space4.\space$ Your formula generates a subset of these.
We begin with Euclid's formula shown here as
$$A=m^2-k^2\quad B=2mk \quad C=m^2+k^2$$
If we let $\space k=1,\space$ we have
$\quad A=m^2-1\quad B=2m\quad C=m^2+1$
This formula also generates trivial triples where one term is zero. A variation of this replaces $\space m\space$ with $\space(2n-1+k)=2n\space$ and produces a formula that generates non-trivial triples for any natural number $n$.
$$\quad A=4n^2-1\quad B=4n\quad C=4n^2+1$$
By inspection, we can see that the middle term will always be even and that $\space C-A=2\space$ for all values of
$\space n.\space$ Algebraically below, we can also "see" that the formula is valid.
\begin{align*}A^2+B^2=&(4n^2-1)^2+(4n)^2\\
=&(16 n^4 - 8 n^2 + 1)+(16n^2)\\
=&(16 n^4 + 8 n^2 + 1)\\
=&(4n^2+1)^2=C^2\\
\end{align*}
| {
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Formation of two families of quadratic equations Question 1:
Find all values of $a,b$ such that roots of $x^2+ax+b=0$ be of the type $(p,p^2)$.
There can be infinitely many pairs of $(a,b)$ such as $a=-(p+p^2),b=p^3$. One may also eliminate $p$ from these two equations: $b^{2/3}+b^{1/3}=-a$ cubing both sides we have $b^2+b-3ab=-a^3$. Next we can relate $b$ to $a$ as $$b=\frac{3a-1\pm\sqrt{(1-3a)^2+4a^3}}{2}$$ to get infinitely many quadratic equations, ordinarily.
The standard and the fate of this equation changes if we ask:
Question 2:Find all pairs of $a$ and $b$ so that if $p$ is a root of $x^2+ax+b=0$ then $p^2$ is also its root.
We can assume that that $p,p^2$ are the roots. When $p^2$ is a root, so $p^4$ should also be a root. Since a quadratic cannot have three roots two of $(p,p^2,p^4)$ need to be equal to determine the allowed values of $p$, then the roots will be $(p,p^2)$.
Case 1: $p=p^2\implies p=0,1$ so roots are $(0,0),(1,1)$ and then $a=0,b=0$; $a=-2,b=1.$
Case 2: $p^2=p^4 \implies p=0, p=\pm 1$ gives new roots as $(-1,1)$. then $a=0, b=-1$.
Case 3: $p=p^4 \implies p=0,1, w, w^2$ give us new roots as $(w,w^2), (w^2,w^4=w)$, giving us $a=1,b=1$. Here $w$ is cube root of unity.
In all there exist only 4 such quadratic equations: $x^2=0, x^2-1=0, x^2-2x+1=0, x^2+x+1=0.$
What could be other rephrasing/solutions of the question 2, stated above?
| This question can be re-stated as:
Find all $(a,b)$ such that $x^2+ax+b=0$ is invariant if the roots are squared.
$y=x^2, x=\sqrt{y}$the the quadratic changes to $$y+a\pm \sqrt{y}+b=0 \implies (y+b)^=a^2y \implies y^2+(2b-a^2)+b^2=0$$
This quadratic should be identical to the given quadratic. Comparing the coeffcients we get $$ \frac{1}{1}=\frac{(2b-a)}{a}=\frac{b^2}{b}.$$
From (i) and (ii), we get $b^2-b\implies b=0,1$
From (i) and (ii) we get $a^2+a-2b=0$
When $b=0$, we get $a=0,-1$
When $b=1$, we get $a=-2,a=1$
Combining we recover four sets of $(a,b)=(0,0),(-1,0),(-2,1),(1,1).$
| {
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Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992)
(i) Show that the condition that the points $P$ $(a\cos A,b\sin A)$ and $Q$ $(a\cos B,b\sin B )$
should subtend a right angle at O is
$$a^2\cos A\cos B+b^2\sin A\sin B=0$$
(ii) Let S be a circle centre $O$ and radius $C$. Find the equation of the tangent to S at
the point $(C\cos t, C\sin t)$.
(iii) If $C = \dfrac{ab}{\sqrt{a^2+b^2}}$, show that the points where a tangent to S cuts the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
subtend a right angle at $O$.
Part (i) can be done by considering gradients of lines from $O$ to each of the points, multiplying them and setting equal to $-1$
Part (ii) gives the result $x\cos t+y\sin t=C$
With part (iii), I have attempted to simply rearrange for $y$ in the tangent equation, subbing into the ellipse equation and trying to solve for $x$ but this results in a huge amount of tedious algebra and I am unable to simplify it properly.
I then attempted to parameterize the ellipse in the form $x=a\cos T, y=b\sin T$ but I am unsure how to go from there to solve for the points. I did attempt to use harmonic addition and this does give some exact solutions in terms of $a,b,t$ but to simplify requires identities for $\sin(\arccos(x))$ and other similar identities like $\arctan\left(\frac{b}{a}\tan t\right)$.
Some useful information may be that we can let the points of intersections be $R(a \cos P,b\sin P)$ and $L(a\cos Q,b\sin Q)$ and substitute these points into the equation for the tangent. Now my problem is taking those equations and getting the equation we want from (i)
| (i) We have $P = (a \cos A, b \sin A)$ and $Q = (a \cos B, b \sin B) $
Since $OP \perp OQ$ , then by using the dot product we get
$ a^2 \cos A \cos B + b^2 \sin A \sin B = 0 $
(ii) A point on the circle with radius $C$ centered at the origin is parameterized as
$ P = C (\cos t, \sin t)$
The tangent vector at $P$ is along the vector $(-\sin t , \cos t ) $ (Perpendicular to the radius vector).
Hence, the parametric equation of the tangent is
$ \ell(t,s) = (C \cos t - s \sin t , C \sin t + s \cos t ) $
(iii) Now the above tangent line is to intersect the ellipse, and we expect two intersection points. Let $P = (a \cos A, b \sin A) $ be the intersection point, then $P$ lies on $\ell$.
Hence
$ a \cos A = C \cos t - s \sin t \hspace{30pt} b \sin A = C \sin t + s \cos t $
Solving for $s$ from the above two equations
$ s = \dfrac{ a \cos A - C \cos t }{-\sin t} = \dfrac{ b \sin A - C \sin t }{\cos t } $
Cross-multiplying yields
$ a \cos A \cos t - C \cos^2 t = - b \sin A \sin t + C \sin^2 t $
So that
$ a \cos A \cos t + b \sin A \sin t = C \hspace{30pt}(1) $
Similarly, if $Q= (a \cos B, b \sin B)$ is the other intersection point, then
$ a \cos B \cos t + b \sin B \sin t = C \hspace{30pt}(2)$
Equations $(1)$ and $(2)$ are linear in $\cos t $ and $\sin t $. Therefore, we can solve for them, and using Cramer's rule (for example), or any other method, we get
$ \cos t = \dfrac{ C b (\sin B - \sin A) }{ ab (\cos A \sin B - \cos B \sin A)} $
$ \sin t = \dfrac{ C a (\cos A - \cos B) }{ ab (\cos A \sin B - \cos B \sin A) } $
Now we can use the fact that $\cos^2 t + \sin^2 t = 1 $ to write
$ C^2 \bigg(b^2 (\sin B - \sin A)^2 + a^2 (\cos A - \cos B)^2 \bigg) = (ab)^2 \bigg( \cos A \sin B - \cos B \sin A \bigg)^2 \hspace{15pt}(3)$
Now, we're given that $ C = \dfrac{ab}{\sqrt{a^2 + b^2} } $
Substituting this into $(3)$,
$b^2 (\sin B - \sin A)^2 + a^2 (\cos A - \cos B)^2 = (a^2 + b^2) \bigg( \cos A \sin B - \cos B \sin A \bigg)^2 \hspace{20pt}(4)$
Expanding both the left and right sides of $(4)$,
$ b^2 (\sin^2 B + \sin^2 A - 2 \sin A \sin B) + a^2 (\cos^2 A + \cos^2 B - 2 \cos A \cos B) = (a^2 + b^2) (\cos^2 A \sin^2 B + \cos^2 B \sin^2 A - 2 \cos A \cos B \sin A \sin B ) $
$ b^2 \sin^2 B (1 - \cos^2 A) + b^2 \sin^2 A (1 - \cos^2 B) + a^2 \cos^2 A (1 - \sin^2 B) + a^2 \cos^2 B (1 - \sin^2 A) - 2 b^2 \sin A \sin B - 2 a^2 \cos A \cos B + 2 (a^2 + b^2) \cos A \cos B \sin A \sin B = 0 $
And this becomes, after using $ \cos^2 A + \sin^2 A = 1 $ and $ \cos^2 B + \sin^2 B = 1 $, and dividing through by $2$:
$ b^2 \sin^2 A \sin^2 B + a^2 \cos^2 A \cos^2 B - a^2 \cos A \cos B - b^2 \sin A \sin B + (a^2 + b^2) \cos A \cos B \sin A \sin B = 0 $
Collecting like terms,
$ (a^2 \cos A \cos B + b^2 \sin A \sin B) ( \sin A \sin B + \cos A \cos B - 1) = 0 $
But $\sin A \sin B + \cos A \cos B - 1 = \cos(A - B) - 1 \ne 0 $ (because $A \ne B$).
