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Find the minimum value of $a^2+b^2$
Let a and b be real numbers for which the equation $x^4 +ax^3 +bx^2
+ax+1=0 \tag1$ has at least one real solution. For all such pairs $(a,b)$, find the minimum value of $a^2 + b^2$.
Using $x + \frac 1 x = y$ in (1):
$y^2 + ay+b-2=0 \tag2$ therefore the first condition is $a^2 - 4b + 8\ge 0$.
The second one, coming from $x^2 -yx + 1=0$, is $y^2 - 4 \ge0$.
The calculus is a mess, so I don't think this is the way to solve it. Does anyone have a smarter idea?
UPDATE
I've corrected (2) following @mathlove suggestion.
| Edit : I should have added a condition that $|-a/2|\ge 2$.
I think that your idea is nice, but note that $(2)$ is incorrect. Using $x+\frac 1x=y$, we have
$$y^2+ay+b\color{red}{-2}=0\tag3$$
So, we have to have
$$a^2-4(b-2)\ge 0\iff b\le \frac{a^2}{4}+2\tag4$$
We want $(3)$ to have at least one real solution $y$ such that $|y|\ge 2$ as you wrote.
Let $f(y):=y^2+ay+b-2$. The condition is represented as
$$\left|-\frac a2\right|\ge 2\quad\text{or}\quad f(-2)\le 0\quad\text{or}\quad f(2)\le 0$$$$\iff |a|\ge 4\quad\text{or}\quad 2-2a+b\le 0\quad\text{or}\quad 2+2a+b\le 0\tag5$$
Hence, drawing $(4)(5)$ in $ab$-plane gives that the minimum value of $a^2+b^2$ is
$$\left(\frac{|2\pm 2\cdot 0+0|}{\sqrt{(\pm 2)^2+1^2}}\right)^2=\color{red}{\frac 45}$$
(note here that $a^2+b^2$ represents the square of the distance from $(0,0)$ to $(a,b)$)
for $(a,b)$ such that $a^2+b^2=\frac 45$ and $b=\pm 2a-2$, i.e. $(a,b)=(\pm 4/5,-2/5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to evaluate sum of fifth powers if sum of $n$th power is known for $n\in\{1,2,3\}$? In a book of mine, I found two problems I couldn't solve. Namely:
Let $a,b,c$ satisfies $a+b+c=3,a^2+b^2+c^2=5, a^3+b^3+c^3=7$. There are three questions:
*
*Find $a^4+b^4+c^4$. There was a solution that the value of the sum of fourth powers is $9$.
But I was unable to solve the following two questions:
*Does the pattern continue? Does $a^5+b^5+c^5=11$?
*Show that there are no solutions for $a,b$, and $c$ in the real numbers, but there are solutions in the complex numbers.
I tried to solve 2. by Sage:
sage: factor((a+b+c)^5-a^5-b^5-c^5)
5*(a^2 + a*b + b^2 + a*c + b*c + c^2)*(a + b)*(a + c)*(b + c)
I have no idea how to do the part 3.
Can anyone show me how to solve 2. and 3.?
| We have
\begin{align}
a^2+b^2+c^2&=(a+b+c)^2-2ab-2bc-2ca\\
5&=9-2(ab+bc+ca)\\
ab+bc+ca&=2\\
(ab+bc+ca)^2&=4\\
a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)&=4\\
a^2b^2+b^2c^2+c^2a^2+6abc&=4\qquad\text{since }a+b+c=3\\
\frac{(a^2+b^2+c^2)^2-(a^4+b^4+c^4)}{2}+6abc&=4\tag{1}
\end{align}
On the other hand, from the identity
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
we have
\begin{align}
7-3abc&=(3)\left(5-2\right)\\
abc&=-\frac{2}3
\end{align}
plugging it into $(1)$ we get
\begin{align}
\frac{5^2-(a^4+b^4+c^4)}{2}+6\left(-\frac{2}3\right)&=4\\
25-(a^4+b^4+c^4)-8&=8\\
a^4+b^4+c^4&=9
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959678",
"timestamp": "2023-03-29T00:00:00",
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Integrating $\frac{x}{1+x^2}$ I need to integrate $\frac{x}{1+x^2}$ and I tried it like this:
$$f(x)=\frac{x}{1+x^2}=\frac{x}{(x+1)(x-1)}=\frac{x+1-1}{(x+1)(x-1)}= \frac{x+1}{(x+1)(x-1)}-\frac{1}{(x+1)(x-1)}=\frac{1}{x-1}-\frac{1}{1+x^2}$$
After I simplified it I integrate it!
$$\int{f(x)}=\int{\frac{1}{x-1}}-\int{\frac{1}{1+x^2}}$$
$\int{\frac{dx}{1+x^2}}$ becomes arctan(x) and I substitute $\int{\frac{dx}{x-1}}$ to $\int{\frac{du}{u}}$ now it becomes $ln(x-1)$ after this been done I get:
$$\int{f(x)}=ln(x-1)-arctan(x)$$
but when I type integrate x/(1+x^2) into wolframAlpha I it tells me the answer is: $$\int{f(x)}=\frac{1}{2}log(x^2+1)$$
where did I go wrong or what should I have done instead ?
Thank you for your help in advance,
Raavgo
| The mistake is at the very beginning: it is not true that $1+x^2 = (1-x)(1+x)$. What you thought about is $1\color{red} - x^2 = (1-x)(1+x)$. (Also, remember that the roots of $1+x^2$ are $\pm \Bbb i$, therefore your factorization is clearly wrong.)
The exercise is much simpler if you notice that $x = \frac 1 2 (1+x^2)'$, therefore
$$\int \frac x {x^2 + 1} \ \Bbb d x = \int \frac {\frac 1 2 (x^2 + 1)'} {x^2 + 1} \ \Bbb d x = \frac 1 2 \log |x^2 + 1| + C ,$$
where $C \in \Bbb R$ is an integration constant.
