Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Complex Modulus Thanks everyone that answered my last question! I had one more question for a different concept. If the modulus of a complex number $z = |z| = \sqrt{(a^2 + b^2)}$,where $a$ and $b$ are the real and imaginary constants of $z$, are you allowed to use the similar property for conjugates and say that the modulus of $(z + 1)$, or $|z + 1| = |z| + |1|$ or would it instead be $\sqrt{(a^2 + b^2 + 1^2)}$?
| $|z+1|=\sqrt{(z+1)(z'+1)}=\sqrt{zz'+z+z'+1}=\sqrt{a^2+b^2+2a+1}$ if $z=a+bi$.
Alternatively, $|z+1|=\sqrt{(z+1)(z'+1)}=\sqrt{(a+1+bi)(a+1-bi)}=\sqrt{(a+1)^2+b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to find the matrix by a given linear transformation and eigenvalue / eigenvector the question I'm trying to resolve is the following:
Given f: R3->R3 and f(1,1,0) = (1,0,0), f(0,1,1) = (1,1,0) and where 3 is an eigenvalue of the eigenvector (1,0,0).
a) Find the associated matrix to f on canonical base R3
Any help would be much appreciated!
| Denote $B= \{(1,1,0), (0,1,1), (1,0,0)\}$ a basis given. Let $C = \{e_1,e_2,e_3\}$ the canonical basis.
Note that $e_1 = 0\cdot (1,1,0) + 0\cdot(0,0,1) + 1(1,0,0), e_2 = 1 \cdot (1,1,0) + 0 \cdot(0,1,1)+ (-1)(1,0,0)$ and $e_3 = (-1)(1,1,0) + 1 \cdot (0,1,1) + 1 \cdot(1,0,0).$
Now, the matrix representation A of the linear map with respect to the canonical basis is given by $A = [f(e_1) \hspace{0.5cm}f(e_2) \hspace{0.5cm} f(e_3)]$.
Note that $3$ is eigevalue of $(1,0,0)$, we have $f(1,0,0) = (3,0,0).$ This is the first column.
So, we have:
$f(e_1) = f(0\cdot (1,1,0) + 0\cdot(0,0,1) + 1(1,0,0)) = 0\cdot f(1,1,0) + 0\cdot f(0,0,1) + 1f(1,0,0) = f(1,0,0) = (3,0,0).$
$f(e_2) = f(1 \cdot (1,1,0) + 0 \cdot(0,1,1)+ (-1)(1,0,0)) = 1 \cdot f(1,1,0) + 0 \cdot f(0,1,1)+ (-1) f(1,0,0)= 1 \cdot (1,0,0) + (-1)(3,0,0) = (-2,0,0).$
$f(e_3) = f((-1)(1,1,0) + 1 \cdot (0,1,1) + 1 \cdot(1,0,0)) = (-1) f(1,1,0) + 1 \cdot f(0,1,1) + 1 \cdot f(1,0,0) = (-1) \cdot(1,0,0) + 1 \cdot (1,1,0) + 1 \cdot (3,0,0) = (3,1,0).$
So $A$ is given by:
$$A = \left(\begin{array}{cc}
3 & -2 & 3\\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $\sum_{M=k-1}^{2 k-1}\left(q C_{M}^{k-1} q^{k-1} p^{M-(k-1)}+p C_{M}^{k} p^{k} q^{M-k}\right)=1$ where $q+p = 1$ How to prove $\sum_{M=k-1}^{2 k-1}\left(q C_{M}^{k-1} q^{k-1} p^{M-(k-1)}+p C_{M}^{k} p^{k} q^{M-k}\right)=1$ ,where $q+p = 1$. And here $C_{m}^{n}=\frac{m !}{n !(m-n) !} $ for $m \geq n$, otherwise $0$. I've tried to expand the sum and considered to use the Taylor series in two variables. But I couldn't get the right answer. Am I thinking in a wrong way?
| Here we seek to prove that
$$\sum_{M=k-1}^{2k-1}
\left({M\choose k-1} (1-x)^k x^{M-(k-1)}
+ {M\choose k} x^{k+1} (1-x)^{M-k}\right) = 1.$$
The LHS is
$$\sum_{q=0}^n {q+n-1\choose n-1} (1-x)^n x^q
+ \sum_{q=1}^n {q+n-1\choose n} x^{n+1} (1-x)^{q-1}
\\ = (1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q
+ x^{n+1} \sum_{q=1}^n {q+n-1\choose n} (1-x)^{q-1}.$$
We have for the first sum
$$ [z^n] \frac{1}{1-z} \sum_{q\ge 0} {q+n-1\choose n-1} x^q z^q
= [z^n] \frac{1}{1-z} \frac{1}{(1-xz)^n}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-z} \frac{1}{(1-xz)^n}.$$
Residues sum to zero and fortunately the residue at infinity is zero
by inspection. Therefore we require the residues at one and at
$z=1/x.$ We start with
$$(-1)^{n+1} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(xz-1)^n}
\\ = \frac{(-1)^{n+1}}{x^n} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-1/x)^n}.$$
The residue at one is
$$\frac{(-1)^{n+1}}{x^n} \frac{1}{(1-1/x)^n}
= \frac{(-1)^{n+1}}{(x-1)^n}.$$
Multiply by the factor in front to get a contribution of $-1.$
For the residue at $z=1/x$ we require
$$\frac{1}{(n-1)!}
\left(\frac{1}{z^{n+1}} \frac{1}{z-1}\right)^{(n-1)}
\\ = \frac{1}{(n-1)!}
\sum_{q=0}^{n-1} {n-1\choose q}
\frac{(-1)^q (n+q)!}{n! z^{n+1+q}}
\frac{(-1)^{n-1-q} (n-1-q)!}{(z-1)^{n-q}}
\\ = (-1)^{n-1} \sum_{q=0}^{n-1} {n+q\choose n}
\frac{1}{z^{n+1+q} (z-1)^{n-q}}.$$
The residue is (evaluate at $z=1/x$)
$$\frac{(-1)^{n+1}}{x^n} (-1)^{n-1}
\sum_{q=0}^{n-1} {n+q\choose n}
x^{n+1+q} \frac{1}{(1/x-1)^{n-q}}
\\ = \sum_{q=0}^{n-1} {n+q\choose n}
x^{q+1} \frac{x^{n-q}}{(1-x)^{n-q}}
\\ = x^{n+1} \sum_{q=0}^{n-1} {n+q\choose n}
\frac{(1-x)^q}{(1-x)^{n}}.$$
Multiply by the factor in front and shift the index to get a
contribution of
$$x^{n+1} \sum_{q=1}^{n} {n+q-1\choose n} (1-x)^{q-1}.$$
Using the fact that residues sum to zero we have shown that
$$(1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q -1
+ x^{n+1} \sum_{q=1}^n {q+n-1\choose n} (1-x)^{q-1} = 0.$$
which is the claim.
Remark. This identity turns out to be an identity by Gosper
which was proved at the following MSE
link. To see this
rewrite as
$$(1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q
+ x^{n+1} \sum_{q=0}^{n-1} {q+n\choose n} (1-x)^{q} = 1.$$
The original is
$$\sum_{q=0}^{m-1} {n-1+q\choose q} x^n (1-x)^q
+ \sum_{q=0}^{n-1} {m-1+q\choose q} x^q (1-x)^m = 1.$$
Now replace $m$ by $n$ and $n$ by $n+1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $ \prod\limits_{k=1}^{n}{\left(1+\frac{1}{k^{3}}\right)}<\mathrm{e} $ How would you prove that $$ \left(\forall n\in\mathbb{N}\right),\ \prod_{k=1}^{n}{\left(1+\frac{1}{k^{3}}\right)}<\mathrm{e} $$
Wolfram|Alpha says its limit would be exactly $ \frac{\cosh{\left(\frac{\sqrt{3}\pi}{2}\right)}}{\pi} $, but I think it might be nicer to solve the problem without trying to compute the infinite product. Any suggestions ?
| You can also use the fact that
$$\ln(1+x)\leq x-\frac{x^2}{3}, x\in\left[0,\frac{1}{2}\right]$$
to conclude
$$\prod\limits_{k=1}^n\left(1+\frac{1}{k^3}\right)\leq2\prod\limits_{k=2}^n e^{\frac{1}{k^3}-\frac{1}{3k^6}}=
2\cdot e^{\sum\limits_{k=2}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)}$$
but
$$\sum\limits_{k=2}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)=
\frac{23}{192}+\sum\limits_{k=3}^n\left(\frac{1}{k^3}-\frac{1}{3k^6}\right)\leq \\
\frac{23}{192}+\int\limits_{2}^{n}\left(\frac{1}{x^3}-\frac{1}{3x^6}\right)dx=
\frac{23}{192}+\frac{59}{480}-\frac{15 n^3-2}{30 n^5}$$
thus
$$\prod\limits_{k=1}^n\left(1+\frac{1}{k^3}\right)<2\cdot e^{\frac{23}{192}+\frac{59}{480}}=\color{red}{2}\cdot e^{\frac{233}{960}}< e^{\color{red}{\frac{3}{4}}+\frac{233}{960}}=e^{\frac{953}{960}}<e$$
| {
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If $x+y^3,x^2+y^2,x^3+y$ are all integers, are $x,y$ both integers?
Let $x,y$ be both real numbers. If $x+y^3,x^2+y^2,x^3+y$ are integers, are $x,y$ both integers?
This question begins with two real numbers while usual number theory tricks are based on the precondition that the variables are integers. Showing $x,y$ are algebraic numbers is easy by observing $x+((x^3+y)−x^3)^3$ is an integer, but how can algebraic numbers help?
| Here's a hint (or at least some useful steps that are too long for a comment): Let $t=xy$. Note that
$$\mathbb Z\ni (x^3+y)(x+y^3)-(x^2+y^2)^2=t^3+t-2t^2=t(t-1)^2$$
and
\begin{align*}
\mathbb Z
&\ni \big[(x^3+y)(x^2+y^2)-(x+y^3)\big]\big[(x+y^3)(x^2+y^2)-(x^3+y)\big]\\
&=t(t-1)^2+(x^2+y^2)^2t(t^2+t-1).
\end{align*}
Therefore $t^3+t^2-t$ is rational (since $x^2+y^2\neq 0$ or we are done). Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence. I need to show that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence, where $x_1 = 2$ and n = 1,2,3,....
I tried to show this with induction, where since $x_1= 2$ then $x_2 = 1.5$, hence the base case satisfies.
Assuming that $x_n \geq x_{n+1}$ , I need to prove that $x_{n+1} \geq x_{n+2}$.
I started with the statement $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$
Also, $x_n \geq x_{n+1}$ then $\frac{1}{2}(x_n + \frac{2}{x_{n+1}}) \geq \frac{1}{2}(x_{n+1} + \frac{2}{x_{n+1}})$
$= \frac{1}{2}(x_n + \frac{2}{x_{n+1}} - \frac{2}{x_n} + \frac{2}{x_n}) \geq x_{n+2} $
$= x_{n+1} - \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq x_{n+2} $
since $- \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq 0$
then I cannot say that $ x_{n+1} \geq x_{n+2} $
How do I proceed forward?
Edit: Since $x_1 = 2$, hence $1 \leq x_1 \leq 2$ holds for base case.
If I assume that $1 \leq x_n \leq 2 $ and show that $1 \leq x_{n+1} \leq 2 $ then by induction we can say that $1 \leq x_n \leq 2 \:\forall n$
I have shown this as:
$x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}, \: now \: since\:1 \leq x_n\leq2\:\Rightarrow x_{n+1} \leq \frac{1}{2}(2 + \frac{2}{1}) = 2$
Also by AM GM inequality, $x_n \geq \sqrt{2} \: \: \forall n$
Now all that I have to show is that $x_n \geq x_{n+1} \: \forall n?$
| Hint:
Actually, you only need to prove that all $x_n$ live in an interval $I$ (to be determined) such that
$$f(x)=\frac12\Bigl(x+\frac2x\Bigr)<x \quad\text{on }I.$$
So you have to determine the variations of the function $f$ and show by induction that $x_n\in I$ for all $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Domain of $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$ $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$
I'm stuck in this one because this can be rewritten as:
$$f(x,y) = \ln(16-x^2-y^2) + \ln(x^2+y^2-4)$$
Yet, the domain of the given function is $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\} \lor \{(16-x^2-y^2<0)\land (x^2+y^2-4<0)\}$. But the domain of the rewritten on is only $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\}$. Which one is the correct one and why does this happen?
| Note that $\{(x,y):16-x^2+y^2<0\}\cap\{(x,y):x^2+y^2-4<0\}=\emptyset$, and hence the two domains are the same in different expressions!
| {
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Limit of $\lim_{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with a > 0 and b > 0 Question:
$\lim_{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$ with $a>0$ and $b>0$.$a$ and $b$ are constant.
| Now that I have solved the problem, I think I should post the answers and thank everyone for their help.
$\lim _{x\to -\infty }\left(\frac{a^x+b^x}{2}\right)^{\frac{1}{x}}$
$=\lim _{x\to -\infty }\left(\frac{1}{2}\right)^{\frac{1}{x}}\cdot \left(a^x+b^x\right)^{\frac{1}{x}}$
$=\lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}$
$\lim _{x\to -\infty }\left(max\left(a^x,b^x\right)\right)^{\frac{1}{x}}\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:-\infty \:}\left(2\cdot max\left(a^x,b^x\right)\right)^{\frac{1}{x}}$
when $x\to -\infty$
,$max\left(a^x,b^x\right)=min\left(a,b\right)^x$
So
$\lim _{x\to -\infty }\left(min\left(a,b\right)^x\right)^{\frac{1}{x}}\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:-\infty \:}\left(2\cdot min\left(a,b\right)^x\right)^{\frac{1}{x}}\:$
$\lim _{x\to -\infty }min\left(a,b\right)\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:-\infty \:}2^{\frac{1}{x}}\cdot min\left(a,b\right)$
Conclusion:
$min\left(a,b\right)\le \lim _{x\to -\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le min\left(a,b\right)$
By the way, I found the following method is also true when $x\to +\infty$.
$\lim _{x\to +\infty }\left(max\left(a^x,b^x\right)\right)^{\frac{1}{x}}\le \lim _{x\to +\infty }\left(a^x+b^x\right)^{\frac{1}{x}}\le \lim \:_{x\to \:+\infty \:}\left(2\cdot max\left(a^x,b^x\right)\right)^{\frac{1}{x}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$.
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.
| Since by Vieta's formulas the sum of the roots is $-2$, there is not much to do here: given that two roots are $\alpha$ and $\alpha^2+2\alpha-4$, the third one has to be $-\alpha^2-3\alpha+2$.
The interesting question, however, is: how do we realize that there are two
roots $\alpha,\beta$ fulfilling $\beta=\alpha^2+2\alpha-4$ ?
Well, the discriminant of the polynomial is $361=19^2$, so all the roots are real and the Galois group over $\mathbb{Q}$ is not $S_3$ but something simpler. By considering the depressed cubic we have
$$ \frac{27}{38\sqrt{19}}\,\underbrace{p\left(\frac{2}{3}(x\sqrt{19}-1)\right)}_{q(x)}=4x^3-3x+\frac{7}{2\sqrt{19}} $$
so by trigonometry we have that a root is given by
$$ \zeta = -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos\Big(\underbrace{\frac{1}{3}\arccos\left(\frac{-7}{2\sqrt{19}}\right)}_{\theta}\Big) $$
and the other roots are given by
$$ -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos(\theta+2\pi/3)\qquad\text{and}\qquad -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos(\theta+4\pi/3) $$
i.e. by
$$ -\frac{2}{3}-\frac{1}{3}\sqrt{19}\cos(\theta)-\frac{1}{\sqrt{3}}\sqrt{19}\sin(\theta)\qquad\text{and}\qquad -\frac{2}{3}-\frac{1}{3}\sqrt{19}\cos(\theta)+\frac{1}{\sqrt{3}}\sqrt{19}\sin(\theta) $$
which are clearly related (to each other, and to $\zeta$) via the Pythagorean theorem $\sin^2\theta+\cos^2\theta=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{align}
I feel I exhausted all identities/properties of binomials without success. Mathematica says it is true, but how to show it?
| For the second one we get
$$\sum_{k=0}^{n-1} (-1)^{n-k-1}
\frac{k+1}{n} {n-1\choose k} {n\choose k+1}
x^k (1-x)^{n-1-k}
\\ = (-1)^{n-1} \sum_{k=0}^{n-1} {n-1\choose k} {n+k-1\choose k}
(-1)^k x^k.$$
The LHS is
$$\sum_{k=0}^{n-1} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
x^k (1-x)^{n-1-k}.$$
The coefficient on $[x^q]$ where $0\le q\le n-1$ of the LHS is
$$\sum_{k=0}^{q} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
[x^{q-k}] (1-x)^{n-1-k}
\\ = \sum_{k=0}^{q} (-1)^{n-k-1}
{n-1\choose k} {n-1\choose k}
(-1)^{q-k} {n-1-k\choose q-k}.$$
Using the RHS we therefore must show that
$${n-1\choose q} {n+q-1\choose q}
= \sum_{k=0}^{q}
{n-1\choose k} {n-1\choose k}
{n-1-k\choose q-k}.$$
Now note that
$${n-1\choose k} {n-1-k\choose q-k} =
\frac{(n-1)!}{k! \times (q-k)! \times (n-1-q)!}
= {n-1\choose q} {q\choose k}.$$
This reduces the claim to
$${n+q-1\choose q} =
\sum_{k=0}^q {n-1\choose k} {q\choose k}$$
which is
$$\sum_{k=0}^q {n-1\choose k} {q\choose q-k}
= [z^q] (1+z)^q \sum_{k=0}^q {n-1\choose k} z^k.$$
Here the coefficient extractor enforces the range and we get
$$[z^q] (1+z)^q \sum_{k\ge 0} {n-1\choose k} z^k
= [z^q] (1+z)^q (1+z)^{n-1} \\ = [z^q] (1+z)^{n+q-1}
= {n+q-1\choose q}$$
as required. This last step can also be done by Vandermonde.
| {
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2-split of $n$ is $\left\{ \lfloor \frac{n}{2} \rfloor,\lceil \frac{n}{2} \rceil \right\}$. What about 3, 4, ...? Clarification: $k$-split of $n$ is an ordered integer sequence $\left\{ a_1,\cdots,a_k \right\}\quad \text{s.t.}$
*
*$0\le a_1\le\cdots\le a_k$
*$a_1+\cdots+a_k=n$
*${\left(a_k-a_1\right)}$ is minimized.
I know that
$$
n = \lfloor \frac{n}{3} \rfloor + \lceil \frac{2n}{3} \rceil,
$$
so I guess 3-split of $n$ is
$$
n = \lfloor \frac{n}{3} \rfloor + \lfloor \frac{\lceil \frac{2n}{3} \rceil}{2} \rfloor+\lceil \frac{\lceil \frac{2n}{3} \rceil}{2} \rceil ?
$$
If so, can this be simplified?
| It is clear your question is satisfied by the following partition:
$$x = \sum_{k=0}^{n-1}\bigg\lfloor\frac{x+k}{n}\bigg\rfloor$$
| {
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Proving $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. I've been asked to prove that $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. Proving that $1+x+\frac{x^2}{2!}>0$ is simple as you just need to consider the discriminant and show that it's less that 0. Showing $1+x+\frac{x^2}{2!} + \frac{x^3}{3!}=0$ only has 1 solution is to just consider the derivative and using the previous proof that $1+x+\frac{x^2}{2!}>0$ has no zero (so it is always increasing).
With this information, how would you be able to prove $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$? I've tried to show that the global minimum (the lone solution to $1+x+\frac{x^2}{2!} + \frac{x^3}{3!}=0$) gives a positive answer, however I have not been successful. How can I show this?
| As you say, $1+x+x^2/2+x^3/6=0$ has a unique solution, (with $x<0$). For this $x$,
$$1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}=\frac{x^4}{4!}>0.$$
| {
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In $\triangle ABC$ with integer sides, $BC$ is the average of the other sides, and $\cos C=AB/AC$. Find the smallest possible area.
In $\triangle ABC$, the length of side $\overline{BC}$ is equal to the average of the other two sides. Also, $\cos C = \frac{AB}{AC}$. Given that all the side lengths are integers, find the smallest possible area of $\triangle ABC$.
I set an altitude, BD, then tried doing $CD=\frac{AD \cdot BC}{AC}$, then solving. But this does not work. What should I do?
| Let $BC=n$, $BA=n-k$ and $AC=n+k$. Then
$$\cos C = \frac{n-k}{n+k}= \frac{ (n+k)^2+n^2 -(n-k)^2}{2n(n+k)}$$
which yields $n=6k$. Thus, the smallest triangle has the integer side lengths 5, 6 and 7, and the corresponding area is $6\sqrt6$.
| {
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How to compute $\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$ I have first written $x^4-2x^2+2=(x^2-1)^2+1$ then I choose $t=x^2-1$. From this I got $$\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$$
From this how to proceed?
| Let
\begin{eqnarray}
I(a)=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx
\end{eqnarray}
Then,
\begin{eqnarray}
I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)}dx\\
&=&\frac{1}{1+a}\int_0^\infty\left(\frac{1}{1+x^2}+\frac{x^2-1}{1+a x^4}\right)dx
=\frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)\\
\end{eqnarray}
Thus,
\begin{align}
&\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt \overset{x=\sqrt t}=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx\\
= & I(1)=\int_0^1 I’(\alpha)d\alpha
= \int_0^1 \frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)d\alpha \\
= & \frac\pi2(\ln2 + 2\coth^{-1} \sqrt2 )
\end{align}
| {
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How to solve this first order ODE right Here is ODE of first order:
$$ y'=- \frac{x^2}{y^3} $$
with $y(0)=1 , y(0)=-1 $
Can my method be the separation method? If I use it, it doesn't work out:
$ \frac{dy}{dx}= -x^2 \frac{1}{y^3} $
$ \leftrightarrow y^3 dy = -x^2 dx $
By integration I get:
$$ \int y^3 dx = \int -x^2 dx $$
$$ \leftrightarrow \frac{1}{4} y^4+c = - \frac{1}{3} x^3+c $$
$$ \leftrightarrow y= \mp \sqrt[4]{ - \frac{4}{3} x^3 +c } $$
Where is my mistake?
| You are correct. If $y\not=0$ then
$$y'=- \frac{x^2}{y^3}\leftrightarrow \int y^3 dy = \int -x^2 dx \leftrightarrow \frac{1}{4} y^4= - \frac{1}{3} x^3+c \leftrightarrow y= \pm \sqrt[4]{ - \frac{4}{3} x^3 +c }. $$
Now for $y(0)=1$, we get
$$ y(x)= \color{blue}{+}\sqrt[4]{1 - \frac{4}{3} x^3},$$
whereas for $y(0)=-1$ we find
$$ y(x)= \color{blue}{-}\sqrt[4]{1 - \frac{4}{3} x^3}.$$
Both solutions, which are symmetric with respect to the line $y=0$, are defined in the half-line $(-\infty,(3/4)^{1/3})$.
| {
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Prove that if $a^{2} + ab + b^{2} | ab(a+b)$ then $\left | a-b \right | > \sqrt[3]{ab}$ This is one of my university homework's problems this week:
Suppose that a and b are two positive integer and $a^{2} + ab + b^{2} | ab(a+b)$ then prove that $\left | a-b \right | > \sqrt[3]{ab}$
I tried to prove that $a^{2} + ab + b^{2} | (a-b)^{3}$ so I can say $a^{2} + ab + b^{2} \leq (a-b)^{3} \Rightarrow ab < (a-b)^{3}$ , but I'm stuck.
sorry for my poor English.
| We need $a \neq b$.
