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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5x + q $$
$$ \text{Since } x + 1 \text{ and } x - 3 \text{ are factors of } f(x), \text{ then } f(-1) = 0 \text{ and } f(3) = 0, \text{ i.e.} $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \color{red}{\leftarrow (1)} \\
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \color{blue}{\leftarrow (2)}
\end{align}
$$ \text{From } \color{red}{(1)}: $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \\
1 + p(-1) + 5(1) + (-5) + q &= 0 \\
1 - p + 5 - 5 + q &= 0 \\
1 - p + q &= 0 \\
q &= p - 1 \color{limegreen}{\leftarrow (3)}
\end{align}
$$ \text{From } \color{blue}{(2)}: $$
\begin{align}
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \\
81 + 27p + 45 + 15 + q &= 0 \\
27p + q + 60 + 81 &= 0 \\
27p + q + 141 &= 0 \\
q &= -27p - 144 \color{orange}{\leftarrow (4)}
\end{align}
$$ \color{orange}{(4)} + \color{limegreen}{(3)}: $$
\begin{align}
-27p - 144 &= p - 1 \\
-27p - p &= 144 - 1 \\
-28p &= 143 \\
28p &= -143 \\
p &= -\frac{143}{28} \\
\therefore p &= -5\frac{3}{28} \color{mediumpurple}{\leftarrow (5)}
\end{align}
$$ \text{Substitute } \color{mediumpurple}{(5)} \text{ into } \color{limegreen}{(3)}: $$
\begin{align}
q &= -5\frac{3}{28} - 1 \\
\therefore q &= -6\frac{3}{28}
\end{align}
The answers were $ p = -5, q = -6 $.
Where did I go wrong?
| Your equation $4$ is not right.
It would be $ q=-27p-141 $
Then you will equate equation $4$ and $3$
$$ -27p-141=p-1 $$
You will get $p$ as $-5$.
Substitute $p$ as $-5$ in equation $3$ and we will get $q$ as $-6$
| {
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"url": "https://math.stackexchange.com/questions/3872075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Compute (if it exists) the one-sided limit of $\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $
Compute (if it exists) the one-sided limit of $$\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $$
I'm finding the question really confusing, especially the one-sided limit of $x$ combined with the series. I've attempted to solve the question the following way using Cesaro summability, but my answer is wrong (the limit exists). Is my choice of method wrong, or have I made some other mistake?
Let $x=1$ so we are looking for $\sum_{k=0}^{\infty} (-1)^k \ k = 0-1 +2-3+4-5+ \dots $The first few partial sums are $0,-1,1,-2,2,-3,3,-4, \dots$ which leads me to think that I should check for Cesaro summability.
Denote $s_n, \ n \geq 0$ as the partial sum. Define $$\sigma_n = \frac{s_0 + s_1 + \dots + s_n}{n+1}$$
Notice that if $n$ is even, then $\sigma_n = 0$. If $n$ is odd, then $\sigma_n = \frac{s_n}{n+1}$. Evaluating the first few $\sigma_n$ gives us $0,-\frac{1}{2}, 0, -\frac{1}{2}, 0,-\frac{1}{2}, 0, -\frac{1}{2}, \dots$. Since $\underset{n \rightarrow \infty}{\lim} \sigma_n$ does not exist, we should check if the series is $(C,2)$ summable.
Define $$\tau_n = \frac{\sigma_0 + \dots + \sigma_n}{n+1}$$
The first few $\tau$:s are $0, -\frac{1}{2}\cdot \frac{1}{2}, -\frac{1}{2}\cdot\frac{1}{3}, -\frac{1}{2}\cdot\frac{1}{4}, \dots$
So $$\tau_n = 0 + \sum_{k=1}^n \left(\frac{1}{1+k}\right)\left( -\frac{1}{2}\right) = 0 + \sum_{k=1}^n -\frac{1}{2+2k}$$
Taking the limit of $\tau$ as $n \rightarrow \infty$ shows that $\tau_n$ diverges, thus $\underset{x \rightarrow 1-}{\lim} \sum_{k=0}^{\infty} (-1)^k \ k \ x^k $ does not exist.
Note: The correct answer is $-\frac{1}{4}$
| Since$$|x|<1\implies\frac1{1+x}=1-x+x^2-x^3+\cdots,$$you have (again, if $|x|<1$)\begin{align}\frac1{(1+x)^2}&=-\left(\frac1{1+x}\right)'\\&=1-2x+3x^2-4x^3+\cdots\\&=\sum_{k=1}^\infty(-1)^{k+1}kx^{k-1}\end{align}and therefore$$|x|<1\implies\frac x{(1+x)^2}=-\sum_{k=0}^\infty(-1)^kkx^k.$$So,$$\lim_{x\to1^-}\sum_{k=0}^\infty(-1)^kkx^k=-\lim_{x\to1^-}\frac x{(1+x)^2}=-\frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n + n + 3$ with $a_{0} = 1$ and $a_{1} = 4$ This is a nonhomogeneous recurrence relation, so there is a homogeneous and a particular solution.
Homogenous:
$a_n - 4a_{n-1} + 3a_{n-2} = 0$
$r^2 - 4r + 3 = 0$
$(r - 3)(r - 1)$
$a_n^h = \alpha(3^n) + \beta(1^n)$
This is where my solution stops because I don't know how to solve the particular solution since it would be $a_n - 4a_{n-1} + 3a_{n-2} = 2^n + n + 3$ and I'm not sure what form it should be. Would it be $A_0(r^n) + A_1(n) + A_2$ where $A_n$ is a constant or not?
I've tried solving it with that form and it ended like this:
$A_0(2^n) + A_1(n) + A_2 - 4(A_0(2^{n-1}) + A_1(n-1) + A_2) + 3(A_0(2^{n-2}) + A_1(n-2) + A_2) = 2^n + n + 3$
After simplifying and dividing $2^{n-2}$:
$A_0(2^n) - 4A_0(2^{n-1}) + 3A_0(2^{n-2}) - 4 = n + 3 + 2A_1(n) + 2A_2 - 2A_1$
And that's where I stop since I don't know what to do next.
Thanks for answering.
| So we have
$$
a_{\,n} - 4a_{\,n - 1} + 3a_{\,n - 2} = 2^{\,n} + n + 3 = q(n)
$$
and the solutions to the homogeneous equations are
$$
3^{\,n} ,\;1
$$
The homogeneous equation has constant coefficients and
$$
q(n) = 2^{\,n} + \left( {n + 3} \right)
$$
is the sum of two terms of the form
$$
c^{\,n} \cdot {\rm polynomial}(n)
$$
Then the theory says that in this case we can look for particular solutions of the form
$$
2^{\,n} \left( {An + B} \right),\quad C\left( {n + 3} \right)^{\,2} + D\left( {n + 3} \right) + E
$$
(method of Undetermined Coefficients).
Since the constant term $E$ is already a homogeneous solution we can omit it and with simple passages we get
$$
A = 0,\;B = - 4,\;C = - 1/4,\;D = - 1
$$
So the solution is
$$
a_{\,n} = \alpha \,3^{\,n} + \beta - 4 \cdot 2^{\,n} - {{\left( {n + 3} \right)^{\,2} } \over 4} - \left( {n + 3} \right)
$$
| {
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"url": "https://math.stackexchange.com/questions/3873779",
"timestamp": "2023-03-29T00:00:00",
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Show that $a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \zeta(2p)a^{2p}$ I'm trying to solve the problem 14.3.9 (Applications of Fourier Series) from Arfken's Mathematical Methods For Physicists:
a) Show that the fourier expansion of $\cos(ax)$ is:
\begin{equation}
\cos(ax) = \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} + \sum_{n=1}^{\infty} \dfrac{(-1)^n}{a^2-n^2} \cos(nx) \right)
\end{equation}
b) From the preceding result show that:
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \zeta(2p)a^{2p}
\end{equation}
where $ \zeta(2p)$ is the riemann zeta function $ \zeta(2p) = \sum_{n=1}^{\infty} \dfrac{1}{n^{2p}}$
I´ve already solved part a), but im stuck on part b), what i did was the following, first i evalueted $\cos(ax)$ at $x=\pi$:
\begin{equation}
\cos(a\pi) = \dfrac{2a\sin(a\pi)}{\pi}\left( \dfrac{1}{2a^2} + \sum_{=1}^{\infty} \dfrac{(-1)^n}{a^2-n^2} \cos(n\pi) \right)
\end{equation}
and after some algebra i ended up with this:
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\left( \dfrac{a^2}{n^2-a^2}\right)
\end{equation}
which is the part i'm stuck, i'm not sure how to relate this last expression with $\sum_{p=1}^{\infty} \zeta(2p)a^{2p}$, i was thinking to use the geometric series and tried something like this:
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^p
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \left(\dfrac{a}{n}\right)^{2p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \dfrac{1}{n^{2p}}a^{2p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \sum_{p=1}^{\infty} \zeta(2p)a^{2p}
\end{equation}
but i get a different result and i don't know in which part i was wrong or if i'm missing something. Any help would be appreciated, thanks.
| We have \begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \dfrac{a^2}{n^2} \left( \dfrac{1}{1-\dfrac{a^2}{n^2}} \right)
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty} \sum_{p=0}^{\infty} \dfrac{a^2}{n^2}\left(\dfrac{a^2}{n^2}\right)^p
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=0}^{\infty} \left(\dfrac{a^2}{n^2}\right)^{p+1}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{n=1}^{\infty}\sum_{p=1}^{\infty} \left(\dfrac{a^2}{n^2}\right)^{p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2\sum_{p=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{n^{2p}}a^{2p}
\end{equation}
\begin{equation}
a\pi\cot{a\pi} = 1-2 \sum_{p=1}^{\infty} \zeta(2p)a^{2p}
\end{equation}
| {
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"url": "https://math.stackexchange.com/questions/3874773",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$
Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$
Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq
\frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$
My Solution.
We write the inequality as
$$\big[2(b+c)-a\big]^2 \big[4\left( {{\mkern 1mu} b + {\mkern 1mu} c} \right){a^2} + 5{\mkern 1mu} \left( {2{\mkern 1mu} c + b} \right)\left( {c + 2{\mkern 1mu} b} \right)a + 3{\mkern 1mu} \left( {b + c} \right)\left( {6{\mkern 1mu} {b^2} + 19{\mkern 1mu} bc + 6{\mkern 1mu} {c^2}} \right)\big]$$
$$+\big[a-2(b+c)\big]\left( {b}^{2}+3bc+{c}^{2} \right) \left( 36{b}^{2}+77bc+36{ c}^{2} \right)+2bc \left( b+c \right) \left( b-c \right) ^{2}\geq 0,$$
which is true.
You can see the text to check here. I'm hoping for alternative proof..
Thank you!
| Suppose $a+b+c=1,$ let $p=a+b+c,\,q=ab+bc+ca,\,r=abc$ then $p \leqslant \frac{1}{3}.$ We write inequality as
$$\frac{5(1-2q+3r)}{q-r} \geqslant \frac{1-2q}{q}+10,$$
or
$$(1+23q)r+2q(2-9q) \geqslant 0. \quad (1)$$
If $0 < q \leqslant \frac{2}{9}$ then $(1)$ is true.
If $\frac{2}{9} < q \leqslant \frac{1}{4},$ from condition we get
$$(a-2b-2c)(b-2c-2a)(c-2a-2b) \geqslant 0,$$
equivalent to
$$r \geqslant \frac{18q-4}{27}.$$
Therefore, we need to prove
$$(1+23q) \cdot \frac{18q-4}{27}+2q(2-9q) \geqslant 0,$$
or
$$\frac{2(9q-2)(1-4q)}{27} \geqslant 0 \quad (\text{true}).$$
If $\frac{1}{4} < q \leqslant \frac 13,$ then
$$\frac{a^2+b^2+c^2}{ab+bc+ca}+10 \leqslant 12.$$
We need to prove
$$5\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right) \geqslant 12.$$
Let $t = \frac{a}{b+c} \geqslant 2,$ using the Cauchy-Schwarz inequality we have
$$\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{(b+c)^2}{a(b+c)+2bc} \geqslant \frac{2}{2t+1}.$$
It's remain to prove that
$$5t + \frac{10}{2t+1} \geqslant 12,$$
or
$$\frac{(10t+1)(t-2)}{2t+1} \geqslant 0.$$
Which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| The Contradiction method also helps here.
Indeed, let $b+c>2,$ $b=kx$, $c=ky$, where $k>0$ and $x+y=2$.
Thus, $$k(x+y)>2,$$ which gives $k>1$ and
$$3=b^2+bc+c^2=k^2(x^2+xy+y^2)>x^2+xy+y^2,$$ which is a contradiction because we'll prove now that $$x^2+xy+y^2\geq3.$$
Indeed, we need to prove that:
$$x^2+xy+y^2\geq3\left(\frac{x+y}{2}\right)^2$$ or
$$(x-y)^2\geq0,$$ which is obvious.
Id est, the assumption that $b+c>2$ was wrong and we are done!
| {
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How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equation is $E=\frac{1}{4}$. Now I thought that for a non-homogeneous equation, where the term $d(n)$ does not depend on the underlying recursion has the form $d(n) = k\mu^n$ and $\mu$ is not a root of the homogeneous equation, the solution is given by
$$a_n = \frac{k\mu^n}{\Phi(\mu)}$$
where $\Phi$ is the characteristic equation of the homogeneous one. In my case $d(n) = \frac{1}{4}\frac{2}{3}^{n}$, so $k=\frac{1}{4}$ and $\mu = \frac{2}{3}$. Thus the solution should be given by
$$a_n = \frac{\frac{1}{4}\frac{2}{3}^n}{\frac{2}{3}-\frac{1}{4}}=\frac{\frac{1}{4}\frac{2}{3}^n}{\frac{5}{12}}=\frac{3}{5}\frac{2}{3}^n$$
However, the solution should be $$\frac{3}{5}\frac{2}{3}^n-\frac{3}{5}\frac{1}{4}^n$$. What did I wrong?
Note: the question arises from another problem, see here
| The telescoping summation helps: $$a_n=\frac{1}{4}a_{n-1}+\frac{1}{4}\left(\frac{2}{3}\right)^{n-1},$$
$$\frac{1}{4}a_{n-1}=\frac{1}{4^2}a_{n-2}+\frac{1}{4^2}\left(\frac{2}{3}\right)^{n-2},$$
$$\frac{1}{4^2}a_{n-2}=\frac{1}{4^3}a_{n-3}+\frac{1}{4^3}\left(\frac{2}{3}\right)^{n-3},$$
$$\cdot$$
$$\cdot$$
$$\cdot$$
$$\frac{1}{4^{n-2}}a_2=\frac{1}{4^{n-1}}a_1+\frac{1}{4^{n-1}}\left(\frac{2}{3}\right)^{1}.$$
Id est, $$a_n=\frac{1}{4^{n-1}}a_1+\frac{1}{4}\left(\frac{2}{3}\right)^{n-1}+...+\frac{1}{4^{n-1}}\left(\frac{2}{3}\right)^{1}=$$
$$=\frac{1}{4^{n-1}}a_1+\frac{\frac{1}{4}\left(\frac{2}{3}\right)^{n-1}\left(\left(\frac{3}{8}\right)^{n-1}-1\right)}{\frac{3}{8}-1}=\frac{a_1}{4^{n-1}}+\frac{2}{5}\left(\left(\frac{2}{3}\right)^{n-1}-\left(\frac{1}{4}\right)^{n-1}\right).$$
| {
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Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \cdot 1} = 9^2 + 7^2 = 130
$$
Therefore, $10|(9^{n+1}+7^{2n})$, is true.
Inductive Step: There exists an integer $k$ such that $(9^{n+1}+7^{2n}) = 10k$. Let $P(k) = 9^{k+1} + 7^{2k}$. Therefore,
$$
P(k+1) = 9^{(k+1)+1} + 7^{2(k+1)} = 9^{k+2} + 7^{2k+2}
$$
$$
P(k+1) = (9^1 \cdot 9^{k + 1}) + (7^2 \cdot 7^{2k})
$$
$$P(k+1) = (2+7)(9^{k+1}) + (7^2 \cdot 7^{2k}) = (2 \cdot 9^{k+1}) + (7 \cdot 9^{k+1}) + (7 \cdot 7 \cdot 7^{2k})
$$
$$
P(k+1) = (2 \cdot 9^{k+1}) + 7(9^{k+1} + 7 \cdot 7^{2k})
$$
| By the binomial theorem, $$9^{n+1}+7^{2n}=(10-1)^{n+1}-(50-1)^n=(10a+(-1)^{n+1})-(50b+(-1)^n)=10a-50b$$ since $n$ and $n+1$ have different parities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ which is not-so-usual, so I decide to write the inequality as
$$
\frac{21 + 2q}{3 + p} +
\frac{21 + 2r}{3 + q} +
\frac{21 + 2p}{3 + r} \geqslant
\frac{69}{4}
$$Where the constraint now is $p + q + r = 3$ and the equality occurs for $p = q = r = 1$. Now we are left to proving that just
$$
2\sum_{cyc}{\frac{q}{3 + p}} +
21\sum_{cyc}{\frac{1}{3 + p}} \geqslant \frac{69}{4}
$$
Now it is sufficient to prove that
$$
\sum_{cyc}\frac{q}{3 + p}\geqslant\frac{3}{4} \quad \textrm{and} \quad \sum_{cyc}\frac{1}{3 + p} \geqslant \frac{3}{4}
$$ The second is true but I can't prove that the first is true.
| we have to prove $$7\sum \frac{1}{1+a}+2\sum_{cyc}\frac{b^2}{b+ab}\ge 69/4$$
but by $AM\ge HM$ $$7\sum \frac{1}{1+a}\ge 63/4$$
also using $ab+bc+ca\le {(a+b+c)}^2/3=1/3$
$$2\sum_{cyc} \frac{b^2}{b+ab}\ge 2\frac{(a+b+c)^2}{ab+bc+ca+a+b+c}\ge 6/4$$
The conclusion is now obvious
| {
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"timestamp": "2023-03-29T00:00:00",
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Upper and Lower semicontinuity and right-left continuity. Let $f: [0,1]\rightarrow\mathbb{R}$ be a function. Is there any relation between upper-lower semicontinuity of $f$ and right-left continuity of $f$?
I mean, is there any implication?
Thanks in advance.
| The only implication is that both left + right continuity and upper + lower semicontinuity are equivalent (and, of course, are equivalent to continuity). Otherwise, we can create a table of counterexamples. Here, we define several real functions of a variable $x$, which possess the given continuity properties at $0$:
$$\begin{array}{|c|c|}
\hline & \text{Left-continuous} & \text{Right-continuous} & \text{Both} & \text{Neither} \\
\hline \text{USC} & \begin{cases} 1 & \text{if }x \le 0 \\ 0 & \text{if } x > 0 \end{cases} & \begin{cases} 0 & \text{if }x < 0 \\ 1 & \text{if } x \ge 0 \end{cases} & - &\begin{cases} 0 & \text{if }x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \\
\text{LSC} & \begin{cases} 0 & \text{if }x \le 0 \\ 1 & \text{if } x > 0 \end{cases} & \begin{cases} 1 & \text{if }x < 0 \\ 0 & \text{if } x \ge 0 \end{cases} & - &\begin{cases} 1 & \text{if }x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \\
\text{Both} & - & - & 0 & - \\
\text{Neither} & \begin{cases} 0 & \text{if }x \le 0 \\ \sin\left(\frac{1}{x}\right) & \text{if } x > 0 \end{cases} & \begin{cases} \sin\left(\frac{1}{x}\right) & \text{if }x < 0 \\ 0 & \text{if } x \ge 0 \end{cases} & - & \begin{cases} \sin\left(\frac{1}{x}\right) & \text{if }x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \\
\hline
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots
Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.
Obviously as soon as you factor this equation to $$(4x^2+8x+a^2)(x^2+ax+1)=0$$ then you have finished.
However, how am I supposed to think of factoring the equation in such a manor? That you are supposed to factor it, is obvious, but how can you find out which are its factors, aside from blind luck?
| Rewrite it as a polynomial in $a$:
$$ a^3 x + a^2 (x^2 + 1) + a ( 4x^3 + 8x^2 ) + (4x^4 + 8x^3 + 4x^2 + 8x) = 0.$$
The factorization almost immediately jumps out by observation:
$$ (a^2 + 4x^2 + 8x ) ( ax + x^2 + 1) = 0 $$
The idea of changing the variable is a common trick. It can be helpful when factoring (e.g. applying the Reminder Factor Theorem), or finding real roots (e.g. apply the quadratic discriminant to another variable).
As an example, try to factorize $ a^4 + b^4 + c^4 - 2a^2b^2 - 2b^2c^2 - 2c^2a^2$.
There are numerous approaches that one can use, so in the spirit of this question, consider it as a quadratic in $a^2$ first.
| {
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Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos\alpha = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.
But the answers are $\frac{\pi}{3}$ and $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?
| Using Principal values
$$-\dfrac\pi3\le\cos^{-1}x-\dfrac\pi3\le\pi-\dfrac\pi3 $$
$$\cos^{-1}x-\dfrac\pi3=\begin{cases} \cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) &\mbox{if } \cos^{-1}x-\dfrac\pi3\ge0\iff x\le\cos\dfrac\pi3 \\
-\cos^{-1}\left(\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2\right) & \mbox{if } x>\cos\dfrac\pi3\end{cases}$$
Now observe that $\dfrac13<\cos\dfrac\pi3=\dfrac12<\dfrac23$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $f(x)=x^2+3x$ is Continuous at $x=a$ I am attempting to give a $\delta,\epsilon$ proof that the following function, $f(x)=x^2+3x$ is continuous at $x=a$, where $a$ is any real number. I have the following:
Given any $\epsilon > 0$, find a $\delta > 0$ such that if $|x-a|<\delta$ then $|(x^2+3x)-((a^2)+3a)|<\epsilon\\ \\$
$\text{We have that:}\\ \\
\quad\quad \begin{align}|(x^2+3x)-((a^2)+3a)| &= |x-a||x+a|+3|x-a| <\epsilon\\
&=|x-a||x+a+3| <\epsilon \end{align}\\$
$\text{Attempt to Bound $|x+a+3|$: Let $\delta \leq 1$}\\ \\
\quad\quad \begin{align} &\Rightarrow |x-a|<1\\
&\Rightarrow -1 <x-a<1\\
&\Rightarrow a-1<x<a+1 ;\text{add $a+3$ in order to obtain $x+a+3$}\\
&\Rightarrow 2a+2<x+a+3<2a+4\\\end{align}$
The issue seems to be my attempt to bound $|x+a+3|$ because I am supposed to reach the following:
$\begin{align}|(x^2+3x)-((a^2)+3a)| &= |x-a||x+a|+3|x-a| <\epsilon\\
&=|x-a||x+a+3|< \epsilon\\
&=|x-a|*(2|a|+3)< \epsilon\\
&=|x-a|< \frac{\epsilon}{2|a|+3}\\ \\
&\text{Choose $\delta=min(1,\frac{\epsilon}{2|a|+3})$} \end{align}\\$
I am not exactly sure how we arrive at $2|a|+3$. An idea that I have is if $2a+2<x+a+3<2a+4$, and $a$ is a real number, then $2a+3$ is in between $2a+2$ and $2a+4$.