(I have used the well-known formula that $\int \frac {f'} f \ \Bbb d x = \log |f| + C$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\geq a+b+c$
If $a,b,c>0$, Then prove that $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\geq a+b+c$$
$\bf{My\; Try::}$ Using Cauchy- Schwarz Inequality
$$\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\geq \frac{a^2+b^2+c^2}{3abc}$$
Now How can i solve after that , Help required, Thanks
| You can actually use Holder's inequality for three variables in the following way:
$$(b+c+a)(c+a+b) \Big(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\Big) \geqslant (a+b+c)^3 $$
From where you get:
$$ \frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} \geqslant a+b+c$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to divide a polynomial by a binomial using factoring when there could be a "remainder" I've learned how to divide a polynomial by a binomial by using the long division trick, and although I get how it works, i'd rather use factoring to get the answer. For example: $(3x^2 + 16x -35)/(x + 7)$ = $(3x^2 + 21x -5x -35)/(x + 7)$ = $((3x^2 + 21x) + (-5x - 35)) / (x + 7)$ = $3x(x + 7) -5(x + 7)) / (x + 7)$ = $(3x - 5)(x + 7)) / (x + 7)$ = $(3x - 5)$. It looks complex but that because I typed every step, it's a lot more simple and easy to understand than the long division trick.
However, I am having trouble factoring out the denominator when there would be a remainder if you did the log division way. For example: $(x^4 - 2x^3 - 42x + 20) / (x - 4)$. I can't find any way to factor out $(x - 4)$. In the long division way, you would put the remainder over the denominator, in this case the answer is $x^3 + 2x^2 +8x - 10 + (-20 / (x - 4))$. Any help would be appreciated, thanks.
| Just like in the long division method, each step you should ask "by what do I need to multiply $(x - 4)$ to get the leading coefficient"? In the beginning, you can't necessarily factor out $(x-4)$ (because then there would be no remainder) but you can factor out $(x-4)$ modulo powers of $x^3$ and reduce your problem to a problem in which the nominator has a lower degree and so on.
More explicitly, you have
$$ \frac{x^4 - 2x^3 - 42x + 20}{x - 4} = \frac{(x-4)(x^3) + (4x^3 - 2x^3 - 42x + 20)}{x - 4} = x^3 + \frac{2x^3 - 42x + 20}{x - 4} \\
= x^3 + \frac{2x^2(x-4) + 8x^2 - 42x + 20}{x - 4} \\
= x^3 + 2x^2 + \frac{8x^2 - 42x + 20}{x-4} = x^3 + 2x^2 + \frac{8x(x - 4) -10x + 20}{x - 4} \\
= x^3 + 2x^2 + 8x - \frac{10x - 20}{x-4} = x^3 + 2x^2 + 8x - \frac{10(x-4) + 20}{x-4} \\
= x^3 + 2x^2 + 8x - 10 - \frac{20}{x-4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove the least postive integer $4$ can define the form $(a-b)(c-d)$ Below the conjecture look is true, but how to prove?
Let $x,y,z,w\in Q^{+}$,and such $xyzw=1$,if postive integer $n=(x-y)(z-w)$,show that:$$n_{\min}=4?$$
because By Now I found this example(other words,I can't find any example is $1,2,3$)
$$4=\left(2-\dfrac{1}{2}\right)\left(3-\dfrac{1}{3}\right)$$
| So you have
$$
\left\{ \begin{gathered}
0 < \text{integer}\,n \hfill \\
\left( {x - y} \right)\left( {z - w} \right) = n \hfill \\
x\,y\,z\,w = 1 \hfill \\
\end{gathered} \right.
$$
which translates into
$$
\left\{ \begin{gathered}
n = \left( {x - y} \right)\left( {z - w} \right) = \frac{{\left( {x - y} \right)\left( {z - w} \right)}}
{{x\,y\,z\,w}} = \left( {\frac{1}
{y} - \frac{1}
{x}} \right)\left( {\frac{1}
{w} - \frac{1}
{z}} \right) \hfill \\
x\,y\,z\,w = 1 \hfill \\
\frac{1}
{x}\frac{1}
{y}\frac{1}
{z}\frac{1}
{w} = 1 \hfill \\
\end{gathered} \right.
$$
Now, apart from the trivial solution $(1,1,1,1)$ which would give $n=0$,
the inversion symmetry that appears above tells you that for $1<x,y$, you can have either
$$
n = \left( {x - \frac{1}
{y}} \right)\left( {y - \frac{1}
{x}} \right) = \frac{{\left( {xy - 1} \right)^2 }}
{{xy}}
$$
or:
$$
n = \left( {x - \frac{1}
{x}} \right)\left( {y - \frac{1}
{y}} \right) = \frac{{\left( {x^2 - 1} \right)\left( {y^2 - 1} \right)}}
{{xy}} = \frac{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {y + 1} \right)\left( {y - 1} \right)}}
{{xy}}
$$
and the conclusion follows easily, just because $n$ is increasing with $x$ and $y$, and there is just to find the first couple $(x,y)$ that renders one of the above expression integral, i.e. $(2,3)$ with the 2nd expression.
| {
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"url": "https://math.stackexchange.com/questions/1963778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $4^{138x + 75} \equiv 4^{15} \pmod{53}$ I want to solve the equation
\begin{equation}
4^{138x + 75} \equiv 4^{15} \pmod{53}
\end{equation}
for $x$. Fermats little theorem yields $4^{52} \equiv 1 \pmod{53}$, but where do I go from here?
| $2^{13}=8192\equiv 30\pmod {53}$, then $2^{26}=900\equiv-1\pmod{53}$. Also, $2^4\not\equiv1\pmod{53}$. Therefore
$$2^n\equiv 1\pmod{53}\iff n\equiv 0\pmod{52}$$
Now,
$$2^{276x+150}\equiv 2^{30}\pmod{53}$$
$$276x+150\equiv 30\pmod {52}$$
$$69x\equiv -30\pmod {13}$$
$$23x\equiv -10\pmod{13}$$
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with the constraint $xy=x+y$ I have to prove that if $x$ and $y$ are two positive numbers such that $xy=x+y$, the following inequality
$$\frac{x}{y^2+4}+\frac{y}{x^2+4}\geq \frac{1}{2}$$
holds. Any help is appreciated, thank you in advance.
| Since $\frac{(x+y)^2}{4}\geq xy=x+y$, we obtain $x+y\geq4$.