WLOG assume $a>b$. Let $d=(a,b)$. Then there exists two coprime positive integers $x> y$ such that $a=dx,b=dy$.
$a^{2} + ab + b^{2} | ab(a+b)$ equivalent to $x^2+xy+y^2 \mid dxy(x+y)$
Since $(x,y)=1$, $x^2+xy+y^2$ and $xy(x+y)$ are coprime. So we must have $x^2+y^2+xy \mid d$. Then $d\geq x^2+y^2+xy>xy$.
$(a-b)^3 > ab \iff d(x-y)^3>xy$. Since $x > y$ then $x-y \geq 1$, combine with above inequality gives the result.
| {
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Schur inequality Show that for all positive real numbers $a$, $b$ and $c$ such that $abc=1$, the inequality $a+b+c+2a^4+2b^4+2c^4\ge \dfrac{3}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right)$ is true
I tried using Schur inequality which gave me $a+b+c+a^4+b^4+c^4\ge a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)$. Then I wanted to show that $a^4+b^4+c^4\ge\dfrac{1}{2}\left(a^2\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b^2\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c^2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right)$ but I'm not sure how to go about that
| We need to prove that
$$\sum_{cyc}(2a^4+a^2bc)\geq\frac{3}{2}\sum_{cyc}a^3(b+c)$$ or
$$\sum_{cyc}(4a^4-3a^3b-3a^3c+2a^2bc)\geq0$$ or
$$3\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(a^4-a^2bc)\geq0,$$ which is true by Schur and Muirhead.
| {
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Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\right]\left[ \frac{d^r}{dx^r}(1-x)^{-\frac{1}{2}} \right].$$
Since we know that $ \frac{d^{2}}{dx^2}(1+x) = 0 $, we can simplify the above sum to the following:
$$ 0+\ldots+\binom{100}{98}\left[\frac{d^{2}}{dx^2}(1+x)\right] \left[ \frac{d^{98}}{dx^{98}}(1-x)^{-\frac{1}{2}} \right]
+\binom{100}{99}\left[\frac{d}{dx}(1+x)\right] \left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] +\binom{100}{100}(1+x) \left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right], $$
which then gives us,
$$ 100\left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] + (1+x)\left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right].$$
This is where I am stucked. I'm not sure resolve those differentials.
| The derivative of $\sqrt{1-x}$ follows a pattern
$$f'(x) = \frac{1}{2}(1-x)^{-\frac{3}{2}}$$
$$f''(x) = \frac{1\cdot 3}{2\cdot 2}(1-x)^{-\frac{5}{2}}$$
$$\vdots$$
$$f^{(n)}(x) = \frac{(2n)!}{4^n n!}(1-x)^{-\frac{(2n+1)}{2}}$$
a product of odds divided by a power of $2$ which makes your derivative
$$\frac{100\cdot 198!}{4^{99}\cdot 99!}(1-x)^{-\frac{199}{2}} + \frac{200!}{4^{100}\cdot 100!}(1+x)(1-x)^{-\frac{201}{2}}$$
| {
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show that limit $\lim_{n\to+\infty}f_{n}(x)=x+3$,if $f_{n+1}(x)=\sqrt{6(1+x)+f_{n}(x^2)}$ let $x$ is give postive real number,if $f_{0}(x)=0,0<x\le\dfrac{1}{2}$, and such $$f_{n+1}(x)=\sqrt{6(1+x)+f_{n}(x^2)}$$
show that
$$\lim_{n\to+\infty}f_{n}(x)=x+3$$
This problem is from AMM 11967(2017),this solution
| Let $g_n(x) = f_n(x) - x - 3$, then $g_0(x) = -(x + 3)$ and it suffices to prove that
$$
\lim_{n\to \infty} |g_n(x)| = 0.
$$
Note that since
$$
\begin{align}
g_{n+1}(x) = f_{n+1}(x) - (x + 3) &= \sqrt{6(1+x) + x^2 + 3 + g_n(x^2)} - (x + 3)\\
&=\frac{g_n(x^2)}{\sqrt{(x+3)^2 + g_n(x^2)} + (x + 3)},
\end{align}
$$
we have
$$
|g_{n+1}(x)| < \frac13|g_{n}(x^2)|.
$$
Therefore,
$$
0 \leq |g_n(x)| < \frac13 |g_{n-1}(x^2)| < \dots < \frac{1}{3^n} |g_0(x^{2^n})| = \frac{|x^{2^n} + 3|}{3^n},
$$
and the result immediately follows from the squeeze theorem.
| {
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How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose that $x,y,z$ are real numbers whose moduli are not equal to $1$.)
I discovered this problem from chapter 3 of Prelude to Mathematics by W. W. Sawyer. Sawyer thought that this problem arose from the sine law: let $a,b,c$ be respectively the lengths of the edges opposite to three vertices $A,B,C$ of a triangle. Define $x=\cos A$ and define $y,z$ analogously. Now equalities $(1)-(3)$ simply relate $a,b$ and $c$ to each other by the cosines of the angles and $(4)$ is just a rewrite of the sine law
$$
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.
$$
However, the algebraic version $(4)$ looks more general. For example, it does not state that $a,b,c$ must be positive or that they must satisfy the triangle inequality.
Sawyer wrote that this isn't a hard problem, but he didn't provide any solution. I can prove $(4)$ using linear algebra. Suppose that $(a,b,c)\ne(0,0,0)$ (otherwise $(4)$ is obvious). Rewrite $(1)-(3)$ in the form of $M\mathbf a=0$:
$$\begin{bmatrix}-1&z&y\\ z&-1&x\\ y&x&-1\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=0.$$
Since $x^2,y^2,z^2\ne1$, $M$ has rank $2$ and $D=\operatorname{adj}(M)$ has rank $1$. Hence all columns of $D$ are parallel to $(a,b,c)^T$ and $\frac{d_{11}}{d_{21}}=\frac{d_{12}}{d_{22}}=\frac{a}{b}$. Since $M$ is symmetric, $D$ is symmetric too. Therefore $\frac{1-x^2}{1-y^2}=\frac{d_{11}}{d_{22}}=\frac{d_{11}d_{12}}{d_{21}d_{22}}=\frac{a^2}{b^2}$, i.e. $\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.
As this problem comes from Hall and Knight's book, I think there should be a more elementary solution. Any ideas?
| It turns out that I solved the equations for the wrong variables. If I rewrite $(1)-(3)$ as
$$\begin{bmatrix}0&c&b\\ c&0&a\\ b&a&0\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}=\begin{bmatrix}a\\ b\\ c\end{bmatrix}$$
and solve for $x,y,z$ instead, I will get the law of cosines, i.e.
$$x=\frac{b^2+c^2-a^2}{2bc}$$
etc.. Therefore $$\frac{a^2}{1-x^2}=\frac{4a^2b^2c^2}{2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)}.$$
As Roman Odaisky has pointed out, this expression can be rewritten as $\frac{a^2b^2c^2}{4s(s-a)(s-b)(s-c)}$, where $s=\frac12(a+b+c)$. By symmetry, $\frac{b^2}{1-y^2}$ and $\frac{c^2}{1-z^2}$ are also equal to the same expression. Geometrically (and according to Heron's formula), this means the common ratio in the law of sines is equal to $\frac{abc}{2T}$ where $T$ is the area of the triangle.
| {
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Proving a general reasult in polynomials. If $f(x) = (x-a)(x-b)(x-c)(x-d) - 1$ where $a, b, c, d$ are distinct integers. Prove that $f(x)$ can't be factorized into integer polynomial with $deg ≥1$
In the above question I proved for three degree polynomial using contradiction. But couldn't use the same for the even degree polynomial.If anyone out there could help me would be of great help.
| By rational root test or Gauss’s Lemma (see the proof there for the lemma over the integers), if $f(x)$ has linear factor over $\mathbb Q$, then $f(x)$ has a factor of the form $x-t,t\in {\mathbb Z}$. But $f(t)=0$ implies $$(t-a)(t-b)(t-c)(t-d)=1,$$ which is impossible, as $t-a,t-b,t-c,t-d$ are nonzero distinct integers.
If $f(x)$ is reducible but of no linear factor, then by Gauss’s Lemma, $$f(x)=(x-a)(x-b)(x-c)(x-d)-1$$ $$=(x^2+Ax+B)(x^2+Cx+D),$$ where $A,B,C,D\in {\mathbb Z}$. It follows that $$f(a)=(a^2+Aa+B)(a^2+Ca+D)=-1$$
$$\Rightarrow a^2+Aa+B=\pm 1.$$ Similarly $$u^2+Au+B=\pm 1,~{\rm for~}u=b,c,d.$$
By the pigeonhole principle (and permuting $a,b,c,d$, $(A,B)$,$(C,D)$ if necessary), one may assume that $$\left\{\begin{array}{c}a^2+Aa+B=1\\ b^2+Ab+B=1\end{array}\right.\Leftrightarrow \left\{\begin{array}{c}a^2+Ca+D=-1\\ b^2+Cb+D=-1\end{array}\right.,$$
where $A,B,C,D$ can be solved as $$A=-(a+b),B=1+ab,C=-(a+b),D=-1+ab.$$ It follows that $$(x-a)(x-b)(x-c)(x-d)-1$$ $$=(x^2-(a+b)x+(1+ab))(x^2-(a+b)x+(-1+ab)). \quad (1)$$ By comparison of coefficient of $x^3$ and the constant term in (1), one has $$a+b=c+d,abcd-1=a^2b^2-1.\quad (2)$$
Case 1: $ab\neq 0\Rightarrow a+b=c+d,ab=cd,$ which shows that $\{a,b\}=\{c,d\},$ a contradiction to the condition that $a,b,c,d$ are distinct.
Case 2: $ab=0$. By symmetry, it suffices to consider the case $a=0$ (so $bcd\neq 0$). Then (2) gives that $b=c+d$, hence from (1), one has $$x(x-(c+d))(x-c)(x-d)-1$$
$$=(x^2-(c+d)x+1)(x^2-(c+d)x-1).\quad (3)$$ By comparison of coefficient of $x$ in (3), one has $$-cd(c+d)=0,$$ which shows that $c+d=0$ (since $cd\neq 0$). This implies that $$0=c+d=a+b=b,$$ contradiction again.
One concludes that if $a,b,c,d$ are distinct integers, then $$f(x)=(x-a)(x-b)(x-c)(x-d)-1$$ is irreducible over $\mathbb Q.$
| {
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Evaluate $\int_0^{\infty} \frac{ dx }{(x^4+c)(x^2+1) } $ Consider $$\int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } $$
where $c \in \mathbb{R}$. Now, it is possible to evaluate this using complex analysis, but I was wondering if there is way to compute this integral without complex analysis.
I was thinking on partial fractions, but there is no easy form to decompose it too.
Here is a way I was thinking: Write $1 = \dfrac{c+x^4 - x^4 }{c}$ to obtain
$$ \frac{1}{c} \int_0^{\infty} \frac{dx}{x^2+1} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)} = \dfrac{ \pi }{2c} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)}$$
I dont know if this is "progress", what do you think?
| $$\dfrac{ 1 }{(x^4+c)(x^2+1) }=\frac{1}{\left(x^2+1\right) \left(x^2-i \sqrt{c}\right) \left(x^2+i \sqrt{c}\right)}$$ Using partial fractions, it is
$$-\frac{1}{2 \left(\sqrt{c}-i\right) \sqrt{c} \left(x^2-i \sqrt{c}\right)}-\frac{1}{2
\left(\sqrt{c}+i\right) \sqrt{c} \left(x^2+i \sqrt{c}\right)}-\frac{i}{2
\left(\sqrt{c}-i\right) \left(\sqrt{c}+i\right) (x-i)}+\frac{i}{2
\left(\sqrt{c}-i\right) \left(\sqrt{c}+i\right) (x+i)}$$ which is not so bad.
| {
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Prove that the value of $\Delta$ is an integer for the given determinant
$\Delta=\begin{vmatrix} \sqrt 6& 2i& 3+\sqrt 6 \\ \sqrt{12}&\sqrt 3 +\sqrt8i &3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &\sqrt 2+ \sqrt {12}i &\sqrt {27}+2i \end{vmatrix}$
taking $\sqrt 6$ out from the first column and performing the following operations
$$R_2\rightarrow R_2 -\sqrt 2 R_1$$
$$R_3 \rightarrow R_3-\sqrt 3 R_1$$
$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$
What should I do next?
| From here
$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$
Pull out a factor of $\sqrt{3}$ from the second row and a factor of $\sqrt{2}$ from the third to get
$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix}$$
Now you can compute it easily by expanding along the first row or by subtracting the second row from the third. Lets subtract.
$$6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 1 & \sqrt{2}i-3 \end {vmatrix} = 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0& 1 &\sqrt 2i-2 \\\ 0& 0 & -1 \end {vmatrix}$$
The determinant is just the product of the diagonal entries, so $6 \cdot 1 \cdot 1 \cdot -1 = -6$.
| {
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Evaluating $\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$ I'm in a little struggle with this limit, can anyone help me, please?
$$\lim_{x \to \frac{\pi}{6}}{(1-2\sin(x))}^{\tan(\frac{\pi}{6}-x)}$$
I tried to use the logarithm to then use L'Hospital's rule but I got stuck here:
$\ln(L)=\lim_{x \to \frac{\pi}{6}}{[\tan(\frac{\pi}{6}-x)\ln(1-2\sin(x))]}$
Thank you!
| Let $f(x) = (1-2\sin x)^{\tan(\frac{\pi}{6}-x)}$, then $f(x) = e^{g(x)}$ with $g(x) = \tan(\frac{\pi}{6}-x) \log (1-2\sin x)$.
$$\begin{align}
\lim\limits_{x \to \frac{\pi}{6}^- } g(x)
&= \lim\limits_{x \to \frac{\pi}{6}^- } \frac{\tan\left(\frac{\pi}{6}-x\right)}{\frac{\pi}{6}-x} \left(\frac{\pi}{6}-x\right)\log \left(1-2\sin x\right)
\\
&\overset{(1)}{=} \lim\limits_{x \to \frac{\pi}{6}^- } \left(\frac{\pi}{6}-x\right)\log \left(1-2\sin x\right)
\\
&=\lim\limits_{x \to \frac{\pi}{6}^- } \frac{ \log (1-2\sin x)}{\frac{1}{\frac{\pi}{6}-x}}
\\
&\overset{\mathrm{H}}{=} \lim\limits_{x \to \frac{\pi}{6}^-} (-2\cos x)\frac{\left(\frac{\pi}{6}-x\right)^2}{1-2\sin x}
\\
&= -\sqrt{3}\lim\limits_{x \to \frac{\pi}{6}^-} \frac{\left(\frac{\pi}{6}-x\right)^2}{1-2\sin x}
\\
&\overset{\mathrm{H}}{=}
-\sqrt{3}\lim\limits_{x \to \frac{\pi}{6}^-}\frac{-2\left(\frac{\pi}{6}-x\right)}{-2\cos x }
\\
&= 0
\end{align}$$
where in $(1)$ I have used $\lim_{y\to0} \frac{\tan y}{y} = 1$ and $H$ denotes the usage of L'Hôpital's rule.
Hence, we conclude that
$$\lim\limits_{x \to \frac{\pi}{6}^-} f(x) = e^0 = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$ For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.
Another proof by $pqr$ method$:$
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4\,{p}^{4}q+8\,{p}^{3}r+27\,{r}^{2} \geqq 0$$
Or$:$ $${\frac { \left( {p}^{4}-5\,{p}^{2}q+6\,pr+4\,{q}^{2} \right)
\left( 7\,{p}^{4}+45\,{p}^{2}q+54\,pr-36\,{q}^{2} \right) }{12{p}^{2}}}
+\,{\frac { \left( {p}^{2}-3\,q \right) \left( 5\,{p}^{2}-3\,q
\right) \left( {p}^{2}-4\,q \right) ^{2}}{12{p}^{2}}} \geqq 0$$
Which is obvious because $p^2 \geqq 3q,\, p^4 -5p^2 q+6pr+4q^2 \geqq 0 \,(\text{Schur degree 4})$
I hope for another proof (without $uvw$!). Thanks for a real lot!
PS$:$ You can get $pqr$'s form more faster by using Maple$,$ see here.
| If $a+b-c<0$ and $a+c-b<0$, we obtain $a<0,$ which is a contradiction.
Thus, it's enough to prove our inequality for $a+b-c>0$, $a+c-b>0$ and $b+c-a>0.$
Now, let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Thus, we need to prove that $$27(x+y)^2(x+z)^2(y+z)^2\geq64xyz(x+y+z)^3,$$ which follows from $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz).$$
Can you end it now?
| {
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"source": "stackexchange",
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Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$
My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we get $BD=BC\cdot \alpha
, \enspace CD=BC\cdot(1-\alpha)\tag{1}$ and also
$$\overrightarrow{AD}=(1-\alpha)\cdot\overrightarrow{AB}+\alpha\cdot \overrightarrow{AC}.$$
By squaring this relation we have that $$AD^2=AB^2(1-\alpha)+AC^2\alpha+BC^2(\alpha^2-\alpha). \tag{$2$}$$
By law of sines we also have $AB=2\sin C$, $AC=2\sin B$ and $BC=2\sin A$.
Now combining with $(1)$ and $(2)$ we may rewrite the desired inequality as follows:
$$((1-\alpha)\sin^2C+\alpha\sin^2B+(\alpha^2-\alpha)\sin^2A)\cdot\alpha^2(1-\alpha)^2\sin^4A\leq \dfrac{2^4}{27^2}.$$
This is where I got stuck. Maybe we could also use the fact that $\sin A=\sin (\pi -B-C)=-\sin(B+C)=-(\sin B\cos C+\sin C\cos B)$ to get rid of $\sin A$?
Thank you in advance!
|
Just some work, that was too big for a comment.
Well, when we have a $\triangle\text{ABC}$:
We know that:
$$
\begin{cases}
\angle\alpha+\angle\beta+\angle\gamma=\pi\\
\\
\frac{\text{a}}{\sin\left(\angle\alpha\right)}=\frac{\text{b}}{\sin\left(\angle\beta\right)}=\frac{\text{c}}{\sin\left(\angle\gamma\right)}\\
\\
\text{a}^2=\text{b}^2+\text{c}^2-2\text{b}\text{c}\cos\left(\angle\alpha\right)\\
\\
\text{b}^2=\text{a}^2+\text{c}^2-2\text{a}\text{c}\cos\left(\angle\beta\right)\\
\\
\text{c}^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(\angle\gamma\right)
\end{cases}\tag1
$$
We also know that the circumradius of that triangle is given by:
$$\text{R}=\frac{\text{a}\text{b}\text{c}}{\sqrt{\left(\text{a}+\text{b}+\text{c}\right)\left(\text{b}+\text{c}-\text{a}\right)\left(\text{a}+\text{b}-\text{c}\right)}}\tag2$$
So, when $\text{R}=1$ we know that:
$$\text{a}\text{b}\text{c}=\sqrt{\left(\text{a}+\text{b}+\text{c}\right)\left(\text{b}+\text{c}-\text{a}\right)\left(\text{a}+\text{b}-\text{c}\right)}\tag3$$
Which is the same as:
$$\text{a}^2\cdot\text{b}^2\cdot\text{c}^2=\left(\text{a}+\text{b}+\text{c}\right)\left(\text{b}+\text{c}-\text{a}\right)\left(\text{a}+\text{b}-\text{c}\right)\tag4$$
| {
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"source": "stackexchange",
"question_score": "2",
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$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula.
Any ideas, hints?
Edit: sorry, there was a mistake in the derivative, I corrected it.
| The product formula for sine says
$$
\frac{\sin(x)}{x}=\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)\tag1
$$
Thus, for $0\lt x\le\pi$
$$
\begin{align}
\frac{\sin(x)}x
&\le1-\frac{x^2}{\pi^2}\tag2\\[3pt]
&\le1-\frac{x^2}{(x+\pi)^2}\tag3\\
x-\sin(x)
&\ge\frac{x^3}{(x+\pi)^2}\tag4
\end{align}
$$
Explanation:
$(2)$: follows from $(1)$
$(3)$: $x\gt0$
$(4)$: subtract from $1$ (which reverses the inequality)
$\phantom{\text{(4):}}$ and multiply by $x$
For $x\gt\pi$,
$$
\begin{align}
(x-\sin(x))(x+\pi)^2
&\ge(x-1)(x+\pi)^2\tag5\\[3pt]
&=x^3+(2\pi-1)x^2+\left(\pi^2-2\pi\right)\!x-\pi^3\tag6\\[3pt]
&\ge x^3\tag7\\
x-\sin(x)
&\ge\frac{x^3}{(x+\pi)^2}\tag8
\end{align}
$$
Explanation:
$(5)$: $\sin(x)\lt1$
$(6)$: expand product
$(7)$: plug in $x\gt\pi$
$(8)$: divide by $(x+\pi)^2$
Inequalities $(4)$ and $(8)$ together show that for $x\gt0$,
$$
x-\sin(x)\ge\frac{x^3}{(x+\pi)^2}\tag9
$$
| {
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"question_score": "3",
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Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x^3+x^6+x^9+\dots$$
$$P_b(x) = 1 +x+x^2+x^3+\dots + x^9 = \frac{1-x^{10}}{1-x}$$
$$P_c(x) = 1+x^2 + x^4 + x^6 + \dots$$
I want to know if my $P_a(x)$ and $P_c(x)$ are stated correctly and how can they be simplified so that I get to $x^{15}$ in an easier way.
Big thanks for your help.
| You don't have to make polynomials and solve it by generating functions,
let $a=3k$, and $c=2j$ for $k,j \in \mathbb{N_0}$
$$3k + 2j + b = 15 \iff 3k+2j=15-b$$
This is a classic linear diophantine equation, and since $\gcd(2,3)=1$, it has solutions. So, all we need to do is find initial solutions to $k,j$ in terms of $b$. An obvious one is
$$j=b \implies3k=15-3b \iff k=5-b$$
So, we know from the properties of linear diophantine equation that
If $(x^*,y^*)$ is a solution to $ax+by=n$, then all integer solutions are in the form of
$$\left(x^*+m\frac{b}{\gcd(a,b)},y^*-m\frac{a}{\gcd(a,b)}\right)$$
for some integer $m$. Applying that to our equation, we get:
$$(k,j) \in \left\{((5-b)+2m,b-3m) \ |\ 0 \le b<10, \ \ b \in \mathbb{N_0},m \in \mathbb{Z}\right\}$$
and one more restriction: $k \ge0 \implies m \ge \frac{b-5}{2}$ and $j \ge0 \implies m \le \frac{b}{3}$
So to find all solutions, you choose $b$ from this set $\{0,1,2,3,4,5,6,7,8,9\}$ and put every integer
$$ \left\lfloor \frac{b-5}{2} \right\rfloor \le m \le\left\lfloor \frac{b}{3} \right\rfloor$$
and the resulting set would be
$$(a,b,c) \in \{(15-3b+6m,b,2b-6m)\}$$
| {
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Find the stronger inequality of $\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geq \frac{1}{\sum ab}$ For $a,b,c>0$ and $a+b+c=1.$ Prove$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}$$
This inequality is easy and there are two nice proof by AM-GM or by C-S also.
