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Asking A Question Why 16 lnx added in the inequality How prove this inequality $a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$
see last answer why did he add 16lnx but lnx doesnt exist
|
This is a well-known trick for some inequalities of the form $f(a) + f(b) + f(c) \ge 0$ under the condition $a, b, c > 0; abc = 1$, assuming that the equality case is $a = b = c = 1$.
The idea is: If we can find a real constant $m$ such that $F(x) := f(x) + m \ln x \ge 0$ for all $x > 0$,
then we have
$0 \le F(a) + F(b) + F(c) = f(a) + f(b) + f(c) + m \ln (abc) = f(a) + f(b) + f(c)$. The desired inequality is proved.
(If we can not find such a $m$, this trick fails.)
How to find $m$?
Since $F(1) = 0$, if $F(x) \ge 0$ for all $x > 0$, then $x = 1$ is the minimum point of $F(x)$ on $x > 0$ which results in $F'(1) = 0$.
From $F'(1) = 0$, we have $f'(1) + m = 0$ and $m = - f'(1)$.
It remains to prove that $f(x) - f'(1)\ln x \ge 0$ for all $x > 0$.
For the problem you refer to,
the inequality is written as $f(a) + f(b) + f(c) \ge 0$
with $f(x) = x^2 + \frac{8}{x} + 1 - 10x$.
We have $f'(1) = -16$.
We need to prove that
$f(x) + 16\ln x \ge 0$ for all $x > 0$. Fortunately, it is true.
There are lots of examples in AoPS and MSE, etc. Here are some.
Problem 1 (jokehim@AoPS): Let $a, b, c > 0$ with $abc = 1$. Prove that
$$\sum_{\mathrm{cyc}} \frac{a^3}{\sqrt{1+a^4}} \ge \sum_{\mathrm{cyc}} \frac{\sqrt2}{a^2+1}.$$
It suffices to prove that, for all $x > 0$,
$$\frac{x^3}{\sqrt{1+x^4}} - \frac{\sqrt2}{x^2+1} - \frac{3}{\sqrt2}\ln x \ge 0.$$
Problem 2: Let $n \ge 3$ be a positive integer.
Let $a, b$ be positive real numbers such that $a^{n+1} + b^{n+1} = 2$.
Prove that $a^n+b^n \ge a^{n-1} + b^{n-1}$.
It suffices to prov that, for all $x \in [0, 2]$,
$$x^n - x^{n - 1} \ge \frac{1}{n+1}(x^{n+1} - 1).$$
Note: The idea is similar, i.e. to find a real constant $m$ such that $x^n - x^{n - 1} \ge m (x^{n + 1} - 1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4415199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Combinatorial argument why is the following below true? $\frac{12!}{2^6 \cdot 3! \cdot 3!} = {12 \choose 6} \cdot 5^2 \cdot 3^2 \cdot 1^2 $
I'm trying to formulate a combinatorial argument and have singled out a case of the two formulas above hoping to see if anyone can explain why the above is true. If so it may help me solve the argument for a general case. Any insight or tips would be much appreciated.
Here is what I know so far from the left side of the equation:
$\frac{12!}{2^6 \cdot 3! \cdot 3!}$ it seems like we are dispersing $12$ distinct objects to two bins of size $3$ but also dividing all the possible subsets of size $6$ (denoted by $2^6$)? Don't know if that is true but this is what I'm inferring right now.
And the right side seems to first pick out a subset of size $6$, and a bit lost from there on what the $5^2 \cdot 3^2 \cdot 1^2$ denotes and how it can be interpreted with the subset of size $6$ using the product rule.
|
Suppose we have $12$ people. We will place six people each in two labeled rooms, $A$ and $B$, and then place the people in each room in three pairs.
There are $\binom{12}{6}$ ways of selecting which six of the twelve people will be placed in room $A$. The rest must be placed in room $B$. Line up the six people in room $A$ in some order, say alphabetically. There are five ways to pair one of the other people in room $A$ with the first person in the line. Remove those people from the line. That leaves four people in that line. There are three ways to pair one of the other three people in the line with the first person remaining in the line. Remove those people from the line. The remaining two people can be paired in one way. Hence, the six people in room $A$ can be placed in three pairs in $5 \cdot 3 \cdot 1$ ways. The six people in room $B$ can be placed in pairs in $5 \cdot 3 \cdot 1$ ways. Hence, the number of ways of placing six people in room $A$ and six people in room $B$, then placing the people in each room in three pairs is
$$\binom{12}{6}(5 \cdot 3 \cdot 1)^2 = \binom{12}{6}5^2 \cdot 3^2 \cdot 1^2$$
There are
$$\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{12!}{2!10!} \cdot \frac{10!}{2!8!} \cdot \frac{8!}{2!6!} \cdot \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{12!}{2^6}$$
ways to make an ordered selection of six pairs of people from the $12$ available people. We will place the first three selected pairs in room $A$ and the remaining three selected pairs in room $B$. However, doing so counts the same three pairs placed in room $A$ in $3!$ ways since the order in which we select the three pairs we place in room $A$ does not matter. Similarly, we have counted the same three pairs of people placed in room $B$ in $3!$ ways. Hence, by making an ordered selection of pairs, we have counted each valid arrangement $3!3!$ ways. Hence, the number of ways of placing six people in room $A$ and six people in room $B$, then placing the people in each room in three pairs is
$$\frac{1}{3!3!}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{12!}{2^6 \cdot 3! \cdot 3!}$$
We have counted the same set of arrangements in two different ways. Equating the two expressions gives the result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is $\lim\limits_{n\to\infty} \prod\limits_{k=1}^n \left(1 + \frac{k+1}{n^2}\right) = \sqrt e$? Mathematica returns these somewhat striking (to me, at any rate) infinite product identities:
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{k+1}{n^2}\right) = \sqrt e$$
and
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{(k+1)^2}{n^3}\right) = \sqrt[3]{e}$$
With larger exponents, checking with software seems to take forever.
I'm making a bit of a leap here, but these results suggest the following closed form for positive integer $p$:
$$\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) = e^{\frac1{p+1}}$$
*
*How does one compute either or both of the first two limits?
*Does the conjecture hold?
In the general case, we have
$$\prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) = \frac{n^{p+1}+(n+1)^p}{n^{p+1}+1} \prod_{k=1}^n \left(1 + \frac{k^p}{n^{p+1}}\right)$$
and the coefficient of the product converges to $1$. From here, I rewrite the product as a sum of logarithms and attempt to rearrange terms to reveal a Riemann sum, but I have had no luck so far. For instance, in the case of $p=1$,
$$\begin{align*}
\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac k{n^2}\right) &= \exp\left(\lim_{n\to\infty}\sum_{k=1}^n \ln\left(1 + \frac k{n^2}\right)\right)\\[1ex]
&= \exp\left(\lim_{n\to\infty}\sum_{k=1}^n \left(\ln\left(n+\frac kn\right) - \ln(n)\right)\right)\\[1ex]
&= \exp\left(\lim_{n\to\infty}\left(\sum_{k=1}^n \ln\left(n+\frac kn\right) - n\ln(n)\right)\right)\\[1ex]
&= \exp\left(\lim_{n\to\infty} n \left(\frac1n \sum_{k=1}^n \ln\left(n+\frac kn\right) - \ln(n)\right)\right)\\[1ex]
\end{align*}$$
|
Consider any real $p > 0$. Note that for any $x>0$, $$ x \ge \ln(1+x) \ge x - x^2.$$ Indeed, at $x=0$ we have equality, and looking at the derivatives of $x,\ln(1+x),x-x^2$ respectivelly, we get $1,\frac{1}{1+x},1-2x$ respectivelly, and since $1 \ge \frac{1}{1+x} \ge 1-2x$ for any $x>0$ the result follows. Hence $$ \ln\prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) = \sum_{k=1}^n \ln\left(1 + \frac{(k+1)^p}{n^{p+1}}\right) $$ can be bounded from both sides by $$ \frac{1}{n}\sum_{k=1}^n \frac{(k+1)^p}{n^p} - \frac{1}{n^2} \sum_{k=1}^n \frac{(k+1)^{2p}}{n^{2p}}\le \ln\prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \le \frac{1}{n}\sum_{k=1}^n \frac{(k+1)^p}{n^p}. $$
Let us firstly deal with $"$linear$"$ factor, i.e $$\frac{1}{n}\sum_{k=1}^n \frac{(k+1)^p}{n^p} = \frac{1}{n} \sum_{k=1}^n \left( \frac{k+1}{n}\right)^p \to \int_0^1 x^pdx = \frac{1}{1+p},$$ where we used the fact that $x \mapsto x^p$ is continuous, to ensure the latter converges to integral via Riemann's Sum Theorem (formally, we need to write $$ \sum_{k=1}^n \left( \frac{k+1}{n}\right)^p = \sum_{k=1}^n \left( \frac{k}{n}\right)^p - \frac{1}{n^p} + \left(\frac{n+1}{n}\right)^p$$ and note that both the second and the last term multiplied by $\frac{1}{n}$ converge to zero. Now, we need to deal with quadratic factor, but we do it really the same way, that is $$ \frac{1}{n^2} \sum_{k=1}^n \frac{(k+1)^{2p}}{n^{2p}} = \frac{1}{n} \cdot \frac{1}{n}\sum_{k=1}^n \left( \frac{k+1}{n}\right)^{2p} \to \frac{1}{\infty} \cdot \int_0^1 x^{2p}dx = 0 \cdot \frac{1}{2p+1} = 0. $$
Hence $$ \ln \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right)$$ is bounded from above by sequence converging to $\frac{1}{1+p}$ and from below by sequence convering to $\frac{1}{1+p}-0 = \frac{1}{1+p}$. Hence by squeeze theorem, $$ \ln \prod_{k=1}^n \left( 1+ \frac{(k+1)^p}{n^{p+1}}\right) \to \frac{1}{1+p}.$$
Since function $x \mapsto e^x$ is continuous, we can take the limit outside and get $$ \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \to \exp\left( \frac{1}{1+p}\right).$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4417359",
"timestamp": "2023-03-29T00:00:00",
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|
Inverse Derivative: Error/Intuition Problem is to find $[f^{-1}(4)]'$ given $f(x)=\frac{x^3+7}{2}$.
Way One: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}$. Therefore, $\sqrt[3]{2x-7}=y$ (i.e. our inverse function). So, $f^{-1}(x)=(2x-7)^{\frac{1}{3}}$. So, $\frac{d}{dx}f^{-1}(x)=\frac{2}{3}(2x-7)^{-\frac{2}{3}}\implies \frac{d}{dx}f^{-1}(4)=\frac{2}{3}.$
Way Two: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}\implies 2x=y^3+7\implies 2=3y^2\frac{dy}{dx}\implies \frac{1}{\frac{3}{2}y^2}=\frac{dy}{dx}$.
Now, this to me seems like the inverse function is $\frac{d}{dx}f^{-1}|_{x=3}=\frac{dy}{dx}|_{x=3}=\frac{1}{\frac{3}{2}y^2}$ when $x=3$ we know $y=f(3)$ which isn't our answer. So, what am I doing wrong here? Can someone explain what is incorrect with this reasoning? I feel like I am missing something when doing these problems.
|
You already know that after switching variables
\begin{align*}
\frac{dy}{dx}=\frac{1}{\frac{3}{2}y^2}\tag{1}
\end{align*}
Since we want to know $\left(f^{-1}(4)\right)^{\prime}$ we obtain again after switching variables from $x=\frac{y^3+7}{2}$ evaluated at $x=4$:
\begin{align*}
4&=\frac{y^3+7}{2}\\
8&=y^3+7\\
y&=1
\end{align*}
putting it into (1) we get
\begin{align*}
\left.\frac{dy}{dx}\right|_{y=1}&=\left.\frac{1}{\frac{3}{2}y^2}\right|_{y=1}
\color{blue}{=\frac{2}{3}}
\end{align*}
in accordance with the first way.
Note: It might be easier to calculate way two without switching variables after all.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the closed form of $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n}}$, where $n\in N$? In my post, I found the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right)^{2}}=\frac{\pi\left(a^{2}+b^{2}\right)}{4 a^{3} b^{3}}$$
Then I want to find the generalized integral using similar technique by using the integral
$$
I(p, q, r)=\int_{0}^{\frac{\pi}{2}} \frac{d x}{p \cos ^{2} x+q \sin ^{2} x+r} \quad \text{ where }p+r> 0 \textrm{ and } q+r>0.
$$
Letting $t=\tan x$ yields
$$
\begin{aligned}
I(p, q, r)&=\int_{0}^{\infty} \frac{d t}{p+qt^{2}+r\left(1+t^{2}\right)}\\
&=\int_{0}^{\infty} \frac{d t}{(q+r) t^{2}+p+r}\\&=\frac{1}{\sqrt{(p+r)(q+r)}} \left.\tan ^{-1}\left(\frac{t\sqrt{q+r} }{\sqrt{p+r}}\right)\right]_{0}^{\infty}\\
&=\frac{\pi}{2 \sqrt{(p+r)(q+r)}}
\end{aligned}
$$
Differentiating it w.r.t. $r$ by $n$ times yields
$$
\begin{aligned}\int_{0}^{\frac{\pi}{2}} \frac{(-1)^{n}n! d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n+1}}=\frac{\pi}{2 }\cdot \frac{\partial^{n}}{\partial r^{n}}\left(\frac{1}{\sqrt{(p+r)(q+r)}}\right) \\
\boxed{\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q\sin ^{2} x+r\right)^{n+1}}=\frac{(-1)^{n} \pi}{2n !} \frac{\partial^{n}}{\partial r^{n}}\left(\frac{1}{\sqrt{(p+r)(q+r)}}\right)}
\end{aligned}
$$
Using the formula, we can evaluate
$$
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{2}}=\frac{-\pi}{2} \frac{\partial}{\partial r}\left(\frac{1}{\sqrt{(p+r)(q+r)}}\right)= \frac{\pi(p+q+2r)}{4\left[(p+r)( q+r)\right]^{\frac{3}{2}}}
$$
For higher derivative, applying Leibniz’s Rule gives
$$
\begin{aligned}\frac{d^{n}}{d r^{n}}\left[(p+r)^{-\frac{1}{2}}(q+r)^{-\frac{1}{2}}\right]&= (-1)^n \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(\frac{1}{2} \right)_k(p+r)^{-\frac{1}{2}-k}\left(\frac{1}{2} \right)_{n-k}(q+r)^{-\frac{1}{2}-n+k} \\&=\frac{(-1)^n}{(q+r)^{n} \sqrt{(p+r)(q+r)}} \sum_{k=0}^{n} \binom{n}{k} \left(\frac{1}{2}\right)_{k}\left(\frac{1}{2}\right)_{n-k}\left(\frac{q+r}{p+r}\right)^{k}\end{aligned}
$$
where $\displaystyle (\alpha)_n=\frac{\Gamma(\alpha+n)}{\Gamma(n)} $.
Now we can conclude that for any natural number $n\geq 2$,
$$
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(p\cos ^{2} x+q\sin ^{2} x+r\right)^{n+1}}\\=\boxed{\frac{\pi}{2n!(q+r)^{n} \sqrt{(p+r)(q+r)}} \sum_{k=0}^{n} \binom{n}{k}\left(\frac{1}{2}\right)_{k}\left(\frac{1}{2}\right)_{n-k}\left(\frac{q+r}{p+r}\right)^{k}}
$$
Question:
Is there any other solution?
|
With $a=p+r$ and $b=q+r$
\begin{aligned}I_n=\int_{0}^{\frac{\pi}{2}} \frac{1}{\left(p \cos ^{2} x+q \sin ^{2} x+r\right)^{n}}dx
=\int_{0}^{\frac{\pi}{2}} \frac{1}{\left(a \cos ^{2} x+b \sin ^{2} x\right)^{n}}dx
\end{aligned}
Integrate, instead
\begin{align}
\sum_{n=1}^\infty I_n t^n=&
\int_0^{\frac{\pi}{2}} \sum_{n=1}^\infty \left(\frac{t}{a \cos^2x+ b \sin^2x}\right)^n dx\\
=&\int_0^{\frac{\pi}{2}}\frac{t}{a\cos^2x+ b\sin^2x-t}dx
=\frac{\pi}{2}\frac{t}{\sqrt{(a-t )(b-t )}}\\
\end{align}
and apply the Taylor series $ \frac{1}{\sqrt{1-x}}=\sum_{i=0}^\infty\frac1{2^{2i}}{2i\choose i}x^i$ to express the result in power of $t^n$
\begin{align}
\sum_{n=1}^\infty I_n t^n
=\>
\sum_{i=0}^\infty\sum_{j=0}^\infty
\frac{\frac{\pi}{2}{2i\choose i} {2j\choose j} t^{i+j+1}}{2^{2i+2j}a^{i+1/2} b^{j+1/2}}
=
\sum_{n=1}^{\infty} \frac{\pi}{2^{2n-1}}\sum_{i+j+1=n} \frac{{2i\choose i}{2j\choose j}}{a^{i+1/2} b^{j+1/2}} t^n
\end{align}
Thus
$$I_n= \frac{\pi}{2^{2n-1}}\sum_{i+j+1=n} \frac{{2i\choose i}{2j\choose j}}{(p+r)^{i+1/2} (q+r)^{j+1/2}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4425783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Is there any other method to compute $\int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y$? After investigating the integral
$$
\int_{0}^{\frac{\pi}{2}} y \ (\cos y) d y
$$
in the post.
I keep on finding the integral with smaller limit
$$
I:=\int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y.
$$
As before, I use the Fourier series of $\ln(\cos y)$
$$
\ln (\cos y)=-\ln 2+\sum_{k=1}^{\infty} \frac{(-1)^{k+1} \cos (2 k y)}{k}
$$
Multiplying it by $y$ followed by integrating from $0$ to $\frac{\pi}{4} $yields
$$
\int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y=-\int_{0}^{\frac{\pi}{4}} y \ln 2 d y+\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \int_{0}^{\frac{\pi}{4}} y \cos (2 k y) dy= -\frac{\pi^{2}}{32} \ln 2+J
$$
Applying integration by parts gives
$$\begin{aligned} J &=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2 k^{2}}\left([y \sin 2 k y]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \sin 2 k y d y\right) \\ &=\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}\left(\frac{\pi}{4} \sin \frac{k \pi}{2}+\left[\frac{\cos 2 k y}{2 k}\right]_{0}^{\frac{\pi}{4}}\right) \\ &=\frac{\pi}{8} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}} \sin \frac{k \pi}{2}+\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3}} \left(\cos \frac{k \pi}{2}-1\right) \\&= \frac{\pi}{2} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2}}+\left[\frac{1}{32} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3}}-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{3}}\right]\\&=\frac{\pi G}{8}- \frac{7}{32}\left(\zeta(3)-2 \sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right)\\&= \frac{\pi G}{8}-\frac{21}{128} \zeta(3),\end{aligned}$$
where $G$ is the Catalan’s Constant.
Now we can conclude that
$$
\boxed{I= \frac{\pi G}{8}-\frac{\pi^{2}}{32} \ln 2-\frac{21}{128} \zeta(3)}
$$
Noting that, $$
\begin{aligned}
\int_0^{\frac{\pi}{4}} x \ln (\sin x) d x+I & =\int_0^{\frac{\pi}{4}} x \ln (\sin x \cos x) d x \\
& =\int_0^{\frac{\pi}{4}}[x \ln (\sin 2 x)-x \ln 2] d x \\
& =\frac{1}{4} \int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\frac{\pi^2}{32} \ln 2 \\
& =\frac{7}{64}\zeta(3)-\frac{\pi^2}{16} \ln 2
\end{aligned}
$$
Hence $$
\boxed{\int_0^{\frac{\pi}{4}} x \ln (\sin x) d x=-\frac{\pi}{8} G-\frac{\pi^2}{32} \ln 2+\frac{35}{128} \zeta(3)}
$$
and
$$
\boxed{\int_0^{\frac{\pi}{4}} x \ln (\tan x)dx=-\frac{\pi}{4} G+\frac{7}{16} \zeta(3)}
$$
Suggestions and alternative methods are highly appreciated.
|
To avoid infinite series, let
$$S=\int_{0}^{\frac{\pi}{4}} y \ln (2\sin y) d y,\>\>\>
C=\int_{0}^{\frac{\pi}{4}} y \ln (2\cos y) d y
$$
and $K=\int_{0}^{\frac{\pi}{2}} y \ln (\tan y) d y $. Then
\begin{align}
S+C=&\int_{0}^{\frac{\pi}{4}} y \ln (2\sin 2y) d y
\overset{2y\to y}=\frac14 \int_{0}^{\frac{\pi}{2}} y \ln (2\sin y)\overset{y\to \frac\pi2-y}{ d y}= \frac18K\\
S-C =& \int_{0}^{\frac{\pi}{4}} y \ln (\tan y) d y
= K - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} y \ln (\tan y) \overset{y\to \frac\pi2-y}{d y}=\frac12K-\frac\pi4G
\end{align}
Evaluate
\begin{align}
K&\overset{t=\tan y}=\int_{0}^{\infty} \frac{\ln t\tan^{-1}t}{1+t^2}dt
= \int_{0}^{1} \int_0^\infty \frac{t\ln t}{(1+t^2 )(1+x^2 t^2)} \overset{t\to \frac1{xt}}{dt}dx\\
&=\int_{0}^{1} \int_0^\infty \frac{-t\ln x}{(1+t^2 )(1+x^2 t^2)} dt\>dx-K= \frac12 \int_0^1\frac{\ln^2x}{1-x^2}dx=\frac78\zeta(3)
\end{align}
Thus, $C= \frac\pi8G -\frac3{16}K$ and
$$\int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y= C-\ln2\int_0^{\frac\pi4}ydy
= \frac{\pi}{8}G-\frac{21}{128} \zeta(3)-\frac{\pi^{2}}{32} \ln 2
$$
|
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"url": "https://math.stackexchange.com/questions/4428738",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
$ \sum_{n=1}^\infty ( n ( \sum_{k=n}^\infty \frac{1}{k^2})^2 - \frac1n) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$
Prove $$\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$$
I decided to write the LHS of the equation as
\begin{align}
\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) &= \sum_{n=1}^\infty n \left( \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac{1}{n^2}\right)\\
&= \sum_{n=1}^\infty n \left(\sum_{k=n}^\infty \frac{1}{k^2} - \frac{1}{n^2}\right)\left( \sum_{k=n}^\infty \frac{1}{k^2} + \frac{1}{n^2}\right)
\end{align}
I'm not sure how to proceed from here. Some hints would be greatly appreciated!
|
One way is to reduce the power of summand by summation by parts.
Take
$$a_n=\bigg(\sum_{k=n}^\infty\frac1{k^2}\bigg)^2-\frac1{n^2},\quad b_n=n,\quad B_n=\sum_{k=1}^nb_n=\frac12n(n+1),$$
we wish to find the closed form of
$$S=\sum_{n=1}^\infty a_nb_n.$$
By summation by parts, with some simplifications,
\begin{align*}
S&=\underbrace{\lim_{n\to\infty}a_nB_n}_{:=L=0}-
\sum_{n=1}^\infty B_n (a_{n+1}-a_n)\\
&=\sum_{n=1}^\infty \frac12\left(\frac1{n^2}+\frac1{n^3}+\frac1{n(n+1)}\right)
+\frac1n\sum_{k=n+1}^\infty\frac1{k^2}+\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\\
&=\frac12(\zeta(2)+\zeta(3)+1)
+\underbrace{\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1{nk^2}}_{:=S'=\zeta(3)}
+\underbrace{\sum_{n=1}^\infty\bigg(\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\bigg)}_{:=S''=1-\zeta(2)}.
\end{align*}
*
*Proof of $L=0$: By Stolz–Cesàro,
$$\lim_{n\to\infty}\frac{\sum_{k=n}^\infty\frac1{k^2}}{\frac1n}
=\lim_{n\to\infty}\frac{\frac1{n^2}}{\frac1{n(n+1)}}=1\quad
\implies\quad \sum_{k=n}^\infty\frac1{k^2}=\frac1n+o\left(\frac1n\right).$$
Thus,
$$\lim_{n\to\infty}a_nB_n=\lim_{n\to\infty}o\left(\frac1{n^2}\right)\frac{n(n+1)}2=0.$$
*Proof of $S'=\zeta(3)$:
\begin{align*}
\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1{nk^2}
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n(n+k)^2}\\
&=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\left(\frac1n+\frac1k\right)\frac1{(n+k)^2}\\
&=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{nk}\int_0^1 x^{n+k-1}\,\mathrm dx\\
&=\frac12\int_0^1\frac1x\sum_{n=1}^\infty\frac{x^n}n\sum_{k=1}^\infty\frac{x^k}k\,\mathrm dx\\
&=\frac12\int_0^1\frac{\ln^2(1-x)}x\,\mathrm dx\\
&=\zeta(3).
\end{align*}
*Proof of $S''=1-\zeta(2)$: We consider the partial sum. Summation by parts gives
$$\sum_{n=1}^\ell\sum_{k=n+1}^\infty\frac1{k^2}
=\ell\sum_{n=\ell+1}^\infty\frac1{k^2}
+\sum_{n=1}^{\ell-1}\frac n{(n+1)^2}.$$
As $\ell\to\infty$,
$$\sum_{n=1}^\infty\bigg(\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\bigg)
=\underbrace{\lim_{\ell\to\infty}\ell\sum_{n=\ell+1}^\infty\frac1{k^2}}_{=1}
+\underbrace{\sum_{n=1}^{\infty}\bigg(\frac n{(n+1)^2}-\frac1n\bigg)}_{=-\zeta(2)}.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$ For $a\in\mathbb R$, I want to evaluate the integral
$$ I = \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx.$$
I tried to integration by parts by considering $\left( \tan^{-1}(x) \right)' = \frac{1}{x^2+1}$, so that
$$I = - \int_{-\infty}^\infty \tan^{-1}(x) \left(\frac{1}{1+x^2} - \frac{1}{1+(x-a)^2}\right).$$
However, I cannot proceed further. Mathematica cannot solve both integrals.
How to evaluate $I$?
|
Too long for a comment
In fact, the evaluation can be done by means of the complex integration method, and this evaluation is rather straightforward. Just one of the possible contours.
$$\int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx=I_1+I_2$$
It is evident that $I_1=0$ (the integrand is odd). As for $I_2$
$$I_2=-\int_{-\infty}^\infty \frac{\tan^{-1}x}{(x+a)^2+1} dx$$
Using $1-ix=\sqrt{1+x^2}e^{-i\tan^{-1}x}\,\Rightarrow\,\ln(1-ix)=\frac{1}{2}\ln(1+x^2)-i\tan^{-1}x\,\Rightarrow\,\boxed{\,\tan^{-1}x=-\Im\ln(1-ix)\,}$
$$I_2=\Im\int_{-\infty}^\infty \frac{\ln(1-ix)}{(x+a)^2+1} dx=\Im\int_{-\infty}^\infty \frac{\ln(1-ix)}{(x+a-i)(x+a+i)} dx$$
Switching to the complex integration and closing the contour by a big circle of radius $R\to\infty$ in the upper half-plane (counterclockwise), taking into consideration that $\ln(1-ix)$ does not have branch points inside the contour, and that there is a singe simple pole at $z=-a+i$
$$I_2=\Im\,2\pi i\underset{z=-a+i}{\operatorname{Res}}\frac{\ln(1-iz)}{(z+a-i)(z+a+i)}=\pi\,\Im\ln(2+ia)=\pi\tan^{-1}\frac{a}{2}$$
$$I=I_1+I_2=\pi\tan^{-1}\frac{a}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $\mathbb{Z}_p[x]/(x^2+x+1)$ is not a field I have to prove the following:
Let $p$ be a prime and denote $\mathbb{Z}_p=\mathbb{Z}/p\mathbb{Z}$. Show that $\mathbb{Z}_p[x]/(x^2+x+1)$ is not a field if and only if $p=3$ or $p \equiv 1 \pmod 3$.
First I consider showing that $x^2+x+1$ is reducible, We may write $x^2+x+1 \equiv 0$ which implies that $x^2 \equiv -x -1$ which implies $x^2 \equiv -x-1 \equiv 2x+2 \pmod 3$. Thus, finally showing that $x^2+x+1 \equiv x^2-2x-2$. Now, $x^2-2x-2$ is reducible. Thus our original polynomial can be written as a reducible polynomial, thus we do not have a field?
Any help would be greatly appreciated.
|
By the quadratic formula, $x^2 + x + 1$ has roots $\frac{-1 \pm \sqrt{-3}}{2}$.
Therefore, we consider when $-3$ has a square root mod $p$. If it does, then $x^2+x+1$ is not prime and $\mathbb{F}_p [x] / (x^2 + x + 1)$ is not a field. Otherwise, $x^2+x+1$ is irreducible and $\mathbb{F}_p [x] / (x^2 + x + 1)$ is a field.
If $ p =3$ then $-3=0$ which has a square root $0$. For $p = 2$, we can check that neither $0$ nor $1$ is a root, so $x^2+x+1$ is irreducible.
Otherwise, use the law of quadratic reciprocity. Let $p \geq 5$ be a prime number. We have $\left( \frac{3}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{-3}{p} \right)$ and $\left( \frac{p}{3} \right) \left( \frac{3}{p} \right) = (-1)^{\frac{p-1}{2}}$. Rearranging, we get $\left( \frac{-3}{p} \right) = \left( \frac{p}{3} \right)\left( \frac{-1}{p} \right)(-1)^{\frac{p-1}{2}}$.
$\left( \frac{-3}{p} \right)$ is what we are trying to find. By Euler's criterion, $\left( \frac{-1}{p} \right)$ is $1$ if $p = 1 \mod 4$ and $-1$ if $p = 3 \mod 4$.
$(-1)^{\frac{p-1}{2}}$ is $1$ if $p = 1 \mod 4$ and $-1$ if $p = 3 \mod 4$. Therefore $\left( \frac{-1}{p} \right)$ and $(-1)^{\frac{p-1}{2}}$ cancel out, so $\left( \frac{p}{3} \right) = \left( \frac{-3}{p} \right)$
By periodicity of the Legendre symbol, $\left( \frac{p}{3} \right)$ is $1$ if $p = 1 \mod 3$, and $-1$ if $p = 2 \mod 3$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4435497",
"timestamp": "2023-03-29T00:00:00",
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|
The number of integer points on the curve $(7x-1)^2+(7y-1)^2=n$ The number of integral solutions to the equation
$$x^2+y^2=n$$
is defined to be $r_2(n)$ and if $n=2^ap_1^{a_1}\dots p_k^{a_k}q^2$ where $p_i\equiv 1\mod 4$ and $q$ is the product of primes which are $3\mod 4$, then
$$r_2(n)=4(a_1+1)\dots(a_k+1).$$
If we restrict $x\equiv y\equiv 1\mod 2$, that is search for solutions to
$$(2x-1)^2+(2y-1)^2=n$$
we find that if $n\equiv 2$ all solutions to the previous one are also solutions to this and otherwise none are. So the number of solutions is $r_2(n)$ or $0$ according to whether $n\equiv 2$ or not.