Therefore,
$ a^2 \cos A \cos B + b^2 \sin A \sin B = 0 $
And from part (i), this shows that the intersection points subtend a right angle at the origin.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Derivative of capital Pi product I wanted to find the derivative of this function at $x=6$
$$y= \prod_{i=1}^{10} (x-i) = (x-1)(x-2) \cdots (x-10) $$
without expanding all of the brackets, so I used the product rule to find a pattern. However, the resulting sum tells me that the derivative is zero at every whole number which is obviously not true. I've been over my solution and I can't see how I've gone wrong. Please could someone highlight where I went wrong? Thank you in advance.
\begin{align*}
\frac{\textit{d}y}{dx} &= (x-2)(x-3) \cdots (x-10) + (x-1) \frac{d}{dx} \biggl((x-2) \cdots (x-10) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1) \frac{d}{dx} \biggl(\prod_{i=2}^{10} (x-i) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\frac{d}{dx} \biggl(\prod_{i=3}^{10} (x-i) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\biggl(\prod_{i=4}^{10} (x-i) \biggr) + \cdots \\
&= \frac{y}{x-1} + \frac{y}{x-2} + \frac{y}{x-3}+\cdots + \frac{y}{x-10} \\
&= \sum_{i=1}^{10} \biggl(\frac{y}{x-i}\biggr)
\end{align*}
| A good strategy for derivatives of products is the logarithmic derivative:
If $f(x)=\prod f_k(x)$ then $\ln f(x)=\sum \ln f_k(x)$ and by taking derivatives on both sides
$$\frac {f’}f =\sum \frac {f_k’}{f_k}$$
Strictly speaking, we need all $f_k(x)>0$, but using the product rule, you can show that $f(x)\ne 0$ is sufficient.
In your example, this takes you very far.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ellipse bounded between two lines and a circle Given two circles with radii $\beta$ and $\beta^{-1}$, where $\beta\geq1$. Also, given two lines $y=x\tan\alpha$ and $y=-x\tan\alpha$, where $\pi/2>\alpha\geq0$. I am interested in all ellipses with center at $(k,0)$ that are bound between these two lines and the larger circle (LHS of Figure). I also assume that ellipses are tangent to points $(\beta^{-1}\cos\alpha, \beta^{-1}\sin\alpha)$ and $(\beta^{-1}\cos\alpha, -\beta^{-1}\sin\alpha)$, and has the following equation:
$$\frac{(x-k)^2}{a^2}+\frac{y^2}{b^2}=1.$$
I assume that $b\geq a$. I found that all such ellipses can be obtained by the following equations:
$$a=\sqrt{\frac{k(k\beta-\cos\alpha)}{\beta}}, \quad b=\sqrt{\frac{k\sin^2\alpha}{\beta\cos\alpha}}.$$
So, by changing $k$ in the range $[\beta^{-1}\cos\alpha, \beta^{-1}\frac{1}{\cos\alpha}]$ I can obtain all such ellipses: desmos.
However, for some combination of $\beta$ and $\alpha$, $k$ cannot reach $\beta^{-1}\frac{1}{\cos\alpha}$, since ellipse will touch larger circle before that (RHS of Figure). Given $\alpha,\beta$ pair, I want to find the largest $k$ allowed, before an ellipse touches the larger circle.
| Suppose the circle with radius $\beta$ and the ellipse touch at a point with co-ordinates $(x,y)$, where $y\neq 0$. Then the gradients of the two curves are equal at that point. For the circle, $x +ym = 0$ and for the ellipse, $b^2(x-k)+a^2ym=0$, where $m$ is the common gradient. This implies that $b^2(x-k)-a^2x=0$ or $(b^2-a^2)x=b^2k$. But also, this point must be on the intersection of the two curves: this gives the quadratic $$(b^2-a^2)x^2-2kb^2x+k^2b^2+a^2\beta^2 - a^2b^2$$ and, substituting for $(b^2-a^2)x$, we get $b^2kx=k^2b^2+a^2\beta^2 - a^2b^2$. Eliminating $x$ between the two expressions gives $$(\beta^2-b^2)(b^2-a^2)-k^2b^2=0.$$ Substituting into this for $a$ and $b$ in terms of $k$ gives $$(\beta^4 \cos^2 \alpha + \sin^2 \alpha)k = \beta^3 \cos \alpha.$$ This will be the maximum value of $k$, unless the $x$ value required is greater than $\beta$, in which case no point on the circle will correspond to it. Thus this condition applies when $(b^2-a^2)\beta \ge b^2k$ or $\beta^2 \cos \alpha \le 1$.
When this condition doesn't apply, there is the possibility that they touch at $(\beta, 0)$, which requires $(\beta-k)^2=a^2$; then using $a$ in terms of $k$ then gives $(2\beta^2 -\cos \alpha)k=\beta^3$ for the maximum $k$.
| {
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Sum of absolute values of coefficients of an equation? If $x =√2+√5$ is a root of $kx^4+mx^3+nx^2+l = 0$ where $k, m, n, l$ are integers
find the value of $|k| + |m| + |n| + |l|$.
Now as all coefficients are integers, so conjugate of x is also a root of the above.
So putting $x=√2+√5$ and $x=√2-√5$ in the above eq one by one and then subtracting the two equations we got.
We get $k.52.√10 + m.22.√5+ n.4.√10+l=0$
Now $m=0$ and $l=0$ for it to become integer.
and $k.52.√10+ n.4.√10=0$
$\implies n=-13k$
Now putting all values in the original equation
we get $kx^4-13kx^2=0 \implies x^2(kx^2-13k)=0 $ and for x not equal to 0
we get $kx^2-13k=0\implies k=0$ is the only integer solution
Hence $k=l=m=n=0$
hence final answer becomes $0$
Is this the correct way to deduce the problem?