Thus, by C-S $\frac{x}{y^2+4}+\frac{y}{x^2+4}\geq\frac{(x+y)^2}{xy^2+x^2y+4(x+y)}=\frac{x+y}{xy+4}=\frac{1}{1+\frac{4}{x+y}}\geq\frac{1}{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Power series representation of $f(x) = \frac{x-1}{x+2}$ What I did so far:
\begin{align*}
\frac{x-1}{x+2}&=(\frac{x-1}{2})\frac{1}{1-(-\frac{x}{2})} \\
&=(\frac{x-1}{2})\sum_{n=1}^\infty (-\frac{x}{2})^n\\
&= (x-1)\sum_{n=1}^\infty \frac{-x^n}{2^{n+1}}
\end{align*}
But i'm stuck there.
| $$f(x)=\frac{x-1}{x+2}=1-\frac{3}{x+2}=1-\frac{3}{2}\frac{1}{1+\frac{x}{2}}\to\\
f(x)=1-\frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{x}{2}\right)^n
$$
There are several ways to represent this exression. If you'd like to have everything under one sum, you can do following:
$$
1=\frac{1}{2}2\to 1=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n+1}\\
f(x)=\sum_{n=0}^{\infty}\left(\left(\frac{1}{2}\right)^{n+1}+\frac{3}{2}(-1)^{n+1}\left(\frac{x}{2}\right)^n\right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$ Given that $x^n - \frac1{x^n}$ is expressible as a polynomial in $x - \frac1x$ with real coefficients only if $n$ is an odd positive integer, find $P(z)$ so that $P\left(x-\frac1x\right) = x^5 - \frac1{x^5}.$
My brain is not working with me today! I don't know how to start this problem. Solutions are greatly appreciated.
| Hint:
$$(x-\frac{1}{x})^5=(x^5-\frac{1}{x^5})-5(x^3-\frac{1}{x^3})+10(x-\frac{1}{x})$$
$$(x-\frac{1}{x})^3=(x^3-\frac{1}{x^3})-3(x-\frac{1}{x})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$ When doing induction should you always try to put your final answer as the "desired " form? For example if: $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ we ought to give the final answer as $$\frac{2(k+1)^{3} + 21(k+1)^{2} + 67(k+1)}{6}?$$
I just expanded both the $\text{LHS}_{k+1}$ and the $\text{RHS}_{k+1}$ to show they were equal after the induction. Like this:
Show that $$\sum^{n}_{k=1}(k+2)(k+4) = \frac{2n^{3} + 21n^{2} + 67n}{6}$$ for all integers $n \geq 1$.
For $n = 1$,
$$\sum^{1}_{k=1}(k+2)(k+4) = 15$$
and
$$\frac{2(1)^{3} + 21(1)^{2} + 67(1)}{6} = 15$$
Assume that it is true for some integer $n = k$, thus $$\sum^{k}_{k=1}(k+2)(k+4) = \frac{2k^{3} + 21k^{2} + 67k}{6}$$ so the $\text{LHS}_{k+1}$ $$\sum^{k+1}_{k=1}(k+2)(k+4) = \sum^{k}_{k=1}(k+2)(k+4) + (k+3)(k+5)$$ $$= \frac{2k^{3} + 21k^{2} + 67k}{6} + \frac{6(k+3)(k+5)}{6}$$ $$=\frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Now the $\text{RHS}_{k+1}$ $$\frac{2(k+1)^{3} + 21(k+1)^{2}+ 67(k+1)}{6} = \frac{2k^{3} + 27k^{2} + 115k + 90}{6}$$ Thus $\text{LHS}_{k+1} = \text{RHS}_{k+1}$ Q.E.D.
| As both members are cubic polynomials, it suffices to show equality for $4$ distinct values of $n$.
$$0=0=\frac{0+0+0}6,\\
0+3\cdot5=15=\frac{2+21+67}6,\\
0+3\cdot5+4\cdot6=39=\frac{2\cdot8+21\cdot4+67\cdot2}6,\\
0+3\cdot5+4\cdot6+5\cdot7=74=\frac{2\cdot27+21\cdot9+67\cdot3}6.\\
$$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverse of $ 3\cdot\frac 57 \mod 7 $ Stuck with the following question: $ 3\cdot\frac 57 \mod 7 $
This can be broken down to $ 3\cdot 5\cdot \frac 17 \mod 7 $ however not sure about the module inverse of $7x = 1 \mod 7$
As per my knowledge we cannot have a inverse, as anything multiplied with $7$ will yield a $0$ when we do a $\mod 7$.
| If we want to find the inverse of $3\cdot 5\cdot \dfrac 1 7 \bmod 7$, we are looking for an $x$ such that $3\cdot 5\cdot \frac 17\cdot x \equiv 1 \pmod 7$. So, we solve as follows
$$
\begin{align*}
3\cdot 5\cdot \frac 17\cdot x &\equiv 1 \bmod 7\\
3\cdot 5\cdot x &\equiv 7 \bmod 7\\
15x &\equiv 0 \bmod 7
\end{align*}
$$
which means $x=0$, so clearly $3\cdot 5\cdot \dfrac 1 7$ does not have an inverse $\bmod 7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $n \in \mathbb{Z}^+$, prove the identity $ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$ Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$
First of all $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$
$$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$
$$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k} \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$
We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$
For the second sum
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$
Integrating both sides from $0$ to $x$, we see that
$$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$
Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$
Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
| $$
\begin{align}
\sum_{k=1}^n\binom{n}{k}\frac{kn^{n-k}}{k+1}
&=n^n\sum_{k=1}^n\binom{n-1}{k-1}\frac1{k+1}\frac1{n^{k-1}}\\
&=n^n\int_0^1\sum_{k=1}^n\binom{n-1}{k-1}\frac{t^k}{n^{k-1}}\,\mathrm{d}t\\
&=n^n\int_0^1t\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{t^k}{n^k}\,\mathrm{d}t\\
&=n^n\int_0^1t\,\left(1+\frac tn\right)^{n-1}\,\mathrm{d}t\\
&=n\int_0^1t\,(n+t)^{n-1}\,\mathrm{d}t\\
&=n\int_n^{n+1}(\color{#C00000}{t}-\color{#00A000}{n})\,t^{n-1}\,\mathrm{d}t\\
&=\color{#C00000}{\frac{n(n+1)^{n+1}-n^{n+2}}{n+1}}-\color{#00A000}{\left(n(n+1)^n-n^{n+1}\right)}\\
&=\frac{n^{n+1}}{n+1}\tag{1}
\end{align}
$$
Subtract the $k=n$ term from $(1)$ and we get
$$
\sum_{k=1}^{n-1}\binom{n}{k}\frac{kn^{n-k}}{k+1}=\frac{n^{n+1}-n}{n+1}\tag{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $ \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x$ I am trying to evaluate the exact value of the following definite integral:
\begin{align} \int^{\frac{91\pi}{6}}_0 |\cos(x)| \, \mathrm{d}x \end{align}
Since $ \int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x$ has symmetry, I did the following:
\begin{align}
\frac{91\pi}{6} \cdot \frac{2}{\pi} = \frac{91}{3}
\end{align}
\begin{align}
\int^{\frac{\pi}{2}}_0 \cos(x) \, \mathrm{d}x = [\sin(x)]^{\frac{\pi}{2}}_0 = 1
\end{align}
\begin{align}
\therefore \text{Bounded Area } = 1 \cdot \frac{91}{3} = \frac{91}{3}
\end{align}
Apparently, the correct answer is $ \frac{61}{2} $ which is very close to my answer. I cannot understand why my answer is wrong. Could someone please advise me?