And if we restrict $x\equiv y\equiv 1\mod 4$, we find that each quadruple of solutions $(\pm x)^2+(\pm y)^2=n$ generates exactly one solution to the new equation. Therefore the number of solutions to $(4x-1)^2+(4y-1)^2=n$ is $\frac{r_2(n)}4$ if $n\equiv 2\mod 4$ and $0$ otherwise.
In general I want to know how many solutions there are for the equation
\begin{equation}\tag{1}
(mx-a)^2+(my-b)^2=n.
\end{equation}
The idea above solves the case $m=2,3,4,5,6$ for all values of $a,b$ but for $m=7$ it doesn't work anymore. This is because $7$ is the first integer for which something like
$$0^1+1^2\equiv 2^2+2^2\mod 7$$
happens. To be precise, $m=7$ is the least so that there are $a,b,c,d$ so that
$$a^2+b^2\equiv c^2+d^2\mod m$$
$$\{a,b\}\neq \{\pm c,\pm d\}$$
for all choices of signs. This means that the number of solutions when $m=7$ is not what you expect. For example the equation $(7x-1)^2+(7y-1)^2=9$ has no solutions, but I expected there to be$\frac{r_2(9)}4=1$. Strangely, the equation $(7x-1)^2+(7y-2)^2=n$ has exactly half as many solutions as I expected for all $n$ up to ten thousand. Since $1^2+2^2=5$, I reasoned that $5$ would behave differently from the other primes, so I excluded its exponent from the product of $r_2(n)$. Therefore I conjecture that the number of solutions to
$$(7x-1)^2+(7y-2)^2=n$$
where $n\equiv 5\mod 7$ and $n=2^a5^bp_1^{a_1}\dots p_k^{a_k}q^2$ is $\frac 12(a_1+1)\dots (a_k+1)=\frac 18r_2\left(\frac n{5^b}\right)$. This product is only odd if $n$ is a square, which is impossible so that doesn't come up.
What I want to know is how many solutions to (1) there are, and in particular the special cases above.
|
For a prime $p\equiv 1\mod 4$, we have a unique (up to sign) representation as a sum of squares $p=a^2+b^2$ and with the identity
$$p(x^2+y^2)=(ax-by)^2+(bx+ay)^2=(ax+by)^2+(-bx+ay)^2$$
we can define the linear transformations
$$\begin{bmatrix}
x\\
y\end{bmatrix}\mapsto\begin{bmatrix}
a & -b\\
b & a\end{bmatrix}\begin{bmatrix}
x\\
y\end{bmatrix}$$
$$\begin{bmatrix}
x\\
y\end{bmatrix}\mapsto\begin{bmatrix}
a & b\\
-b & a\end{bmatrix}\begin{bmatrix}
x\\
y\end{bmatrix}.$$
Call those matrices $M_+$ and $M_-$. Matrices of this form commute and their product is also of this form. We can classify all primes by the residues of $a,b\mod 7$. There are 9 possibilities up to sign.
\begin{array}{|c|c|}
\hline
p\mod 7& (a,b) \\ \hline
1 & (0,1),(2,2)\\ \hline
2 & (0,3),(1,1) \\ \hline
3 & (1,3) \\ \hline
4 & (0,2),(3,3) \\ \hline
5 & (1,2) \\ \hline
6 & (2,3) \\ \hline
\end{array}
So for an $n=2^ap_1^{a_1}\dots p_k^{a_k}q^2$ write each $p_i=a_i^2+b_i^2$ and define matrices $M_{i+}, M_{i-}$ as above. Each of the representations of $n$ as a sum of squares corresponds to a product
$$(M_{1+}^{t_1}M_{1-}^{a_1-t_1})(M_{2+}^{t_1}M_{2-}^{a_2-t_2})\dots (M_{k+}^{t_k}M_{k-}^{a_k-t_k})=\begin{bmatrix}
\pm x & \mp y\\
\pm y & \pm x\end{bmatrix}$$
where $0\leq t_1\leq a_1, \dots 0\leq t_k\leq a_k$. Notice that there are $4$ choices of sign and $(a_1+1)\dots(a_k+1)$ choices of $t_i$, precisely $r_2(n)$ choices in total.
This means that to find how many solutions there are to $x^2+y^2=n$ subject to $x\equiv\alpha\mod 7$, $y\equiv\beta$ we just need to classify the prime factors of $n$ into one of the nine classes and figure out how many ways there are to choose the $t_i$ such that
$$(M_{1+}^{t_1}M_{1-}^{a_1-t_1})(M_{2+}^{t_1}M_{2-}^{a_2-t_2})\dots (M_{k+}^{t_k}M_{k-}^{a_k-t_k})\equiv\begin{bmatrix}\alpha&-\beta\\ \beta&\alpha\end{bmatrix}\mod 7.$$
There are only $2\cdot 9=18$ types of matrices in this product, and they have finite order, and they commute, so it must be possible to calculate how many solutions there are as a function of the $a_i$ alone.
In other words if
$$n=2^a(p_{11}^{a_{11}}p_{12}^{a_{12}}\dots p_{1k_1}^{a_{1k_1}})(p_{21}^{a_{21}}p_{22}^{a_{22}}\dots p_{2k_1}^{a_{2k_2}})\dots (p_{91}^{a_{91}}p_{92}^{a_{92}}\dots p_{9k_1}^{a_{9k_9}})q^2$$
where each $p_{ij}$ has a representation as a sum of squares of the type $i$ modulo 7, then the number of solutions to $(7x+\alpha)^2+(7y+\beta)^2=n$ can be calculated as a function of the $a_{ij}$ alone and does not depend on the primes themselves. This is kind of the answer I wanted for the question, but I wish it was less computationally intensive.
|
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|
How to calculate this limit $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $? $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $
I tried L'Hopital, but it gets complicated.
I tried also this:
$$ \lim\limits_{x\rightarrow0}\frac{1-(1+x^2)\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
$$ \lim\limits_{x\rightarrow0}\frac{\sin^2{x}-x^2\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
, but I don't know what to do from here.
|
Another way.-Let $f(x)=\dfrac{1-\sqrt{1+x^2}\cos(x)}{x^4}$. Since $f(x)=f(-x)$ and is well defined for $|x|\gt0$ by prolongement by continuity we can put $f(0)=a$ where $a$ is precisely the asked limit. One has therefore
$$1-\sqrt{1+x^2}\cos(x)=ax^4$$ Now we have
$$1-(1+\frac{x^2}{2}-\frac{x^4}{8}+\cdots)(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots)=ax^4$$ The first coefficient in LHS is for $x^4$ and its value is equal to $\dfrac14+\dfrac18-\dfrac{1}{24}=\dfrac13$ so we have
$$\frac{x^4}{3}-\frac{x^6}{12}+\cdots=ax^4$$ from which the searched limit is $=\dfrac 13$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the matrix of a linear transformation from $\Bbb R^{2\times2}$ to $P_2$ Let $T:\Bbb R^{2\times2}\rightarrow P_2$ be a linear transformation, where $\Bbb R^{2\times2}$ is the vector space of all $2\times2$ matrices and $P_2$ is the vector space of all polynomials up to second order.
We consider the basis $B$ for $R^{2×2}$,
$$B = \begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix},\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix},{\begin{bmatrix}
0 & 0\\
1 & 0
\end{bmatrix},\begin{bmatrix}0 & 0\\
0 & 1\end{bmatrix}}, $$
and the basis $C$ for the vector space $P_2$,
$$\ C = \begin{pmatrix}
1 & t & t^{2}
\end{pmatrix}.$$
The transformation $T$ is defined by
$$T\begin{pmatrix}
a & b \\ c &d\end{pmatrix} = (c + d) + (b −c)t + (2a + b)t^{2}.$$
How do you find the matrix representation of $T$ relative to the chosen bases $B$ and $C$?
I think the first step would be to take the transformation $T$ of each basis vector $B_1 = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$, $B_2=\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$, $B_3=\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}$, $B_4 = \begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}$. If that's correct, what's the intuitive idea behind this? Why would I want to see how the basis vectors transform?
|
You need to find out how basis vectors transform under $T$:
$$
\begin{split}
T\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix} &= 2t^2 \\
T\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} &= t + t^2 \\
T\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix} &= 1-t \\
T\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} &= 1.
\end{split}
$$
Then, the columns of the matrix of the transformation, are the transformed basis vectors, i.e.
$$T=\begin{pmatrix}
0 & 0 & 1 & 1\\
0 & 1 &-1 & 0\\
2 & 1 & 0 & 0
\end{pmatrix}.
$$
|
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"language": "en",
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|
Finding the sum of finite geometric series I'm doing the following summation $\sum_{l=k}^{n}2^l$
$\sum_{l=k}^{n}2^l = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$
$S_n=a_1\dfrac{1-r^n}{1-r} \therefore S_n=2^k\dfrac{1-(2)^n}{1-2} = 2^{k+n}-2^k$
But my final result seems to be incorrect compared to the one obtained within the calculator
Am i doing something wrong when using the formula above?
|
The following summation is
$$\sum_{l=k}^{n}2^l = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$$
can be written as
$$\sum_{l=0}^{n-k}2^{l+k} = 2^k + 2^{k+1} + 2^{k+2} + \ldots+ 2^{n-1} + 2^{n}$$
by shifting the index $k$-times, which is a series with $n−k+1$ terms
$S_{n-k+1}=a_1\dfrac{1-r^{n-k+1}}{1-r} \therefore S_{n-k+1}=2^k\dfrac{1-(2)^{n-k+1}}{1-2} = 2^{n+1}-2^k$
This result seems to be correct compared to the one obtained within the calculator.
|
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|
Solve $\tan^2(x)+\tan(x)=2$ for $0\leq x\leq 2\pi$ I am trying to solve the trigonometric equation
$$\tan^2(x)+\tan(x)=2$$ for $0\leq x\leq 2\pi$. At first glance, I try to rearrange the trigonometric equation into something more manageable such that
\begin{align}
\frac{\sin^2(x)}{\cos^2(x)}+\frac{\sin(x)}{\cos(x)}&=2 \\
\sin^2(x)+\sin(x)\cos(x)=&2\cos^2(x).
\end{align}
I do not see how to progress from here. I have also tried using the Pythagorean identity $\tan^2(x)=\sec^2(x)-1$, but this did not seem to help. Any suggestions are appreciated.
|
Let $u = \tan(x)$
It follows that
$$\tan^2(x) + \tan(x) = 2 \iff u^2 + u - 2 = 0 \iff (u-1)(u + 2) = 0$$
$$\therefore \tan(x) = -2 \; \text{or} \; \tan(x) = 1$$
$$\therefore x = \pi n - \tan^{-1}(2)\; \text{or} \; x = \frac{\pi}{4} + \pi n, \; n \in \mathbb{Z}$$
In the given domain $0 \le x \le 2\pi$, we have
$$x = \boxed{\pi - \tan^{-1}(2)} \approx 2.034$$
$$x = \boxed{2\pi - \tan^{-1}(2)} \approx 5.176$$
$$x = \boxed{\frac{\pi}{4}}$$
$$x = \boxed{\frac{5\pi}{4}}$$
$\blacksquare$
|
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|
Difficulty evaluating $\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$ In evaluating
$$\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$$ I did not have a situation where the polynomial under the square root has lower degree. Neither trigonometric substitutions nor variable changes help, at least how I apply them.
I would like at least to know how to evaluate a similar integral, because I already spent a lot of time on this one.
|
Substitute $ x =2^{3/2}\tan y$, along with $a=2^{7/6}$
\begin{align}
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{2}\int_0^\frac\pi2\frac{\sec y}{1 + (a\tan y)^3} dy
=\frac12 (I_1 + I_2)
\end{align}
where
\begin{align}
I_1=& \int_0^\frac\pi2\frac{\sec y}{1 + a\tan y} dy
=\frac2{\sqrt{1+a^2}}\tanh^{-1} \frac{\sqrt{1+a^2}}{1+a}\\
I_2= &\int_0^\frac\pi2\frac{\sec y \ (2-a\tan y)}{1 -a\tan y+a^2\tan^2 y}\overset{y\to \frac\pi2 -y}{dy}\\
=& \ \frac12 \int_0^\frac\pi2\frac{\sec y \ (2-a\tan y)}{1 -a\tan y+a^2\tan^2 y}+\frac{\csc y \ (2-a\cot y)}{1 -a\cot y+a^2\cot^2 y}\ dy\\
=& \int_0^\frac\pi2 \frac{2[a^3+a^2+a-2)\sin2y-4a^2+2a](\sin y +\cos y)}{(a^4-a^2+1)\cos4y+4a(a^2+1)\sin2y-(a^4+7a^2+1)} \ \overset{t=\sin y -\cos y}{dy}\\
=& \int_{-1}^1 \frac{(a^3+a^2+a-2)t^2-(a-2)(a^2-a+1)}
{(a^4-a^2+1)t^4 -2(a^4-a^3-a^2-a+1)t^2+(a^2-a+1)^2}dt\\
=& \ \frac{\sqrt2p_-}{\sqrt{q_-}}\bigg(\frac\pi2+\tan^{-1}\frac{s_-}{\sqrt{q_-}}\bigg)
- \frac{\sqrt2p_+}{\sqrt{q_+}}\coth^{-1}\frac{s_+}{\sqrt{q_+}}
\end{align}
with
\begin{align}
&s_\pm=\sqrt{a^4-a^2+1}\pm(a^2-a+1)\\
&p_\pm=\frac12\bigg( \frac{a^3+a^2+a-2}{\sqrt{a^4-a^2+1}}\pm (a-2)\bigg)\\
&q_\pm= \sqrt{a^4-a^2+1}\ (a^2-a+1)\pm (a^4-a^3-a^2-a+1)
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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|
Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.
Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}=\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p},$$
then
\begin{align*} \frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p}}\sim \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \frac{k^p}{e^p}}= e^p\cdot \frac{k}{\sqrt[k]{a_1a_2\cdots a_k}}. \end{align*}
This perhaps can not work.
|
This is similar to robjohn's solution and it is based on the solution of this page: http://www.math.org.cn/forum.php?mod=viewthread&tid=28918
By Cauchy-Schwarz, we have
$$
\sum_{k=1}^n a_k k^p \sum_{k=1}^n \frac k{a_k} \geq \left(\sum_{k=1}^n k^{\frac{p+1}2}\right)^2\geq \left( \frac{n^{\frac{p+3}2}}{\frac{p+3}2}\right)^2=\frac{4n^{p+3}}{(p+3)^2}.
$$
Then
$$
\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}4 \frac1{n^2}\sum_{k=1}^n \frac k{a_k}.
$$
Summing over $n$, we have
$$
\sum_{n=1}^{\infty}\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}4 \sum_{n=1}^{\infty}\frac1{n^2}\sum_{k=1}^n \frac k{a_k}.
$$
Interchanging order of the summation on the right side, we have
$$
\sum_{n=1}^{\infty}\frac1{n^2}\sum_{k=1}^n \frac k{a_k}=\sum_{k=1}^{\infty} \frac k{a_k} \sum_{n=k}^{\infty} \frac1{n^2}\leq \sum_{k=1}^{\infty} \frac k{a_k} \left( \frac1{k^2} + \frac1k\right)\leq \sum_{k=1}^{\infty} \frac 2{a_k}.
$$
Hence,
$$
\sum_{n=1}^{\infty}\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}2 \sum_{k=1}^{\infty} \frac1{a_k}.
$$
|
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"url": "https://math.stackexchange.com/questions/4453465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Remarquable identities $f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}$ Let $n$ be an integer, and
\begin{equation}
f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}
\end{equation}
\begin{equation}
g(n) = \frac{(bc)^n}{(a-b)(a-c)} + \frac{(ac)^n}{(b-a)(b-c)} + \frac{(ab)^n}{(c-a)(c-b)}
\end{equation}
We have the following impressive identities, for all $a,b,c$,
\begin{align}
f(0) &= 0 \\
f(1) &= 0 \\
f(2) &= 1 \\
f(3) &= a+b+c \\
f(4) &= a^2 + b^2 + c^2 + ab + ac + bc \\
f(5) &= a^3 + b^3 + c^3 + a^2b + a^2c + b^2c + ab^2 + ac^2 + bc^2 \\
f(6) &= a^4 + b^4 + c^4 + a^3b + a^3c + b^3c + ab^3 + ac^3 + bc^3 + a^2bc + ab^2c + abc^2 +a^2b^2 + a^2c^2 + b^2c^2 \\
\\
g(0) &= 0 \\
g(1) &= 1 \\
g(2) &= ab + ac + bc \\
g(3) &= a^2b^2 + a^2c^2 + b^2c^2 + a^2bc + ab^2c + abc^2 \\
g(4) &= a^3b^3 + a^3c^3 + b^3c^3 + a^3b^2c + a^3bc^2 + a^2b^3c + ab^3c^2 + a^2bc^3 + ab^2c^3 + a^2b^2c^2
\end{align}
which I have verified by plugging the expressions into Wolfram Alpha. It seems that the general form should be, for $n > 2$.
\begin{align}
f(n) &= \sum_{i+j+k = n-2}a^ib^jc^k \\
g(n) &= \sum_{\substack{i+j+k = 2(n-1)\\1\leq i,j,k \leq n-1}}a^ib^jc^k
\end{align}
The questions are :
*
*How to demonstrate the statements for $n$ general using induction. Intuitively, we should use induction, however I do not see the induction step.
*Could we demonstrate the general case without using induction ? There is a link, between these formulas and Vandermondt matrices (see below), would there be a nice demonstration using matrices ?
=======================================================================
I arrived at such identities when working with partial fractions decomposition,
and after some related work I realized that the Vandermondt matrices where almost the inverse of the matrices which appear when we do partial fractions decomposition,
Then I realised that the Vandermondt matrices have very nice inverse :
\begin{equation}
\begin{pmatrix}
1&1 \\
a&b
\end{pmatrix}
\begin{pmatrix}
-\frac{b}{(a - b)} & \frac{1}{(a - b)} \\
-\frac{a}{(b - a)} & \frac{1}{(b - a)} \\
\end{pmatrix}
= I_{2}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&1&1 \\
a&b&c \\
a^2&b^2&c^2
\end{pmatrix}
\begin{pmatrix}
\frac{b c}{(a - b) (a - c)} & -\frac{b + c}{(a - b) (a - c)} & \frac{1}{(a - b) (a - c)} \\
\frac{a c}{(b - a) (b - c)} & -\frac{a + c}{(b - a) (b - c)} & \frac{1}{(b - a) (b - c)} \\
\frac{a b}{(c - a) (c - b)} & -\frac{a + b}{(c - a) (c - b)} & \frac{1}{(c - a) (c - b)}
\end{pmatrix}
= I_{3}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&1&1&1 \\
a&b&c&d \\
a^2&b^2&c^2&d^2 \\
a^3&b^3&c^3&d^3
\end{pmatrix}
\begin{pmatrix}
-\frac{bcd}{(a - b) (a - c)(a-d)} & \frac{bc + cd + bd}{(a - b) (a - c)(a-d)} &-\frac{b+c+d}{(a - b) (a - c)(a-d)} & \frac{1}{(a - b) (a - c)(a-d)}\\
-\frac{a cd}{(b - a) (b - c)(b-d)} & \frac{ac + ad + cd}{(b - a) (b - c)(b-d)} & -\frac{a + c + d}{(b - a) (b - c)(b-d)}& \frac{1}{(b - a) (b - c)(b-d)}\\
-\frac{a bd}{(c - a) (c - b)(c-d)} & \frac{ab + ad + bd}{(c - a) (c - b)(c-d)} & -\frac{a + b + d}{(c - a) (c - b)(c-d)}&\frac{1}{(c - a) (c - b)(c-d)}\\
-\frac{a bc}{(d - a) (d - b)(d-c)} & \frac{ab + ac + bc}{(d - a) (d - b)(d-c)} & -\frac{a + b + c}{(d - a) (d - b)(d-c)}&\frac{1}{(d - a) (d - b)(d-c)}
\end{pmatrix}
= I_{4}
\end{equation}
The identities $f(0),f(1), f(2)$ are the last column of the inverse equation for 3-dim matrices. However, it seems that such matrix argument is not sufficient to prove the case for $n$ general, and that many similar identities (the other places in the matrices) should exist.
|
Using the method that @RonaldBlaak used, we see that
$$\begin{align}
\sum_{n\ge0}z^ng(n)&=\sum_{n\ge0}\frac{(bcz)^n}{(a-b)(a-c)}+\sum_{n\ge0}\frac{(acz)^n}{(b-a)(b-c)}+\sum_{n\ge0}\frac{(abz)^n}{(c-a)(c-b)}\\
&=\frac{1}{(a-b)(a-c)(1-bcz)}+\frac{1}{(b-a)(b-c)(1-acz)}+\frac{1}{(c-a)(c-b)(1-abz)}\\
&=\frac{z}{(1-bcz)(1-acz)(1-bcz)}\\
&=z\left(\sum_{i\ge0}(bcz)^i\right)\left(\sum_{j\ge0}(acz)^j\right)\left(\sum_{k\ge0}(abz)^k\right)\\
&=z\sum_{n\ge0}z^n\sum_{i+j+k=n}(bc)^i(ac)^j(ab)^k\\
&=\sum_{n\ge0}z^{n+1}\sum_{i+j+k=n}a^{j+k}b^{i+k}c^{i+j}\\
&=\sum_{n\ge0}z^{n+1}\sum_{i+j+k=n}a^{n-i}b^{n-j}c^{n-k}\\
&=\sum_{n\ge1}z^n(abc)^{n-1}\sum_{i+j+k=n-1}a^{-i}b^{-j}c^{-k},
\end{align}$$
thus showing that $g(0;a,b,c)=0$ and $$g(n;a,b,c)=\sum_{i+j+k=n-1}a^{n-1-i}b^{n-1-j}c^{n-1-k}=(abc)^{n-1}\sum_{i+j+k=n-1}a^{-i}b^{-j}c^{-k},$$
which lends itself to the relation, as was pointed out in the comments,
$$g(n;a,b,c)=(abc)^{n-1}f\left(n+1;\frac1a,\frac1b,\frac1c\right).$$
Nicely enough, this is still a polynomial :)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\ {$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$} Here is the following question:
Solve: $\sec(2x) \ge\sec(x) , x\in [0,\pi]$\
{$\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}$}
Note: This is part b of a question where in part a, I was asked to solve: $\cos(2x)=\cos(x), x\in[0,2\pi]$
For part a, I got my answers as $x = 0,\frac{2\pi}{3},\frac{4\pi}{3},2\pi$
Part a aided me as I was able to get part b to the following form:
*
*$\cos(x)>=\cos(2x)$
However, here is where I am confused. Are the answers just $x>=0, x>=\frac{2\pi}{3}$ by copying the signs?
|
Recall that $\sec(x) = \frac{1}{\cos x}$ and $\cos(2x) = 2 \cos^2(x) - 1$. So, in terms of $\cos(x)$, you have:
$$\frac{1}{2 \cos^2(x) - 1} \ge \frac{1}{\cos(x)}$$
For convenience, let $c = \cos(x)$.
$$\frac{1}{2 c^2 - 1} \ge \frac{1}{c}$$
Now, cross-multiply. Note that because $-1 \le c \le 1$, $2c^2 - 1$ is always positive. But $c$ can be either positive or negative. (It can't be zero, because then the division on the RHS is undefined.) If it's positive, then:
$$c \ge 2c^2 - 1$$
$$0 \ge 2c^2 - c - 1$$
$$0 \ge (2c + 1)(c - 1)$$
Since $0 < c \le 1$, $2c + 1$ is positive, and $c - 1$ is nonpositive. Thus, the inequality is true for all $0 < c \le 1$.
OTOH, if $c < 0$, then we need to flip the inequality sign in all the steps above, so $$0 \le (2c + 1)(c - 1)$$
Since $-1 \le c < 0$, $2c + 1$ is positive, and $c - 1$ is strictly negative, so $(2c + 1)(c - 1)$ is negative, and the inequality is not satisfied.
Therefore, the original inequality is true if $0 < \cos(x) \le 1$. Since the domain of $x$ is restricted to $[0, \pi] ∖ \{ \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4} \}$, this means that $x \in (0, \frac{\pi}{2}) ∖ {\frac{\pi}{4}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4456112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
|
Since the inequality is both homogeneous and cyclic, assume that $a = \max(a, b, c, d) = 1$.
It suffices to prove that, for all $b, c, d \in (0, 1]$,
$$\frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} \le 3.$$
To this end, first, we have
\begin{align*}
\frac{1}{\sqrt{1+b^2}}+\frac{b}{\sqrt{b^2+c^2}}
&= \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2b}{\sqrt{(1 + b^2)(b^2 + c^2)}}}\\
&\le \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2b}{b + bc}}\\
&= \sqrt{\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2}{1 + c}}\\
&\le 1 + \frac14 \left(\frac{1}{1+b^2} + \frac{b^2}{b^2+c^2} + \frac{2}{1 + c}\right)\\
&\le 1 + \frac14\left(\frac{2}{1 + c} + \frac{2}{1 + c}\right)\\
&= 1 + \frac{1}{1 + c} \tag{1}
\end{align*}
where we have used Cauchy-Bunyakovsky-Schwarz inequality,
and $\sqrt{u} \le 1 + \frac{u}{4}$,
and $\frac{2}{1+c} - \frac{1}{1+b^2} - \frac{b^2}{b^2+c^2} = \frac{(1-c)(b^2-c)^2}{(1+c)(b^2+1)(b^2+c^2)}\ge 0$.
Second, we have
\begin{align*}
\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+1}} &= \sqrt{\frac{c^2}{c^2+d^2}
+ \frac{d^2}{d^2+1} + \frac{2cd}{\sqrt{(c^2+d^2)(d^2+1)}}}\\
&\le \sqrt{\frac{c^2}{c^2+d^2}
+ \frac{d^2}{d^2+1} + \frac{2cd}{cd + d}}\\
&= \sqrt{\frac{c^2}{c^2+d^2}
+ \frac{d^2}{d^2+1} + \frac{2c}{c + 1}}\\
&\le \frac12 + \frac12 \left( \frac{c^2}{c^2+d^2}
+ \frac{d^2}{d^2+1} + \frac{2c}{c + 1}\right)\\
&\le \frac12 + \frac12 \left( \frac{1}{1+d^2}
+ \frac{d^2}{d^2+1} + \frac{2c}{c + 1}\right)\\
&= \frac12 + \frac12 \left( 1 + \frac{2c}{c + 1}\right)\\
&= 1 + \frac{c}{c + 1} \tag{2}
\end{align*}
where we have used Cauchy-Bunyakovsky-Schwarz inequality,
and $\sqrt{u} \le \frac12 + \frac12 u$.
Using (1) and (2), the desired result follows.
We are done.
|
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"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the area of the shaded region. Given, $MT=2$ and $AC=8$. Calculate the area of the shaded region.
$\triangle BMT_(notable)\implies (a, 2a, a\sqrt5)\\
\therefore MT = 2, BT=4\\
\triangle ABF \sim \triangle MBT \implies k = \frac{MT}{BT }=\frac{1}{2} =\frac{AF}{BF}\therefore AF =2.2 = 4, BF = 2.4 = 8 \\
S\triangle AMC = S\triangle MCB$
Missing a detail,,,
I managed to solve
|
$MT=2, BT = 4, BM = 2\sqrt5\\
\triangle AGC \sim \triangle MTB\\
\frac{GC}{TB}=\frac{AC}{MB}\\
\frac{GC}{4}=\frac{8}{2\sqrt5}\\
∴GC=\frac{16}{\sqrt5}\\
S_{\triangle MBC} =\frac{1}{2}.MB⋅GC=\frac{1}{2}\cdot 2\sqrt5 \cdot \frac{16}{\sqrt5}=16$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to express a function into powers of $(x-1)$ and $(y-2)$ using Taylor's formula? Use Taylor's formula to express the following in powers of $(x-1)$ and $(y-2)$:
$f(x,y)=x^3 + y^3 + xy^2$
Solution:
$f(1,2)=1 +8 + 4=13$
$f_x (1,2) = 3 + 4=7$
$f_y (1,2) = 12 + 4=16$
$f_{xx} (1,2) = 6$
$f_{yy} (1,2)= 12 + 2=14$
$f_{xy} (1,2)= 4 = f_{yx} (1,2)$
$f_{xxx} (1,2)= 6$
$f_{yyx} (1,2)= 2 = f_{xyy} (x,y) = f_{yxy} (1,2)$
$f_{yyy} (1,2)= 6$
Hence,
$x^3 +y^3 + xy^2 = 13 +7(x-1) + 16(y-2) + \frac{1}{2!} \left[ 6(4)(x-1)^2 +2(x-1)(y-2) + 14(y-2)^2\right] +\frac{1}{3!} \left[ 6(x-1)^3 - 3(0)(x-1)^2 (y-2) + 3(2)(x-1)(y-2) - 6(y-2)^3\right] $
Is the solution correct and how to compute for the remainder if it is applicable? Another question is, is the Taylor's formula the same equation with the Taylor expansion and/or the one stated by the Taylor's Theorem?
|
Your approach is fine, although it should be somewhat revised. A way to check your calculation is recalling that when expanding the function at $x_0=1$ and $y_0=2$ we can replace in $f=f(x,y)$:
\begin{align*}
x&=(x-x_0)+x_0=(x-1)+1\\
y&=(y-y_0)+y_0=(y-2)+2
\end{align*}
We obtain
\begin{align*}
f(x,y)&=x^3+y^3+xy^2\\
&=\left((x-1)+1\right)^3+\left((y-2)+2\right)^3+\left((x-1)+1\right)\left((y-2)+2\right)^2\\
&=(x-1)^3+3(x-1)^2+3(x-1)+1\\
&\qquad+(y-2)^3+6(y-2)^2+12(y-2)+8\\
&\qquad+(x-1)(y-2)^2+4(x-1)(y-2)+4(x-1)\\
&\qquad+(y-2)^2+4(y-2)+4\\
&=13+7(x-1)+16(y-2)\\
&\qquad+\frac{1}{2!}\left(\color{blue}{6}(x-1)^2+\color{blue}{8}(x-1)(y-2)+14(y-2)^2\right)\\
&\qquad+\frac{1}{3!}\left(6(x-1)^3+6(x-1)(y-2)^2\color{blue}{+6}(y-2)^3\right)\tag{1}
\end{align*}
Since the expression (1) is already $f=f(x,y)$, the remainder term is zero.