One other way is below
$x =√2+√5 $
$⇒ x−√2 = √5 $
$⇒ (x−√2)^2 = (√5)^2 $
$⇒ x^2+2−2√2x =5 $
$⇒ (x^2 − 3) = 2√2x $
$⇒ (x^2 − 3)^2 = (2√2x)^2 $
$⇒ x^4 − 6x^2 + 9 = 8x^2 $
$⇒x^4 − 14x^2 + 9 = 0 $
$⇒ k = 1, m = 0, n = −14, l = 9 $
$⇒ |k| + |m| + |n| + |l| =1 + 0 + 14 + 9 = 24
$
| That doesn't sound right. Note that
$$x^2 = (\sqrt{2} + \sqrt{5})^2 = 7 + 2\sqrt{10}.$$
Thus
$$(x^2-7)^2 = 40.$$
It follows that $x$ is a root of the polynomial
$$p(x) := x^4 - 14x^2 + 9,$$
in which case you would have $|k|+|m|+|n|+|l| = 24$. That said, clearly every integer multiple of $p$ also works, so your question is not well-posed.
| {
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Prove $\frac{3n^5 + 20n^3 + 7}{2n^5 - 1}$ converges to $\frac{3}{2}$ by definition Let $a_{n+1} = \frac{3n^5 + 20n^3 + 7}{2n^5 - 1}$, $l = \frac{3}{2}$. Prove that $a_n$ converges to $l$.
I used the absolute value of $a_{n} - l$, to obtain $\frac{40n^3 + 17}{4n^5-2}$. I then obtained these inequalities:
$\frac{40n^3 + 17}{4n^5-2} < \frac{40n^3 + 17}{4n^5-2n^5} = \frac{40n^3 + 17}{2n^5} < \frac{40n^3 + 17n^3}{2n^5} = \frac{57n^3}{2n^5} < \frac{57}{2n^2} < \frac{57}{2n}$, which holds for all natural numbers $n\gt1$.
I then did the usual steps letting $N$ be any natural number greater than $\frac{57}{2\epsilon}$. Does this seem correct ?
Any help ???
| I tried a different way...
Let $\epsilon>0$ and consider the formal definition of limit for sequences. Then, we want to find an $N(\epsilon)$ such that for all $n>N(\epsilon)$,
$$\left|\frac{3n^5+20n^3+7}{2n^5-1}-\frac{3}{2}\right|<\epsilon\tag{*}$$
i.e.,
$$\left|\frac{40n^3+17}{2(2n^5-1)}\right|<\epsilon$$
is satisfied. Assuming, $N(\epsilon)\geq 1$, then $2n^5-1>0$ and then,
$$\frac{40n^3+17}{2(2n^5-1)}<\epsilon\implies \frac{10}{\epsilon n^2}+\frac{17}{4\epsilon n^5}+\frac{1}{2n^5}<1.$$
We may assume that $\frac{10}{\epsilon n^2}<\frac{1}{3}$, $\frac{17}{4\epsilon n^5}<\frac{1}{3}$, $\frac{1}{2n^5}<\frac{1}{3}$ or $n>\sqrt\frac{30}{\epsilon}$, $n>\sqrt[5]\frac{51}{4\epsilon}$, $n>\sqrt[5]\frac{3}{2}$.
Now, let $$N(\epsilon)=\max\left\{1, \sqrt\frac{30}{\epsilon}, \sqrt[5]\frac{51}{4\epsilon}, \sqrt[5]\frac{3}{2} \right\}.$$
Then, $(*)$ is satisfied for all $N>N(\epsilon)$.
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How to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without L'Hospital's Rule? I am asked to solve $\lim\limits_{x\to 1} \frac{1-\cos(\sin(x^3-1))}{x^3-1}$ without using L'Hospital's Rule.
I'm not sure how to go about it.
There is an indeterminate form $(\frac{0}{0})$ at $x=1$ and L'Hospital's Rule seems like the best course of action.
I tried to multiply by $\frac{1+\cos(\sin(x^3-1))}{1+\cos(\sin(x^3-1))}$ to get $\lim\limits_{x\to 1} \frac{1-\cos^2(\sin(x^3-1))}{x^3-1}\cdot\frac{1}{1+\cos(\sin(x^3-1))} = \lim\limits_{x\to 1} \frac{\sin(\sin(x^3-1))}{x^3-1}\cdot \sin(\sin(x^3-1))\cdot\frac{1}{1+\cos(\sin(x^3-1))}$
I'd try to get a limit of the form $\lim\limits_{u\to 0} \frac{\sin(u)}{u}$ because I know that that limit is $1$, but the problem is that sine is composed with itself.
Any help would be appreciated!
| By standard limits, since as $x \to 1 \iff x^3-1 \to 0$, we have that
$$\frac{1-\cos(\sin(x^3-1))}{x^3-1}=\frac{1-\cos(\sin(x^3-1))}{\sin^2(x^3-1)}\cdot \frac{\sin(x^3-1)}{x^3-1}\cdot \sin(x^3-1)\to \frac12 \cdot 1 \cdot 0 =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Volume of sum of a cube and a sphere Let $$C=\{(x, y, z)\in \mathbb{R}^3:|x|\leq1, |y|\leq 1, |z|\leq 1\}$$ be a cube and $$S=\{(x, y, z)\in \mathbb{R}^3:x^2+y^2+z^2\leq 1\}$$ be a sphere. Let us define $$C+S=\{c+s:c\in C, s\in S\}.$$Find the volume of $C+S$.
The main requirement is to determine the limits of $x, y$ and $z$ for the set $C+S$. After evaluating them, I get $$-2\leq x\leq 2$$ $$-1-\sqrt{1-x^2}\leq y\leq 1+\sqrt{1-x^2}$$ $$-1-\sqrt{1-x^2-y^2}\leq z\leq 1+\sqrt{1-x^2-y^2}.$$ This means the volume of $C+S$ is $$\int_{-2}^2\int_{-1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\int_{-1-\sqrt{1-x^2-y^2}}^{1+\sqrt{1-x^2-y^2}}dz\, dy\, dx.$$
Is this the correct approach? Please help. Also, if this is the correct approach, then how to evaluate the integral $$ 2\int_{-2}^2\int_{-1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}}\sqrt{1-x^2-y^2} dy\, dx.$$
| Intuitively $C+S$ consists of
*
*Eight eighth-spheres of radius $1$ (purple), with total volume $\frac43\pi$
*$12$ quarter-cylinders of height $2$ and radius $1$ (blue), with total volume $3\cdot2\pi=6\pi$
*Six $2×2×1$ cuboids (white) with total volume $24$
*a central cube (pink) of side $2$ which is $C$ and has volume $8$
Hence the total volume is $32+\frac{22}3\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
limits of 2 variables with trigonometric terms I'm trying to determine the following limit
$$ \lim_{(x,y) \to (0,0)} \frac{x^2-\sin(x^2y^2)+y^2}{x^2+\sin(x^2y^2)+y^2}$$
I tried to use polar coordinates and then Taylor and got
$$ \lim_{r \to 0} \frac{2-2 \cdot \cos^2(\theta) \cdot \sin^2(\theta) \cdot \cos(r^2\cos^2(\theta)\sin^2(\theta))}{2+2 \cdot \cos^2(\theta) \cdot \sin^2(\theta) \cdot \cos(r^2\cos^2(\theta)\sin^2(\theta))}$$
Which does not say anything I guess? How can I solve this type of questions?