SOS also help here$:$
$$\text{LHS}-\text{RHS}=\frac{3\Big[\sum\limits_{cyc} (ab+bc-2ca)^2 + (ab+bc+ca) \sum\limits_{cyc} (a-b)^2 \Big]}{2 \prod (ab+2c^2 +2c)}+\frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)} \geqq 0$$
By SOS$,$ I can only found this stronger with same condition$:$ $$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\geqq \frac{1}{ab+bc+ca}+\frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)}$$
But it's is very easy so I wanna to find another harder version for it$?$
Thanks for a real lot!
| An improvement of your last inequality is the following: For $a, b, c > 0$ with $a+b+c=1$,
\begin{align}
&\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\\
\ge\ & \frac{1}{ab+bc+ca} + 4\cdot \frac{\prod (a-b)^2}{(ab+bc+ca) \prod (ab+2c^2 +2c)}.\tag{1}
\end{align}
$4$ is the best constant in the sense that we cannot replace it with a constant strictly larger than $4$.
(1) is verified by Mathematica. One may find a step-by-step proof of it.
The following inequality is also true: For $a, b, c > 0$ with $a+b+c=1$,
\begin{align}
&\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}\\
\ge\ & \frac{1}{ab+bc+ca} + \frac{27}{32}[(a-b)^2+(b-c)^2+(c-a)^2].\tag{2}
\end{align}
$\frac{27}{32}$ is the best constant in the sense that we cannot replace it with a constant strictly larger than $\frac{27}{32}$. (2) is verified by Mathematica. One may find a step-by-step proof of it.
In general, we first find a function $f(a, b, c)\ge 0$ with $f(1/3, 1/3, 1/3)=0$, then determine
$$\alpha = \inf_{a, b, c > 0; \ a+b+c=1}\frac{\dfrac{1}{ab+2c^{2}+2c}+\dfrac{1}{bc+2a^{2}+2a}+\dfrac{1}{ca+2b^{2}+2b} - \dfrac{1}{ab+bc+ca}}{f(a,b,c)}.$$
Then the following inequality is true: For $a, b, c > 0$ with $a+b+c=1$,
$$\frac{1}{ab+2c^{2}+2c}+\frac{1}{bc+2a^{2}+2a}+\frac{1}{ca+2b^{2}+2b}
\ge \frac{1}{ab+bc+ca} + \alpha f(a,b,c).\tag{3}$$
| {
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Modular calculation high exponent? I want to show, that $5^{96}\equiv -1 \pmod{193}$, without using the formula for quadratic residue.
So far I have :
$5^{96}\equiv 5^{4\cdot24} \equiv 625^{24}\equiv 46^{24}\equiv 186^{12}\equiv -7^{12}\equiv 7^{12}\equiv 7^{3\cdot4}\equiv 150^4\equiv -43^4\equiv 43^4\equiv 112^2\equiv -81^2\equiv 3^8\\\
$
I think Euler's totient doesn't help, as $\varphi(193)=192>96=\frac{192}{2}$ or can I write this ?
$5^{96}\equiv 5^{192-96} \equiv 5^{-96}\equiv(5^{-1})^{96}\equiv 116^{96}\equiv -77^{96} \pmod{193} $
What am I doing wrong ? What am I missing ? Thanks in advance.
| Let $x=5^{96}$. Then in the field $\Bbb F_{193}$ we have
$$
1=5^{\phi(193)}=(5^{96})^2=x^2,
$$
so that $(x-1)(x+1)=0$ in $\Bbb F_{193}$. Since a field has no zero divisors, we must have either $x=1$ or $x=-1$.
But $5^4=625=46$, so that $5^8=46^2=186$ and $5^{16}=49$,
$5^{32}=85$, so that
$$
5^{96}=(5^{32})^3=85^3=-1.
$$
Your calculation is correct, too, since
$$
3^8=6561=-1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$ Prove $\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(a + b + c)$
So, using AM-GM, or just pop out squares under square roots we can show:
$$\sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ac + a^2} \ge \sqrt{3}(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}),$$
i.e. we need next to show that $(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge (a + b + c)$, but i don't know how to do it.
Any help appreciated
| Also, we can use SOS here:
$$\sum_{cyc}\sqrt{a^2+ab+b^2}-\sqrt3(a+b+c)=\sum_{cyc}\left(\sqrt{a^2+ab+b^2}-\frac{\sqrt3(a+b)}{2}\right)=$$
$$=\sum_{cyc}\frac{4(a^2+ab+b^2)-3(a+b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}=\sum_{cyc}\frac{(a-b)^2}{2\left(2\sqrt{a^2+ab+b^2}+\sqrt3(a+b)\right)}\geq0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find inverse of $[x+1]$ in factor ring $\mathbb{Q}[x]/\left\langle x^3-2 \right\rangle$ Find inverse of $[x+1]$ in factor ring $\mathbb{Q}[x]/\left\langle x^3-2 \right\rangle$. I remember that I need to use the extended Euclidean algorithm, but it has been some time, so I am a bit rusty. Thanks in advance!
Edit: I tried it and, by solving $(x+1)(ax^2+bx+c)=1$, with $x^3=2$, i got the solution for the inverse: $[x^2/3 -x/3 + 1/3]$, is the method and result correct?
| Extended Euclidean algorithm is exactly right. Maybe
$x^3 -2 = (x^2 -x +1) (x+1) - 3 $
$ x+1 \, \, = (-x/3)(-3) + 1$. This, we may write as $(-x/3)(-3) = (x+1) -1$
So that finally, $(x^3 - 2) (-x/3) = (x^2 -x +1) (x+1) (-x/3) - 3 (-x/3) $
The right side further becomes $(x^2 -x +1) (x+1) (-x/3) + (x+1) - 1 = (x+1) \big[p(x) \big] - 1 $. Thus $$ (x^3 -2) + 1 = (x+1)p(x) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $ \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$, find $|a_1|$
If $a_1a_2 = 1, a_2a_3 = 2, a_3a_4 = 3 \cdots$ and $\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$. Find $|a_1|$
I could conclude that $\displaystyle \lim_{n \to \infty}a_n$ must be $\infty$. But couldn't get any further. Any help is welcome.
| By the Wallis product
\begin{align*}
\frac{2}{\pi } & = \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\frac{{(2n - 1) \cdot (2n + 1)}}{{2n \cdot 2n}}} = \mathop {\lim }\limits_{N \to + \infty } \prod\limits_{n = 1}^N {\frac{{a_{2n - 1} a_{2n} a_{2n + 1} a_{2n + 2} }}{{a_{2n} a_{2n + 1} a_{2n} a_{2n + 1} }}} \\ & = \mathop {\lim }\limits_{N \to + \infty } \left( {\prod\limits_{n = 1}^N {\frac{{a_{2n - 1} }}{{a_{2n} }}} \prod\limits_{n = 1}^N {\frac{{a_{2n + 2} }}{{a_{2n + 1} }}} } \right) = \mathop {\lim }\limits_{N \to + \infty } \left( {\prod\limits_{n = 1}^N {\frac{{a_{2n - 1} }}{{a_{2n} }}} \prod\limits_{n = 2}^{N + 1} {\frac{{a_{2n} }}{{a_{2n - 1} }}} } \right) \\ & = \frac{{a_1 }}{{a_2 }}\mathop {\lim }\limits_{N \to + \infty } \frac{{a_{2N + 2} }}{{a_{2N + 1} }} = \frac{{a_1 }}{{a_2 }}.
\end{align*}
Thus,
$$
\frac{2}{\pi } = \frac{{a_1 }}{{a_2 }} = \frac{{a_1^2 }}{{a_1 a_2 }} = a_1^2 ,
$$
i.e.,
$$
\left| {a_1 } \right| = \sqrt {\frac{2}{\pi }} .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Surface integral of piecewise volume boundary? How should I go about solving the surface integral
$$\iint_{\partial V} 2(x^2+y^2) \, dS,$$ where $V$ is the region bounded by the paraboloid $$z=\frac12-x^2-y^2$$ and the cone $$z^2=x^2+y^2?$$
I have done surface integrals before, but I am unsure how to proceed with finding the boundary of the solid $V.$
| The outward surface normals for the two surfaces are given by the gradients of the expressions:
$$z+x^2+y^2-\frac{1}{2}=0 \implies \vec{n} = \frac{1}{\sqrt{3-4z}}\langle 2x, 2y, 1\rangle$$
$$x^2+y^2-z^2 = 0 \implies \vec{n} = \frac{1}{2\sqrt{2}|z|}\langle 2x, 2y, -2z \rangle$$
and consider the two vector fields
$$F_1 = \sqrt{3-4z} \langle x, y , 0 \rangle $$
$$F_2 = 2\sqrt{2} |z| \langle x, y, 0 \rangle$$
The integral of the first one on the upper surface and the second one of the second surface gives us the integrand in the problem. If we consider the disk created by the two surfaces' plane of intersection, the integral on that disk for both of those vector fields is $0$. Thus we can use the divergence theorem to calculate the integral:
$$\iint_{\partial V} 2(x^2+y^2)\:dS = \iiint_{V_1} 2\sqrt{3-4z}\:dV + \iiint_{V_2} 4\sqrt{2} z \:dV$$
The easiest way to do these integrals is in cylindrical coordinates
$$\int_0^{2\pi} \int_0^{\frac{-1+\sqrt{3}}{2}} \int_{\frac{-1+\sqrt{3}}{2}}^{\frac{1}{2}-r^2} 2r\sqrt{3-4z}\:dz\:dr\:d\theta = \frac{2\pi}{3} \int_0^{\frac{-1+\sqrt{3}}{2}} r\left(5-2\sqrt{3}\right)^{\frac{3}{2}} - r\left(4r^2+1\right)^{\frac{3}{2}} \:dr = \frac{\pi\left(5-2\sqrt{3}\right)^2\left(\sqrt{3}-1\right)^2}{12} - \frac{\pi\left(5-2\sqrt{3}\right)^{\frac{5}{2}}}{30}$$
$$\int_0^{2\pi} \int_0^{\frac{-1+\sqrt{3}}{2}} \int_0^z 2\sqrt{2} zr\:dr\:dz\:d\theta = 2\sqrt{2}\pi \int_0^{\frac{-1+\sqrt{3}}{2}} z^3\:dz = \frac{\pi\left(\sqrt{3}-1\right)^4}{16\sqrt{2}}$$
and the final answer is the sum of these two.
| {
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If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$ While trying to prove it, I proved the following two inequalities:
$a^4+b^4+c^4\ge abc(a+b+c)$ and
$(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$
The later one, on some simplification gives
$a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$
But we can't claim that $ab+bc+ca\ge a+b+c$ for all positive $a, b, c.$ So this doesn't help. So am not quite sure how to approach the inequality in question. Please suggest.. Thanks in advance. (BTW can we use Cauchy-Schwarz's inequality? I tried but couldn't think of a proper choice for two triplets.)
| Using “Titu's Lemma” (also called “Engel's form” of the Cauchy-Schwarz inequality) you have for $a, b, c > 0$
$$
\frac{a^3b + b^3c + c^3a}{abc} = \frac{a^2}{c} + \frac{b^2}{a} +\frac{c^2}{b}
\ge \frac{(a+b+c)^2}{c+a+b} = a+b+c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given $x\cdot x =x$ Prove that it is a commutative ring I have seen some threads related to this, but I want to know whether my approach is correct.
Given $R$ a ring and that $\forall$ $x \in R, x\cdot x=x$. Then prove
that it is a commutative ring.
My try:
Considering $$(x+y)\cdot(x+y)=x\cdot x+x\cdot y+y\cdot x+y\cdot y$$ we get
$$x+y=x+y+x\cdot y+y\cdot x$$
using Cancellation laws:
$$x\cdot y=-(y\cdot x)$$
$\implies$
$$x\cdot y=y\cdot (-x)$$
We have $$x\cdot x=x$$
$$(-x)\cdot (-x)=-x$$
But $$(-x)\cdot (-x)=x$$
So $$x=-x.$$
Hence $$x\cdot y=y\cdot x$$
Is this a valid proof?
| Your proof looks fine! Just for fun, here is a another one using the same ideas.
Let $a,b \in R$. We have that \begin{align}
(a+b)(a+b) &= a\cdot a + a \cdot b + b \cdot a + b \cdot b\\
&= a + a \cdot b + b \cdot a + b\\
&\overset{*}{=} a + b
\end{align}
and using the Cancellation Law in $*$ gives us that $a\cdot b = -(b \cdot a)$, so $b \cdot a = -(a \cdot b)$. But then
\begin{align}
(a\cdot b)\cdot (a\cdot b)&= a \cdot (b \cdot a) \cdot b\\
&= a \cdot (-(a \cdot b)) \cdot b\\
&=a \cdot ((-a)\cdot b) \cdot b\\
&= (a \cdot (-a))\cdot (b \cdot b)\\
&=(-a)\cdot b\\
&= -(a \cdot b)\\
&= a\cdot b
\end{align}
(where the last equality follows from $(a\cdot b)\cdot (a \cdot b) = a \cdot b$) and so $a\cdot b = - (a \cdot b)$ together with $b \cdot a = -(a \cdot b)$ give us that $a \cdot b = b \cdot a$. But $a$ and $b$ were chosen arbitrarily, so $R$ is indeed a commutative ring.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in this form: $\sqrt{16x^2 - 3^2}$ so I know to use $x = 3\sec u$:
$$
\begin{align}
& \int \frac{\sqrt{16x^2 - 9}}{x} \, dx \\
= {} & \int \frac{\sqrt{16x^2 - 3^2}}{x} \, dx \\
= {} & \int \frac{\sqrt{16(3\sec u)^2 - 3^2}}{3\sec u} 3\sec u\tan u \, du \\
= {} & \int \frac{(\sqrt{16(3\sec u)^2 - 3^2)}(3\sec u\tan u)}{3\sec u} \, du \\
= {} & \int \sqrt{(16(3\sec u)^2 - 3^2)}(\tan u) \, du
\end{align}
$$
This doesn't seem to make it easy. However, using a calculator online, it suggests I instead use $x = \dfrac{3}{4}\sec{u}$ which simplifies the integral to a crisp $\int 3\tan^2 u \, du$.
My question is, how did the calculator get $a = \dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
| I am confused by the suggestion to use trigonometric substitution, since $$\frac{\sqrt{16x^2 - 9}}{x} = 16 x \frac{\sqrt{16x^2 - 9}}{16x^2},$$ and the substitution $$u^2 = 16x^2 - 9, \quad 2u \, du = 32 x \, dx$$ yields $$\begin{align*}
\int \frac{\sqrt{16x^2 - 9}}{x} \, dx
&= \int \frac{u}{u^2 + 9} u \, du \\
&= \int 1 - \frac{9}{u^2 + 9} \, du \\
&= u - 3 \tan^{-1} \frac{u}{3} + C \\
&= \sqrt{16x^2 - 9} - 3 \tan^{-1} \frac{\sqrt{16x^2 - 9}}{3} + C.
\end{align*}$$
Trigonometric substitution certainly works, but in such cases, we can certainly avoid it.
| {
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"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of complex roots' fractions According to this:
If $\omega^7 =1$ and $\omega \neq 1$ then find value of
$\displaystyle\frac{1}{(\omega+1)^2} +
\frac{1}{(\omega^2+1)^2} +
\frac{1}{(\omega^3+1)^2} +
... + \frac{1}{(\omega^6+1)^2}=?$
First I try like
$\displaystyle\frac{1}{\omega+1} +
\frac{1}{\omega^2+1} +
\frac{1}{\omega^3+1} +
... + \frac{1}{\omega^6+1} = 3
$
I have done distribution them and finally got the solution $\dfrac{5}{3}$
However, this is, without a doubt, a time-consuming way.
Can someone please suggest easier way to solve this one.
| Using $\omega^7=1$ the second sum computes to
$$
\frac{3(\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1)}{\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1}=3
$$
For the first sum I do not obtain $5/3$. I obtain
$$
3\cdot\frac{5\omega^6+ 2\omega^5 + 11\omega^4 - 4\omega^3 + 11\omega^2 + 2\omega + 5}{9\omega^6 + 10\omega^5 + 7\omega^4 + 12\omega^3 + 7\omega^2 + 10\omega + 9}=-9,
$$
because $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find value of $\cot(\theta-\alpha)$ Let $\theta$ and $\alpha$ be the solutions to $$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=\dfrac{3\cos2x+6}{10}$$
find the value of $\cot(\theta-\alpha)$, where $\dfrac{\pi}{2}<\theta<\alpha<\dfrac{3\pi}{2}$
My attempt :
Let $A=\dfrac{x}{2}$, $$\dfrac{\sin^2A}{1-\cot A}+\dfrac{\cos^2A}{1-\tan A}=\dfrac{3\cos 4A+6}{10}$$
$$\dfrac{\sin A\cos A-\sin^4A-\cos^4A}{2\sin A\cos A-\cos^2A-\sin^2A}=\dfrac{12\cos^2A-3}{10}$$
$$\dfrac{\sin A\cos A+\cos^2A-\sin^2A}{2\sin A\cos A-1}=\dfrac{12\cos^3A-3}{10}$$
Are there any better ways for approaching this problem? or I've come the right way.
| Note that
$$\dfrac{\sin^2\dfrac{x}{2}}{1-\cot\dfrac{x}{2}}+\dfrac{\cos^2\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}=
\frac{\sin^3\frac x2 -\cos^3\frac x2}{\sin\frac x2 - \cos\frac x2}
=1+\frac12\sin x
$$
which, along with $\frac{3\cos2x+6}{10}= \frac{9-6\sin^2 x}{10} $, leads to
$$6\sin^2 x +5\sin x +1=0$$
Then, $\sin\alpha =-\frac12$, $ \sin \theta =-\frac13$ and correspondingly $\tan\alpha = \frac1{\sqrt3}$, $\tan\theta = \frac1{2\sqrt2}$
$$\cot(\theta- \alpha) = \frac{1+\tan \theta \tan\alpha }{\tan \theta - \tan \alpha}
=-\frac{8\sqrt2+9\sqrt3}{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate $\iint\frac{dxdy}{(1+x^2+y^2)^2}$ over a triangle
Calculate
$$\iint\frac{dxdy}{(1+x^2+y^2)^2}$$ over the triangle $(0,0)$, $(2,0)$, $(1,\sqrt{3})$.
So I tried changing to polar coordinates and I know that the angle is between $0$ and $\frac{\pi}{3}$ but I couldn't figure how to set the radius because it depends on the angle.
| Let us have a solution based on an alternative idea. We consider on the triangle $T$ the one-form
$$
\omega=\frac 12\cdot \frac {x\; dy - y\; dx}{1+x^2+y^2}\ .
$$
Then
$$
\begin{aligned}
2d\omega
&=
\frac\partial{\partial x}\left(\frac x{1+x^2+y^2}\right)
dx\wedge dy
+
\frac\partial{\partial x}\left(\frac {-y}{1+x^2+y^2}\right)
dy\wedge dx
\\
&=\frac 2{(1+x^2+y^2)^2}\; dx\wedge dy\ .
\end{aligned}
$$
We apply Stokes now. We parametrize the boundary of $T$ using the maps
*
*$t\to(t,0)$ for $t$ from $0$ to $2$, and there will be no contribution because of $y=0$,
*$t\to(2-t,t\sqrt 3)$ for $t$ from $0$ to $1$,
*$t\to(t,t\sqrt 3)$ for $t$ from $1$ to $0$, and there will be no contribution, because $x\; dy-y\; dx$ becomes $t\;(t\sqrt 3)'\; dt -(t\sqrt 3)\; t'\; dt$,
and compute explicitly:
$$
\begin{aligned}
&\int_{\partial T}
\frac {x\;dy}{1+x^2+y^2}
=
\int_0^2\frac {t\cdot 0'\; dt}{1+t^2+0^2}
\\
&\qquad\qquad\qquad
+
\int_0^1\frac {(2-t)\; (t\sqrt 3)'\; dt}{1+(2-t)^2+3t^2}
+
\int_1^0\frac {t\; (t\sqrt 3)'\; dt}{1+t^2+3t^2}
\ ,
\\[3mm]
&\int_{\partial T}
\frac {y\;dx}{1+x^2+y^2}
=
\int_0^2\frac {0\cdot t'\; dt}{1+t^2+0^2}
\\
&\qquad\qquad\qquad
+
\int_0^1\frac {t\sqrt 3\; (2-t)'\; dt}{1+(2-t)^2+3t^2}
+
\int_1^0\frac {t\sqrt 3\; t'\; dt}{1+t^2+3t^2}
\ ,
\\[3mm]
&\iint_T\frac {dx\; dy}{(1+x^2+y^2)^2}=
\iint_T d\omega
\\
&\qquad=
\int_{\partial T} \omega
\\
&\qquad
=\frac 12\int_0^1
\frac {(2-t)\cdot(t\sqrt 3)'-(t\sqrt 3)\; (2-t)'}{1+(2-t)^2+3t^2}
\; dt
\\
&\qquad=\frac {\sqrt 3}2\int_0^1
\frac {(2-t)+t}{(2t-1)^2+2^2}
\; dt
=\color{blue}{\frac {\sqrt 3}2\arctan\frac 12}\ .
\end{aligned}
$$
(Note: All details are included for didactical reasons, now please remove all details to have a two lines computation, given the formula for $d\omega$ and the cancellations on the first and third line path parametrizing $\partial T$.)
A sage numerical check using Fubini...
sage: var('x,y');
sage: f = 1 / (1 + x^2 + y^2)^2
sage: assume(x>0)
sage: assume(x<2)
sage: J1 = integral( integral(f, y, 0, x *sqrt(3)), x, 0, 1)
sage: J2 = integral( integral(f, y, 0, (2-x)*sqrt(3)), x, 1, 2)
sage: (J1+J2).n()
0.401530607798613
sage: ( sqrt(3)/2*atan(1/2) ).n()
0.401530607798613
| {
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"url": "https://math.stackexchange.com/questions/3709310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$
Now if $(x+4y+3)(x+2y)=2$, then either $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\text{ or }\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\text{ or }\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\text{ or }\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}.$$
We have $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\iff (x,y)=(6,-2), \\\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\iff (x,y)=(3,-1), \\\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\iff (x,y)=(0,-1),\\\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}\iff (x,y)=(3,-2).$$
Now since, all the four pairs $(6,-2),(3,-1),(0,-1),(3,-2)$ satisfies the integer equation $(x+4y+3)(x+2y)=2$, thus we can conclude that $(x+4y+3)(x+2y)=2\iff (x,y)=(6,-2),(3,-1),(0,-1),(3,-2).$
Hence, we can conclude that the integer equation $x^2+6xy+8y^2+3x+6x=2$ is satisfied if and only if $(x,y)=(6,-2),(3,-1),(0,-1),(3,-2)$, and we are done.
Is the solution correct and rigorous enough? And, I am always confused while solving equations regarding the usage of the if and only if arguments, which I feel is very necessary in order to have a complete and rigorous solution, but I rarely find it's usage in any book while solving equations of any kind. So, is it necessary? Also, is there a better solution than this?
| Below is a different solution. It's not as easy as other methods to which this equation might be amenable, but it generalizes nicely to degree two polynomial Diophantine equations in two variables and has a nice geometric interpretation. The general idea is to parameterize all rational solutions and then restrict them to integers. Here we go.