Blue marked parts should be revised in your calculcation.
|
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|
Proof verification that $ \lim_{n\to \infty} \frac{n^2+n-1}{n^2 + 2n +2}=1$ EDIT: I've had some problems uploading this question today as I initially used the mobile verision, hence the quite absurd first proof if you saw it. Here is the full one:
We do this using the epsilon-N definition of the limit of :
$$\forall \varepsilon>0 ,\hspace{1mm} \exists N>0 \hspace{1mm}\text{s.t}\hspace{2mm} n\geq N \implies |a_{n}-L|<\varepsilon$$
Now,
\begin{align}
|a_{n}-L| & = \left| \frac{n^2+n-1}{n^2 + 2n +2}-1 \right| \\ &=\left| \frac{-n-3}{n^2 + 2n +2} \right|
\end{align}
We now consider the definition of $|x|$ for q quick moment, where we have that:
$$|x|=\left\{
\begin{aligned}
&x \hspace{2mm}\text{if} \hspace{2mm}x\geq0\\
-&x \hspace{2mm}\text{if} \hspace{2mm}x<0
\end{aligned} \right\}
$$
Since we can split the absolute value into the numerator and denominator separately, we have that:
$$|-n-3|=n+3 \hspace{2mm} \text{if} \hspace{2mm} -n-3<0 \iff -3<n$$
However, in our proof, we require $N>0$, thereby $n>0$, meaning that $|-n-3|=n+3$. A similar argument shows that $|n^2 +2n +2|=n^2+2n+2$ for $n>0$. Thereby, for $n>0$:
\begin{align}
|a_{n}-L| &=\frac{n+3}{n^2 + 2n +2} \\
&\leq \frac{n+3}{n^2+2n}
\end{align}
For $n\geq 3$,
\begin{align}|a_n-L| &\leq \frac{n+3}{n^2+2n} \\
&\leq \frac{2n}{n^2+2n} \\
&= \frac{2}{n+2} \\
&\leq \frac{2}{n} < \varepsilon
\end{align}
Thus, to complete our proof, we choose $N=\max\{{\frac{2}{\varepsilon},3}\}$.
Does this look good?
|
HINT
As a suggestion, you do not need to split into two cases. Suppose that $n\geq n_{\varepsilon}$. Then one gets that
\begin{align*}
\left|\frac{n^{2} + n - 1}{n^{2} + 2n + 2} - 1\right| & = \frac{n + 3}{n^{2} + 2n + 2} = \frac{n + 3}{(n + 1)^{2} + 1} \leq \frac{3n + 3}{(n + 1)^{2}} \leq \frac{3}{n} \leq \frac{3}{n_{\varepsilon}}
\end{align*}
Can you take it from here?
|
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|
Why am I getting wrong solution to the system $2x+6y-3z=10$, $5x+2y-1z=12$? Suppose we have this system of equations:
$$\begin{align}
2x+6y-3z&=10\tag{1} \\
5x+2y-1z&=12\tag{2}
\end{align}$$
Why does solving as follows give an incorrect solution?
$$\begin{align}
2x+6y-3z-10&=5x+2y-1z-12 \\
-3x+4y-2z&=-2\tag{3}
\end{align}$$
According to this equation, $(0,0,1)$ is a solution. However, when I plug this into the two starting equations, the result is inconsistent.
What is wrong with the logic I used in finding the solution?
By the way, I was only looking for a single solution of the two starting equations, not a general solution. I know how to use matrices to solve the system, but I am still curious as to what is wrong with the approach I demonstrated above.
|
\begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align}
Matrix form: $AX=b$
$A=\begin{pmatrix} 2 & 6&-3\\5&2&-1\end{pmatrix}$
$X=\begin{pmatrix}x\\y\\z\end{pmatrix}
$
$b=\begin{pmatrix} 10\\12\end{pmatrix}$
Solve : $[A|b]=\begin{pmatrix} 2 & 6&-3&|10\\5&2&-1&|12\end{pmatrix}$
Then you get the solution :
$X=\{c\begin{pmatrix} 0\\1\\2\end{pmatrix}+\begin{pmatrix} 2\\1\\0\end{pmatrix}:c\in\Bbb{R} \}$
Now back to the original problem :
$-3x+4y-2z=-2$
represent a plane inside the $3$ dimensional space.But the solution of the original system lies on a line.
Also the line lies on the plane $-3x+4y-2z=-2$.
Hence any solution of the original system are also the solution of $-3x+4y-2z=-2$.
\begin{align}
2x+6y-3z&=10 \tag 1 \\
5x+2y-1z&=12 \tag 2
\end{align} $$\overset{(1')=(1)-(2)}\downarrow$$
\begin{align}
-3x+4y-2z&=-2 \tag {1'} \\
5x+2y-z&=12 \tag {2'}
\end{align}
Now solve the system $(1') , (2')$ and you will get the original solutions as both systems are equivalent system.This is the way how elimination works!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Geometry problem inspired by Babylonian tablets (positive matrix factorization) Consider the following problem: given positive real numbers $a,b,c>0$, find positive real numbers $x,y,z,w >0$ such that
$$x^2+y^2 =a$$
$$z^2+w^2=c$$
and
$$xz+wy=b$$
This has an obvious geometric interpretation: the sum of the areas of two squares is known as $a$, the sum of the areas of two other squares is know, and the area of the rectangles formed by one side of the first square and one side from third plus the area of one side of the second square and one side of the fourth square is known as $b$. Can we find these sides? Here is an example, $(a,b,c)=(3,5,11)$ is solved with $(x,y,z,w)=(1, \sqrt{2}, 3, \sqrt{2}).$ This was inspired by geometric presentations of problems on BM 013901, but they are simpler, like to find positive $x,y$ such that $x^2+y^2=a$ and $x+y=c$ and IM 052304 is a list of problems by their first line and has ones like “I sum areas of 4 square sides”, etc.
It seems like this should be solvable with elementary methods but I am at a loss. I tried to look at numerical examples to get a feel for any methods that might be generalizable. All I’ve tried so far is, you can solve for, say, $y$ in the third equation and substitute this into the first, do some simplification and you’ll get a quadratic equation in $x$ with coefficients depending on $w,z,b,a$.
Also, rephrased in linear algebra this is equivalent finding positive entry-wise $B$ such that $A=BB^T$ where $A$ is symmetric, positive entrywise, and PSD, and where both are $2\times 2$ matrices. I’ve seen on mathoverflow and various papers on arxiv that this decomposition fails, in general, for $5\times 5$ matrices with positive entry-wise replaced nonnegative entry-wise. But I can’t seem to find explicit decomposition methods for $n< 5$. From what I’ve seen, if you can write each entry as an inner product of linear dependent vectors then you can rotate them into the positive orthant of the plane. But I don’t see how find such vectors in general for this $2\times 2$ case. Apologies if this is elementary, my linear algebra is not my forte.
|
quite a mess. Given positive definite quadratic form $a x^2 \pm 2bxy + c y^2 $ so that $ac>b^2,$ take $r > 0,$ with
$$ 0 < r < \min \left( \frac{b}{a}, \frac{c}{b} \right) $$
Then define $$ p = \frac{c-br}{b-ar} $$
We see that $p > 0.$ It follows from $a r^2 - 2 b r + c > 0$ that $p > r > 0. $
$$
\left(
\begin{array}{rr}
p&-1 \\
-r&1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
a&b \\
b&c \\
\end{array}
\right)
\left(
\begin{array}{rr}
p&-r \\
-1&1 \\
\end{array}
\right) =
\left(
\begin{array}{cc}
ap^2-2bp+c&0 \\
0&ar^2-2br+c \\
\end{array}
\right)
$$
Alright, the determinant of $ \left(
\begin{array}{rr}
p&-r \\
-1&1 \\
\end{array}
\right) $ is $p-r>0$ and its inverse is
$ \frac{1}{p-r}\left(
\begin{array}{rr}
1&r \\
1&p \\
\end{array}
\right) $
with all positive elements
$$
\frac{1}{(p-r)^2}
\left(
\begin{array}{rr}
1&1 \\
r&p \\
\end{array}
\right)
\left(
\begin{array}{cc}
ap^2-2bp+c&0 \\
0&ar^2-2br+c \\
\end{array}
\right)
\left(
\begin{array}{rr}
1&r \\
1&p \\
\end{array}
\right) =
\left(
\begin{array}{rr}
a&b \\
b&c \\
\end{array}
\right)
$$
LATER: I forgot this; the diagonal matrix
$D= \left(
\begin{array}{cc}
ap^2-2bp+c&0 \\
0&ar^2-2br+c \\
\end{array}
\right) $ has a real diagonal square root by simply taking the (positive) square roots of the diagonal elements. Call that matrix $E$, so that $EE = E^T E = D$ If we now name $W = \frac{1}{p-r} E\left(
\begin{array}{rr}
1&r \\
1&p \\
\end{array}
\right) $ which has positive elements, we see
$$ W^T \; W = \left(
\begin{array}{rr}
a&b \\
b&c \\
\end{array}
\right) $$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Finding a representation of an element of Fibonacci sequence using power series. I need help finding the power series of $f\left(x\right)=\frac{-1}{x^{2}+x-1}$ probably for $x=0$.
I got a sequence defined as $a_{0}=1,\ a_{1}=1\ \ a_{n}=a_{n-1}+a_{n-2}$ for $n\ge 2$. I found out that the radius of convergence of the series $f\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n}$ is bigger than $\frac{1}{2}$. Afterwards I found for all $|x|<\frac{1}{2}$ that $$f\left(x\right)=\frac{-1}{x^{2}+x-1}.$$ And now I think I have to find out that the power series of $\frac{-1}{x^{2}+x-1}$ is $$\sum_{n=0}^{\infty}\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}}{\sqrt{5}}x^{n}$$ and then I can conclude from the similarities of power series that $a_{n}=\frac{\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}}{\sqrt{5}}$.
I started by getting to this equality: $$f\left(x\right)=\frac{1}{\left(x-\frac{-1-\sqrt{5}}{2}\right)\left(x-\frac{-1+\sqrt{5}}{2}\right)}\\=\frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{-1+\sqrt{5}}{2}}-\frac{1}{x-\frac{-1-\sqrt{5}}{2}}\right).$$ I wanted to show out that $$\frac{1}{x-\frac{-1+\sqrt{5}}{2}}=\sum_{n=0}^{\infty}\left(\frac{1+\sqrt{5}}{2}\right)^{n}x^{n}, \\ \frac{1}{x-\frac{-1-\sqrt{5}}{2}}=\sum_{n=0}^{\infty}\left(\frac{1-\sqrt{5}}{2}\right)^{n}x^{n},$$ but I can't manage to do that, I tried expanding the Taylor series around $0$ but it's not working for me.
Any hints or help would be great. Thank you!
|
$$\frac{1}{x-\frac{-1+\sqrt{5}}{2}}=\frac{2}{1-\sqrt{5}}\frac{1}{1-x\cdot\frac{2}{\sqrt{5}-1}}$$
Now let’s use rationalising/conjugacy technique: $$\frac{2}{\sqrt{5}-1}=\frac{2(\sqrt{5}+1)}{(\sqrt{5})^2-1^2}=\frac{\sqrt{5}+1}{2}$$
So you get: $$-\frac{1+\sqrt{5}}{2}\cdot\frac{1}{1-x\cdot\frac{1+\sqrt{5}}{2}}$$
Can you proceed with a geometric series?
Two notes: You’ve dropped a minus sign in your factorisation of $f$, and you’ve dropped a power - those two series you ask about should feature $\varphi^{n\color{red}{+1}}x^n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the $z$ coordinate of center of mass for $C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$ The Problem
Let $C$ be the cone
$$C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$$
Assume that $C$ has a constant mass density and find the z coordinate of the center of mass.
The work I have done:
\begin{align*}
M_{xy}&=\iint_{R}\int_{\sqrt{x^2+y^2}}^{1} z \,\delta \,dx \,dy \,dx\\
&=\iint_{R} \delta \,\frac{z^2}{2} \Bigg|_{\sqrt{x^2+y^2}}^{1} \,dy \,dx\\
&=\iint_{R} \delta \,\Big(\frac{1}{2}-\frac{\sqrt{x^2+y^2}}{2} \Big) \,dy \,dx
\end{align*}
Switching to polar coordinates now and pulling out the constants:
\begin{align*}
&=\frac{\delta}{2}\int_{0}^{2\pi}\int_{0}^{1} \big(r-r^3 \big) \,dr \,d\theta\\
&=\frac{\delta}{2} (\frac{\pi}{2})=\frac{\pi}{4}\,\delta\,.\\
\\
M&=\iint_{R}\int_{\sqrt{x^2+y^2}}^{1} \,\delta \,dx \,dy \,dx\\
&=\iint_{R} \delta (1-\sqrt{x^2+y^2}) \,dy \,dx
\end{align*}
Switching again to polar:
\begin{align*}
&=\delta \int_{0}^{2\pi} \int_{0}^{1} (1-r) \,r \,dr \,d\theta\\
&=\delta \int_{0}^{2\pi} \frac{1}{6} d\theta = \frac{\pi}{3}\,\delta
\end{align*}
Therefore:
$$\bar{z}=\frac{M_{xy}}{M} = \frac{\pi\delta}{4}\cdot\frac{3}{\pi\delta}=\frac{3}{4}$$
Is my answer correct?
Edit: I also know that I could have switched to cylindrical from the beginning. If I did, would the integral I need to evaluate be $$M=\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} \,r \,dz \,dr\,d\theta$$
and
$$M_{xy}=\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} \,rz \,dz \,dr\,d\theta\,?$$
|
This is a cone of base radius $1$, and height $1$. The center of mass formula is
$ \overline{z} = \displaystyle \dfrac{ \iiint z dV } {\iiint dV} $
The numerator is
$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^3)dz = \dfrac{\pi}{2} $
The denominator is
$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^2)dz = \dfrac{2\pi}{3} $
Hence,
$ \overline{z} = \dfrac{ \dfrac{1}{2} }{ \dfrac{ 2}{3}} = \dfrac{3}{4} $
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that $a^3+b^3+c^3+2abc\leq 2$
Let $a,b,c \in [0,1[$ such that $a+b+c=2$. Prove that
$$a^3+b^3+c^3+2abc\leq 2$$
My attempt:
put $a=x^{\frac{2}{3}}$,$b=y^{\frac{2}{3}}$,$c=z^{\frac{2}{3}}$
\begin{align*}
&x^{2}+y^{2}+z^{2}+2x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}} \\
&\quad \leq x^2+y^2+z^2+\frac{2}{3}(x^{2}+y^{2}+z^{2})\\
&\quad =\frac{5}{3}(x^{2}+y^{2}+z^{2}),
\end{align*}
because $\frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}$. And
\begin{align*}
\frac{5}{3}(x^{2}+y^{2}+z^{2})
&= \frac{5}{3}(x^{\frac{1}{3}} \cdot x^{\frac{5}{3}} + y^{\frac{1}{3}} \cdot y^{\frac{5}{3}} + z^{\frac{1}{3}} \cdot z^{\frac{5}{3}}) \\
&\leq \frac{5}{3} \sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}\sqrt{x^{\frac{2}{3}} + y^{\frac{2}{3}} + z^{\frac{2}{3}}}\\
&= \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}} + y^{\frac{10}{3}} + z^{\frac{10}{3}}}
\end{align*}
Now I need to find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$ \frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2. $$
I have two question:
*
*Is my attempt correct? (I mean if my algebraic manipulation is true.)
*Can you find a good frame to $\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}$, or $\sqrt{a^5+b^5+c^5}$ for make
$$\frac{5\sqrt{2}}{3}\sqrt{x^{\frac{10}{3}}+y^{\frac{10}{3}}+z^{\frac{10}{3}}}\leq 2 \quad ?$$
If you have an other method, you can post it, but it will be nice if you can complete my attempt.
|
New proof:
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
&a^3+b^3+c^3 + 2abc\\
\le\,& a^2 + b^2 + c^2 + 2abc\\
=\,& (a + b)^2 - 2ab + c^2 + 2abc\\
=\,& (a + b)^2 - 2ab(1 - c) + c^2\\
\le\,& (a + b)^2 - 2(a + b - 1)(1 - c) + c^2 \\
=\,& (2 - c)^2 - 2(1-c)(1-c) + c^2 \\
=\,& (2 - c)^2 - (1-c)^2 + c^2 - (1-c)^2\\
=\,& 1 \cdot (3-2c) + (2c-1)\cdot 1\\
=\,& 2.
\end{align*}
We are done.
Old proof:
WLOG, assume that $c = \min(a,b,c)$.
Using $0 \le (1 - a)(1 - b) = 1 - a - b + ab$, we have $ab \ge a + b - 1$.
We have
\begin{align*}
&a^3 + b^3 + c^3 + 2abc\\
=\,& (a + b)^3 - 3ab(a + b) + c^3 + 2abc \\
=\,& (a + b)^3 - ab(3a + 3b - 2c) + c^3\\
\le\,& (a + b)^3 - (a + b - 1)(3a + 3b - 2c) + c^3\\
=\,& c^2 - c + 2\\
\le\,& 2.
\end{align*}
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Directional Derivative in $\mathbb{R}^2$
Definition Let $f: \mathbb{R}^n \to \mathbb{R}$ and $u \in \mathbb{R}^n$ be a unit vector. The directional derivative of $f$ in the direction of $u$ is
$$D_uf(j) = \displaystyle\lim_{t \to 0} \frac{f(j+tu) - f(j)}{t}$$
provided that this limit exists.
I am preparing for my summer exit exams in my grad program. I saw the following problem in my notes, but with no solution. So I attempted it myself. I am looking for solution verification. Please correct me if I have made any mistakes along the way.
Example Suppose $f(x,y) = x^2+3xy+4y^2$ and $j =(2,1)$ and $u = \langle \frac{3}{5}, - \frac{4}{5} \rangle$. To find $D_uf(j)$, we find the limit.
$$D_uf(j) = \displaystyle\lim_{t \to 0} \frac{f(j+tu) - f(t)}{t} = \displaystyle\lim_{t \to 0} \frac{f((2,1)+ t(\frac{3}{5}, \frac{-4}{5}))-f(2,1)}{t}$$
$$= \displaystyle\lim_{t \to 0} \frac{f(2+\frac{3}{5}t, 1 - \frac{4}{5}t)-f(2,1)}{t}$$
$$= \displaystyle\lim_{t \to 0} \frac{(2+\frac{3}{5}t)^2+3(2+\frac{3}{5}t)(1-\frac{4}{5}t) +4(1-\frac{4}{5}t)^2-[2^2-3(2)(1)+4(1)^2]}{t}$$
$$= \displaystyle\lim_{t \to 0} \frac{( \frac{1337}{100}t^2+\frac{23}{5}t+14) - 2}{t} = \displaystyle\lim_{t \to 0} \frac{\frac{1337}{100}t^2+\frac{23}{5}t+12}{t}$$
$$= \displaystyle\lim_{t \to 0} 13.37t + \displaystyle\lim_{t \to 0} \frac{23}{5} + \displaystyle\lim_{t \to 0} \frac{12}{t} = \frac{23}{5} + \infty$$
How do I get rid of this pesky $\frac{12}{t}$? Perhaps I have made an error somewhere.
EDIT / UPDATE:
This only works if we have $f(x,y) = x^2-3xy+4y^2$, where the second term is negative instead of positive. Doing so yields
$$D_uf(j) = \displaystyle\lim_{t \to 0} \frac{f(j+tu) - f(t)}{t} = \displaystyle\lim_{t \to 0} \frac{f((2,1)+ t(\frac{3}{5},\frac{-4}{5}))-f(2,1)}{t}$$
$$= \displaystyle\lim_{t \to 0} \frac{f(2+\frac{3}{5}t, 1 - \frac{4}{5}t)-f(2,1)}{t}$$
$$= \displaystyle\lim_{t \to 0} \frac{(2+\frac{3}{5}t)^2-3(2+\frac{3}{5}t)(1-\frac{4}{5}t) +4(1-\frac{4}{5}t)^2-[2^2-3(2)(1)+4(1)^2]}{t}$$
$$= \displaystyle\lim_{t \to 0} \frac{\frac{109}{25}t^2 -t +2 - [2]}{t} = \displaystyle\lim_{t \to 0} = \displaystyle\lim_{t \to 0} \Big( \frac{109}{25}t - 1 \Big) = -1$$
|
Maybe you can recognize that
$Df(\mathbf{x}_0)[\mathbf{u}]
=g'(0)$ with the scalar-valued function
$g(t) = f(\mathbf{x}_0 + t\mathbf{u} )$.
Denote $\mathbf{x}= \mathbf{x}_0 + t\mathbf{u}$.
It is simple to show using chain rule that
$g'(t) = \nabla_\mathbf{x}f(\mathbf{x}_0 + t\mathbf{u}):\mathbf{u}$
from which you can deduce
$$
Df(\mathbf{x}_0)[\mathbf{u}]= g'(0)
= \nabla_\mathbf{x}f(\mathbf{x}_0):\mathbf{u}
$$
In your application
$$
\nabla_\mathbf{x}f(\mathbf{x}_0)
=
\begin{pmatrix}
2x_0+3y_0 \\
8y_0+3x_0
\end{pmatrix}
=
\begin{pmatrix}
7 \\
14
\end{pmatrix}
$$
The directional derivative is
$\frac15 (7\cdot 3-14\cdot 4)=-7$
Here the dot colon indicates the inner product between vectors.
|
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|
Doubt regarding validity of answer in a mod equation question
$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$
Find the set of values of $x$.
The answer given is $[0,\frac12]\cup [2,3]$.
What I don't get is how is the solution a range? I'm getting the solution as $0,\frac 12, 2 ,3, \frac 25$. That makes sense to me as the first expression is always positive, $$\because x^2-2x+1+1=(x-1)^2+1> 0$$ and the other two expressions give two cases which on being solved gives 4 solutions in total as it's a quadratic.
$$x^2-2x+2-|(x-2)(2x-1)|-|(x-0)(x-3)|=0$$
Using wavy curve on the two expressions inside mod, we get that they are negative for $x\in (\frac 12,2)$ and $x\in [0,3]$ respectively. So, we can conclude that if the first expression is negative then the second expression is negative too as $[\frac 12,2]$ is in $[0,3]$. So we can get 3 cases:
Both are positive, or both are negative, or the first one is positive and the second one is negative.
Solving for each gives the solutions which I said I've received earlier in the post.
So where is the range coming from? Am I interpreting the mod operator wrong? Or is my book wrong?
|
$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$
Since $x^2-2x+2 = x^2-2x+1+1=(x-1)^2 +1 > 0$, $|x^2-2x+2|-|2x^2-5x+2|= x^2-2x+2-|2x^2-5x+2|=|x^2-3x|$
then
$2x^2-5x+2=(2x-1)(x-2)=0$, $x=\frac{1}{2}$, and $x=2$
$|2x^2-5x+2|= 2x^2-5x+2$ when $-\infty \leq x \leq\frac{1}{2}$ or $2 \leq x \leq\infty$
$|2x^2-5x+2|= -(2x^2-5x+2)$ when $\frac{1}{2}< x < 2$
and
$x^2-3x=x (x-3)=0$ then $x=0$ or $x=3$
$|x^2-3x|=x^2-3x$ when $-\infty \leq x \leq 0$ or $3 \leq x \leq\infty$
$|x^2-3x|=-(x^2-3x)$ when $0 < x < 3$
After arranging intervals:
$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$ will be
$1.$ $x^2-2x+2-2x^2+5x-2=x^2-3x$, when, $-\infty \leq x \leq 0$ or $3 \leq x \leq\infty$, then, $x=0$ or $x=3$
$2.$ $x^2-2x+2-2x^2+5x-2=-x^2+3x$, when, $0 < x \leq \frac{1}{2}$ or $2 \leq x < 3$ , then, $0=0$ ,so, solution is $\forall x$ in these intervals.
$3.$ $x^2-2x+2+2x^2-5x+2=-x^2+3x$, when, $\frac{1}{2} < x < 2 $ , then, $2x^2-5x+2=0$, $x=\frac{1}{2}$, and $x=2$ ,so, no solution.
Finally
$0 \leq x \leq \frac{1}{2}$ or $2 \leq x \leq3$
|
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|
find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2}{3(yx^3+zy^3+xz^3)+x^3+y^3+z^3}=\dfrac{3}{yx^3+zy^3+xz^3+1}$ and I don't know what to do next (I think my approach is not right)
|
@SuzuHirose's use of standard Lagrange methods indeed finds an extremum. Here's a color-coded plot of the target function on the cubic surface to show that:
The minima occur when $x = y = z = 1$, etc., as seen in the figure, where blue is small.
|
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"url": "https://math.stackexchange.com/questions/4474396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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|
How to prove $s > \frac{11}{2}t-3t\ln t$ in this problem The problem is:
Given $f(x)=\frac{\ln x}{x}$, line $l$ is the tangent of curve
$y=f(x)$ at $(t,f(t))$, and intersects the curve at another point
$(s,f(s))$ where $s<t$.
(1) Find the range of $t$;
(2) (i) Prove $\ln x\le 1 +\frac{1}{e}(x-e)-\frac{1}{2e^2}(x-e)^2+\frac{1}{3e^3}(x-e)^3$;
$\quad$ (ii) Prove $s>\frac{11}{2}t-3t\ln t$.
Now I have solved the problem (1) and (2)(i), but I can't figure out the proof of (2)(ii).
In problem (1) we can get $t\in(e^{3/2},+\infty)$, and the intersection is the null point of the function $F(x)=\frac{\ln x}{x}-\frac{1-\ln t}{t^2}(x-t)-\frac{\ln t}{t}$.
Therefore, we have: $\color{red}{\frac{\ln s}{s}-\frac{1-\ln t}{t^2}(s-t)-\frac{\ln t}{t}=0}$, where $s<t$ and $t>e^{3/2}$. Then we intend to prove $s>\frac{11}{2}t-3t\ln t$.
Maybe the inequality in (2)(i) is helpful for enlarging and reducing the inequality, but I can't find how to use it properly.
|
For (2)-(ii):
From question (1), we have $t > \mathrm{e}^{3/2}$.
We only need to consider the case when $\frac{11}{2}t - 3t\ln t > 0$, i.e. $t < \mathrm{e}^{11/6}$.
Let
$$G(x) := x F(x) = \ln x - \frac{1-\ln t}{t^2}(x-t)x - \frac{x\ln t}{t}.$$
We have $G(s) = 0$.
We have
$$G'(x) = \frac{(t - x)(t + 2x - 2x\ln t)}{xt^2}.$$
Let $x_0 = \frac{t/2}{\ln t - 1}$.
Since $t > \mathrm{e}^{3/2}$,
we have $x_0 < t$.
We have $G'(x) > 0$ on $(0, x_0)$.
Let $x_1 = \frac{11}{2}t - 3t\ln t$.
Since $t > \mathrm{e}^{3/2}$, we have $x_1 < x_0$.
We can prove that $G(x_1) < 0$ (see the remarks at the end).
Since $G'(x) > 0$ on $(0, x_0)$,
using $G(x_1) < 0$ and $0 < x_1 < x_0$ and $G(s) = 0$,
we have $x_1 < s$.
We are done.
Remarks:
Using $\frac{x_1}{t} = \frac{11}{2} - 3\ln t$, we have
\begin{align*}
G(x_1) &= \ln x_1 - \frac{1-\ln t}{t^2}(x_1-t)x_1 - \frac{x_1\ln t}{t} \\
&= \ln t + \ln(11/2 - 3\ln t) - (1 - \ln t)\cdot (11/2 - 3\ln t - 1)(11/2 - 3\ln t)\\
&\qquad - (11/2 - 3\ln t)\ln t\\
&= y + \ln(11/2 - 3y) - (1 - y)\cdot (11/2 - 3y - 1)(11/2 - 3y) - (11/2 - 3y)y \\
&= \ln(11/2 - 3y) + 9y^3 - 36y^2 + \frac{201}{4}y - \frac{99}{4}
\end{align*}
where $y = \ln t \in(3/2, 11/6)$.
Let
$$g(y) := \ln(11/2 - 3y) + 9y^3 - 36y^2 + \frac{201}{4}y - \frac{99}{4}.$$
We have
$$g'(y) = - \frac{81(2y - 3)^3}{4(11 - 6y)}.$$
Thus, $g'(y) < 0$ on $(3/2, 11/6)$.
Also, $g(3/2) = 0$.
Thus, $g(y) < 0$ on $(3/2, 11/6)$.
Thus, $G(x_1) < 0$.
We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4475136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find expression of $c_n$, where $c_n = a_n + b_n$ Given the recurrence relation $a_{n+2} = 3a_{n+1} + 6a_n$ and $b_{n+2} = b_{n+1} + b_n$ I am supposed to find an expression of the recurrence relation for $c_n := a_n + b_n$. I tried to find some form of linear dependence to obtain a recurrence relation for $c_n$ but this did not let me finish.
Appreciate your help!
|
Using that $\,a_{n+2} - 3a_{n+1} - 6a_n = 0\,$ and $\,b_{n+2} = b_{n+1} + b_n\,$:
$$
\require{cancel}
\begin{align}
c_{n+2}-3c_{n+1}-6c_n &= \cancel{(a_{n+2}-3a_{n+1}-6a_n)} + (b_{n+2}-3b_{n+1}-6b_n)
\\ &= -2b_{n+1} - 5 b_n
\end{align}
$$
Then, using that $\,b_{n+2} - b_{n+1} - b_n = 0\,$:
$$
\begin{align}
(c_{n+4} - 3c_{n+3}-6c_{n+2}&) - (c_{n+3}-3c_{n+2}-6c_{n+1}) - (c_{n+2}-3c_{n+1}-6c_n)
\\ &=\; -2\cancel{(b_{n+3}-b_{n+2}-b_{n+1})} - 5\bcancel{(b_{n+2}-b_{n+1}-b_{n})}
\\ &= 0
\end{align}
$$
$$
\iff c_{n+4} - 4c_{n+3}-4c_{n+2}+9 c_{n+1}+ 6 c_n = 0
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4475812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Estimate the condition number in the second norm of the matrix An Let $A_{n}$ be a matrix of size $n$ for $n \geq 1$, and a structure:
$$
A_{n}=\left[\begin{array}{cccccccc}
\sqrt{21} & 1 & 0 & 0 & \ldots & 0 & 0 & 0 \\
0 & \sqrt{21} & 1 & 0 & \ldots & 0 & 0 & 0 \\
0 & 0 & \sqrt{21} & 1 & \ldots & 0 & 0 & 0 \\
0 & 0 & 0 & \sqrt{21} & \ldots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \ldots & \sqrt{21} & 1 & 0 \\
0 & 0 & 0 & 0 & \ldots & 0 & \sqrt{21} & 1 \\
0 & 0 & 0 & 0 & \ldots & 0 & 0 & \sqrt{21}
\end{array}\right]
$$
Estimate the condition number in the second norm of the matrix $A_{n}$.
My solutions is as follows
$$
\|A\|_{2}=\sqrt{\lambda_{\max }\left(A^{*} A\right)}=\sigma_{\max }(A) .
$$
where $\sigma_{\max }(A)$ represents the largest singular value of matrix $A$. So
$$
\begin{aligned}
&\left\|A_{n}\right\|_{2}=\sqrt{21} \\
&\left\|A_{n}^{-1}\right\|_{2}=\frac{1}{\sqrt{21}}
\end{aligned}
$$
Condition number
$$
\mu=|| A_{n} \| \cdot|| A_{n}^{-1}||=1
$$
I want to check if this is correct and I'm used to only solving this for symmetric matrices
|
As you indicated, you essentially need to estimate the eigenvalues of $A^*A$.