| We have that for $xy=0$ the limit is equal to $1$, for $xy\neq 0$ we have
$$ \frac{x^2-\sin(x^2y^2)+y^2}{x^2+\sin(x^2y^2)+y^2}=1-2\frac{\sin(x^2y^2)}{x^2y^2}\frac{x^2y^2}{x^2+\sin(x^2y^2)+y^2}\to 1$$
indeed $\frac{\sin(x^2y^2)}{x^2y^2}\to1$ and
$$0\le \left|\frac{x^2y^2}{x^2+\sin(x^2y^2)+y^2}\right|\le \frac{x^2y^2}{x^2+y^2}$$
with
$$\frac{x^2y^2}{x^2+y^2}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration type problem Let $f(x)$ and $g(x$) be continuous, positive function such that $f(–x) = g(x) – 1$,$f(x)=\frac{g(x)}{g(-x)}$ and $\int\limits_{ - 20}^{20} {f\left( x \right)dx} = 2020$
then the value of is $\int\limits_{ - 20}^{20} {\frac{{f\left( x \right)}}{{g\left( x \right)}}dx} $ is
(A) 1010
(B) 1050
(C) 2020
(D) 2050
My approach is as follow
GIven $f\left( { - x} \right) = g\left( x \right) - 1\& f\left( x \right) = \frac{{g\left( x \right)}}{{g\left( { - x} \right)}}$
$\frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{1}{{g\left( { - x} \right)}}\& f\left( { - x} \right) = g\left( x \right) - 1 \Rightarrow f\left( x \right) + 1 = g\left( { - x} \right) \Rightarrow \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{1}{{f\left( x \right) + 1}}$
$\int\limits_{ - 20}^{20} {\frac{{f\left( x \right)}}{{g\left( x \right)}}dx} \Rightarrow \int\limits_{ - 20}^{20} {\frac{1}{{f\left( x \right) + 1}}dx} $
Not able to approach from here
| So firstly we make the computation
$$\frac{f(x)}{g(x)}=\frac{1}{g(-x)}=\frac{1}{1+f(x)}.$$
Now notice that
$$\int_{-20}^{20}\frac{f(x)}{g(x)}~\mathrm{d}x=\int_{-20}^{20} \frac{\mathrm{d}x}{1+f(x)}=\int_{-20}^{0} \frac{\mathrm{d}x}{1+f(x)}+\int_{0}^{20} \frac{\mathrm{d}x}{1+f(x)}=\int_{0}^{20} \frac{\mathrm{d}x}{1+f(-x)}+\int_{0}^{20} \frac{\mathrm{d}x}{1+f(x)}=\int_0^{20}\frac{2+f(x)+f(-x)}{1+f(x)+f(-x)+f(x)f(-x)}~\mathrm{d}x.$$
However using the fact that $f(x)=\frac{g(x)}{g(-x)}$, we can easily see that
$$f(x)f(-x)=1,$$
and so
$$\int_{-20}^{20} \frac{f(x)}{g(x)}~\mathrm{d}x=\int_0^{20}\frac{2+f(x)+f(-x)}{2+f(x)+f(-x)}~\mathrm{d}x=\int_0^{20}\mathrm{d}x=20.$$
So either there is a mistake in my computation, or none of the answers are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can we prove $\lim\limits_{n\to \infty} \sqrt[n]{n^2+n}=1$ using Theorem 1 from Ch. 20 of Spivak's Calculus? In Spivak's Calculus there is the following theorem (I am paraphrasing)
Ch. 22, Theorem 1 If a function $f$ is defined in an open interval containing $c$ except perhaps at $c$, with $$\lim\limits_{x\to
> c}f(x)=l\tag{1}$$ then
For every sequence ${a_n}$ such that
*
*each $a_n$ is in the domain of $f$
*each $a_n\neq 0$
*$\lim\limits_{n\to\infty} a_n=c$
the sequence $f(a_n)$ satisfies
$$\lim\limits_{n\to\infty} f(a_n)=l\tag{2}$$
Conversely, if for every sequence $a_n$ satisfying the three
conditions above there is a sequence $f(a_n)$ satisfying $(2)$, then
$(1)$ is true.
Can apply this theorem directly to prove that
$$\lim\limits_{n\to \infty} \sqrt[n]{n^2+n}=1$$
?
Here is the proposed argument
$$\sqrt[n]{n^2+n}=e^{\frac{1}{n}\log{(n^2+n)}}$$
$$a_n=\frac{\log{(n^2+n)}}{n}$$
$$\lim\limits_{n\to \infty} a_n=0$$
Let $f(x)=e^x$.
Then $\lim\limits_{x\to 0} f(x)=1$.
Since each $a_n$ is in the domain of $f$, which is all numbers, each $a_n\neq 0$ and $$\lim\limits_{n\to \infty} a_n=0$$, we can infer that the sequence $f(a_n)$ satisfies
$$\lim\limits_{x\to\infty} f(a_n)=\lim\limits_{n\to \infty} \sqrt[n]{n^2+n}=1$$
| Without the use of theorems then:
\begin{align}
\sqrt[n]{n^2 + n} &= e^{\ln(n^2 + n)/n} = 1 + \frac{\ln(n^2 + n)}{n} + \frac{\ln^{2}(n^2 + n)}{2 \, n^2} + \mathcal{O}\left( \frac{\ln^3(n)}{n^3} \right) \\
&= 1 + \frac{2 \, \ln(n)}{n} + \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right) + \mathcal{O}\left(\frac{1}{n^2}\right)
\end{align}
An alternate view is:
\begin{align}
\sqrt[n]{n^2 + n} &= e^{\ln(n^2 + n)/n} = e^{\ln\left(1 + \frac{1}{n}\right)/n} \times e^{2 \, \ln n/n} \\
&= e^{\ln\left(1 + \frac{1}{n}\right)/n} \, \left( 1 + \frac{2 \, \ln n}{n} + \frac{2^2 \, \ln^2 n}{2 \, n^2} + \mathcal{O}\left(\frac{\ln^3 n}{n^3}\right) \right)
\end{align}
Since both $\ln\left(1 + \frac{1}{n}\right)$ and $\frac{1}{n}$ tend to zero as $n \to \infty$ then $\text{exp}\left( \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)\right) \to 1$ as $n \to \infty$. This leaves
$$ \lim_{n \to \infty} \sqrt[n]{n^2 + n} = \lim_{n \to \infty} \left( 1 + \frac{2 \, \ln n}{n} + \frac{2^2 \, \ln^2 n}{2 \, n^2} + \mathcal{O}\left(\frac{\ln^3 n}{n^3}\right) \right). $$
Since $ \ln n < n$ then $\frac{\ln n}{n} \to 0$ as $n \to \infty$ then
$$ \lim_{n \to \infty} \sqrt[n]{n^2 + n} = 1.$$
Otherwise the problem proposer has provided a solution by use of a theorem which produces the same desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
what is the projection of point in 3d plane If the coordinate $T(3,4,3) ,B(6,0,0),C(6,0,6)$ and $P(2,0,0)$ in $3$ -dimensions. What is the coordinate of the foot altitude of $P$ with respect to plane $TBC$.
My-Progress:
I am trying to use the equation of plane before resulting
$4x+3y=24$. What I need to do next or this step is unnecessary.
| Write the vector parametric equation of the plane. This is
$r(t,s) = T + t (B - T) + s (C - T) $
Substituting
$r(t,s) = (3,4,3) + t (3, -4, -3) + s(3, -4, 3) $
So the two vectros spanning the plane are $\mathbf{v_1} = [3, -4, -3]^T $ and $\mathbf{v_2} = [3, -4, 3]^T$.
Define the $3 \times 2 $ matrix $A$ as follows
$ A = [\mathbf{v_1, v_2} ] = \begin{bmatrix} 3 && 3 \\ -4 && -4 \\ -3 && 3 \end{bmatrix} $
Then the projector onto the plane is
$P = A (A^T A)^{-1} A^T $
Such the projection of a point $\mathbf{p}$ onto the plane is given by
$ \text{Proj}(\mathbf{p}) = T + P (\mathbf{p} - T) $
Let's compute the projection matrix $P$. We have
$ (A^T A)^{-1} = \dfrac{1}{900} \begin{bmatrix} 34 && - 16 \\ -16 &&34 \end{bmatrix} $
Therefore,
$P = \begin{bmatrix} 0.36 && -0.48 && 0 \\ -0.48 && 0.64 && 0 \\ 0 && 0 && 1 \end{bmatrix} $
And finally, the projection of $Q (2, 0, 0)$ is
$ \text{Proj}(Q) = (3,4,3) + P (Q - (3,4,3) ) = (3,4,3) + P (-1, -4, -3) \\= (3,4,3) + ( 1.56, -2.08, -3 ) = (4.56, 1.92, 0) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$
If $x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$, find the value of $x-\dfrac{1}{x}$.