| for $k\in \mathbb Z$
$\int_{\frac{\pi}{2}+2k\pi}^{\frac{3\pi}{2}+2k\pi}(-cos(x))dx+$
$\int_{\frac{3\pi}{2}+2k\pi}^{\frac{5\pi}{2}+2k\pi}cos(x)=4$
so,
$I=\int_0^{\frac{\pi}{2}}cos(x)dx+4\times7+$
$\int_{15\pi-\frac{\pi}{2}}^{15\pi}(-cos(x))dx+\int_{15\pi}^{15\pi+\frac{\pi}{6}}(-cos(x))dx=1+28+1+\frac{1}{2}=\frac{61}{2}$
| {
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"url": "https://math.stackexchange.com/questions/1971315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What we can infer from equating Taylor series for $(1+x)^n$ and binomial theorem for $(1+x)^n$? $(1+x)^n=1+\frac{(n)ln(1+x)}{1!}+\frac{(n^2)(ln(1+x))^2}{2!}+\frac{(n^3)(ln(1+x))^3}{3!}+\cdots$
$(1+x)^n=1+nx+(^nC_2){x^2}+(^nC_3){x^3+\cdots}$
Here, equating both the series doesn't make any sense of progress to get any sort of result since the former one contains natural logs of $(1+x)$ and the later one has factorial values attached to the variable $x$. So, comparing the coefficients looks like a tedious and cumbersome task. But I was still wondering if we can extract any result from the comparison of the two forms?
| As said in comment, what you did wrong,was to consider that $$(1+x)^n= e^{n\, \ln(1+x)}=e^y$$ and you did apply the expansion of $$e^y=1+y+ \frac{y^2}{2!}+ \frac{y^3}{3!}+\cdots$$ and making $y=n\ln(1+x)$ $$(1+x)^n=1+\frac{n\ln(1+x)}{1!}+\frac{n^2 \ln^2(1+x)}{2!}+\frac{n^3 \ln^3(1+x)}{3!}+\cdots\tag 1$$ which is your first expression.
However, you can still recover using Taylor expansion of $\ln(1+x)$ since $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ $$\ln^2(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)=x^2-x^3+O\left(x^4\right)$$ $$\ln^3(1+x)=x^3+O\left(x^4\right)$$ Replacing in $(1)$, you then get $$(1+x)^n=1+n x+\left(\frac{n^2}{2}-\frac{n}{2}\right)
x^2+\left(\frac{n^3}{6}-\frac{n^2}{2}+\frac{n}{3}\right) x^3+O\left(x^4\right)$$ which, after simplifications, leads to $$(1+x)^n=1+n x+\frac{1}{2} n(n-1) x^2+\frac{1}{6} n(n-1)(n-2) x^3+O\left(x^4\right)$$ which is your second expression.
You took a long way $\cdots$ but it works !
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence formula for a series $x_n$ -- very stuck!!! The Problem:
$x_n$ is defined as $$\sum_{k=0}^\infty \frac{(n+k)!}{k!(2n+2k)!}(\frac{1}{2})^{2k} $$
and satisfies the recurrence: $x_n-2(2n+1)x_{n+1}=x_{n+2}$
But I cannot show that this is the case...
My Attempt:
$$LHS=\sum_{k=0}^\infty\frac{(n+k)!}{k!(2n+2k)!}(\frac{1}{2})^{2k}-2(2n+1)\frac{(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{(n+k)!(2n+1+2k)(2n+2+2k)}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}-2(2n+1)\frac{(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+k)!(n+1+k)(2n+1+2k)-2(2n+1)(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+1+k)!( (2n+1+2k)-(2n+1))}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+1+k)!( 2k)}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+2+k)!( 2k)}{k!(2n+2+2k)!(n+2+k)}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{(n+2+k)!}{(k-1)!(2n+2+2k)!(2n+4+2k)}(\frac{1}{2})^{2(k-1)-1}$$
But from this point, I have no idea how to obtain $x_{n+2}$ from this...
| We have that $\frac{1}{(2n+2k)!}$ is the coefficient of $z^{n+k}$ in $\sum_{m\geq 0}\frac{z^m}{(2m)!}=\cosh(\sqrt{z})$.
On the other hand, by stars and bars we have that $\frac{1}{4^k}\binom{n+k}{k}$ is the coefficient of $z^k$ in
$$ \sum_{m\geq 0}\binom{n+m}{m}\frac{z^m}{4^m}=\frac{1}{(1-z/4)^{n+1}} $$
or the coefficient of $z^{-k}$ in $\left(1-\frac{1}{4z}\right)^{-1-n}$, so that
$$ \sum_{k\geq 0}\frac{(n+k)!}{4^k k!(2n+2k)!} = n!\cdot[z^n]\frac{\cosh(\sqrt{z})}{\left(1-\frac{1}{4z}\right)^{n+1}}=n!\cdot [z^{2n}]\frac{\cosh z}{\left(1-\frac{1}{4z^2}\right)^{n+1}}. $$
By using Cauchy integral formula and the residue theorem, we may check that
$x_n$ is related with the Fourier series of $e^{-\cos\theta}$ over $(-\pi,\pi)$. In particular, we have:
$$ x_n = \sum_{k\geq 0}\frac{(n+k)!}{4^k k!(2n+2k)!} = \frac{\sqrt{\pi}}{2}\cdot I_{\frac{2n-1}{2}}\left(\frac{1}{2}\right) $$
where $I_\nu$ is a modified Bessel function of the first kind. The recurrence relation for the $x_n$ sequence then follows from the integral representation of such function, the cosine addition formulas and integration by parts, as exploited here. Notably, given such recurrence relation, we also have that $x_n$ is a linear combination of $\cosh\frac{1}{2}$ and $\sinh\frac{1}{2}$ for any $n\in\mathbb{N}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify rational fraction
$$\large \frac {bx(a^2x^2+2a^2y^2+b^2y^2)+ay(a^2x^2+2b^2x^2+b^2y^2)}{bx+ay}$$
My attempt:
$$\dfrac{ba^2x^3+2a^2bxy^2+b^3xy^2+a^3x^2y+2b^2x^2ay+b^2ay^3}{bx+ay}$$
I multiplied it out like this, then got stuck...