Fortunately, in this case it is easy to compute $A^*A$ explicitly.
$$
A^*A=\left[\begin{array}{cccccccc}
22 & \sqrt{21} & 0 & 0 & \ldots & 0 & 0 & 0 \\
\sqrt{21} &22 & \sqrt{21} & 0 & \ldots & 0 & 0 & 0 \\
0 & \sqrt{21} &22 & \sqrt{21} & \ldots & 0 & 0 & 0 \\
0 & 0 & \sqrt{21} &22 & \ldots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \dots &22 & \sqrt{21} & 0 \\
0 & 0 & 0 & 0 & \ldots & \sqrt{21} &22 & \sqrt{21} \\
0 & 0 & 0 & 0 & \ldots & 0 & \sqrt{21} &22
\end{array}\right]
$$
You can now bound the eigenvalues of this matrix using the Gershgorin circle theorem, which tells you that they must be in $[22-2\sqrt{21},22+2\sqrt{21}]$.
Now,
$$
22-2\sqrt{21} = (\sqrt{21} - 1)^2, \text{ and}\\
22+2\sqrt{21} = (\sqrt{21} + 1)^2.
$$
Therefore, an estimate for the condition number of $A$ is
$$
\frac{\sqrt{21} + 1}{\sqrt{21} - 1}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4478168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$
$$\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$$
I completed the square:
$\int \dfrac{dx}{(1+x)\sqrt{2-(x-1)^2}}$
And then substituted $\sqrt 2\sin θ = x-1$ which gives
$\int \dfrac{dθ}{\sqrt2 \sinθ + 2}$
But now I'm stuck. Can someone please help?
|
You may continue with $\theta=\frac\pi2+t$
\begin{align}\int \frac{1}{\sqrt2 \sin\theta+ 2}d\theta
= &\int \frac{1}{\sqrt2 \cos t+ 2}dt= \int \frac{1}{2\sqrt2 \cos^2\frac t2+ 2-\sqrt2}dt\\
=&\int \frac{2d(\tan\frac t2)}{(2-\sqrt2)\tan^2\frac t2+(2+\sqrt2)}=\sqrt2\tan^{-1}\frac{\tan\frac t2}{\sqrt2+1}+C
\end{align}
Or, simply integrate as follows
$$\int \frac{dx}{(1+x)\sqrt{1+2x-x^2}}
= \int \frac{d(\frac{x}{1+x})}{\sqrt{1-2(\frac{x}{1+x})^2}}= \frac1{\sqrt2}\sin^{-1} \frac{\sqrt2x}{1+x}+C
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4478268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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|
Find the maximum of $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$,where$a,b,c>0$ $a,b,c>0$, find the maximum of :
$$\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$$
I try to find the minimum of $\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}=\frac{4a+1}{\sqrt{a}}\cdot\frac{9a+b}{\sqrt{ab}}\cdot\frac{4b+c}{\sqrt{bc}}\cdot\frac{9c+1}{\sqrt{c}}
=\left(4\sqrt{a}+\frac{1}{\sqrt{a}}\right)\left(9\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\left(4\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}\right)\left(9\sqrt{c}+\frac{1}{\sqrt{c}}\right) \geq 4 \times 6\times 4\times 6$
when I try to use AM-GM inequality, I find I can't take equality in the above $4$ parentheses.so I only find a upper bound of the expression.: $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)} < \frac{1}{576}$
|
Let $Q = \dfrac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}= \dfrac{abc}{(b+9a)(4a+1)(4b+c)(9c+1)}$. Apply Cauchy-Schwarz inequality twice: $Q \le \dfrac{abc}{(2\sqrt{a}\cdot \sqrt{b}+1\cdot 3\sqrt{a})^2\cdot(2\sqrt{b}\cdot 3\sqrt{c}+\sqrt{c}\cdot 1)^2} = \left(\dfrac{\sqrt{b}}{(2\sqrt{b}+3)(6\sqrt{b}+1)}\right)^2= \left(\dfrac{t}{(2t+3)(6t+1)}\right)^2$, with $t = \sqrt{b}$. And apply AM-GM inequality:$\left(\dfrac{t}{(2t+3)(6t+1)}\right)^2= \left(\dfrac{t}{12t^2+20t+3}\right)^2= \dfrac{1}{\left(12t+\dfrac{3}{t}+20\right)^2}\le \dfrac{1}{\left(20+2\sqrt{12t\cdot \dfrac{3}{t}}\right)^2}=\dfrac{1}{32^2}=\dfrac{1}{1024}$. Thus $Q_{\text{max}} = \dfrac{1}{1024}$. This maximum value occurs when: $12t = \dfrac{3}{t}\implies t = \dfrac{1}{2}\implies \sqrt{b} = \dfrac{1}{2}\implies b = \dfrac{1}{4}$. Also: $\dfrac{2\sqrt{b}}{3\sqrt{c}}=\dfrac{\sqrt{c}}{1}\implies 3c=2\sqrt{b}=2\cdot\dfrac{1}{2} = 1\implies c = \dfrac{1}{3}.$, and $\dfrac{2\sqrt{a}}{\sqrt{b}}=\dfrac{1}{3\sqrt{a}}\implies 6a=\sqrt{b}=\dfrac{1}{2}\implies a = \dfrac{1}{12}.$ In summary, the maximum of $\dfrac{1}{1024}$ achieved when $(a,b,c) = \left(\dfrac{1}{12}, \dfrac{1}{4}, \dfrac{1}{3}\right)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4478545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Calculating complex integral with Residue - where is my fault? \begin{align}\int_0^{2\pi} \frac{\cos(x)}{13+12\cos(x)} dx & = \displaystyle\int_0^{2\pi} \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\
& = \cdots \\
&= -i\displaystyle\int_0^{2\pi} \frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)z}dz \\
&= 2\pi i (-i) \operatorname{Res}\left(\frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)z};0\right) \\
&=2\pi \lim_{z\to0}\frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac{3}{2}\right)}\\
&=2\pi \frac{0^2+1}{\left(0+\frac{2}{3}\right)\left(0+\frac{3}{2}\right)}\\
&=2\pi
\end{align}
But Wolfram Alpha says it's $\frac{4\pi}{15}$.
What am I doing wrong ?
|
You forget the residue at $-2/3$, because the change of variable is $z=e^{it}$, so the integral in $z$ is not from $0$ to $2\pi$, is in the unit circle, that includes inside $-2/3$.
$\begin{align}\displaystyle\int_0^{2\pi} \frac{cos(x)}{13+12cos(x)} dx & = \displaystyle\int_\gamma \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\
& = \dots \\
&= -i\displaystyle\int_\gamma \frac{z^2+1}{(3z+2)(2z+3)2z}dz \\
&= 2\pi i (-i) (Res(f;0)+Res(f;2/3)) \\
&=2\pi (\lim_{z\to0}\frac{z^2+1}{(3z+2)2(2z+3)}+\lim_{z\to\frac{-2}{3}}\frac{z^2+1}{2z(2z+3)3})\\
&=-\frac{4\pi}{15}
\end{align} $
Another mistake was in the factorization of the denominator:
$$12z^3+26z^2+12z\not=z(z+2/3)(z+3/2)$$
though the roots are the same
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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|
Computing $D_v^2(f)(x_0)$, if $f(x,y)=x^3+x+y^2-2y+1$, $v=\frac{1}{5}(3,4), x_0=(\frac{1}{3},1)$
Compute $D_v^2(f)(x_0)$, if $f(x,y)=x^3+x+y^2-2y+1$, $v=\frac{1}{5}(3,4), x_0=(\frac{1}{3},1)$ using the formula $D_v^2(f)(x_0)=v^TH_f(x_0)v$.
So, $v^T= \left[
\begin{array}{cc}
\frac{3}{5}\\
\frac{4}{5}
\end{array}
\right] $
and I found also Hessian matrix to be: $ \left[
\begin{array}{cc}
2&0\\
0&6
\end{array}
\right] $ and $v=(\frac{3}{5}, \frac{4}{5})$ but matrices cannot be multiplied in this way. Would appreciate if somebody could help.
|
You're almost there, but are misunderstanding the convention. Actually, $$v=\begin{bmatrix} \frac{3}{5} \\ \frac{4}{5}\end{bmatrix}$$ and
$$v^T=\begin{bmatrix} \frac{3}{5} & \frac{4}{5}\end{bmatrix}.$$
|
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"url": "https://math.stackexchange.com/questions/4482564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate $\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}$ . Problem:
Evaluate
$$
\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}.
$$
My question:
This is a question from a previous complex analysis qualifying exam that I am trying to work through. I know that the straightforward formula for calculating this residue would be
$$
\frac{1}{4!} \lim_{z\to 0} \frac{d^4}{dz^4} \left(z^5 \cdot \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)} \right)
$$
However, I'm guessing there is a better method in this case than taking four derivatives or trying to turn this into a Laurent series. Can someone point me in the right direction here?
|
Let $$f(z) = \frac{(z^6-1)^2}{z^5(2z^4 - 5z^2 + 2)}, \quad g(z) = \frac{(z^3-1)^2}{2z^2 - 5z + 2},$$ so that we have $f(z) = g(z^2)/z^5$. So by computing the series expansion of $g$ about $0$, we can obtain the Laruent expansion of $f$, since $g$ has no singularity at $0$.
To this end, observe $$2z^2 - 5z + 2 = (2z-1)(z-2) = 2(1-2z)(1-z/2),$$ hence $$\begin{align}
g(z) &= \frac{1}{6}(z^3 - 1)^2 \left( \frac{4}{1-2z} - \frac{1}{1-z/2} \right) \\
&= \frac{1}{6} (z^3 - 1)^2 \sum_{k=0}^\infty 4(2z)^k - (z/2)^k \\
&= \frac{1}{6} (z^6 - 2z^3 + 1) \sum_{k=0}^\infty (2^{k+2} - 2^{-k}) z^k.
\end{align}$$
We specifically need the coefficient of $z^2$ in this expansion, since this yields the coefficient of $z^4$ in $g(z^2)$, which in turn is the coefficient of $1/z$ in the Laurent expansion of $f$. This corresponds to the choice $k = 2$: $$g(z) = \cdots + \frac{1}{6}(2^4 - 2^{-2}) z^2 + \cdots.$$ Thus the required coefficient is $21/8$, which is also the desired residue of $f$ at $0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is independent of $n$.
Taking
$$f(n)=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
I could prove that
$$f(n+25)=f(n)$$
But, I have no idea how to show
$$f(n+k)=f(n)\\\forall k\in \{1,2,\dots 24\}$$
Of course, we can check these finite number of cases by force. But is there any other elegant method to handle this?
|
Here is a variation which is also related to Bill Dubuque's hint. We want to show that
\begin{align*}
f(n)&=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor\\
&=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor+4
\end{align*}
is independent from $n$. Since
\begin{align*}
f(0)&=\left\lfloor\frac{13}{25}\right\rfloor-\left\lfloor\frac{-\left\lfloor\frac{-17}{25}\right\rfloor}{3}\right\rfloor+4\\
&=0-0+4=4
\end{align*}
we claim the following is valid for all non-negative integers $n$:
\begin{align*}
\color{blue}{\left\lfloor\frac{8n+13}{25}\right\rfloor=\left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor}\tag{1}
\end{align*}
We start with the right-hand side of (1) and obtain
\begin{align*}
\color{blue}{\left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor}
&=\left\lfloor\frac{-\left\lfloor\frac{-24n-17}{25}\right\rfloor}{3}\right\rfloor
=\left\lfloor\frac{\left\lceil\frac{24n+17}{25}\right\rceil}{3}\right\rfloor\tag{2.1}\\
&=\left\lfloor\frac{\left\lfloor\frac{24n+41}{25}\right\rfloor}{3}\right\rfloor
=\left\lfloor\frac{24n+41}{75}\right\rfloor\tag{2.2}\\
&=\left\lfloor\frac{24n+39}{75}\right\rfloor\tag{2.3}\\
&\,\,\color{blue}{=\left\lfloor\frac{8n+13}{25}\right\rfloor}
\end{align*}
and the claim follows.
Comment:
*
*In (2.1) we bring $n$ into the inner floor symbol and use the rule
$$\lfloor -x\rfloor=-\lceil x\rceil$$
*In (2.2) we use the rule
\begin{align*}
\left\lceil \frac{n}{m} \right\rceil =\left \lfloor \frac{n+m-1}{m} \right\rfloor
\end{align*}
and also the nested division rule
\begin{align*}
\left\lfloor\frac{\left\lfloor x/m\right\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{mn}\right\rfloor
\end{align*}
*In (2.3) we can check for $0\leq n<25$ the validity of $\left\lfloor\frac{24n+41}{75}\right\rfloor=\left\lfloor\frac{24n+39}{75}\right\rfloor$. We can alternatively note the difference between these two expressions inside the floor symbols is $\frac{2}{75}$. So, different natural numbers would occur, iff there is a natural number $m$ with
\begin{align*}
\frac{24n+39}{75}=m-\frac{1}{75}\qquad\quad\text{and}\qquad\quad\frac{24n+41}{75}=m+\frac{1}{75}
\end{align*}
This is equivalent with the existence of a solution of the linear congruence relation
\begin{align*}
24n+41&\equiv 1\pmod{75}\\
24n&\equiv 35\pmod{75}\\
\end{align*}
But this has no solution, since
\begin{align*}
35&\not\equiv 0\pmod{\gcd(24,75)}\\
35&\not\equiv 0\pmod{3}\\
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Equation of line passing through two given lines (3D) A line from the origin meets the line $x-2\over 1$ = $y-1\over-2$ = $z+1\over1$ And $(x-\frac83)\over2$ = $y+3\over-1$ = $z-1\over1$ at P and Q respectively. If the length $PQ = d$ , then $(d)^2$ is
My attempt :
I assumed general points on the given lines, thus introducing two variables, say a and b. Then I found out the direction ratio of the required line in terms of a and b, (by subtracting the pairs of each of the three points). The required line would be a line with the already found out DR and passing through the assumed point on line 1(in terms of a). I wrote the expression for a general point on the required line by introducing a third variable , say c , and then equating each of the variables to zero as the line passes through origin. 3 equation and three variables.
But solving this is very lengthy. After which we would have to apply distance formula as well. I am looking for a smarter and shorter approach,
Thank you
This is also my first question on this site
|
Let point $P$ be on the first line, then
$P = P_0 + t V $
And let point $Q$ be on the second line, then
$Q = Q_0 + s W $
Since we want the line extending from $P$ to $Q$ to pass through the origin, then it follows that $P$ is on the plane that passes through the origin and contains $Q(s)$. Similarly, $Q$ is on the plane that passes through the origin and contains the line $P(t)$.
There is a single plane that passes through the origin and contains the first line. The equation of this plane is
$ ( P_0 \times V ) \cdot P = 0 $
We can find where the second line intersects this plane, as follows
$ (P_0 \times V) \cdot ( Q_0 + s W ) = 0 $
From this,
$ s = - \dfrac{ Q_0 \cdot ( P_0 \times V ) }{ W \cdot (P_0 \times V ) }$
And this will give us the value of $s$, thus the required Q is
$ Q = Q_0 + s W $
similarly, the value of $t$ where the first line intersects the plane passing through the origin and containing the second line is the solution of the following equation
$ (Q_0 \times W) \cdot ( P_0 + t V ) = 0 $
From this
$ t = - \dfrac{ P_0 \cdot (Q_0 \times W) }{ V \cdot (Q_0 \times W) } $
Now we have $t$ and $s$, so the distance squared between $P$ and $Q$ is
$d^2 = ( P_0 + t V - Q_0 - s W) \cdot (P_0 + t V - Q_0 - s W) $
Numerically, we first compute
$ N_1 = P_0 \times V = (2,1,-1) \times (1,-2,1) = (-1, -3 , -5 ) $
$ N_2 = Q_0 \times W = (8/3, -3, 1 ) \times (2,-1,1) = (-2, -2/3, 10/3 ) $
Therefore,
$ s = - \dfrac{ (8/3, -3, 1) \cdot (-1, -3, -5) }{ (2, -1, 1) \cdot (-1, -3, -5) } = - \dfrac{ -8/3 + 9 - 5 }{-2 + 3 - 5 } = \dfrac{1}{3} $
And
$ t = - \dfrac{ (2, 1, -1) \cdot (-2, -2/3, 10/3) }{ (1, -2, 1) \cdot (-2, -2/3, 10/3 ) } = - \dfrac{ ( -12 - 2 - 10 ) }{ -6 + 4 + 10 } = \dfrac{24}{8} = 3 $
Now
$ d^2 = \| (2, 1, -1) - (8/3, -3, 1) + 3 (1, -2, 1) - \dfrac{1}{3} (2, -1, 1) \|^2 $
This simplifies to
$ d^2 = \| (5/3 , -5/3 , 2/3) \|^2 = \dfrac{25+25+4}{9} = \dfrac{ 54 }{9} = 6 $
|
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|
Find $f:\mathbb{R} \to \mathbb{R}$ which satisfies $f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y$.
Find $f:\mathbb{R} \to \mathbb{R}$ which satisfies $f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y$.
Let $P(x, y): f(x)^2+2f(x)f(y)+y^2=\big(x-f(y)\big)^2+4f(x)y.$
$P(0, 0): f(0)^2-2f(0)f(0)=\big(0-f(0)\big)^2.$
$\therefore -f(0)^2=f(0)^2, f(0)=0.$
$P(x, 0): f(x)^2=x^2.$
$\therefore x^2+2f(x)f(y)+y^2=x^2-2xf(y)+y^2+4f(x)y.$
$\Rightarrow 2f(y)(f(x)+x)=4f(x)y, \ f(y)(f(x)+x)=2f(x)y.$
How can I disprove that $f(x)= \begin{cases} x & \text{if ****} \\ -x & \text{if ****}\end{cases}$?
|
You have shown that $\lvert f(x) \rvert = |x|$
and
$$ f(y)(f(x)+x)=2f(x)y $$
for all $x$ and $y$.
Consider whether $f(1)$ is $1$ or $-1$.
If $f(1) = 1$, then (plugging in $y = 1$)
$$ 1 \cdot (f(x) + x) = 2 f(x) \cdot 1, $$
so $f(x) = x$ for all $x$.
If $f(1) = -1$, then (plugging in $x = y = 1$)
$$ -1 \cdot (-1 + 1) = 2 \cdot (-1) \cdot 1, $$
which is false.
Therefore, the only function that satisfies the original equation is $f(x) = x$.
|
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|
Check convergence of series I want to check the convergence of the following series $$\sum_{n=0}^{\infty}\left (\frac{3+n^3}{4-n+n^2-3n^3}\right )^n \ \\ \sum_{n=0}^{\infty}\left (\frac{3+n^2}{4+2n^2}\right )^n \\ \sum_{n=1}^{\infty}\left (\sqrt{2n}-\sqrt{n}\right ) $$
I have done the following :
$$\sum_{n=0}^{\infty}\left (\frac{3+n^3}{4-n+n^2-3n^3}\right )^n \rightarrow \lim_{n\rightarrow \infty}\left |\frac{3+n^3}{4-n+n^2-3n^3}\right |=\left |-\frac{1}{3}\right |=\frac{1}{3}<1 \text{ Convergence } \\ \sum_{n=0}^{\infty}\left (\frac{3+n^2}{4+2n^2}\right )^n \rightarrow \lim_{n\rightarrow \infty}\left |\frac{3+n^2}{4+2n^2}\right |=\frac{1}{2}<1 \text{ Convergence }\\ \sum_{n=1}^{\infty}\left (\sqrt{2n}-\sqrt{n}\right ) \rightarrow \sqrt{2n}-\sqrt{n} = \sqrt{2}\sqrt{n}-\sqrt{n} =(\sqrt{2}-1)\cdot \sqrt{n} \geq \frac{1}{4}\sqrt{n} \\ \text{ since the series of } \sqrt{n} \text{ diverges we get that also the original series diverges}$$ Is everything correct ?
|
The limit of the general term is, after your work, $\lim_{n\to\infty}(\sqrt2-1)\sqrt n=\infty$.
Thus since it doesn't go to zero, the series diverges (sometimes called the divergence test).
|
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|
show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\
\iff & 1-a^2\ge b^2-ab \\
\iff &(1-a)(1+a)\ge b(b-a) \tag{*}
\\ \iff & 1+a\ge b-a
\end{align}
which is obvious. Q.E.D.
But I can't understand how we get last inequality from $(*)$.
|
Let $0\le a , b \le 1$
We prove that $$0\le \frac a{1+b}+\frac b{1+a} \le 1$$
Let $$F=\frac a{1+b}+\frac b{1+a}$$
Then $F\ge 0$, note that
$$\frac{a}{1+b} \le \frac{a}{a+b}$$
and
$$\frac{b}{1+a}\le \frac{b}{b+a}$$
Adding the last two we get
$$F\le 1$$
Equality holds when $$a=1=b.$$
|
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|
(Absolute) convergence of $\sum_{k=1}^n (-1)^k \frac{3^k+2}{4^k -6}$ We want to check if the following sequence converges (absolutely):
$$\sum_{k=1}^n (-1)^k \frac{3^k+2}{4^k -6}$$
Since we have $(-1)^k$, we can use the Leibniz criterion.
We have to prove that $a_n$ is a null sequence and that it is monotonically decreasing.
$$\lim_{k \to \infty} \frac{3^k+2}{4^k-6} = \lim_{k \to \infty} \frac{(\frac{3}{4})^k + \frac{2}{4^k}}{1 - \frac{6}{4^k}} = 0$$
To prove that it is monotonous decreasing, we have to show $a_n \geq a_{n+1}$
$$a_n \geq a_{n+1} \\ \Leftrightarrow \frac{3^k+2}{4^k-6} \geq \frac{3^{k+1} +2}{4^{k+1}-6} $$
But how do we continue from here?
|
As Lionel Ricci's comment indicates, rather than trying to use the Leibniz criterion (note that, as described in Leibniz Criterion, this criterion (also called the alternating series test) only indicates the series converges, not that it necessarily converges absolutely), instead use that, for $k \ge 2$, we have $4^k - 6 \gt 4^{k-1}$ and $3^k + 2 \lt 3^{k+1}$. Thus,
$$\frac{3^k+2}{4^k-6} \lt \frac{3^{k+1}}{4^{k-1}} = \frac{3^2(3^{k-1})}{4^{k-1}} = 3^2\left(\frac{3}{4}\right)^{k-1} \tag{1}\label{eq1A}$$
Therefore,
$$\sum_{k=1}^{\infty}\left|(-1)^k\left(\frac{3^k+2}{4^k-6}\right)\right| \lt \frac{5}{2} + \sum_{k=2}^{\infty}9\left(\frac{3}{4}\right)^{k-1} = \frac{5}{2} + \frac{9\left(\frac{3}{4}\right)}{1 - \frac{3}{4}} = \frac{59}{2} \tag{2}\label{eq2A}$$
|
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|
Express $|\pi - \frac{23}{7}|$ without the absolute value symbol
Express $\left|\pi - \dfrac{23}{7}\right|$ without the absolute value symbol.
I know I have to check if $\pi - \dfrac{23}{7}$ is greater than (or equal to) zero, but how can I do it analytically (without a calculator)?
I know that $\pi \gt 3=\dfrac{21}{7}$ but how to compare $\pi$ with $\dfrac{23}{7}$?
|
I will not use any approximations in this answer.
Consider the definite integral $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt.$$ Simply expand the numerator using binomial formula and reduce the numerator in terms of the denominator. I’ll skip a few steps for the sake of brevity: $$\int_0^1\frac{t^4(1-t)^4}{1+t^2}dt=\int_0^1\left(-4t^5+t^6+t^4+\frac{4t^6}{1+t^2} \right)dt$$$$= \int_0^1\left(-4t^5+t^6+5t^4-\frac{4t^4}{1+t^2} \right)dt= \int_0^1\left(-4t^5+t^6+5t^4-4t^2+\frac{4t^2}{1+t^2} \right)dt $$$$= \int_0^1\left(-4t^5+t^6+5t^4-4t^2+4-\frac{4}{1+t^2} \right)dt$$$$=\bbox[5px, border:2px solid red]{\frac{22}{7}-\pi.}$$ This is a very nice expression containing both $\dfrac{22}{7}$ and $\pi$.
Now, note that the function $\displaystyle f(t)= \frac{t^4(1-t)^4}{1+t^2}$ is ALWAYS positive for all $t\in (0,1)$. This means that the integral $\displaystyle\int_0^1 \frac{t^4(1-t)^4}{1+t^2} dt$ is also strictly positive. Thus, we get, $$\bbox[5px, border:2px solid gold]{\frac{22}{7}-\pi>0\implies \frac{22}{7}>\pi.}$$
Hence we finally arrive at the desired conclusion: $$\pi<\frac{22}{7}<\frac{23}{7}.$$
Thus, we can write $\Bigg |\pi-\dfrac{23}{7}\Bigg|=\dfrac{23}{7}-\pi$.
|
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How does one show this complex expression equals a natural number? We have:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3}+ \frac{7}{3 \left(\frac{10}{3^{3/2}}i-3\right)^{1/3}}=2$$
This comes from solving the cubic equation of $x^3-7x+6=0$ which factors as $(x-2)(x-1)(x+3)=0$
We can simplify the problem into finding just one part of this, namely:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3} = 1 + \frac{2}{\sqrt{3}} i$$
Now, is there any general method (which doesn't involve factorising a cubic!) in which we get from the LHS of the above equation to the RHS? Or do we simply say that the cubic formula fails in this case and we have to resort to trial-and-error factorisation to find the result?
In which case, if there is a general method, could we not express this method as another cubic formula circumvents the intermediate complex steps?
BTW, I am only interested in cases in which the solutions of the cubic are rational or real algebraic solutions (without complex sub-parts). So we might be able to use this fact in a general method.
(Some might say that even finding $27^{1/3}$ is trial-and-error in a way, since we could try numbers 1,2,3... to see which one works. But we shall ignore and just say that finding the cube root of an integer is "allowed"!)
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Write your complex numbers in polar. Then you have
$$
\left(-3+i\frac{10}{3\sqrt{3}}\right)^{1/3}=\left[\frac{7\sqrt{7}}{3\sqrt{3}}\exp\left(i\pi-i\sin^{-1}\frac{10}{7\sqrt{7}}\right)\right]^{1/3} = \frac{\sqrt{7}}{\sqrt{3}}\exp\left(\frac{i\pi}{3} - \frac{i}{3}\sin^{-1}\frac{10}{7\sqrt{7}}\right).$$
Now honestly simplifying that $\sin^{-1}$ would require solving a cubic equation. However, since we know our answer should come out to $\pi/3 - \sin^{-1}(2/\sqrt{7}) = \sin^{-1}(1/\sqrt{28})$, we can cheat and calculate
$$
\sin\left(3\sin^{-1}\frac{1}{2\sqrt{7}}\right) = \frac{3}{2\sqrt{7}}-\frac{4}{56\sqrt{7}} = \frac{10}{7\sqrt{7}},
$$
allowing us to miraculously guess the answer. Thus, $\pi/3 - \sin^{-1}(10/\sqrt{343})/3 = \sin^{-1}(2/\sqrt{7})$, and we have
$$
\frac{\sqrt{7}}{\sqrt{3}}\exp\left(\frac{i\pi}{3} - \frac{i}{3}\sin^{-1}\frac{10}{7\sqrt{7}}\right) = \frac{\sqrt{7}}{\sqrt{3}}\exp\left(i\sin^{-1}\frac{2}{\sqrt{7}}\right) = \frac{\sqrt{7}}{\sqrt{3}}\left(\frac{\sqrt{3}}{\sqrt{7}} + i\frac{2}{\sqrt{7}}\right) = 1+i\frac{2}{\sqrt{3}}.
$$
|
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|
Missing solutions from complex cubic root From Cardano's work in the 1600s, we have this famous example of a cubic polynomial equation:
$$x^3-15x-4=0.$$
Plugging the coefficients into the Cardano-Tartaglia formula, we get an expression for the solutions that reduces to:
$$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}\\=\sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}.$$
We are here dealing with cubic roots of imaginary numbers, which are known to provide three solutions. In fact, all three solutions to this particular equation are real.
Now to my question. I could replace the imaginary numbers under the radicals with their polar forms and solve for all three solutions by equating arguments and modulii, respectively, from the subsequent cubic equations that can be extracted. But, when trying out a purely algebraic route, only one solution pops out with the other two solutions still "hidden":
$$x=\sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}\\=\sqrt[3]{(2+i)^3}+\sqrt[3]{(2-i)^3}\\=2+i+2-i\\ =4$$
In this latter approach, where does the issue arrise that causes two solutions to not appear?
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$x = \sqrt[3]{2+i\sqrt{121}}+\sqrt[3]{2-i\sqrt{121}}$
This is misleading, because a non-zero complex number has $3$ cube roots, and the above makes it look like $x$ could take $9$ values between the different combinations of the two cube roots.
The better way to write the cubic formula is $x = \sqrt[3]{2+i\sqrt{121}} + \dfrac{5}{\sqrt[3]{2+i\sqrt{121}}}$ where the two cube roots are always chosen to be the same. ( The $5$ in the numerator comes from $\left|2 + i \sqrt{121}\right|^2 = 2^2 + 121 = 125 = 5^3$. )
$\sqrt[3]{2+i\sqrt{121}}\;$ [...] $\; =\sqrt[3]{(2+i)^3}\;$ [...] $\; = 2+i\;$ [...]
$(2+i)$ is indeed one of the cube roots of $\sqrt[3]{2+i\sqrt{121}}$, and the other two are $\omega(2+i)$ and $\omega^2(2+i)$ where $\omega = \dfrac{-1 \pm i \sqrt{3}}{2}$ is a primitive cube root of unity. Then the three solutions are:
$$
x = \omega^k(2+i) + \frac{5}{\omega^k (2+i)} = \omega^k(2+i) + \overline{\omega}^k (2 - i) = 2 \text{Re}\left(\omega^k (2+i)\right) \;\;\Big|\;\; k = 0, 1, 2
$$
|
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pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $()^4$.
I tried to rewrite $8x^4=(2x)^4\times\frac{1}{2}$, but It wasn't helpful :
$$\left(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\right)((2x)^4+(3y)^4+(4z)^4)\ge (x+y+z)^4$$
So, is there any nice transform to apply C-S inequality?
or should I apply Lagrange Multiplier?
|
Apply AM-GM inequality repeatedly three times:
$8x^4+\dfrac{243}{2}=8x^4 + \dfrac{81}{2}+\dfrac{81}{2}+\dfrac{81}{2} \ge 4\sqrt[4]{8x^4\cdot \dfrac{81}{2}\cdot \dfrac{81}{2}\cdot \dfrac{81}{2}}=4\sqrt[4]{(27x)^4}= 4|27x|\ge 27(4x)$.
$27y^4 + 81 = 27y^4+ 27+27+27 \ge 4\sqrt[4]{27y^4\cdot 27\cdot 27\cdot 27}=4|27y| \ge 27(4y)$.