Here's what I've tried so far.
$$\begin{align}
\left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\
\rightarrow \quad \left(x-\dfrac{1}{x}\right)^3&=108+76\sqrt{2}-3\left(x-\dfrac{1}{x}\right) \\u:=x-\dfrac{1}{x} \quad\rightarrow \quad u^3+3u-108-76\sqrt{2}&=0
\end{align}$$
Got stuck here since I didn't know how to solve this cubic equation.
I also tried factorizing $x^3-\dfrac{1}{x^3}$.
$$\begin{align}x^3-\dfrac{1}{x^3}&=\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(\left(x+\dfrac{1}{x}\right)^2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+1\right)\left(x+\dfrac{1}{x}-1\right)
\end{align}$$
Again, I didn't know what I could do with this.
| I thought it might be helpful with answering now.
\begin{align}
\newcommand{w}{\omega}
& x^3-\frac{1}{x^3}=108+76\sqrt{2}. \\
& 108+76\sqrt{2} = \left(x-\frac 1 x\right)\left(x^2+1+\frac 1 {x^2}\right) \\
& = \left(x-\frac 1 x \right)^3+3x\cdot \frac 1 x \left(x-\frac 1 x\right). \\
\ \\
& u^3+3u=108+76\sqrt{2}. \\
& 108+76\sqrt{2}= 4(27+19\sqrt{2})=4(5+3\sqrt{2})(1+\sqrt{2})^2. \\
& (5+3\sqrt{2})^2+3=46+30\sqrt{2} \neq 12+8\sqrt{2}. \\
& \vdots \\
& (3+2\sqrt{2})^2+3=20+12\sqrt{2}=4(5+3\sqrt{2}). \\
\therefore \; & u=3+2\sqrt{2}.
\end{align}
Substituting $x-\dfrac 1 x=3+2\sqrt{2}$, it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Show that $|\sin(n\pi\sqrt{5})| \ge \frac{c}{n}$ for some $c>0$ Consider the sequence $\{x_n\}$ defined by
$$
x_n = |\sin (n\pi\sqrt{5})|.
$$
Show that there exists a real number $c>0$ such that $x_n \ge \frac{c}{n}$ for all $n \ge 1$.
I have been given two hints:
*
*On $\left[0,\frac{\pi}{2}\right]$, one has $\sin{t} \ge t - \frac{t^3}{6}$. (This is an elementary argument in calculus.)
*For any integers $p$ and $q$ with $p^2+q^2 \ne 0$, one has
$$
|p\sqrt{5}-q| \ge \frac{1}{p\sqrt{5}+q}.
$$
(This corresponds to the fact that $\sqrt{5}$ is irrational.)
For this question, given fixed $n$, I will consider finding $m$ such that $|n\sqrt{5}-m|\le\frac{1}{2}$ (such $m$ always exists, because $\left[n\sqrt{5}-\frac{1}{2},n\sqrt{5}+\frac{1}{2}\right]$ is an interval of length $1$, at least one integer will fall in it), so that
$$
|\sin(n\pi\sqrt{5})|=|\sin(n\pi\sqrt{5}-m\pi)|=|\sin(n\sqrt{5}-m)\pi|=\sin|(n\sqrt{5}-m)\pi|.
$$
Since then $|(n\sqrt{5}-m)\pi| \in \left[0,\frac{\pi}{2}\right]$, we may use these two hints. But I think I am stuck here.
By the first hint,
$$
\begin{aligned}
\sin|(n\sqrt{5}-m)\pi| &\ge |(n\sqrt{5}-m)\pi| - \frac{|(n\sqrt{5}-m)\pi|^3}{6} \\
&\ge \frac{1}{(n\sqrt{5}+m)\pi} - \frac{|(n\sqrt{5}-m)\pi|^3}{6}
\end{aligned}
$$
But I do not think this inequality brings me any close to the desired conclusion, for the existence of $n^3$ term on the right hand side.
Alternatively, I also think about applying the second hint immediately:
$$
\begin{aligned}
\sin|(n\sqrt{5}-m)\pi| & \ge \sin \frac{\pi}{n\sqrt{5}+m} \\
& \ge \frac{\pi}{n\sqrt{5}+m} - \frac{\pi^3}{6(n\sqrt{5}+m)^3}
\end{aligned}
$$
This approach looks more promising but it is still not obvious that I will deduce the desired result.
Is it possible to deduce the desired result from one of my approaches (most likely the second)? Or maybe I should try a even finer approach? I hope further steps should not be a bunch of lengthy fractions. Thanks in advance!
| From your last result, we have
$$
\begin{align}
\sin|(n\sqrt{5}-m)\pi| & \ge \sin \frac{\pi}{n\sqrt{5}+m} \\
& \ge \frac{\pi}{n\sqrt{5}+m} - \frac{\pi^3}{6(n\sqrt{5}+m)^3} \\
&= \alpha\frac{\pi}{n\sqrt{5}+m} + \left((1-\alpha)\frac{\pi}{n\sqrt{5}+m}- \frac{\pi^3}{6(n\sqrt{5}+m)^3} \right) \tag{1}
\end{align}
$$
We will choose $\alpha \in(0,1)$ such that
$$(1-\alpha)\frac{\pi}{n\sqrt{5}+m}- \frac{\pi^3}{6(n\sqrt{5}+m)^3} \ge 0 \tag{2}$$
We have
$$\begin{align}
(2) &\Longleftrightarrow 6(n\sqrt{5}+m)^2\ge\frac{\pi^2}{1-\alpha}\\
&\Longleftrightarrow \alpha \le 1-\frac{1}{6}\left(\frac{\pi}{n\sqrt{5}+m} \right)^2\\ \tag{3}
\end{align}$$
From $|n\sqrt{5}-m|\le\frac{1}{2}$, we deduce that $m \ge n\sqrt{5} -\frac{1}{2}$, then the right hand side of $(3)$ is
$$
RHS(3) \ge 1-\frac{1}{6}\left(\frac{\pi}{2\sqrt{5}n-\frac{1}{2}} \right)^2 \ge 1-\frac{1}{6}\left(\frac{\pi}{2\sqrt{5}-\frac{1}{2}} \right)^2 >\frac{1}{4}
$$
We can choose $\alpha = \frac{1}{4}$. Return back to $(1)$, we have
$$
\sin|(n\sqrt{5}-m)\pi| > \frac{1}{4}\frac{\pi}{n\sqrt{5}+m} > \frac{1}{4}\frac{\pi}{2\sqrt{5}n + \frac{1}{2}} > \frac{1}{4}\frac{\pi}{3\sqrt{5}n}
$$
So, we can choose $c = \frac{1}{4}\frac{\pi}{3\sqrt{5}}$ and for all $n$
$$\sin|(n\sqrt{5})\pi| \ge \frac{1}{4}\frac{\pi}{3\sqrt{5}} \frac{1}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is $\ln(-1)$ equal to $\pi i$, undefined, or both? Calculating $\ln(-1)$ with ...
*
*My calculator: Error (maybe "undefined")
*Advanced calculator: $\pi i$
| answer
It's a bit more complex:
Actually, all complex solutions are $\ln\left( -1 \right) = \pi \cdot \mathrm{i} + 2 \cdot k \cdot \pi \cdot \mathrm{i} = \left( \pi + 2 \cdot k \cdot \pi \right)\cdot \mathrm{i}$ with $k \in \mathbb{Z}$ aka there are infinitely many complex solutions in complex.