| First, let's pull out the $bx$ and $ay$ in the numerator:
\begin{eqnarray}
bx(a^2 x^2 &+& 2a^2 y^2 + b^2 y^2) + ay(a^2 x^2 + 2b^2 x^2 + b^2 y^2)\\
&=& (bx+ay)(a^2 x^2 + b^2 y^2) + 2a^2 b x y^2 + 2 a b^2 x^2 y\\
&=& (bx+ay)(a^2 x^2 + b^2 y^2) + ay(2a b x y) + bx(2a b x y)\\
&=& (bx+ay)(a^2 x^2 + 2abxy + b^2 y^2).
\end{eqnarray}
Now divide out the $bx + ay$ and factor to get
$$a^2 x^2 + 2abxy + b^2 y^2 = (ax + by)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving Inequalities similar to Nesbitt's Is it possible to use inequalities like Cauchy-Schwarz or QM-AM-GM-HM to find the minimum value of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}$ for $a,b,c\gt0$?
From just trying different values, the minimum seems to be $11\over6$,
but how would one prove this? I tried setting
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}=S$, giving $S+6=(a+b+c)(\frac{1}{a+b}+\frac{1}{a+c}+\frac{2}{a+b})$, but I'm not sure how to proceed from here, or if this is even the right first step.
| Yes of course! We can use C-S.
For $a=b=1+\sqrt2$ and $c=1$ we get a value $2\sqrt2-1$.
By C-S $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{2kc^2}{kac+bkc}\geq$$
$$\geq\frac{(a+b+\sqrt{2k}c)^2}{ab+ac+ab+bc+kac+kbc}$$
The equality occurs for $\frac{1}{b+c}=\frac{1}{a+c}=\frac{1}{\sqrt{\frac{k}{2}}(a+b)}$, which gives $k=1$.
Id est, it remains to prove that $(a+b+\sqrt2c)^2\geq2(2\sqrt2-1)(ab+ac+bc)$,
which is $2c^2-2(\sqrt2-1)(a+b)c+a^2+b^2-4(\sqrt2-1)ab\geq0$,
for which it's enough to prove that
$$(\sqrt2-1)^2(a+b)^2-2\left(a^2+b^2-4(\sqrt2-1)ab\right)\leq0$$
which is $(a-b)^2\geq0$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Confused about Erdos-Gallai theorem Consider the sorted degree sequence $1,3,3,3$.
Erdos-Gallai states that for $k$ in $1 \leq k \leq n$:
$$\sum_{i=1}^{k} d_i \leq k(k-1) + \sum_{j=k+1}^n min(d_j,k)$$.
And that $\Sigma d_i$ must be even, which it is.
$k=1$ gives the inequality $1 \leq 1(0) + (1 + 1 + 1)$, or $1 \leq 3$, which passes.
$k=2$ gives the inequality $1 + 3 \leq 2(1) + (2 + 2)$, or $4 \leq 6$, which passes.
$k=3$ gives the inequality $1 + 3 + 3 \leq 3(2) + (3)$, or $7 \leq 9$, which passes.
$k=4$ gives the inequality $1 + 3 + 3 + 3 \leq 4(3)$, or $10 \leq 12$, which passes.
But it's easy to see that the degree sequence $1, 3, 3, 3$ is agraphical. What am I missing here?
| As hardmath commented, my ordering was backwards. Erdos-Gallai states that the degree sequence must be ordered largest degree first; that is, the sequence must be $3,3,3,1$.
So we have:
$k=1$ gives $3 \leq 1(0) + (1 + 1 + 1)$, or $3 \leq 4$, which passes.
$k=2$ gives $3 + 3 \leq 2(1) + (2 + 1)$, or $6 \leq 5$, which fails.
Then we're done, but for good measure:
$k=3$ gives $3 + 3 + 3 \leq 3(2) + 1$, or $9 \leq 7$, which fails.
$k=4$ gives $3 + 3 + 3 + 1 \leq 4(3)$, or $10 \leq 12$, which passes.
Thanks hardmath!
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry equation $\tan2x+3 \sec x +3=0$ for $0\leq x\leq 360$ Solve the trigonometry equation for $0\leq x\leq 360$
$$\tan2x+3 \sec x +3=0$$
I've no idea how should I start. Is there any identity I've to use? Hope someone can show out the working and explain for it. Thanks in advance.
| Clearly, $\cos2x\cos x\ne0$
Multiply both sides of the given equation $$\cos x\sin2x+3(1+\cos x)\cos2x=0$$
$$0=2\sin x\cos^2x+3(1+\cos x)\cos2x$$
$$=4\sin\dfrac x2\cos\dfrac x2\cos^2x+3\cdot2\cos^2\dfrac x2\cos2x$$
$$=2\cos\dfrac x2\left(2\sin\dfrac x2\cos^2x+3\cos\dfrac x2\cos2x\right)$$
If $\cos\dfrac x2=0,\dfrac x2=(2n+1)90^\circ\implies x=(2n+1)180^\circ\equiv180^\circ\pmod{360^\circ}$
Else $2\sin\dfrac x2\cos^2x+3\cos\dfrac x2\cos2x=0$
$-2\tan\dfrac x2=\dfrac{3(2\cos^2x-1)}{\cos^2x}$
Use $\cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$ which unfortunately leaves us with a bi-quadratic equation in $\tan\dfrac x2$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the equation is correct I now the Lagrange identity: $$\sqrt{a+\sqrt b}=\sqrt\frac{{a+\sqrt{a^2-b}}}{2}+\sqrt\frac{{a-\sqrt{a^2-b}}}{2}$$
but i didnt know how to prove that the equation
$$\sqrt[3]{2+\sqrt 5}+\sqrt[3]{2-\sqrt 5}=1$$
I do not think that should be used above identity. I tried to resolve the draw as irrational but appears more complicated, so please help me to solve.