$64z^4 + \dfrac{243}{4} = 64z^4 + \dfrac{81}{4}+\dfrac{81}{4}+\dfrac{81}{4} \ge 4\sqrt[4]{64z^4\cdot \dfrac{81}{4}\cdot \dfrac{81}{4}\cdot \dfrac{81}{4}}=4|27z| \ge 27(4z)$.
Now add these three inequalities side by side, we obtain:
$8x^4+27y^4+64z^4 + \dfrac{81\cdot 13}{4} \ge 27(4x+4y+4z) = 27\cdot 13\implies 8x^4+27y^4+64z^4 \ge \dfrac{27\cdot 13}{4}=\dfrac{351}{4}$. This is also the minimum value of the expression $8x^4+27y^4+64z^4$. Equality occurs when: $8x^4 = \dfrac{81}{2}, 27y^4 = 27, 64z^4 = \dfrac{81}{4}\implies x = \dfrac{3}{2}, y = 1, z = \dfrac{3}{4}$
|
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How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then decomposed it into partial fractions and ended up with
$40+ \sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}}$. I know the answer is 40 from WolframAlpha, however I don't know how to go about showing that the sum with the complex numbers is 0.
How should I proceed?
|
Notice that :
$$S = \sum_{k = 1}^{80} \dfrac{k^2}{81^2 + 2 k^2 - 2 \times 81 \times k} = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2}$$
and :
$$\sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} = \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2}$$
which allows us to deduce that :
$$2 S = \sum_{k = 1}^{80} \dfrac{k^2}{(81 - k)^2 + k^2} + \sum_{k = 1}^{80} \dfrac{(81 - k)^2}{k^2 + (81 - k)^2} = \sum_{k = 1}^{80} \dfrac{k^2 + (81 - k)^2}{k^2 + (81 - k)^2} = \sum_{k = 1}^{80} 1 = 80$$
Hence :
$$S = 40$$
|
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Radius of convergence opposite result I am trying to find the radius of convergence of the series $\sum_{n=1}^\infty{\frac{(n!)^3 \cdot x^n}{(3n)!}}$. I know that it should be $|x|<27$, but i get the exactly the opposite result ($27<|x|$). I am using the ratio test to determine it (e.g. $\rho = \lim_{n\to\infty} \left|\frac{a_n}{a_{n+1}}\right|$).
My calculations are as follows:
$$
\lim_{n\to\infty} \left|\frac{\frac{(n!)^3 \cdot x^n}{(3n)!}}{\frac{((n+1)!)^3 \cdot x^{n+1}}{(3(n+1))!}} \right|
= \left| \lim_{n\to \infty} \frac{(n!)^3 \cdot x^n}{(3n)! \cdot \frac{((n+1)!)^3 \cdot x^{n+1}}{(3(n+1))!}} \right|
=\lim_{n \to \infty} \left| \frac{\frac{(n!)^3 \cdot x^n}{\frac{((n+1)!)^3 \cdot x^{n+1}}{(3(n+1))!}}}{(3n)!} \right|
= \lim_{n\to\infty} \left| \frac{\frac{(n!)^3 \cdot x^n \cdot (3(n+1))!}{((n+1)!)^3 \cdot x^{n+1}}}{(3n)!} \right|
= \lim_{n\to \infty} \left| \frac{(n!)^3 \cdot x^n \cdot (3(n+1))!}{((n+1)!)^3 \cdot x^{n+1} \cdot (3n)!} \right|
= \lim_{n\to\infty} \left| \frac{(n!)^3 \cdot (3(n+1))!}{((n+1)!)^3 \cdot x \cdot (3n)!} \right|
= \lim_{n\to\infty} \left| \frac{(n!)^3 \cdot (3n+3)!}{((n+1)!)^3 \cdot x \cdot (3n)!} \right|\\
= \lim_{n\to\infty} \left| \frac{(n!)^3 \cdot (3n+3)\cdot(3n+2)\cdot(3n+1)}{((n+1)!)^3 \cdot x} \right|
= \lim_{n\to\infty} \left| \frac{(3n+3)\cdot(3n+2)\cdot(3n+1)}{(n+1)^3 \cdot x} \right|\\
= \lim_{n\to\infty} \left| \frac{27 n^3 + 54 n^2 + 33 n + 6}{n^3x + 3 n^2x + 3 nx + 1x} \right|\\
\overset{\cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}}}{=} \lim_{n\to\infty} \left| \frac{27 + \frac{54}{n} + \frac{33}{n^2} + \frac{6}{n^3}}{x + \frac{3x}{n} + \frac{3x}{n^2} + \frac{x}{n^3}} \right|\\
\Rightarrow \frac{27}{|x|} < 1 \Rightarrow 27 < |x|
$$
Where did I go wrong? Did I somehow mess up the direction of the inequality?
|
There shouldn't be any $x$'s. Then you will get $r=\lim\dfrac{ \mid a_n\mid}{\mid a_{n+1}\mid}=27$.
That's you use the $a_n$ in $$\sum a_nx^n$$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating $\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$ I need to find the summation
$$S=\sum_{r=0}^{1010} \binom{1010}r \sum_{k=2r+1}^{2021}\binom{2021}k$$
I tried various things like replacing $k$ by $2021-k$ and trying to add the 2 summations to a pattern but was unable to find a solution. For more insights into the question, this was essential to solve a probability question wherein there were 2 players, A and B. A rolls a dice $2021$ times, and B rolls it $1010$ times. We had to find the probability of A having number of odd numbers more than twice of B. So if B had $r$ odd numbers, A could have odd numbers from $2r+1$ to $2021$, hence the summation. I can get the required probability by dividing this by $2^{2021+1010}$.
|
For $N=1010$ or others
\begin{equation}
S_N\equiv \sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
=
\end{equation}
\begin{equation}
\sum_{r=0}^N\binom{N}{r}\sum_{k=2r+1}^{2N+1}\binom{2N+1}{k}
+
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2N+1}\binom{2N+1}{k}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
\sum_{r=0}^N\binom{N}{r}2^{2N+1}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
2^N 2^{2N+1}
-
\sum_{r=0}^N\binom{N}{r}\sum_{k=0}^{2r}\binom{2N+1}{k}
\end{equation}
\begin{equation}
=
2^{3N+1}
-
% \sum_{k=0}^{2N}\sum_{r=k}^{N}
\sum_{r=0}^{N}\sum_{k=0}^{2r}
\binom{N}{r}\binom{2N+1}{k}
\end{equation}
substituing $r'=N-r$
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k=0}^{2N-2r'}
\binom{N}{N-r'}\binom{2N+1}{k}
\end{equation}
and substituting $k'=2N+1-k$
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k'=2N+1}^{2N+1-(2N-2r')}
\binom{N}{N-r'}\binom{2N+1}{2N+1-k'}
\end{equation}
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=N}^{0}\sum_{k'=2N+1}^{1+2r'}
\binom{N}{r'}\binom{2N+1}{k'}
\end{equation}
\begin{equation}
=
2^{3N+1}
-
\sum_{r'=0}^{N}\sum_{k'=2N+1}^{1+2r'}
\binom{N}{r'}\binom{2N+1}{k'}
=
2^{3N+1}-S.
\end{equation}
Adding $S$ to both sides gives $2S=2^{3N+1}$, therefore
\begin{equation}
S=2^{3N}.
\end{equation}
|
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"url": "https://math.stackexchange.com/questions/4508840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Parameterizing the parabola $9x^2 +y^2-6xy+4x-4y+1=0$
Find parametrization of curve given by equation:
$$9x^2 +y^2-6xy+4x-4y+1=0$$
My attempt:
Let's notice that \begin{split} 9x^2 +y^2-6xy+4x-4y+1=0 & \iff (3x)^2 -6xy + y^2 +4x -4y +1=0\\
& \iff (3x-y)^2 +6x -2y + 1 -2x -2y=0\\
& \iff (3x-y)^2 +2(3x -y) + 1^2 -2x -2y=0\\
& \iff (3x-y+1)^2-2(x+y)=0.\end{split}
So this look like some crazy parabola, right? But I don't know how to derive parametric equation of this?
|
One possible parametrization can be obtained by setting $u = 3x-y+1$ and $v = 2(x + y)$, so that
$$
\begin{cases}
u = 3x - y + 1 \\
v = 2(x + y)
\end{cases} \Longleftrightarrow
\begin{cases}
x = \frac{1}{8} \left(2u + v - 2 \right)\\
y = \frac{1}{4} \left(-2u + 3v + 2 \right)
\end{cases}
$$
and $9x^2 +y^2-6xy+4x-4y+1=0 \Leftrightarrow u^2 = v$. New parabola can be trivially parametrized as $(u, v) = (t, t^2)$ for $t \in \mathbb{R}$, which leads to
$$
\begin{cases}
x = x(t) = \frac{1}{8}\left(2t + t^2 - 2 \right), \\
y = y(t) = \frac{1}{8}\left(-2t + 3t^2 + 2 \right), \\
t \in \mathbb{R}.
\end{cases}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$
The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests a promising method of finding a polynomial close to $\arctan$
*
*divide $1$ by $1+t^2$, to obtain a polynomial plus a remainder:
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}\tag{1}$$
How exactly do we perform this division of $1$ by $1+t^2$ to obtain $(1)$?
|
I haven't read the book but if the author has already introduced polynomial long division or synthetic division, then either one can be used to evaluate even "weird-looking" fractions like
$$\frac 1{1+t^2}$$
For division of polynomials where the denominator has a degree greater than one, then we neglect the highest order term and invert the signs of the lower order terms all the way down to the constant. That is, since the denominator in this instance is $t^2+1$, ignore the highest order term $t^2$ and invert the signs, meaning our vertical column will be $0$ (coefficient of $t$) and $-1$ (constant term).
To synthetically divide the two terms, we first rewrite the numerator in terms of higher order powers. Since there is only a constant term, this becomes really trivial.
$$1=1+\sum\limits_{k=0}^n0t^k=1+0t+0t^2+0t^3+\cdots+0t^n$$
Our numerator goes in the top row of our division table and our denominator along the left-hand side. For this example, I will use the case of $n=8$ but you can easily extend this to any $n$ by adding more zeros in the numerator.
\begin{array}{c|ccccccccc} & t^0 & t & t^2 & t^3 & t^4 & t^5 & t^6 & t^7 & t^8\\ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & & & & & & & & & &\\-1 & & & & & & & & & &\\\hline\end{array}
The first step is to bring down the first term from our numerator like so
\begin{array}{c|ccccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\color{blue}{0} & & \color{blue}{\star} & & & & & & & &\\-1 & & & & & & & & & &\\\hline & \color{blue}{1}\end{array}
Next, multiply the two blue terms together and put the result where the $\color{blue}{\star}$ is positioned (second column). Finally, sum the second column vertically and write the sum in the second position. Your table should now look like
\begin{array}{c|ccccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\color{blue}{0} & & 0 & \color{blue}{\star} & & & & &\\\color{red}{-1} & & & \color{red}{\star} & & & & & &\\\hline & \color{red}{1} & \color{blue}{0}\end{array}
In a similar manner as before, multiply the two blue numbers together and place the result in the third column where $\color{blue}{\star}$ is, and repeat with the red numbers and $\color{red}{\star}$. Sum the third column and write the result in the third column below the horizontal line.
\begin{array}{c|ccccccccc} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\\color{blue}{0} & & 0 & 0 & \color{blue}{\star} & & & &\\\color{red}{-1} & & & -1 & \color{red}{\star} & & & & &\\\hline & 1 & \color{red}{0} & \color{blue}{-1}\end{array}
Continue this process until you reach the end. The final result should look something like this:
\begin{array}{c|ccccccccc} & t^0 & t & t^2 & t^3 & t^4 & t^5 & t^6 & t^7 & t^8\\ & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\0 & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\-1 & & & -1 & 0 & 1 & 0 & -1 & 0 & 1\\\hline & 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & \color{red}{1}\end{array}
The bottom row of numbers is the quotient, starting off with the constant term and increasing in power as we move across towards the right. The final term highlighted in red is the remainder of the division. In this case, we have that
$$\frac 1{1+t^2}=1-t^2+t^4-t^6+\frac {t^8}{1+t^2}$$
If we continue, then for a general $n$, it seems that
$$\frac 1{1+t^2}=1-t^2+t^4-t^6+\cdots+\frac {(-1)^{n+1}t^{2n+2}}{1+t^2}$$
Next step would be to use induction to prove that this formula holds true for all $n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43$$ Or in general, where there are $n$ variables can we say that their sum (in the same format as above) is no less than $\frac{n}{n-1}$$?$
All the variables used here are positive real numbers.
My thoughts:
I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.
|
This answer is a generalization of the method used in the first answer given above.
Problem:
Prove that:
$$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$
Proof:
Using AM-GM inequality we have,
$$\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
\implies &\sum_{1\le i\le n} x_i^2\ge \frac 2{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
\implies &\left(\sum_{1\le i\le n} x_i\right)^2\ge\left( 2+\frac 2{n-1}\right)\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
&\qquad\qquad\quad =\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\end{align}$$
Finally, applying Cauchy-Schwars, we obtain
$$\begin{align}\left(\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\right)\left(\sum_{1\le j\le n}^n \left({x_j}{\sum_{1\le i\le n} x_i-x_j}\right)\right)&\ge \frac{ \left(\sum_{1\le j\le n} x_i\right)^2}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\
&\ge \frac {\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\
&=\boxed {\frac{n}{n-1}.}\end{align}$$
|
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"url": "https://math.stackexchange.com/questions/4511844",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Find the radius of the circle cutting sides of equilateral triangle Consider the figure below. The triangle is equilateral. Find the radius of the circle.
My try:
I have dropped perpendiculars from the center of the circle to respective chords naming them $x,y,z$ and named the other line segments as $m,n,p,q,s,t$ as shown below.
By Viviani's theorem we have
$$x+y+z=\frac{\sqrt{3}a}{2}$$ Where $a$ is the side length of the triangle.
By Chords theorem we have:
$$\begin{aligned}
m(m+5) &=t(t+3) \\
q(q+4) &=s(s+3) \\
p(p+4) &=n(n+5) \\
m+n+5 &=p+q+4=s+t+3=a
\end{aligned}$$
Also by Pythagoras theorem if $R$ is the radius of the circle we have:
$$R^{2}-x^{2}=\frac{9}{4}, \quad R^{2}-y^{2}=4 ,\quad R^{2}-z^{2}=\frac{25}{4}$$
I guess this is a very tedious task? Any better ways?
|
If you set up a Cartesian coordinate system with its origin at the center of the circle and $BC$ as a horizontal line beneath it, then the coordinates of the chord endpoints work out to:
$$G = \left( - \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$T = \left( \frac{5}{2}, - \sqrt{R^2 - \frac{25}{4}} \right)$$
$$N = \left(1 + \frac{\sqrt{3}}{2}\sqrt{R^2 - 4}, \frac{1}{2}\sqrt{R^2 - 4} - \sqrt{3} \right)$$
$$M = \left(-1 + \frac{\sqrt{3}}{2}\sqrt{R^2 - 4}, \frac{1}{2}\sqrt{R^2 - 4} + \sqrt{3}\right)$$
$$E = \left(\frac{3}{4} - \frac{\sqrt{3}}{2} \sqrt{R^2 - \frac{9}{4}}, \frac{1}{2}\sqrt{R^2 - \frac{9}{4}} + \frac{3\sqrt{3}}{4} \right)$$
$$F = \left(-\frac{3}{4} - \frac{\sqrt{3}}{2} \sqrt{R^2 - \frac{9}{4}}, \frac{1}{2}\sqrt{R^2 - \frac{9}{4}} - \frac{3\sqrt{3}}{4}\right)$$
From these, we can find equations for the lines that form the three sides of the triangle.
$$\overline{AB} = \overline{MN} : y = -\sqrt{3}x + 2\sqrt{R^2 - 4}$$
$$\overline{AC} = \overline{EF} : y = \sqrt{3}x + 2\sqrt{R^2 - \frac{9}{4}}$$
$$\overline{BC} = \overline{GT} : y = -\sqrt{R^2 - \frac{25}{4}}$$
From the intersections between those three lines, we find the triangle's vertices.
$$A = \overline{AB} \cap \overline{AC} = \left( \frac{1}{\sqrt{3}}\sqrt{R^2 - 4} - \frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{9}{4}}, \sqrt{R^2 - 4} + \sqrt{R^2 - \frac{9}{4}} \right)$$
$$B = \overline{AB} \cap \overline{BC} = \left( \frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{25}{4}} + \frac{2}{\sqrt{3}}\sqrt{R^2 - 4}, -\sqrt{R^2 - \frac{25}{4}} \right)$$
$$C = \overline{AC} \cap \overline{BC} = \left(-\frac{1}{\sqrt{3}}\sqrt{R^2 - \frac{25}{4}} - \frac{2}{\sqrt{3}}\sqrt{R^2 - \frac{9}{4}}, -\sqrt{R^2 - \frac{25}{4}} \right)$$
And by applying the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to any two of these points, the common side length $a$ of the equilateral triangle is:
$$a = \frac{2}{\sqrt{3}}\left( \sqrt{R^2 - \frac{9}{4}} + \sqrt{R^2 - 4} + \sqrt{R^2 - \frac{25}{4}} \right)$$
But this is just what OP already found by Vivani's theorem, and adds no new information.
As indicated by @math for entertainment's answer, there isn't a unique solution by $R$. We need one more piece of information to find it.
Assuming that information is the side length $a$, then we'll need to solve the above radical equation for $R$.
|
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|
How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$? I can't use l'hopital, so change of variable is the only way.
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$
|
We have: $\dfrac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}=\dfrac{\dfrac{x^2}{\sqrt{x^2+p^2}+p}}{\dfrac{x^2}{\sqrt{x^2+q^2}+q}}=\dfrac{\sqrt{x^2+q^2}+q}{\sqrt{x^2+p^2+p}}\to \dfrac{2q}{2p} = \dfrac{q}{p}$ as $x \to 0$.
|
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|
Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$
This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve the problem. I've tried the following: $$\begin{align} 4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ})&=16\sin^2(12^{\circ})\cos^2(12^{\circ})+8\sin^2(12^{\circ})
\cos(12^{\circ})\\
\\
&=8\sin^2(12^{\circ})\cos(12^{\circ})\Big(2\cos(12^{\circ} ) + 1\Big)\end{align}$$
As you can see, I was trying to simplify the expression so that it'll contain only $\sin(12^{\circ})$ and $\cos(12^{\circ})$, since I thought by unifying the angles I would have a bigger chance of solving it. However, I couldn't find a way to make any further progress. Can someone show me the way?
|
From DeMoivre's Theorem:
$$(\cos(12°) + i \sin(12°))^5 = \cos(60°) + i\sin(60²) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
For brevity, let $c = \cos(12°)$ and $s = \sin(12°)$.
$$(c + is)^5 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
$$c^5 + 5isc^4 - 10s^2c^3 - 10is^3c^2 + 5s^4c + is^5 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
Splitting this equation into separate real and imaginary parts gives:
$$c^5 - 10s^2c^3 + 5s^4c = \frac{1}{2}$$
$$5sc^4 - 10s^3c^2 + s^5 = \frac{\sqrt{3}}{2}$$
But we can use the Pythagorean identity $s^2 + c^2 = 1$ to write each equation in terms of only one variable.
$$c^5 - 10(1-c^2)c^3 + 5(1-c^2)^2c = \frac{1}{2}$$
$$5s(1-s^2)^2 - 10s^3(1-s^2) + s^5 = \frac{\sqrt{3}}{2}$$
Or, after expanding and combining terms:
$$16c^5 - 20c^3 + 5c - \frac{1}{2} = 0$$
$$16s^5 - 20s^3 + 5s - \frac{\sqrt{3}}{2} = 0$$
Now, just plug the coefficients into the Quintic Formula, and...oh, wait, there isn't one.
Fortunately, we can use the Rational Root Theorem to determine that $c = \frac{1}{2}$ is a solution to the first question. This can't be the actual value of $\cos(12°)$ because $\cos$ is decreasing on $[0, 180°]$ and so $\frac{\sqrt{3}}{2} = \cos(30°) < \cos(12°) < \cos(0) = 1$, but it does mean that we can factor out $c - \frac{1}{2}$ to reduce the equation to a quartic.
$$16c^4 + 8c^3 - 16c^2 - 8c + 1 = 0$$
Which AFAICT doesn't factor, but maybe we don't need to find an exact value for $c$, as the value that we ultimately want is:
$$4\sin^2(24°)+4\sin(24°)\sin(12°)$$
$$= 4(2cs)^2 + 4(2cs)s$$
$$=16c^2s^2 + 8cs^2$$
$$=16c^2(1-c^2) + 8c(1-c^2)$$
$$=16c^2 - 16c^4 + 8c - 8c^3$$
$$=- 16c^4 - 8c^3 + 16c^2 + 8c$$
$$=-(16c^4 + 8c^3 - 16c^2 - 8c)$$
$$=-(-1)$$
$$=1$$
Gee, it sure was lucky that the target expression happened to work out to be almost the same polynomial we found earlier.
|
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|
Show for $0 < A < B, \int_{-\infty}^\infty \frac{A^2 - A + B^2 - B + (A + B - 2AB)\cos(v) + v^2}{(A^2-2AB\cos(v)+B^2+v^2)^2}dv > 0$. The integral in question doesn't have a closed form solution as far as I know, but it does converge. The denominator is clearly positive, and the numerator of the integrand is always positive, except in a very narrow situation, which I will describe:
We have that $A^2 + B^2 - 2AB\cos(v) \geq (B-A)^2 > 0$. Furthermore, we have $\cos(v) \geq 1 - \frac{v^2}{2}$ by the Taylor formula, so we also have $(A + B)(\cos(v) - 1) + v^2 \geq (A + B)(- \frac{v^2}{2}) + v^2 = v^2(1 - \frac{A+B}{2})$. So only in the limited scenario where $\frac{A+B}{2} > 1$ could the integrand be negative. Now I'm stuck trying to prove this would not be enough to force the entire integral to be negative.
This inequality is part of a conjecture I have, so there is a chance it could be wrong!
|
If $A + B - 2AB \le 0$, then
$$\mathrm{NUM} \ge A^2 - A + B^2 - B + (A + B - 2AB) = (A - B)^2 > 0.$$
If $A + B - 2AB > 0$ and $A + B \ge 2$, then
$$\mathrm{NUM}
\ge A^2 - A + B^2 - B - (A + B - 2AB) = (A + B)(A + B - 2) \ge 0 .$$
If $A + B - 2AB > 0$ and $A + B < 2$, using $\cos v \ge 1 - v^2/2$, then
\begin{align*}
\mathrm{NUM} &\ge A^2 - A + B^2 - B + (A + B - 2AB)(1 - v^2/2) + v^2\\
&= \frac12(2AB + 2 - A - B)v^2 + (A - B)^2\\
&\ge 0.
\end{align*}
|
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|
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to $a+b+c=1$ and $a,b,c>0$
Minimize $f(a,b,c) = \dfrac{1}{a²+b²+c²} +\dfrac{1}{9abc}$ subject to
$a+b+c=1$ and $a,b,c>0$
My attempt:
Since $f(a,b,c) \ge 0$, then we can minimizing the addends individually. Furthermore, minimizing fractions is the same as maximizing their denominators.
Apply the Lagrange Multiplier on ${a²+b²+c²}$ with the contraints $\implies a=b=c=\frac{1}{3}.$
Similarly, apply the Lagrange Multiplier on ${9abc}$ with the contraints $\implies a=b=c=\frac{1}{3}.$
Therefore, the minimum of $f\!\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)=6.$
I'm looking for a better solution not using the Lagrange Multiplier.
|
Let's try this with AM-GMs. First note that as $(x+y+z)^2\geqslant 3(xy+yz+zx)$,
$$(ab+bc+ca)^2 \geqslant 3abc(a+b+c)=3abc $$
$\begin{align*}
\implies 9abc(a^2+b^2+c^2) & \leqslant 3(a^2+b^2+c^2)(ab+bc+ca)^2 \\
&\leqslant 3\left(\frac{a^2+b^2+c^2+2(ab+bc+ca)}3 \right)^3 \\
&=\frac19(a+b+c)^6=\frac19
\end{align*}$
Hence
$$f \geqslant \frac2{\sqrt{(a^2+b^2+c^2)(9abc)}}\geqslant \frac2{1/3}=6$$
with equality iff $a=b=c=\frac13$.
|
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|
Limit of $u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k !$ I encountered a sequence $(u_n)_{n \in \mathbb{N}} $ defined as
$$ u_{n}=\frac{1}{n !} \sum_{k=0}^{n} k ! $$
And I wonder what is the limit. It seems to be 1 but even Wolfram Alpha cannot figure it out.
My first idea was to write $$ \frac{k!}{n!} = \frac{1}{(k+1)...(n-1)n}$$ and
$$ (k+1)^{n-k} \leq (k+1)...(n-1)n \leq n^{n-k} $$
EDIT : I found a solution but Yanko found something more elementary.
My solution :
\begin{align}
u_n &= 1 + \frac{1}{n} + \frac{1}{n(n-1)} + \frac{1}{n(n-1)(n-2)} + ... + \frac{1}{n!} \\
&= 1 + \frac{1}{n} + (n-1) \times o(\frac{1}{n^2}) \\
&= 1 + \frac{1}{n} + o(\frac{1}{n}) \\
\end{align}
Thus all terms converge towards 0 except the first one, and the limit is 1.
|
Write it as
$$\frac{1}{n!} \sum_{k=0}^n k! = \frac{n!}{n!} + \frac{(n-1)!}{n!}+...+\frac{1!}{n!} = 1+\frac{1}{n}+\frac{1}{n(n-1)}+...+\frac{1}{n!}$$
Now this is clearly greater or equal to $1$. On the other hand the last $(n-1)$ terms (all except the first two) are smaller than $\frac{1}{n(n-1)}$ so we have an upper bound of
$$1+\frac{1}{n} + \frac{1}{n} = 1+\frac{2}{n}$$
which converges to $1$.
Use the Sandwich theorem.
|
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Argue on elementary grounds that the average value of $x/y$ exceeds 1 if x and y are chosen randomly between 1 and 2 Textbook problem:
Two numbers, first $x$ and then $y$, are chosen at random between $1$ and $2$. What is the average value of the quotient $\frac{x}{y}$? Can you argue on elementary grounds that the answer must exceed 1?
My answer: If $x$ has been picked, then the average of the quotient is
$$
\int_1^2 \frac{x}{y}\,dy = x\ln(2)
$$
Averaging over possible values of $x$ yields
$$
\int_1^2 x\ln(2)\, dx = \frac{3}{2}\ln(2) \approx 1.03
$$
But this is not an argument based on elementary grounds I don't think.
Interpreting the average value of a function over an interval as being the height of the rectangle with base on the interval and with the same area as the area under the function over the interval leads me to the following argument.
The average value of $x$ is $3/2$. The quotient $x/y$ then ranges from $3/4$ to $3/2$. We can underestimate the area under this curve by splitting it up into a rectangle with length $1$ and height $3/4$ and a right triangle on top of the rectangle whose hypotenuse is the tangent line to the curve $\frac{3}{2y}$ at the point $(3/2,1)$. This gives a triangle with height $7/12$ and base $7/8$.
So an underestimate for the area under the curve would be $\left(1\cdot \frac{3}{4}\right) + \left(\frac{1}{2}\cdot \frac{7}{8}\cdot \frac{7}{12}\right) = \frac{193}{192}$.
Here is an illustration of my quotient function with $x = 3/2$, the tangent line, and the rectangle:
Since the base of the rectangle with the same area on $[1,2]$ has length $1$ its height must be at least $\frac{193}{192}$ to match the area under the graph. Hence, the height, or average value, exceeds $1$.
Question: Does this seem to be an argument the author could be looking for instead of integrating?
This is a single variable calculus text. The section is ''The Average Value of a Function". Note that not until the next chapter is probability introduced.
Update: I have obtained a copy of the author's own solutions manual. Here is the official solution:
To argue on elementary grounds that the answer must exceed 1, note that the outcomes $a/b$ and $b/a$ must occur equally often, so that the answer is certainly greater than the minimum over all $a$ and $b$ in $[1,2]$ of the average of these two numbers. Now this average is
$$
\frac{1}{2}\left[\frac{a}{b} + \frac{b}{a}\right] = \frac{a^2 + b^2}{2ab}
$$
and this is always greater than 1 because $a^2 + b^2 - 2ab = (a-b)^2 > 0$.
|
We can argue as follows. The probability distribution is symmetric w.r.t. interchanging $x$ and $y$. This means that we can compute the expectation value by restricting $x$ to be larger than or equal to $y$ and computing the expectation value of $f(x,y) = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}\right)$ over this modified probability distribution.
The fact that the expectation value exceeds $1$ then follows from the fact that the function $f(x,y)$ is larger than or equal to $1$:
$$f(x,y) - 1 = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}-2\right) = \frac{x^2+y^2-2xy}{2xy} = \frac{(x-y)^2}{2 xy}\geq 0$$
The function $f(x,y)$ is then equal to $1$ when $x = y$ which occurs on a subset of measure zero, so the expectation value is clearly larger than $1$.
|
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|
$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $ Find all x's
It is given that $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $$ Find all possible values of $x$.
What I did: $$\frac{1}{\log_2(x-2)}+\frac{1}{\log_2(x+2)}=\frac{5}{6} $$ $$\log_{(x-2)}2 +\log_{(x+2)}2 =\frac{5}{6}$$
What is the next ? I stuck in here..
I am looking for algebraic approach. The answer is $+6,-6$
|
Ah you meant $\log_2((x-2)^2)$ not $(\log_2(x-2))^2$? That wasn't clear to me. Are you sure that its +- 6? Wolframalpha gives completely different answers, no matter whether I use $\log_2((x-2)^2)$ or $(\log_2(x-2))^2$.
This might help you a bit, but don't expect really nice answers unless you've made a mistake in the phrasing of your question.
$$\frac{1}{\log_2((x-2)^2)} + \frac{1}{\log_2((x+2)^2)} = \frac{5}{12}$$
$$\frac{1}{2\log_2(x-2)} + \frac{1}{2\log_2(x+2)} = \frac{5}{12}$$
$$\frac{1}{\log_2(x-2)} + \frac{1}{\log_2(x+2)} = \frac{5}{6}$$
|
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|
Simplifying $\sqrt{a^{4 n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}$ as far as possible, using only positive integers So far here is my steps but I am unsure of how to do the final step
$$\sqrt{a^{4 n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}$$
$$\left(a^{4 n}\right)^{\frac{1}{2}} \cdot \frac{3 a}{a^{n}} \cdot a^{1+n} \cdot a \cdot \frac{(2 a)^{3 n}}{2 a}
$$
by canceling the a's we therefore get
$$ \frac{a^{2 n} \cdot 3 \cdot 2^{3 n} \cdot(a)^{3 n}}{2} $$
How would you then express this in all positive integers
|
Just to clarify the comments, I have a full worked solution provided.
\begin{align}
&\quad\sqrt{a^{4n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}\\
&=\left(a^{4n}\right)^{\frac{1}{2}}\cdot3\cdot\frac{1}{a^n}\cdot a^{1+n}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=a^{\frac{4n}{2}}\cdot3\cdot a^{-n}\cdot a^{n+1}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=3\cdot a^{2n}\cdot a^{-n}\cdot a^{n+1}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=3\cdot2^{3n-1}\cdot a^{2n-n+(n+1)+(3n-1)}\\
&=3\cdot2^{3n-1}\cdot a^{2n-n+n+1+3n-1}\\
&=3\cdot2^{3n-1}\cdot a^{5n}\\
&=\frac{3}{2}a^{5n}\cdot2^{3n}
\end{align}
|
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|
How can we simplify the limit of the angle at a vertex of a traingle as the triangle gets arbitrarily small? (the limit involves the law of cosines) Let $f$ be a mapping $\mathbb{R} \implies \mathbb{R}$
Let $\forall x, r \in \mathbb{R}$ let $T(x, r)$ be the triangle whose three vertices are the following points:
*
*$(x-r, f(x-r))$
*$(x+0, f(x+0))$
*$(x+r, f(x+r))$
Let $\theta(x, r)$ be the the interior angle at point $(x, f(x))$ on triangle $T(x, r)$.