"Advanced calculators" like Wolfram|Alpha know this sometimes and show this general solution.
reason
This is because of zheRelationship between exponential functions and trigonometric functions $\exp\left( x \cdot \mathrm{i} \right) = \cos\left( x \right) + \sin\left( x \right) \cdot \mathrm{i}$ (see Euler's formula) and the periodicity of the $\operatorname{cis}\left( x \right)$ function or of the $\sin\left( x \right)$ function and $\cos\left( x \right)$ function to $2 \cdot \pi$:
$$
\begin{align*}
\ln\left( -1 \right) &= x \cdot \mathrm{i}\\
x \cdot \mathrm{i} &= \ln\left( -1 \right) \quad\mid\quad \exp\left( ~~ \right)\\
\exp\left( x \cdot \mathrm{i} \right) &= \exp\left(\ln\left( -1 \right)\right)\\
\operatorname{cis}\left( x \right) &= -1\\
\cos\left( x \right) + \sin\left( x \right) \cdot \mathrm{i} &= -1 \quad\mid\quad \text{Since } -1 \text{ is not imaginary, the } \sin(x) = 0 \text{.}\\
\cos\left( x \right) + 0 \cdot \mathrm{i} &= -1\\
\cos\left( x \right) + 0 &= -1\\
\cos\left( x \right) &= -1 \quad\mid\quad \arccos\left( ~~ \right)\\
\arccos\left(\cos\left( x \right)\right) &= \arccos\left( -1 \right) + 2 \cdot k \cdot \pi\\
x &= \arccos\left( -1 \right) + 2 \cdot k \cdot \pi\\
x &= \pi + 2 \cdot k \cdot \pi\\
\\
\Rightarrow \ln\left( -1 \right) &= \left( \pi + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i} = \pi \cdot \mathrm{i} + 2 \cdot k \cdot \pi \cdot \mathrm{i}\\
\end{align*}
$$
You can write the genral $\ln\left( z \right)$ as
$\ln\left( z \right) = \ln\left( |z| \cdot e^{\left( \arg(z) + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i}} \right) = \ln\left( |z| \right) + \ln\left( e^{\left( \arg(z) + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i}} \right) = \ln\left( |z| \right) + \left( \arg(z) + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i}$.
but it's a bit more complex
Since the complex numbers $\mathbb{C}$ are not the only hypercomplex numbers $\mathbb{C}^{*}$ that can have an imaginary unit $\mathrm{i}$ with $\mathrm{i}^{2} = -1$, but also an infinite number of other hypercomplex numbers, e.g. the quaternions $\mathbb{H}$ (imaginary units $\mathrm{i}$, $\mathrm{j}$ and $\mathrm{k}$ with $\mathrm{i}^{2} = \mathrm{j}^{2} = \mathrm{k}^{2} = \mathrm{i} \cdot \mathrm{j} \cdot \mathrm{k} = -1$), the octonions $\mathbb{O}$, sedenions $\mathbb{S}$, ..., and you can apply Euler's formula to any hypercomplex number with an imaginary unit $\mathrm{i}_{k}$ with $\mathrm{i}_{k}^{2} = -1$, which means that there are again an infinite number of solutions.
| {
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"url": "https://math.stackexchange.com/questions/4560293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given $A,B$ are $n \times n$ matrices, $(A+B)^2=A+B,r(A+B)=r(A)+r(B)$. Prove: $A^2=A,B^2=B$ Given $A,B$ are $n \times n$ matrices, $(A+B)^2=A+B,r(A+B)=r(A)+r(B)$.
Prove: $A^2=A,B^2=B$
I don't know how to use $r(A+B)=r(A)+r(B)$
| It seems like you are working on your linear algebra worksheet(very similar to mine).@Burno B 's answer is beautiful,but there's also a tricky method just using the properties of the rank of matrix.
We know that the elementary row (or column) transformation doesn't change the rank of a matrix,so we have
$$
\begin{pmatrix}
A&0&0\\
0&B&0\\
0&0&I-A-B
\end{pmatrix}
\sim
\begin{pmatrix}
A&0&0\\
0&B&0\\
A&B&I-A-B
\end{pmatrix}
\sim
\begin{pmatrix}
A&0&A\\
0&B&B\\
A&B&I
\end{pmatrix}
$$
$$
\sim
\begin{pmatrix}
A-A^2&-AB&0\\
-BA&B-B^2&0\\
A&B&I
\end{pmatrix}
\sim
\begin{pmatrix}
A-A^2&-AB&0\\
-BA&B-B^2&0\\
0&0&I
\end{pmatrix}
$$
and
$$
r(A)+r(B)+r(I-A-B)=r\begin{pmatrix}
A-A^2&-AB\\
-BA&B-B^2
\end{pmatrix}+r(I)=r\begin{pmatrix}
A-A^2&-AB\\
-BA&B-B^2
\end{pmatrix}+n
$$
Also,
$$
(A+B)^2=(A+B)\implies r(A+B)+r(I-A-B)=n
$$
$$
r(A+B)=r(A)+r(B)\implies r(A)+r(B)+r(I-A-B)=n
$$
which implies
$$
r\begin{pmatrix}
A-A^2&-AB\\
-BA&B-B^2
\end{pmatrix}=0
$$
so
$$
A^2=A,B^2=B,AB=BA=0
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 1
} |
Prove $\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} -\dfrac{1}{x+y} - \dfrac{1}{y+z} -\dfrac{1}{x+z} +\dfrac{1}{x+y+z}$ is a convex function My problem: Prove $g(x,y,z)=\dfrac{1}{x} +\dfrac{1}{y} +\dfrac{1}{z} -\dfrac{1}{x+y} - \dfrac{1}{y+z} -\dfrac{1}{x+z} +\dfrac{1}{x+y+z}$ with $x,y,z >0$ is a convex function.
I tried to use the definition of convex function. Moreover, i tried to use the differentiating but i still get stuck with them.
Can you help me to solve this problem.
| Hints.
If $f(x, y, z): \Omega \to \mathbb{R}$ is twice differentiable and the Hessian is positive semidefinite in the entire domain $\Omega$, then $f(x, y, z)$ is convex.
$\Omega$ must be assumed to be convex too.
If the Hessian has a negative eigenvalue at a point $(x_0, y_0, z_0) \in\Omega$, then the function is not convex.
The domain in your case is convex, since $x>0$, $y>0$, $z>0$
Can you proceed?
THe Hessian Matrix reads:
$$\left(
\begin{array}{ccc}
\frac{2}{x^3}+\frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3}-\frac{2}{(x+z)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+z)^3} \\
\frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+y)^3}+\frac{2}{y^3}-\frac{2}{(y-z)^3} & \frac{2}{(x+y+z)^3}+\frac{2}{(y-z)^3} \\
\frac{2}{(x+y+z)^3}-\frac{2}{(x+z)^3} & \frac{2}{(x+y+z)^3}+\frac{2}{(y-z)^3} & \frac{2}{(x+y+z)^3}-\frac{2}{(x+z)^3}-\frac{2}{(y-z)^3}+\frac{2}{z^3} \\
\end{array}
\right)$$
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
Olympiad-like way to solve the equation $(x^2-4)(x^2+6x+6)=x^2-1$?
Solve the equation:
$$(x^2-4)(x^2+6x+6)=x^2-1$$
I found this question from math olympiad textbook for beginners. But there is no specific hint for the solution.
Is there any faster way to solve this equation?
I see that the given equation is equivalent to
$$(x-2)(x+2)(x^2+6x+6)=(x-1)(x+1)$$
But, I don't see how can I proceed.
After expanding I got
$$x^4+6x^3+x^2-24x-23=0$$
Now, I need factorisation. But factoring doesn't seem like the good track to me.