Any help would be greatly appreciated!
| Use the sequential criterion for continuity at $a$. Let $(a_n)_{n=1}^\infty$ be a real sequence with limit $a$. Now, it suffices to show $\lim_{n\to\infty}f(a_n)=f(a)$.
$$\lim_{n\to\infty}f(a_n)=\lim_{n\to\infty}a_n(a_n+3)=\lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}(a_n+3)=a(a+3)=f(a)$$
| {
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
And I have no idea what to do next.
Any help would be much appreciated
| Let $C$ be the solutions. Since $C = -C$ so we can look for solutions with non negative real part.
Note that $2 = |2i| = |z+i - (z-i)| \le |z+i|+|z-i|$.
Let $f(x) = |x+iy+i| + |x+iy-i|$ and note that $f$ is continuous everywhere, differentiable for $ x > 0$ and $f'(x) >0$ for such $x$. It follows from the mean value theorem that $f(x) > f(0) \ge 2$ for $x>0$. Hence it follows that $x=0$.
It is not hard to see that $|iy+i| + |iy-i| = |y+1|+|y-1| = \max(2,2|y|)$, hence
we must have $|y| \le 1$, that is $C \subset [-i,i]$.
It is straightforward to verify that $[-i,i] \subset C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove using the $\epsilon-\delta$ definition that $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$ for any $a > 0$, $a \in \mathbb{R}$. Scratch Work
Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$. Then for every $\epsilon > 0$, we aim to find a $\delta > 0$ such that
$$
0 < |x-a| < \delta \Rightarrow \left|\sin \sqrt{x} - \sin \sqrt{a}\right| < \epsilon
$$
Note that
\begin{align*}
\sin x &\le x \\
\cos x &\le 1
\end{align*}
for every $x \ge 0$.
Also
\begin{align*}
|x-a| &= |\sqrt x - \sqrt a| \cdot |\sqrt x + \sqrt a|
\end{align*}
Proof
Choose $\delta = \min\{a, \epsilon \sqrt {a}\}$. We have
\begin{align*}
0 < |x-a| < \delta &\Rightarrow \left|\sin \sqrt{x} - \sin \sqrt{a}\right| \\
&= \left|2\cos\left(\frac {\sqrt{x} + \sqrt{a}} 2\right) \cdot \sin\left(\frac {\sqrt{x} - \sqrt{a}} 2 \right) \right| \\
&\le 2 \cdot 1 \cdot \frac {\left|\sqrt {x} - \sqrt {a} \right|} 2 \\
&= \frac {|x-a|} {|\sqrt x + \sqrt a|} \\
&\le \frac 1 {\sqrt a} \cdot \epsilon \sqrt a \\
&= \epsilon
\end{align*}
I am more concerned with my workings which led me to the proof. Is there any mistake in my manipulations? Any advice would be greatly appreciated!
| I agree with user837206’s comment: your post is correct.
I suggest some tweaks:
You start with “Suppose $\lim_{x \to a} \sin \sqrt{x} = \sin \sqrt{a}$.” That’s the statement you are trying to prove. It’s just some notes, not the actual proof, so it’s not technically wrong. But make sure you are clear about what you have already proved and what you still need to prove.
You follow with “for every $\epsilon > 0$, we aim to find a $\delta > 0$ …”. I would say we can find such a $\delta > 0$.
Turning to the actual proof, you can strengthen most of your inequalities:
For every $x > 0$, $\sin x < x$, so
$$\left| 2 \cos\left(\frac{\sqrt{x} + \sqrt{a}}{2}\right) \sin\left(\frac{\sqrt{x} - \sqrt{a}}{2}\right) \right| < 2 \cdot 1 \cdot \frac{\left|\sqrt{x} - \sqrt{a}\right|}{2}.$$
By using either of these inequalities:
\begin{align*}
|x - a| < \delta \leq a &\implies x > 0 \\
&\implies \frac{1}{\left| \sqrt{x} + \sqrt{a} \right|} < \frac{1}{\sqrt{a}} \tag{1} \\
|x - a| < \delta \leq \epsilon\sqrt{a} &\implies |x - a| < \epsilon\sqrt{a} \tag{2}
\end{align*}
We can make this inequality strict:
$$\frac{|x - a|}{\left| \sqrt{x} + \sqrt{a} \right|} < \frac{1}{\sqrt{a}} \cdot \epsilon\sqrt{a}.$$
Actually, do we really need the absolute value in $\left| \sqrt{x} + \sqrt{a} \right|$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $
Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b = 0$, $\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \overrightarrow c } \right) = 0$ and $2\left| {\overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$. If $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $, then find all possible values of $\mu $.
My approach is as follows
Given $\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \overrightarrow c } \right) = 0\& 2\left| {\overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$
$\overrightarrow u = \left( {\overrightarrow b - \overrightarrow a } \right);\overrightarrow v = \left( {\overrightarrow b + \overrightarrow c } \right)$
$2\left| {\overrightarrow v } \right| = \left| {\overrightarrow u } \right| \Rightarrow 4\left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow c } \right|}^2} + 2\overrightarrow b .\overrightarrow c } \right) = \left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow a } \right|}^2} - 2\overrightarrow a .\overrightarrow c } \right)$
$\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c \Rightarrow \overrightarrow b - \overrightarrow u = \mu \overrightarrow b + 4\left( {\overrightarrow v - \overrightarrow b } \right)$
$ \Rightarrow \overrightarrow b .\overrightarrow v - \overrightarrow u .\overrightarrow v = \mu \overrightarrow b .\overrightarrow v + 4\left( {\overrightarrow v .\overrightarrow v - \overrightarrow b .\overrightarrow v } \right) \Rightarrow 5\overrightarrow b .\overrightarrow v = \mu \overrightarrow b .\overrightarrow v + 4{\left| {\overrightarrow v } \right|^2}$\
$ \Rightarrow 5\overrightarrow b .\left( {\overrightarrow b + \overrightarrow c } \right) = \mu \overrightarrow b .\left( {\overrightarrow b + \overrightarrow c } \right) + 4{\left| {\overrightarrow v } \right|^2} \Rightarrow 5\overrightarrow b .\overrightarrow b + 5\overrightarrow b .\overrightarrow c = \mu \overrightarrow b .\overrightarrow b + \mu \overrightarrow b .\overrightarrow c + 4{\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 8\overrightarrow b .\overrightarrow c $
Not able to proceed from here onward
| As $\vec a \perp \vec b$ like $\hat i \perp \hat j$, it's strategic to represent
$$ \vec c = \frac{1}{4}\vec a - \frac{\mu}{4}\vec b$$
Remaining two conditions are,
$$ (\vec b - \vec a)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)=0$$
and
$$ (\vec b - \vec a)\cdot(\vec b - \vec a) = 2^2(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)\cdot(\frac{1}{4}\vec a + \frac{4-\mu}{4}\vec b)$$
which should give you a quadratic in $\mu$.
| {
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A simple binomial inequality seeking its proof from the book? Let $k\in \mathbb{N}$ and $1\leq a \leq k$ be fixed. Imagine we have two groups $G_{1}, G_{2}$ of $k$ indistinguishable elements each forming two sequences with indices $1, \dots , k$. Choose the element with index $a$ in $G_{1}$ and the element with index $b$ in $G_{2}$. Consider the expression
$$f(k,a,b) = \binom{a+b-2}{a-1}\binom{2k-b-a}{k-a},$$
which counts the number of ways to merge the elements before the two chosen elements and after them into two sequences.
Example: $k=5$, $a=3$, $b=4$:
$ - - \circleddash - -$
$ - - - \circleddash -$
The number of ways to merge the set of elements before the two chosen is $\binom{2+3}{2}$ and to merge the set of elements after the two chosen is $\binom{2+1}{1}$. In total, we get $\binom{2+3}{2}.\binom{2+1}{1}$.
Show that
$$\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}\geq min\left(\binom{2k-1-a}{k-a},\binom{a+k-2}{a-1}\right),$$
i.e.,
$$f(k,a,b)\geq min\left(f(k,a,1),f(k,a,k)\right).$$
In other words, when $k$ and $a$ are fixed, $f$ reaches its minimum at one of the endpoints $b=1$ or $b=k$.
An ugly proof by expansion of the binomial coefficients exist, but
does anyone know an elegant proof?
| By expanding the binomials, it can be shown directly that
$$
\frac{f(k,a,b)}{f(k,a,b-1)} = \frac{\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}}{\binom{a+b-3}{a-1}\binom{2k-b-a+1}{k-a}} \\
= \frac{a+b-2}{b-1} \frac{k-b+1}{2k-a-b+1} = (1 + \frac{a-1}{b-1})(1 - \frac{k-a}{2k-a-b+1})
$$
Interpreting this ratio as a function of a continuous $b$, it is straightforward to calculate that the ratio becomes $1$ for $b^* = \frac{ak - 1}{k-1}$.
Since $ b < 2k - a +1$, both terms in the last product always fall with $b$. It follows that the ratio is smaller than $1$ for $ \frac{ak - 1}{k-1} < b$ and hence $f(k,a,b)$ is falling in that range of $b$, so the minimum is obtained at the "endpoint", for the highest $b$. Conversely, for $ \frac{ak - 1}{k-1} > b$ the ratio is larger than $1$ and hence $f(k,a,b)$ is rising in that range of $b$, so the minimum is obtained at the other "endpoint", for the lowest $b$.
Together, this proves the claim that $f(k,a,b)\geq \min\left(f(k,a,1),f(k,a,k)\right).$ $\qquad \Box$
| {
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Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.
(2) Assuming that the condition hold for $n=k.$
(3) Proving that it holds for $n=k+1$
$$3^{4(k+1) + 2} + 1 = 3^{4k + 6} + 1 = 3^4 \cdot 3^{4k+2} + 1$$
Since we assumed that $5 \mid 3^{4k+2} + 1$ we have that
$$3^4 \cdot 3^{4k+2} + 1 = 3^4 \cdot 5t, \text{ where $t \in \Bbb Z$}.$$
Thus $5 \mid 3^{4(k+1) + 2} + 1$.
Is this a valid proof? I'm not entirely sure I'm correct with this...
| An alternative strategy for this kind of problem is to consider the difference between consecutive terms.
Let $f(n)=3^{4n+2}+1$. Then $f(n+1)-f(n)=80 \cdot 3^{4 n + 2}$ is a multiple of $5$.
The claim follows by induction since $f(0)=10$ is a multiple of $5$.
(Actually, this proves that $f(n)$ is always a multiple of $10$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4}$. I noticed that
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\sqrt{n^3} \Big(\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n-1}}\Big) $$
That doesn't seem to lead to $-\frac{1}{4}$ when $n\to \infty$ . I tried another form of the original expression:
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)$$
If I multiply by the conjugate, we obtain
$$2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
Now that doesn't seem to be of any use either. Any ideas?
| Dividing both the numerator and the denominator by $\sqrt{n^3}$, you have
\begin{eqnarray}
&&\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}\\
&=&\frac{2}{\big(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}+2\big)\big( \sqrt{1-\frac1{n^2}}+1\big)}.
\end{eqnarray}
Now you can take the limit to get the result.
| {
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Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can from here find all x such that $x \in \mathbb{Z}$ ??
i did use the hint for $y = 2x+1$ so that $x =\frac{y-1}{2}$
made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$
$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$
I can not understand with the help of the clues, can some 1 some formal proof ?
| $2x+1|x^3-3x+2\iff 2x+1|8(x^3-3x+2)$
$\iff 2x+1|8(x^3-3x+2)-(2x+1)^3=-12x^2-30x+15$
$\iff 2x+1|3(2x+1)^2-12x^2-30x+15=-18x+18$
$\iff 2x+1|9(2x+1)-18x+18=27$
$\iff x\in \{-14,-5,-2,-1,0,1,4,13\}.$
| {
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Suppose $x$ and $y$ are unequal real numbers
Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$
For the above question, I got $x$ not equal to $-1,0,1$ and $y = 1/x$.
Then, I tried finding $x-y$ using $\big((x+y)^2-4xy)\big)^{\frac{1}{2}}$ and then solving for $x$ and $y$.
I think I over complicated it and there must be some easier way. Please help me out.
| Tricky one.
Let $x+y = u$, $x-y=v$.
$$ xy=1 \Rightarrow u^2-v^2 =4$$
and cubing original equation :$$ \sqrt[3]{\dfrac{u}{v}} + \sqrt[3]{\dfrac{v}{u}} = u $$
gives
$$ \dfrac{u}{v} + \dfrac{v}{u} + 3u = u^3 $$
$$ \Rightarrow u^2 + v^2 + 3u^2v = u^4v$$
Substitute $u^2=4+v^2$ ,
$$(4+v^2) + v^2 + 3(4+v^2)v = (4+v^2)^2v $$
directly yields
$$ \boxed{v^5 + 5v^3 -2v^2 + 4v =4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3902929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Distance between the curves from a particular point The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to
(A) 1
(B) 2
(C) 3
(4) 4
My approach is as follow Curve $x^2+4x+y^2$ represent a circle with centre $(-2,0)$ and radius $2$.
The parametric equation of the circle is $-2cos\theta+2$ and $2sin\theta$
$D^2=(u-2)^2+v^2$
Also $D^2=(u+2cos\theta-2)^2+(v-2sin\theta)^2$
How we will proceed from here
| The distance between the circle and $P(u,v)$ is the distance between $P$ and the centre minus the radius. So we have$$\left|\sqrt{(u+2)^2+v^2}-2\right|=\sqrt{(u-2)^2+v^2}$$Squaring,$$\begin{align*}&(u+2)^2+v^2+4-4\sqrt{(u+2)^2+v^2}=(u-2)^2+v^2\\\implies&2u+1=\sqrt{(u+2)^2+v^2}\\\implies&4u^2+4u+1=(u+2)^2+v^2\\\implies&u^2-v^2/3=1.\end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cubic Formula Doesn't Seem to Work for $x^3+3x^2-50x-52=0$ So I'll jump right into it. I've derived the following formula set for solving cubics of the form $ax^3+bx^2+cx+d=0$:
$$z_{k\pm}=\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}\pm\sqrt{\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+4b^3d}{108a^4}}}$$
$$x=z_{k\pm}-\frac{\left(\frac{-b^2}{3a^2}+\frac{c}{a}\right)}{3z_{k\pm}}-\frac{b}{3a}$$
The idea is that you solve for the 6 $z$-values defined by $k=0, 1, 2$ and each of these $k$-values has a positive and negative result based upon the selection for the $\pm$ sign. Also note that this derivation is heavily based upon this article (http://math.sfsu.edu/smith/Documents/Cubic&Quartic.pdf). The article dictates that these 6 $z$-values translate to only 3 distinct $x$-values (since cubics only have 3 solutions).
This all went smoothly and I was able to successfully show the viability of the formula set by solving the cubic $-2x^3+3x^2-x+5=0$. The formula was successful in capturing all 3 $x$-values without any issues.
However, upon further inspection of the formula, I have noticed an issue when the radicand of the square root is negative. This would make the radicand of the cube root imaginary, thus leaving $z$ with no real solutions; hence, $x$ would have no real solutions when $\frac{4ac^3+27a^2d^2-18abcd-b^2c^2+b^3d}{108a^4}\lt0$. Of course this cannot be true since every cubic has at least one real solution.
To test this theory I tried to solve
$$x^3+3x^2-50x-52=0$$
using the formula. This cubic does indeed make the radicand of the square root negative and happens to have three real roots: $-1, -1-\sqrt{52}, -1+\sqrt{52}$. However, upon solving the formula for these values, $z_{0+}$ and $z_{0-}$ were complex and $z_{1+}$ gave a domain error on my TI-84+ calculator. My reference article provides some insight into dealing with a negative radicand, but it doesn't make sense to me. Any help is greatly appreciated!
| For negative radicands, the 6 $z$-values translates into 3 distinctive $x$-values as well.
Take the example $x^3+3x^2-50x-52=0$ given in the post, $z$‘s are evaluated as
$$z_{k\pm} =\sqrt[3]{\pm \sqrt{-\frac{53^3}{27}}}= \pm\sqrt{\frac{53}3}e^{i \frac{1+4\pi k}6},\>\>\>k=0,1,2$$
and the corresponding $x$’s are
$$x_{k\pm} = z_{k\pm} +\frac{53}{3z_{k\pm}}-1
= z_{k\pm} +{z_{k\mp}}-1=2Re(z_{k\pm})-1
$$
Explicitly, the 6 $x$-values are
\begin{align}
x_{0+} & = 2\sqrt{\frac{53}3}\cos\frac\pi6-1=\sqrt{53}-1\\
x_{0-} & = -2\sqrt{\frac{53}3}\cos\frac\pi6-1=-\sqrt{53}-1\\
x_{1+} & = 2\sqrt{\frac{53}3}\cos\frac{5\pi}6-1=-\sqrt{53}-1\\
x_{1-} & = -2\sqrt{\frac{53}3}\cos\frac{5\pi}6-1=\sqrt{53}-1\\
x_{2+} & = 2\sqrt{\frac{53}3}\cos\frac{3\pi}2-1=-1\\
x_{2-} & =- 2\sqrt{\frac{53}3}\cos\frac{3\pi}2-1=-1\\
\end{align}
of which, only 3 values, $-1$, $\sqrt{53}-1$ and $-\sqrt{53}-1$, are distinctive, as expected.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.
I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.
| Consider the identity $$(2k+1)^2=4k^2+4k+1$$ which relates the sum of the odd squares to the sum of the squares.
The extra terms $4k+1$ are easily dealt with with a telescoping,
$$(k+1)^2-k^2=2k+1.$$
Altogether,
$$1^2+3^2+ \dots +(2n+1)^2=4\frac{n(n+1)(2n+1)}6+2(n+1)^2-1-n.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| $5^{6n}-2^{2n}=125^{2n}-2^{2n}$
$=(125-2)\left(125^{2n-1}+125^{2n-2}2+125^{2n-3}2^2+\cdots+125^22^{2n-3}+125\cdot2^{2n-2}+2^{2n-1}\right)$
is divisible by $3$, because $125-2$ is.
Now substitute $m=n+1$ to show that $5^{6m+6}-2^{2m+2}$ is divisible by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that $\left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n\leq \frac{x}{2}$ using induction I was asked to show that
For every $n\in\mathbb{N}$ and $x\in [0,1]$ the following is true:
\begin{equation*}
\left(1-x+\frac{x^2}{2}\right)^n-(1-x)^n\leq \frac{x}{2}.
\end{equation*}
However, I found a mistake in my proof and now I don't know how to proceed.
This was my attempt:
The case when $n=1$ is true because when $x\in [0,1]$, we have $x^2\leq x$, and so
\begin{equation}
1-x+\frac{x^2}{2}-1+x=\frac{x^2}{2}\leq \frac{x}{2}.
\end{equation}
Suppose now that the statement holds for some $k\in \mathbb{N}$,
\begin{equation}
\left(1-x+\frac{x^2}{2}\right)^k-(1-x)^k\leq \frac{x}{2}.
\end{equation}
Knowing that $0\leq x\leq 1$, we can see that $1-1\leq x$, which implies $1-x\leq 1$. Additionally, we can see that $2x\geq x$ and so $x^2\leq x$,
\begin{align}
x\geq \frac{x}{2}\geq \frac{x^2}{2}&\Rightarrow x\geq \frac{x^2}{2}\\
&\Rightarrow 0\geq \frac{x^2}{2}-x\\
&\Rightarrow 1\geq \frac{x^2}{2}-x+1
\end{align}
Thus,
\begin{align}
{ \left(1-x+\frac{x^2}{2}\right)^{k+1}-(1-x)^{k+1}}&= \left(1-x+\frac{x^2}{2}\right)^k\left(1-x+\frac{x^2}{2}\right)-(1-x)^k(1-x) \\
&\leq \left(1-x+\frac{x^2}{2}\right)^k-(1-x)^k\\
&\leq \frac{x}{2}
\end{align}
The statement holds for $k+1$.
The mistake is the part where I wrongly use
\begin{equation}
-(1-x)^k(1-x)\leq -(1-x)^k.
\end{equation}
But I don't know how to proceed now.
Any hint or advice is appreciated. Thank you.
| Rearrange the induction hypothesis so it reads $\left(1-x+\dfrac {x^2}2\right)^k \le \dfrac x2 + (1-x)^k$.
Now the induction step can go as follows:
\begin{align}\left(1-x+\frac {x^2}2\right)^{k+1} - (1-x)^{k+1} &\le \left(1-x+\frac {x^2}2\right)\left(\frac x2 + (1-x)^k\right)-(1-x)^{k+1}\\
&=\frac x2-\frac {x^2}2+\frac{x^3}4+(1-x)^{k+1}+\frac{x^2}2(1-x)^k-(1-x)^{k+1}\\
&=\frac x2 - \frac{x^2}2\left(1-\frac x2-(1-x)^k\right)\\
&\le \frac x2-\frac{x^2}2\left(1-\frac x2 - (1 - x)\right)\\
&=\frac x2 - \frac{x^3}4 \le \frac x2
\end{align}
The part where we eliminated the exponent $k$ holds since $(1-x)^k \le (1-x)$ for $x \in [0,1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $\alpha$ be a root of $(x^2-a)$ and $\beta$ be a root of $(x^2-b)$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$. QUESTION: Let $K$ be a field of characteristic different of $2$. Let $F$ be a splitting field for $(x^2-a)(x^2-b)\in K[x]$. Let $\alpha$ be a root of $x^2-a$ and $\beta$ be a root of $x^2-b$. Provide conditions over $a$ and $b$ to have $F=K(\alpha+\beta)$.
MY ATTEMPT:
Let $\alpha=\sqrt{a}$, $\beta=\sqrt{b}$ and $\gamma=\alpha+\beta$.
First of all, we have $F=K(\alpha, \beta)$ due to the definition of splitting field.
Defining $K(\alpha+\beta)=K(\gamma)$.