Previously, thank you for the helping.
| Let $a=\sqrt[3]{2+\sqrt 5}$, $b=\sqrt[3]{2-\sqrt 5}$. By inspection $a^3+b^3=4$ and $a \cdot b=-1$.
Let $c=a+b$. Then: $$4 = a^3+b^3=(a+b)(a^2 - a \cdot b + b^2) = (a+b)\left((a+b)^2 - 3 a \cdot b\right) = c^3 + 3 c $$
But the equation $c^3 + 3 c - 4 = 0$ has $1$ as the unique real root, so $c=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which number is greater, $2^\sqrt2$ or $e$?
Claim: $\color{red}{2^\sqrt2<e}$
Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus
$$ f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln x \implies e^{x-1} > x^{\sqrt{x}} $$
Putting $x=2$, we obtain
$$e>2^\sqrt2$$
What do you think about this approach? Is there an easier way?
| Approximating the integral by the area of a trapezoid we get for $x>1$ $$\log(x)=\int_1^x\frac{\mathrm{d}t}{t}<(x-1)\frac{1+\frac1x}2=\frac{x^2-1}{2x}$$ So $$\sqrt{2}\log(2)=2\sqrt{2}\log(\sqrt{2})<1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\left(2-\frac 1r\right)x^r+1\leq (x+1)^r$ for $x\in (0,1)$ and $r\in (1,2)$ Let $x\in (0,1)\ \&\ r\in (1,2)$.
How to show that $$\left(2-\frac 1r\right)x^r+1\leq (x+1)^r$$
| Let $f(x) = (x+1)^r - \left(2 - \frac{1}{r}\right)x^r$ where $x$ and $r$ satisfies your conditions.
We have $f^{'}(x) = r(x+1)^{r-1} - \left(2 - \frac{1}{r}\right)rx^{r-1}$
We have $f^{'}(x) = 0 \leftrightarrow r(x+1)^{r-1} = \left(2 - \frac{1}{r}\right)rx^{r-1}$
Therefore $x = \frac{1}{\left( 2-\frac{1}{r}\right)^{\frac{1}{r-1}} - 1}.$
Define $x_0:= \frac{1}{\left( 2-\frac{1}{r}\right)^{\frac{1}{r-1}} - 1}$ and note that $f^{'}(x) >0$ if $x > x_0$ and $f^{'}(x) >0$ if $x < x_0$. Thus $x_0 $is the global minimum of $f.$
A direct computation leads to
$$f(x_0)= \frac{\left( 2 - \frac{1}{r}\right)^{\frac{r}{r-1}} - \left( 2 - \frac{1}{r}\right)}{\left(\left( 2 - \frac{1}{r}\right) - 1\right)^r}$$
Since $2-\frac{1}{r} >1$ and $\frac{r}{r-1}> 1$ we have $f(x_0)>0$
Therefore
$$ \left( 2 - \frac{1}{r}\right)x^r \leq (x+1)^r $$
I think that you only need to consider $x>0$ instead of $x \in (0,1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic approximation of Catalan Numbers The nth Catalan number is :
$$C_n = \frac {1} {n+1} \times {2n \choose n}$$
The problem 12-4 of CLRS asks to find :
$$C_n = \frac {4^n} { \sqrt {\pi} n^{3/2}} (1+ O(1/n)) $$
And Stirling's approximation is:
$$n! = \sqrt {2 \pi n} {\left( \frac {n}{e} \right)}^{n} {\left( 1+ \Theta \left(\frac {1} {n}\right) \right)} $$
So, the nth catalan number becomes :
$$C_n = \frac {2n!}{(n+1)(n!)^2} $$
That, after applying Stirling's approximation becomes:
$$C_n = \left( \frac {1}{1+n} \right) \left( \frac {4^n}{\sqrt{\pi n}} \right) \frac {1}{\left( 1+\Theta \left(1/n\right) \right)}$$
And then, it becomes hopeless. The Asymptotic bound comes in the denominator, not in the numerator.
What should be done now?
Any help appreciated.
Moon
| In this answer, it is shown that
$$
\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}\tag{1}
$$
Using the fact that as $n\to\infty$,
$$
\begin{align}
(n+a)^b
&=n^b\left(1+\frac an\right)^b\\
&=n^b\left(1+O\left(\frac1n\right)\right)\tag{2}
\end{align}
$$
we get
$$
\binom{2n}{n}=\frac{4^n}{\sqrt{\pi n}}\left(1+O\left(\frac1n\right)\right)\tag{3}
$$
Furthermore
$$
\frac1{n+1}=\frac1n\left(1+O\left(\frac1n\right)\right)\tag{4}
$$
Therefore,
$$
\frac1{n+1}\binom{2n}{n}=\frac{4^n}{\sqrt{\pi}n^{3/2}}\left(1+O\left(\frac1n\right)\right)\tag{5}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $x+y+z$ divides $x^2+y^2+z^2$ where x,y,z are consecutive terms in a geometric series The natural numbers $x,y,z$ are consecutive terms in a geometric series. Proves that $x+y+z$ divides $x^2+y^2+z^2$.
$x = ar\\
y = ar^2 = xr\\
z = ar^3 = xr^2$
So..
$\begin{align}
x + y + z & = x + xr + xr^2 \\
& = x(1+r+r^2)\end{align}$
$\begin{align}
x^2 + y^2 + z^2 & = x^2 + x^2r^2 + x^2r^4\\
& = x^2(1+r^2+r^4)\end{align}$
I can see that x is a factor for both but I've no idea where to go from here
| An alternate solution is to observe that
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)$$
and
$$xy+xz+yz=x^2r(1+r+r^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c}$, then find $a,b,c$.
If
$$\int_1^ \infty \frac {x^3+3}{x^6(x^2+1)} \, \mathrm d x=\frac{a+b\pi}{c} $$
then find $a, b, c$.