Let $\Theta(x)$ be the limit of $\theta(x, r)$ as $r$ approaches $0$.
Question: How can we simplify the following formula?
$\begin{align} \Theta(x) &= \lim \limits_{r \to 0} \theta(x, r) \\
&= \lim \limits_{r \to 0} \cos^{-1}\begin{pmatrix} \dfrac{-(r^{2}) + a^{2}(x,r) + b^{2}(x, r)}{2*a(x,r)*b(x,r)} \end{pmatrix} \\
\end{align}$
where
$\qquad \qquad a(x, r) = \sqrt{r^{2}+ \begin{pmatrix}f(x)-f(x-r)\end{pmatrix}^{2}} $
$\qquad \qquad b(x, r) = \sqrt{r^{2}+ \begin{pmatrix}f(x)-f(x+r)\end{pmatrix}^{2}}$
The law of cosines was used here.
A Remark
If $f$ is differentiable then $\forall x \in \mathbb{R}, Θ(x) = 180º$
However, I am interested in calculating the angle when $f$ represents a polygonal-chain
|
Define $A = \frac{f(x) - f(x - r)}{r}$ and $B = \frac{f(x + r) - f(x)}{r}$. If $f$ is differentiable, then $A$ and $B$ both converge to $f'(x)$ as $r \rightarrow 0$. But this generalization allows for piecewise functions that "turn the corner" at $(x, f(x))$. Anyhow, if we make the substitutions
$$f(x - r) = f(x) - rA$$
$$f(x + r) = f(x) + rB$$
in your definitions of functions $a$ and $b$, then these functions simplify to:
$$a(x, r) = r \sqrt{1 + A^2}$$
$$b(x, r) = r \sqrt{1 + B^2}$$
Plug these into your formula for $\cos(\theta)$.
$$\cos(\theta(x, r)) = \dfrac{-r^{2} + a^{2}(x,r) + b^{2}(x, r)}{2 a(x,r) b(x,r)}$$
$$= \dfrac{-r^{2} + r^2(1 + A^2) + r^2(1 + B^2)}{2 r \sqrt{1 + A^2} r \sqrt{1 + B^2}}$$
$$= \dfrac{r^2 + r^2A^2 + r^2B^2}{2 r^2 \sqrt{(1 + A^2)(1 + B^2)}}$$
$$= \dfrac{1 + A^2 + B^2}{2\sqrt{(1 + A^2)(1 + B^2)}}$$
If $f$ is differentiable at $x$, then $A = B = f'(x)$, and the above expression simplifies to $1 - \frac{1}{2(1+f'(x)^2)}$. Otherwise,
$$\Theta(x) = \cos^{-1}\left(\dfrac{1 + f'_-(x)^2 + f'_+(x)^2}{2\sqrt{(1 + f'_-(x)^2)(1 + f'_+(x)^2)}}\right)$$
Where $f'_-(x)$ and $f'_+(x)$ denote left and right derivatives.
|
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|
Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$$
converges absolutely, conditionally or diverges, and was wondering whether my proof is correct or not.
I began by noticing that
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} -...$$
$$= 1- \frac{1}{2} + \frac{1}{2\times4} - \frac{1}{2 \times 4 \times 6}+...+\frac{1}{(2n)\times(2n-2)... \times2} -...$$
In other words, each $n$th term is the inverse of the product of all even natural numbers, and thus the initial sum is simply $ 1 + \sum a_n$ with
$$a_n := (-1)^n \frac{1}{\prod_{j=1}^n2j}$$
We can show that
$$\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty} \frac{\prod_{j=1}^{n}2j}{\prod_{j=1}^{n+1}2j}=\lim_{n\to\infty} \frac{1}{2(n+1)} = 0 < 1$$
and therefore the series converges absolutely according to the ratio test.
Is this proof correct?
|
Even easier, we can explicitly evaluate the sum. As you noticed, if for $n \ge 0$ $$a_n = \frac{(-1)^n}{(2n+1)!} \prod_{k=1}^n (2k+1),$$ then $$\begin{align}
a_n &= \frac{(-1)^n}{(2n+1)!} \prod_{k=1}^n (2k)(2k+1) \left(\prod_{k=1}^n (2k) \right)^{-1} \\
&= \frac{(-1)^n}{(2n+1)!} (2n+1)! \left( 2^n \prod_{k=1}^n k \right)^{-1} \\
&= \frac{(-1/2)^n}{n!}.
\end{align}$$
In fact, you did all of the above except the last step.
Therefore, $$\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty \frac{(-1/2)^n}{n!} = e^{-1/2}.$$
If we just want to establish convergence, we can easily bound the sum by noting
$$|a_n| \le \frac{1}{n!}$$
for all $n$.
|
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|
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x^2\left(4+\frac{2}{x}+\frac{1}{x^2}-a^2-\frac{2ab}{x}-\frac{b^2}{x^2}\right)}{x^2\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
Applying limit on the numerator and denominator
$$\frac{\lim_{x \to \infty}\left(4-a^2+\frac{2-2ab}{x}+\frac{1-b^2}{x^2}\right)}{\lim_{x \to \infty}\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
It can be seen that the denominator tends to $0$
|
It should be a question to find a and b.
We need to eliminate $x^2$ term, otherwise limit is zero, so $a=\pm2$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x\left(2-2ab+\frac{1-b^2}{x}\right)}{x\left(\sqrt{\frac{4x^2+2x+1}{x^2}}+a+\frac{b}{x}\right)}=\frac{2-2ab}{2+a}=-\frac{1}{2}$$
Reject $a = -2$
When $a=2, b=1$
|
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|
For $0For $0<t\leq 1$ show
$$ \text{ln}(t)\geq \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
On a side note, I've been taking this summer class on inequalities. The professor told us that his main source are all sorts of math olympiads. Given that I'm gonna take Calc 3 next semester and Real Analysis afterwards, are those types of inequalities really as essential as my prof suggests? As you can tell, I'm not really capable of proving most of them by myself yet...
|
Fact 1: If $x \in (0, 1]$, then
$$\ln x \ge \frac{(x - 1)(1 + x^{1/3})}{x + x^{1/3}}.$$
(See [1], page 272. This inequality is due to Karamata.)
By Fact 1, it suffices to prove that
$$\frac{(t - 1)(1 + t^{1/3})}{t + t^{1/3}}
\ge \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
or
$$\frac{1 + t^{1/3}}{t + t^{1/3}}
\le \frac{1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right).$$
Letting $t = u^3$, it suffices to prove that, for all $u\in (0, 1]$,
$$\frac{1 + u}{u^3 + u}
\le \frac{1}{2u^3+2}\left(1+\sqrt{\frac{2u^6+5u^3+2}{u^3}}\right)$$
or
$$\frac{2u^4 + u^3 + u + 2}{u^3 + u} \le \sqrt{\frac{2u^6+5u^3+2}{u^3}}$$
or
$$\left(\frac{2u^4 + u^3 + u + 2}{u^3 + u}\right)^2
\le \frac{2u^6+5u^3+2}{u^3}$$
or
$$\frac{2(u^2 + u + 1)(u^2 - u + 1)(1 + u)^2(1 - u)^4}{u^3(u^2 + 1)^2} \ge 0$$
which is true.
We are done.
Reference.
[1] D. S. Mitrinovic, P. M. Vasic, “Analytic Inequalities,” 1970.
|
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finding integral of $ \int \frac{1}{\sin x + \sqrt{3} \cos x}\ dx $ In $ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx $, If I multiply and divide by $1/2$ I get $$ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx ,$$
then I can write $ 1/2=\sin (\frac{\pi}{6}) $ and $ \sqrt{3}/2=\cos (\frac{\pi}{6}) $ and apply $\cos(a-b)=\cos a \cos b + \sin a \sin b$,
I'll get $ \int\frac{1}{\cos(\frac{\pi}{6} - x)}\ dx $, I get $ \int \sec(\frac{\pi}{6} - x)\ dx $, now we know $ \int \sec(x)\ dx = \log|\tan(\frac{\pi}{4}+\frac{x}{2})| + c $ ,
so I finally get $$ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx = \frac{1}{2}\log|\tan(\frac{\pi}{4}+\frac{\pi}{6}-\frac{x}{2})| + c $$
However if at this step $ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx $
I take $ 1/2=\cos (\frac{\pi}{3}) $ and $ \sqrt{3}/2=\sin (\frac{\pi}{3})$, then what I get is $ \int\frac{1}{\sin(\frac{\pi}{3} + x)}\ dx $, which is equal to $ \int \csc(\frac{\pi}{3} + x)\ dx $, now we know $ \int \csc(x)\ dx = \log|\tan(\frac{x}{2})| + c $
so I finally get $$ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx = \frac{1}{2}\log|\tan(\frac{\pi}{6}+\frac{x}{2})| + c $$
why answer is different, I have rechecked and could not find any mistake
|
Both of your methods have a mistake in them. In your first method, you should have
$$\frac{\sin\left(\frac{\pi}{6}\right)}{\sin\left(\frac{\pi}{6}\right)\sin\left(x\right)+\cos\left(\frac{\pi}{6}\right)\cos\left(x\right)} = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}-x\right)}.$$
Similarly, your second method should have
$$\frac{\cos\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)\sin\left(x\right)+\sin\left(\frac{\pi}{3}\right)\cos\left(x\right)} = \frac{\cos\left(\frac{\pi}{3}\right)}{\sin\left(\frac{\pi}{3}+x\right)}.$$
I'm guessing you forgot about the numerator, right? I think you can figure out the rest from there. Integrating both of them should result in the same answer, probably differing by a constant (I didn't try it out).
|
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In how many ways can the ice hockey team be formed?
A ice hockey team consists of 1 goaltender, 2 defencemen and 3 forwards. The coach has available to him: 3 goaltenders, 7 defencemen, 10 forwards and 4 players that can play both position of a defenceman and a forward. In how many ways can the team be formed?
If there would not be players that can play two positions the solution would be:
$$\binom{3}{1}\cdot \binom{7}{2}\cdot \binom{10}{3} = 7560$$
ways. So my question is how do I take the 4 players that can play two different positions into account in the calculation? Obviously if the four players has been chosen to be defencemen, they can't be chosen to be forwards.
|
Given the explanation here
Combination problem - picking a basketball team with restrictions one could expect two solutions, 50916 and 42273.
\begin{array}{c|c}
play & rest & \\
\hline
1 + g.G & 1 + G + \frac {G^2}{2!} + \frac {G^3}{3!} \\
1 + d.D + d^2\frac {D^2}{2!} & 1 + D + \frac {D^2}{2!} + \cdots + \frac {D ^7}{7!} \\
1 + f.F +f^2\frac {F ^2}{2!} + f^3\frac {F^3}{3!} & 1 + F + \frac {F ^2}{2!} + \cdots + \frac {F^{10}}{10!} \\
option \ A), \ B) & 1 + V + \cdots +\frac {V^4}{4!}\\
\hline
\end{array}
Option $A)$
$$1 + (f+d) \cdot V + (f+d)^2 \frac {V^2}{2!}+(f+d)^3 \frac {V^3}{3!} +(f+d)^4 \frac {V^4}{4!}$$
Option $B)$
$$1 + (f+d) \cdot V + (f^2+fd+d^2) \frac {V^2}{2!}+(f^3+ffd+fdd+ d^3)\frac {V^3}{3!} +(f^4+fffd+ffdd+fddd + d^4) \frac {V^4}{4!}$$
The GF/egf is the product of the eight factors in the table.
we are now interested in the coefficient of
$$ g \cdot d^2 \cdot f^3 \cdot \frac { G^3 }{3!}\frac { D^7 }{7!}\frac { F^{10} }{10!}\frac { V^4 }{4!}$$
|
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|
Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $
Let $x, y, z > 0$. Prove that
$$\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1.$$
I encountered this problem today and thought that it was straightforward, and to be honest it still might be I just can't figure it out for some reason.
I approached this problem using the famous Cauchy-Schwarz inequality. I applied the inequality to every denominator. For example for the first fraction:
$$(1+xy+xz)(1 + \frac{1}{x}+ \frac{1}{x}) \ge (1+y+z)^2$$
After doing this for every fraction I switched the denominator for every one of them with the left side of the Cauchy-Schwarz inequality. Reducing each fraction we are left with
$$\frac{1}{1+2/x} + \frac{1}{1+2/y} + \frac{1}{1+2/z} \ge 1 $$
I am not sure how to continue from here. I tried to use Titu's lemma but it didn't work. I will be very happy if someone can help me out here.
Edit: My Cauchy was wrong.Thanks to all who contributed :)
|
By the Cauchy-Schwarz inequality,
$$
(1+xy+xz)(1+y/x+z/x)
\\
=\left(1,\sqrt{xy},\sqrt{xz}\right)^2
\left(1,\sqrt{ y/x},\sqrt{z/x}\right)^2\geq(1+y+z)^2
$$
which we can rearrange to
$$
{1+xy+xz\over(1+y+z)^2}\geq{x\over x+y+z}.
$$
The rest is smooth.
Credit
|
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|
How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact answer. My working is the following:
$$y=\sqrt{\frac{1-x}{1+x}}$$
This can be written as $$y=\frac{(1-x)^\frac{1}{2}}{(1+x)^\frac{1}{2}}$$
Using the quotient rule we get $$\frac{dy}{dx}=\frac{(1+x)^\frac12\frac{d(1-x)^\frac12}{dx}-(1-x)^\frac12\frac{d(1+x)^\frac12}{dx}}{1+x}$$
Hence $$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}$$
What is the next step?
|
Continuing from your second-last step:
$$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{\sqrt{1+x}^2}{2(1+x)\sqrt{1-x}\sqrt{1+x}}-\frac{\sqrt{1-x}^2}{2(1+x)\sqrt{1+x}\sqrt{1-x}}$$
$$=\frac{-(1+x)-(1-x)}{2(1+x)\sqrt{1-x^2}}=\frac{-1}{(1+x)\sqrt{1-x^2}}$$
|
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|
How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations and substituting into the first. None of these methods worked. How do I do this?
|
If $b, c, d$ are integers,
then $c$ can only be
$1$ or $-1$.
In either case,
$b+d=-5$.
From the last,
$(b,d)$ is one of
$(1,6), (2,3),
(3,2), (6,1)$,
or these replaced by their negatives.
The only ones that work are
$(-2,-3)$ and $(-3,-2)$.
If $c=1$, then $b-d=-1$ so
$b=-3$, $d=-2$.
If $c=-1$, then $b-d=1$ so
$b=-2$, $d=-3$.
|
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|
How to use little-o notation to solve $\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}$ with Taylor polynomials? Consider the limit
$$\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}\tag{1}$$
Let me preface my question by saying that I am studying the Chapter of Spivak's Calculus entitled "Approximation by Polynomial Functions", in which Taylor's Theorem is introduced. Infinite series have not been introduced yet.
I've just read the Introduction chapter of the book Asymptotic Methods in Analysis by N.G. de Bruijn to see the definitions used for the O-symbol and o-symbol.
The goal here is to solve the limit $(1)$ by using Taylor's Theorem. I asked a separate question where I show my (prolix) solution to the limit.
In that question, one person provided a much shorter answer that utilized little-o notation.
I'd like to understand a bit better how to use such notation.
Here is that solution
$$\begin{align}\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}&=\lim\limits_{x\to 0} \frac{\left ( x^2-\frac{x^6}{3!}+... \right )-\left ( x-\frac{x^3}{3!}+... \right )^2}{\left ( x^2-\frac{x^6}{3!}+... \right ) \left (x-\frac{x^3}{3!}+... \right )^2}\\&\tag{2}=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+... \right ) \left (x^2+... \right )}\\&\tag{3}=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^4}\\
&=\frac13.\end{align}$$
*
*What happened to little-o in the numerator between $(2)$ and $(3)$? At first sight it would seem we have $o(x^4)-o(x^4)$, but as far as I can tell the entity $o(x^4)$ can't be used in a subtraction like this (ie we can't just cancel them, can we?).
*Why wasn't little-o used in the denominator?
*More generally, what exactly (if possible in painstaking detail) do we do to get from $(2)$ to $(3)$?
|
*
*$o(x^4)$ means "any function $f(x)$ such that $\frac{f(x)}{x^4}\to0$" (here: when $x\to0$). So $o(x^4)-o(x^4)$ means "any difference of any two such functions" (which may be $\ne$), and such a difference is again some $o(x^4)$.
*My lazy writing was only to imitate the style of Spivak's solution you reported in your previous question. More rigorously, this becomes
$$\begin{align}\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}&=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x-\frac{x^3}{3!}+o(x^3)\right )^2}{\left ( x^2+o(x^2)\right ) \left (x+o(x)\right )^2}\\&\tag{2}=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+o(x^2)\right ) \left (x^2+o(x^2)\right )}\\&\tag{3}=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^4}\\
&=\frac13.\end{align}$$
*Using the notation $o(1)$ for "any function $f$ which tends to $0$", we have $o(x^4)=x^4o(1)$, hence$$\begin{align}\tag{2}\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+o(x^2)\right ) \left (x^2+o(x^2)\right )}&=\lim\limits_{x\to 0} \frac{\frac{2x^4}{3!}+o(x^4)}{x^4+o(x^4)}\\&=\lim\limits_{x\to 0} \frac{\frac{2x^4}{3!}(1+o(1))}{x^4(1+o(1))}\\&=\lim\limits_{x\to 0} \frac{\frac{2x^4}{3!}}{x^4}\lim\limits_{x\to 0} \frac{1+o(1)}{1+o(1)}\\&\tag{3}=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^4}.\end{align}$$
But this can be written more shortly, using the notion of asymptotic equivalence.
|
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|
Find numbers divisible by 6 Find the number of all $n$, $1 \leq n \leq 25$ such that $n^2+15n+122$ is divisible by 6.
My attempt. We know that:
\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}
But $n^2+3n+2=(n+1)(n+2)$, then $n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$, now we have
\begin{align*}
n(n^2+15n+122) & \equiv n(n+1)(n+2)\pmod{6} \\
n^3+15n^2+122n & \equiv 0 \pmod{6}
\end{align*}
I have done this and I think I have complicated the problem even more.
|
You were almost there with your first line. You have $n^2+3n+2=(n+1)(n+2)\text{ mod 6}$
$(n+1)$ and $(n+2)$ are two consecutive numbers so one of them is even. That gives you that this polynomial is divisible by $2$ for all $n$.
If $n$ is either congruent to $1$ or $2$ mod $3$ then $(n+2)$ or $(n+1)$, respectively, is divisible by $3$. Therefore all non-multiples of $3$ are solutions to this problem.
|
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|
How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$? How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$?
I'm trying to solve $(2x - 4y + 6)dx + (x+y-3)dy = 0$. The two lines intersect at $(1,2)$. After substituting to $(\tau = x-1, \mu = y-2)$ and then to $(\tau, u = \mu/\tau)$ I get $-\dfrac{d\tau}{\tau} = \dfrac{(u+1)du}{u^2 - 3u +2}$ and then I integrate. On the left I have $- \ln |x-1| + C$. On the right, as the denominator is the same as $(u-3/2)^2 - 1/4$ I substitute with $z = (u-3/2)$ and finally get $-2\ln |z + 1/2| + 3 \ln |z - 1/2| + C \equiv -2 \ln |\frac{y}{x} - 1| + 3\ln |\frac{y}{x} -2| + C $. After raising to powerof $e$ I get $|x-1| = \dfrac{(\frac{y}{x} - 1)^2}{|\frac{y}{x} - 2|^3}*e^C \equiv \dfrac {|x|(y-x)^2}{|y-2x|^3}*e^C$ but I'm not able to obtain the form in the answer key.
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You first set, correctly, $u=\fracμτ=\frac{y-2}{x-1}$ but then in the transformation back you used $u=\frac yx$. This is not compatible.
|
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|
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Firstly, $\sin x - 1=0 \implies x= \frac{\pi}{2}$
Next, $(\sqrt{2} \cos x +1)=0 \implies x= \frac{3\pi}{4}$
Why is this solution wrong? Is there more than one $x$ value for $\sin x - 1=0$ or $(\sqrt{2} \cos x +1)=0$ in the given interval? How do I know that?
|
Your overlooked the fact that there are two values of $x \in [0, 2\pi]$ such that $\sqrt{2}\cos x + 1 = 0$.
As we can see from the sine graph, there is only one value of $x \in [0, 2\pi]$ such that $\sin x = 1$.
A number $x$ is a solution of the equation $\sqrt{2}\cos x + 1 = 0 \iff \cos x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$. As we can see from the cosine graph, since
$$-1 < -\frac{\sqrt{2}}{2} < 1$$
there are two values of $x$ in the interval $[0, 2\pi]$ such that $\cos x = -\frac{\sqrt{2}}{2}$.
The interval $[0, 2\pi]$ contains the coterminal angles $0$ and $2\pi$, so there are two values of $x$ such that $\cos x = 1$, namely $0$ and $2\pi$.
However, if we were to instead consider the interval $[0, 2\pi)$, the equations $\sin x = 1$ and $\sin x = -1$ each have one solution, while the equation $\sin x = y$ has two solutions whenever $-1 < y < 1$. Similarly, in the same interval, the equations $\cos x = 1$ and $\cos x = -1$ each have one solution, while the equation $\cos x = y$ has two solutions whenever $-1 < y < 1$.
Next, consider the diagram below.
When does $\sin\theta = \sin\varphi$?
Two angles have the same sine if the $y$-coordinates of the points where their terminal sides intersect the unit circle are equal. Clearly, this is true if $\theta = \varphi$. By symmetry, it is also true when $\theta = \pi - \varphi$. Moreover, it is also true for any angles coterminal with these angles. Hence, the general solution of the equation $\sin\theta = \sin\varphi$ is
\begin{align*}
\theta & = \varphi + 2k\pi, k \in \mathbb{Z} & \theta = \pi - \varphi + 2m\pi, m \in \mathbb{Z}
\end{align*}
Now, consider the equation $\sin x - 1 = 0$.
\begin{align*}
\sin x - 1 & = 0\\
\sin x & = 1
\end{align*}
A particular solution of this equation is
$$x = \arcsin(1) = \frac{\pi}{2}$$
Thus, the general solution of this equation is
\begin{align*}
x & = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z} & x & = \pi - \frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}\\
& & & = \frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}
\end{align*}
Since we want a solution in the interval $[0, 2\pi]$, we must take $k = 0$ and $m = 0$, which yields the unique solution
$$x = \frac{\pi}{2}$$
When is $\cos\theta = \cos\varphi$?
Two angles have the same cosine if the $x$-coordinates of the points where their terminal sides intersect the unit circle are equal. Clearly, this is true if $\theta = \varphi$. By symmetry, it is also true if $\theta = -\varphi$. Moreover, any angles coterminal with these angles have the same cosine. Thus, the general solution of the equation $\cos\theta = \cos\varphi$ is
\begin{align*}
\theta & = \varphi + 2k\pi, k \in \mathbb{Z} & \theta = -\varphi + 2m\pi, m \in \mathbb{Z}
\end{align*}
Now, consider the equation $\sqrt{2}\cos x + 1 = 0$.
\begin{align*}
\sqrt{2}\cos x + 1 & = 0\\
\sqrt{2}\cos x & = -1\\
\cos x & = -\frac{1}{\sqrt{2}}\\
\cos x & = -\frac{\sqrt{2}}{2}
\end{align*}
A particular solution of this equation is
$$x = \arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$$
Thus, the general solution of the equation $\sqrt{2}\cos x + 1 = 0$ is
\begin{align*}
x & = \frac{3\pi}{4} + 2k\pi, k \in \mathbb{Z} & x & = -\frac{3\pi}{4} + 2m\pi, m \in \mathbb{Z}
\end{align*}
Since we seek solutions in the interval $[0, 2\pi]$, we must take $k = 0$ and $m = 1$, which yields the solutions
\begin{align*}
x & = \frac{3\pi}{4} & x & = 2\pi - \frac{3\pi}{4} = \frac{5\pi}{4}
\end{align*}
|
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|
find a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}$
Let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. Order the (real) solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$ with $x\ge 1$ from smallest to largest by $x_1,x_2,\cdots$. Provide a closed form formula for $\sum_{k=1}^n \frac{1}{x_{2k}^2 - x_{2k-1}^2}.$
First, I'm not even sure why there are infinitely many solutions to $\sqrt{\lfloor x\rfloor \lfloor x^3\rfloor} + \sqrt{\{x\}\{x^3\}} = x^2$. One inequality that might be useful is the Cauchy-Schwarz inequality. There's also the AM-GM inequality. We have $ac + bd \leq \sqrt{a^2+b^2}\sqrt{c^2+d^2}$ for all real numbers $a,b,c,d$ where $ac,bd\ge 0$ with equality iff $(a,b),(c,d)$ are proportional vectors in $\mathbb{R}^2$. Observe that $\lfloor x\rfloor$ always has the same sign as $x$, so $\lfloor x\rfloor \lfloor x^3\rfloor$ has the same sign as $x^4 \ge 0$. It might be useful to substitute $\{x\} = x-\lfloor x\rfloor$ into the original equation and simplify the result somehow. Also, it could be possible to write the given sum as a telescoping sum.
|
Using Cauchy-Bunyakovsky-Schwarz inequality, we have
$\frac{\{x\}}{\lfloor x\rfloor}
= \frac{\{x^3\}}{\lfloor x^3\rfloor}$
or
$$\lfloor x^3\rfloor = x^2 \lfloor x\rfloor. \tag{1}$$
Clearly, $x=1, 2, \cdots$ are solutions of (1).
Let $k \in \mathbb{Z}_{>0}$.
Consider the solutions of (1) in $(k, k + 1)$ i.e. $k < x < k + 1$. We have $\lfloor x \rfloor = k$.
Let $u = k(x^2 - k^2)$.
We have $x = k\sqrt{1 + u/k^3}$
and $x^3 = (k^3 + u)\sqrt{1 + u/k^3}$.
We have
$x^2\lfloor x \rfloor = k^3 + u$.
Thus, $u$ is an positive integer.
(1) If $u \ge 2$, we have
$$x^6 - (k^3 + u + 1)^2
= \frac{(u-2)k^6 + [2(u-2)^2 + 6(u-2) + 3]k^3 + u^3}{k^3} > 0$$
which results in
$\lfloor x^3\rfloor \ge k^3 + u + 1 > k^3 + u = x^2 \lfloor x\rfloor$.
(2) If $u = 1$ i.e. $x = k \sqrt{1 + 1/k^3}$, we have $x^3 = (k^3 + 1)\sqrt{1 + 1/k^3}$.
It is easy to prove that
$\lfloor x^3 \rfloor = k^3 + 1$.
Also, $x^2 \lfloor x\rfloor = k^3 + 1$. Thus, $x = k \sqrt{1 + 1/k^3}$
is a solution of (1).
Thus, we have $x_{2k-1} = k$ and $x_{2k} = k\sqrt{1 + 1/k^3}$ for $k = 1, 2, \cdots$.
|
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|
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2+B(x-3)(x+2) + C(x-3)}{(x-3)(x+2)} = \frac{Ax^2 + 4Ax + 4A + Bx^2 - Bx - 6B +Cx - 3C}{(x-3)(x+2)}$
Looking at the numerator: $x^2(A+B) + x(4A-B+C) + (4A-6B-3C)$
So, comparing coefficients:
$A+B=1$,
$4A-B+C=0$
$4A-6B-3C=0$
I am struggling to solve these 3 equations to find A,B,C
|
There is another way of finding $A$, $B$ and $C$ which is more efficient than forming and solving simultaneous equations.
Starting with the identity $$x^2=A(x+2)^2+B(x-3)(x+2)+C(x-3)$$
Choose values of $x$ to make the brackets zero.
So, putting $x=2$ gives $$4=C(-5)\implies C=-\frac45$$
Putting $x=3$ gives $$9=A(25)\implies A=\frac{9}{25}$$
Now that you’ve run out of convenient values of $x$ you can put $x=0$ or just look at coefficients, such as $x^2$:
$$1=A+B\implies B=\frac{16}{25}$$
|
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The equation of the line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ intersecting the lines $9x+y+z+4=0 =5x+y+3z$ & $x+2y-3z-3=0=2x-5y+3z+3$ The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$ is
(A) $7x + y + 2z + 2 = 0 = x - 2y + z + 1$
(B) $7x - y + 2z + 2 = 0 = x + 2y + z + 1$
(C) $x + 7y + z + 1 = 0 = 2x - y + 2z + 2$
(D) None of these
I am solving but only putting the general equation
Let the line be represented by $\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4} \Rightarrow {L_m}:\overrightarrow r = a\hat i + b\hat j + c\hat k + s\left( {2\hat i + 3\hat j + 4\hat k} \right) \Rightarrow {L_m}:\overrightarrow r = \overrightarrow {{c_m}} + s\overrightarrow {{t_m}} $
The equation of the line $9x + y + z + 4 = 0 = 5x + y + 3z \Rightarrow {L_2}$
On calculating we get ${L_2}:\frac{x}{{ - 1}} = \frac{{y + 6}}{{11}} = \frac{{z - 2}}{{ - 2}} \Rightarrow \overrightarrow r = 0\hat i - 6\hat j + 2\hat k + t\left( { - \hat i + 11\hat j - 2\hat k} \right) \Rightarrow {L_2}:\overrightarrow r = \overrightarrow {{c_2}} + t\overrightarrow {{t_2}} $
The equation of the line $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3 = {L_3}$
On calculating we get ${L_3}:\frac{x}{1} = \frac{{y - 1}}{1} = \frac{{z - \frac{2}{3}}}{1} \Rightarrow \overrightarrow r = 0\hat i + \hat j + \frac{2}{3}\hat k + u\left( {\hat i + \hat j + \hat k} \right) \Rightarrow {L_3}:\overrightarrow r = \overrightarrow {{c_3}} + u\overrightarrow {{t_3}} $
It is getting more and more complicated as need to find $\left( {\overrightarrow {{c_2}} - \overrightarrow {{c_m}} } \right).\left( {\overrightarrow {{t_m}} \times \overrightarrow {{t_2}} } \right) = 0$ & $\left( {\overrightarrow {{c_3}} - \overrightarrow {{c_m}} } \right).\left( {\overrightarrow {{t_m}} \times \overrightarrow {{t_3}} } \right) = 0$
Can this be solved by some other method
|
Answer D since none of the lines in A, B, C is parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}.$ Indeed, solving the pairs of associated homogeneous equations, we find that:
*
*(A) is directed by the vector $(-1,1,3)$
*(B) by $(1,1,-3)$
*(C) by $(1,0,-1).$
If the question was more open, I would have solved it the following way:
find $(a,b,c)\in L_3$ such that $(a,b,c)+t(2,3,4)\in L_2$ for some $t\in\mathbb R$ (write the 4 equations in the 4 unknowns $a,b,c,t$ and eliminate $t$ to get 3 equations in 3 unknowns). I found $(a,b,c)=(-3,-3,-4)$ (and $t=\frac65$).
|
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|
$S_{n}$ is the sum of every third element in the $n$th row of the Pascal triangle, beginning on the left with the second element. Find $S_{100}$. Problem:
Let $S_{n}$ be the sum of every third element in the $n$th row of the Pascal triangle, beginning on the left with the second element. Find the value of $S_{100}$.