Let's show that $K(\alpha, \beta)\subset K(\gamma)$:
*
*From $\gamma=\alpha+\beta$ follows that
\begin{align*}
\gamma^2&=(\alpha+\beta)^2\\
&=\alpha^2+2\alpha\beta+\beta^2\\
&=(\sqrt{a})^2+2\sqrt{a}\sqrt{b} +(\sqrt{b})^2\\
&=(a+b)+2\sqrt{a}\sqrt{b}\qquad (*)\\
\end{align*}
*Now we are going to show that $\sqrt{b}\in K(\gamma)$
Indeed, multiplying both sides in $(*)$ by $\sqrt{b}$ we have:
$\gamma^2\sqrt{b}=(a+b)\sqrt{b}+2\sqrt{a}(\sqrt{b})^2$. Then $$\sqrt{b}=\frac{2b\sqrt{a}+(a+b)\sqrt{b}}{\gamma^2}\in K(\gamma)$$
*
*Similarly, $\sqrt{a}\in K(\gamma)$, this is
$\gamma^2\sqrt{a}=(a+b)\sqrt{a}+2(\sqrt{a})^2\sqrt{b}$, then
$$\sqrt{a}=\frac{(a+b)\sqrt{a}+2a\sqrt{b}}{\gamma^2}\in K(\gamma)$$
MY DOUBT:
I guess there is no conditions over $a$ and $b$ such that $\alpha=\sqrt{a}$ and $\beta=\sqrt{b}$, however I'm not sure. And I don't know how to connect this with the hypothesis that $K$ has characteristic different of two. Would you help me please?
| Once you know that $[ K( \alpha,\beta ) : K ]=4$, with basis $\{1,\alpha,\beta,\alpha\beta\}$, you can proceed as follows:
$$
\begin{pmatrix} 1 \\ \gamma \\ \gamma^2 \\ \gamma^3 \end{pmatrix}
=
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ a+b & 0 & 0 & 2 \\ 0 & a+3b & 3a+b & 0
\end{pmatrix}
\begin{pmatrix} 1 \\ \alpha \\ \beta \\ \alpha\beta \end{pmatrix}
$$
The matrix has determinant $4(b-a)$ and so is invertible iff $a\ne b$ since the characteristic of $K$ is not $2$. Therefore, $\{1,\gamma,\gamma^2,\gamma^3\}$ is also a basis and so generates the same space, that is, $K( \alpha,\beta ) = K(\gamma)=K( \alpha + \beta )$.
Bottom line: The main condition is that $[ K( \alpha,\beta ) : K ]=4$, or equivalently, $\beta \not\in K( \alpha)$.
This approach does not work in characteristic $2$ because $[K(\gamma):K]\le 2$ since $\gamma^2 = a+b \in K$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.
| The set $E$ is a ball of radius ${1\over2}$, centered at $\left({1\over2},0,0\right)$. The integrand
$$f(x,y,z):={yz\over x^2+y^2+z^2}$$
satisfies $f(x,-y,z)\equiv-f(x,y,z)$. This means that $f$ is odd with respect to the symmetry plane $y=0$ of $E$. It is then obvious that the requested integral has value $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &= 1 \\
a^2 + \frac{1}{2a^2} &= 1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &= 1 \\
1 \pm i + 4b^2 &= 2 \\
\pm i + 4b^2 &= 1 \\
4b^2 &= 1 \pm i \\
2b &= \sqrt{1 \pm i} \\
b &= \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}
Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?
| With a couple of corrections, your answer seems consistent:
$z=\dfrac{1\pm i}2\implies a=\color{red}\pm\sqrt{\dfrac{1\pm i}2}$
$\pm i+4b^2=1\implies 4b^2=1\color{red}\mp i\implies b=\color{red}\pm\dfrac{\sqrt{1\color{red}\mp i}}{2}$
| {
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"question_score": "4",
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How do I prove that if $n$ is greater or equal to $4$, then $(2n)!$ is greater than or equal to $10^n$? I used the method of induction to prove this.
For the basis step, when $n=4$, the statement is $[2(4)]!≥10^4$ which is the same as $40320≥10000$ which is true.
Next, for the inductive step, we show $\mathrm{S}_k$ implies $\mathrm{S}_{k+1}$ for some integer $n=k≥1$.
By assuming that $(2k)!≥10^k$ is true. Therefore, we need to show $(2k+2)!≥10^{k+1}$ for $n=k+1$. First, $10^{k+1}=10\cdot10^k$. By the induction hypothesis, $(2k)!≥10^k$, so $10\cdot(2k)!≥10\cdot10^k$.
Since $k≥4$, we know that $2(k+1)≥10$. So, $2(k+1)\cdot(2k)!≥10\cdot10^k$ by substitution. This simplifies to $(2(k+1))!≥10\cdot10^k$ which equals $(2(k+1))!≥10^{k+1}$ which we needed to show.
I know my expansion from $2(k+1)(2k)!$ to $(2(k+1))!$ is wrong, but I don't know what the correct way to get to $(2(k+1))!$ from $2(k+1)(2k)!$.
| With induction but more clear:
You have proved the base case, so let's continue.
Let $(2n)!\ge 10$. Then:
$$(2(n+1))!=(2n+2)!=(2n+2)(2n+1)(2n)!=(4n^2+6n+2)(2n)!\ge (4n^2+6n+2)\cdot 10^n$$
Clearly $4n^2+6n+2 \ge 10$ for $n \ge 1$ so:
$$(2(n+1))! \ge (4n^2+6n+2)\cdot 10^n \ge 10\cdot 10^n=10^{n+1}$$
$$(2(n+1))! \ge 10^{n+1}$$
And we have proved the inductive step $\blacksquare$
Just to clarify, you were actually almost there. We know that $2(k+1)\cdot (2k)! = (2k+2) \cdot (2k)! \le (2k+2)(2k+1) \cdot (2k)! = (2(k+1))!$
If $2(k+1)\cdot (2k)! \ge 10^{k+1}$ and $(2(k+1))! \ge 2(k+1)\cdot (2k)!$ then $(2(k+1))! \ge 10^{k+1}$
| {
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Recursive relation induction problem: $c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$ This is part of problem 18 of chapter 13 of Michael Spivak's Calculus:
Define $c_n = \int_{0}^{1}x^n dx$
(b) Prove: $$c_n = \frac{1}{2^n} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~n} {n \choose k}c_k$$
(c) Prove $c_n = \frac{1}{n+1}$ from this (without calculating the integral).
Note the actual question in the book is more than this, I am just condensing it to the parts I'm stuck on. I was able to do (b), but not (c). It seems to me to be another induction problem, but I am going in circles. So I assume it's true for $\forall n \leq p$, i.e. $ \forall n \leq p : c_n = \frac{1}{n+1}$. Then for $c_{p+1}$, we have:
$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k \text{ even}}^{0~ \leq ~k~ \leq ~{p+1}} {p+1 \choose k}c_k$$
Now already there's some technical issues as to whether the last term in the sum will be $c_{p+1}$ or $(p+1)c_p$, depending on whether $p+1$ is even or odd. So I split up the formula into those 2 separate cases, like so:
$$c_{n} = \frac{1}{2^{n} - 1} \cdot \sum_{k~ =~ 0}^{k ~= ~n/2-1} {n \choose 2k}c_{2k} ~~~: ~~\text{for n = even}$$
$$c_{n} = \frac{1}{2^{n}} \cdot \sum_{k~ =~ 0}^{k ~= ~(n-1)/2} {n \choose 2k}c_{2k} ~~~~~~: ~~\text{for n = odd}$$
Ok so we'll probably have to do 2 separate proofs by induction, one for when $p$ is odd, one when $p$ is even. Also we'll need to prove both the $n = 0$ and $n = 1$ base cases, but that's trivial, no problems there.
Let's try when $p$ is even. So we assume $p$ is even and $ \forall n \leq p : c_n = \frac{1}{n+1}$. Then $p+1$ is odd, so:
$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}c_{2k}$$
$$~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1}$$
And now I am stuck.
I need to somehow show $\frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1} = \frac{1}{p+2}$. I tried using the identity ${n+1 \choose k} = {n \choose k} + {n \choose k-1}$, which gives us:
$$c_{p+1} = \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} {p+1 \choose 2k}\frac{1}{2k+1}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \sum_{k~ =~ 0}^{k ~= ~p/2} \Bigr[{p \choose 2k} + {p \choose 2k-1}\Bigr]\frac{1}{2k+1}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{1}{2^{p+1}} \cdot \biggr[c_p ~+~ \Bigr[ \sum_{k~ =~ 0}^{k ~= ~p/2} {p \choose 2k-1}\frac{1}{2k+1}\Bigr]\biggr]$$
But now how to deal with this new summation that runs on odd numbers; $2k - 1$, instead of the even numbers; $2k$, that we have in our definition of $c_p$ (which we need to incorporate somehow)? I tried to transform the ${p \choose 2k - 1}$ to ${p \choose 2k - 2}$ or ${p \choose 2k}$, using the identity ${n \choose k} = \frac{n - k + 1}{k}{n \choose k - 1}$ but the coefficients didn't cancel and looked very ugly so now I have no ideas.
Any?
| Note that $$\sum_{k \ge 0} a_{2k} = \sum_{k \ge 0} \frac{1+(-1)^k}{2}a_k.$$
Taking $a_k=\binom{p+1}{k}\frac{1}{k+1}$ yields
\begin{align}
\frac{1}{2^{p+1}}\sum_{k\ge 0} \binom{p+1}{2k}\frac{1}{2k+1}
&=\frac{1}{2^{p+1}}\sum_{k\ge 0} \frac{1+(-1)^k}{2} \binom{p+1}{k}\frac{1}{k+1} \\
&=\frac{1}{2^{p+1}(p+2)}\sum_{k\ge 0} \frac{1+(-1)^k}{2} \binom{p+2}{k+1} \\
&=\frac{1}{2^{p+2}(p+2)}\left(\sum_{k\ge 0} \binom{p+2}{k+1} - \sum_{k\ge 0} (-1)^{k+1} \binom{p+2}{k+1}\right) \\
&=\frac{((1+1)^{p+2}-(p+2)) - ((1-1)^{p+2}-(p+2))}{2^{p+2}(p+2)} \\
&=\frac{2^{p+2}}{2^{p+2}(p+2)} \\
&=\frac{1}{p+2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$
Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
{\rm d}t=\dfrac{\pi }{\sqrt{2}}$.
We can obtain
$$\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t=\lim_{x\to 1}\int_0^1\frac{x\sqrt{xu}f(xu)}{\sqrt{f(x)-f(xu)}}{\rm d}u,$$
but how to go on ?
| Finally I was able to handle this problem. It is just tedious calculation. Let $u=\frac{f(t)}{f(x)}$ and then
\begin{eqnarray}t&=&1-\frac{(1-x)(x-1+\sqrt{(1-x)^2+8u(1+x)}}{2u(1+x)},\\
dt&=&-\frac12\frac{ (x-1) \left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)}{u^2(1+x)\sqrt{(1-x)^2+8u(1+x)}}du.
\end{eqnarray}
So
\begin{eqnarray}
&&\lim_{x \to 1}\int_0^x\frac{\sqrt{t}f(t)}{\sqrt{f(x)-f(t)}}{\rm d}t\\
&=&\lim_{x \to 1}\int_{\frac1{f(x)}}^1\frac{\sqrt{t}\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
}{u \sqrt{(1-u)
(x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}{\rm d}u\\
&=&\int_0^1\frac{1}{\sqrt{2u(1-u)}}du\\
&=&\frac{\pi}{\sqrt 2}.
\end{eqnarray}
Here
$$ \lim_{x \to 1}\frac1{f(x)}=0, \lim_{x \to 1}t=1, $$
and it turns out that when $x=1$, the integrand
$$ \frac{\left(4 u (x+1)+(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)\right)^2
\sqrt{\frac{(x-1) \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)}{2 u (x+1)}+1}}{u \sqrt{(1-u)
(x+1) \left(8 u (x+1)+(x-1)^2\right)} \left(\sqrt{8 u (x+1)+(x-1)^2}+x-1\right)^2}\bigg|_{x=1}=\frac{1}{\sqrt{2u(1-u)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Nicomachus theorem proof - what did I do wrong? The exercise asked me to proof
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$...$
I formulate the equation as
(1)$$a^3 = \sum_{i=0}^{a-1} (a-1)a+1+2i$$
Proof base case:
$$1^3=\sum_{i=0}^{1-1} (1-1)(1)+1+2i=1$$
Proof (a+1) case:
(2) $$(a+1)^3=a^3+3a^2+3a+1$$
(3) $$(a+1)^3=\sum_{i=0}^{(a+1)-1}((a+1)-1)(a+1)+1+2i=\sum_{i=0}^{a}(a)(a+1)+1+2i$$
factor out last summation term, so number of iterations match (1):
$$\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+(a)(a+1)+1+2a=\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+a^2+3a+1$$
rearrange the summation term, so part of equation match (1):
$$\sum_{i=0}^{a-1}[(a)(a-1)+1+2i]+\sum_{i=0}^{a-1}(2a)+a^2+3a+1$$
substitute summation with (1):
(4) $$a^3+\sum_{i=0}^{a-1}(2a)+a^2+3a+1=a^3+(2a)(a-1)+a^2+3a+1=a^3+3a^2+a+1$$
From (2) I got $a^3+3a^2+3a+1$ but from (3) & (4) I got $a^3+3a^2+a+1$
My result with induction is off by 2a. Can someone please help me to understand where I did wrong?
| There are simpler ways to prove this relation.
The sum $\displaystyle S(n)=\sum\limits_{i=0}^n i=\frac{n(n+1)}{2}$ is well known and easy to prove, for instance
$\begin{align}2S(n)&=\bigg(1+2+\cdots+(n-1)+n\bigg)+\bigg(n+(n-1)+\cdots+2+1\bigg)\\&=\bigg((n+1)+(n+1)+\cdots+(n+1)\bigg)\\&=n(n+1)\end{align}$
Then notice $(a-1)a+1$ does not depend on the summation index $i$, thus we can factor it out of the sum and multiply by the number of terms.
$\begin{align}\sum\limits_{i=0}^{a-1}((a-1)a+1+2i)&=((a-1)a+1)\times a+2\sum\limits_{i=0}^{a-1}i\\&=(a^3-a^2+a)+2S(a-1)\\&=(a^3-a^2+a)+(a-1)a\\&=a^3\end{align}$
Now to come back to your solution, it is a bit confusing because in (2) you start from your result, then go back to your starting point and try to match both...
In any case, this method involves also factorizing constant values out of the sum so the previous method is more straightforward in this regard.
$\begin{align}(a+1)^3&=(a^3)+(3a^2+3a+1)
\\&=\sum\limits_{i=0}^{a-1}((a-1)a+1+2i)+(3a^2+3a+1)&\text{induction hypothesis P(a)}
\\&=\sum\limits_{i=0}^{a-1}((a+1)a+1+2i)-\underbrace{\sum\limits_{i=0}^{a-1}(2a)}_{-2a^2}+(3a^2+3a+1)&\text{shift formula by 2a, remove the excess}
\\&=\sum\limits_{i=0}^{a-1}((a+1)a+1+2i)+(a^2+3a+1)&\text{simplify constant sum}
\\&=\sum\limits_{i=0}^{a}((a+1)a+1+2i)-\underbrace{((a+1)a+1+2a)}_{a^2+3a+1}+(a^2+3a+1)&\text{add/remove last term of sum}
\\&=\sum\limits_{i=0}^{a}((a+1)a+1+2i)&\text{after cancelling terms P(a+1) is verified}\end{align}$
You pretty much got all of these ideas, so mostly it is a question of organizing your calculations in a way that is less prone to mistakes (negative sign forgotten or missed term, ...).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that determinant is zero iff $x= 2, 3$
Prove that
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix}=0\Leftrightarrow x= 2, 3$$
I see that if determinant is zero, two lines must equal, in this problem, there are between line 1 and line 2, 3, but how can I prove there is no other solutions ?? And what for $x^{3}, x^{4}, etc$ instead ?? Thks
| Why not to expand the determinant and find all the solutions?
$$\begin{vmatrix}
1 & x & x^{2}\\
1 & 2 & 2^{2}\\
1 & 3 & 3^{2}
\end{vmatrix} =
\begin{vmatrix}
1 & x & x^{2}\\
0 & 2-x & 2^{2}-x^2\\
0 & 1 & 5
\end{vmatrix} =
\begin{vmatrix}
2-x & 2^{2}-x^2\\
1 & 5
\end{vmatrix}$$
$$=10-5x-(4-x^2) = x^2-5x+6=(x-2)(x-3)$$
Which equals to zero iff $x=2,3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Elliptic Curve Discriminant Here is my attempted approach to prove that the discriminant $\triangle = 4a^3+27b^2$ of an elliptic curve in the form of $y^2 = x^3 + ax +b$ is zero. I have a problem at the end which doesn't bring me to the expected conclusion. Below is the process, please let me know the error in the process. I try to approach this from the definition of non-singularity (part of the definition of an elliptic curve), which is equivalent to the statement: the equation $y^2 = x^3 + ax +b$ is differentiable everywhere on the graph. From that definition, I try to derive the derivative of $y$ in respect to $x$ through implicit differentiation:
\begin{align}
y^2 &= x^3+ax+b\\ \label{ref1}
2yy' &= 3x^2+a\\
y' &= \frac{3x^2+a}{2y}
\end{align}
If the graph is singular, then $y'$ does not exist, in other words:
\begin{align}
2y &= 0\\
3x^2+a &\neq 0
\end{align}
I replace $y = 0$ in the equation of elliptic curve which yields
\begin{equation}
0 \ = \ x^3 + ax + b
\end{equation}
I apply Cardano's method, who realize that the form could be represented as
\begin{equation}
(\alpha-\beta)^3 + 3\alpha\beta(\alpha-\beta) = \alpha^3 - \beta^3
\end{equation}
in which
\begin{align}
\alpha\beta &= \frac{a}{3} \\
\alpha^3 - \beta^3 &= -b
\end{align}
By substituting $\alpha = \frac{a}{3\beta}$ in the second equation from above, I obtain
\begin{equation}
(\frac{a}{3\beta})^3 - \beta^3 = -b
\end{equation}
I further simplify this by considering $\beta^3$ as a whole, i.e.
\begin{align}
\frac{a^3}{27}-\beta^6 &= -b\beta^3\\
(\beta^3)^2 - b\beta^3 -\frac{a^3}{27} &= 0
\end{align}
By quadratic formula,
\begin{equation}
\beta^3 = \frac{b\pm \sqrt{b^2+\frac{4a^3}{27}}}{2} = \frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}
\end{equation}
From $\alpha^3 - \beta^3 = -b$, I get
\begin{equation}
\alpha^3 = \beta^3 - b = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}
\end{equation}
From $3x^2+a \neq 0$:
\begin{equation}
3\left(\left(-\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3} - \left(\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3}\right)^2 + a \neq0
\end{equation}
The equation above is from $3(\alpha-\beta)^2+a \neq 0$. By De Moivre's Formula there are two other equations, $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$ and $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})^2-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$. These result in several representations of b in terms of a and one integer solution, $a=0,b=0$ shouldn't exist at the same time, $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$. These two meet my expectation, $\triangle = 0$. However, there are other four representations that do not meet my expectation $b \neq \pm \sqrt{\frac{a^3}{54}\pm\frac{5ia^3}{6\sqrt{3}}}$ and $b \neq \pm \sqrt{\frac{a^3}{54}\mp\frac{5ia^3}{6\sqrt{3}}}$. I also try to see what I could get from \begin{equation}
3\left(\left(-\frac{b}{2}\right)^\frac{1}{3} - \left(\frac{b}{2}\right)^\frac{1}{3}\right)^2 + a \neq0
\end{equation}
which yields that $a=0,b=0$ shouldn't exist at the same time, or $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$ these satisfy $\triangle = 0$ and also a weird pair of $b \neq \pm \frac{ia^{3/2}}{12\sqrt{3}}$, which doesn't even satisfy $\triangle = 0$. Could someone explain this?
| When $3x^2 + a \ne0$ and $2y = 0$, the curve is still nonsingular, its just that the slope of the tangent line is infinite.
The singularity occurs when
$$3x^2 + a = 0 \text{ and } 2y =0$$
simultaneously i.e.
$a = - 3x^2$ so $0 = x^3 + (-3x^2)x + b$
giving
$$ b = 2x^3$$
this is what leads to the equation
$$27b^2 + 4a^3 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2} \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?
| Note that\begin{align}\int_1^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx&=-\int_1^0\frac{1+y^{-2}}{1+y^{-4}}\left(-\frac1{y^2}\right)\,\mathrm dy\\&=\int_0^1\frac{1+y^2}{1+y^4}\,\mathrm dy\end{align}and that therefore\begin{align}\int_0^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx&=\int_0^1\frac{1+x^2}{1+x^4}\,\mathrm dx+\int_1^\infty\frac{1+x^2}{1+x^4}\,\mathrm dx\\&=2\int_0^1\frac{1+x^2}{1+x^4}\,\mathrm dx.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Apply product and chain rule to differentiate with respect to one variable I have the following:
$$
u = \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}} \phi\left(\frac{x}{t}\right)
$$
I need to find $u_t$
This equation was found by solving the characteristic equation:
$$
\frac{dx}{xt} = \frac{dt}{t^2} = \frac{du}{-\left(\frac{1}{4}x^2 + \frac{1}{2}t\right)u}
$$
solving the equation from the first two yields:
$$
r = \frac{x}{t}
$$
solving from the second two yields:
$$
v = u\sqrt{t}e^{\frac{x^2}{4t}}
$$
we let $v = \phi(r)$
this gives us:
$$
u\sqrt{t}e^{\frac{x^2}{4t}} = \phi\left(\frac{x}{t}\right)
$$
that is where the equation for $u$ comes from.
Here is what I have:
\begin{align}
u_t & = \frac{-1}{2t^{\frac{3}{2}}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) + \frac{x^2}{4t^2}\frac{1}{\sqrt{t}}e^\frac{-x^2}{4t}\phi\left(\frac{x}{t}\right) + \left( \frac{-x}{t^2} \right) \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi'\left(\frac{x}{t}\right) \\
& = \frac{-u}{2t} + \frac{x^2u}{4t^2} - \left( \frac{x}{t^2} \right) \frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi'\left(\frac{x}{t}\right) \\
\end{align}
However the solution given has:
$$
u_t = \frac{-u}{2t} + \frac{x^2u}{4t^2} - \phi'\left( \frac{x}{t} \right) \left(\frac{xu}{t^2}\right)
$$
I can't figure out what I am doing wrong before I look at $u_{xx}$ (I already tried that and again, am making the same mistake).
| Your mistake is in the second term. It should be
$$\frac{x^2}{4t^{2}}\frac{1}{\sqrt{t}}e^{\frac{-x^2}{4t}}\phi\left(\frac{x}{t}\right) = \frac{x^2}{4t^{2}}u$$
You just forgot the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Height of a cube edge from the floor
A cube $ABCD.EFGH$ has side length $2a$ cm. Point $A$ is lifted $a$ cm from the floor, point $C$ is still on the floor, point $B$ and point $D$ are on the same height from the floor. What is the height of point $E$ from the floor?