Now, using partial fractions I calculated
$$a = 62-10\ln (2) \qquad\qquad b = -15 \qquad\qquad c = 20$$
but it took me more than 45 minutes to do all the work. The question was asked in an MCQ exam where only 4-5 minutes are available. I am probably missing something which can help to solve it. Thanks!
Note it was asked in an exam for students of grade 12 so I basically don't know very complex integrations thus I am searching for integration via elementary functions only.
| First, break the integral in two
$$ \int_0^{\infty} \frac{ x^3 + 3 }{x^6(x^2+1)} = \int_1^{\infty} \frac{dx}{x^3(x^2+1)} + 3 \int_1^{\infty} \frac{dx}{x^6(x^2+1)} $$
First, we integrat the second integral. Notice
$$ \int\limits_1^{\infty} \frac{(1 +x^2 - x^2) dx }{x^6(x^2+1)} = \int\limits_1^{\infty} \frac{dx}{x^6} - \int\limits_1^{\infty} \frac{ d x}{x^4(x^2+1)} = \frac{1}{5} - \int\limits_1^{\infty} \frac{dx}{x^4} + \int\limits_1^{\infty} \frac{ dx }{x^2 ( x^2+1)} =$$
$$ = \frac{1}{5} - \frac{1}{3} + \int\limits_1^{\infty} \left( \frac{1}{x^2} - \frac{1}{1+x^2} \right) dx = -\frac{2}{15} + \frac{ \pi }{2} +1 $$
Now, for the first integral, use same trick
$$ \int\limits_1^{\infty} \frac{ (1 + x^2 - x^2)dx}{x^3(x^2+1)} = \int_1^{\infty} \frac{dx}{x^3} - \int_1^{\infty} \frac{ dx }{x(x^2+1)} = \frac{1}{2} - \int_1^{\infty} \frac{ dx }{x(x^2+1)}$$
Using $x = \tan t$, we solve the last integral easily,
$$ \int_1^{\infty} \frac{ \sec^2 t dt }{tan t \sec^2 t } = \int_1^{\infty} \frac{ \cos t dt }{\sin t} = \ln ( \sin t ) = \ln ( \frac{ x }{\sqrt{x^2+1}}) = \frac{1}{2} \ln \left( \frac{x^2 }{x^2+1} \right) \bigg|_1^{\infty} = \frac{1}{2} \ln (1/2)$$
Thus,
$$ \int_0^{\infty} \frac{ x^3 + 3 }{x^6(x^2+1)} = \frac{-2}{15} + \frac{\pi}{2} + 1 + \frac{1}{2} - \frac{1}{2} \ln(1/2) = \boxed{ \frac{30 \ln 2+ 81 + 30 \pi }{60}}$$
| {
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"answer_id": 1
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Prob. 11, Chap. 3, in Baby Rudin: If $a_n > 0$ and $\sum a_n$ diverges, then how do we show that $\sum \frac{a_n}{1+a_n}$ too diverges? Here's Prob. 11, Chap. 3, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Suppose $a_n > 0$, $s_n = a_1 + \cdots + a_n$, and $\sum a_n$ diverges.
(a) Prove that $\sum \frac{a_n}{1+ a_n}$ diverges. [ I have no clue of how to prove this!]
(b) Prove that $$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq 1- \frac{s_N}{s_{N+k}}$$ [I can show this.] and deduce that $\sum \frac{a_n}{s_n}$ diverges. [How to?]
(c) Prove that $$ \frac{a_n}{s_n^2} \leq \frac{1}{s_{n-1}} - \frac{1}{s_n}$$ and deduce that $\sum \frac{a_n}{s_n^2}$ converges. [This I can show, I think.]
(d) What can be said about (the convergence or divergence of) $$\sum \frac{a_n}{1+ n a_n} \ \ \ \mbox{ and } \ \ \ \sum \frac{a_n}{1+ n^2 a_n}?$$ [ How to answer this?]
I would prefer those answers that use only the machinary developed by Rudin himself upto this point in the book.
Here's what I can show:
Since $a_n > 0$, we have $0 < s_{N+1} < \cdots < s_{N+k}$ and so
$$ \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \geq \frac{a_{N+1} + \cdots + a_{N+k}}{s_{N+k}} = 1- \frac{s_N}{s_{N+k}}.$$
As $a_n > 0$, so, for all $n = 2, 3, 4, \ldots$, we have $0 < s_{n-1} < s_n$ and therefore $$ \frac{a_n}{s_n^2} = \frac{s_n - s_{n-1}}{s_n^2} \leq \frac{s_n - s_{n-1}}{s_n s_{n-1}} = \frac{1}{s_{n-1}} - \frac{1}{s_n},$$ and hence $$ 0 \leq \sum_{k=1}^n \frac{a_k}{s_k^2} = \frac{1}{a_1} + \sum_{k=2}^n \frac{a_k}{s_k^2} \leq \frac{1}{a_1} + \sum_{k=2}^n \left( \frac{1}{s_{k-1}} - \frac{1}{s_k} \right) = \frac{1}{a_1} + \frac{1}{s_1} - \frac{1}{s_n} \to \frac{2}{a_1} + 0 $$ as $n \to \infty$ because $a_n > 0$ and $\sum a_n$ diverges and hence $s_n = a_1 + \cdots + a_n \to \infty$ as $n \to \infty$. So if $\lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} $ exists [ but how to show this?}, then we must have $$ 0 \leq \lim_{n\to\infty} \sum_{k=1}^n \frac{a_k}{s_k^2} \leq \frac{2}{a_1}.$$
Am I right?
| Consider two cases: either there are infinitely many $k$ such that $a_k \geq 2$, or there are only finitely many such $k$.
In the first case, the series $\sum_k \frac{a_k}{1+a_k}$ has infinitely many terms which are at least $\frac23$, so it diverges.
In the second case, except for finitely many $k$, we have that
$$
\frac{a_k}{1+a_k} \geq \frac{a_k}3,
$$
so the series also diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ We know $ \alpha $ and $ \beta $ are roots of $ax^2+bx+c = 0 $.
also $S_{n-1} = \alpha^{n-1} + \beta^{n-1}$ , $S_{n} = \alpha^{n} + \beta^{n}$ and $S_{n+1} = \alpha^{n+1} + \beta^{n+1}$.