First we notice that $(0,-1)$ works. This will be helpful because we can use it to generate all other rational solutions. Note that for $x=0,$ the only solutions are $y=-1,\frac{1}{4}.$ Note that if $(p,q)$ is a rational solution other than $(0,-1)$ and $\left(0,\frac{1}{4}\right),$ then the line through $(p,q)$ and $(0,-1)$ is non-vertical and has slope $\frac{q+1}{p},$ which is rational. The main idea is the converse: If there is a non-vertical line through $(0,-1)$ with a rational slope $m,$ then it intersects the curve of real solutions at a rational point. We can prove it as follows: The equation of the line is $y=mx-1.$ By substituting it into the curve's equation and simplifying, we get $$(1+6m+8m^2)x^2 -(3+10m)x=0.$$ Since $x\ne 0,$ we get $$(1+6m+8m^2)x =3+10m.$$ The only ways that $1+6m+8m^2=0$ are $m=-\frac{1}{4},-\frac{1}{2}.$ Each of these cause the left side to disappear while making the right side non-zero, so we don't have to worry about division by zero when we produce $$x=\frac{3+10m}{1+6m+8m^2},$$ which is rational. So the rational solutions for $x\ne 0$ are parameterized by $$(x,y)=\left(\frac{3+10m}{1+6m+8m^2},mx-1\right)$$ over all rational $m.$ We want to find out which of there are not merely rational, but integers.
Isolating $m$ in $$x=\frac{3+10m}{1+6m+8m^2}$$ using the quadratic formula yields $$m=\frac{5-3x\pm\sqrt{(x-3)^2+2^4}}{8x}.$$ So we want to find all cases where $(x-3)^2+2^4=z^2$ for some integer $z.$ By difference of squares, we want to solve the Diophantine equation $$(z-x+3)(z+x-3)=2^4.$$ Via casework, we get the solutions $$(x,z)=(6,5),(3,4),(0,5),(0,-5),(3,-4),(6,-5).$$ Omitting the cases where $x=0,$ we can find all the corresponding values of $$m=\frac{5-3x\pm z}{8x}$$ and subsequently all the values of $y.$ There turn out to be some overlaps in these cases, but anyway, the solutions are $$(6,-2),(3,-1),(3,-2),(0,-1),$$ where the last one is the one that we initially guessed.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Does this double integral exists? I've calculated $\iint_D\frac{y}{x^2+y^2}\,dA$ where D is bounded by $y=x$, $y=2x$, $x=2$, in this way:
$$\iint_D\frac{y}{x^2+y^2}\,dA=\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx=\cdots = \ln \frac{5}{2}.$$
However, I wonder if the fact that the $\frac{y}{x^2+y^2}$ is not bounded on D invalidates all the calculations and in fact the double integral does not exist. All the theorems that I consult have as a hypothesis that the integrand is bounded on the region of integration and hence my doubt regarding this double integral. Can anyone help me? Does this integral exist?
| $$\int \frac{ydy}{x^2+y^2} = \frac{1}{2} \int \frac{d(y^2+x^2)}{x^2+y^2} =\frac{1}{2} \ln(x^2+y^2) + C$$
So
$$\int_0^2\int_x^{2x}\frac{y}{x^2+y^2}\,dy\,dx= \frac{1}{2} \int_{0}^{2}\ln(x^2+y^2)|_{x}^{2x} dx$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$M = \int_0^{\pi/2} \frac{\cos x}{(x+2)} dx$ and $N = \int_0^{\pi/4} \frac{\sin x \cos x}{(x+1)^2} dx$, then value of $M - N$ is? Options are :
(A) $\pi\quad$ (B) $\frac{\pi}{4}\quad$ (C) $\frac{2}{\pi + 4}\quad$ (D) $\frac{2}{\pi - 4}$
I have solved it using Taylor's expansion of the numerators.
E.g, $\cos x = \cos ((x+2) -2) = \cos(x+2) \cos 2 + \sin(x+2) \sin 2$, then using Taylor's formula expanded $\cos(x+2)$ and $\sin(x+2)$ to get polynomials of $(x+2)$, which is in the denominator.
But due to the denominator, I am getting results in terms of log, which does not match any of the options given.
| Using Integration by Parts, we can write N as
$$N = -\frac{\sin x\cos x}{x+1}\vert_0^{\frac{\pi}{4}}+\int_0^\frac{\pi}{4}\frac{\cos^2x-\sin^2x}{x+1}dx=-\frac{2}{\pi+4}+\int_0^{\frac{\pi}{4}}\frac{\cos2x}{x+1}dx$$
In the second integral, let $x \to 2x \implies dx \to 2dx$
$$\therefore N=-\frac{2}{\pi+4}+\int_0^\frac{\pi}{2}\frac{\cos x}{x+2}dx=-\frac{2}{\pi+4}+M$$
$$\therefore M-N=\frac{2}{\pi+4}$$
P.S Thanks to Claude Leibovici for the corrected expression.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$ $$\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$$
I will give my solution, which, to my surprise, turned out to be erroneous:
$S_n=\frac{a_1+a_n}{2}n=\frac{\frac{n}{1+n^2}+\frac{n}{n^2+n^2}}{2}n=\frac{1}{4}\frac{3n^2+1}{1+n^2}$
$\lim_{n\to\infty}S_n=\frac{3}{4}$
But the correct answer is $\frac{\pi}{4}$
I do not understand how to get the exact answer.
| Divide the general term $\dfrac{n}{k^2+n^2}$ by $n^2$ we have : $\dfrac{1}{n}\cdot \dfrac{1}{1+ \frac{k^2}{n^2}} \to \displaystyle \int_{0}^{1} \dfrac{dx}{1+x^2}= \dfrac{\pi}{4}$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the unit digit of: $\left\lfloor{10^{20000}\over 100^{100}+3}\right\rfloor$
Find the unit digit of:
$$\left\lfloor{10^{20000}\over 100^{100}+3}\right\rfloor$$
I am completely clueless about how to deal with such huge powers.
I noticed that the numerator is $({100^{100}})^{100}$, which brings some sort of similarity with the denominator, but I couldn't get anything useful from it, I tried to add and subtract terms in order to reduce the power:
Let $x = 100^{100}$
${x^{100}-3^{100} + 3^{100}\over x+3 }$
Now the first two terms combined are divisible by $x+3$ , but it still didn't work out to solve for the unit digit.
Could someone please teach me how to solve for the unit digit?
Thanks !
| $$\frac{x^{100}-3^{100} + 3^{100}}{ x+3 }$$
"Now the first two terms combined are divisible by $x+3$". In fact, you have done the critical step!
$$\frac{x^{100}-3^{100} + 3^{100}}{ x+3 }= \frac{x^{100}-3^{100}}{ x+3 }+\frac{3^{100}}{ x+3 }$$
Since $\dfrac{3^{100}}{ x+3 }=\dfrac{3^{100}}{ 100^{100}+3 }<1$, we see that the unit digit wanted is the same as the unit digit of $\dfrac{x^{100}-3^{100}}{ x+3 }$.
The unit digit of $x^{100}-3^{100}$ is 9 since the unix digit of $x=100^{100}$ is $0$ and the unit digit of $3^{100}=(81)^{25}$ is $1$. The unit digit of $x+3$ is $3$.
So the unit digit of $\dfrac{x^{100}-3^{100}}{ x+3 }$ is $\dfrac{\cdots9}{\cdots3} = 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding Inverse of a 4*4 matrix Find the inverse of the following matrix
$$
\begin{array}{c}
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
\text { where } c_{0}=\frac{1+\sqrt{3}}{4 \sqrt{2}}, c_{1}=\frac{3+\sqrt{3}}{4 \sqrt{2}}, c_{2}=\frac{3-\sqrt{3}}{4 \sqrt{2}} \text { and } c_{3}=\frac{1-\sqrt{3}}{4 \sqrt{2}}
\end{array}
$$
I had thought of replacing the values with trigonometric values like Sin 15 , cos 15 .... But it seems very lengthy. Any shorter approach?
| Posting as an answer as it the MathJax for the matrix will not fit in a comment. Mathematica gives a rather nasty expression for the inverse of the matrix:
$$
\left(
\begin{array}{cccc}
\frac{1}{560} \left(8 \sqrt{6}-29\right) & \frac{1}{560} \left(99-8 \sqrt{6}\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}+99\right) \\
\frac{1}{560} \left(8 \sqrt{6}+99\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}-99\right) & \frac{1}{560} \left(29-8 \sqrt{6}\right) \\
\frac{1}{560} \left(99-8 \sqrt{6}\right) & \frac{1}{560} \left(8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}+99\right) & \frac{1}{560} \left(-8 \sqrt{6}-29\right) \\
\frac{1}{560} \left(-8 \sqrt{6}-29\right) & \frac{1}{560} \left(8 \sqrt{6}+99\right) & \frac{1}{560} \left(29-8 \sqrt{6}\right) & \frac{1}{560} \left(8 \sqrt{6}-99\right) \\
\end{array}
\right)
$$
Showing this result analytically seems a daunting task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is the permutation matrix P of PLU decomposition unique? Let $A$ be a square matrix. Then there exists a permutation matrix $P$ such that $A=PLU$, where $L$ is a lower triangular matrix and $U$ is an upper triangular matrix. To further ensure the uniqueness, we assume that the main diagonal of $L$ (or $U$) to be 1. So, the question is, is the permutation matrix unique, i.e., can we find another $P'\ne P$ such that $A=PLU=P'L'U'$ where $L', U'$ are still triangular matrix? If yes, what is the condition for the uniqueness?
| No, the permutaion is not unqiue. Here is an example:
$$
\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix} =\underbrace{\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{L}\underbrace{\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{U}\\
\underbrace{\begin{pmatrix}
0 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix}}_{P}\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 1
\end{pmatrix} =\begin{pmatrix}
1 & 1 & 0\\
1 & 0 & 0\\
0 & 0 & 1
\end{pmatrix} =\underbrace{\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{L}\underbrace{\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}}_{U}
$$
where $\displaystyle P$ is the permutation matrix that switches first and second row. It follows that we can find a different permutation matrix but has a different LU-decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Proof of the following approximation for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant1+x+x^2$ I was going through the text Introduction to Algorithms by Cormen et. Al. where I came across the following approximation...
for $|x| \leqslant 1$ we have $1+x \leqslant e^x \leqslant 1+x+x^2$
Now I could prove the left side of the inequality and it holds for all real $x$. I proved using Mean Value Theorem( Taylor Series).
Let $f(x)= e^x$
So by Mean Value Theorem we have,
$$f(x) = f(0) + x.f'(\vartheta x) = e^0 + x.e^{\vartheta x} , 0< \vartheta <1 $$
$$ 0< \vartheta < 1 $$
Case 1:
(if $x \geqslant 0$)
$$\begin{aligned}\implies &0 \leqslant \vartheta .x \leqslant x\\\implies& e^0 \leqslant e^{\vartheta .x} \leqslant e^x\\\implies &x.1 \leqslant x.e^{\vartheta .x} \leqslant x.e^x\\\implies& 1+x \leqslant 1+x.e^{\vartheta .x} \leqslant 1+x.e^x\end{aligned}$$
$$\implies 1+x \leqslant f(x) \leqslant 1+x.e^x \ldots\tag 1$$
Case 2:
(if $x \lt 0$)
$$\begin{aligned}\implies &0 \gt \vartheta .x \gt x\\\implies& e^0 \gt e^{\vartheta .x} \gt e^x\\\implies& x.1 \lt x.e^{\vartheta .x} \lt x.e^x\\\implies& 1+x \lt 1+x.e^{\vartheta .x} \lt 1+x.e^x\end{aligned} $$
$$\implies 1+x \lt f(x) \lt 1+x.e^x \ldots\tag 2$$
So from cases $(1)$ and $(2)$ we have that $e^x \geqslant 1+x\ \forall x\in\Bbb R$ and hence for $|x| \leqslant 1$ .
Now my attempt to prove the right inequality.
Here again I attempt using Taylor Series,
$$f(x) = f(0) + x.f'(0) + \frac{x^2}{2!}.f''(\theta .x) = e^0 + x.e^0+ \frac{x^2}{2!}.e^{\vartheta .x} , 0< \vartheta <1 $$
$$\implies f(x) = 1 + x + \frac{x^2}2.e^{\vartheta .x} $$
(if $0\leqslant x \leqslant 1$)
$$\begin{aligned}\implies& 0 \leqslant \theta .x \leqslant x\\\implies& e^0 \leqslant e^{\vartheta .x} \leqslant e^x\\\implies& \frac{x^2}2.1 \leqslant \frac{x^2}2.e^{\vartheta .x} \leqslant \frac{x^2}{2}.e^x\\\implies& 1+ x +\frac{x^2}{2} \leqslant 1+x+\frac{x^2}{2}.e^{\vartheta .x} \leqslant 1+x+\frac{x^2}2.e^x\end{aligned}$$
$$\implies 1+ x +\frac{x^2}2\leqslant f(x) \leqslant 1+x+\frac{x^2}2.e^x\ldots\tag 3$$
Now,
$$\begin{aligned}0\leqslant x \leqslant 1 \implies& 1 \leqslant e^x \leqslant e \implies \frac12 \leqslant \frac{e^x}2 \leqslant \frac{e}2(\approx 1.4)\\\implies &1+x+\frac12.x^2 \leqslant 1+x+\frac{e^x}2.x^2 \leqslant 1+x+\frac{e}2x^2 \lt 1+x+ 1.4 x^2\end{aligned}$$
How to show a more tighter bound for f(x)? I don't get it. Maybe I made some mistake.
(if $-1\leqslant x \lt 0$)
$$\begin{aligned}\implies 0 \gt \vartheta .x \gt x\\\implies& e^0 \gt e^{\vartheta .x} \gt e^x\\\implies&\frac{x^2}{2}.1 \gt \frac{x^2}2.e^{\vartheta .x} \gt \frac{x^2}2.e^x\\\implies &1+ x +\frac{x^2}{2} \gt 1+x+\frac{x^2}2.e^{\vartheta .x} \gt 1+x+\frac{x^2}2.e^x\\\implies &1+ x +\frac{x^2}{2} \gt f(x) \gt 1+x+\frac{x^2}2.e^x\end{aligned}$$
$$\implies 1+x+\frac{x^2}2.e^x \lt f(x) \lt 1+ x +\frac{x^2}2 \lt 1+ x +x^2 \ldots\tag 4$$
| For $0\le x\le1$, $e^x=1+x+\sum_{n\ge2}\frac{x^n}{n!}\le1+x+\sum_{n\ge2}\frac{x^2}{2^{n-1}}=1+x+x^2$. For $-1\le x<0$, write $y:=-x$ so we want to prove $0<y\le1\implies e^{-y}\le1-y+y^2$. Indeed, $e^{-y}=1-y+\sum_{n\ge2}\frac{(-1)^ny^n}{n!}\le1-y+\sum_{n\ge2}\frac{y^2}{2^{n-1}}=1-y+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$. Prove: $\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{3}{2}$ with $a, b, c$ are positive real numbers.
Let $a \le b \le c$:
\begin{align*}
\sqrt{\dfrac{bc}{a(3b+a)}} = \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{c}{3b+a}} &\ge \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{c}{3c+c}} = \dfrac{1}{2} \sqrt{\dfrac{b}{a}} && (1)
\\
\sqrt{\dfrac{ac}{b(3c+b)}} = \sqrt{\dfrac{a}{b}}\sqrt{\dfrac{c}{3c+b}} &\ge \sqrt{\dfrac{a}{b}}\sqrt{\dfrac{c}{3c+c}} = \dfrac{1}{2}\sqrt{\dfrac{a}{b}} && (2)
\\
\sqrt{\dfrac{ab}{c(3a+c)}} \ge \sqrt{\dfrac{ab}{c(3c+c)}} &\ge \dfrac{1}{2} \dfrac{\sqrt{ab}}{c} && (3)
\end{align*}
With $(1)+(2)+(3)$, we have:
$$
\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{1}{2}\left(\sqrt{\dfrac{b}{a}} + \sqrt{\dfrac{a}{b}} + \dfrac{\sqrt{ab}}{c} \right).
$$
And I can not find the method to finish this problem. I am trying to think another method.
| After replacing $a$ at $\frac{1}{a},$ $b$ at $\frac{1}{b}$ and $c$ at $\frac{1}{c}$ we need to prove that:
$$\sum_{cyc}\frac{a}{\sqrt{c(3a+b)}}\geq\frac{3}{2}.$$
Now, by Holder $$\left(\sum_{cyc}\frac{a}{\sqrt{c(3a+b)}}\right)^2\sum_{cyc}ac(3a+b)\geq(a+b+c)^3.$$
Id est, it's enough to prove that
$$4(a+b+c)^3\geq9\sum_{cyc}(3a^2c+abc),$$ which is smooth:
Let $\{a,b,c\}=\{x,y,z\}$ where $x\geq y\geq z$.
Thus, by Rearrangement and AM-GM we obtain:
$$9\sum_{cyc}(3a^2c+abc)=27(a\cdot ac+b\cdot ba+c\cdot cb+xyz)\leq$$
$$\leq27(x\cdot xy+y\cdot xz+z\cdot yz+xyz)=27y(x+z)^2=108y\left(\frac{x+z}{2}\right)^2\leq$$
$$\leq108\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4(x+y+z)^3=4(a+b+c)^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric inequalities in a triangle What is the proof of
$~~\cos^2 A+\cos^2 B+\cos^2 C \leq 1 ~~$
in an acute triangle ?
This will be of help in finding the answer (if such exists) to finding the minimal T in any ∆ ABC when
$$T \geq \sin^k A+ \sin^k B+ \sin^k C~ ,~~~~ k \geq 3$$
| We have
\begin{align}
\cos^2A+\cos^2B+\cos^2C&=\frac12(1+\cos2A)+\frac12(1+\cos2B)+\cos^2C\\
&=\frac12(2+\cos2A+\cos2B)+\cos^2C\\
&=\frac12(2+2\cos(A-B)\cos(A+B))+\cos^2C\\
&=1+\cos(A-B)\cos(\pi-C)+\cos^2C\\
&=1-\cos(A-B)\cos C+\cos^2C\\
&=1-\cos C(\cos(A-B)-\cos(\pi-(A+B)))\\
&=1-\cos C(\cos(A-B)+\cos(A+B))\\
\cos^2A+\cos^2B+\cos^2C&=1-2\cos A\cos B\cos C\\
\implies\cos^2A+\cos^2B+\cos^2C&\leq1\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i'm learning by myself limits since my teachers doesn't respond any email and they are not making lectures, just pdf's... I'm trying to do my best, help pls). Also is there any good book or suggestion to learn limits .
\begin{align*}&\lim_{b \to a} \dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2} \cdot \dfrac{(a-\sqrt{a^2-b^2})}{(a-\sqrt{a^2-b^2})} \\=&
\lim_{b \to a} \dfrac{a[a^2-(a^2-b^2)]-b^2(a-\sqrt{a^2-b^2})}{a[(a^2-\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
= &\lim_{b \to a} \dfrac{-ab^2-ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
=&\lim_{b \to a} \dfrac{-2ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
\end{align*}
| Notice: \begin{align}a \cdot (a + \sqrt{a^2-b^2}) - b^2 &= (a^2-b^2) +a\sqrt{a^2-b^2} \\&=
\sqrt{a^2-b^2}(\sqrt{a^2-b^2} +a)\end{align}
and \begin{align}a \cdot (a - \sqrt{a^2-b^2}) - b^2 &= (a^2-b^2) -a\sqrt{a^2-b^2} \\ &=
\sqrt{a^2-b^2}(\sqrt{a^2-b^2} -a)\end{align}
So \begin{align} \lim_{b \to a} \dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}= \lim_{b \to a} \dfrac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}-a} \end{align}
Now it should be easy...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\sum\limits_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$? I am trying to prove this. I used the telescoping method, but the problem is I need the first fraction to be $\frac{1}{4(n+1)}$ and I also tried to relate it to alternating Harmonic series which didn't work. Any hint would be greatly appreciated.$$\sum_{n=1}^{\infty}\left ( \frac{1}{4n+1}-\frac{1}{4n} \right )=\frac{1}{8}\left ( \pi-8+6\ln{2} \right )$$
| This series can be evaluated with some generating function techniques. It will be related to the well-known generating function
$$ H(x) = \sum_{n=1}^\infty \frac{1}{n} x^n $$
which has closed form found by:
$$ H'(x) = \sum_{n=1}^\infty x^{n-1} = \sum_{n=0}^\infty x^n = \frac{1}{1-x} $$
$$ H(x) = \int\! \frac{dx}{1-x} = \ln(1-x) + C = \ln(1-x) $$
since $H(0)=0$ requires $C=0$. For this problem, we'll want $\ln$ to represent the principal branch of the complex function.
But the function $f$ involves only subsequences of $1/n$, not every term. A standard technique for dealing with subsequences in generating functions is a selector function for roots of unity:
$$ \sigma_k(x) = \frac{1}{k} \sum_{\ell=0}^{k-1} x^{\ell} $$
which has the property that for every $m \in \mathbb{Z}$,
$$ \sigma_k \! \left(e^{2 \pi m i/k}\right) = \begin{cases}
1 & \quad\mathrm{if}\ m \equiv 0 \pmod{k} \\
0 & \quad\mathrm{if}\ m \not\equiv 0 \pmod{k} \end{cases} $$
To match the series terms, note that
$$ \forall m\in \mathbb{Z}: \sigma_4(i^{m-1}) - \sigma_4(i^m) =
\begin{cases} 1 & \quad\mathrm{if}\ m \equiv 1 \pmod{4} \\
-1 & \quad\mathrm{if}\ m \equiv 0 \pmod{4} \\
0 & \quad\mathrm{otherwise} \end{cases} $$
So define
$$ f(x) = \sum_{m=4}^\infty \frac{\sigma_4(x^{m-1}) - \sigma_4(x^m)}{m} $$
and the desired infinite sum will be $f(i)$.
$$ f(x) = -1 + \sum_{m=1}^\infty \frac{\sigma_4(x^{m-1}) - \sigma_4(x^m)}{m} $$
$$ f(x) = -1 + \sum_{m=1}^\infty \frac{1}{4m} (1 + x^{m-1} + x^{2m-2} + x^{3m-3} - 1 - x^m - x^{2m} - x^{3m}) $$
$$ f(x) = -1 + \frac{1}{4} \sum_{m=1}^\infty \left[ (x^{-1} - 1)\frac{x^m}{m} +
(x^{-2} - 1)\frac{x^{2m}}{m} + (x^{-3} - 1)\frac{x^{3m}}{m} \right] $$
$$ f(x) = -1 + \frac{1}{4}(x^{-1} - 1)\ln(1-x) +
\frac{1}{4}(x^{-2} - 1)\ln(1-x^2) +
\frac{1}{4}(x^{-3} - 1)\ln(1-x^3) $$
Then the infinite sum is
$$ f(i) = -1 + \frac{-1-i}{4} \ln(1-i) - \frac{2}{4} \ln 2 + \frac{-1+i}{4}\ln(1+i) $$
Since $1-i = \sqrt{2}\,e^{-\pi i/4}$ and $1+i = \sqrt{2}\,e^{\pi i/4}$,
$$ f(i) = -1 + \frac{-1-i}{4}\left(\frac{1}{2} \ln 2 - \frac{\pi i}{4}\right)
- 2 \ln 2 + \frac{-1+i}{4}\left(\frac{1}{2} \ln 2 + \frac{\pi i}{4}\right) $$
Multiplying out and collecting like terms cancels all the imaginary parts. (If they didn't cancel, we'd know something was wrong with the calculations, since the original sum is clearly real!) So finally, the sum is
$$ f(i) = -1 + \frac{\pi}{8} - \frac{3}{4} \ln 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a^2-2a+b^2-2b+c^2-2c=0$
Adding 3 to both sides we get
$a^2-2a+1+b^2-2b+1+c^2-2c+1=3$
This can be simplified to the following
$(a-1)^2+(b-1)^2+(c-1)^2=3$
Therefore,
$0\leq(a-1)^2,(b-1)^2,(c-1)^2\leq3$
From here we can deduce the range of values that a,b,c can take as
$1-\sqrt{3}\leq a,b,c\leq1+\sqrt{3}$
I don't know know if this helps to answer the question.
|
\begin{align} x+y+z&=k \tag{1}\label{1},\\ x^2+y^2+z^2&=2k\tag{2}\label{2}. \end{align}
Expressing $y,z$ in terms of $x,k$ gives
\begin{align}
y,z&=
\tfrac12\,(k-x\pm\sqrt{-k^2+2 x k-3x^2+4k})
,
\end{align}
so we must have
\begin{align}
-k^2+2 x k-3x^2+4k\ge0
,
\end{align}
which leads to expression of $k$ in terms of $x$
\begin{align}
k(x)&=x+2+\sqrt{4+4x-2x^2}
,\\
k'(x)&=
\frac{\sqrt{4+4x-2x^2}+2-2x}{\sqrt{4+4x-2x^2}}
,
\end{align}
$k'(x)=0$ at $x=2$.
\begin{align}
k_{\max}&=k(2)=6
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$ using L'Hopital $$\displaystyle \lim_{ x\to 0} \frac{\tan x - \sin x}{x^3}$$
$$ \displaystyle \lim_{ x\to 0} \frac{\sec^2x - \cos x}{3x^2}$$
$$ \displaystyle \lim_{x\to 0} \frac{2\cos^{-3}x \sin x + \sin x}{6x}$$
Is it indeed complicated using LHopital, how do I continue?
| God knows why do you want to use L'Hopital rule only. Either you can use brute force Differentiation or you can simply put aside $\cos x$ as others have pointed out, because it is not causing any problem in the limit.