I know that the substitution $x=y-\frac {b}{4a}$ in $ax^4+bx^3+cx^2+dx+e=0$ can work. But, this gets us a lot of more work. Because the original equation was not given that way.
| We want to make the right-hand side a constant. Therefore, we subtract $x^2-4$ from both sides:
$$
\begin{align}(x^2-4)(x^2+6x+6)-(x^2-4)=(x^2-1)-(x^2-4)\end{align}
$$
Thus we obtain:
$$(x^2-4)(x^2+6x+5)=3$$
The following steps will lead us to the solution:
$$
\begin{align}
&(x-2)(x+2)(x+1)(x+5)=3\\
\implies &\color{#E2062C}{(x-2)(x+5)}\color{#0000CD}{(x+1)(x+2)}=3\\
\implies &(\color{#E2062C}{x^2+3x}-10)(\color{#0000CD}{x^2+3x}+2)=3\end{align}
$$
Then letting $x^2+3x=u$, you have:
$$
\begin{align}&(u-10)(u+2)=3\\
\implies &u^2-8u-23=0.\end{align}
$$
Finally, you can use the quadratic formula to complete the solution.
| {
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} |
Sum of $(1/n^2)\sum_{k=2}^{n/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}$ in the large $n$ limit. This summation appears in studying circulant matrices. Consider an $n-$dimensional circulant matrix with with entries $c_1=c_{n-1}=1$ and $c_0=-2$ (all other coefficients beign $0$). This represents, for example, a random walk on a closed ring of $n$ states. The following calculation can be viewed as the average decay time for all modes (notice that the denominator of the series' terms correspond to the negative eigenvalues of the circulant matrix, $-\lambda_{k}$. Prefactor $1/n^2$ is a normalization constant). From the symmetry of the eigenvalues set one can distinguish two cases: $n$ odd or even.
If $n$ is odd,
$$\frac{1}{n^2}\sum_{k=2}^{(n+1)/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}\,.$$
If $n$ is even,
$$\frac{1}{n^2}\sum_{k=2}^{n/2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}\,.$$
I've attempted to approximate this summation by using the continuous limit but I get the wrong values when comparing to numerical evaluation.
The goal is to find an expression for the large $n$ behavior.
| Similar to @Svyatoslav's second answer
$$\frac 1 {n^2}\sum_{k=2}^{\frac n 2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}=\frac 1 {2n^2}\sum_{k=2}^{{\frac n 2}}\frac{1}{\sin ^2\left(\frac{\pi (k-1)}{n}\right) }=\frac 1 {2n^2}\sum_{k=1}^{{\frac n 2}-1}\frac{1}{\sin ^2\left(\frac{\pi k}{n}\right) }$$
Using
$$\frac{1}{\sin ^2(t)}=\sum_{m=-1}^\infty a_m\,t^{2m}$$ where the $a_m$ form tha apparently unknown sequence
$$\left\{1,\frac{1}{3},\frac{1}{15},\frac{2}{189},\frac{1}{675},
\frac{2}{10395},\frac{1382}{58046625},\frac{4}{1403325},\frac{361
7}{10854718875},\cdots\right\}$$
$$\frac{1}{\sin ^2\left(\frac{\pi k}{n}\right) }=\sum_{m=-1}^\infty a_m\,\frac{\pi^{2m}}{n^{2m}} k^{2m}$$
$$\sum_{k=1}^{\frac n 2-1} k^{2m}=H_{\frac{n}{2}-1}^{(-2 m)}=\zeta (-2 m)-\zeta \left(-2 m,\frac{n}{2}\right)$$ where appear generalized harmonic numbers, the zeta function and the Hurwitz zeta function.
Using all the above gives (apparently at least)
$$\frac 1 {n^2}\sum_{k=2}^{\frac n 2}\frac{1}{1-\cos\left(\frac{2\pi(k-1)}{n}\right)}=\frac 1{12} \left(1-\frac 1{n^2}\right)+O\left(\frac{1}{n^{16}}\right)$$
Let $n=10^p$ and compute the absolute difference
$$\left(
\begin{array}{cc}
p & \Delta_p \\
1 & 2.5\times 10^{-3} \\
2 & 2.5\times 10^{-5} \\
3 & 2.5\times 10^{-7} \\
4 & 2.5\times 10^{-9} \\
5 & 2.5\times 10^{-11} \\
\end{array}
\right)$$
I suppose that comments are not needed.
| {
"language": "en",
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"answer_count": 2,
"answer_id": 1
} |
Estimating area below a hyperbola : showing $\log(3)\gt 1$ In a recent answer of mine, I needed to find an explicit constant rational number $c$ such that $\log(3)\geq c \gt 1$ without using a calculator.
I proceeded using a Riemann sum :
$$\log(3)=\int_{1}^{3}\frac{dt}{t}=\sum_{k=1}^8 \int_{1+\frac{k-1}{4}}^{1+\frac{k}{4}}\frac{dt}{t}\geq \sum_{k=1}^8\frac{1}{4+k}=\frac{28271}{27720} $$
If we look at the complexity of the last step in this computation,
we have a sum of $8$ rational numbers, so this makes $3\times 7$ multiplications, $7$ additions, and $7$ fraction reductions.
Since there are estimates with smaller numbers, such as $\log(3)\gt \frac{12}{11}$, I wonder if there are simpler (calculator-free) proofs.
Perhaps by taking a suitable tangent below the hyperbola to evaluate the area ?
| Solution 1 The Taylor series for $$\log \left(\frac{1+x}{1-x}\right) = 2 \operatorname{artanh} x$$ is $2\left(x + \frac{1}{3} x^3 + \frac{1}{5} x^5 + \cdots\right)$, with all coefficients nonnegative, so for positive $x$, $$2 \operatorname{artanh x} > 2 x + \frac{x^3}{3} ,$$ and in particular $$\log 3 = 2 \operatorname{artanh} \frac{1}{2} > 2 \left[ \frac{1}{2} + \frac{1}{3}\left(\frac{1}{2}\right)^3 \right] = \frac{13}{12} > 1 .$$
Solution 2 Alternatively, evaluating a Dalzell-style integral gives
$$0 < \int_0^\sqrt{2} \frac{x^2 (1 - x)^2}{1 + x^2} = \log 3 - \left(2 - \frac{2}{3} \sqrt{2}\right),$$
so it suffices to find a rational number between $1$ and $2 - \frac{2}{3} \sqrt{2}$, or, by rearranging, a number between $2 \sqrt{2}$ and $3$, but rearranging shows that $\frac{17}{6}$ suffices for the former, and correspondingly $1 < \frac{19}{18} < \log 3$. Probably a more clever choice of integral can reduce the amount of subsequent computation.
Remark Incidentally, direct computation shows that the (signed) area over the interval $[1, 3]$ and under the tangent line to the curve $x y = 1$ at $(a, \frac{1}{a})$ is $\frac{4 (a - 1)}{a^2}$, which has a maximum of $1$ (at $a = 2$), hence without modification this method cannot yield an explicit rational between $1$ and $\log 3$.
| {
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"url": "https://math.stackexchange.com/questions/4573589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Uniform convergence of this series I'm studying the convergence of this series on $ \mathbb{R}_{\geqslant 0}$:
$$ \sum_{n=1}^\infty \frac{\sqrt{nx}}{1+n^2x} $$
It's clear that series converges pointwise, since it's $0$ at $x=0$, otherwise it's equal to:
$$ \frac{1}{\sqrt{x}} \sum_{n=1}^\infty \frac{\sqrt{n}}{\frac{1}{x}+n^2}\leqslant\frac{1}{\sqrt{x}} \sum_{n=1}^\infty \frac{\sqrt{n}}{n^2}=\frac{1}{\sqrt{x}} \sum_{n=1}^\infty \frac{1}{n^{\frac{3}{2}}}$$.
The function $\frac{\sqrt{nx}}{1+n^2x}$ has a maximum at $\frac{1}{n^2},$ and if we sum those values we get the sum $\sum \frac{1}{2\sqrt{n}}$, which diverges, so total convergence fails.
Uniform convergence holds on $[1, \infty)$:
$$ \sum_{n=k}^\infty \frac{\sqrt{nx}}{1+n^2x} = \frac{1}{\sqrt{x}} \sum_{n=k}^\infty \frac{1}{n^{\frac{3}{2}}} \leqslant \sum_{n=k}^\infty \frac{1}{n^{\frac{3}{2}}} \to 0 \ \text{ as } \ k \to \infty$$
But I believe that uniform convergence fails on $[0,1]$, but I can't prove why. Does anyone know how to show this?
| Let's substitute $x = \delta^2$. Then the series becomes
\begin{align*}
\sum_{n=1}^{\infty} \frac{\sqrt{n\delta^2}}{1 + n^2\delta^2}
&= \sum_{n=1}^{\infty} \frac{\sqrt{n\delta}}{1 + (n\delta)^2} \sqrt{\delta}
= \frac{1}{\sqrt{\delta}} \sum_{n=1}^{\infty} \frac{\sqrt{n\delta}}{1 + (n\delta)^2} \delta.