I'm sorry for my bad English, but I try to illustrate it as follows.
The left cube is the original one while the right side is the cube after we lifted the point $A$. In my mind, to find the height of point $E$ from the floor is to find the length of line $EN$. We can use Pythagorean theorem $EN^2=ME^2-MN^2$. But how to find the length of line $ME$ and $MN$?
| You can also use similar triangles. Let $P$ be the point on $MN$ directly below $A$, and let $Q$ be the point of intersection of $EN$ and line $AC$. Since $\angle CAM$ is a right angle, we have that
$$\triangle CAP \sim \triangle AMP \sim \triangle EAQ$$
$CA \,$ is clearly $\, 2a \sqrt{2}\, $ so $\, PC = a\sqrt{7}\,.$ Then
$$\begin{align}
\frac{AC}{PC} &= \frac{EA}{EQ} \\
\\
EQ &= \frac{EA \cdot PC}{AC} \\
\\
&= \frac{(2a)(a\sqrt{7})}{2a\sqrt{2}} \\
\\
&= a \sqrt {\frac{7}{2}} \\
\\
&= \frac{a}{2} \sqrt{14}
\end{align}$$
and so the height of $E$ is
$$\begin{align}
EQ + QN &= \frac{a}{2} \sqrt{14} + a \\
\\
&= \boxed{ a \left( \frac{\sqrt{14}}{2} + 1 \right) }
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove combinatorically - $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$ $\sum_{k=0}^{n} 2^{k}\binom{n}{k} = 3^{n}$
I have no idea which story to form, I thought about $n$ students assuming n=4
picking 4 students for a committee with 3 roles. since each students has 3 options.
then on the other side.. $2^{0}\binom{4}{0} + 2^{1}\binom{4}{1} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{3}+2^{4}\binom{4}{4}$ = $2^{0}\binom{4}{4} + 2^{1}\binom{4}{3} +2^{1}\binom{4}{2} + 2^{3}\binom{4}{1}+2^{4}\binom{4}{0}$
but im just stuck here...
| The "binomial theorem" says that $(x+ y)^n= \sum_{i=0}^n \begin{pmatrix}n\\ i\end{pmatrix} x^i y^{n-i}$. If x= 2 and y= 1 that becomes $(2+ 1)^n= \sum_{i=0}^n\begin{pmatrix}n \\i \end{pmatrix} (2^i)(1^{n-i})$ or $3^n= \sum_{i=0}^n \begin{pmatrix}n \\ i\end{pmatrix} 2^i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$
*$\left(1- \frac {x+1}{4x+3}\right)^x$
*$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$
*$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$
*$e^{\infty} = \infty$
answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!
| Now @Infinity_hunter has explained your error, note that $\frac34-\frac{3x+2}{4x+3}=\frac{1}{4(4x+3)}>0$, so $0<\left(\frac{3x+2}{4x+3}\right)^x<\left(\frac34\right)^x$ proves the limit is $0$ by squeezing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Proving inequality given some conditions I want to show that $-x^2y^2-y^4+4y^2 \geq 0$, if $x^4+2y^2 \leq 4$ and $-x^2y^2-y^4+4y^2 \leq 0$, if $x^4+2y^2 \geq 32$. Is there a factoring trick that helps to prove this explicitly?
| *
*For the first question you need to prove
$$
-2x^2-2y^2+8 \geq 0
$$
with the condition $-2y^2 \geq x^4 -4$. This condition implies $x^2 < 2$ since otherwise it can never be fulfilled. Combining the two gives the sharper $
-2x^2+x^4+4 \geq 0
$ or $(x^2-1)^2 +3\ge 0$, done.
*For the second question you need to prove
$$
-2x^2-2y^2+8 \leq 0
$$
with the condition $-2y^2 \leq x^4 -32$. This condition needs to be applied only for $x^2 < \sqrt{32}$ since otherwise it is fulfilled anyway. So for this case, we have $
-2x^2+x^4-24 \leq 0
$ or $(x^2-1)^2 \le 25$, or $x^2 \le 6$, but this is fulfilled with $x^2 < \sqrt{32} < 6$.
Consider the other case $x^2 \ge \sqrt{32}$, then we see directly that $
-2x^2-2y^2+8 \leq -2x^2 +8 \leq - 2 \sqrt{32} +8 < 0
$
| {
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"url": "https://math.stackexchange.com/questions/3953249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other. Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.
I'm not sure how to solve this at all, maybe Vieta's formulas for quadratics can help.
| $f(x) = x^2 +bx+ 9;\;g(x) = x^2 +ax+c$
The sum of the roots of $f(x)$ is $s=-b$ and their product is $p=9$
The sum the roots of $g(x)$ is $-a$ and their product is $c$
So we must have $a=-b;\;c=9$
$f(x)=x^2+bx+9;\;g(x)=x^2-bx+9$
$h(x)=f(x)+g(x)=2x^2+18=0\to x=\pm 3i$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability to get 5 successes in at most 7 trials The night before the party, you plan to rehearse the song you are to sing and would like to sing it perfectly 5 times before sleeping. Since you are a great singer, the probability of you singing it perfectly on each attempt is 5/6. Suppose your attempts are mutually independent, what is the probability that you'll rehearse the song at most 7 times?
Here's my approach:
\begin{align*} \text{Probability to get 5 successes in 5 tries} = 4C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^0\dfrac{5}{6} &\approx 0.402\\ \text{6 tries} = 5C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^1\dfrac{5}{6} &\approx 0.334\\ \text{7 tries} = 6C4\left(\dfrac{5}{6}\right)^4\left(\dfrac{1}{6}\right)^2\dfrac{5}{6} &\approx 0.167\\\\ \text{Probability to get 5 successes in at most 7 tries} &= 0.903 \end{align*}
Was my approach correct? May I also know what type of Probability Distribution is suitable to model this problem?
| If the problem is asking for the $5^{th}$ success in the $7^{th}$ trial then your approach is incorrect, continue reading for the correct solution. If the question is like the body "$5^{th}$ success in $7$ trials then what you did is correct in terms of choosing the binomial distribution, not the arithmetic, I think the arithmetic mistakes you did show some misunderstanding of the binomial distribution. You did the following:
\begin{align*}
7C5\left(\dfrac{5}{6}\right)^5\left(\dfrac{1}{6}\right)^2 &\approx 0.2344\\
+\ 7C6\left(\dfrac{5}{6}\right)^6\left(\dfrac{1}{6}\right)^1 &\approx 0.3907\\
+\ 7C7\left(\dfrac{5}{6}\right)^7\left(\dfrac{1}{6}\right)^0 &\approx 0.2791\\
&= 0.9042
\end{align*}
However, you have two things wrong: your combinations and powers.
Your first equation seems correct to me (in case you meant $5$ successes in $7$ trials), for the second equation what you did gives the probability of $6$ heads in $7$ trials, but in your question you should calculate the probability of $5$ heads in $5$ trials + the probability of $5$ heads in $6$ trials+ the probability of $5$ heads in $5$ trials, so the second combination should be:
$\binom{6}{5}\cdot (\frac{5}{6})^5 \cdot \frac{1}{6}$
We used this combination as you have $5$ successes that can happen anywhere during $6$ trials.
Note that you should always have $(\frac{5}{6})^5$ since you always have $5$ successes, however, you should have $(\frac{1}{6})^{n-5}$ where $n$ is the total number of trials. Similarly, your third equation should be the one of $5$ success in $5$ trials which is:
$\binom{5}{5}\cdot (\frac{5}{6})^{5} \cdot (\frac{1}{6})^0=(\frac{5}{6})^5$
I will leave the numerical answers to you.
Moving to the second interpretation of the question, when you want to find the probability that the $k^{th}$ success happens in the $n^{th}$ trial you use the negative binomial distribution.
For example. the probability that the $5^{th}$success happens in the $6^{th}$ trial is:
$\binom{5}{4} \cdot (\frac{5}{6})^5 \cdot \frac{1}{6}$
We use the following combination as the first four successes can happen anywhere in the first five trials but the last success is for sure in the fifth trial so we don't include it in the combination as it has a fixed position at the end, can you continue the rest of the cases?
| {
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$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.
Case 2: If $b^2 - 4ac$ is not a perfect square, $b = \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$
Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)
So the equation is not quadratic and $c = 0$.
$a + bp^\frac{1}{3} = 0$
$-\dfrac{a}{b} = p^\frac{1}{3}$
This is a contradiction and hence $b = 0$ and $a = 0$
Is there any other way to solve this?
| Here's another way. If $a+bq+cq^2=0$, then $a=-(bq+cq^2)$, hence $a^2=b^2q^2+2bcq^3+c^2q^4$. So if $q^3=p$, then we have two equations:
$$\begin{align}
a+bq+cq^2&=0\\
(a^2-2bcp)-c^2pq-b^2q^2&=0
\end{align}$$
Multiplying both sides of the first equation by $b^2$ and the both side of the second equation by $cp$, and then adding the resulting equations, we have
$$(ab^2+a^2c-2bc^2p)+(b^3-c^3p)q=0$$
Now if $q$ is irrational (i.e., if $p$ is not a perfect cube) and the other variables are rational, then we must have $ab^2+a^2c-2bc^2p=b^3-c^3p=0$. But $b^3-c^3p=0$ for a non-cube $p$ implies $b=c=0$, in which case the original equation, $a+bq+cq^2=0$, implies $a=0$.
| {
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Two variables function limit In an exercise of a limit of a function of two variables in the solution I read this inequality:
$$ \frac{x^2y^2}{(x^2+y^2)^\frac{3}{2}} \le \frac{1}{2}\sqrt{x^2+y^2} $$
how did they arrive at this result?
| That inequality is equivalent to$$x^2y^2\leqslant\frac12\sqrt{x^2+y^2}(x^2+y^2)^{3/2}=\frac{(x^2+y^2)^2}2.$$Besides,$$(x^2+y^2)^2-2x^2y^2=x^4+y^4\geqslant0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$ $$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$
I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ where $a_1,a_2,a_3\in \{a,b,c,d,e\}$.
|
Thanks to @CalvinLin for pinpointing that we cannot "assume WLOG" that the five variables are sorted as $a \le b \le c \le d \le e$.
We can still assume that the maximum element among them is $e$ $^{(*)}$. Then, by comparing $\color{red}{\text{red}}$ variables, we observe
*
*$$3d = (\color{red}e + a + b)^3 \ge (a + b + \color{red}c)^3 = 3e \\ \implies d \ge e \overset*\implies \boxed{d=e} \tag{1}$$
*
*$$3a = (b + c + \color{red}d)^3 \ge (\color{red}a + b + c)^3 = 3e \\\implies a \ge e \overset*\implies \boxed{a = e} \tag{2}$$
*
*$$3a = (\color{red}b + c + d)^3 \le (c + d + \color{red}e)^3 = 3b \\\implies b \ge a = e \overset*\implies \boxed{b = e} \tag{3}$$
*
*$$3c = (d + e + \color{red}a)^3 \ge (\color{red}c + d + e)^3 = 3b \\ \implies c \ge b \implies \boxed{c = b} \tag{4}$$
Finally, $(1)\land(2)\land(3)\land(4) \implies \boxed{a = b = c = d = e}$.
Also, @AlbusDumbledore provided an equivalent solution in AoPS
| {
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Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!
| One can use trigonometric functions to avoid squaring both sides to get a degree 4 polynomial. Clearly $x>1$. Let $x=\sec t$, $t\in(0,\frac{\pi}{2})$. Then the equation becomes
$$ \sec t+\csc t=\frac{35}{12} $$
or
$$ \frac{\sin t+\cos t}{\sin t\cos t}=\frac{35}{12}. $$
Squaring both sides gives
$$ \frac{1+\sin(2t)}{\sin^2(2t)}=\frac{35^2}{24^2} $$
which is equivalent to
$$ 35^2\sin^2(2t)-24^2\sin(2t)-24^2=0 $$
or
$$ (25\sin(2t)-24)(49\sin(2t)+24)=0. $$
Since $\sin(2t)>0$, one has
$$ \sin(2t)=\frac{24}{25} $$
which gives
$$ \cos(2t)=\pm\frac{7}{25}. $$
So
$$ \cos t=\sqrt{\frac{1+\cos(2t)}{2}}=\frac{3}{5} \text{ or } \frac{4}{5} $$
and hence
$$ x=\frac{5}{3} \text{ or }\frac{5}{4}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.
Solution
We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$
Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$
Hence, the answer is $31a^4$.
The original exam
| A very nice question.
You can think in this way: if we denote
$$x+y = s\\
x^3 + y^3= t\\
x^4 + y^4 = u$$
$s$, $t$, $u$ depend on only two parameters $x$, $y$, so they have only two degrees of freedom. Therefore, there must be some relation between $s$, $t$, $u$. It can be obtained by eliminating $x$, $y$ from the above equation. The relation obtained will be a polynomial one. I used WA and got this relation
$$s^6 - 8 s^3 t - 2 t^2 + 9 s^2 u=0$$
You can check it directly by substituting $s=x+y$, $t=x^3 + y^3$, $u=x^4 + y^4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of integration terms If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then
(A) $I>\ln99$
(B) $I<\ln99$
(C) $I<\frac{49}{50}$
(D) $I>\frac{49}{50}$
The official answer is B and D
My approach is as follow
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $$
$$\frac{A}{x} + \frac{B}{{x + 1}} = \frac{{A\left( {x + 1} \right) + Bx}}{{x\left( {x + 1} \right)}} = \frac{{x\left( {A + B} \right) + A}}{{x\left( {x + 1} \right)}}$$
$$A = k + 1\text{ and } B = - \left( {k + 1} \right)$$
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\left( {\frac{{k + 1}}{x} - \frac{{k + 1}}{{x + 1}}} \right)} dx} \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\int\limits_k^{k + 1} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx} $$
$$I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left. {\ln\frac{x}{{x + 1}}} \right|_k^{k + 1}} \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{k + 1}}{{k + 2}} - \ln\frac{k}{{k + 1}}} \right)}$$ $$\Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{{{\left( {k + 1} \right)}^2}}}{{k\left( {k + 2} \right)}}} \right)} $$
Not able to approach from here
| You have
$$\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx \le \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x} = \int_1^{99} \frac{dx}{x} = \ln 99$$
hence B is true. Also
$$\begin{aligned}\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx &\ge \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x+1} \\
&\ge \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{100}\\
&=\frac{98}{100}= \frac{49}{50}
\end{aligned}$$
hence D is also true.
| {
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"timestamp": "2023-03-29T00:00:00",
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Intuition on the concept of bounding a sum For example, in calculating the following limit
$$\lim_{n\to\infty}\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}$$
I know (because my professor wrote) that one of the methods of solving this limit is via the sandwich rule since we know the lower and upper bound
$$\frac{3n}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{3n}{\sqrt{n^2+1}}$$
But I don't understand why it's so, because I would have guessed it would be
$$\frac{1}{\sqrt{n^2+3n}}<\sum_{k=1}^{3n}\frac{1}{\sqrt{n^2+k}}<\frac{1}{\sqrt{n^2+1}}$$
instead
| A single term of the series obeys the inequality $$\frac{1}{\sqrt{n^2 + 3n}} < \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{\sqrt{n^2 + 1}}$$ for any integer $k \in \{1, 2, \ldots, 3n\}$, but the sum does not. Instead, you have to sum each of the expressions accordingly:
$$\sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + 1}},$$ and because the leftmost and rightmost sums are independent of $k$, we get
$$\frac{3n}{\sqrt{n^2 + 3n}} < \sum_{k=1}^{3n} \frac{1}{\sqrt{n^2 + k}} \le \frac{3n}{\sqrt{n^2 + 1}}.$$
| {
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Find $a$ such that $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions
Find $a$ so that equation $\log _{2}^{2}x-{{\log }_{\sqrt{2}}}x=a-\sqrt{a+{{\log }_{2}}x}$ has exactly $2$ solutions.
My approach: Since that $$\log_{2}(x)=\frac{\ln(x)}{\ln(2)} \quad \text{and} \quad \log_{\sqrt{2}}(x)=\frac{\ln(x)}{\ln(\sqrt{2})}$$
then, we can re-write the equation as $$\left(\frac{\ln(x)}{\ln(2)}\right)^{2}-\left( \frac{\ln(x)}{\ln(\sqrt{2})}\right)^{2}=a-\sqrt{a+\frac{\ln(x)}{\ln(2)}}$$
Let $t:=\ln(x)$ so the equation above we can re-write in terms of $t$ as $$\left(\frac{1}{\ln(2)}t\right)^{2}-\left( \frac{1}{\ln(\sqrt{2})}t\right)^{2}=a-\sqrt{a+\frac{1}{\ln(2)}t}$$
Then, the idea that I was thinking is squaring both sides of the equation, but there appears an equation of degree 4 that is difficult to solve. How could I continue from there? or maybe a simpler approach?
New approach: Using the point out of @Ross Millikan and since that $$\log_{\sqrt{2}}(x)=\frac{\log_{2}(x)}{\log_{2}(\sqrt{2})}=2\log_{2}(x)$$
and let $t:=\log_{2}(x)$ so the equation can be we-write in terms of $t$ as $$t^{2}-2t=a-\sqrt{a+t} \implies t^{2}-2t\color{blue}{+1}=a-\sqrt{a+t}\color{blue}{+1} \implies (t-1)^{2}=a+1-\sqrt{a+t}$$
or maybe $$ t^{2}-2t=a-\sqrt{a+t} \overset{y^{2}=a+t}{\implies} (y^{2}-a)^{2}-2(y^{2}-a)=a-y^{2}$$
solving a little the expression above we can see that $$y^{4}-2y^{2}a+a^{2}-2y^{2}+2a-a+y^{2}=0$$
how can I solve this?
| Consider the function $$f(x)=\left(\log_2 x\right)^2-\log_{\sqrt 2} x-a+\sqrt{a+\log_2 x},$$ defined for all $x\ge 2^{-a}.$ Then we find that $$f'(x)=\frac{1}{2x\log_e 2\cdot \sqrt{a+\log_2 x}}\left(4\log_2 x\sqrt{a+\log_2 x}-4\sqrt{a+\log_2 x}+1\right).$$ Letting $M=\log_2 x,$ we see that $f'(x)=0$ whenever $16(M+a)(M-1)^2=1,$ or in other words when $$g(M)=16M^3+16(a-2)M^2+16(1-2a)M-1=0,$$ which is to say, when $\log_2 x =m,$ for some real number $m$ a root of the cubic in $M.$ This happens either once or thrice.
Now for exactly two roots of the original equation we must have a unique solution to the cubic. We find that the discriminant of $g'(M)$ is given by $$\Delta=16^2(a-2)^2-3\times 16\times 16(1-2a)=16^2(a+1)^2>0$$ for all real values of $a.$ Thus to ensure a unique solution we must have that $$g\left(\frac{a-2+|a+1|}{3}\right)g\left(\frac{a-2-|a+1|}{3}\right)>0.$$ This gives us a preliminary condition on $a.$
Now for the values of $a$ satisfying the inequality above we have ensured that $f'(x)$ vanishes exactly once, namely whenever $x=2^m,$ where $m=m(a)$ is the unique solution of $g(M)=0.$ Finally, since for $x\to +\infty,$ we have that $f(x)>0$ and for $x=2^{-a}$ we have that $f(x)=a(a+1),$ it follows that if $a=-1$ or $a=0$ we already have one root, and if these values of $a$ satisfy the previous condition then we have found two values of $a.$ On the other hand if $-1<a<0$ then there cannot be two roots if $a$ also satisfies the previous condition.
This leaves us with the case when $a<-1$ or $a>0,$ so that $f(2^{-a})>0.$ Thus in this case there are also exactly two roots provided the previous condition is satisfied and also the minimum value $f(2^m)$ is negative. This gives us a further condition on $a,$ namely that $(m-1)^2-(1+a)+\sqrt{a+m}<0.$
In summary we must have that $a$ satisfies the following conditions:
(1) $g\left(\frac{a-2+|a+1|}{3}\right)g\left(\frac{a-2-|a+1|}{3}\right)>0,$
and exactly one of the following:
(2) $a=-1$ or $a=0,$
or
(3) $a<-1$ or $a>0$ and $(m-1)^2-(1+a)+\sqrt{a+m}<0.$
Given these, then $f(x)=0$ has exactly two roots.
PS. Note that the conditions are to be selected as (1) and (2) or (1) and (3).
| {
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How many variables in the multinomial expansion have an exponent that's different from 2? Multinominal: $(a + b + c + d)^{10}$
Question: How many variables in the multinomial expansion have an exponent that's different from 2?
I want to solve it using generating functions.
So we can write: $t_1 + t_2 + t_3 + t_4 = 10$, When $t_i$ cannot use $x^2$.
That equals to the generating function:
$$
(1+x+x^3+x^4...)^4 = (1+x+\frac{x^3}{1-x})^4
$$
But now what? How can I keep expending this till I find the coefficient of $x^{10}$?
| You can rewrite $$(1+x+x^3+x^4...)^4=(1+x)^4(1+x^3+x^5+x^7+...)^4$$
$$=(x^4+4x^3+6x^2+4x+1)(1+x^3+x^5+x^7+x^9+...)^4$$
Now first look at the $x^4$ term: the only way to add exponent up to $10$ is $4+3+3=10$, therefore we have ${4\choose 2}=6$ ways.
Next look at $4x^3$ term, the only way is $3+7=10$, therefore we have $4{4\choose 1}=16$ ways.
Next look at $6x^2$ term, the only way is $2+3+5=10$, therefore $6{4\choose 1}{3\choose 1}=72$ ways.
Next look at $4x$ term, we have $1+9=1+3+3+3=10$, therefore $4({4\choose 1}+{4\choose 3})=32$ ways.
Next look at $1$ term, we have $0+3+7=0+5+5=10$, therefore ${4\choose1}{3\choose1}+{4\choose 2}=18$ ways.
Finally add them all up, $6+16+72+32+18=144$ ways.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$
Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.
I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.
This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?
| The case $x_1=x_2=...=x_n=1$ is trivial, and for $n >2$, is the only case when equality holds. We prove this below.