However, a creative way to do this limit-
Let $L= \displaystyle \lim_{x\to 0 } \frac{\tan x - \sin x}{x^3}$
Now, let $x=3\theta$ , as $ x \rightarrow 0$ , $\theta \rightarrow 0$
$$L= \displaystyle \lim_{\theta \to 0 } \frac{\tan (3 \theta) - \sin(3\theta)}{(3\theta)^3} \\ = \displaystyle \lim_{\theta \to 0 } \dfrac{\frac{3 \tan(\theta)- \tan^3 (\theta)}{1-3\tan^2(\theta)} -(3\sin(\theta)-4\sin^3(\theta))}{27\theta^3} $$
Take the LCM and arrange the terms:
$$ L= \displaystyle \lim_{\theta \to 0} \dfrac{3\tan \theta -3\sin \theta - \tan^3 \theta +4\sin^3 \theta +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3(1-3\tan^2 \theta) }$$
Note that $(1-3\tan^2 \theta)$ is just $1$ as $\theta \rightarrow 0$, So, we can shift the limit on it so separate it out. (I would appreciate if someone could write that in a better way, I can't)
This limit is now limited to
$$ L= \displaystyle \lim_{\theta \to 0} \dfrac{3(\tan \theta -\sin \theta) +4\sin^3 \theta - \tan^3 \theta +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3} $$
Do you feel the Deja Vu?
$$L = \frac{3L}{27} +\displaystyle \lim_{\theta \to 0} \dfrac{4\sin^3 \theta - \tan^3 \theta +9 \sin \theta \tan^2 \theta -12\sin^3\theta \tan^2 \theta }{27 \theta^3} \\ \implies 24L= 4 \displaystyle \lim_{\theta \to 0} \frac{\sin^3 \theta}{\theta^3} - \displaystyle \lim_{\theta \to 0} \frac{\tan^3\theta}{\theta^3} + 9 \displaystyle \lim_{\theta \to 0} \frac{\sin \theta \tan^2 \theta }{\theta^3 } -12 \displaystyle \lim_{\theta \to 0} \frac{\sin^3 \theta \tan^2 \theta}{\theta^3} $$
Hence, $ L = \frac{ 4 -1 +9}{24} = \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Factoring $ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$
Factor $$ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a$$
I used the factor theorem to get factors
$$f(a, b, c)=(a-b)(b-c)(a-c)\;g (a, b, c)$$
for some polynomial $g (a, b, c)$.
How can I continue using this method?
(sorry for the previously messed up question I'm new to this website and didn't fully understand the guidelines).
| @JohnBentin's method is the slickest, but if you don't want to check one example to get the scaling factor, let $x:=b^2-a^2,\,y:=c^2-b^2$ so your sum is$$\begin{align}abx+bcy-ca(x+y)&=a(b-c)x+c(b-a)y\\&=a(b-c)(b-a)(b+a)+c(b-a)(c-b)(c+b)\\&=(b-a)(c-b)(c(c+b)-a(b+a))\\&=(b-a)(c-b)(c^2+bc-ab-a^2)\\&=(b-a)(c-b)(c-a)(a+b+c).\end{align}$$In fact, his method helps you guess some components of this calculation as you go, so it's not as taxing as it looks.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Remainder when divided by $7$ What would be the remainder when
$12^1 + 12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.
| In the comments, you recognized that $12^1+12^2+12^3+\cdots+12^{100}$
$\equiv \underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\underbrace{5+4+6+2+3+1}+\,5+4+6+2\pmod7.$
Note that the sum over each brace is a multiple of $7$, and you're almost done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{1+x-1+x}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{2x}$$
*
*if i use $x=\sin^2\theta$
$$\int \frac{(\sqrt{1+\sin^2\theta}+\cos\theta)}{2\sin^2\theta}\sin2\theta\ d\theta=\int (\sqrt{1+\sin^2\theta}+\cos\theta)\cot\theta d\theta$$
*if i use $x=\tan^2\theta$
$$\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta=\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{\sin\theta\cos\theta} d\theta$$
Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.
Thanks
| With the change of variable $x=\sin 2t$, we have
$$\sqrt{1+x}-\sqrt{1-x}=\sqrt{\cos^2t+2\cos t\sin t+\sin^2t}-\sqrt{\cos^2t-2\cos t\sin t+\sin^2t}=2\sin t.$$
Then
$$\int\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int\frac{2\cos2t}{2\sin t}dt=\int\left(\frac1{\sin t}-2\sin t\right)dt
\\=\text{arcoth}(\cos t)+2\cos t+C.$$
From the biquadratic equation
$$4\cos^2t\,(1-\cos^2t)=x^2$$ you draw $\cos t$ as a function of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Olympiad inequality proof issue Prove that $(a^2+b^2)^2\geq(a+b+c)(a+b-c)(b+c-a)(c+a-b)\ \forall \ a,b,c\in\mathbb{R^+} $.
I, forgetting to consider whether $a_1$ and $a_2$ are strictly non-negative (don't think they are), found a proof (almost) using the AM-GM inequality with $a_{1}=(a+b+c)(a+b-c)=(a+b)^2-c^2$ and $a_2=(b+c-a)(c+a-b)=c^2-(a-b)^2$, leading to (after manipulation):
$4a^2b^2 \geq (c^2-(a-b)^2)((a+b)^2-c^2)=(a+b+c)(a+b-c)(b+c-a)(c+a-b)$
$4a^2b^2\leq(a^2+b^2)^2$, clearly, so initial inequality proven.
However, I just realised AM-GM only holds for non-negative reals, and my $a_1$ and $a_2$ are both non-negative only if $b-a\leq c\leq b+a$.
Is there another way to prove this using AM-GM, or an extension of the same idea that covers the cases where $c\gt b+a \gt b-a$ or $b+a \gt b-a \gt c$? Also, why does the proof so cleanly yield the result?
| There is also the following way.
Since for $\prod\limits_{cyc}(a+b-c)\leq0$ the inequality is true,
it's enough to prove our inequality for $\prod\limits_{cyc}(a+b-c)>0.$
Now, if $a+b-c<0$ and $a+c-b<0$ so $a+b-c+a+c-b<0,$
which is a contradiction.
Thus, it's enough to assume that $z=a+b-c>0$, $y=a+c-b>0$ and $x=b+c-a>0$
and we need to prove that
$$\left(\left(\frac{y+z}{2}\right)^2+\left(\frac{x+z}{2}\right)^2\right)^2\geq(x+y+z)xyz$$ or
$$(x^2+y^2+2z^2+2yz+2xz)^2\geq16(x+y+z)xyz,$$ which is true by AM-GM twice:
$$(x^2+y^2+2z^2+2yz+2xz)^2\geq(2xy+2z^2+2yz+2xz)^2=$$
$$=4(xy+z(x+y+z))^2\geq4\left(2\sqrt{xyz(x+y+z)}\right)^2=16(x+y+z)xyz.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{b}{b-1}\right)}}$ for all $b>2$. While solving a bigger problem, I've reduced it to an inequality $$\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{\color{red}b}{b-1}\right)}}$$for $b>2$, which looks plausible when looking at the plots. I've tried using Jensen inequality here with $x\mapsto x^{b-1}$, but not much luck.
I've also checked that the inequality works empirically with Wolfram Alpha.
Yes, this is written correctly, opposite to what some of the people in the comments below are trying to claim. I'm surprised this even needs claryfying.
| This is a partial answer, at least. I'm planning to keep thinking about this tomorrow, but maybe someone else can step in and finish what I have. The argument is almost complete, but I have other things to work on right now.
First we log both sides to get the equivalent inequality
$$
b \log \left ( 1+2^{b^{\frac{1}{b-1}}-1} \right )
\overset{?}{<}
\log \left ( 1+2^{b^{\frac{b}{b-1}}-1} \right )
$$
Then we factor out the dominant term of each log, and separate to get
$$
b
\left [
\log \left ( 2^{b^{\frac{1}{b-1}}-1} \right )
+
\log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
\right ]
\overset{?}{<}
\log \left ( 2^{b^{\frac{b}{b-1}}-1} \right )
+
\log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )
$$
Then we apply some log rules and rearrange
$$
b^{\frac{b}{b-1}} \log(2)
-
b \log(2)
+
b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
\overset{?}{<}
b^{\frac{b}{b-1}} \log(2)
-
\log(2)
+
\log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )
$$
We can cancel the first term of each side, and swap the second terms to make them positive
$$
\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
\overset{?}{<}
b \log(2) + \log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )
$$
Now, looking at the left hand side, notice $b^{\frac{1}{b-1}} \to 1$ from above. So we get the following honest upper bound on the left hand side
$$
\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
<
\log(2) + b \log(2)
$$
In fact, this inequality is not very tight -- It turns out for $b > 4$, we have
$$\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < b \log(2)$$
but my only proof (so far) is desmos:
Looking at the right hand side, notice $b^{\frac{b}{b-1}} \to \infty$, so
$2^{1-b^{\frac{b}{b-1}}} \to 2^{1-\infty} \to 0$, so it's a good thing that
$b \log(2)$ eventually dominates the left hand side!
In summary, by actually proving the tighter inequality $\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < b \log(2)$, which I think is a reasonable goal, we can show that your desired inequality holds for $b > 4$ (really $3.384$).
But we're currently using $0$ as our lower bound for
$\log \left ( 1 + 2^{1-b^{\frac{b}{b-1}}} \right )$. One can see from the graph that the desired inequality holds from $2$ to $3.384$, so if that pleases you then we're done. If not, then slightly more work is needed, but I haven't spent much time thinking about this case yet.
I hope this helps ^_^
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?
| hint
Multiply the first term by $ b-c$ to get in the numerator
$$b^3+c^2b-a^2b-b^2c-c^3+a^2c$$
and the others, by permutation
$$c^3+a^2c-b^2c-c^2a-a^3+b^2a$$
$$a^3+b^2a-c^2a-a^2b-b^3+c^2b$$
The result is $ -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to prove that $f(x)-x f'(x)\neq 0$? I have this function:
$f(x)=\cosh ^{-1}\left(\frac{4 a^2 x^2+\left(a^2 x^2-1\right)^2 \cosh (2 \pi x)}{\left(a^2 x^2+1\right)^2}\right),$
where $0<x<\frac{1}{a}$ and $a$ is a positive real number. I want to prove that $\;f(x)-x f'(x)>0$, or at least to prove that $f(x)-x f'(x)\neq 0$. Is there any way to prove this?
| Partial answer
Remarks: It appears that $h''<0$ on $(0, \frac{1}{a})$, if $a \ge 1$ (I think that proof is not very difficult, according to the expression of $h''$ below).
The difficult case is $a < 1$ in which $h''$ can be positive.
Let us prove that if $a \ge 2$, then $f(x) - xf'(x) > 0$ on $(0, \frac{1}{a})$.
We have $f = \ln (1 + h^2 + h\sqrt{h^2 + 2})$ where
$$h(x) = \frac{(1 - a^2x^2)(\mathrm{e}^{\pi x} - \mathrm{e}^{-\pi x})}{(a^2x^2 + 1)\sqrt{2}}.$$
Note: $h(x) > 0$ for each $x$ in $(0, \frac{1}{a})$.
Let $F(x) = f(x) - xf'(x)$. We have $F(0) = 0$.
It suffices to prove that $F'(x) > 0$ on $(0, \frac{1}{a})$.
We have
$$F'(x) = -xf''(x) = \frac{x}{(h^2+2)^{3/2}}[h(h')^2 - (h^2+2)h''].$$
It suffices to prove that $h'' < 0$ on $(0, \frac{1}{a})$.
We have
$$h'' = -\frac{\mathrm{e}^{2\pi x} - 1}{(a^2x^2+1)^3\mathrm{e}^{\pi x}\sqrt{2}}
\left(P \frac{\mathrm{e}^{2\pi x} + 1}{\mathrm{e}^{2\pi x} - 1} + C
\right)$$
where
$$P = 8\pi a^4 x^3 + 8\pi a^2 x,$$
$$C = \pi^2 a^6 x^6+\pi^2 a^4 x^4-\pi^2 a^2 x^2-12 a^4 x^2-\pi^2+4 a^2.$$
Since $\frac{\mathrm{e}^{2\pi x} + 1}{\mathrm{e}^{2\pi x} - 1}
\ge \frac{\mathrm{e}^{2\pi/a} + 1}{\mathrm{e}^{2\pi/a} - 1} \ge \frac{a}{2\pi} + \frac{1}{2}$
for $0 < x < \frac{1}{a}$ (easy to prove), it suffices to prove that
$P (\frac{a}{2\pi} + \frac{1}{2}) + C > 0$ or
\begin{align}
&\pi^2 a^6 x^6 + \pi^2 a^4 x^4 - \pi^2 a^2 x^2 - 12 a^4 x^2 - \pi^2 + 4 a^2\\
&\qquad\qquad + (4 \pi a^4 x^3 + 4 \pi a^2 x) + (4 a^5 x^3 + 4 a^3 x)> 0.
\end{align}
Since $4 \pi a^4 x^3 + 4 \pi a^2 x \ge 8\pi a^3 x^2$ and $4 a^5 x^3 + 4 a^3 x \ge 8a^4 x^2$ by AM-GM, it suffices to prove that
$$\pi^2 a^6 x^6 + \pi^2 a^4 x^4 - \pi^2 a^2 x^2 - 12 a^4 x^2 - \pi^2 + 4 a^2 + 8\pi a^3 x^2 + 8a^4 x^2
> 0$$
or
$$[\pi a^4 x^4+2 \pi a^2 x^2+\pi+ 2 a (1 - a^2x^2)] (\pi a^2 x^2-\pi+2 a) > 0$$
which is clearly true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can i solve $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$ How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$
My work:
I substituted $x=\tan\theta$, $dx=\sec^2\theta d\theta $
integral becomes
$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$
$\int \dfrac{\tan\theta(\tan^2\theta+1)+\tan \theta-7}{\sec\theta}\sec^2\theta d\theta$
$\int (\tan\theta(\sec^2\theta)+\tan \theta-7)\sec\theta d\theta$
$\int \tan\theta\sec^3\theta\ d\theta+\int \sec\theta \tan \theta\ d\theta-7\int \sec\theta d\theta$
$\int \tan\theta\sec^3\theta+\sec\theta -7\ln|\sec\theta+\tan\theta|+C$
I got stuck here in solving first part of above integral. I can't see the way to solve it. please help me solve it by substitution or other method. thanks
| You're almost there.
$$\int \tan \theta \sec^3 \theta d \theta = \int \frac {\sin \theta}{\cos^4 \theta} d \theta = -\int \frac{du}{u^4}=\frac 13 u^{-3} = \frac {\sec^3 \theta}{3}+ C$$
Since $x= \tan \theta, x^2+1=\sec^2 \theta$, so
$$\frac{\sec^3 \theta}{3} + C = \frac{\sqrt{(x^2+1)^3}}{3}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sums of powers of cosines and sines shifted by $2\pi/3$ I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi/3)\sin^2(x+2\pi/3)+\cos^2(x+4\pi/3)\sin^2(x+4\pi/3) &= 3/8,\\
\cos^4(x)\sin^4(x)+\cos^4(x+2\pi/3)\sin^4(x+2\pi/3)+\cos^4(x+4\pi/3)\sin^4(x+4\pi/3) &= 9/128,
\end{split}
$$
and of course the most elementary
$$
\cos(x)+\cos(x+2\pi/3)+\cos(x+4\pi/3)=0.
$$
The last identity admits a rather intuitive interpretation in terms of unitary complex numbers centered about the origin. My questions are:
*
*Do the other identities admit similar more or less intuitive interpretations as well?
*Do such identities have names?
*Not all powers and combinations produce a constant; What is the general form of the expressions that do?
Context: The first two identities came up while calculating the elastic response of a two-dimensional truss (a planar lattice of nodes connected with springs) that is invariant by rotations of order 3, in which case $x$ describes the orientation of the truss. We know that such trusses must exhibit an isotropic response and that justifies, in a rather convoluted manner, that these expressions must be constants. The other expressions I found by trial and error. I am looking for a satisfying, non-brute-force, non-too-group-theoretic, explanation.
| Yes, any polynomial identity involving $\cos(mx + c)$ and $\sin(mx+c)$ for various constants $c$ and integers $m$ can be written in the form
$R(z) = 0$ where $z = e^{ix}$ and $R$ is a rational function involving the $e^{ic}$. For this to be true, the numerator of $R(z)$ must simplify to the polynomial $0$.
For example, let's take
$$ \cos^2(x) + \cos^2(x+2\pi/3) + \cos^2(x+4\pi/3)=3/2 $$
Expressed in terms of $z = e^{ix}$, this becomes
$$ \frac{z^2}{4} + \frac{1}{2} + \frac{1}{4z^2} + \frac{z^2}{4} e^{4\pi i/3} + \frac{1}{2} + \frac{1}{4 z^2} e^{-4\pi i/3} + \frac{z^2}{4} e^{8\pi i/3} + \frac{1}{2} + \frac{1}{4 z^2} e^{-8\pi i/3} = \frac{3}{2} $$
which simplifies to
$$ \left(1 + e^{4\pi i/3} + e^{8\pi i/3}\right) \frac{z^2}{4} +
\left(1 + e^{-4\pi i/3} + e^{-8\pi i/3}\right) \frac{1}{4 z^2} = 0 $$
and that is true, as we verify by showing $$1 + e^{4\pi i/3} + e^{8\pi i/3} = 0$$
and $$ 1 + e^{-4\pi i/3} + e^{-8\pi i/3} = 0$$
Note that if $w = e^{4\pi i/3}$, the first is $1 + w + w^2 = (1-w^3)/(1-w)$, and
$w^3 = e^{4\pi i} = \left(e^{2\pi i}\right)^2 = 1$. Similarly for the second.
EDIT: For question 3, you basically want to know what polynomial identities are satisfied by the $e^{ic}$. If there is only one $c$, then $e^{ic}$ must be an algebraic number, and all polynomial identities it satisfies are multiples of its minimal polynomial. For example, if $c = 2 m \pi/n$
with $m$ and $n$ coprime, then the minimal polynomial is the cyclotomic polynomial $C_n(w)$. Things can be more complicated if there are several different $c$.
EDIT: For example, the $6$'th cyclotomic polynomial is $C_6(w) = w^2 - w + 1$, and its roots are $e^{2\pi i k/6}$. where $k$ and $6$ are coprime, i.e. $e^{\pi i/3}$ and $e^{- \pi i/3}$.
We might take $$(z+1/z)(w - 1 + 1/w) = z w + \frac{1}{zw} - z - \frac{1}{z} + \frac{z}{w} + \frac{w}{z}$$
which with $w = \exp(i\pi/3)$ and $z = \exp(ix)$ becomes
$$ 2 \cos(x+\pi/3) - 2 \cos(x) + 2 \cos(x-\pi/3) = 0 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does not yield the correct answer, yet evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}$$
gives the correct answer
Im at a loss here, why am I not getting the correct answer by finding the difference of the sum between the two bounds?
| Let the terms of the sum be $a_n$. You need to find:
$$a_{n+1}+a_{n+2}+\cdots+a_{2n}=\\
(a_1+\cdots+a_{n}+a_{n+1}+\cdots+a_{2n})-(a_1+\cdots+a_n)=\\
S_{2n}-S_n$$
In your first method, you are subtracting the term $a_{n+1}$ and losing it.
Addendum: Note the correct formula to use is:
$$S_n=\sum_{k=1}^n k(k+1)=\frac{2n^3+6n^2+4n}{6}$$
Now consider the difference:
$$\sum_{r=n+1}^{2n} r(r+1)=S_{2n}-S_n=\\
\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}=\\
\frac{7}{3}n^3+3n^2+\frac{2}{3}n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {xyz}$$ Cubing both sides and multiplying by $8$, $$\frac{8a^3}{27}>8xyz$$
Also, by $AM>GM$, $$(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)>8xyz$$
Now, how do I find the relation between $(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)$ and $\frac{8a^3}{27}$?
| You can proceed like this: $$(a-x)(a-y)(a-z) \leqslant \left( \dfrac{(a-x)+(a-y)+(a-z)}{3} \right) ^3 =\dfrac{8}{27} a^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To prove that (as I believe) a particular pair of diophantine equations has no solutions in positive integers. The diophantine equation $$a^2+b^2+ab=d^2$$ has positive integer solutions. If, in addition,$$ a^2+b^2=c^2$$ I suspect that it does not. Is there an easy way of seeing this?
| \begin{cases}
\begin{split}
a^2+b^2+ab&=d^2\\
a^2+b^2&=c^2\\
\end{split}
\end{cases}
Parametric of solution of second equation is given below.
$a = m^2-n^2$
$b = 2mn$
$c = m^2+n^2$
Substitute above ${a,b}$ to first equation, then we get
$$d^2 = m^4+2nm^3+2m^2n^2-2n^3m+n^4$$
Let $x=\frac{m}{n}, y=\frac{d}{n^2}$, the we get
$$y^2 = x^4+2x^3+2x^2-2x+1$$
Above quartic equation can be transformed to elliptic curve below.
$Y^2 = X^3-X^2-9X+9$ with
$x = \frac{2X}{Y+X-3}, y = \frac{6Y-18+X^3+9X}{(Y+X-3)^2}.$
We can get rational points of above elliptic curve using Magma calculator below.
For example, we get $(X,Y)=(\frac{13}{4}, \frac{15}{8}).$
From $x = \frac{2X}{Y+X-3}$ and $x=\frac{m}{n}$, we get $(m,n)=(52,17).$
$(m,n)=(52,17)$ gives positive solution $(a,b,c,d)=(2415, 1768, 2993, 3637).$
Thus there is a positive solution $(a,b,c,d).$
If $n>m$ or $mn<0$ there is no positive solution $(a,b,c,d).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756405",
"timestamp": "2023-03-29T00:00:00",
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find function $f$ such that $f(x)=xf(x-1)$ and $f(1) = 1$
Find function $f$ such that $f(x)=xf(x-1)$ and $f(1) = 1$.
I can prove that there is just one function as $f$ (see Proof1).
I know that there exists a pi function $\Pi(z) = \int_0^\infty e^{-t} t^z\, dt$ that fits as $f$ so it is the only solution.