My work:
For brevity, I worded the problem differently but it is the same as the one discussed in this question, however, my doubt is not the same. I understood the problem and I was able to conjecture that $$S_{100} = \frac{2^{100} - 1}{3}$$
Then, the author of the book where I found the problem says that this can be easily proved through induction but I can't figure out how and would appreciate any help.
Thanks in advance.
Edit: Huge thanks to Rob Pratt for his solution, but I finally came up with the proof by induction I was looking for. I am posting it here in case anyone ever has the same doubt I had.
Assume that for all $k \le n$ such that $k \equiv 4 \mod{6}$ (As well, assume $n \equiv 4 \mod{6}$) $$S_{k,0} = S_{k,1} = S_{k,2} - 1 \text{ and } S_{k,1} = \frac{2^k - 1}{3}$$ Now, the following facts follow from the definition of the Pascal triangle, for any $m \in \mathbb{N}$
$$\begin{array} SS_{m,0} = S_{m-1,0} + S_{m-1,2} \\ S_{m,1} = S_{m-1,0} + S_{m-1,1} \\ S_{m,2} = S_{m-1,1} + S_{m-1,2} \end{array}$$
Knowing the above we can deduce the following
$$S_{n+6,1} = S_{n+5,0} + S_{n+5,1} = S_{n+4,0} + S_{n+4,2} + S_{n+4,0} + S_{n+4,1} = 2(S_{n+3, 0} + S_{n+3,2}) + S_{n+3,1} + S_{n+3,2} + S_{n+3,0} + S_{n+3,1} = 3(S_{n+2,0}+S_{n+2,2}) + 2(S_{n+2,0}+S_{n+2,1}) + 3(S_{n+2,1} + S_{n+2,2}) = 5(S_{n+1,0}+S_{n+1,2}) + 6(S_{n+1,1} + S_{n+1,2}) + 5(S_{n+1,0}+S_{n+1,1}) = 10(S_{n,0} + S_{n,2}) + 11(S_{n,1} + S_{n,2}) + 11(S_{n,0} + S_{n,1}) = 21S_{n,0} + 22S_{n,1} + 21S_{n,2}$$
Now let's use our induction hypothesis,
$$S_{n+6,1} = 21S_{n,0} + 22S_{n,1} + 21S_{n,2} = 21S_{n,1}+ 22S_{n,1}+21(S_{n,1} + 1) = 64S_{n} + 21 = \frac{2^{n+6} - 64}{3} + 21 = \frac{2^{n+6}-1}{3}$$
So our proof is concluded.
|
Let $\omega =\exp(2\pi i/3)$ be the primitive cube root of unity. Because $$\frac{1+\omega^k+\omega^{2k}}{3}=\begin{cases}1&\text{if $3 \mid k$}\\0&\text{otherwise}\end{cases}$$
we have $$\sum_k a_{3k} = \sum_k a_k \frac{1+\omega^k+\omega^{2k}}{3}.$$
Now take $$a_k=\binom{n}{k+1}$$ and apply the binomial theorem to each of the resulting three sums to obtain
\begin{align}
S_n
&= \sum_{k \ge 0} \binom{n}{3k+1} \\
&= \sum_{k \ge 0} \binom{n}{k+1}\frac{1+\omega^k+\omega^{2k}}{3} \\
&= \frac{1}{3}\sum_{k \ge 0} \binom{n}{k+1}
+ \frac{1}{3}\sum_{k \ge 0}\binom{n}{k+1} \omega^k
+ \frac{1}{3}\sum_{k \ge 0}\binom{n}{k+1} \omega^{2k} \\
&= \frac{1}{3}\sum_{k \ge 1} \binom{n}{k}
+ \frac{1}{3\omega}\sum_{k \ge 1} \binom{n}{k}\omega^k
+ \frac{1}{3\omega^2}\sum_{k \ge 1} \binom{n}{k}(\omega^2)^k \\
&= \frac{1}{3}(2^n-1)
+ \frac{1}{3\omega}((1+\omega)^n-1)
+ \frac{1}{3\omega^2}((1+\omega^2)^n-1) \\
&= \frac{1}{3}(2^n-1)
+ \frac{\omega^2}{3}((-\omega^2)^n-1)
+ \frac{\omega}{3}((-\omega)^n-1).
\end{align}
The last two terms cancel when $n\equiv 4 \pmod6$, leaving $(2^n-1)/3$.
|
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|
Whose probability of winning is higher? There are $10$ blue marbles and $5$ red marbles in a black bag. Two players $A$ and $B$ take turns taking non-refundable (without replacement) $1$ marble each time (A takes first). The game ends when someone gets a red ball and that person loses. Whose probability of winning is higher?
Please any help. I honestly don't know how to approach this one. Thank you.
|
You can calculate the winning probability of the person to move recursively as a function of the number of blue balls in the bag, since as long as the game is still going, the number of red balls will always be $\ 5\ $. If $\ w_n\ $ is the probability that the person to draw will win when there are $\ n\ $ blue balls in the bag, then
\begin{align}
w_0&=0\ \ \text{ and}\\
w_n&=\frac{n\big(1-w_{n-1}\big)}{n+5}
\end{align}
for $\ n\ge1\ $. This gives the probabilities listed in the following table
\begin{array}{c|c|}
n&w_n\\
\hline
1&\frac{1}{6}\\
\hline
2&\frac{2}{7}\left(1-\frac{1}{6}\right)=\frac{5}{21}\\
\hline
3&\frac{3}{8}\left(1-\frac{5}{21}\right)=\frac{2}{7}\\
\hline
4&\frac{4}{9}\left(1-\frac{2}{7}\right)=\frac{20}{63}\\
\hline
5&\frac{5}{10}\left(1-\frac{20}{63}\right)=\frac{43}{126}\\
\hline
6&\frac{6}{11}\left(1-\frac{43}{126}\right)=\frac{83}{231}\\
\hline
7&\frac{7}{12}\left(1-\frac{83}{231}\right)=\frac{37}{99}\\
\hline
8&\frac{8}{13}\left(1-\frac{37}{99}\right)=\frac{496}{1287}\\
\hline
9&\frac{9}{14}\left(1-\frac{496}{1287}\right)=\frac{113}{286}\\
\hline
10&\frac{10}{15}\left(1-\frac{113}{286}\right)=\frac{173}{429}\\
\hline
\end{array}
Thus, for the game given, with $\ 10\ $ blue marbles in the bag, the probability that the player going first wins is $\ \frac{173}{429}\approx0.403\ $. Before the game starts, the probability that the player going second will win is $\ \frac{256}{429}\approx0.597\ $, and so he or she is more likely to win than the player who goes first.
For what it's worth, you can obtain the expression
$$
w_n=\frac{(-1)^nn!}{(n+5)!}\sum_{j=1}^n\frac{(-1)^j(j+4)!}{(j-1)!}
$$
for the solution of the recursion for $\ w_n\ $. However, using this expression to calculate $\ w_n\ $ is no more efficient than using the recursion directly.
|
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|
Number of tangential quadrilaterals with distinct sidelengths chosen in set $\{1,2,\ldots, 8\}$
There are $8$ line segments of length $1,2,3,\ldots,8$ $\textrm{units}$. How many quadrilaterals can be made from these line segments such that circles can be inscribed in the quadrilaterals made$?$
I applied Pitot theorem. I first fixed on side as $8$ then found all possible combinations. Then, I fixed $7$ and counted all possible combinations excluding those which include $8$ and so on. I am listing all the combinations that I found.
$(8,1,7,2),(8,1,6,3),(8,1,5,4),(8,2,7,3),(8,2,6,4),(8,3,7,4),(8,3,6,5),(8,4,7,5)$
$(7,1,6,2),(7,1,5,3)(7,2,6,3),(7,2,5,4),(7,3,6,4),(7,4,6,5),(6,1,5,2),(6,1,4,2)$
$(6,2,5,3),(6,3,5,4),(5,1,4,2),(5,2,4,3),(4,1,2,3)$
I found $21$ combinations in all. Is my method and my answer correct$?$ Is there any other elegant way to arrive to the correct answer$?$
Any help is greatly appreciated. There is a similar question this on this site, stating the same problem but the solution given there doesn't give any answer or any elegant method.
|
A good way to proceed is to take the twenty-eight possible pairs of side lengths and classify them according to the sum of the lengths:
$\{1,2\}$
$\{1,3\}$
$\{1,4\},\{2,3\}$
$\{1,5\},\{2,4\}$
$\{1,6\},\{2,5\},\{3,4\}$
$\{1,7\},\{2,6\},\{3,5\}$
$\{1,8\},\{2,7\},\{3,6\},\{4,5\}$
$\{2,8\},\{3,7\},\{4,6\}$
$\{3,8\},\{4,7\},\{5,6\}$
$\{4,8\},\{5,7\}$
$\{5,8\},\{6,7\}$
$\{6,8\}$
$\{7,8\}$
Now select any pair which will represent two opposite sides and join it with another pair having the same sum (which will be the other pair of opposing sides). Clearly there is no match for sums of $3,4,14,15$. For the rest of the cases, if you can choose from $m$ pairs with a given sum you have $\binom{m}{2}=m(m-1)/2$ choices for the two opposing side-pairs that make up the quadrilateral. Thus one choice with sums of $5$, one with sums of $6$ (both $m=2$), three apiece with sums of $7$ and $8$ (both $m=3$), etc., and the sum of these all these combination counts will be $22$ as indicated in the comments.
|
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|
Evaluate $\int_0^1 \cos^{-1} x\ dx$ by first finding $\frac{d}{dx}(x\cos^{-1} x)$ Question
Evaluate $$\int_0^1 \cos^{-1} x\ dx$$ by first finding the value of $$\frac{d}{dx}(x\cos^{-1} x).$$
My Working
As the question said to evaluate $$\frac{d}{dx}(x\cos^{-1} x),$$ I used the product rule to differentiate. We first have to let $u=x$ and $v=\cos^{-1} x$, so
\begin{align}
& \quad\frac{d}{dx}(x\cos^{-1} x)\\
&=u^\prime v+v^\prime u\\
&=1\cdot \cos^{-1}x + x\cdot\frac{-1}{\sqrt{1-x^2}}\\
&=\cos^{-1}x-\frac{x}{\sqrt{1-x^2}}
\end{align}
Unfortunately, after that, I have no idea about how to proceed, but I think that we should go somewhere from
$$\int\left[\frac{d}{dx}(x\cos^{-1}x)\right]\ dx+\int\frac{x}{\sqrt{1-x^2}}\ dx=\int \cos^{-1}x???$$
Thank you for your help!
|
After seeing the helpful comments, I can start using $u$-substitution for the solution. Let $u=1-x^2$, then
\begin{align}
\frac{du}{dx}&=-2x\\
du&=-2x\ dx\\
\int_0^1\frac{x}{\sqrt{1-x^2}}&=-\frac{1}{2}\int^1_0\frac{-2x}{\sqrt{1-x^2}}\ dx\\
&=-\frac{1}{2}\int^1_0 \frac{1}{\sqrt{u}}\ du\\
&=-\frac{1}{2}\int^1_0 u^{-\frac{1}{2}}\ du\\
&=-\frac{1}{2}\left[2u^\frac{1}{2}\right]^1_0\\
&=-\frac{1}{2}\left[2(1-x^2)^\frac{1}{2}\right]^1_0\\
&=-\frac{1}{2}\left(2\sqrt{1-1^2}-2\sqrt{1-0^2}\right)\\
&=-\frac{1}{2}\cdot-2\\
&=1\\
\int^1_0\left[\frac{d}{dx}(x\cos^{-1}x)\right]\ dx&=\left[x\cos^{-1}x\right]^1_0\\
&=0\\
\int^1_0\cos^{-1}x\ dx&= 0+1=1
\end{align}
|
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|
Prove that $\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)=-1$ The question is from The number of solutions of $ax^2+by^2\equiv 1\pmod{p}$ is $ p-\left(\frac{-ab}{p}\right)$
And I wonder how to prove the following equation although someone gives the trick:
\begin{split}
\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)=-1
\end{split}
Here is the trick:using$\left(\frac{y^*}{p}\right)$ instead of$\left(\frac{y}{p}\right)$
$y^*$ is the inverse of $y$ modulo $p$
|
Here is my answer:
We can replace $\left(\frac{y}{p}\right)$ with $\left(\frac{y^*}{p}\right)$ such that:
\begin{split}
\left(\frac{y}{p}\right) = \left(\frac{y^*}{p}\right).
\end{split}
We can easily verify it by using primitive root.
When $y = 0$ there is no inverse, so we ignore the first term of $\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)$.
So we have:
\begin{split}
&\sum_{y^*=1}^{p-1}\left(\frac{y^*}{p}\right)\left(\frac{y+d}{p}\right)\\
&=\sum_{y^*=1}^{p-1}\left(\frac{1+y^*d}{p}\right)\\
&=\sum_{y=1}^{p-1}\left(\frac{1+yd}{p}\right).
\end{split}
Because we know that:
\begin{split}
\sum_{y=0}^{p-1}\left(\frac{1+yd}{p}\right) = 0
\end{split}
Legendre symbol: Showing that $\sum_{m=0}^{p-1} \left(\frac{am+b}{p}\right)=0$
So it lacks the first term, that is $\left(\frac{1}{p}\right) = 1$, and we can get:
\begin{split}
&\sum_{y=0}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)\\
&=\sum_{y=1}^{p-1}\left(\frac{y}{p}\right)\left(\frac{y+d}{p}\right)\\
&=-1.
\end{split}
There is a better way I think:
\begin{split}
\left(\frac{y(y+d)}{p}\right)=\left(\frac{y(y+dyy^\ast)}{p}\right)=\left(\frac{y^2(1+dy^\ast)}{p}\right)=\left(\frac{1+dy^\ast}{p}\right)
\end{split}
from$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$
|
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|
If integers $s$, $x$, $y$, $z$ are such that $s=5xy+5yz+5xz=2x^2+2y^2+2z^2$, then $10s$ is a perfect square I'm having a problem with a question:
Consider integer $s$ and integers $x,y,z$ such that
$$s = 5xy+5yz+5xz \quad\text{and}\quad s= 2x^{2} + 2y^{2} + 2z^{2}$$ Prove that $10s$ is a perfect square.
I'm pretty sure it has something in common with
$$(x+y+z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy+yz+xz)$$ e.g. if I somehow (prob. by writing $10s$ as $4s + 6s$ etc.) show $10s$ equals $4, 9$ or $16$ times $(x+y+z)^{2}$, then I'm done, yet I have no clue how to show this.
Another idea was to check modulo, and I found at least one of $x,y,z$ is divisible by $5$ and at least two of $x,y,z$ are even still don't know how to use it.
|
$(xy+yz+xz)=\frac 15 s$ and $x^2 + y^2 + z^2 = \frac 12s$
Now $(x^2 + y^2 + z^2) + 2(xy + yz+xy)= (x+y+z)^2$
so $\frac 12 s + \frac 25 s = \frac 9{10}s=(x+y+z)^2$
So $10s = \frac {100}9(x+y+z)^2 = (\frac {10}3(x+y+z))^2$.
Dang.
So close!
But if we can prove $x+y+z$ is divisible by $3$ we will be okay. Can we?
Well, it must. We have $\frac 9{10}s = (x+y+z)^2$ so $9x = 10(x+y+z)^2$ is an integer. So $9\mid 10(x+y+z)^2$ so $9|(x+y+z)^2$ and $3|x+y+z$. Hurray! we are done.
|
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|
Variable substitution with 3 variables I'm trying to determine the following triple integral
$$
\iiint_{K}\left[\left(x - a\right)^{2} +
\left(y - b\right)^{2} + \left(z - c\right)^2\right]{\rm d}x\,{\rm d}y\,{\rm d}z,\quad
K = \left\{\left(x,y,z\right): x^{2} + y^{2} + z^{2}
\leq 1\right\}.
$$
I tried the following variable substitution
$$\begin{cases}
x = r\sin\left(\theta\right)\cos\left(\varphi\right) + a
\\[1mm]
y = r\sin\left(\theta\right)\sin\left(\varphi\right) + b
\\[1mm]
z = r\cos\left(\theta\right) + c
\end{cases}$$
with
$$
0<r<1 \quad , \quad 0< \theta < \pi \quad , \quad
0 < \varphi < 2\pi
$$
and I'm getting $4\pi/5$, which is wrong, the answer should be $4\pi/5 + 4\pi\left(a^{2} + b^{2} + c^{2}\right)/3$, but I can't really find where I missed up !.
|
By using your change of coordinates and the given limits you are not evaluating the given integral, but $\iiint_K (x^2+y^2+z^2) dxdydz$.
The correct limits are different and not so easy to find.
On the other hand, by expanding the squares, we split the integral into three terms,
$$\begin{align}
I&=\iiint_K ((x-a)^2+(y-b)^2+(z-c)^2) dxdydz \\
&=\iiint_K (x^2+y^2+z^2) dxdydz
-2\iiint_K (ax+by+cz) dxdydz\\
&\qquad+(a^2+b^2+c^2) \iiint_K 1 dxdydz
\end{align}$$
and the integral in the middle is zero by symmetry (for instance
$\iiint_K ax dxdydz=0$ because the integrand function is $x$-odd and $K$ is symmetric with respect to the plane $x=0$).
Finally, using the spherical coordinates (centered at the origin),
$$I=2\pi\int_{r=0}^1\int_{\phi=0}^\pi r^2\cdot r^2\sin(\phi) drd\phi+0+(a^2+b^2+c^2)|K|=\frac{4\pi}{5} +\frac{4\pi}{3}(a^2+b^2+c^2).$$
|
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|
Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$x^3+y^3 = 4(x+y)$
$x^3+y^3 = 8$
Then:
$8 = 4(x+y)$
$x+y = 2$
However if we replace that value in $(x+y)^2$ we have:
$(x+y)^2 = 2^2 =4$
$(x+y)^2 = x^2+y^2+2xy = 5 + 2 = 7$
As you can see $4 \neq 7$, what is happening?
|
We can also take a geometric interpretation of the problem. The curves corresponding to the equations
$$ xy \ = \ 1 \ \ \ , \ \ \ x^2 \ + \ y^2 \ = \ 5 \ \ \ , \ \ \ x^3 \ + \ y^3 \ = \ 8 $$
all have symmetry about the line $ \ y \ = \ x \ $ (only the first two, however, has symmetry about the origin). This suggests an approach in which we look at intersections of these curves with lines through the origin $ \ y \ = \ mx \ \ . \ $ (We do not need to consider a "horizontal" or "vertical" line, since the first curve has the coordinates axes as asymptotes.) If there are value(s) of $ \ m \ $ which "work" for all three curves, then we have located their mutual intersections.
With this substitution, we have
$$ m·x^2 \ = \ 1 \ \ \ , \ \ \ (1 + m^2)·x^2 \ = \ 5 \ \ \ , \ \ \ (1 + m^3)·x^3 \ = \ 8 \ \ . $$
For the first two curve equations, we obtain
$$ (1 + m^2)·\frac{1}{m} \ \ = \ \ 5 \ \ \Rightarrow \ \ m^2 - 5m + 1 \ \ = \ \ 0 \ \ \Rightarrow \ \ m \ \ = \ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \ . $$
(As $ \ \frac{2}{5 \ +\ \sqrt{21}} \ = \ \frac{2·(5 \ - \ \sqrt{21})}{25 - 21} \ = \ \frac{5 \ - \ \sqrt{21}}{2} \ \ , \ $ we see that these lines are indeed symmetrical about $ \ y \ = \ x \ \ . \ $ These slopes also tell us that the circle and rectangular hyperbola intersect only in the first and third quadrants.)
However, in the third equation, this gives us
$$ (1 + m^3)·\left( \frac{1}{m} \right)^{3/2} \ \ \stackrel{?}{=} \ \ 8 \ \ \Rightarrow \ \ (1 + m^3) \ \ \stackrel{?}{=} \ \ 8· m^{3/2} \ \ $$
$$ \Rightarrow \ \ \left( \ 1 \ + \ \left[ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \right]^3 \ \right) \ \ \stackrel{?}{=} \ \ 8· \left[ \ \frac{5 \ \pm \ \sqrt{21}}{2} \ \right]^{3/2} $$
$$ \Rightarrow \ \ \ 8 \ + \ \ ( \ 5^3 \ \pm \ 3·5^2·\sqrt{21} \ + \ 3·5·21 \ \pm \ 21^{3/2} \ ) \ \ \stackrel{?}{=} \ \ 16·\sqrt2· ( \ 5 \ \pm \ \sqrt{21} \ )^{3/2} $$
$$ \Rightarrow \ \ \ 8 \ + \ \ ( \ 440 \ \pm \ 96·\sqrt{21} \ \ ) \ \ \stackrel{?}{=} \ \ 16·\sqrt2· ( \ 5 \ \pm \ \sqrt{21} \ )^{3/2} $$
[then, squaring both sides]
$$ \Rightarrow \ \ \ ( \ 448 \ \pm \ 96·\sqrt{21} \ )^2 \ \ \stackrel{?}{=} \ \ 512· ( \ 448 \ \pm \ 96·\sqrt{21} \ ) \ \ \Rightarrow \ \ \ 448 \ \pm \ 96·\sqrt{21} \ \ \neq \ \ 512 \ \ . $$
So we may conclude that the pairwise intersections of the three curves do not all fall on the same radial lines from the origin, and thus there is no solution to the given system of non-linear equations.
Interestingly, the second and third curves nearly meet tangentially. The line $ \ y \ = \ x \ $ intersects the circle at $ \ x^2 \ = \ \frac52 \ \Rightarrow \ \left(\sqrt{\frac52} \ , \ \sqrt{\frac52} \ \approx \ 1.5811 \right) \ \ $ and intersects the "cubic curve" at $ \ x^3 \ = \ 4 $ $ \Rightarrow \ \left(\sqrt[3]{4} \ , \ \sqrt[3]{4} \ \approx \ 1.5874 \right) \ \ , \ $ so the intersections of these two curves (in the first quadrant -- there is a second "diagonally symmetric" pair in the second and fourth quadrants ) lie on lines with reciprocal slopes close to $ \ 1 \ \ . \ $ (Those slopes are approximately $ \ m \ \approx \ 1.197 \ \ , \ \ m \ \approx \ 0.835 \ \ . \ ) $
|
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"url": "https://math.stackexchange.com/questions/4571660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find for what values of a and b in R the limit exists (No De L'Hopital) I was given this exercise in my math course at university.
The question is to find, without using De l'Hopitals and other methods which may use derivates and similars, for what values of $a$ and $b$ in $\mathbb{R}$ the following statement is true.
$$
f(x)= \left\{\begin{matrix} \dfrac{\sin x}{x} & x > 0 \\ ax + b & x \leq 0 \\ \end{matrix}\right. $$ $$ \lim_{x \to 0^+} \dfrac{f(x) - ax - b}{x} = 0$$
I started by writing it as:
$$ \lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x} -ax - b}{x} = 0$$ because, due to the fact that we are approaching from $x > 0, \, f(x) = \dfrac{\sin x}{x}$
Then I multiplied $\frac{\sin x}{x}$ by $\frac{\sin x}{\sin x}$ in order to get:
$$ \lim_{x \to 0^+} \dfrac{\dfrac{\sin^2x}{x\sin x} -ax - b}{x} = 0$$
From there:
$$ \lim_{x \to 0^+} \dfrac{\dfrac{\sin^2x}{x\sin x} -ax - b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos^2x}{x\sin x}-ax-b}{x} = 0 $$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos x}{x} \times \dfrac{1+\cos x}{\sin x}-ax-b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos x}{x} \times \dfrac{1+\cos x}{x} \times \dfrac{x}{\sin x} - ax - b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1 - \cos x}{x} \times \dfrac{1}{x} \times \left(1 + \cos x\right) \times \dfrac{x}{\sin x} -ax - b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos x}{x^2} \times \left(1+ \cos x\right) \times \dfrac{x}{\sin x} -ax - b}{x} = 0$$
Which, substituting the value of $x$ to the value it is approaching, results in:
$$ \dfrac{\dfrac{1}{2}\times\left(1+1\right)\times 1 - a \times 0 - b}{x} = 0 $$
$$ \Longleftrightarrow \dfrac{1 - b}{x} = 0 $$
Which should resolve to the I.F. $$\dfrac{0}{0}$$ when $b=1$
So to me it seems like, with the calculations I've done, I can't reach an answer about which value of $a$ and $b$ make the statement true
|
Assuming the limit exists:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x} -ax - b}{x} = \lim_{x \to 0^+} \dfrac{\sin x -ax^2 - bx}{x^2}=\lim_{x \to 0^+} \left(\dfrac{\sin x- bx}{x^2}-a\right)
\end{equation}
Now consider:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\sin x- bx}{x^2}
\end{equation}
If $b\neq1$:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\sin x- bx}{x^2}=\lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x}- b}{x}= \pm \infty
\end{equation}
If $b=1$, we have:
$$ \forall x \in (0, \pi/2), \sin x \leq x \leq \tan x$$
So:
\begin{equation}
\dfrac{\sin x - \tan x}{x^2} \leq \dfrac{\sin x - x}{x^2} \leq 0.
\end{equation}
Now, considering that:
$$ \sin x - \tan x = \tan x(\cos x - 1) = - \dfrac{\tan x \sin^2 x}{1+\cos x} $$
and
$$\dfrac{\sin x}{x} \to 1 \quad \text{as} \quad x\to 0$$
we have
$$ \lim_{x \to 0} \frac{\sin x - \tan x}{x^2} = - \lim_{x \to 0} \frac{\tan x}{1+\cos x}\left(\frac{\sin x}{x}\right)^2 = 0. $$
Applying the squeezing theorem, we conclude:
$$ \lim_{x\to0^+} \frac{\sin x - x}{x^2} = 0. $$
Since
$$ \lim_{x \to 0^+} \dfrac{f(x) - ax - b}{x} = 0$$
and
$$ \lim_{x \to 0^+} \left(\dfrac{\sin x- bx}{x^2}-a\right)=\lim_{x \to 0^+} \dfrac{f(x) - ax - b}{x} = 0$$
we must have $a=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determining probability of a draw in simple coin flipping game Consider a 2-player game where each player flips a coin every round. If a player flips heads, they get a point for that round. Either both players get a point, only one, or none. The game ends when either of the players get to 10 points. If the game ends with both players getting 10 points, then a draw occurs. What is the probability of getting a draw?
I solved a simpler version of this problem (players only needed 3 points to win) using a Markov chain, where each game state is the score. For the simpler case I have these states:
0-0, 1-0, 1-1, 2-0, 2-1, 2-2, 3-X(win), 3-3
Players are indistinguishable from one another (2-1 and 1-2 are functionally the same) and winning scores are indistinguishable except for ties (3-0, 3-1, 3-2 are all wins for a player, 3-3 is a tie)
Then I got this Markov matrix:
$M =\begin{bmatrix} 0.25&0&0&0&0&0&0&0 \\ 0.5&0.25&0&0&0&0&0&0 \\ 0.25&0.25&0.25&0&0&0&0&0 \\ 0&0.25&0&0.25&0&0&0&0 \\ 0&0.25&0.5&0.25&0.25&0&0&0 \\ 0&0&0.25&0&0.25&0.25&0&0 \\ 0&0&0&0.5&0.5&0.5&1&0 \\ 0&0&0&0&0&0.25&0&1 \end{bmatrix}$
Calculating probabilities:
$\lim_{k \to \infty}M^k\begin{bmatrix}1\\0\\0\\0\\0\\0\\0\\0\end{bmatrix} = \begin{bmatrix}0\\0\\0\\0\\0\\0\\\frac{70}{81}\\\frac{11}{81}\end{bmatrix}$
So the draw probability in this case is $\frac{11}{81}$.
The Markov matrix for this is an 8x8, but if I were to do the same for the 10 point game, I would get a (1+2+...+10)+2=57x57 matrix, which is a bit impractical to handle.
Are there more optimized methods for this type of problem?
|
Here is a semi-solution, for general $ k $. I try and compute for $ n = 3 $. Unfortunately, it looks like I have a different value. Of course, there may be some calculation mistake. For $ k = 10$, you will need to do some calculations to finish it off.
The number of coin tosses until you get $ k $ heads follow a negative Binomial distribution with parameters $ k $ and $ p $. Here we are taking the number of tosses (not the number of failures) as the variable and a fair coin, i.e., $ p = 1/2$. Player 1 wins if the negative binomial with parameters $k$ and $1/2$ (number of tosses to get $k$ heads), associated with her coin tosses, is smaller than the corresponding negative binomial with parameters $k$ and $1/2$, associated with the coin tosses of player 2 and vice versa. Thus, the game will be a draw if and only if both negative Binomial random variables assume the same value. Since the coin tosses of player 1 and player 2 are independent, clearly the negative Binomial random variables are independent. Thus, we have the required probability as
\begin{equation*}
\alpha_k = \mathbb{P} ( X_1 = X_2 )
\end{equation*}
where $ X_1, X_2 $ are independent and both having negative Binomial distribution with parameters $ k $ and $ 1/2 $.
Writing down the pmf, we have
\begin{equation*}
\alpha_k = \mathbb{P} ( X_1 = X_2 ) = \sum_{ n = k}^{\infty} \binom{n-1}{k-1} \frac{1}{2^n} \binom{n-1}{k-1} \frac{1}{2^n} = \sum_{ n = k}^{\infty} \binom{n-1}{k-1}^2 \frac{1}{4^n} .
\end{equation*}
Now, to sum this we use the expansion of $ (1-x)^{-k} $ for $ k \geq 1 $ (here $k = 10 $). Note that
\begin{equation*}
(1-x)^{-k} = \sum_{ n = 0}^{ \infty} \binom{ n+k-1}{ n } x^n = \sum_{ n = 0}^{ \infty} \binom{ n+k-1}{ k-1 } x^n.