Suppose the $x_i$s are not all $1$. Then, there exists $i,j \in \{1,2,..,n\}, i \neq j$, such that $x_i<1<x_j$. Replace the pair $(x_i \ ,x_j)$ with $(x_i'\ ,x_j')$ , such that:
$$x_i'=1, \ x_j'=x_ix_j.$$
$x_i'$ and $x_j'$ have the same product as $x_i$ and $x_j$. Consider the sum
$$\dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} = \dfrac{x_i+x_j+2(n-1)}{(n-1)^2 + (n-1)(x_i+x_j) + x_ix_j}.$$
It would be ideal if the above sum was strictly smaller than $\dfrac{1}{n-1+x_i'} + \dfrac{1}{n-1+x_j'} = \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j}.$ (By our replacement, we increase the number of $x_i$s which are equal to $1$, while strictly increasing the L.H.S. of the given inequality and preserving the given constraint. Eventually, all the $x_i$s will equal $1$, and the inequality becomes an equality.) Attempting to proceed along this line,
\begin{align}
& \dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} < \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j} \\
& \iff \dfrac{1}{n-1+x_i} - \dfrac{1}{n} < \dfrac{1}{n-1+x_ix_j} - \dfrac{1}{n-1+x_j} \\
& \iff \dfrac{1-x_i}{n(n-1+x_i)} < \dfrac{x_j-x_ix_j}{(n-1+x_ix_j)(n-1+x_j)} \\
& \iff \dfrac{1}{n(n-1+x_i)} < \dfrac{x_j}{(n-1+x_ix_j)(n-1+x_j)} \\
& \iff (n-1+x_ix_j)(n-1+x_j) < nx_j(n-1+x_i) \\
& \iff x_ix_j^2 - x_ix_j + nx_j - x_j + n^2-2n+1 - n^2x_j + nx_j < 0 \\
& \iff n^2(1-x_j) - x_ix_j(1-x_j) - 2n(1-x_j) + (1-x_j) < 0 \\
& \iff n^2-x_ix_j -2n + 1 >0 \\
& \iff x_ix_j < (n-1)^2.
\end{align}
Unfortunately, it is possible to have $x_ix_j \geq (n-1)^2$. Fortunately, this issue can be circumvented. W.L.O.G. let $x_1 \leq x_2 ... \leq x_n , x_1 < 1 < x_n$. There are $2$ cases:
Case $1$:
$x_1x_n \geq (n-1)^2 \Rightarrow \dfrac{x_1+x_n+2(n-1)}{(n-1)^2 + (n-1)(x_1+x_n) + x_1x_n } \leq \dfrac{x_1+x_n+2(n-1)}{2(n-1)^2 + (n-1)(x_1+x_n)} = \dfrac{1}{n-1}.$ Since $\dfrac{1}{n-1+x_n}$ is positive, we conclude that $\dfrac{1}{n-1+x_1} < \dfrac{1}{n-1}$. But the largest term in the sum $ \displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i}$ is $\dfrac{1}{n-1+x_1}$. Thus,
$$\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} \leq \dfrac{1}{n-1} + \displaystyle \sum_{i=2}^{n-1} \dfrac{1}{n-1+x_i} < \dfrac{1}{n-1} + \dfrac{n-2}{n-1} =1. $$
Case $2$: $x_1x_n < (n-1)^2 \Rightarrow \dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_n} < \dfrac{1}{n} + \dfrac{1}{n-1+x_1x_n}$. Letting $x_1'=1, x_n'=x_1x_n$ and $x_j'=x_j \ \forall \ j \notin \{1,n\},$ we have $\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} < \displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i'}.$ And we are done by a repeated application of this "smoothing" procedure! (At each stage, W.L.O.G. re-order the variables $x_i$s from smallest to largest, since the previous round of "smoothing" disrupted the original order.) If, at any stage, the product of the smallest term with the largest term is greater than or equal to $(n-1)^2$, we are done by Case $1$. Otherwise, we continually increase the L.H.S. and eventually end up with all the $x_is$ equal to $1$, giving the equality case L.H.S. $=1$.
| {
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Difficult Cauchy Problem Let $a, b, c>0$ such that $a^{2}+b^{2}+c^{2}=3 a b c$. Prove the following inequality:
$$
\frac{a}{b^{2} c^{2}}+\frac{b}{c^{2} a^{2}}+\frac{c}{a^{2} b^{2}} \geq \frac{9}{a+b+c}
$$
I tried using the fact that $a^2 + b^2 + c^2 = 3abc$ but I could only think of one case where $a=b=c$. I also tried using knowing what is $a^2 + b^2 + c^2$ in terms of $a$, $b$ and $c$ to try and get $a+b_c$ in terms of them using the sum of squares where $(a+b)^2 - 2ab = a^2 + b^2$ but I reached a dead end where I could not simplify any further.
I think assuming $a=b=c$ did get me somewhere but since it is a proof problem I cannot assume but rather have to prove that $a=b=c$. I cannot figure that bit out.
I know Cauchy but very little bit, so I do not now if it can be applied too well here but I think it is a right approach.
| Proposed inequality can be rewriten like this:$$(a+b+c)(a^3+b^3+c^3)\geq 9a^2b^2c^2$$
By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3) \geq (a^2+b^2+c^2)^2$$ and we are done!
| {
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How can we make it easy to solve quadratics without the Quadratic Formula? For example ,
$$5x^2 - 7x-2=0$$
Then, we only need to find is when
$$5x^2-7x=2$$
$$x(5x-7)=2$$
Since that when we get $2-2=0$.
Is there any way we can find them easily without solving quadratics using formula?
I am just getting very enthusiastic in this topic.
| Unfortunately, your strategy does not help that much because it is very difficult to solve
$$
x(5x-7)=2
$$
without simply returning to the original equation. This is because there is a $2$ on the RHS. The reason we want to be able to write a quadratic in the form
$$
(x-p)(x-q)=0
$$
is so that we can exploit the zero-product property: if $a \times b=0$, then $a=0$ or $b=0$. Once we have written an equation in the form $(x-p)(x-q)=0$, we can still instantly deduce that $x=p$ or $x=q$. This is something we can't do when we have something like
$$
x(5x-7)=2 \, .
$$
Instead, I would suggest dividing through by $5$, which yields
$$
x^2 - \frac{7}{5}x - \frac{2}{5} = 0 \, .
$$
Then, add $2/5$ to both sides:
$$
x^2 - \frac{7}{5}x = \frac{2}{5} \, \tag{*}\label{*} .
$$
The LHS is almost a perfect square. Note that
$$
\left(x-\frac{7}{10}\right)^2=x^2-\frac{7}{5}x+\frac{49}{100} \, .
$$
If we subtract $49/100$ from this equation, we get
$$
\left(x-\frac{7}{10}\right)^2-\frac{49}{100}=x^2-\frac{7}{5}x \, .
$$
Hence, the equation $\eqref{*}$ can be rewritten as
$$
\left(x-\frac{7}{10}\right)^2-\frac{49}{100}=\frac{2}{5} \, .
$$
Add $49/100$ to both sides:
$$
\left(x-\frac{7}{10}\right)^2 = \frac{2}{5}+\frac{49}{100} = \frac{89}{100} \, .
$$
Take the square root of both sides:
$$
x-\frac{7}{10} = \pm\sqrt{\frac{89}{100}} \, .
$$
Add $7/10$ to both sides:
$$
x = \frac{7}{10} \pm \sqrt{\frac{89}{100}} \, ,
$$
and we are done. This method of solving quadratics is known as completing the square. Unlike factorisation, it can be used to solve any quadratic. In fact, the quadratic formula comes from completing the square on the general quadratic equation
$$
ax^2 + bx + c = 0 \, .
$$
Please let me know if you have any questions.
| {
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Find the density function of $Y = X^2 - 1$ for a piecewise-linear r.v. $X$ A density function $f$, of a continuous random variable $X$, is given by:
$$
f(x) = \left\{
\begin{array}{lc}
\frac{1}{2} & \quad 0 < x < 1 \\
\frac{x}{3} & \quad 1 \leq x < 2 \\
\ 0 & \text{else}
\end{array}
\right.
$$
Find the density function of the random variable $Y = X^2 - 1.$
I'm struggling with this problem can someone explain me how to proceed with this kind of problems?
| First note that Y can take values between -1 and 3. If we find the cumulative probability for Y - let's call it $G(y) = p(Y \le y)$ - then its derivative will be the density function.
For $0 < x < 1$, $y = x^2 - 1$, so $-1 < y < 0$. For $y$ in this range,
$p(Y \le y) = p(X^2 - 1 \le y) = p(X \le \sqrt{y+1}) = \frac12 \sqrt{y+1}$
The last step follows from the observation that $p(X<x) = \frac12x$ for $0 < x <1$
Now consider $1 \le x \lt 2$ where $0 \le y < 3$.
$p(Y \le y) = p(Y \le 0) + p(0 < Y < y)$
We know $p(Y \le 0)$ from the first part, ie $\frac12 \sqrt{0+1} = \frac12$
$p(0 < Y < y) = p(0 < X^2 - 1 < y) = p(1 < X < \sqrt{y+1})$
$$=\int_1^\sqrt{y+1}\frac{x}{3}dx$$
$$=\left[\frac{x^2}{6}\right]_1^\sqrt{y+1}$$
$$=\frac{y}6$$
So $G(y) = \frac12 + \frac{y}6 = \frac {y+3}{6}$
Putting it all together:
$$G(y) = \left\{
\begin{array}{lc}
\frac{1}{2}\sqrt{y+1} & \quad -1 < y < 0 \\
\frac{y+3}{6} & \quad 0 \leq y < 3 \\
\ 0 & \text{else}
\end{array}
\right.$$
If you find the derivative of this function with respect to y, you will have the density function required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving an inequality with n positive real numbers Is there any alternative way of proving
$$\sum_{k=1}^n\frac{a_k-1}{a_k+1}\geq\frac{\prod_{k=1}^n a_k -1}{\prod_{k=1}^na_k+1}$$ for any positive real numbers $a_1,a_2,\cdots,a_n,$ greater than or equal to 1, other than the one given in one of the the answers; An inequality associated with sum of n real numbers?
| We write the inequality as
$$\sum_{i=1}^n\frac{1}{a_i+1}\leq \frac{n-1}{2}+\frac{1}{a_1a_2\cdots a_n+1}. \quad (1)$$
For $n=1$ the inequality become
$$\frac{1}{a_1+1}+\frac{1}{a_2+1} \leqslant \frac{1}{2} + \frac{1}{a_1a_2+1}, \quad (2)$$
or
$$\frac{(a_1-1)(a_2-1)(a_1a_2-1)}{2(a_1+1)(a_2+1)(a_1a_2+1)} \geqslant 0.$$
Which is true.
Now, let $x=a_1a_2 \cdots a_k,$ suppose $(1)$ is true for $n=k,$ we get
$$\sum_{i=1}^k\frac{1}{a_i+1} \leqslant \frac{k-1}{2}+\frac{1}{x+1}. \quad (3)$$
We will show that this is also true for $n=k+1.$
Indeed, setting $y = a_{k+1},$ use $(3)$ and $(2)$ we have
$$\sum_{i=1}^{k+1}\frac{1}{a_i+1} = \sum_{i=1}^k\frac{1}{a_i+1} + \frac{1}{y+1} \leqslant \frac{k-1}{2}+\frac{1}{x+1} + \frac{1}{y+1}$$
$$\leqslant \frac{k-1}{2}+\frac{1}{2}+\frac{1}{xy+1} = \frac{k}{2}+\frac{1}{x_1x_2\cdots x_{k+1}+1}.$$
Done.
| {
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Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$
$$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$
$$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$
$$= (250\sqrt{2009} - 1259)^5$$
However, the answer given is $32$, so there could have been more simplifications.
As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?
| Your working leads to the answer. Here is the correction -
$a = \frac{1 + \sqrt{2009}}{2}$
$a^2 - 3 = \frac{999 + \sqrt{2009}}{2}$
$a (a^2 - 3) = 752 + 250 \sqrt {2009}$
$500 (a + 1) = 250 ( \sqrt {2009} + 3)$
$a(a^2-3) - 500(a+1) = 2$
So $(a^3 - 503a - 500)^5 = 32$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$ Find all incongruent solutions to $x^2 \equiv 23 \pmod{77}$
Using the Chinese Remainder Theorem, I obtained $x^2 \equiv 23 \pmod{7}$ and $x^2 \equiv 23 \pmod{11}$.
For $x^2 \equiv 23 \pmod{7}$, the answers are:
\begin{align}
x \equiv 3 \pmod{7} \\
x \equiv 4 \pmod{7}
\end{align}
For $x^2 \equiv 23 \pmod{11}$, the answers are:
\begin{align}
x \equiv 1 \pmod{11} \\
x \equiv 10 \pmod{11}
\end{align}
But the solutions for $x^2 \equiv 23 \pmod{77}$ are:
\begin{align}
x \equiv 10 \pmod{77} \\
x \equiv 32 \pmod{77} \\
x \equiv 45 \pmod{77} \\
x \equiv 67 \pmod{77}
\end{align}
Why are my answers not the same as the solutions? Can someone help me with this? Thanks!
| To get the given solution, combine congruences in your solution with the Chinese remainder theorem: $3\bmod7\land1\bmod11\iff45\mod77$, etc.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster? I have to calculate this integral:
\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with}
\quad a \in \mathbb{R} \end{align}
My attempt:
Using, trigonometric substitution
\begin{align}
\tan \theta &= \frac{x}{a}\\ \Longrightarrow \ x&=a \tan \theta\\ \Longrightarrow \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta
\end{align}
Thus,
\begin{align}
\int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}}
\end{align}
My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help
| $$x^2 \sqrt{a^2+x^2};\;\, x\to a \sinh u$$
$$dx=\cosh u\,du$$
$$\int x^2 \sqrt{a^2+x^2}\,dx=\int (a^2 \sinh ^2 u)(a \cosh u )\sqrt{a^2 \sinh ^2 u+a^2}\,du=$$
$$=a^4\int \sinh^2 u\cosh^2 u\,du=\frac{a^4}{4}\int\sinh^2 2u\,du=\frac{a^4}{8} \int (\cosh 4 u-1) \, du=$$
$$=\frac{1}{8} a^4 \left(\frac{1}{4} \sinh 4 u-u\right)+C=\frac{1}{8} \left(x \sqrt{a^2+x^2} \left(a^2+2 x^2\right)-a^4 \text{arcsinh}\left(\frac{x}{a}\right)\right)+C$$
Useful formulas
$\cosh^2 u -\sinh^2 u=1$
$\sinh 2u=2\sinh u\cosh u$
$\cosh 4u =\sinh ^2 2u +\cosh ^2 2u=2\sinh^2 2u+1\to \sinh^2 2u = \frac{1}{2}(\cosh4u - 1)$
$x=a\sinh u\to u=\text{arcsinh}\left(\frac{x}{a}\right)$
$\sinh 4u = 2\sinh 2u \cosh 2u = 4\sinh u\cosh u(\cosh^2 u+ \sinh^2 u)$
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Write the equation $f(x,y)=0$ that describes the set $S$ Let $A=(1,1)$, $B=(1, -2)$ and $C=(5,-5)$.
Write the equation $f(x,y)=0$ with $f\in \mathbb{R}[x,y]$ that describes the set $S$ of points that have same distance of $A$ as to $BC$.
$$$$
I have done the following:
Let $P=(x,y)$ be a point of $S$.
The it must be $d(P,A)=d(P,BC)$.
The line through $B$ and $C$ is $$y+2=\frac{-2+5}{1-5}(x-1) \Rightarrow y+2=-\frac{3}{4}(x-1) \Rightarrow 4y+8=-3x+3 \Rightarrow 3x+4y+5=0$$
From $d(P,A)=d(P,BC)$ we get $$\sqrt{(x-1)^2+(y-1)^2}=\frac{|3x+4y+5|}{\sqrt{3^2+4^2}}$$ right?
Do we get from that the desired equation?
| We have
$$\sqrt{(x-1)^2+(y-1)^2}=\frac{|3x+4y+5|}{\sqrt{3^2+4^2}} \iff 25(x-1)^2+25(y-1)^2-(3x+4y+5)^2=0.$$
| {
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Finding maximum radius of inscribed circle I am working through a pure maths text book as a hobby. I am having difficulty with the second half of this problem:
The equal sides AB and AC of an isosceles triangle have length a, and the angle A is denoted by $2 \theta$. Show that the radius of the inscribed circle of the triangle is given by $\frac {a \sin\theta \cos \theta}{1 + \sin \theta}$
Show that, if a is constant and $\theta$ varies between $0^\circ$ and $90^\circ$, the maximum value of r occurs when $\sin \theta = \frac{(\sqrt 5 - 1)}{2}$
I have solved the first part of the question. Now I need to differentiate $\frac {a \sin\theta \cos \theta}{1 + \sin \theta}$
Using the quotient formula I get this to be:
$\frac{dr}{d\theta} = \frac{(1 + \sin \theta)(a\cos^2\theta - a\sin^2\theta) - (a\sin \theta \cos\theta)\cos \theta}{(1 + \sin\theta)^2}$
But I cannot get this to equal the answer given.
| HINT:
Let $$\frac{dr}{d\theta}=0$$
and then rewrite the numerator completely in terms of $\sin \theta$.
If you need any more help please don't hesitate to ask :)
EDIT:
As you correctly found, the numerator of the derivative is equal to
$$a(1-2\sin^2\theta-\sin^3\theta)$$
Hence, $$a(1-2\sin^2\theta-\sin^3\theta)=0$$
as when a fraction is equal to $0$ the denomiantor plays no part in making the fraction equal to $0$; the numerator must be equal to $0$. Now, assuming $a\ne0$, we can divide through by $a$ to obtain
$$1-2\sin^2\theta-\sin^3\theta=0\implies \sin^3\theta+2\sin^2\theta-1=0$$
Now we can simply verify that $\sin\theta=\frac{\sqrt5 -1}{2}$ is a solution the equation above, and we are done.
Alternatively, we can find this solution directly. Notice that substituting $\sin\theta=-1$ gives us $0$; hence by the factor theorem $(\sin\theta+1)$ is a factor of $\sin^3\theta+2\sin^2\theta-1$. Hence,
$$\sin^3\theta+2\sin^2\theta-1=(\sin\theta+1)(\sin^2\theta+\sin\theta-1)=0$$
Solving for the second factor we obtain the required solution. Notice that the other $2$ solutions are outside of the range of allowed solutions for $\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $ find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$
Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\frac{c}{a} $
So I can express $b=-a(\alpha+\beta)$
$c=a.\alpha\beta$
Once I substitute for b and c in the equation I can get, $$\alpha\beta x^2+(\alpha+\beta)(\alpha\beta+1)x+(\alpha\beta+1)^2=0$$
I want to know whether there is any different approach other than this method?
Any hint is higly valued. thank you!
| Let $r_1$ and $r_2$ be the roots of the equation
$$
a c x^{2}-b(c+a) x+(c+a)^{2}=0
$$
Then $$
\begin{aligned}
r_{1}+r_{2} &=\frac{b(c+a)}{a c}=\frac{b}{a}+\frac{b}{c}\\&=-(\alpha+\beta)-\frac{\alpha+\beta}{\alpha \beta} \\
&=-(\alpha+\beta)-\left(\frac{1}{\beta}+\frac{1}{\alpha}\right)\\&=-\left(\alpha+\frac{1}{\beta}\right)-\left(\beta+\frac{1}{\alpha}\right)
\end{aligned}
$$
and
$$
\begin{aligned}
r_{1} r_{2} &=\frac{(c+a)^{2}}{a c}=\frac{c^{2}+2 a c+a^{2}}{a c} \\
&=\frac{c}{a}+2+\frac{a}{c} \\
&=\alpha \beta+2+\frac{1}{\alpha \beta} \\
&=\left[-\left(\alpha+\frac{1}{\beta}\right)\right]\left[-\left(\beta+\frac{1}{\alpha}\right)\right]
\end{aligned}
$$
Hence the required roots are $\displaystyle -\left(\alpha+\frac{1}{\beta}\right) $ and $ \displaystyle -\left(\beta+\frac{1}{\alpha}\right) .$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2] I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.
I tried multiple ways to understand why this assumption was made, but I can't figure it out. My first assumption was that since $m \in S$ it's true that:
$m > 3^{m/3}$
and by consequence:
$m^3 > 3^{m}$
but it still does not prove $3^m < m$. Any help is very appreciated.
| The well ordering principle comes into play in trying to find the first $n$ where $n^3 > 3^n$.
We know $1^3 < 3^1$.
But can there be any $n^3 > 3^n$?
If so, there must be a first $n$ where $n^3 > 3^n$. But if $n$ is the first then it must be that $(n-1)^3 \le 3^{n-1}$.
Now hopefully we will be able to show $(n-1)^3 \le 3^{n-1}$ while $n^3 > 3^n$ can't ever happen which means we can never have a first case where $k^3 \le 3^k$ is not true, which means $k^3 \le 3^k$ always will be true.
So multiplying both sides by $3$ we have$3(n-1)^3 \le 3^n$. But we also have $3^n < n^3$. So $3(n-1)^3 \le 3^n < n^3$.
And $3(n^3 - 3n^2 + 3n -1) < n^3$.
Well now simplify that and
$2n^3 - 9n^2 + 9n - 3< 0$ so
$2n^3 +9n < 9n^2 + 3$.
$2n + \frac 9n < 9 + \frac 3{n^2}$.
So $2n < 2n + \frac 9n < 9 + \frac 3{n^2} $.
If $n \ge 2$ then $\frac 3{n^2} < 1$ and so we either have $n =1$ or $2n < 9+\frac 3{n^2}< 9+1 = 10$. In any event we must have $n < 5$ or in other words, $n=1,2,3,$ or $4$.
So we test that $n=1,2,3,4$ and get $1^3 < 3^1; 2^3 =8 < 9 =3^2; 3^3 = 3^3; 4^3=64 < 81 = 3^4$. so none of those are the first exception.
So the first exception can not exist.
And if there can't be a first value, there can't be any value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3979260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $$
I have solved this integral before by substitution and change in the limits of integration, but in this chapter the book asks to solve it by differentiation under the integral sign. I have tried several ways of solving this, but the only one that i thought it was leading me somewhere was:
$$f(a) = \int_0^1 \frac{\ln(x+a)} {x^2 + 1} \, \mathrm{d}x $$
So that:
$$f'(a) = \int_0^1 \frac{1} {(x+a)(x^2 + 1)} \, \mathrm{d}x $$
Then i tried to separate this last integral by partial fractions, my result on this was:
$$\frac {1} {(x+a)(x^2 + 1)} = \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1}\right)$$
And the integral reduces to:
$$f'(a) = \int_0^1 \frac{1} {a^2 + 1} \left(\frac {1} {x+a} - \frac{x-a} {x^2+1} \right) \, \mathrm{d}x $$
Then this last expression evaluates to:
$$f'(a) = \frac{1} {a^2 + 1} (\ln(a+1) - \ln(a) - \ln(4)+ \frac{π}{4} a)$$
Then integrating from 0 to 1 with respect to $a$ we will get:
$$f(1) - f(0) = \int_0^1 \frac{\ln(a+1)} {a^2 + 1} \, \mathrm{d}a - \int_0^1 \frac{\ln(a)} {a^2 + 1} \, \mathrm{d}a $$
(The last two terms of $f'(a)$ cancel each other after the integration so i didn't wrote them)
But then the two integrals on the right hand side are equal to $f(1) - f(0)$ so the differentiation under the integral led nowhere.