My problem is, although I know the answer to be $\Pi(z)$ I can not reach it myself. I mean how can I find $f$ without knowing the answer before? how can I reach $\Pi(z)$ from the equation $f(x)=xf(x-1)$ and $f(1) = 1$?
Proof1
suppose $f(x)$ and $g(x)$ are two solutions to $f(x)=xf(x-1)$ and $f(1) = 1$
$$
\frac{f(x)}{g(x)} = h(x) \Rightarrow \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g^2(x)} = h'(x) \\
\Rightarrow \frac{f(x-1) + xf'(x-1)}{g(x)} - \frac{f(x)}{g(x)}\frac{g(x-1) + xg'(x-1)}{g(x)} = h'(x) \\
\Rightarrow \frac{f(x-1)}{g(x)}+\frac{xf'(x-1)}{g(x)} - \frac{f(x)}{g(x)}(\frac{g(x-1)}{g(x)}+\frac{ xg'(x-1)}{g(x)}) = h'(x) \\
\Rightarrow \frac{1}{x}\frac{f(x)}{g(x)}+\frac{xf'(x-1)}{xg(x-1)} - \frac{f(x)}{g(x)}(\frac{1}{x}+\frac{xg'(x-1)}{xg(x-1)}) = h'(x) \\
\Rightarrow \left [\frac{1}{x}\frac{f(x)}{g(x)} - \frac{f(x)}{g(x)}\frac{1}{x}\right ] + \left [\frac{xf'(x-1)}{xg(x-1)} - \frac{f(x)}{g(x)}\frac{xg'(x-1)}{xg(x-1)}\right ] = h'(x) \\
\Rightarrow 0 + \left [\frac{\not{x}f'(x-1)}{\not{x}g(x-1)}\frac{g(x-1)}{g(x-1)} - \frac{\not{x}f(x-1)}{\not{x}g(x-1)}\frac{\not{x}g'(x-1)}{\not{x}g(x-1)}\right ] = h'(x) \\
\Rightarrow \frac{f'(x-1)g(x-1) - g'(x-1)f(x-1)}{g^2(x-1)} = h'(x) \\
\Rightarrow h'(x-1) = h'(x) \\
$$
now we know that $h(x)$ must be a line. But $h(x) = h(x+n)$ $^{h(x+1) = \frac{f(x+1)}{g(x+1)} = \frac{{x +1}f(x)}{{x +1}g(x)} = \frac{f(x)}{g(x)} = h(x)}$. So $h(x)$ must be a horizontal line so it is a constant. so:
$$
\frac{f(x)}{g(x)} = c, \frac{f(1)}{g(1)} = 1 \Rightarrow \frac{f(x)}{g(x)} = 1 \Rightarrow f(x) = g(x)
$$
| No, it is not true. Given any solution $f$ and any function $g$ that is periodic with period $1$ (i.e. $g(x+1) = g(x)$) and $g(1)=1$, then $h(x) = f(x) g(x)$ also satisfies your equations. For example, you could take $g(x) = 1 + \sin(2\pi x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real root of $r(x) = x^3+12x^2+8x+1,$ I was able to prove that all roots of the original sextic can be expressed in radicals. The process is as follows:
*
*Divide $p(x+iy)$ by $r(x)$ to get $\frac{1}{8}x^3 + \frac{3}{16}x^2 + x\left(\frac{7}{32}-\frac{15y^2}{8}\right) + \left(\frac{95}{32}-\frac{15y^2}{16}\right) + \frac{R(x,y)}{p(x)}$ where $R(x,y) = A(y)x^2 + B(y)x + C(y)$ and $A(y) = 15y^4 - \frac{15y^2}{4} - \frac{201}{16}, B(y) = 15y^8 - 30y^6 + 12y^4 + \frac{75y^2}{8} - \frac{767}{32}, C(y) = -y^6+5y^4-\frac{193y^2}{16}-\frac{63}{32}.$
*The equation $R(x_0, y_0) = 0$ is a quartic in $y_0^2,$ which we can solve exactly to obtain $y_0^2$ and hence $y_0.$
*Polynomial division reduces $p(x)$ to a quartic, and now we apply the quartic formula again to find the other $4$ roots.
However, I don't want to perform the rest of the computations. Is there a cleaner way to use the observation that $r(x_0) = 0,$ perhaps in the realm of abstract algebra?
| $$\frac{x^9+6x^6+31x^3-1}{x^3-3x^2+4x-1}$$
| {
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Prove that $\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$ for all $n \ge 1.$ The sequence $a_n$ is defined by $a_1 = \frac{1}{2}$ and $a_n = a_{n - 1}^2 + a_{n - 1}$ for $n \ge 2.$
Prove that $\frac{1}{a_1 + 1} + \frac{1}{a_2 + 1} + \dots + \frac{1}{a_n + 1} < 2$ for all $n \ge 1.$
| You have
$$\frac{a_k}{a_{k-1}}=a_{k-1}+1$$ and
$$a_k - a_{k-1}=a_{k-1}^2$$
Hence,
$$\sum_{k=1}^n\frac 1{a_k+1}= \sum_{k=1}^n\frac{a_{k}}{a_{k+1}}= \sum_{k=1}^n\frac{a_{k}^2}{a_{k}a_{k+1}}$$
$$= \sum_{k=1}^n\frac{a_{k+1}-a_k}{a_{k}a_{k+1}}=\sum_{k=1}^n\left(\frac{1}{a_{k}} - \frac 1{a_{k+1}}\right)$$
I leave the last step up to you.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-n\pi i-\frac{\pi i}{2}+\frac{\pi i}{4}}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-i\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}+\frac{1}{\cos\left(n\pi +\frac{\pi}{4}\right)+i\sin\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\sqrt{2}}\frac{1}{\frac{1}{\sqrt{2}}(\cos n\pi -\sin n\pi )+\frac{i}{\sqrt{2}}(\cos n\pi +\sin n\pi)}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\cos n\pi +i\cos n\pi}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{i^{2n}+i^{2n+1}}-i^{n^2}\end{align}$$
How should I proceed?
| Let $f=\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}$. Then
$$\begin{align}f&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-(2n+1)\frac{\pi i}{2}+\frac{\pi i}{4}}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-i\left(n\pi +\frac{\pi}{2}-\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\end{align}$$
and
$$\begin{align}\operatorname{Re}f&=\frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}+\cos\left(n\pi +\frac{\pi}{4}\right)\right)-\cos\frac{n^2\pi}{2}\\&=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}}{2}+\frac{1}{\sqrt{2}}\left(\cos n\pi -\sin n\pi \right)\right)-\cos \frac{n^2\pi}{2}\\&=\frac{1}{2}+\frac{1}{2}\cos n\pi -\cos\frac{n^2\pi}{2}\\&=\frac{1+(-1)^n}{2}-\cos \frac{n^2\pi}{2}.\end{align}$$
Since $\frac{1+(-1)^n}{2}=0$ for odd $n$ and $\frac{1+(-1)^n}{2}=1$ for even $n$ and $\cos\frac{n^2\pi}{2}=0$ for odd $n$ and $\cos\frac{n^2\pi}{2}=1$ for even $n$, $\operatorname{Re}f=0$ for $n\in\mathbb{N}$.
Similarly for $\operatorname{Im}f$:
$$\operatorname{Im}f=\frac{1-(-1)^n}{2}-\sin\frac{n^2\pi}{2}.$$
Since $\frac{1-(-1)^n}{2}=1$ for odd $n$ and $\frac{1-(-1)^n}{2}=0$ for even $n$ and $\sin\frac{n^2\pi}{2}=1$ for odd $n$ and $\sin\frac{n^2\pi}{2}=0$ for even $n$, $\operatorname{Im}f=0$ for $n\in\mathbb{N}$. Therefore $f=0$ for all $n\in\mathbb{N}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Rozenberg, but I couldn't find a solution to it by myself or on this site, so decided to ask.
I tried my best with fallacy-
$$4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \Rightarrow 4 \ge a+b+c+d$$ with the constraint given. Proceeding-
$$(a+b+c+d)\left({a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a}\right) \ge (a+b+c+d)^2 $$
$$\Rightarrow {a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge a+b+c+d$$
That means, I need $a+b+c+d\ge 4$ to complete the proof but instead got $4 \ge a+b+c+d$ !
$2.$Let $a, b, c$ be positive real numbers such that $abc = 1$.
Show that
$${1\over b(a+b)}+{1\over c(b+c)}+{1\over a(c+a)}\ge \frac{3}{2}.$$
$3.$If a, b, c and d are positive real numbers such that $a + b + c + d = 4$. Prove that
$$ {a \over 1+b^2c}+{b \over 1+c^2d}+{c \over 1+d^2a}+{d \over 1+a^2b} \ge 2. $$
The #2 is credited to the Zhautykov Olympiad 2008 and there is no need for revealing my attempts as I've no idea what to do, Lastly,
This is a doubt, not an question that definitely has an answer, but if the doubt comes out to have an answer, nevertheless, this is a problem.
$4.$ Prove without Induction:
$$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} + ...+\sqrt{a_n^2+b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$
| The third problem.
By C-S and AM-GM twice we obtain:
$$\sum_{cyc}\frac{d}{1+a^2b}=\sum_{cyc}\frac{d^2}{d+a^2bd}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(a+a^2bd)}=$$
$$=\frac{16}{4+ab(ad+bc)+cd(bc+ad)}=\frac{16}{4+(ab+cd)(ad+bc)}\geq$$
$$\geq\frac{16}{4+\left(\frac{ab+cd+ad+bc}{2}\right)^2}=\frac{16}{4+\left(\frac{(a+c)(b+d)}{2}\right)^2}\geq\frac{16}{4+\left(\frac{\left(\frac{a+c+b+d}{2}\right)^2}{2}\right)^2}=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a diagonal matrix and a symmetric positive semidefinite matrix Let the following symmetric matrix
$A = \left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\
-a_{12} & a_{12}+a_{23} & -a_{23} \\
-a_{13} & -a_{23} & a_{13}+a_{23} } \right]$,
where each $a_{ij}$ is a nonnegative real number, and hence, $A$ is a symmetric positive semidefinite matrix, because:
$
x^T A x = (a_{12}+a_{13}) x_{1}^2 +
(a_{12}+a_{23}) x_{2}^2 +
(a_{13}+a_{23}) x_{3}^2
- 2( a_{12} x_1 x_2 + a_{13} x_1 x_3 + a_{23} x_2 x_3 ) \\
= a_{12}(x_1 - x_2)^2 + a_{13}(x_1 - x_3)^2 + a_{23}(x_2 - x_3)^2 \geq 0 $
On the other hand:
Let $B = A + D =
\left[ \matrix{a_{12}+a_{13} & -a_{12} & -a_{13} \\
-a_{12} & a_{12}+a_{23} & -a_{23} \\
-a_{13} & -a_{23} & a_{13}+a_{23} } \right] +
\left[ \matrix{d_{11} & & \\
& d_{22} & \\
& & d_{33} } \right]$,
where $d_{ii} \in \mathbb{R}$, that is, $D$ is a diagonal matrix with positive and negative real numbers.
What are the limits of $d_{ii}$ so that B is still a positive semidefinite matrix?
| Note that $A$ is positive semidefinite is not positive definite. In other words, $A$ is positive semidefinite and singular.
First, note that if all three $d_{ii}$ are positive, then $A + D$ will necessarily fail to be positive semidefinite.
So, we must suppose that at least on of these is zero. Without loss of generality, suppose that $d_{33} = 0$. If both $d_{11},d_{22}$ are non-zero, then we find that $A + D$ can only be positive semidefinite if $A$ is diagonal with $a_{33} = 0$. In this case, this only occurs when $A = 0$.
For the remaining case, we must suppose that one of $d_{11},d_{22}$ are zero. Without loss of generality, suppose that $d_{22} = 0$. In other words, we are looking for the maximal value of $d_{11}$ for which
$$
A - \pmatrix{d_{11} & 0 & 0\\0 &0 &0\\0 &0&0}
$$
is positive semidefinite. As is shown here, for instance, this matrix will be positive definite for $d_{11} \leq 1/A^+_{11}$, where $A^+_{11}$ denotes the $1,1$ entry of the Moore-Penrose pseudoinverse $A^+$ of $A$.
As we established above, it is sufficient to consider the case in which at most one of the $d_{ii}$ are non-zero, since $A + D$ will fail to be positive semidefinite otherwise. We compute (via matlab)
$$
A^+ = \left(\begin{array}{ccc} \frac{4\,b_{1}+b_{2}+b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{1}+2\,b_{2}-b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{1}-b_{2}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)}\\ -\frac{2\,b_{1}+2\,b_{2}-b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & \frac{b_{1}+4\,b_{2}+b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{2}-b_{1}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)}\\ -\frac{2\,b_{1}-b_{2}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & -\frac{2\,b_{2}-b_{1}+2\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} & \frac{b_{1}+b_{2}+4\,b_{3}}{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)} \end{array}\right)
\\ =
\frac{1}{9(b_1b_2 + b_1 b_3 + b_2 b_3)}\sum_{i=1}^3b_i\,(-2e_i + e_{i+1} + e_{i+2})(-2e_i + e_{i+1} + e_{i+2})^T,
$$
where in the above we take $b_{1} = a_{23}, b_2 = a_{13},b_3 = a_{12}$. In the second line, $e_i$ denotes the $i$th standard basis vector and addition in the indices is modulo $3$.
We see then, for instance, that the maximal $d_{11}$ for which $A + D$ is positive semidefinite is given by
$$
d_{11} = \frac{9\,\left(b_{1}\,b_{2}+b_{1}\,b_{3}+b_{2}\,b_{3}\right)}{4\,b_{1}+b_{2}+b_{3}}.
$$
Similar conditions hold for $d_{22},d_{33}$.
| {
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$6$-digit permutation problem How many $6$-digit different numbers greater than $400000$ can be formed from the digits $2$, $2$, $5$, $6$, $7$, $7$, $8$?
Note: $22$ and $77$ are allowed in the number. For instance, $622775$ is allowed.
My approach: $5$ ways to choose the first digit, $6$ ways to choose the second, $5$ ways to choose the third, $4$ ways to choose the fourth, $3$ ways to choose the fifth and $2$ ways to choose the sixth. So
the total arrangement is $5\times 6\times 5\times 4\times 3\times 2 = 3600$. But with the $2,2, 7, 7$ coming up, I think there is more do be done with it. I tried dividing $3600$ by $(2! \times 2!)$. I tried this with a small number of digits and it gave the wrong answer. I am stuck. Can anyone help me with the answer or hint. Thanks
| Since the number must exceed $40000000$, the leading digit may be $5$, $6$, $7$, or $8$. Of these digits, only $7$ may be repeated. Therefore, it makes sense to handle the case in which the leading digit is $7$ separately from the others.
Since there are a total of seven available digits, six of them must be used. Since those digits are $2, 2, 5, 6, 7, 7, 8$, there must be at least one repeated digit, but both $2$ and $7$ may appear twice.
The leading digit is $5$, $6$, or $8$: The leading digit can be chosen in three ways.
Both $2$ and $7$ appear twice: Select the leading digit from among $5$, $6$, or $8$. If both $2$ and $7$ appear twice, then we must choose two of the remaining five positions for the $2$s and two of the remaining three positions for the $7$s. We then have two choices for the final open position, namely one of the two numbers from among $5$, $6$, or $8$ which we did not choose as the leading digit. Hence, there are
$$\binom{3}{1}\binom{5}{2}\binom{3}{2}\binom{2}{1}$$
such numbers.
Exactly one of $2$ or $7$ appears twice: Select the leading digit from among $5$, $6$, or $8$. Choose whether $2$ or $7$ appears twice. Choose which two of the remaining five positions are occupied by that number. Each of the remaining three positions must be filled with one of the three distinct digits that remain, which can be done in $3!$ ways. For example, if the leading digit is $5$ and $2$ appears twice, the remaining digits are $6$, $7$, $7$, and $8$. However, since only $2$ appears twice, $7$ can only appear once. Thus, the final three positions would have to be filled with $6$, $7$, and $8$. Hence, there are
$$\binom{3}{1}\binom{2}{1}\binom{5}{2}3!$$
such numbers.
The leading digit is $7$: The remaining digits are $2, 2, 5, 6, 7, 8$. Therefore, $2$ appears either once or twice.
The digit $2$ appears twice: If $2$ appears twice, we must choose two of the five remaining positions for the $2$s. We then choose which three of the remaining four distinct digits $5, 6, 7, 8$ appear in the remaining three positions, and arrange them in those positions. There are
$$\binom{5}{2}\binom{4}{3}3!$$
such numbers.
The digit $2$ appears once: If $2$ appears once, we must fill the remaining five positions with the distinct digits $2, 5, 6, 7, 8$, which can be done in $5!$ ways.
Total: Since the cases described above are mutually exclusive and exhaustive, the number of six-digit numbers greater than $400000$ which can be formed using the digits $2, 2, 5, 6, 7, 7, 8$ is
$$\binom{3}{1}\binom{5}{2}\binom{3}{2}\binom{2}{1} + \binom{3}{1}\binom{2}{1}\binom{5}{2}3! + \binom{5}{2}\binom{4}{3}3! + 5! = 900$$
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} |
Prove that $\binom{n}{0}\binom{n+1}{n} +\binom{n}{1}\binom{n}{n-1} +\binom{n}{2}\binom{n-1}{n-2} +\cdots +\binom{n}{n}\binom{1}{0} = 2^{n-1}(n+2)$
Prove the below:
$$\binom{n}{0}\cdot\binom{n+1}{n} +\binom{n}{1}\cdot\binom{n}{n-1} +\binom{n}{2}\cdot\binom{n-1}{n-2} +\cdots +\binom{n}{n}\cdot\binom{1}{0} = 2^{n-1}\cdot(n+2)$$
Attempt:
Consider $$(1+x)^{n} = \binom{n}{0} + \binom{n} {1}x + \binom{n} {2}x^² + \cdots + \binom{n} {n} x^n$$
The series in the question is the coefficient of $x^n$ in
$$\binom{n} {0}(1+x)^{n+1}+\binom{n} {1}x(1+x)^{n} +\binom{n} {2}x^2(1+x)^{n-1 }+ \cdots + \binom {n} {n} x^n (1+x)$$
Knowing the above, how can I rewrite the coefficient of $x^n$, as that is usually the key to solving such problems?
| Hint:
$$\binom nr\cdot\binom{n+1-r}{n-r}=\binom nr(n+1-r)=(n+1)\binom nr-r\binom nr$$
For $r\ge1,$ $$r\binom nr=n\binom{n-1}{r-1}$$
Now put $m=n,n-1$ one after another in $$(1+1)^m=\sum_{r=0}^m\binom mr$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward .
Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it .
Can anyone help me ? Some hints or suggestions to this problem will be appreciated !!
| After square completion, there is a simple solution just using the triangle inequality for the Euklidean distance:
Using $(x-1)^2 + (y+1)^2 =4$ we get
$$x^2+y^2= \left(\left|\binom xy\right|\right)^2 \stackrel{\triangle-ineq.}{\leq} \left(\left|\binom{x-1}{y+1}\right| + \left|\binom{-1}{1}\right|\right)^2= (2+\sqrt 2)^2=6+4\sqrt2$$
Now, just add $\sqrt{32}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Interpreting $a^2b \div \frac13a^2b^3$. The maths book I'm using shows:
$$ a^2b \div \frac13a^2b^3 $$
Which would be something like:
$$ a^2 \cdot b ÷ \frac13 \cdot a^2 \cdot b^3 $$
My understanding of order of operations it would equal:
$$ 3a^4b^4 $$
However in the book the second term is evaluated before the division (switching it to multiplication) and equaling:
$$ a^2b \cdot \frac{3}{a^2b^3} = \frac{3a^2b}{a^2b^3} = \frac{3}{b^2} $$
If a multiplication or division signs shows between terms, should that be taken as multiplying or dividing one term by the other?
| In this form
$$a^2b \div \frac13a^2b^3$$
my interpretation is to consider the two terms as a whole, that is
$$\left(a^2b\right) \div \left(\frac13a^2b^3\right)=\frac{3a^2b}{a^2b^3}=\frac 3 {b^2}$$
Otherwise, in this form
$$a^2 \times b ÷ \frac13 \times a^2 \times b^3$$
whithout parenthesis my interpretation would be
$$a^2 \times \left(b ÷ \frac13\right) \times a^2 \times b^3=3a^4b^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Inequality with a High Degree Constraint This question-
Suppose that $x, y, z$ are positive real numbers and $x^5 + y^5 + z^5 = 3$. Prove that $$ {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3 $$
The inequality has a high degree constraint which can convert a $5$-degree polynomial to a $0$-degree term and makes it difficult.
On trying C-S to manage-
$$ \left({x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3}\right)\left(x^5 + y^5 + z^5\right) \ge 9 \Rightarrow
\left(x^2y+y^2z+z^2x\right)^2\geq9 \Rightarrow x^2y+y^2z+z^2x\geq3 $$Still gives a third degree inequality and not a useful fifth degree.
How can I do it and solve the problem?
| A solution due to AoPS user @realquarterb:
By AM-GM, we have $1 + 19 x^{100/19} \ge 20 x^5$.
By AM-GM, we have $10\frac{x^4}{y^3} + 3x^{10} + 6x^5y^5 \ge 19x^{100/19} \ge 20x^5 - 1$.
Thus, we have $10 \sum_{\mathrm{cyc}} \frac{x^4}{y^3} + 3(x^5+y^5+z^5)^2 \ge 20(x^5+y^5+z^5) - 3$. (Q. E. D.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Show that the span of the $\{x_1, x_2, x_3, x_4 \}$ is $\{(a, b,c, d): 2a - b = 0, 2a - 3c - d = 0\}$. Show that the span of the $\{x_1, x_2, x_3, x_4 \}$ is $\{(x, y,z, w): 2x - y = 0, 2x - 3z - w = 0\}$ where $x_1 = (1,2,1, -1), x_2 = (2,4,1,1), x_3 = (-1,-2,-2,-4) \text{ and } x_4 = (3,6,2,0)$.
My approach:
We have to solve for $c = (c_1, c_2, c_3, c_4)$ in the following system.
$$\vec x = c_1\vec v_1 + c_2\vec v_2 + c_3\vec v_3 + c_4\vec v_4$$
$$\left[\begin{array}{cccc|c}
1 & 2 & -1 & 3& x\\
2 & 4 & -2 & 6 & y\\
1 & 1 & -2 & 2 & z\\
-1 & 1 & -4 & 0 & w\\
\end{array}\right]$$
$$\left[\begin{array}{cccc|c}
1 & 2 & -1 & 3& x\\
0 & 0 & 0 & 0 & 2x-y\\
0 & 1 & 1 & 1 & x-z\\
0 & 0 & 1 & 0 & {2x -3z - w \over 8}\\
\end{array}\right]$$
In order to have a solution $2x-y = 0$.
Hence according to me, the span should be $\{(x, y,z, w): 2x - y = 0\}$
I don't get why the other condition ie. $2x - 3z - w = 0$ is part of the span. Could someone please explain to me why it has to be included.
| I suspect it should be $x_3 = (-1,-2,-2,4)$. Then the correct reduced matrix is
$$\left[\begin{array}{cccc|c}
1 & 2 & -1 & 3& x\\
2 & 4 & -2 & 6 & y\\
1 & 1 & -2 & 2 & z\\
-1 & 1 & 4 & 0 & w\\
\end{array}\right] \sim \left[\begin{array}{cccc|c}
1 & 2 & -1 & 3& x\\
0 & 0 & 0 & 0 & 2x-y\\
0 & 1 & 1 & 1 & x-z\\
0 & 0 &0 & 0 & {2x -3z - w}\\
\end{array}\right]$$
so a solution exists if and only if
$$2x-y=0, \quad 3x-3z-w=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $S$ is the unit square with opposite corners $(0, 0)$ and $(1, 1)$, show $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2})$ Let $S$ be the unit square with opposite corners $(0, 0)$ and $(1, 1)$. Show $$\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}=\log(1+\sqrt{2}).$$
What I tried:
Since $\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}$ converges in $S$. So the integral is convergent independent of choice of ${D_n}$, the let $D_n$ be square with opposite corners $(1/n, 1/n)$ and $(1, 1)$.