\end{equation*}
Now, set
\begin{equation*}
f_k (x) = x^{k-1} (1-x)^{-k}
\end{equation*}
and observe that
\begin{equation*}
f_k (x) = x^{k-1} (1-x)^{-k} = \sum_{ n = 0}^{ \infty} \binom{ n+k-1}{ k-1 } x^{n+k-1} = \sum_{ n = k }^{ \infty} \binom{ n-1}{ k-1 } x^{n-1}.
\end{equation*}
Differentiate both sides $(k-1)$ times with respect to $ x $ and taking the derivative inside the sum (can be done since the power series converges absolutely for $|x| < 1 $), we have
\begin{equation*}
f_{k}^{(k-1)} (x) = \sum_{ n = k}^{ \infty} \binom{ n-1}{ k-1 } (n-1)(n-2) \dotsm (n-k+1) x^{n-k}.
\end{equation*}
Now, dividing by $ (k-1)! $ and multiplying by $ x^{k} $, we have
\begin{align*}
\frac{x^{k}}{(k-1)!} f_{k}^{(k-1)} (x) & = \sum_{ n = k}^{ \infty} \binom{ n-1}{ k-1 } \frac{ (n-1)(n-2) \dotsm (n-k+1) }{ (k-1)!} x^{n} \\
& = \sum_{ n = k}^{ \infty} \binom{ n-1}{ k-1 } \binom{ n-1}{ k-1 } x^{n} \\
& = \sum_{ n = k}^{ \infty} \binom{ n-1}{ k-1 }^2 x^{n} .
\end{align*}
Thus, we have
\begin{equation*}
\alpha_k = \frac{x^{k}}{(k-1)!} f_{k}^{(k-1)} (x) \biggl|_{ x = 1/4} .
\end{equation*}
For $ k = 3 $, we have
\begin{equation*}
\frac{x^{3}}{(3-1)!} f_3^{(2)} (x) = \frac{x^{3}}{2} \Biggl[ 2 (1-x)^{-3} - 12 x (1-x)^{-4} + 12 x^2 (1-x)^{-5} \Biggr].
\end{equation*}
We have
\begin{align*}
\alpha_3 & = \frac{x^{3}}{(3-1)!} f_3^{(2)} (x) \biggl|_{ x = 1/4} \\
& = \frac{ 1}{2 \cdot 4^3} \Biggl[ 2 \times\Bigl( \frac{3 }{4} \Bigr)^{-3} - 12 \times \frac{1}{ 4} \times \Bigl( \frac{3 }{4} \Bigr)^{-4} + 12 \times \frac{1}{ 4^2} \times \Bigl( \frac{3 }{4} \Bigr)^{-5} \Biggr] \\
& = \frac{ 1}{ 2 \cdot 4^3} \Biggl[ 2 \times \Bigl( \frac{4 }{3} \Bigr)^{3} - 3 \times \Bigl( \frac{4 }{3} \Bigr)^{4} + 3 \times \frac{1}{ 4} \times \Bigl( \frac{4 }{3} \Bigr)^{5} \Biggr] \\
& = \frac{ 1}{2 \cdot 4^3} \times \frac{4^3}{ 3^4} \Bigl[ 2 \times 3 - 3 + 4 \Bigr] = \frac{ 7 }{ 162 }.
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/4577755",
"timestamp": "2023-03-29T00:00:00",
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|
Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x^2} d x}_K+ \underbrace{\int_0^1 \frac{\ln (1-x)}{1+x^2} d x}_L
\end{aligned}
Denotes the Catalan’s constant by $G$.
By my post, $$J=\frac{\pi}{2}\ln2-G$$
Dealing with the last $2$ integrals, we use a powerful substitution $x=\frac{1-t}{1+t} ,$ then $dx=-\frac{2dt}{(1+t)^2}.$
$$
\begin{aligned}
K&=\int_1^0 \frac{\ln 2-\ln (1+t)}{\frac{2+2 t^2}{(1+t)^2}} \frac{-2 d t}{(1+t)^2} \\
&=\ln 2 \int_0^1 \frac{d t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2}
\end{aligned}
$$
Hence $$K=\frac{\pi}{8} \ln 2 $$
$$
\begin{aligned}
L=& \int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\
&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t . \\
&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G
\end{aligned}
$$
Combining them to get
$$
\begin{aligned}
I &=\left(\frac{\pi}{2} \ln 2-G\right) +\frac{\pi}{8} \ln 2 +\left(\frac{\pi}{8} \ln 2-G\right)\\
&=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
I do want to know if it can be solved by any other elegant methods. Your comments and methods are highly appreciated.
|
Substitute $x=\tan t$. Then
$$1-x^4=4x^2\frac{1-x^2}{1+x^2}\left(\frac{1+x^2}{2x}\right)^2
=4\tan^2t\ \frac{\cos 2t}{\sin^2 2t}
$$
and
\begin{aligned}
&\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x \\
=& \int_0^{\pi/4}\ln8+2\ln \tan t +\ln (2\cos 2t)
-2\ln (2\sin 2t) \ d t\\
=&\ \frac{3\pi}4\ln2 - 2G
\end{aligned}
where $\int_0^{\pi/4} \ln \tan t\ dt=-G$ and
$\int_0^{\pi/4}\ln (2\cos 2t)dt
= \int_0^{\pi/4}\ln (2\sin 2t) d t=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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|
Linear approximation of $\sqrt[7]{e}$ I need to find a linear approximation $\sqrt[7]{e}$. I know that for $x_0 = a + \Delta x$
$$
f(x_0) \approx f(a)+f'(a)\cdot \Delta x
$$
Thus my inital function is $f(x) = \sqrt[7]{x}$ and $e = 1 + (e - 1)$, so
$$
f(e) \approx \sqrt[7]{1} + \frac{1}{7}1^{\frac{-6}{7}}\cdot(e - 1) = \frac{6}{7} + \frac{e}{7} \approx 1.2455.
$$
However, I can't say that I'm entirely pleased by the result. Even though one could argue that $f(x) = \sqrt[7]{x}$ grows in such slow manner that approximating $f(e)$ by $f(1)$ is good enough for linear approximation, I'm not convinced. So I started to wonder: does there exist a function $g(x): f(g(x)) = \sqrt[7]{g(x)}$ and $f(g(c)) = \sqrt[7]{g(c)}= \sqrt[7]{e}$ that could provide a more precise approximation?
|
$[k,k]$ Padé approximants $P_k$ of $e^{\frac{1}{n}}$ are not bad
$$P_1=\frac{2 n+1}{2 n-1}\qquad P_2=\frac{12 n^2+6 n+1}{12 n^2-6 n+1} \qquad P_3=\frac{120 n^3+60 n^2+12 n+1}{120 n^3-60 n^2+12 n-1}$$ and so on are more then decent. For $n=7$ they generate
$$\left(
\begin{array}{ccc}
k & P_k & \log_{10} \left(\left| P_k-e^{\frac{1}{7}}\right| \right)\\
1 & \frac{15}{13} & -3.55 \\
2 & \frac{631}{547} & -7.02 \\
3 & \frac{44185}{38303} & -10.86 \\
4 & \frac{4330761}{3754241} & -14.95 \\
5 & \frac{545720071}{473072669} & -19.24 \\
6 & \frac{84045221695}{72856945267} & -23.68 \\
\end{array}
\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4583585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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|
Show that the equation $X^2=J_n$has no solution if $n\geq2$, if $J_n$ is the jordan block of size $n$ with zeros on diagonal. More clearly:
$$
J_n=
\begin{pmatrix}
0 & 1 & 0 &\cdots & 0 & 0 \\
0 & 0 & 1 &\cdots & 0 & 0 \\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0 & 0 & 0 &\cdots & 1 & 0 \\
0 & 0 & 0 &\cdots & 0 & 1 \\
0 & 0 & 0 &\cdots & 0 & 0
\end{pmatrix}
$$
The solutions manual says: $X$ is nilpotent, thus $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$ and $X^2$ cannot have rank $n - 1$.
I understand that $X$ is nilpotent, but I do not understand how that implies $\operatorname{dim}(\operatorname{ker} X^2) \geq 2$.
|
If $X^2$ has rank $n-1$, then $n-1\ge\operatorname{rank}(X)\ge\operatorname{rank}(X^2)=n-1$. Therefore $\dim XV=\dim X^2V=n-1$, where $V=\mathbb F^n$. However, as $X^2V=X(XV)\subseteq X(V)=XV$, we must have $X^2V=XV$. It follows that $X^kV=XV\ne0$ for every integer $k>0$, which is a contradiction to the assumption that $X$ is nilpotent.
By the way, one may prove that the equation $X^2=J_n$ is not solvable without considering the rank of $X^2$: if $J_n=X^2$, then $X^{2n}=J_n^n=0$. Hence $X$ is nilpotent. Thus we must have $X^n=0$, because $X$ is $n\times n$. But then $0=X^nX^{n-2}=X^{2n-2}=J_n^{n-1}\ne0$, which is a contradiction.
|
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"url": "https://math.stackexchange.com/questions/4587913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Show that scalene triangle $\triangle ABC$ is a right-triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$ As the title suggests, this is a college entrance exam practice problem from Japan, it is as follows:
Given a scalene triangle $\triangle ABC$, prove that it is a right triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$
I found this problem pretty interesting, and after some thinking, I found a way to solve it, and I'll show my attempt here. I want to know, are there any other/better ways to solve this? Or is there anything about my solution that can be improved? Please let me know!
Here's my attempt:
Some have that: $$\sin(A)\cos(A)=\sin(B)\cos(B)$$
From Law of Sines we know that:
$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R$$
And from Law of Cosines:
$$\cos(B)=\frac{a^2+c^2-b^2}{2ac}$$ $$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$
Therefore:
$$(\frac{a}{2R})(\frac{b^2+c^2-a^2}{2bc})=(\frac{b}{2R})(\frac{a^2+c^2-b^2}{2ac})$$
Dividing and multiplying by $a$ on the left side and by $b$ on the right side gives us:
$$\frac{(a^2)(b^2+c^2-a^2)}{4Rabc}=\frac{(b^2)(a^2+c^2-b^2)}{4Rabc}$$
$$a^2b^2+a^2c^2-a^4=a^2b^2+b^2c^2-b^4$$
$$(a^2-b^2)c^2-(a^2-b^2)(a^2+b^2)=0$$
$$(a+b)(a-b)(-a^2-b^2+c^2)=0$$
Now, obviously the case $a=-b$ cannot be true. We also know that $a=b$ does not work in this particular case since this triangle is scalene, therefore we're left with:
$$-a^2-b^2+c^2=0$$
$$a^2+b^2=c^2$$
And this proves that the triangle is right angled
|
Your method is correct.
If you can use trigonometric identities, proceed as follows:
$2\sin(A)\cos(A)=2\sin(B)\cos(B)$
$\sin(2A)=\sin(2B)$
Thus $2A$ and $2B$ are either equal or supplementary, so correspondingly $A$ and $B$ are equal or complementary. But equality of angles implies the triangle is isosceles, a contradiction; so $A$ and $B$ must be complementary from which the third angle ($\angle C$) has to measure $180°-90°=90°$ and thus the triangle is right.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4588868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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|
Evaluate $\int_0^1\arcsin^2(\frac{\sqrt{-x}}{2}) (\log^3 x) (\frac{8}{1+x}+\frac{1}{x}) \, dx$ Here is an interesting integral, which is equivalent to the title
$$\tag{1}\int_0^1 \log ^2\left(\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}\right) (\log ^3x) \left(\frac{8}{1+x}+\frac{1}{x}\right) \, dx = \frac{5 \pi ^6}{1134}-\frac{22 \zeta (3)^2}{5}$$
Question: how to prove $(1)$?
Using power expansion of $\log^2(\sqrt{x+1}-\sqrt{x})$, we can derive an equivalent form of $(1)$:
$$\tag{2}\sum _{n=1}^{\infty } \frac{1}{n^2 \binom{2 n}{n}} \left(8 \sum _{j=1}^n \frac{(-1)^j}{j^4}+\frac{(-1)^n}{n^4}\right)=-\frac{22 \zeta (3)^2}{15}-\frac{97 \pi ^6}{34020}$$
Letting $\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}=\sqrt{u+1}$ gives
$$\tag{3}\int_0^{\phi} \log^2 (1+u) \log^3\left(\frac{u^2}{1+u}\right) \frac{(u+2)(9 u^2+u+1)}{u (u+1) (u^2+u+1)} du = \frac{10 \pi ^6}{567}-\frac{88 \zeta (3)^2}{5}$$
with $\phi = (\sqrt{5}+1)/2$. But all these variations look equally difficult.
Any idea is welcomed.
|
Too long for comment: Let
$$I_1 = \int_0^1\log^2\left(\sqrt{\frac x4+1}+\sqrt{\frac x4}\right) \log^3(x) \, \frac{dx}x \\
I_2 = \int_0^1 \log^2\left(\sqrt{\frac x4+1}+\sqrt{\frac x4}\right) \log^3(x) \, \frac{dx}{x+1}$$
so the desired integral is $I=I_1+8I_2$.
Taking a cue from Ali Shadhar's solution to a related problem, we may be able to handle $I_1$ via the chain of substitutions
$$u = \log\left(\sqrt{\frac x4}+\sqrt{\frac x4+1}\right) =\operatorname{arsinh}\left(\sqrt{\frac x4}\right) \to v=e^u \to w=\sqrt{v+1}$$
Then
$$\begin{align*}
I_1 &= 16\int_0^{\log(\phi)}u^2\log^3(2\sinh(u))\coth(u)\,du \\[1ex]
&= 16\int_1^\phi\log^2(v)\log^3\left(\frac{v^2-1}v\right)\frac{v^2+1}{v^2-1}\,\frac{dv}v \\[1ex]
&= 2\int_0^\phi \log^2(w+1) \left[\log^3(w)-\frac32\log^2(w)\log(w+1)\right.\\
&\qquad\qquad\qquad\qquad\qquad\left.+\frac34\log(w)\log^2(w+1)-\frac18\log^3(w+1)\right]\left(\frac2w-\frac1{w+1}\right) \, dw \\[1ex]
&= 4\int_0^\phi \frac{\log^3(w)\log^2(w+1)}w \, dw - 4\int_0^\phi \frac{\log^2(w)\log^3(w+1)}w \, dw \\
&\qquad + \frac32 \int_0^\phi \frac{\log(w)\log^4(w+1)}w\,dw - \frac15 \int_0^\phi \frac{\log^5(w+1)}w \, dw + K
\end{align*}$$
where
$$K = - \frac23 \log^3(\phi) \log^3(\phi+1) + \frac34 \log^2(\phi) \log^4(\phi+1) - \frac3{10} \log(\phi) \log^5(\phi+1) + \frac1{24}\log^6(\phi+1)$$
Fully expanding the penultimate integrand produces some other integrals with denominators $y+1$, but they can each be done by parts to get absorbed into the integrals and constant $K$ shown here.
With $f(x)=\frac{\log^a(x) \log^b(x+1)}x$, let
$$\mathcal K(a,b) = \int_0^\phi f(x) \, dx, \quad \mathcal L(a,b) = \int_0^1 \cdots ,\quad \mathcal M(a,b) = \int_{\frac1\phi}^1 \cdots, \quad \mathcal N(a,b) = \int_0^{\frac1\phi} \cdots$$
We find
$$\begin{align*}
\mathcal K(a,b) &= \int_0^\phi \frac{\log^a(x) \log^b(x+1)}x \, dx \\[1ex]
&= \int_0^1 \frac{\log^a(x)\log^b(x+1)}x \, dx + \int_{\frac1\phi}^1 \frac{\log^a(x) \left(\log(x+1)-\log(x)\right)^b}x \, dx \\[1ex]
&= \mathcal L(a,b) + \sum_{\beta=0}^b (-1)^{a+\beta} \binom b\beta \mathcal M(a+\beta,b-\beta)
\end{align*}$$
so we can rewrite $I_1$ as several "simpler" integrals,
$$\begin{align*}
I_1 &= 4 \mathcal K(3,2) - 4 \mathcal K(2,3) + \frac32 \mathcal K(1,4) - \frac15 \mathcal K(0,5) + K \\[1ex]
&= -4 \mathcal L(2,3) + 2 \mathcal M(2,3) + 3 \mathcal M(3,2) + 2 \mathcal M(4,1) - \frac32 \mathcal M(5,0) + \frac32 \mathcal N(1,4) + 4 \mathcal N(3,2) - \frac15 \mathcal K(0,5) + K
\end{align*}$$
If Ali's answer is any indication, this should resolve into something involving polylogarithms of $\phi$ and zetas. I suppose zeta-free terms vanish in the end by virtue of a polylog ladder. Not sure if $I_2$ can be approached in the same way.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $1^3 + 2^3 + ... + 2004^3$ is divisible by $2005$. This is a first year high school problem. There's no series or any other more "advanced" math involved. There should just be a way to factor $2005$ out of the sum.
Judging by similar problems I assume that it is, in general, true that, for a natural number $n>2$, $n$ divides $1^3 + 2^3 + \cdots + (n-1)^3$.
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Here's maybe a more 9th-grade-friendly approach, assuming you know how to factorise a sum of two cubes:
$\begin{eqnarray} S & = & 1^3 + 2^3 + \ldots + 2003^3 + 2004^3 \\
& = & (1^3 + 2004^3) + (2^3 + 2003^3) + \ldots + (1002^3 + 1003^3) \\
& = & (1 + 2004)(1^2 + 1 \times 2004 + 2004^2) + (2 + 2003)(2^2 + 2 \times 2003 + 2003^2) + \ldots + (1002 + 1003)(1002^2 + 1002 \times 1003 + 1003^2) \\
& = & 2005\left(1^2 + 1 \times 2004 + 2004^2 + 2^2 + \ldots) \right) \end{eqnarray}$
Note that this trick only works because we have an even number of terms so we can pair them up, so for example $1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225$ which is not divisible by 6.
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{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
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Observe $\sqrt {1-x^2}≥0$ implies that, $x≥0.$
Thus, under the restriction $0≤x≤\frac {\sqrt 2}{2}$ we have:
$$\begin{align}x\sqrt{1-x^2}&=\sqrt{1-x^2}-x\\
\left(x\sqrt {1-x^2}\right)^2&=1-2x\sqrt {1-x^2}\end{align}$$
Letting $x\sqrt {1-x^2}=u$, we have:
$$\begin{align}&u^2=1-2u\\
\implies &u^2+2u-1=0\end{align}$$
This leads to:
$$\begin{align}&u=\sqrt{1-x^2}-x\\
\implies &ux=u-x^2,\,x≠0\\
\implies &x^2+ux-u=0.\end{align}$$
Finally, note that all possible roots of the original equation must satisfy the inequality $0≤x≤\frac {\sqrt 2}{2}.$
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"timestamp": "2023-03-29T00:00:00",
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|
Integration of Gauss ${}_{2}F_{1}$ hypergeometric function The indefinite integral representation of Gauss hypergeometric function is
$$\int {{}_2{F_1}\left( {a,b;c;z} \right)dz = \frac{{c - 1}}{{\left( {a - 1} \right)\left( {b - 1} \right)}}} {}_2{F_1}\left( {a - 1,b - 1;c - 1;z} \right)$$
However I don't understand how to find the following definite integral:
$$\int\limits_{ - \infty }^\infty {\left( {1 + \frac{{{z^2}}}{{{n_2}{{\cos }^2}\left( \theta \right)}}} \right){}_2{F_1}\left( {a,b;\frac{1}{2};\frac{{{\gamma ^2}}}{{4{\alpha _1}{\alpha _2}}}} \right)dz}$$
where;
$$\gamma = \frac{{2yz\sin \left( \theta \right)}}{{\sqrt {{n_1}{n_2}} {{\cos }^2}\left( \theta \right)}},\quad {\alpha _1} = 1 + \frac{{y^2}}{{{n_1}{{\cos }^2}\left( \theta \right)}},\quad {\alpha _2} = 1 + \frac{{z^2}}{{{n_2}{{\cos }^2}\left( \theta \right)}}$$
I would appreciate it if anyone had a solution to this problem. Thank you in advance.
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Consider the integral in the form
$$ I = \int_{-\infty}^{\infty} (1 + p \, t^2) \, {}_{2}F_{1}\left(a, b; c; \frac{q^2 \, t^2}{4 \, \alpha \, (1 + p \, t^2)} \right) \, dt. $$
Converting the ${}_{2}F_{1}$ into its series form then
$$ I = \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b)_{n}}{n! \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha}\right)^n \, J_{n} $$
where $J_{n}$ is the integral
$$ J_{n} = \int_{-\infty}^{\infty} \frac{t^{2 n} \, dt}{(1 + p \, t^2)^{n-1}}. $$
Since
\begin{align}
\int_{-\infty}^{\infty} f(t^2) \, dt &= \int_{-\infty}^{0} f(t^2) \, dt + \int_{0}^{\infty} f(t^2) \, dt \\
&= 2 \, \int_{0}^{\infty} f(t^2) \, dt \hspace{5mm} \text{let} \, t = -u \, \text{in the first integral} \\
&= \int_{0}^{\infty} f(u) \, u^{-1/2} \, du \hspace{5mm} \text{where} \, t = \sqrt{u} \, \text{was used}.
\end{align}
With this then $J_{n}$ becomes
$$ J_{n} = \int_{0}^{\infty} u^{n-1/2} \, (1 + p \, u)^{1-n} \, du.$$
Comparing this to the hypergeometric function ${}_{2}F_{1}$ integral form, namely,
$$ \int_{0}^{\infty} t^{c-b-1} \, (1+t)^{c-a} \, (1-z+t)^{-a} \, dt = \frac{\Gamma(b) \, \Gamma(c-b)}{\Gamma(c)} \, {}_{2}F_{1}(a, b; c; z).$$
In the case $c = a$ this reduces to
\begin{align}
\int_{0}^{\infty} t^{a-b-1} \, (1-z+t)^{-a} \, dt &= \frac{\Gamma(b) \, \Gamma(a-b)}{\Gamma(a)} \, {}_{2}F_{1}(a, b; a; z) \\
&= \frac{\Gamma(b) \, \Gamma(a-b)}{\Gamma(a)} \, (1-z)^{-a}.
\end{align}
Now,
\begin{align}
J_{n} &= \int_{0}^{\infty} u^{n-1/2} \, (1 + p \, u)^{1-n} \, du \\
&= p^{1-n} \, \int_{0}^{\infty} u^{n-1/2} \, \left(\frac{1}{p} + t\right)^{-(n-1)} \, du \\
&= p^{1-n} \, \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(n + \frac{1}{2}\right)}{\Gamma(n-1)} \, \left(\frac{1}{p}\right)^{3/2} \\
&= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \frac{\left(\frac{1}{2}\right)_{n}}{(n-2)! \, p^{n}}.
\end{align}
Returning to $I$:
\begin{align}
I &= \sum_{n=0}^{\infty} \frac{(a)_{n} \, (b)_{n}}{n! \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha}\right)^n \, J_{n} \\
&= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, (a)_{n} \, (b)_{n}}{(n-2)! \, (1)_{n} \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha \, p}\right)^n \\
&= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \sum_{n=2}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, (a)_{n} \, (b)_{n}}{(n-2)! \, (1)_{n} \, (c)_{n}} \, \left(\frac{q^2}{4 \, \alpha \, p}\right)^n \\
&= \frac{\Gamma\left(- \frac{3}{2}\right) \, \Gamma\left(\frac{1}{2}\right)}{\sqrt{p}} \, \frac{\left(\frac{1}{2}\right)_{2} \, (a)_{2} \, (b)_{2}}{(1)_{2} \, (c)_{2}} \, \sum_{n=0}^{\infty} \frac{\left(\frac{5}{2}\right)_{n} \, (a+2)_{n} \, (b+2)_{n}}{n! \, (3)_{n} \, (c+2)_{n}} \, \left(\frac{q^2}{4 \, \alpha \, p}\right)^{n+2} \\
I &= \frac{\pi \, (a)_{2} \, (b)_{2}}{2 \, (c)_{2}} \, \left(\frac{q^2}{4 \, \alpha \, p} \right)^{2} \, {}_{3}F_{2}\left(\frac{5}{2}, a+2, b+2; 3, c+2; \frac{q^2}{4 \, \alpha \, p} \right).
\end{align}
Associating the constants will lead to the desired result.
For the case of $c = \frac{1}{2}$ the given result has a nice reduction as seen by
\begin{align}
I_{c = 1/2} &= \int_{-\infty}^{\infty} (1 + p \, t^2) \, {}_{2}F_{1}\left(a, b; \frac{1}{2}; \frac{q^2 \, t^2}{4 \, \alpha \, (1 + p \, t^2)} \right) \, dt \\
&= \frac{\pi \, (a)_{2} \, (b)_{2} \, q^4}{24 \, (\alpha \, p)^2} \, {}_{2}F_{1}\left(a+2, b+2; 3; \frac{q^2}{4 \, \alpha \, p} \right).
\end{align}
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"timestamp": "2023-03-29T00:00:00",
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|
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!
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Here is another way to solve for $\cos(C)$, by area calculations, done in 2 ways.
From Bob Dobb' answer, we have:
$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = \sqrt{\frac{128}{5}}$
$\displaystyle Δ = \frac{a\,b\,c}{4\,R} \qquad \qquad \quad
→ Δ^2 = \frac{125}{32}\,c^2$
Using my own triangle area formula:
$\displaystyle z = \frac{c^2 - (a-b)^2}{4} = \frac{c^2-1}{4}$
$\displaystyle Δ = \sqrt{(a\,b-z)\,z} \qquad → Δ^2 = \frac{-81+82\,c^2-c^4}{16}$
Area squared calculations, done 2 ways, must match:
$\displaystyle \frac{125}{32}\,c^2 - \frac{-81+82\,c^2-c^4}{16} = \frac{(c^2-6)(c^2-\frac{27}{2})}{16} = 0$
$c^2=6 \qquad → \cos(C) = \frac{5^2 + 4^2 - 6}{2×5×4} = \frac{7}{8}$
$c^2=\frac{27}{2} \quad\; → \cos(C) = \frac{5^2 + 4^2 - \frac{27}{2}}{2×5×4} = \frac{11}{16}$
It is interesting we have 2 solutions for $\cos(C)\;$!
If we pick one solution for $\cos(C)$, the other is $\cos(A-B)\;$!
$\displaystyle \cos(A-B) = \frac{7}{8} \quad → \cos(C) = \frac{11}{16}$
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"timestamp": "2023-03-29T00:00:00",
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|
Computing Denominator of Bayes theorem - independently vs conditionally Lets say that you have n independent variables (assume n is even) $x_1$ to $x_n$,
Let $X = [x_1, x_2, x_3, ...., x_n]$
We have another variable $y$ that can only take two values $0$ and $1$ such that
$P(y=0)=P(y=1)=0.5$,
Given that $P(x_i=1|y=1) = a$, $P(x_i=1|y=0)=b$ for all $i$ from $1$ to $n$
Find $P(y=1|X)$ where $X = [1,0,1,0,....,1,0]$.
If we use Baye's theorem, we get
$$
P(y=1|X) = \frac{P(X|y=1)P(y=1)}{P(X)}
$$
Now to compute the denominator $P(X)$,
Option 1 (compute independently)
should I write it as $P(X) = P(x_1=1)P(x_2=0)P(x_3=1)...P(x_{n}=0)$
To solve this, I first find $P(x_i=1) = P(x_i=1|y=1)P(y=1) + P(x_i=1|y=1)P(y=1)$
This gives us,
$P(x_i=1)=(a+b)/2$
and $P(x_i=0)=1-(a+b)/2$
So the final answer using this approach is
$P(X) = ((a+b)/2)^{n/2} (1-(a+b)/2)^{n/2}$
OR
Option 2 (compute conditionally)
should I write it as $P(X) = P(X|y=1)P(y=1) + P(X|y=0)P(y=0)$
To solve this we have,
$P(X|y=1) = P(x_1=1|y=1)P(x_2=0|y=1)...P(x_n=0|y=1)$
$P(X|y=1) = a(1-a)a(1-a)...(1-a)$
Similarly
$P(X|y=0) = b(1-b)b(1-b)...(1-b)$
This gives us the final answer as
$P(X) = a^{n/2}(1-a)^{n/2}0.5 + b^{n/2}(1-b)^{n/2}0.5$
Answers from the above two options are clearly different, which one is correct?
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You're asking which of two options you should pick, but I think both of those ideas will be important here!
For convenience I'll assume $n$ is even. If it's odd you need to handle the end of the products carefully since $X$ will have more $1$s than $0$s.
First, for any $k$ we have
$$
\begin{align}
P(x_k = 0) &= P(x_k = 0 | y = 0) P(y=0) + P(x_k=0|y=1) P(y=1) = \frac {1-b} 2 + \frac {1-a} 2 = \frac{2-a-b}2
\end{align}$$
and
$$P(x_k = 1) = P(x_k = 1 | y = 0) P(y=0) + P(x_k=1|y=1) P(y=1) = \frac b 2 + \frac a 2 = \frac{a+b}2.$$
Now we'll use those to compute $P(X)$:
$$
\begin{align}
P(X) &= P(x_1=1)P(x_2=0)P(x_3=1)...P(x_{n}=0) \\
&= \left( \frac{a+b}{2} \right) \left( \frac{2-a-b}{2} \right) \left( \frac{a+b}{2} \right) \cdots \left( \frac{2-a-b}{2} \right) \\
&= \frac 1 {2^n} (a+b)^{n/2} (2-a-b)^{n/2}.
\end{align}
$$
Side note: You initially said your problem was "Find $P(X | y=1)$" but then in the next line you started trying to compute $P(y=1|X)$. I'm kind of assuming the problem really said to compute $P(y=1|X)$, which means your Bayes' Theorem approach is the right idea. But just in case I'll say: if you really did want to compute $P(X | y=1)$ then that wouldn't be a Bayes' Theorem situation at all; you could just write $P(X | y=1) = P(x_1=1|y=1) P(x_2=0|y=1) P(x_3=1|y=1) \cdots P(x_n=0|y=1)$ and go from there.
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How to prove $\sum_{n=1}^{+\infty}\dfrac{1}{(n+1)\sqrt[e]{n}}$ converges to a number $I can prove that $\sum_{n=1}^{+\infty}\dfrac{1}{(n+1)\sqrt[e]{n}}$ converges, but don't know how to transform the $\sqrt[e]{n}$ item to prove it less than $e$, could someone help/hint me?
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We use $\int x^{-1-1/e} dx = -ex^{-1/e} + C$.
$$\sum_{n=1}^{\infty} \frac{(1+1/n)^{1/e}}{(n+1)^{1+1/e}} = \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} + \frac{1}{4 \times 3^{1/e}} + \sum_{n=4}^{\infty} \frac{(1+1/n)^{1/e}}{(n+1)^{1+1/e}} $$
$$ \le \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} + \frac{1}{4 \times 3^{1/e}} + (\frac{5}{4})^{1/e} \int_4^{\infty} x^{-1-1/e} dx $$
$$ = \frac{1}{2} + \frac{1}{3 \times 2^{1/e}} + \frac{1}{4 \times 3^{1/e}} + (\frac{5}{16})^{1/e} e \approx 2.697177 < e$$
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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.