Do i need some other approach? Or did i made any mistake?
Any help is appreciated.
| Even if I think that @Tavish's solution is the most efficient, wht you did must work.
$$f(a) = \int_0^1 \frac{\log(x+a)} {x^2 + 1} \,dx$$
$$f'(a) = \frac{1} {a^2 + 1} \left(\log(a+1) - \log(a) - \ln(4)+ \frac{\pi}{4} a\right)$$ Now, you must use $a^2+1=(a+i)(a-i)$ to make
$$ \int f'(a)\,da=\frac{1}{8} \pi \log \left(a^2+1\right)+\frac{1}{2} i \text{Li}_2(-i
a)-\frac{1}{2} i \text{Li}_2(i a)-$$ $$\frac{1}{2} i
\text{Li}_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) (a+1)\right)+\frac{1}{2} i
\text{Li}_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) (a+1)\right)-$$ $$\frac{1}{2} i
\log ((1+i)-(1-i) a) \log (a+1)+\frac{1}{2} i \log (-(1+i) (a+i)) \log
(a+1)-$$ $$\log (a) \tan ^{-1}(a)-\frac{1}{2} \log (2) \tan ^{-1}(a)$$
$$ \int_0^1 f'(a)\,da=\frac{1}{2} (2 C-i (\text{Li}_2(1-i)-\text{Li}_2(1+i))+\pi \log (2))-\frac{1}{2} i
\left(\text{Li}_2\left(\frac{1}{2}+\frac{i}{2}\right)-\text{Li}_2\left(\frac{1}{
2}-\frac{i}{2}\right)\right)=C+\frac{1}{8} \pi \log (2)$$ Since
$$\int_0^1 \frac{\log(x)} {x^2 + 1} \,dx=-C$$ then the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove a well-known inequality using Cauchy-Schwarz or AM-GM For $a;b>0$ and $ab \geq 1$ we have a well-known inequality:
$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2} \geq \dfrac{2}{1+ab}$
Which is equivalent to: $\dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)} \geq 0$ (true)
But is there a solution for this inequality by Cauchy-Schwarz or AM-GM? Thank you.
| Again not AM-GM or C-S (unless you count $a^2+b^2\geq2ab$) but I thought this was kinda sweet:
\begin{align*}
\frac{1}{1+a^2}+\frac{1}{1+b^2}=1-\frac{(ab)^2-1}{1+a^2+b^2+(ab)^2}&\geq1-\frac{(ab)^2-1}{1+2ab+(ab)^2}\\
&=1-\frac{ab-1}{ab+1}\\
&=\frac{2}{1+ab}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Simplified expression for centers of period-three Mandlebrot bulbs The Mandelbrot set contains three regions (two bulbs and a cardioid) with periods of three. These regions each contain a fixed point which is a root of the expression $x^3 + 2x^2 + x + 1$.
The fixed point at the center of the period-three cardioid has the value
$$-\frac{1}{3} \left(2 + \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \sqrt[3]{\frac{25 + 3 \sqrt{69}}{2}}\right)$$
This simplifies to $-ρ^2$, where $ρ$ is the plastic constant equal to $\sqrt[3]{\frac{9 + \sqrt{69}}{18}} + \sqrt[3]{\frac{9 - \sqrt{69}}{18}}$.
The fixed points at the centers of the two period-three bulbs are
$$-\frac{2}{3} + \frac{1 + i \sqrt{3}}{6} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \frac{1 - i \sqrt{3}}{3 \sqrt[3]{2}^2 \sqrt[3]{25 - 3 \sqrt{69}}}$$ and $$-\frac{2}{3} + \frac{1 - i \sqrt{3}}{6} \sqrt[3]{\frac{25 - 3 \sqrt{69}}{2}} + \frac{1 + i \sqrt{3}}{3 \sqrt[3]{2}^2 \sqrt[3]{25 - 3 \sqrt{69}}}$$
The real part of these two points simplifies to $\frac{ρ^2 - 2} {2}$. The imaginary parts are approximately $\pm$ 0.744861766619744236593170... $i$, but I can't figure out a closed-form expression for that. I'm hopeful that I can find something in terms of $ρ$, since that worked for the real part and for the cardioid point, but so far I've been stumped.
| Given that I know the real root of the function $x^3 + 2x^2 + x + 1$ is $-ρ^2$, I can factor that out to get $(x+ρ^2)(x^2 + (2 - ρ^2) x + (2 - ρ^6 + 3ρ^4 - 3ρ^2))$.
Using the quadratic formula on that second term, I get $\frac{ρ^2 ± \sqrt{-4 + 8ρ^2 - 11ρ^4 + 4ρ^6} - 2} {2}$.
Because $ρ^3 = ρ + 1$ and $ρ^4 = ρ^2 + ρ$, the term inside the square root simplifies to $ρ^2 - 3ρ$. That is negative, so factoring out $ί$ gives $\sqrt{3ρ - ρ²} ί$.
Rearranging, that gives me the two complex roots as $\frac{ρ^2 - 2} {2} ± \frac{\sqrt{3ρ - ρ²}} {2} ί$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $
How to go further
| By AM-GM $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge 2(\sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} )=2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$
Now using $x^2+y^2+z^2\ge xy+yz+zx$ $$\begin{align*}2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 2(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}})=2(\sqrt{a}+\sqrt{b}+\sqrt{c})& \\ \ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}\end{align*}$$ as $abc=1$ we see that $k=3$
Equality when $a=b=c=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Time period of a nonuniform oscillator I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the variables for $\eqref{eq1}$ to get
$$\frac{1}{\omega-a\sin(\theta)}d\theta=dt.$$
I know that for the time to increase from $0$ to $T$, then the oscillator has gone from $0$ to $2\pi$. Therefore
$$\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta=\int_0^Tdt$$
and so
$$T=\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta.$$
This is where the fun begins. I use the Weierstrass substitution of $u=\tan\left(\frac{\theta}{2}\right)$ to evaluate this which changes my integral to
$$T=\int_{\theta=0}^{\theta=2\pi}\frac{2}{\omega u^2+\omega-2ua}du.$$
I take the factor of $2$ outside of the integral and complete the square of the quadratic in the denominator to arrive at
$$T=2\int_{\theta=0}^{\theta=2\pi}\frac{1}{\omega\left(\left(u-\frac{a}{\omega}\right)^2+1-\frac{a^2}{\omega^2}\right)}du.$$
I factor out $\frac{1}{\omega}$ from the integral, and use the substitutions $x=u-\frac{a}{\omega}$ and $y=1-\frac{a^2}{\omega^2}$ to make this
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{1}{x^2+y}dx.$$
I then use the substitution $x=\sqrt{y}\tan(v)$, meaning $x^2=y\tan^2(v)$ and $dx=\sqrt{y}\sec^2(v)dv$. This makes my integral
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{\sqrt{y}\sec^2(v)}{y\tan^2(v)+y}dv.$$
I take out a factor of $\frac{\sqrt{y}}{y}$ and find the following:
$$T=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\tan^2(v)+1}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\sec^2(v)}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}1 dv=\frac{2\sqrt{y}}{\omega y}\left[v\right]^{\theta=2\pi}_{\theta=0}.$$
Then I start the process of re-inserting my many substitutions. First $v=\tan^{-1}\left(\frac{x}{\sqrt{y}}\right)$, then $x=u-\frac{a}{\omega}$, then $u=\tan\left(\frac{\theta}{2}\right)$, and finally $y=1-\frac{a^2}{\omega^"}$. This gets me
$$T=\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}\left[\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)\right]^{2\pi}_{0}.$$
I can simplify this slightly. Firstly,
$$\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}=\frac{2}{\omega\sqrt{1-\frac{a^2}{\omega^2}}}=\frac{2}{\omega\sqrt{\frac{\omega^2-a^2}{\omega^2}}}=\frac{2}{\sqrt{\omega^2-a^2}},$$
and secondly
$$\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)=\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right).$$
So now I have
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$
I feel like I'm so very close to $\eqref{eq2}$ now, but this is where I become stuck. I know that $\tan\left(\frac{2\pi}{2}\right)=\tan\left(\frac{0}{2}\right)=0$, which means when I evaluate the square brackets between $0$ and $2\pi$ they just cancel out and I'm left with
$$T=0.$$
Could someone please tell me where I have gone wrong, or what I am missing here? Been stuck on this one all day to no avail!
| Taking on board the comment of @Hans Ludmark, I believe I have found the answer but I'm not entirely sure if it is rigorous. Considering the discontinuity at $\theta=\pi$ I now re-examine the equation
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$
I have split the brackets to be evaluated into two. I now write this as
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{\pi}_{0}+\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{\pi}\right),$$
where the bracket on the left is to the left of $\theta=\pi$ and the bracket on the right is to the right of $\theta=\pi$.
I then write this as
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^--\tan^{-1}\left(\frac{\omega\tan\left(\frac{0}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)+\tan^{-1}\left(\frac{\omega\tan\left(\frac{2\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)-\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^+\right)$$
where a '$-$' superscript denotes the term on the left of $\pi$ and a '$+$' superscript denotes the term on the right of $\pi$. The terms $$-\tan^{-1}\left(\frac{\omega\tan\left(\frac{0}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)$$ and $$+\tan^{-1}\left(\frac{\omega\tan\left(\frac{2\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)$$ cancel since $\tan\left(\frac{0}{2}\right)=\tan\left(\frac{2\pi}{2}\right)=0$ This leaves
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^--\tan^{-1}\left(\frac{\omega\tan\left(\frac{\pi}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)^+\right)$$
For the term on the left of $\pi$, we can take the limit to find
$$\lim\limits_{x\to\frac{\pi}{2}^-}\tan(x)=\infty,$$
and for the term on the right of $\pi$ we take the limit to find
$$\lim\limits_{x\to\frac{\pi}{2}^+}\tan(x)=-\infty.$$
This reduces our equation to
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\tan^{-1}(\infty)-\tan^{-1}(-\infty)\right).$$
Taking limits again we find that
$$\lim\limits_{x\to\infty}\tan^{-1}(x)=\frac{\pi}{2}$$
and
$$\lim\limits_{x\to-\infty}\tan^{-1}(x)=-\frac{\pi}{2}.$$
Hence:
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left(\frac{\pi}{2}--\frac{\pi}{2}\right)=\frac{2\pi}{\sqrt{\omega^2-a^2}}.$$
Could someone let me know if this is a sound method?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$
To find the domain:
$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$
$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$
$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$
$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$
The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$
But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.
Thanks
| Look at the plot below (first picture). The function can be inverted only in the interval where it is bijective.
Take the derivative
$$f'(x)=\frac{4}{3} \sin \left(\frac{1}{3} (2 x+1)\right) \cos \left(\frac{1}{3} (2 x+1)\right)=\frac{2}{3} \sin \left(\frac{2}{3} (2 x+1)\right)$$
$$f'(x)=0\to \sin \left(\frac{2}{3} (2 x+1)\right)=0$$
In the interval we are interested in, this happens when
$$\frac{2}{3} (2 x+1)=0;\;\frac{2}{3} (2 x+1)=\pi$$
which gives $$x_1=-\frac12;\;x_2=\frac{3 \pi }{4}-\frac{1}{2}$$
We will invert $f(x)$ in the interval $[x_1,x_2]$.
Set $y=f(x)$ and get
$$\sin ^2\left(\frac{1}{3} (2 x+1)\right)=y\to \sin \left(\frac{1}{3} (2 x+1)\right)=\sqrt{y}$$
then
$$\frac{1}{3} (2 x+1)=\arcsin\sqrt y\to 2x+1=3\arcsin\sqrt y\to x=\frac{1}{2} \left(3 \arcsin\left(\sqrt{y}\right)-1\right)$$
Only now we swap $x$ and $y$ to get
$$f^{-1}(x)=\frac{1}{2} \left(3 \arcsin\left(\sqrt{x}\right)-1\right)$$
The domain of $f^{-1}(x)$ is the range of $f(x)$ that is $[0,1]$.
Look at the second picture below. The inverse function graph is the symmetric of the function $f(x)$ wrt the line $y=x$.
$$...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$ I saw this twitter post that reads:
Find all the pairs of positive integers $(x,y)$ satisfying $$ x^2 + y^2 - 5xy + 5 = 0 . $$
I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's interesting is that there are (at least?) 12 general forms that looks like they all resemble the solutions to Pell's equation.
For example, the first general solution presented reads:
$$ \begin{array} { r c l }
x &=&
\dfrac1{42} \Big [ 21 \left(55 - 12\sqrt{21} \right)^n
- \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 21 \left(55 + 12\sqrt{21} \right)^n
+ \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big ] \\ \phantom0 \\
y &=&
\dfrac1{42} \Big [ 63 \left(55 - 12\sqrt{21} \right)^n
- 13 \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 63 \left(55 + 12\sqrt{21} \right)^n
+ 13 \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big] \end{array} $$ for $n = 0,1,2,3\ldots $.
Plugging $n=0,1,2,3$ gives the first few solutions
$$(x,y) = (1,3), (67,321), (7369,35307), (810523,3883449)
$$
I tried converting these solutions to the form of $X^2 - DY^2 = A$ or something similar to Pell's equation but I got nothing.
Is it a coincidence that this innocuous-looking quadratic Diophantine equation is actually a caboodle of Pell's equation in disguise? If so, how can we derive them all?
| This sort of thing has a clear description in terms of Conway's topograph; I find it more convenient to use equivalent form $u^2 + uv - 5 v^2.$ The outcome for the original problem is sequences (note that the rule deals with every other element). For instance,
$5\cdot 9 - 2 = 43$ and $5 \cdot 14 -3 = 67$
$$ x_{n+4}= 5 x_{n+2} - x_n,$$
$$ y_{n+4}= 5 y_{n+2} - y_n,$$
There are two interleaved subsequences.
$$
\begin{array}{c}
1&1&2&3&9&14&43&67&206&321&987 &1538&4729& \ldots \\
3&2&1&1&2&3&9&14&43&67&206&321&987&\ldots \\
\end{array}
$$
Running the indices backwards leads to different solutions, but they are just transpositions of the ones above.
Let's see, given $u^2 + uv - 5 v^2 = -5$ and $x=u+3v, y=v$ gives $x^2 - 5 xy + y^2 = -5$ and the reverse holds as well
Apparently I drew one of these in 2016
Examples in previous answers:
Generate solutions of Quadratic Diophantine Equation
diagrams
Another quadratic Diophantine equation: How do I proceed?
How to find solutions of $x^2-3y^2=-2$?
Generate solutions of Quadratic Diophantine Equation
Why can't the Alpertron solve this Pell-like equation?
Finding all solutions of the Pell-type equation $x^2-5y^2 = -4$
If $(m,n)\in\mathbb Z_+^2$ satisfies $3m^2+m = 4n^2+n$ then $(m-n)$ is a perfect square.
how to solve binary form $ax^2+bxy+cy^2=m$, for integer and rational $ (x,y)$ :::: 69 55
Find all integer solutions for the equation $|5x^2 - y^2| = 4$
Positive integer $n$ such that $2n+1$ , $3n+1$ are both perfect squares
Maps of primitive vectors and Conway's river, has anyone built this in SAGE?
Infinitely many systems of $23$ consecutive integers
Solve the following equation for x and y: <1,-1,-1>
Finding integers of the form $3x^2 + xy - 5y^2$ where $x$ and $y$ are integers, using diagram via arithmetic progression
Small integral representation as $x^2-2y^2$ in Pell's equation
Solving the equation $ x^2-7y^2=-3 $ over integers
Solutions to Diophantine Equations
How to prove that the roots of this equation are integers?
Does the Pell-like equation $X^2-dY^2=k$ have a simple recursion like $X^2-dY^2=1$?
http://math.stackexchange.com/questions/1737385/if-d1-is-a-squarefree-integer-show-that-x2-dy2-c-gives-some-bounds-i/1737824#1737824 "seeds"
Find all natural numbers $n$ such that $21n^2-20$ is a perfect square.
Is there a simple proof that if $(b-a)(b+a) = ab - 1$, then $a, b$ must be Fibonacci numbers? 1,1,-1; 1,11
To find all integral solutions of $3x^2 - 4y^2 = 11$
How do we solve pell-like equations?
Diophantine equation $x^2 + xy − 3y^2 = 17$ <1,1,-3>
| {
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"url": "https://math.stackexchange.com/questions/3991684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Area of shaded triangular region from absolute value functions So, I got this question from my teacher. I've tried solving it but to no avail. I cannot work out a method either, which can help me solve this.
So, it gives us two absolute value functions
$g(x) = 4|x-3|+3$
$f(x) = -6|x-3|+9$
There are two points, A & B as shown in the figure given below.
I don't understand, when I find the points of intersection by solving
$g(x) = f(x)$
then what are those $x$ values. Are they the coordinates of point A and B?
And what would be the next steps leading to the complete solution of this question.
I'd appreciate if anyone points out the steps or just solves it for my ease.
Question FIGURE
| From your diagram, we can see that we must use $A,B$, and the vertex of $f(x)$ to calculate the area of our triangle. $A, B$ are the intersection points of $f(x)$ and $g(x)$.
Thus, we set $g(x) = f(x)$ to solve for our intersection points, and we get $4|x-3| +3 = -6|x-3|+9$. Simplifying, we get $10|x-3| = 6$, so $|x-3| = 0.6$. The two values that have an absolute value of $0.6$ are $0.6 $ and $-0.6$. Thus, we have $x-3 = 0.6$ or $x-3 = -0.6$. Solving for $x$, in each case, we get $x=2.4, 3.6$.
We plug these values into either $f(x)$ or $g(x)$, it doesn't matter because both functions are equal at our intersection points. So, we have $g(2.4) = 4|2.4-3| +3 = 5.4$ and $g(3.6) = 4|3.6-3|+3=5.4$. Thus, our intersection points are $(2.4,5.4)$ and $(3.6,5.4)$.
Now, we have to find the vertex of $f(x) = -6|x−3|+9$. Looking at our diagram, we see that the vertex of $f(x)$ is when it is the biggest. Since we are subtracting $-6|x−3|$ in $f(x)$, we try to make it as small as possible to make $f(x)$ as big as possible. Since absolute value is never negative, the smallest we can make it is $0$. We achieve $-6|x−3|=0$ when $x=3$. So, the vertex of $f(x)$ has $x$-coordinate $3$, and the $y$-coordinate is just $f(3) = 9$ because we made $-6|x−3|=0$. Thus, the vertex of $f(x)$ is $(3,9)$.
So, we have all the points of our triangle $(2.4,5.4), (3.6,5.4), (3,9)$. The base of our triangle is between the points $(2.4,5.4)$ and $(3.6,5.4)$ and the distance between them is $3.6-2.4 = 1.2$. So $b=1.2$. The height of our triangle is the distance from $(3,9)$ to our base. The $y$-coordinate of $(3,9)$ is $9$, and the $y$-coordinate of our base is $5.4$. The difference between the $y$ is $3.6$, so height $h = 3.6$. Use area of triangle formula $A =bh/2$ to get $A = 1.2*3.6/2 =2.16$. Thus, the area of the shaded traingle is $2.16$.
| {
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"url": "https://math.stackexchange.com/questions/4001563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$. Why is this inequality true? The proof to the generalized harmonic series has the following inequality:
$$\sum_{k = 1}^{\infty}\frac{1}{k^c} < 1 + \sum_{k = 1}^{\infty}\big(\frac{1}{2^\omega}\big)^k$$
with $c > 1$ and $c = 1 + \omega$, $\omega > 0$.
I am aware that the generalized harmonic series can be proven alternatively. But I don't see the above inequality. Why is that true? I tried to list out a few of the first terms and compare them term-by-term.
LHS: $ 1 + \frac{1}{2\cdot2^{\omega}} + \frac{1}{3\cdot3^{\omega}} + \frac{1}{4\cdot4^{\omega}} + \frac{1}{5\cdot5^{\omega}} + \dots$
RHS: $ 1 + \{\frac{1}{2^{\omega}} + \frac{1}{2^{2\omega}} + \frac{1}{2^{3\omega}} + \frac{1}{2^{4\omega}} + \dots\} = 1 + \{\frac{1}{2^{\omega}} + \frac{1}{4^{\omega}} + \frac{1}{8^{\omega}} + \frac{1}{16^{\omega}} + \dots\}$
Starting from $k = 5$, however, the term of LHS is not necessarily smaller than that of RHS.
| Group by packets of size $2^n$:
$$\sum_{k=1}^{+\infty}\frac 1 {k^c}=\sum_{n=0}^{+\infty}\left(\sum_{m=2^n}^{2^{n+1}-1}\frac 1 {m^c}\right)< \sum_{n=0}^{+\infty}\frac{2^n}{(2^n)^c}=\sum_{n=0}^{+\infty}\frac 1 {2^{\omega n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Min $P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1}$
Given $x,y,z$ are positive numbers such that $$x^2y+y^2z+z^2x=3$$
Find the minium of value: $$P=\frac{x^5y}{x^2+1} +\frac{y^5z}{y^2+1} +\frac{z^5x}{z^2+1} $$
My Attempt:
$$P= x^3y+y^3z+z^3x - \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1} \right)$$
$$x^2+1 \geq 2x \Rightarrow \left( \frac{x^3y}{x^2+1}+ \frac{y^3z}{z^2+1}+\frac{z^3x}{z^2+1}\right)\leq \frac{1}{2}\left(x^2y+y^2z+z^2x\right)=\frac{3}{2}$$
$$P \geq x^3y+y^3z+z^3x - \frac{3}{2}$$
I was stuck when I was finding the minium of $x^3y+y^3z+z^3x$
Help me, thank you so much
| There's no minimum of $P$ at all. Consider an intersection of $x^2y+y^2z+z^2x = 3$ and $y = x^{-4}$, i.e. $x^{-2} + x^{-8}z + z^2x = 3$ or $x^9 z^2 + z + x^6 - 3x^8 = 0$. We can find from here that there exists a positive solution
$$
z = \dfrac{-1 + \sqrt{1+12x^{17}-4x^{15}}}{2x^9}.
$$
at least for $x$ big enough. Moreover,
$$
\lim_{x \to +\infty}z\sqrt{x} = \sqrt{3}.
$$
Hence
$$
\lim_{x \to +\infty}\left[ \dfrac{x^5 y}{x^2+1} + \dfrac{y^5 z}{y^2+1} + \dfrac{z^5 x}{z^2+1}\right] = \lim_{x \to +\infty}\left[\dfrac{y^5 z}{y^2+1} + \dfrac{z^5 x}{z^2+1}\right] = \lim_{x \to +\infty}\dfrac{z^5 x}{z^2+1} = \lim_{x \to +\infty}\dfrac{3^{\frac{5}{2}}x^{-\frac{5}{2}} x}{3 x^{-1}+1} = 0.
$$
So $P$ does not have a minimal value on the set $(x,y,z) \in (0, +\infty) \times (0, +\infty) \times (0, +\infty)$ (because $P$ is obviously positive on it).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$ I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$
We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.