Then
\begin{align}
\iint_S \frac{y^2}{\sqrt{(x^2+y^2)^3}}&=\int^1_{1/n}\int^1_{1/n}\frac{y^2}{\sqrt{(x^2+y^2)^3}}
\\&=\int^1_{1/n}y^2dy\int^1_{1/n}\frac{1}{(x^2+y^2)^{2/3}}dx
\\&=\int^1_{1/n}y^2\frac{x}{y^2\sqrt{y^2+x^2}}|^1_{1/n}dy
\\&=\log(y+\sqrt{1+y^2}|^1_{1/n}-\frac{1}{n}\log(y+\sqrt{y^2+(1/n)^2})|^1_{1/n}
\\&=\log(1+\sqrt{2})-0-0+\frac{1}{n}\log\left(\frac{1}{n}+\sqrt{\frac{1}{n}^2+\frac{1}{n}^2}\right)
\end{align}
But $\frac{1}{n}\log(\frac{1}{n}+\sqrt{\frac{1}{n}^2+\frac{1}{n}^2})$ seem to go to infinity.
| Under the variable interchange $x\leftrightarrow y$ we have that
$$I = \iint\limits_{[0,1]^2} \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}dA = \iint\limits_{[0,1]^2} \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}dA$$
and adding the two integrals gets us
$$2I = \iint\limits_{[0,1]^2} \frac{x^2+y^2}{(x^2+y^2)^{\frac{3}{2}}}dA \implies I = \frac{1}{2} \iint\limits_{[0,1]^2} \frac{1}{\sqrt{x^2+y^2}}dA$$
By symmetry we can do the integral in polar coordinates
$$I = 2\cdot\frac{1}{2}\int_0^1 \int_0^x \frac{1}{\sqrt{x^2+y^2}}dA = \int_0^{\frac{\pi}{4}}\int_0^{\sec\theta}\:dr\:d\theta$$
$$ = \log|\sec\theta+\tan\theta|\Biggr|_0^{\frac{\pi}{4}} = \log(1+\sqrt{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $A_n A_{n+1}=A_{n}+2 A_{n+1}$ to disagree with a question Solving $$A_n A_{n+1}=A_{n}+2 A_{n+1} \tag1$$ to disagree with a question attached below:
Let us re-write *1) as
$$A_{n+1}(A_n-2)=A_n \tag 2$$
Let $A_n-2=B_n/B_{n-1}$ in (2) to simplify and to get
$$B_{n+1}-2B_{n-1}+B_n=0 \tag3$$
Let $B_n=x^n$, we get $x^2+x-2=0 \implies x=1,-2.$
We get $$B_n=P +Q (-2)^n \tag4$$. Finally we get
$$A_n=\frac{P+Q(-2)^n}{P+Q(-2)^{n-1}}+2=\frac{3P}{P+Q(-2)^{n-1}}\tag5$$
The solution of (1) given by (5) is algebraic of exponential but not not periodic.
Can one justify the attached question containing the recurrence relation (1)?
Edit:
Let $Q/P=R$, to write $$A_n=\frac{3}{1+R(-2)^{n-1}}.$$
| This is a Ricatti recurrence:
$\begin{align*}
w_{n + 1}
= \frac{a w_n + b}{c w_n + d}
\end{align*}$
with $c \ne 0$ and $a d - b c \ne 0$. There are several ways to solve them. Brand "A Sequence Defined by a Recurrence Relation", AMM 62:7 (1955), pp 489-492 goes as follows.
Define:
$\begin{align*}
y_{n + 1}
&= \alpha - \frac{\beta}{y_n} \\
\alpha
&= a + d \\
\beta
&= a d - bc
\end{align*}$
Replacing $y_n = x_{n + 1} / x_n$ gives now:
$\begin{align*}
x_{n + 2} - \alpha x_{n + 1} + \beta x_n
&= 0
\end{align*}$
We need two starting values, pick $x_0 = 1$ for convenience giving $x_1 = y_0$, and you are set.
Another road is to recognize the recurrence as a Möbius tranformation:
$\begin{align*}
w_{n + 1}
&= \frac{a w_n + b}{c w_n + d}
\end{align*}$
It turns out those compose just like $2 \times 2$ matrices multiply, so if you define:
$\begin{align*}
M
&= \pmatrix{a & b \\
c & d} \\
M^n
&= \pmatrix{a^{(n)} & b^{(n]} \\
c^{(n)} & d^{(n)}}
\end{align*}$
then:
$\begin{align*}
w_n
&= \frac{a^{(n)} w_0 + b^{(n)}}{d^{(n)} w_0 + d^{(n)}}
\end{align*}$
Yet another way is given by Mitchell "An Analytic Ricatti Solution for Two-Target Discrete-Time Control", Journal of Economic Dynamics and Control 24:4 (2000), pp 615-622.
Define the auxilliary sequence:
$\begin{align*}
x_n
&= \frac{1}{1 + \eta w_n}
\end{align*}$
to get:
$\begin{align*}
x_{n + 1}
&= \frac{(d \eta - c) x_n + c}
{b \eta^2 - (a - d) \eta - c) x_n + a \eta + c}
\end{align*}$
Picking $\eta$ so that $b \eta^2 - (a - d) \eta - c = 0$, this is a linear recurrence. Bonus is that it is first order, so it can be solved even if the coefficients aren't constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x+\frac{1}{y+\frac{1}{z+\frac13}}=\frac{118}{51}$ find $xyz$. I'm having a difficult time figuring out how to solve this algebra problem.
If $$x+\cfrac 1 {y+\cfrac 1 {z+\cfrac 1 3}}=\frac{118}{51}$$
find $xyz$.
| I will presume that $x,y,z$ are intended to be positive integers.
\begin{align}
\frac{118}{51} & = 2 + \frac{16}{51} = 2 + \frac 1 {\left( \frac{51}{16} \right)} = 2 + \cfrac 1 {3 + \cfrac 3 {16}} \\[15pt]
& = 2 + \cfrac 1 {3 + \cfrac 1 {\left( \frac{16} 3 \right)}} = 2 + \cfrac 1 {3 + \cfrac 1 {5 + \cfrac 1 3}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Challenging limit: $\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$ Here is a challenging limit proposed by a friend:
$$\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$$
and he claims that the closed form for this limit is really pleasant.
I am not good at limits so I am not going to show any work and just leave it to those who find it interesting.
Addendum: A similar problem proposed by the same person:
$$\lim_{\alpha\to0^{+}}\left(\frac{2}{3\alpha^3}-\frac{4\pi}{3\alpha}\int_1^\infty\frac{x\cosh(\pi\alpha x)}{\sinh^2(\pi\alpha x)\sqrt{x^2-1}}dx\right)$$
| Let $f \colon (0,\infty) \to \mathbb{R},$
\begin{align}
f(\alpha) &= \frac{1}{2\alpha} - \int \limits_1^\infty \frac{\mathrm{d} x}{\sinh(\pi \alpha x) \sqrt{x^2 -1}} \stackrel{x = \cosh(t)}{=} \frac{1}{2 \alpha} - \int \limits_0^\infty \frac{\mathrm{d} t}{\sinh(\pi \alpha \cosh(t))} \\
&= \frac{1}{2 \alpha} - \int \limits_0^\infty \left[\frac{1}{\pi \alpha \cosh(t)} + 2 \pi \alpha \cosh(t) \sum \limits_{k=1}^\infty \frac{(-1)^k}{\pi^2 k^2 + \pi^2 \alpha^2 \cosh^2(t)}\right] \mathrm{d} t \\
&\!\!\!\!\!\!\!\!\stackrel{u = \alpha \sinh(t)}{=} \frac{2}{\pi} \int \limits_0^\infty \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^2 + \alpha^2 + u^2} \, \mathrm{d} u \, .
\end{align}
Here we have used the pole expansion of $\operatorname{csch}$ and the elementary integral $\int_0^\infty \operatorname{sech}(t) \, \mathrm{d} t = \frac{\pi}{2}$. Since the partial sums of the remaining alternating series (with terms decreasing in absolute value) are bounded by the first term, i. e. the integrable function $u \mapsto \frac{1}{1 + \alpha^2 + u^2}$ , the dominated convergence theorem allows us to interchange summation and integration. We obtain
$$ f(\alpha) = \frac{2}{\pi} \sum \limits_{k=1}^\infty (-1)^{k-1}\int \limits_0^\infty \frac{\mathrm{d} u}{k^2 + \alpha^2 + u^2} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + \alpha^2}} $$
for $\alpha > 0$. The series on the right-hand side converges uniformly on $\mathbb{R}$ (see this question), so it defines a continuous function of $\alpha$ on $\mathbb{R}$. In particular, as predicted by Claude Leibovici in the comments,
$$ \lim_{\alpha \to 0^+} f(\alpha) = \lim_{\alpha \to 0^+} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + \alpha^2}} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + 0^2}} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k} = \log(2)$$
holds. The additional problem can be solved by noting that
\begin{align} \frac{2}{3 \alpha^3} - \frac{4\pi}{3 \alpha} \int \limits_1^\infty \frac{x \cosh(\pi \alpha x)}{\sinh^2(\pi \alpha x) \sqrt{x^2-1}} \, \mathrm{d} x &= - \frac{4}{3 \alpha} f'(\alpha) = \frac{4}{3} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{(k^2 + \alpha^2)^{3/2}} \\
&\!\!\!\stackrel{\alpha \rightarrow 0^+}{\longrightarrow} \frac{4}{3} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^3} = \frac{4}{3} \operatorname{\eta}(3) = \operatorname{\zeta}(3) \, .
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3783000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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An uncertainty for $a^3+b^3+c^3-3abc$ I have a doubt regarding $a^3+b^3+c^3-3abc$.
For factoring, it is easy that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$ as $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
But on using AM-GM inequality, we see-
$$ {a^3+b^3+c^3\over 3}\geq abc \Rightarrow a^3+b^3+c^3\ge3abc $$
AM-GM inequality ensures that equality holds if and only if all variables are equal.
So, equality holds if and only if $a=b=c$, which is trivial.
But we see by factoring that equality holds also if $a+b+c=0$. As the inequality is nothing more than a mere AM-GM, so equality should hold where the inequality ensures us it holds. But it also holds for $a+b+c=0$.
How is it possible and if it is, how can I find such equality cases?
| When $a+b+c=0$, then $a,b,c$ all of $a,b,c$ are not positive, So AM_GM will not be used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3786879",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Use infinite series to evaluate $\lim_{x \rightarrow \infty} (x^3 - 5x^2 + 1)^{\frac{1}{3}} - x$ Use infinite series to evaluate $\displaystyle\lim_{x \rightarrow \infty} (x^3 - 5x^2 + 1)^{\frac{1}{3}} - x$
I know the limit is $-\dfrac{5}{3}$ after rationalizing the expression, but I don't know how to prove it using Taylor series. Could someone give me any hints? I prefer hints to complete solutions.
| Taylor series ... Take
$$
(x^3-5x^2+1)^{1/3} = x\left(1-\frac{5}{x} + \frac{1}{x^3}\right)^{1/3}
$$
Taylor series for $1-5z+z^3$ near $z=0$ is $1 - \frac{5}{3}z + o(z)$ as $z \to 0$. So as $x \to \infty$,
\begin{align}
(x^3-5x^2+1)^{1/3} - x &= x\left(1-\frac{5}{x} + \frac{1}{x^3}\right)^{1/3} - x
\\ &= x\left(1 - \frac{5}{3x} + o(1/x)\right) - x
= -\frac{5}{3} + o(1) \to -\frac{5}{3} .
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Obtain sum of a sequence from sum of its odd terms. I would like to compute the sum
$$
\sum_{k=1}^\infty \frac{1}{k^4}
$$
by using the Fourier series of $f(x)=|x|$ over $(-\pi,\pi)$. Coefficients $b_k$ are all $0$ because $f$ is even. Doing the integration stuff, I obtained:
$$
a_0 = \pi
$$
and
$$
a_k = \frac{2}{k^2}\bigg((-1)^k-1\bigg)
$$
for $k>0$. The Parseval's equality gives:
$$
\frac{a_0^2}{2} + \sum_{k=1}^\infty (a_k^2+b_k^2)= \frac{1}{\pi}\int_{-\pi}^{\pi}f^2dx
$$
which gives
$$
\frac{\pi^2}{2} + \sum_{k=1}^\infty \frac{4}{\pi^2k^4}(2-2(-1)^k) = \frac{2}{3}\pi^2
$$
which simplifies to
$$
\sum_{k=1}^\infty \frac{1}{k^4} - \sum_{k=1}^\infty \frac{(-1)^k}{k^4} = \frac{\pi^4}{48}
$$
which basically says:
$$
\sum_{k=0}^\infty \frac{1}{(2k+1)^4}=\frac{\pi^4}{96}
$$
any idea how to obtain the sum from there?
| Observe that what you have is that $2\sum_{k=0}^{\infty} \frac 1{(2k+1)^4}=\frac {\pi^4}{48}$. Calling $\sum_{k=0}^{\infty} \frac 1{k^4}=S$ you have that $\sum_{k=0}^{\infty} \frac 1{(2k)^4}=\frac 1{16} S$ and finally you have $S-\frac 1{16}S=\frac 12 \frac {\pi^4}{48}$ from which $S=\frac {\pi^4}{90}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded, monotone, and find its limit Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_1 = 0 \leq 2$. Assume that the sequence is bounded for $n = k$. Then,
\begin{align*}
0 \leq x_k &\leq 2 \\
\vdots \\
\text{lower bound } \leq x_{k + 1} &\leq \text{upper bound}
\end{align*}
I am thrown off by the term $x_{n + 2}$ in the recursive formula and I can't see the algebra to produce the above steps without getting $x_{n + 2}$ in the expression of the upper / lower bound.
Thank you.
Update:
I have added this to the prove:
We have $0 \leq x_1 = 0 \leq 2$ and $0 \leq x_2 = 0 \leq 2$. Assume that the sequence is bounded for $k+1$,
\begin{align*}
0 &\leq x_{k + 1} \leq 2 \\
0 &\leq x_k + x_{k+1} \leq 4 \\
0 &\leq x_k + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq \frac{1}{6} x_{k} + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq x_{k+2} \leq 4
\end{align*}
Therefore, by the principle of mathematical induction, the sequence is bounded.
Is this valid?
| Observe that $x_1 = 0$, $x_2 = 0$, $x_3 = 1$, $x_4 = \frac{4}{3}$. We can prove by induction that $x_n <2$ for all $n$. Suppose that the inequality is true for $x_1, x_2,\ldots, x_{n+1}$. Then
$$
x_{n + 2} = \frac{1}{3}x_{n + 1} + \frac{1}{6}x_n + 1 < \frac{2}{3} + \frac{2}{6} + 1 = 2.
$$
Now we show that the sequence is monotonically increasing. Suppose that $x_1 \leq x_2 \leq x_3 \leq \ldots \leq x_{n+1}$ holds for some $n\geq 2$. Then
$$
x_{n + 2} - x_{n + 1} = \frac{1}{3}(x_{n + 1} - x_n ) + \frac{1}{6}(x_n - x_{n - 1} ) \geq 0.
$$
Thus $x_n$ is bounded from above and increasing, hence it is convergent. Its limit $x$ must satisfy
$$
x = \frac{1}{3}x + \frac{1}{6}x + 1,
$$
i.e., we must have $x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\lim_{x\to\infty} \frac{1}{x} = 1$ is false $$\lim_{x\to\infty} \frac{1}{x} = 1$$
Given $\epsilon > 0$
$$\left|\frac{1}{x} - 1\right| < \epsilon.$$
Rewrite it as
$$-\epsilon < \frac{1}{x} - 1 < \epsilon$$
$$-\epsilon + 1< \frac{1}{x} < \epsilon + 1$$
If epsilon is very small, then on both sides we are getting value close to $1$, but the function gets closer to zero, hence both sides false.
If $\epsilon$ is big, then on the right side we are getting big positive value, but with $ n \in (0,1)$ the function gets bigger too. Hence right side fails.
Is this a sound proof? And if yes, how would I rewrite it with math symbols?
| To show that the statement is false, find an $\epsilon$ which does not have a corresponding $x^\star$, such that whenever $x \geq x\star$, $| \frac{1}{x} - 1 | < \epsilon$ does not hold. Suppose $L = 1$ and let $\epsilon = \frac{1}{2}$.
Consider when $x^\star \geq 1$, then
\begin{align*}
| \frac{1}{x} -1 | \geq \frac{1}{2}
\end{align*}
whenever $x \geq 2$. Consider when $x^\star <1$, then
\begin{align*}
|\frac{1}{x} -1 | \geq \frac{1}{2}
\end{align*}
whenever $x \geq 2$.
Hence, there does not exist an $x^\star$ for $\epsilon = \frac{1}{2}$. Therefore, it must certainly be false that the limit is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I solve the recurrence relation:$F\left(n\right)=F\left(n-1\right)+2F\left(n-2\right),\:F\left(0\right)=1,F\left(1\right)=1$ I've been trying to solve it using Power Series and Partial Fractions but got stuck with.
| Let $F(n) = x^n$. Then we have $x^n - x^{n-1} - 2x^{n-2} = x^{n-2}(x^2 - x - 2) = 0$. So in order to have solution(s) in the form of $Cx^n$ we should set $$x^2 - x - 2 = (x-2)(x+1)= 0 \implies x = 2 , -1$$ Combining these solutions, the general solution is $F(n) = a_12^n + a_2(-1)^n$. Using initial conditions we have $$a_1 + a_2 = 1$$ And $$2a_1 -a_2 = 1$$ Which leads to $F(n) = \frac{2}{3}2^n+\frac{1}{3}(-1)^n$. We can check that easily using WA. It's really like solving the differential equation $y'' - y' - 2y = 0$.
Edit: Actually we can also use unilateral $\mathcal{Z}-$transform (which is similar to Laplace transform). It's defined by $$X(z) = \sum_{n = 0}^{\infty}x[n]z^{-n}$$We can prove that $\mathcal{Z}\{x[n-1]\} = z^{-1}X(z) + x[-1]$ which leads to $\mathcal{Z}\{x[n-2]\} = z^{-2}X(z) + z^{-1}x[-1] + x[-2]$. So we have $$X(z) - (z^{-1}X(z) + x[-1]) - 2(z^{-2}X(z) + z^{-1}x[-1] + x[-2]) = 0$$ It's easy to compute $x[-1]$ and $x[-2]$ by extrapolating $$x[1] = x[0] + 2x[-1] \implies x[-1] = 0$$ $$x[0] = x[-1] + 2x[-2] \implies x[-2] = \frac{1}{2}$$
The result is $$X(z) = \frac{1}{1 - z^{-1} - 2z^{-2}}$$
Apply partial fraction decomposition $$\frac{1}{1 - z^{-1} - 2z^{-2}} = \frac{-1}{(z^{-1} + 1)(2z^{-1} - 1)} = \frac{1}{3}(\frac{1}{z^{-1} + 1}) - \frac{2}{3}(\frac{1}{2z^{-1} -1}) = \frac{1}{3}(\frac{1}{1 -(-1)z^{-1}}) + \frac{2}{3}(\frac{1}{1- (2)z^{-1}})$$
Also $\mathcal{Z}\{a^n u[n]\} = \frac{1}{1-az^{-1}}$ when $|z| \gt |a|$. Applying inverse transform to both sides $$x[n] = \frac{1}{3}(-1)^n u[n] + \frac{2}{3}2^n u[n]$$
Which is the same answer as previous.
| {
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How to find the sum of this geometric serie: $ 3+ \sqrt3 + 1 + ...$ I am trying to find the sum of this geometric series but can't find it:
$ 3+ \sqrt3 + 1 + ...$
The solution I get is:
$S=\frac{3(3+\sqrt{3})}{4}$
but the answer key shows:
$S=\frac{3\sqrt{3}(\sqrt{3}+1)}{2}$
This exercice is from a book called Pre-Calculus in a Nutshell. I could solve the other geometric series but this questio has a square root and I must be making a mistake when simplifying.
Here are the steps I took to find my solution, maybe you can see where it goes wrong?
The sum of a geometric serie is
$(S) = \frac{a}{1-r}$ when |r| < 1
$3*r=\sqrt{3}$
Therefore:
$r=\frac{\sqrt{3}}{3}$
$|\frac{3}{\sqrt{3}}|<1 $ so I can use that formula
$a=3$
which gives me
$S=\frac{3}{1-\sqrt{3}}$
simplifying I get:
$S=\frac{3*(1+\sqrt{3})}{1-\sqrt{3}*(1+\sqrt{3})}$
$S=\frac{9+3\sqrt{3}}{1- (-3)}$
Simplifying more:
$S=\frac{3(3+\sqrt{3})}{4}$
| Indeed the common ratio was incorrect, it should be $\frac1{\sqrt3}=\frac{\sqrt3}{3}$
\begin{align}
S &= \frac{3}{1-\frac{\sqrt3}{3}}=\frac{9}{3-\sqrt3}=\frac{9(3+\sqrt3)}{9-3} =\frac{3(3+\sqrt3)}2 = \frac{3\sqrt3(\sqrt3+1)}{2}
\end{align}
Edit: Alternative working:
$$S=\frac{3}{1-\frac1{\sqrt3}}=\frac{3\sqrt{3}}{\sqrt3-1}=\frac{3\sqrt3(\sqrt3+1)}{2}$$
| {
"language": "en",
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Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$ My task was to prove the question above over real variables.
I thought that this minor inequality should help-
$$ 3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2). $$
which is true.
By this inequality, the original inequality is converted to-
$$ (1 - ab)^2 (1 - c)^2 + (ab-c)^2 + abc \geq 0 $$
This proves the inequality for $abc\geq 0$.
I want to prove this Inequality for $abc\lt0$. But I couldn't find a solution for $abc\lt0$.
Any extensions for $abc\lt0$ are thankfully accepted.
| Here is a straightforward solution based on quadratic inequalities.
For simplicity, denote $A=1-a+a^2$ and $B=1-b+b^2$. We need to show that
$$
3AB(1-c+c^2) \geq 1+abc +a^2b^2c^2.
$$
This is equivalent to showing that
$$
(3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1) \geq 0.
$$
If we regard the left-hand side above as a quadratic function of $c$,
$$
f_{A,B,a,b}(c)= (3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1),
$$
it suffices to show $f_{A,B,a,b}(c)\geq 0$ for any real $a,b,c$. From the fact
$A=1-a+a^2 \geq \frac 3 4 a^2$ and $B\geq \frac 3 4 b^2$, we know the leading coefficient of $f_{A,B,a,b}$ is strictly positive, i.e., $3AB - a^2b^2 >0$. Now it remains to show the discriminant of $f_{A,B,a,b}$ is non-positive. Namely,
$$
(3AB +ab)^2 -4(3AB-a^2b^2)(3AB-1) \leq 0,
$$
or equivalently,
$$
4AB + 2ABab + 4ABa^2b^2 \leq a^2b^2 +9A^2B^2.
$$
By AM-GM inequality, we have
$$
2ABab \leq a^2b^2 + A^2B^2.
$$
Therefore, it suffices to show
$$
4AB + 4ABa^2b^2 \leq 8A^2B^2,
$$
or equivalently,
\begin{equation}\begin{split}
1+a^2b^2 \leq& 2AB \\
=& 2(1-a+a^2)(1-b+b^2), \\
\end{split}\end{equation}
which is also used in Vasc's solution provided by Michael Rozenberg.
| {
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$3^{123} \mod 100$
The Question:
Evaluate $3^{123}\mod 100$
My Attempt
So initially I attempted to list the powers of 3 and find a pattern of the last two digits - which, despite much painful inspection did not yield an obvious useful pattern.