\end{align*}
However, thinking of the last series as a Riemann sum, it is reasonable to expect that
$$ \sum_{n=1}^{\infty} \frac{\sqrt{n\delta}}{1 + (n\delta)^2} \delta \quad \xrightarrow{\delta \to 0^+} \quad \int_{0}^{\infty} \frac{\sqrt{x}}{1 + x^2} \, \mathrm{d}x > 0 $$
and hence the overall quantity will diverge as $\delta \to 0^+$. This claim can be verified relatively easily by a simple comparison between the "Riemann sum" and the actual integral.
This means that the sum is unbounded on $(0, \infty)$, and this cannot happen for a uniform limit of bounded functions. Therefore the sum does not converge uniformly.
| {
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Integration of $\int_0^{1/4} \frac{x^3}{ \sqrt{1-9x^2}} dx$ using trigo substitution
Evalute $\int_0^{1/4} \frac{x^3}{ \sqrt{1-9x^2}} dx$ using trigo substitution
Because $\sqrt{a^2 - b^2x^2} => x = a/b \sin \theta$
So, Let $x= \frac{1}{3} \sin \theta$
$dx = \frac{1}{3} \cos \theta$
Elimination of roots:
$\sqrt{1-9x^2} = \sqrt{1- \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta $
The limit of integration is $x=0$ to $x=1/4$
Therefore, the new limit of integration is:
$0 \le \theta \le 0.8481$ This is in the 1st quadrant and therefore $\cos \theta$ is positive
Actual substitution:
$\int_0^{1/4} \frac{x^3}{ \sqrt{1-9x^2}} dx = \int_0^{0.8481} \frac{\frac{1}{9} \sin^3 \theta}{\cos \theta} (\frac{1}{3} \cos \theta)d\theta =\int_0^{0.8481} \frac{\sin^3 \theta}{27} d\theta$
However, the calculated value of the definite integral is incorrect. I was told that my trigo substitution is correct. I cannot find which step I made a careless in.
| Let
$$ I = \int_{0}^{1/4} \, \frac{x^3 \, dx}{\sqrt{1- 9 \, x^2}}. $$
Make the substitution $3 \, x = \sin\theta$ to obtain the following.
Let $ a = \sin^{-1}(1/4)$.
\begin{align}
I &= \int_{0}^{a} \, \frac{\sin^3\theta}{3^3} \, \frac{1}{\sqrt{1-\sin^2\theta}} \, \frac{\cos\theta \, d\theta}{3} \\
&= \frac{1}{3^4} \, \int_{0}^{a} \frac{\sin^3\theta \, \cos\theta \, d\theta}{\cos\theta} \\
&= \frac{1}{3^4} \, \int_{0}^{a} \sin^3\theta \, d\theta \\
&= \frac{1}{3^4 \, 4} \, \int_{0}^{a} ( 3 \, \sin\theta - \sin(3\theta) ) \, d\theta \\
&= \frac{1}{3^5 \, 4} \left[ \cos(3\theta) - 9 \, \cos\theta \right]_{0}^{a} \\
&= \frac{1}{3^5 \, 4} \, \left[8 + \cos\left(3 \, \sin^{-1}\left(\frac{3}{4}\right)\right) - 9 \, \cos\left(\sin^{-1}\left(\frac{3}{4}\right)\right) \right] \\
&= \frac{1}{3^5 \, 4} \, \left[ 8 - \frac{5 \, \sqrt{7}}{16} - 9 \, \frac{\sqrt{7}}{4} \right] \\
&= \frac{1}{3^5 \, 4} \, \left(8 - \frac{41 \, \sqrt{7}}{16} \right).
\end{align}
A straight integration reveals
\begin{align}
I &= \int_{0}^{1/4} \, \frac{x^3 \, dx}{\sqrt{1- 9 \, x^2}} \\
&= - \frac{1}{3^5} \, [ (9 x^2 + 2) \, \sqrt{1 - 9 x^2} ]_{0}^{1/4} \\
&= \frac{1}{3^5 \, 4} \, \left( 8 - \frac{41 \, \sqrt{7}}{16} \right)
\end{align}
as expected.
| {
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Long, complicated implicit differentiation, finding $y'$ Find $y''$ where $y^2 + xy = \ln (x+1)$
Using implicit differentiation,
$\frac{dy}{dx} = \frac{1}{2y+x} (\frac{1}{x+1} -y)$
Now finding the second derivative of this function, using the product rule,
$\frac{d^2y}{dx^2} = (\frac{-2 \frac{dy}{dx} +1}{(2y+x)^2} \times \frac{1}{x+1} - y) + (\frac{1}{2y+x} \times (\frac{-1}{(x+1)^2} - \frac{dy}{dx}))$
This looks very different from what the answer key is telling me:
$\frac{d^2y}{dx^2} = \frac{-1}{2y+x}(\frac{1}{(1+x)^2} + 2\frac{dy}{dx} + 2(\frac{dy}{dx})^2)$
| To follow the path you took, applying the Product Rule to
$$ \frac{dy}{dx} \ \ = \ \ \frac{1}{2y+x} · \left(\frac{1}{x+1} -y \right) \ \ = \ \ (2y + x)^{-1} \ · \ \left(\frac{1}{x+1} -y \right) $$
[which means needing to bring in the Chain Rule as well] produces
$$ \frac{d^2y}{dx^2} \ \ = \ \ \left[ \ -(2y + x)^{-2} \ · \ \left(2·\frac{dy}{dx} \ + \ 1 \right) \ \right] \ · \ \left( \frac{1}{x+1} \ - \ y \right) $$ $$ + \ \ (2y + x)^{-1} \ · \ \left[ \ \frac{-1}{(x+1)^2} \ - \ \frac{dy}{dx} \ \right] \ \ , $$
so what you have to this point appears to be correct, but could be simplified (somewhat). The person writing the answer key "extracted" a common factor among the terms to obtain
$$ \frac{d^2y}{dx^2} \ \ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \ (2y + x)^{-1} · \left(2·\frac{dy}{dx} \ + \ 1 \right) · \left( \frac{1}{x+1} \ - \ y \right) \ \right] \ \ , $$
which can be re-arranged as
$$ \frac{d^2y}{dx^2} \ \ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \ \left(2·\frac{dy}{dx} \ + \ 1 \right) · \underbrace{(2y + x)^{-1} · \left( \frac{1}{x+1} \ - \ y \right)}_{dy/dx} \ \right] \ \ , $$
in which we notice that two of the factors in the last term are the expression you found for the first derivative, permitting us to write
$$ \frac{d^2y}{dx^2} \ \ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \ \left(2·\frac{dy}{dx} \ + \ 1 \right) · \frac{dy}{dx} \ \right] $$
$$ = \ \ -(2y + x)^{-1} \ · \ \left[ \ \frac{1}{(x+1)^2} \ + \ \frac{dy}{dx} \ + \ 2·\left(\frac{dy}{dx}\right)^2 \ + \ \frac{dy}{dx} \ \right] \ \ , $$
leading to the given answer.
[My personal preference in finding higher derivatives by implicit differentiation is that demonstrated by copper.hat. You may need to find a simplified expression for each successive derivative, but it depends on the expression being differentiated. It is generally a good idea to keep in mind that original expression and what you found for the derivatives, as they may be useful in making calculational simplifications.
Also, you should be more careful about your use of parentheses: you have two places in your result that could cause confusion in making the calculation. ]
| {
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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.