But I have a question: what should I do if the problem is for 4 variables, eg:
Let $a;b;c;d \geq 0$ and $a+b+c+d=3$, prove: $ab^2+bc^2+cd^2+da^2 \leq 4$
Thank you.
| If we fix $b$ and $d$ and write $a=x, c=s-x$ where $s=3-b-d\ge 0$ then
$$f(x) = ab^2+bc^2+cd^2+da^2 = b^2 x+b(s-x)^2+d^2(s-x)+dx^2, x \in [0,s]$$
is quadratic with leading coefficient $b+d \ge 0$ so its maximum is obtained at either $x=0 (a=0) $ or $x=s (c=0)$.
Similarly if we fix $a$ and $c$, then setting either $b=0$ or $d=0$ will increase $ab^2+bc^2+cd^2+da^2$. Now WLOG we can assume $c=d=0$. Then we need to prove
$$ab^2 \le 4 \text { where } a+b=3$$
which is obvious via AM-GM:
$$\frac 14 ab^2 = a \cdot \frac b2 \cdot \frac b2 \le \left( \frac{a+\frac b2 + \frac b2}{3} \right)^3 = 1$$
Remark 1: We can easily generalize to: if $a_i\ge 0, \sum_{i=1}^n a_i = 3$ then $S=\sum_{i=1}^n a_i a_{i+1}^2 \le 4$. Note that $a_{n+1}$ is defined as $a_1$.
For any non-adjacent $a_i$, $a_j$ we can show that if we fix everything else, then $S$ obtains maximum at either $a_i=0$ or $a_j=0$ so we are left with two adjacent $a_k, a_{k+1}\ge 0$ while every other terms are zeros. Then we apply AM-GM and we are done.
Remark 2: It's easy to see that your inequality becomes equality if and only if $(a_1, \cdots, a_n)=(1,2,0, \cdots,0)$ or one of its cyclic permutations.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Puzzling solution for $1/x = 3 - 2\sqrt{x}$ What is the solution for
$$\frac{1}{x} = 3 - 2\sqrt{x}$$
When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.
However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.
What am I missing?
| Rewrite the equation as
$$
2\sqrt{x}=3-\frac{1}{x}
$$
This shows that there is an implied condition, besides $x>0$ necessary in order that the equation makes sense: indeed you need
$$
3-\frac{1}{x}\ge0
$$
which becomes, taking into account that $x>0$, $3x-1\ge0$, so $x\ge1/3$.
Now you can safely square and get
$$
4x=9-\frac{6}{x}+\frac{1}{x^2}
$$
and therefore
$$
4x^3-9x^2+6x-1=0
$$
This factors as $(x-1)^2(4x-1)=0$. Since $1/4<1/3$, this is a spurious solution that must be discarded. Hence the only solution is $x=1$.
| {
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"url": "https://math.stackexchange.com/questions/4015535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Smallest possible area for triangle I'm solving the following question:
The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?
This is my attempt at solving the problem:
Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$
Let's find the tangent of the ellipse:
$2x^2 + y^2 = 1$
$4x + 2y \dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{-2x}{y}$
Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:
$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$
So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$
Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.
So now area is $1/2 a.b = 1/2 a^2$
Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.
I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ? Unfortunately, the book I'm using doesn't provide
the answer to this question.
|
Outline : We will write down the equation of tangent at arbitrary point $P$ of ellipse. Let it intersect the coordinate axes in $A$ and $B$. We identify the legs of the right triangle formed can be identified as $x$- and $y$- intercepts of the tangent line. So we find the smallest area of triangle minimizing the product of the intercepts.
Let $P=(h,k)$. Then tangent at $P$ is $2xh+ky=1$ which can be written as
$$\frac{x}{1/(2h)}+\frac{y}{1/k}=1$$
Area $=\frac{1}{2}\cdot \frac{1}{(2h)}\cdot \frac{1}{k}\cdot$ where $2h^2+k^2=1$
By AM-GM,
$$\frac{2h^2+k^2}{2} \ge \sqrt{2h^2\cdot k^2}$$
$$\Rightarrow \frac{1}{hk} \ge \frac{2\sqrt{2}}{2h^2+k^2}=2\sqrt{2}$$
$$\therefore A_{\text{min}}=\frac{1}{4hk}=\frac{2\sqrt{2}}{4}=\frac{1}{\sqrt{2}}$$
Equality occurs at $2h^2=k^2$ which can be used to determine coordinates of optimal $P$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$
But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.
| Hint:
Let $\dfrac{x+y}{xy}=k$
$$(xy)^3=3(xy)^2+(x+y)^3-3xy(x+y)$$
$$\iff(xy)^3(1-k^3)-3(xy)^2+3xy(kxy)=0$$
$$\iff(xy)^3(1-k^3)+3(xy)^2(k-1)=0$$
As $xy\ne0$ $$(xy)(1-k^3)=3(1-k)$$
Clearly, $k=1$ is a solution.
Otherwise, $$xy=\dfrac3{1+k+k^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.
| Hint:
$$(a+b)(ax^2+by^2)-(ax+by)^2=\cdots=ab(x-y)^2\ \ \ \ (1)$$
$$(a+b)(ax^3+by^3)-(ax+by)(ax^2+by^2)=\cdots=ab(x-y)^2(x+y)\ \ \ \ (2)$$
On division, we get $x+y$
$$(a+b)(ax^3+by^3)-(ax^2+by^2)^2=\cdots=abxy(x-y)^2\ \ \ \ (3)$$
$(1)/(3)$ will give $xy$
$$(a+b)(ax^4+by^4)-(ax+by)(ax^3+by^3)=\cdots=ab(x-y)^2((x+y)^2-xy)\ \ \ \ (4)$$
$(4)/(3)$ will give $$\dfrac{23(ax^4+by^4)-79\cdot691}{23(691)-217^2}=\dfrac{xy}{(x+y)^2-xy}$$
Now replace the values of $x+y$ and $xy$
| {
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"source": "stackexchange",
"question_score": "4",
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If $x+y+z \geq xyz$ Prove that $x^2 + y^2 + z^2 \geq xyz$. $x,y,z \in \mathbb R$
My attempt:
Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So :
$$x^2 + y^2 + z^2 \geq x+y+z $$
But this is actually not always true, take the case when $ 0<x,y,z <1$ And you will see that this is not working.
Maybe I’m wrong.
| I guess you guys are too fast and / or assume too much prior knowledge. Let's digest it.
We need to be careful since all numbers are reals. Hence AM-GM and square root methods don't work.
First, note that if $xyz < 0$, then there is nothing to prove, since in this case $x^2 + y^2 + z^2 \geq 0 \geq xyz$ holds in all cases. So we need to consider $xyz \ge 0$ only.
We want to establish the following train of inequalities:
$$
(x^2+y^2+z^2)^2 \ge (xy+yz+zx)^2 \ge {3xyz(x+y+z)} \ge {3} (xyz)^2 = 3 |xyz|^2
$$
The first one follows from
$$
x^2+y^2+z^2 - ( xy+yz+zx) = \frac12 \Big[ (x-y)^2+(y-z)^2+ (z-x)^2 \Big] \ge 0
$$
The second one follows from
$$
(xy+yz+zx)^2 - 3xyz(x+y+z) = \frac12 \Big[(xy-yz)^2+(zx-xy)^2+(yz-zx)^2 \Big] \ge 0
$$
For the third one, for $xyz > 0$, then this follows from the question's condition: $x + y+ z \ge xyz$.
So indeed it holds for all reals:
$$
x^2+y^2+z^2 \ge \sqrt{3} |xyz| \ge \sqrt{3} \cdot xyz
$$
This is also the tightest inequality, since we observe that equality holds for the question's condition and for this solution at $x=y=z = \sqrt{3}$. (River Li asked for that constant $k = \sqrt{3}$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of $\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}$ I found some combinatorial identities in my old notebooks, but I cannot recall how I derived them. Can anyone help provide the most elementary/elegant proofs for the following identity? Specifically, the connection between the summation and either of the two closed-form expressions.
$$
\sum_{k=0}^n{ \binom{2k}{k} 2^{2n-2k} } = (2n+1) \binom{2n}{n} = \frac{n+1}{2} \binom{2(n+1)}{n+1}
$$
Some specific cases written out for some simple visualization: ($n=0$ through $n=3$)
$$
\begin{align}
\binom{0}{0} 2^{0} &= 1\binom{0}{0} = \frac{1}{2}\binom{2}{1} \\
\binom{0}{0} 2^{2} + \binom{2}{1} 2^{0} &= 3\binom{2}{1} = \frac{2}{2}\binom{4}{2} \\
\binom{0}{0} 2^{4} + \binom{2}{1} 2^{2} + \binom{4}{2} 2^{0} &= 5\binom{4}{2} = \frac{3}{2}\binom{6}{3}\\
\binom{0}{0} 2^{6} + \binom{2}{1} 2^{4} + \binom{4}{2} 2^{2} + \binom{6}{3} 2^{0} &= 7\binom{6}{3} = \frac{4}{2}\binom{8}{4}
\end{align}
$$
EDIT: I should specify that I have already found an inductive proof, but it is not satisfying to me. I am specifically looking for a bijective/double-counting proof ideally.
| A Couple of "Common" Series
We have the Geometric Series
$$
\sum_{k=0}^\infty 4^kx^k=\frac1{1-4x}\tag1
$$
and the Generating Function for the Central Binomial Coefficients
$$
\sum_{k=0}^\infty\binom{2k}{k}x^k=\frac1{\sqrt{1-4x}}\tag2
$$
Equation $(2)$ is the Generalized Binomial Theorem applied to $\binom{-1/2}{k}=\left(-\frac14\right)^k\binom{2k}{k}$, which is shown in this answer.
Using the Cauchy Product
We get the product of the preceding functions by taking half the derivative of $(2)$ and adjusting the index:
$$
\sum_{k=0}^\infty\frac{k+1}2\binom{2k+2}{k+1}x^k=\frac1{\sqrt{1-4x}^3}\tag3
$$
The Cauchy Product Formula then says that
$$
\begin{align}
\sum_{k=0}^n\binom{2k}{k}4^{n-k}
&=\frac{n+1}2\binom{2n+2}{n+1}\tag4\\[3pt]
&=(2n+1)\binom{2n}{n}\tag5
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$
| I thought it might be instructive to present an approach that relies on straightforward contour integration. To that end, we now proceed.
Let $I$ be the integral given by
$$I=\int_0^\infty \frac{\log(x)}{x^3-1}\,dx$$
Now, moving to the complex plane, we analyze the contour integral $J$ given by
$$J=\oint_C \frac{\log^2(z)}{z^3-1}\,dz$$
where $C$ is the classical keyhole contour with a semi-circular deformation at $z=1+i0^-$. That is, $J$ is given by
$$\begin{align}
J&=\int_0^\infty \frac{\log^2(x)}{x^3-1}\,dx-\text{PV}\int_0^\infty\frac{(\log(x)+i2\pi)^2}{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I+4\pi^2\text{PV}\int_0^\infty \frac1{x^3-1}\,dx+i\frac{4\pi^3}3\\\\
&=-i4\pi I-\frac{4\pi^3}{3\sqrt3}+i\frac{4\pi^3}3\tag1
\end{align}$$
Rearranging $(1)$ reveals that the integral of interest can by written in terms of $J$ as
$$I=i\frac1{4\pi}J+i \frac{\pi^2}{3\sqrt 3}+\frac{\pi^2}{3}\tag2$$
Now, applying the residue theorem we see that $\frac{iJ}{4\pi}$ is given by
$$\begin{align}
\frac{iJ}{4\pi}&=-\frac12 \left(\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i2\pi/3}\right)+\text{Res}\left(\frac{\log^2(z)}{z^3-1}, z=e^{i4\pi/3}\right)\right)\\\\
&=-\frac{4\pi^2}{27}\left(\frac54+i\frac{3\sqrt{3}}{4}\right)\tag3
\end{align}$$
Using $(3)$ in $(2)$ yields the coveted result
$$I=\frac{4\pi^2}{27}$$
as expected!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 2
} |
Bounding the difference between consecutive terms in the sequence $a_n=(1+1/n)^n$ I'm pretty sure the following estimate holds for $C>1$, but I can't prove it:
$$a_n:=\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1}\leq\frac{C}{n^2}$$
For context, I was looking for some such bound in order to prove that $na_n\to0$.
I tried using the fact that
$$\left(1+\frac{1}{n}\right)^n\leq\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}+\cdots+\color{red}{\frac{1}{n!}}:=S_n$$
because, by some algebra with the binomial theorem,
$$\left(1+\frac{1}{n}\right)^n=\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}\color{blue}{\left(1-\frac{1}{n}\right)}+\cdots+\color{red}{\frac{1}{n!}}\color{blue}{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)}$$
where the blue factors are all less than one.
But I also need a good lower bound on the terms $(1+n^{-1})^{n}$, and I can't see how to pick the right lower bound to get, say,
$$a_n\leq S_n-\text{lower bound applied to the term $\left(1+\frac{1}{n-1}\right)^{n-1}$}\leq\frac{C}{n^2}$$
| Suppose that $p\geq 1$. Then
$$
\left( {1 + \frac{1}{p}} \right)^p = \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right) \ge \exp \left( {1 - \frac{1}{{2p}}} \right) \ge e\left( {1 - \frac{1}{{2p}}} \right)
$$
and
\begin{align*}
&\left( {1 + \frac{1}{p}} \right)^p \le \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right)\le \exp \left( {1 - \left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right)} \right) \\ & \le e\left( {1 - \left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right) + \frac{1}{2}\left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right)^2 } \right) \le e\left( {1 - \frac{1}{{2p}} + \frac{{11}}{{24p^2 }}} \right).
\end{align*}
These inequalities follow from the fact that for all $x>0$,
$$
x - \frac{{x^2 }}{2} < \log (1 + x) < x - \frac{{x^2 }}{2} + \frac{{x^3 }}{3},\quad e^{ - x} < 1 - x + \frac{{x^2 }}{2}.
$$
In the last step, I re-arranged the terms and observed that when multiplied by $p^2$, the sum of the terms beyond $-\frac{1}{2p}$ tend monotonically from below to $\frac{11}{24}$. Thus, for all $n\geq 2$,
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^n - \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1} & \le e - \frac{e}{{2n}} + \frac{{11}}{{24n^2 }} - e + \frac{e}{{2(n - 1)}}
\\ &
= \frac{e}{2}\frac{1}{{n(n - 1)}} + \frac{{11}}{{24n^2 }} \le \frac{{24e + 11}}{{24}}\frac{1}{{n^2 }} < \frac{4}{{n^2 }} .
\end{align*}
Addendum: By the mean value theorem
\begin{align*}
&\left( {1 + \frac{1}{n}} \right)^n - \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1} = \left( {1 + \frac{1}{\xi }} \right)^\xi \left( {\log \left( {1 + \frac{1}{\xi }} \right) - \frac{1}{{\xi + 1}}} \right)
\\ &
\le e\left( {\log \left( {1 + \frac{1}{\xi }} \right) - \frac{1}{{\xi + 1}}} \right) \le \frac{e}{{2\xi ^2 }} \le \frac{e}{{2(n - 1)^2 }},
\end{align*}
where $n-1<\xi<n$ is a suitable number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
| Alternative approach that avoids Calculus:
square both sides.
LHS squared is $(2x-1) - 2\sqrt{x(x-1)}$
RHS squared is $(2x+1) - 2\sqrt{x(x+1)}$
So, taking LHS - RHS, the question reduces to whether
$-2 + 2\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 0.$
This is equivalent to asking whether
$\sqrt{x}[\sqrt{x+1} - \sqrt{x-1}] > 1.$
Edit
Stealing a caveat from Quanto's answer, note that when $x\geq 1$, the LHS above is clearly positive.
Again squaring both sides, the problem reduces to asking whether
$x[2x - 2\sqrt{x^2 -1}] > 1.$
This is answered by applying the scalar of $[2x + 2\sqrt{x^2 -1}],$ which is a positive scalar,
to both sides, so that the problem is transformed to asking whether
$$x[2x - 2\sqrt{x^2 -1}][2x + 2\sqrt{x^2 -1}] > [2x + 2\sqrt{x^2 -1}].\tag1$$
Since the LHS of equation (1) above simplifies to $4x$,
and since it is clear that $2\sqrt{x^2 - 1} < 2x$
the problem is resolved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
A Stirling number identity representing the second-order Eulerian numbers. Graham, Knuth, and Patashnik give in
CMath
a thorough introduction to the Stirling numbers.
On table 250 and table 251, they compile two pages of
Stirling number identities.
Of course, there are many more such identities. We give here one
which is not included and which we could not find anywhere else
(which means little), but Knuth would perhaps like.
They can be proved with the techniques from CMath. Maybe someone
enjoys showing us the proof?
As in CMath, the Stirling cycle numbers are unsigned.
{n, k} and [n, k] are defined for 0 <= k <= n, and for this
range the following identity holds:
$$ \sum \limits_{j\,=\,0}^{k} (-1)^{k-j}\binom{2n+1}{k-j}
{n+j \brace j}
= \sum \limits_{j\,=\,0}^{n-k} (-1)^j \binom{2n+1}{j}
{n+m \brack m} $$
For better readability we use the shortcut $ m = n-k-j+1$.
References to the OEIS are A132393, A048993, and A340556.
| We seek to show that with $0\le k\le n$ the following identity holds:
$$\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j} =
\sum_{j=0}^{n-k} (-1)^j {2n+1\choose j} {2n-k-j+1\brack n-k-j+1}.$$
We will start with the LHS. The chapter 6.2 on Eulerian Numbers
of Concrete Mathematics by Knuth et al. proposes the formula
$${n\brace m} = (-1)^{n-m+1} \frac{n!}{(m-1)!} \sigma_{n-m}(-m)$$
where $\sigma_n(x)$ is a Stirling polynomial and we have the identity
$$\left(\frac{1}{z} \log\frac{1}{1-z}\right)^x
= x \sum_{n\ge 0} \sigma_n(x+n) z^n.$$
We get
$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^x
= x \sigma_{n-m}(x+n-m)$$
and hence
$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}
= -n \sigma_{n-m}(-m)$$
which implies that for $n\ge m\ge 1$
$$\bbox[5px,border:2px solid #00A000]{
{n\brace m} = (-1)^{n-m} \frac{(n-1)!}{(m-1)!}
[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}.}$$
This gives for the LHS
$$\sum_{j=1}^k (-1)^{k-j} {2n+1\choose k-j}
(-1)^n \frac{(n+j-1)!}{(j-1)!} [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-j}
\\ = (-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
[w^{k-1}] (1+w)^{2n+1}
\\ \times \sum_{j=1}^k {n+j-1\choose n} (-1)^{j-1} w^{j-1}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-j+1}.$$
Now the coefficient extractor in $w$ enforces the upper limit of the
sum and we may extend $j$ to infinity, getting
$$(-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
[w^{k-1}] (1+w)^{2n+1}
\frac{1}{(1+w/(\frac{1}{z} \log\frac{1}{1-z}))^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
[w^{k-1}] (1+w)^{2n+1}
\frac{1}{(w+\frac{1}{z} \log\frac{1}{1-z})^{n+1}}.$$
Continuing,
$$ (-1)^{n-k+1} n! [z^n]
[w^{n+k}] (1+w)^{2n+1}
\frac{1}{(1+\frac{1}{w} \frac{1}{z} \log\frac{1}{1-z})^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
[w^{n+k}] (1+w)^{2n+1}
\sum_{q\ge 0} {n+q\choose n} (-1)^q \frac{1}{w^q}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^q
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=n+k}^{2n+1} {2n+1\choose j}
{n+j-(n+k)\choose n} (-1)^{j-(n+k)}
\\ \times
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j-(n+k)}
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} n!
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j} [z^{n+j}]
\left(\log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} n!
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j}
\\ \times \frac{j!}{(n+j)!} \times (n+j)! [z^{n+j}]
\frac{1}{j!} \left(\log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1}
{2n+1\choose j+n+k} (-1)^j {n+j\brack j}
\\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1}
{2n+1\choose 2n-j+1} (-1)^{n-k-j+1} {2n-k-j+1\brack n-k-j+1}
\\ = \sum_{j=0}^{n-k+1}
{2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$
The Stirling number is zero for $j=n-k+1$ and we get at last
$$\sum_{j=0}^{n-k}
{2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$
This is the RHS and we have the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
computing the limit $\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$ I'm trying to compute the following limit and would greatly appreciate your heartening feedback on my solution.
The limit:
$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$
My steps in deriving the solution:
Preliminary identities:
*
*$\sec \theta = \frac{1}{\cos \theta}$
*$\tan \theta = \frac{\sin \theta}{\cos \theta}$
$\frac{1}{\cos \theta}-\frac{\sin\theta}{\cos\theta} = \frac{1-\sin\theta}{\cos\theta} = \frac{1-\sin^2\theta}{(1+\sin\theta)\cos\theta} = \frac{\cos^2\theta}{\cos\theta}\cdot \frac{1}{1+\sin\theta} = \frac{\cos\theta}{1+\sin\theta} = \frac{0}{1+1}$
When $\theta \to \frac{\pi}{2}$ then $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$
The answer being:
$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta) = 0$
| Your proof is fine. Here's another:$$\sec\theta,\,\tan\theta\to\infty\implies\sec\theta+\tan\theta\to\infty\implies\sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta}\to0.$$
| {
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"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
What did I do wrong with this integral problem?