So I then attempted to simplify this and use Euler's Generalisation of Fermat's Theorem to solve this:
The theorem states: $a^{\phi(n)} \equiv 1 \pmod{n}$
So:
$3^{123}\mod 100$
= $3^{41^3}\mod 100$
= $(3^{40} \times 3^1)^3\mod 100$
I think I'm ok up to that point. Now, $\phi(100) = 40$
So am I right in the following?
$(3^{40} \times 3^1)^3\mod 100$ $\cong$ $(1 \times 3^1)^3\mod 100$
= $3^3\mod 100$
= 27.
Am I correct?
Thanks!
| The OP began by looking for a pattern but stated
that
...despite much painful inspection did not yield an obvious useful pattern.
You can use some light theory to actually predict the form and structure of the pattern.
Observe that if $a \in \{0,2,4,6,8\}$ and $b \in \{1,3,7,9\}$ and
$\quad 3 \times (10 a + b) \equiv 10 \,a' + b' \pmod{100} \text{ with } a',b' \in \{0,1,2,3,4,5,6,7,8,9\}$
then in fact $a' \in \{0,2,4,6,8\}$ and $b' \in \{1,3,7,9\}$.
This is our main (theoretical) pattern and
$\quad 3^1 \equiv 03 \pmod{100}$
$\quad 3^2 \equiv 09 \pmod{100}$
$\quad 3^3 \equiv 27 \pmod{100}$
$\quad 3^4 \equiv 81 \pmod{100}$
$\quad\text{-------------------------}$
$\quad 3^5 \equiv 43 \pmod{100}$
It easy to verify that the units digit will move
$\quad 3 \mapsto 9 \mapsto 7 \mapsto 1$
inside each of these four cycles.
Considering that $3$ is a unit, we can argue that one of these $4$-cycles will end on
$$\quad 01 \quad \text{the multiplicative identify}$$
and that no repetition is possible until the identify is reached.
Since the tens digit can only cycle over the set $\{0,2,4,6,8\}$, there are at most five of these $4$-cycles that have to be calculated.
Calculating the $2^{nd}$ $4$-cycle:
$\quad 3^5 \equiv 43 \pmod{100}$
$\quad 3^6 \equiv 29 \pmod{100}$
$\quad 3^7 \equiv 87 \pmod{100}$
$\quad 3^8 \equiv 61 \pmod{100}$
$\quad\text{-------------------------}$
Calculating the $3^{rd}$ $4$-cycle:
$\quad 3^9 \equiv 83 \pmod{100}$
$\quad 3^{10} \equiv 49 \pmod{100}$
$\quad 3^{11} \equiv 47 \pmod{100}$
$\quad 3^{12} \equiv 41 \pmod{100}$
$\quad\text{-------------------------}$
Calculating the $4^{th}$ $4$-cycle:
$\quad 3^{13} \equiv 23 \pmod{100}$
$\quad 3^{14} \equiv 69 \pmod{100}$
$\quad 3^{15} \equiv 07 \pmod{100}$
$\quad 3^{16} \equiv 21 \pmod{100}$
$\quad\text{-------------------------}$
At this point we really don't have to calculate the $5^{th}$ $4$-cycle since we know it has to be the last one.
We can now use the fact that
$\tag 1 3^{20} \equiv 1 \pmod{100}$
and work out the remaining details for the OP's question.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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What function $f(x)$ would this limit, involving 3 integrals be equal to? What function $f(x)$ would the limit
$$\lim_{n\rightarrow{0^+}}\left(-\int_{-\infty}^{-x-n}\frac{xK_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da+\int_{-x+n}^{x-n}\frac{xK_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-\int_{x+n}^\infty\frac{xK_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da\right)$$
be equal to?
I tried doing numerical integration to get an idea of what the function would be, but didn't get to try enough of a variety of rectangle lengths or spreads of $a$ to see if it was really converging on anything. Also I know that simply subtracting $\infty$ from $\infty$ is undefined, which is the reason I expressed it as a limit.
Also $K_n(x)$ denotes the modified bessel function of the second kind.
| The final answer is $$2\ \text{Chi}(x)\cosh(x) - 2\ \text{Ci}(x)\cos(x) - 2\ \text{Shi}(x)\sinh(x) - 2\ \text{Si}(x)\sin(x)+ \pi\sin(x)$$ for $x > 0$. $\text{Ci}(x)$ is the cosine integral, $\text{Si}(x)$ is the sine integral, $\text{Chi}(x)$ is the hyperbolic cosine integral, and $\text{Shi}(x)$ is the hyperbolic sine integral.
$x$ can be factored out to get:
$$x\lim_{n\rightarrow{0^+}}\left(-\int_{-\infty}^{-x-n}\frac{K_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da+\int_{-x+n}^{x-n}\frac{K_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-\int_{x+n}^\infty\frac{K_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da\right)$$
Since the first integral is equal to the third, and the integrand in the second integral is even, this simplifies to:
$$x\lim_{n\rightarrow{0^+}}\left(2\int_{0}^{x-n}\frac{K_1\left(\sqrt{x^2-a^2}\right)}{\sqrt{x^2-a^2}}da-2\int_{x+n}^\infty\frac{K_1\left(\sqrt{a^2-x^2}\right)}{\sqrt{a^2-x^2}}da\right)$$
Using the integral definition $\big($formula $(5)\big)$ of $K_1(z)$, this becomes $$x\lim_{n\rightarrow{0^+}}\left(2\int_{0}^{x-n}\int_0^{\infty}
\frac{\cos(t)}{(t^2+x^2-a^2)^{3/2}}dtda-2\int_{x+n}^\infty\int_0^{\infty}
\frac{\cos(t)}{(t^2+a^2-x^2)^{3/2}}dtda\right)$$
Then switching the order of integration, this becomes $$x\lim_{n\rightarrow{0^+}}\left(2\int_0^{\infty}\int_{0}^{x-n}
\frac{\cos(t)}{(t^2+x^2-a^2)^{3/2}}dadt-2\int_0^{\infty}\int_{x+n}^\infty
\frac{\cos(t)}{(t^2+a^2-x^2)^{3/2}}dadt\right)$$
This simplifies to $$x\lim_{n\rightarrow{0^+}}\left(2\int_0^{\infty}\frac{\cos(t)(x-n)}{(t^2+x^2)\sqrt{2nx-n^2+t^2}}dt-2\int_0^{\infty}\cos(t)\left(\frac{1}{t^2-x^2}-\frac{n+x}{(t^2-x^2)\sqrt{n^2+t^2+2nx}}\right)dt\right)$$
Combining everything together into one integral yields $$2x\lim_{n\rightarrow{0^+}}\left(\int_0^{\infty}\cos(t)\left(\frac{x-n}{(t^2+x^2)\sqrt{2nx-n^2+t^2}}-\frac{1}{t^2-x^2}+\frac{n+x}{(t^2-x^2)\sqrt{n^2+t^2+2nx}}\right)dt\right)$$
Since this converges uniformly, $n = 0$ can be plugged in to get $$2x\left(\int_0^{\infty}\cos(t)\left(\frac{x}{t(t^2+x^2)}-\frac{1}{t^2-x^2}+\frac{x}{t(t^2-x^2)}\right)dt\right)$$
The integrand can be simplified $$2x\left(\int_0^{\infty}\cos(t)\left(\frac{x}{t(t^2+x^2)}-\frac{1}{t(t+x)}\right)dt\right)$$
Using integration by parts makes this $$f(x) = \int_{0}^{\infty}\ln\left(1+\frac{2xt}{x^{2}+t^{2}}\right)\sin\left(t\right)dt$$
Differentiating under the integral yields $$f'(x) = \int_0^{\infty}\left( \frac{2}{t+x}-\frac{2x}{t^2+x^2} \right) \sin(t) dt$$
Evaluating (with Mathematica), yields that $$f'(x) = 2\text{Ci}(x)\sin(x) + 2\text{Chi}(x)\sinh(x) - 2\text{Shi}(x)\cosh(x) - 2\text{Si}(x)\cos(x) + \pi\cos(x)$$ where $\text{Ci}(x)$ denotes the cosine integral, $\text{Si}(x)$ denotes the sine integral, $\text{Chi}(x)$ denotes the hyperbolic cosine integral, and $\text{Shi}(x)$ denotes the hyperbolic sine integral.
The antiderivative can be computed analytically, but it was somewhat long, so I will just skip over it.
Then since $f(0) = 0$, $f(x) = \int_0^x f'(u)du$, which evaluates to $$2\ \text{Chi}(x)\cosh(x) - 2\ \text{Ci}(x)\cos(x) - 2\ \text{Shi}(x)\sinh(x) - 2\ \text{Si}(x)\sin(x)+ \pi\sin(x)$$
I checked this numerically and it seems to hold for all positive $x$.
| {
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If $P(x)$ is a polynomial of degree three in $x$, and $y^2 = P(x)$, show that $\frac{D(y^3D^2y)}{y^2}$ is constant If $P(x)$ is a polynomial of degree three in $x$, and $y^2 = P(x)$, show that
$$\frac{D(y^3D^2y)}{y^2}$$
is a constant, where D denotes the derivative operator.
I have tried expressing the expression above in terms of $P$ and its derivatives (in the hopes of showing the derivative is $0$) but I couldn't manage to do that.
| First consider
$$Dy = \frac{1}{2y}Dy^2$$
and
$$D^2y = D(\frac{1}{2y}Dy^2) =\frac{-1}{2y^2}(Dy)(Dy^2) + \frac{1}{2y}D^2y^2 = -\frac{1}{4y^3}(Dy^2)^2 + \frac{1}{2y}D^2y^2 $$
Therefore, the numerator in your expression is
\begin{align}
D(y^3D^2y) &= D(-\frac{1}{4}(Dy^2)^2 + \frac{y^2}{2}D^2y^2)
\\
&=-\frac{1}{2}(Dy^2)(D^2y^2) + \frac{1}{2}(Dy^2)(D^2y^2) + \frac{y^2}{2}D^3y^2
\\
&=\frac{y^2}{2}D^3y^2
\end{align}
then your original expression becomes
$$\frac{D(y^3D^2y)}{y^2} = \frac{1}{2}D^3 y^2$$
If $y^2 = ax^3+bx^2+cx+d$, then
$$\frac{D(y^3D^2y)}{y^2} = 3a$$
| {
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Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+ca$ and $r = abc$.
$$ LHS = 1 + p^2 - 2q + q^2 - 2pr + r^2 \geq 2q \Rightarrow 1 + p^2 + q^2 + r^2 \geq 4q + 2pr $$
It's quite well-known that $p^2\geq 3q$ and $q^2\geq 3pr$. So,
$$ 1 + 3q + 3pr + r^2 \geq 4q + 2pr \Rightarrow 1 + pr + r^2 \geq q $$
But I don't know how to prove it. It can also be seen that $a\ge b\ge c$, but I can't exploit symmetry.
Any help is thankfully welcome.
| Proceeding along your approach:
It suffices to prove that $1 + p^2 - 2q + q^2 - 2pr + r^2 \ge 2q$.
Since $p^2 \ge 3q, q^2 \ge 3pr$, it suffices to prove that
$1 + 3q - 2q + q^2 - \frac{2}{3}q^2 \ge 2q$
or $1 - q + \frac{q^2}{3} \ge 0$
or $\frac{1}{3}(q - \frac{3}{2})^2 + \frac{1}{4} \ge 0$ which is true. We are done.
| {
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When its possible to write a number $a+b\sqrt{c}$ as the square of a sum? I was teaching my high school students how to find the roots of 2nd degree equations like the following:
$\frac{x^2}{2}+\sqrt{3}x-\sqrt{2}=0$
In this case, using the formula we get:
$x=-\sqrt{3}\pm\sqrt{3-2\sqrt{2}}$
To simplify I writted $3-2\sqrt{2}$ as $(\sqrt{2}-1)^2$, so I get:
$x=-\sqrt{3}\pm(\sqrt{2}-1)$
In this case was simple to write $3-2\sqrt{2}$ as $(\sqrt{2}-1)^2$, there are some harder cases to find this factorization.
My question is when it's possible to write $a+b\sqrt{c}$ as $(m+n)^2$,
and how to find these numbers $m$ and $n$?
| This is not as hard as it seems.
[I will slightly alter your notation, and instead of $(m+n)^2$ I will use $(m + n\sqrt{c})^2$.]
$$ (m + n\sqrt{c})^2 = a + b \sqrt{c} $$
Expanding and collecting everything on the left-hand side gives us:
$$ n^2 c + (2mn - b) \sqrt{c} + (m^2 - a) = 0$$
We are given $a, b, c \in \mathbb{Z}$, and likewise we want $m, n \in \mathbb{Z}$.
Well, the first and third terms are integers, and the second one isn't (unless $c$ is a perfect square -- otherwise we could have used simpler methods). This means that the coefficient of $\sqrt{c}$ must be equal to $0$. This gives us the new equation $mn = \frac{b}{2}$. So, $b$ must be even for this to work.
The two remaining terms also need to sum to 0. This means $n^2 c + m^2 = a$.
So, let's sum up what we have so far:
*
*$b$ must be even.
*$mn = \frac{b}{2}$
*$n^2 c + m^2 = a$
Unless $b = 0$, in which case we have a much simpler problem, we have that $n \neq 0$. This means that we can safely divide by $n$ , and get:
*
*$m = \frac{b}{2n}$
We can substitute that into our last equation, which gives us:
*
*$n^2 c + \frac{b^2}{4n^2} = a$
If we let $N = n^2$, we can multiply by $N$ and rewrite all this as:
*$N^2 c + \frac{b^2}{4} = a$
So, we get: $N^2 = \frac{a}{c} - \frac{b^2}{4c}$.
This means:
$$N^2 = \frac{4a - b^2}{4c}$$
The right-hand side must be positive (so $4a > b^2$) and it must be a perfect square.
So, the constraint you were looking for are:
*$b$ is even.
*$4a > b^2$
*$\frac{4a - b^2}{4c}$ is a perfect square.
[The first constraint is not strictly necessary, if you don't mind working with fractions instead of integers.]
For your example, we get $$\frac{4\times 3 - 2^2}{4 \times 2} = \frac{8}{8} = 1 = 1^2$$
So, we know $n = \pm \sqrt{N}$, and $m = \pm \frac{b}{2\sqrt{N}}$. We can safely choose one sign for $n$ and take the corresponding sign for $m$ (the other option would just give us $(-m -n\sqrt{c})$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate $\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$? The wolframalpha gives the answer $0$:
Wolframalpha culculation
I tried like this:
Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$,then:
$\lim \limits_{(x,y)→(0,0)} \frac{(x^2+y^2)^2}{xy}$ = $\lim \limits_{r→0} \frac{r^4}{r^2 \sin(\theta)\cos(\theta)}$ = $\lim \limits_{r→0} \frac{2r^2}{\sin(2\theta)} =0$
but it seems wrong when $\theta =0$ and the limit $\lim \limits_{r→0} \frac{2r^2}{\sin(2\theta)}$ may not exist.
So how to calculate the limit?
| We have that for $x=y=t\to 0$
$$\frac{(x^2+y^2)^2}{xy}=\frac{4t^4}{t^2}=4t^2 \to 0 $$
but for $x=t\to 0$ and $y=t^3$
$$\frac{(x^2+y^2)^2}{xy}=\frac{(t^2+t^6)^2}{t^{4}}=1+2t^4+t^8 \to 1 $$
therefore the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $
But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$
My Approach #2
$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$
Which also seems to be a dead end.
Any ideas on how to evaluate this?
| $$\lim _{x\to 2}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right) = \lim _{x\to 2}\left(\sqrt{x+2}+\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right) = \lim _{x\to 2}\left(\sqrt{x+2}+\frac{\sqrt{x-2}}{\sqrt{x}+\sqrt{2}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Find the sum of series without differentiation Given a series $\sum_{i > 0}\frac{i^2}{z^i}$, and $\sum_{i > 0}\frac{i}{z^i} = \frac{z}{(z - 1)^2}$ I need to find the sum
My method does not require differentiation but there is a difficulty.
Let $S = \frac{1^2}{z} + \frac{2^2}{z^2} + \frac{3^2}{z^3} + ... + \frac{i^2}{z^i}$
Let $zS = 1 + \frac{2^2}{z} + \frac{3^2}{z^2} + ... + \frac{i^2}{z^{i - 1}}$
Thus, $zS - S = 1 + \frac{2^2 - 1^2}{z} + \frac{3^2 - 2^2}{z^2} + ... + \frac{i^2 - (i - 1)^2}{z^{i - 1}} - \frac{i^2}{z^i}$
Thus, $(z - 1)S = 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}$
My question is how can I proceed? The numerator of each term is not 1 so I cannot use any formula to calculate the sum.
| Let
\begin{eqnarray*}
S=\sum_{i=1}^{\infty} i^2 x^i.
\end{eqnarray*}
Now multiply by $(1-3x+3x^2-x^3)$ and note that for $i \geq 2$
\begin{eqnarray*}
(i-3)^2-3(i-2)^2+3(i-1)^2-i^3=0.
\end{eqnarray*}
Examine the lower order terms more carefully, and we have
\begin{eqnarray*}
(1-3x+3x^2-x^3)S=x(1+x)
\end{eqnarray*}
giving the well known formula
\begin{eqnarray*}
S=\sum_{i=1}^{\infty} i^2 x^i =\frac{x(1+x)}{(1-x)^3}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $ \int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $ How can I evaluate
$$ \int_{0}^{1} \frac{\ln(x+\sqrt{1-x^2})}{\sqrt{1+x^2}} \, \mathrm{d}x $$
U-substitution has not worked for me. Integration by parts, Differentiation under integral sign, Mathematica is not coming up with a solution either.
Is there a closed form for this integral?
Thank you kindly for your help and time.
| To calculate this integral I'll use series expansion $$\frac 1 {\sqrt{1+x^2}}=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n}$$ for $|x|\le1$
$$I=\int_{0}^{1}\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}x^{2n} \ln\left(x+\sqrt{1-x^2}\right)dx$$
By dominated convergence
$$I=\sum_{n=0}^{\infty }\frac {(-1)^n} {2^{2n}}\binom{2n}{n}
\int_{0}^{1} \ln\left(x+\sqrt{1-x^2}\right)x^{2n} dx$$
Let $$ J=\int_{0}^{1} \ln\left(x+\sqrt{1-x^2}\right)x^{2n}$$
Now, Let $x=\cos\theta$
$$\implies J=\int_{0}^{\fracπ2}\ln\left(\cos\theta+\sin\theta\right)\left(\cos^{2n}\theta\right) (\sin\theta) d\theta$$
$$ \implies J=\frac12 \int_{0}^{\fracπ2}\ln\left(1+\sin2\theta\right)\left(\cos^{2n}\theta\right)\left(\sin\theta\right) d\theta$$
$$ \implies J=\frac12 \int_{0}^{\fracπ2}\left(\cos^{2n}\theta\right) \left(\sin\theta\right) \sum_{k=1}^{\infty }(-1)^{k-1}\frac {\sin^k 2\theta}{k} d\theta$$
$$ \implies J=\frac12 \int_{0}^{\fracπ2}\left(\cos^{2n}\theta\right) \left(\sin\theta\right) \sum_{k=1}^{\infty }(-1)^{k-1}\frac {2^k \left(\sin^k \theta \right)\left(\cos^k\theta\right)}{k} d\theta$$
By dominated convergence
$$J= \sum_{k=1}^{\infty }\frac {(-1)^{k-1} 2^{k-1}}{k}\int_{0}^{\fracπ2}\left(\cos^{2n+k}\theta\right) \left(\sin^{k+1}\theta\right) d\theta$$
Using $$\int_{0}^{\fracπ2}\left(\sin^m\theta\right) \left(\cos^n\theta\right)d\theta=\frac{\Gamma\left(\frac{n+1}2\right)
\Gamma\left(\frac{m+1}2\right)}{2
\Gamma\left(\frac{m+n+2}2\right)}$$
$$J=\sum_{k=1}^{\infty }\frac{(-1)^{k-1} 2^{k-2}}{k}\frac{\Gamma\left(\frac{k+2}2\right)
\Gamma\left(\frac{2n+k+1}2\right)}{
\Gamma\left(\frac{2n+2k+3}2\right)}$$
On substituting $J$ in orignal integral, we get
$$I=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}\frac {(-1)^{(n+k-1)}}{2^{(2n-k+2)}k}\binom{2n}{n}\frac{\Gamma\left(\frac{k+2}2\right)
\Gamma\left(\frac{2n+k+1}2\right)}{
\Gamma\left(\frac{2n+2k+3}2\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Given the O.D.E. $2yy''=1+(y')²$ by using the transformation $y'=z, y''=z·(dz/dy)$ find the solutions.Why cases for y are not taken? By using this transformation we have:
$$\dfrac {2z\;dz}{z^2+1}=\dfrac {dy}y$$
Then by integrating the book comes to the conclusion that:
$z=±\sqrt {c_1y-1}$ , where $c_1$ is a non zero arbitrary constant.
How do we know that $c_1y-1 \ge 0$ in order to apply a root? Why doesn't it examine cases for $y$?
| It's an autonomous equation.
We need to solve the ODE: $yy''=1+(y')^{2}$.
Now, treating $y$ as the independent variable, let $z=y'$ wich gives $y''=z\frac{dz}{dy}$, so $$2y\frac{dz}{dy}z=z^{2}+1$$Solving for $\frac{dz}{dy}$, we can see that $$\frac{dz}{dy}=\frac{z^{2}+1}{2yz} \overbrace{\implies}^{\div \frac{z^{2}+1}{2z}} \frac{2\frac{dz}{dy}z}{z^{2}+1}=\frac{1}{y} \implies \int \frac{2\frac{dz}{dy}z}{z^{2}+1} dy = \int \frac{1}{y} dy \implies \ln(z^{2}+1)=\ln(y)+c_{1}.$$
Solving for $z$, we can see that $$z=-\sqrt{e^{c_{1}}y-1} \quad \vee \quad z=\sqrt{e^{c_{1}}y-1}.$$
Now, sustituing back for $z=\frac{dy}{dx}$, we can see that $$\frac{dy}{dx}=-\sqrt{c_{1}y-1} \quad \vee \quad \frac{dy}{dx}=\sqrt{c_{1}y-1}.$$For $\frac{dy}{dx}=-\sqrt{c_{1}y-1}$, we can obtain that $$y=\frac{1}{c_{1}}+\frac{1}{4}(x-c_{2})^{2}c_{1}$$also for $\frac{dy}{dx}=\sqrt{c_{1}y-1}$, we can obtain that $$y_{x}=\frac{1}{c_{1}}+\frac{1}{4}(x+c_{2})^{2}c_{1}$$Finally, collecting solutions and simplify the arbitrary constanst we can see that $$\boxed{y(x)=\frac{1}{c_{1}}+\frac{1}{4}(x+c_{2})^{2}c_{1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Fractions in Questions and Answers
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