Find the value of the following definite integral
$$\int_0^{\pi/2} \sqrt{1 - \sqrt{3}\sin 2x + 2\cos^2x} \, dx$$
*
*My attempt:
$$\int_0^{\pi/2} \sqrt{1 - \sqrt{3}\sin 2x + 2\cos^2x} \, dx$$
$$= \int_0^{\pi/2} \sqrt{2 - \sqrt{3}\sin 2x + \cos2x} \, dx$$
$$= \int_0^{\pi/2} \sqrt{2 + 2\cos \left(2x + \dfrac{\pi}{3}\right)} \, dx$$
$$= \sqrt{2} \int_0^{\pi/2} \sqrt{1 + \cos \left(2x + \dfrac{\pi}{3}\right)} \, dx$$
Let $u = 1 - \cos \left(2x + \dfrac{\pi}{3}\right) \implies du = 2\sin \left(2x + \dfrac{\pi}{3}\right) \, dx = 2\sqrt{2u - u^2} \, dx \implies dx = \dfrac{1}{2\sqrt{2u - u^2}} \, du$
The integral become:
$$\sqrt{2}\int_{1/2}^{3/2} \dfrac{\sqrt{2-u}}{2\sqrt{2u - u^2}} \, du$$
$$= \sqrt{2}\int_{1/2}^{3/2} \dfrac{1}{2\sqrt{u}} \, du$$
$$= \sqrt{2} \cdot \left(\sqrt{u}\right) \Biggr|_{1/2}^{3/2} = \sqrt{3} - 1$$
However, using my calculator and other online calculators, they're all giving $3 - \sqrt{3}$, which is different from mine.
So what did I do wrong?
A small realization I found out is that:
$$\text{(Calculator answer) = $\sqrt{3} \cdot$ (My answer)}$$
| $$\int_0^{\pi/2}\sqrt{1-\sqrt{3}\sin(2x)+2\cos^2(x)}dx$$
then as you did we know that:
$$\cos(2x)=2\cos^2(x)-1\Rightarrow 2\cos^2(x)=\cos(2x)+1$$
so we have:
$$\int_0^{\pi/2}\sqrt{2-\sqrt{3}\sin(2x)+\cos(2x)}dx$$
try substituting $u=2x\Rightarrow dx=du/2$ so:
$$\int_0^\pi\sqrt{2-\sqrt{3}\sin(u)+\cos(u)}\frac{du}2$$
now set up
$$\sqrt{3}\sin(u)+\cos(u)=A\cos(u+\alpha)$$
solving gives $2\cos\left(x-\frac\pi3\right)$
now we have:
$$\frac1{\sqrt{2}}\int_0^\pi\sqrt{1-\sin\left(u-\frac\pi3\right)}du$$
letting $v=u-\frac\pi3\Rightarrow dv=du$ we get:
$$\frac1{\sqrt{2}}\int\limits_{-\pi/3}^{2\pi/3}\sqrt{1-\sin v}\,dv$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove an identity by induction I am working on the following problem:
$\text{For $n$ $\ge$ 1, Prove the following:}$
$n(1+x)^{n-1} = \sum_{k=1}^{n} k {n\choose k}x^{k-1} $
$\text{Deduce that: } \sum_{k=1}^{n}k{n \choose k}=n2^{n-1} $
I followed the proof by induction as follows:
$\text{Base Case: let n=1 , both RHS and LHS will evaluate to 1}$
$\text{Assume true for n=l. Now consider n=l+1}$
$(l+1)(1+x)^l=\sum_{k=1}^{l+1}k {l+1 \choose k} x^{k-1}$
$(l+1)(\sum_{k=0}^{l} {l \choose k} x^k)=\sum_{k=1}^{l+1}k {l+1 \choose k} x^{k-1}$
After this, I am kinda lost. any ideas on how to continue the proof?
| Well, $(n+1)(1+x)^n = (1+x)\frac{n+1}n\cdot n(1+x)^{n-1}= (1+x)\frac{n+1}n\cdot \sum\limits_{k=1}^nk{n\choose k}x^{k-1}=$
$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1}+ x\sum\limits_{k=1}^nk{n\choose k}x^{k-1})=$
$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1} + \sum\limits_{k=1}^nk{n\choose k}x^{k})=$
$\frac {n+1}n(\sum\limits_{k=1}^nk{n\choose k}x^{k-1} + \sum\limits_{k=2}^{n+1}(k-1){n\choose k-1}x^{k-1})=$
$\frac {n+1}n(1{n\choose 1}x^0 +\sum\limits_{k=2}^n[k{n\choose k}+(k-1){n\choose k-1}]x^{k-1} + n{n\choose n}x^n)=$
$\frac {n+1}n\cdot n + \frac {n+1}n\sum\limits_{k=2}^n[k\frac {n!}{k!(n-k)!}+(k-1)\frac {n!}{(k-1)!(n-k+1)!}]x^{k-1} +\frac {n+1}n\cdot nx^n)=$ ...
$(n+1) + \frac {n+1}n\sum\limits_{k=2}^n[\frac {n!(n-k+1)}{(k-1)!(n-k+1)!}+\frac {n!(k-1)}{(k-1)!(n-k+1)!}]x^{k-1} + (n+1)x^n =$
$k{n+1\choose k}x^{k-1}|_{k=1} +\frac {n+1}n\sum\limits_{k=2}^n\frac{n!n}{{(k-1)!(n-k+1)!}}x^{k-1} +k{n+1\choose k}x^{k-1}|_{k=n+1}=$
$k{n+1\choose k}x^{k-1}|_{k=1} +\sum\limits_{k=2}^n\frac{(n+1)!}{{(k-1)!(n-k+1)!}}x^{k-1} +k{n+1\choose k}x^{k-1}|_{k=n+1}=$
$k{n+1\choose k}x^{k-1}|_{k=1} +\sum\limits_{k=2}^nk\frac{(n+1)!}{{k!(n-k+1)!}}x^{k-1} +k{n+1\choose k}x^{k-1}|_{k=n+1}=$
$\sum\limits_{k=1}^{n+1}k{n+1\choose k}x^{k-1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}
It checks out numerically, for $n=1,\dots,7$ that
\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}
so \eqref{0} seems to follow the pattern
\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}
for some function $f$.
Items \eqref{5} and \eqref{7} look promising
as they agree to $f(n)=2n\cot^2(\tfrac\pi{n})$,
but the other fail on that.
Edit:
Also, it looks like
\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}
and
\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}
with the same $f$.
Edit
Thanks to the great answer by @Quanto,
the function $f$ can be defined as
\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}
$\endgroup$
| The close-form result can be expressed as
$$\color{blue}{ \int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx = -nG+\frac\pi2 n \ln 2
+ \pi \sum_{k=1}^{[\frac n2]}\ln \cos\frac{(n+1-2k)\pi}{4n} }
$$
as shown below. Note that
\begin{align}
I_n =
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
\overset{x\to\frac1x} == \frac12 J_n - nG
\end{align}
where $ \int_1^\infty\frac{\ln x}{1+x^2} \,dx=G$ and
$$J_n =\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx $$
Substitute
$$1+x^{2n} = \prod_{k=1}^{n}(1+e^{i\pi\frac{n+1-2k}n }x^2)
$$
and use the known result
$\int_0^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= \pi\ln(1+a^{\frac12})
$ to integrate
\begin{align}
J_{n}& =\int_0^\infty\frac{dx}{1+x^2} \sum_{k=1}^{n}
\ln (1+e^{i\pi\frac{n+1-2k}n }x^2)
= \pi\sum_{k=1}^{n} \ln (1+e^{i\pi\frac{n+1-2k}{2n} })\\
&=n \pi \ln 2 + 2\pi\sum_{k=1}^{[\frac n2]} \ln \cos\frac{(n+1-2k)\pi}{4n}
\end{align}
where the symmetry of the sequence is recognized in the last step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
Solve in the set of integers $2^x +5^x = 3^x + 4^x$
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x + 4^x$$
With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$
So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.
If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).
Any easier solution?
| For $x\geq2$ we have $3^{x-2},4^{x-2}\leq5^{x-2}$, and so from $5^2=3^2+4^2$ it follows that
$$2^x+5^x>5^x=5^{x-2}(3^2+4^2)=5^{x-2}3^2+5^{x-2}4^2\geq3^x+4^x.$$
For $x\leq-2$ we have $3^{x+2},4^{x+2}\leq2^{x+2}$, and so from $2^{-2}>3^{-2}+4^{-2}$ it follows that
$$2^x+5^x>2^x=2^{x+2}(3^{-2}+4^{-2})=2^{x+2}3^{-2}+2^{x+2}4^{-2}\geq3^x+4^x.$$
Then it suffices to check that the equality holds for $x=0$ and $x=1$, but not for $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:
*
*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.
*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2z^2 = 3xyz
$$
*Since (from 1.) the left hand side (LHS) is strictly positive, then $xyz \gt 0$
*Exploring the LHS of the equation from 1.:
$$
2x^2y^2+2x^2z^2+2y^2z^2 = (xy-xz)^2+(xy-yz)^2+(xz-yz)^2+2xyz(x+y+z)
$$
therefore
$$
x^2y^2+x^2z^2+y^2z^2 \geq xyz(x+y+z)
$$
*Replacing LHS of the last inequation by $3xyz$ (according to 2.) gives:
$$3xyz \geq xyz(x+y+z)$$ and since (from 3.) $xyz$ is positive, we can divide both sides by $xyz$, obtaining: $$x+y+z \leq 3$$
*$xyz$ can be positive in two cases:
Case I. $x, y, z$ are all positive. In this case $x \geq 1$, $y \geq 1$, $z \geq 1$,
so
$$
x+y+z \geq 3
$$
but according to 5. this must be an equation. Therefore all inequations for $x$, $y$ and $z$ must turn into equations, so $x=y=z=1$.
Case II. One of $x, y, z$ is positive, two others are negative. Without loss of generality we can assume $x \gt 0$, $y \lt 0$, $z \lt 0$
...
| ... then write $m = -y, n = -z$. Then $x,m,n$ are positive integers satisfying:
$$\frac{x(-m)}{-n} + \frac {x(-n)}{(-m)} + \frac {(-m)(-n)}{x} = 3$$
or
$$\frac {xm}{n} + \frac {xn}{m} + \frac {mn}{x} = 3$$
which is just our original equation. Using our previous analysis, $x=m=n=1$. Hence $y=z=-1$...
A shortcut to your analysis is to use the AM$\ge$GM inequality: for positive $x,y,z$,
$$\frac {xy}{z} + \frac {xz}{y} + \frac {yz}{x} \ge 3\sqrt[3]{xyz} \ge 3$$
as $xyz \ge1$. Equality holds iff $x=y=z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Challenge: find angle x in the figure via logical approach Background: this problem was from social app, but I cannot find original source. This is an interesting problem that I have not seen similar ones. It is a hard problem, cannot be solved with angle chasing. I worked out via analytic geometry programming, but cannot figure out a logical approach. Is there a logical approach to solve it? May need help from some theorems that I am not familiar with.
Problem: There are 5 known angles in the figure. Find the angle x. Is there an approach to solve it through deduction?
| Label the bottom two points (starting from left) with $A$, $B$, and the top points (starting from left) with $C$, $D$, $E$. Let $F$ be the intersection of $AD$ and $BC$. Let $\omega$ be the circumcircle of $ABF$. Let $G$ be the second common point of $\omega$ and $AE$. Let $H$ be the second common point of $BD$ and $\omega$. Finally, let $I$ be the point on the ray $BA^\to$ such that $AG=AI$.
Since $AI=AG$, we have $\angle AIG = \frac 12 \angle BAG = \frac 12 \cdot 44^\circ = 22^\circ$. Then we find that $\angle GBI = \angle FBI + \angle GBF = 49^\circ + \angle GAF = 49^\circ + 30^\circ = 79^\circ$, and $\angle IGB = 180^\circ - \angle BIG - \angle GBI = 180^\circ - 22^\circ - 79^\circ = 79^\circ$. Hence $IBG$ is isosceles and $IB=IG$.
Let $\Omega$ be the circle with center $I$ and radius $IB=IG$. Let $C'\in\Omega$ be such that the chord $BC'$ passes through $F$. Note that $\angle GIC' = 2 \angle GBC' = 2\angle GBF = 2\cdot 30^\circ = 60^\circ$, hence triangle $GIC'$ is equilateral. In particular, $C'I=C'G$. Since $AI=AG$, it follows that $AC'$ bisects the angle $GAI$. Hence $\angle GAC' = \frac 12 \angle GAI = \frac 12 \cdot (180^\circ - \angle BAG) = 90^\circ - \frac 12 \cdot 44^\circ = 90^\circ - 22^\circ = 68^\circ = \angle GAC$, and therefore $C'$ lies on the line $AC$.
Moreover, $\angle C'BI = 90^\circ - \frac 12 \angle BIC' = 90^\circ - \frac 12(\angle BIG + \angle GIC') = 90^\circ - \frac 12 \cdot (22^\circ + 60^\circ) = 49^\circ = \angle CBI$, and therefore $C'$ lies on the line $BC$.
Hence $C'$ is the intersection of $AC$ and $BC$, hence $C'$ coincides with $C$.
Next, we shall prove that $C, G, H$ are collinear. Indeed, $\angle CGH = \angle CGA + \angle AGH = \angle AIC + (180^\circ - \angle HBA) = \angle GIC + \angle BIG + 180^\circ - 82^\circ = 60^\circ + 22^\circ + 180^\circ - 82^\circ = 180^\circ$.
Applying Pascal theorem for (degenerated) hexagon $FBBHGA$ we see that $BE$ is tangent to $\omega$. Therefore $\angle EBD = \angle BAH = \angle BAF - \angle HAF = 74^\circ - \angle HBF = 74^\circ - 33^\circ = 41^\circ$.
Remark: This is a special case of the following phenomenon:
The proof presented above can be adapted to the general case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4042623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $a,b\in \mathbb R^+$, $ |a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}}$ Prove $a+b\leq 2$
Question: If $a,b\in \mathbb R^+$, $|a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}},$ prove $a+b≤2$.
I figured out that $a+b\leq \frac {1}{\sqrt{a}}+\frac {1}{\sqrt{b}}$, but I am not sure how to prove that $a+b\leq 2$ after doing this.
Can anyone help me?
| Just my 2 cents to the Albus' answer, without using power mean. We can assume $0<a\leq b$ ($a=0$ can be checked easily) then $$\frac{(a+b)^2}{4}\leq a^3+b^3 \iff \frac{\left(1+\frac{b}{a}\right)^3}{4}\leq 1+\left(\frac{b}{a}\right)^3 \tag{1}$$
Now
$$f(x)=\frac{\left(1+x\right)^3}{4} -1-x^3=(1+x)\left(\frac{(1+x)^2}{4}-(1-x+x^2)\right)
=\\
(x+1)\left(\frac{-3+6x-3x^2}{4}\right)=-\frac{3}{4}(x+1)(x-1)^2$$
We can see that $f(x)\leq 0$ for $x\geq1$, which explains $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two circles inscribed in a semicircle Two circles are inscribed in a semi circle. Given the areas of the shaded triangles, what's the radius of the semicircle?
(Note there's a similar but different question here)
|
Let $EM=a$ be tangential to both circles at the point $E$ and the angle $\theta$. Then, the area $[ABE] = 33= a^2\sin\theta$ and
\begin{align}
[ABCD]&=153 =[AOD]+[BOC]=[COD] \\
& = \frac12Rr_1\cos\alpha + \frac12Rr_2\cos\beta + \frac12R^2\sin(\alpha+\beta)
\end{align}
Substitute $\sin\alpha = \frac{r_1}{R-r_1}$, $\sin\beta= \frac{r_2}{R-r_2}$ into above equation and, after simplifing
\begin{align}
153=aR\left( \frac1{(1-\frac{r_1}{R})(1-\frac{r_2}{R})}-1\right)\tag1
\end{align}
Note that $r_1 = a \tan\frac{\theta}2$, $r_2 = a \cot\frac{\theta}2$ and $AB=AO +OB$, or
\begin{align}
2a & = \sqrt{(R-r_1)^2-r_1^2} + \sqrt{(R-r_2)^2-r_2^2}
= \sqrt{R^2-2aR\tan\frac{\theta}2}+ \sqrt{R^2-2aR\cot\frac{\theta}2}
\end{align}
which reduces to
$$R = \frac a{\sqrt2-\csc\theta}$$
Then, along with $a^2=33\csc\theta$, substitute $r_1$, $r_2$ and $R$ derived above into (1) to obtain the equation for $\theta$
$$\frac{153}{33} = \frac{\csc\theta}{\sqrt2-\csc\theta}\left(\frac1{3-4\sqrt2\csc\theta+3\csc^2\theta} -1\right)
$$
Solve to get the valid solution $\csc\theta = \frac{17}{12\sqrt2}$ and, in turn, the radius
$$R = \frac{\sqrt{33\csc\theta}}{\sqrt2-\csc\theta} = \frac67\sqrt{187\sqrt2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a formula for (0.5)²+(1)²+(1.5)²+(2)²+...+(25)² (1/4)(1³) + (1/9)(1³+2³) + (16)(1³+2³+3³) + ... + (1/2704)(1³+2³+3³+...+51³) = ?
Initially, I factored it as 1³ (1/2² + 1/3² + ... + 1/52²) + 2³ (1/3² + ... + 1/52²) + 3³ (1/4² +...+ 1/52²) + ... + 51³ (1/52²), but got stuck to split them as separate series.
Alternatively, I observed this pattern
1³/4 + (1³/9 + 2³/9) + (1+8+27)/16 + 100/25 + ... + (1+2+3+...+51)/52
= (1/2)² + (3/3)² + (6/4)² + (10/5)² + ... + ((1+...+51)/52)²
= (1/2)² + (2/2)² + (3/2)² + (4/2)² + (5/2)² ... + (51/2)²
= (1/2)² (1+2²+3²+...+51²)
= (1/2)² (1/6)(51)(2(51)+1)(51+1)
= (1/2)² (1/6)(51)(103)(52)
= (1/2)² (17)(103)(26)
= 45526/4
= 11381.25
Does this seem correct?
| The sum can be expressed as:
$$
\frac1{2^2}\cdot 1^3+\frac1{3^2}\cdot (1^3+2^3)+\frac1{4^2}\cdot (1^3+2^3+3^3)+\cdots +\frac1{52^2}\cdot (1^3+2^3+\cdots +51^3)=$$
$$\sum_{i=2}^{52} \frac1{i^2} \sum_{j=1}^{i-1} j^3\stackrel{(1)}{=}\\
\sum_{i=2}^{52} \frac1{i^2}\cdot (\frac{(i-1)i}{2})^2=\sum_{i=2}^{52} \frac{(i-1)^2}4=\frac14\sum_{i=1}^{51} i^2\stackrel{(2)}{=}\cdots$$
where:
$$\sum_{i=1}^{n} i^3=(\frac{n(n+1)}{2})^2 \quad \quad (1)$$
$$\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6} \quad \quad (2)$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $a+b+c=3,$ show $a+b+c\geq ab+bc+ac$ by using AM-GM A am having issues with a problem that states: let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Show that $$a^bb^cc^a\le1$$
As this was an example problem, a solution path was provided but I am confused as to why the solution works. It writes:
$$1=\frac{a+b+c}{3}$$
$$\ge\frac{ab+bc+ca}{a+b+c}$$
$$\ge (a^bb^cc^a)^{\frac{1}{a+b+c}}$$
I understand how weighted AM-GM could show the second inequality but I can't seem to understand how one can show $a+b+c \ge ab+bc+ca$ which is what the first inequality implies.
Btw: at this point in the text, they have only proved AM-GM and weighted AM-GM. I can think of a way to show this using Chebyshev's Inequality but I was hoping to try to understand what the text meant by this specific solution path.
| Hint
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$
but, by $MA\ge MG$: $a^2+b^2\ge 2ab, a^2+c^2\ge 2ac, b^2+c^2\ge 2bc$, what gives you
$$a^2+b^2+c^2\ge ab+ac+bc.$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trying to find if this trigonometry problem is wrong or not I know That we have a triangle ABC with $$A=90° $$ $$ b+c=18 $$ $$ \cot(B)=4\sin^2(C) $$ $$ 0<B,C<90 $$
But , upon calculating i end up a = 18, b = 18, c = 0 which should not be possible
| $$\frac{c}{b}=\cot(B)=4\sin^2(C)=\frac{4c^2}{a^2}$$
$$a^2=4bc=b^2+c^2$$
Now use $c=18-b$ and get:
$$b^2-18b+54=0$$
$$b=9\pm3\sqrt{3}$$
Therefore, $a=6\sqrt{6}$; $b,c=9\pm3\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How/why $\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}$ turn into (negative fraction) $-\frac{x+a}{x²a²}$? I have that the solution of $\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)$ is $-\frac{x+a}{x²a²}$, but when I try to solve it I get (positive) $\frac{x+a}{x²a²}$.
The way I've been getting it is by making the denominators equals $\left(\frac{1}{x²(x-a)}\frac{a²}{a²}\right)-\left(\frac{1}{a²(x-a)}\frac{x²}{x²}\right)$ which gives $\frac{a²-x²}{x²a²(x-a)}$ that by factoring the nominator is $\frac{(a-x)(a+x)}{x²a²(x-a)}$, allowing me to cancel $\frac{(a-x)}{(a-x)}$, finishing as $\frac{x+a}{x²a²}$.
Where it should've become negative?
OBS: I don't know if matters but it's all part of a limits problem that goes like $\lim_{x\to a}\left(\frac{1}{x²(x-a)}-\frac{1}{a²(x-a)}\right)=-\frac{x+a}{x²a²} = \frac{-2}{a³}$, but my problem is just where it became negative as it's suppose to become.
| $$\frac{1}{x^2(x - a)} - \frac{1}{a^2(x - a)}$$
$$\frac{1}{(x-a)}\left(\frac{1}{x^2} - \frac{1}{a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{a^2 - x^2}{x^2a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{(a + x)(a - x)}{x^2a^2}\right)$$
$$\frac{1}{(x-a)}\left(\frac{(a + x) -1(x - a)}{x^2a^2}\right)$$
$$\frac{-(a + x)}{x^2a^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
| The series can also be written as
$$
\begin{align}
\overbrace{\ \sum_{k=1}^\infty\frac1{3^k}\ }^{\substack{\text{each power}\\\text{of $3$}}}+\overbrace{\ \sum_{k=1}^\infty\frac1{9^k}\ }^{\substack{\text{even powers}\\\text{of $3$}}}
&=\frac{\frac13}{1-\frac13}+\frac{\frac19}{1-\frac19}\\
&=\frac12+\frac18\\[6pt]
&=\frac58
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
| Integrate by parts
\begin{align}
\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx
&= \int_0^\infty \frac{\ln x}{2x}\>d(\frac{x^2}{x^2+1})\\
&\overset{ibp}= \frac12\int_0^\infty \underset{=\ 0}{\frac{\ln x}{x^2+1}dx }- \frac12\int_0^\infty \frac{1}{x^2+1}dx
= -\frac\pi4
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
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