How we can find relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ ?
Note : We know that relation has $a$ , $b$ and $c$.
| \begin{align*}
\alpha^n+\beta^n &=
(\alpha+\beta)(\alpha^{n-1}+\beta^{n-1})-
\alpha \beta (\alpha^{n-2}+\beta^{n-2}) \\
&= (\alpha+\beta) S_{n-1}-\alpha \beta S_{n-2} \\
S_{n} &= -\frac{b}{a}S_{n-1}-\frac{c}{a}S_{n-2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
When does $A[x,y,z]^T=[y,2x,0]^T$ have a nonzero solution? $$A[x,y,z]^T=[y,2x,0]^T$$ has a nonzero solution
if (and only if?)
$$\det(A-B) = 0$$
where
$$B = \begin{bmatrix}
0 & 1 & 0\\
2 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
Is that right?
| Note that $$B\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} = \begin{pmatrix}
0 & 1 & 0\\
2 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
y\\
2x\\
0
\end{pmatrix}.$$ Thus
$$A\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =\begin{pmatrix}
y\\
2x\\
0
\end{pmatrix}\iff (A-B)\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}.$$ Now, the homogenous system $$ (A-B)\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$$ has a non-trivial solution iff $\det(A-B)=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the improper integral I was asked a question, to evaluate the improper integral:
$$\int \limits_{0}^{\infty}\frac{x^{1/3}}{1+x^2}dx$$
I am using complex analysis to solve this. Consider some small radius r, large radius R, and small angle η. I am considering
the integral along γ1 + γR + γ2 + γr, where γR is the circle of radius R, omitting points with argument
between −η and η, γr is clockwise around the circle of radius r, omitting points with argument between
−η and η, γ1 is along the ray of argument η, from γr to γR, and γ2 is along the ray of argument −η,
from γR to γr.
I think the integral along γR and γr is 0. I estimated them by ML inequality and the integral along those tend to 0. But I am not able to evaluate the integral along γ1 and γ2.
I think I can employ the residue theorem for them. But I do not know what the index is.
Any help is appreciated.
| Consider the contour integral
\begin{align}
\int_C f(z)\ dz=\int_{C}\frac{z^{1/3}}{1+z^2}\ dz
\end{align}
where $C=L_1+C_R +L_2 +C_\epsilon$ is given by
where
\begin{align}
z^{1/3} = \exp\left(\frac{1}{3}\log z \right)
\end{align}
is given by the branch of the logarithm with $-\frac{\pi}{2}<\theta \le \frac{3\pi}{2}$.
It's not hard to see that $f(z)$ has a simple pole at $z=i$ and analytic everywhere else on the strictly upper half plane. Hence by Cauchy's theorem, we have that
\begin{align}
\int_{L_1} \frac{z^{1/3}}{1+z^2}\ dz + \int_{C_R}\frac{z^{1/3}}{1+z^2}\ dz+ \int_{L_2}\frac{z^{1/3}}{1+z^2}\ dz+\int_{C_\epsilon}\frac{z^{1/3}}{1+z^2}\ dz = 2\pi i \operatorname{Res}_{z= i} f(z) = \pi i^{1/3} = \pi e^{i\pi/6}.
\end{align}
Let us simplify each integral. Observe
\begin{align}
\int_{L_1} \frac{z^{1/3}}{1+z^2}\ dz = \int^R_{\epsilon} \frac{x^{1/3}}{1+x^2}\ dx
\end{align}
and
\begin{align}
\int_{L_2} \frac{z^{1/3}}{1+z^2}\ dz =&\ \int^{-\epsilon}_{-R} \frac{x^{1/3}}{1+x^2}\ dx .
\end{align}
Next, observe that
\begin{align}
\left|\int_{C_R}\frac{\exp\left(\frac{1}{3}\log|z|+i\frac{\theta}{3} \right)}{1+z^2}\ dz\right|\le \int_{C_R} \frac{R^{1/3}}{R^2-1}\ |dz| \le C \frac{R^{1+1/3}}{R^2-1}\rightarrow 0
\end{align}
as $R \rightarrow \infty$. Lastly, we have that
\begin{align}
\left|\int_{C_\epsilon} \frac{z^{1/3}}{1+z^2}\ dz\right|\le \frac{C\epsilon^{1/3+\epsilon}}{1-\epsilon^2}\Rightarrow 0
\end{align}
as $\epsilon\rightarrow 0$.
Thus, it follows
\begin{align}
\int^\infty_{-\infty} \frac{x^{1/3}}{1+x^2}\ dx = \pi e^{i\pi/6}.
\end{align}
Edit: This post was made when the original question was to evaluate
\begin{align}
\int^\infty_{-\infty} \frac{x^{1/3}}{1+x^2}\ dx.
\end{align}
The reader should note that this integral doesn't have a unique answer. You could get what I have shown or you could also get what robjohn have gotten. The ambiguity comes from the interpretation of $x^{1/3}$, which is not as simple as it looks. In particular, we need to first determine the meaning of $(-1)^{1/3}$. Most students in high school or even freshmen in college are taught that $(-1)^{1/3} = -1$ since $(-1)^3 = -1$, which is correct since a first course in calculus is usually restricted to the real numbers.
However, once students have learned a bit about the complex numbers then they should start to re-evaluate what $(-1)^{1/3}$ actually means. The first thing they should do is to look at any pre-calculus books (in particular, they should look at the pre-calculus book they had learned from) to see the definition of $a^b$. It shouldn't come as a surprise that most of the books only define $a^b$ for $a>0$.
So how should we define $(-1)^{1/3}$? The general definition used in complex analysis for raising a complex number $z$ to a complex power $a$ is defined as follow
\begin{align} z^a = \exp\left(a \log z \right) = \exp\left( a\ast
[\log|z|+i \arg \theta]\right),
\end{align}
i.e. you need to make a choice for your range of $\theta$ (in other words, you need to choose a branch of the logarithm) otherwise $z^a$ will be a multi-valued function on the complex plane, which mean it is not a function.
Why is it not a function? Let us look at the example $(-1)^{1/3}$. By definition
\begin{align}
(-1)^{1/3} = \exp\left(\frac{1}{3}\log(-1) \right) = \exp\left(\frac{i}{3}\arg (-1) \right).
\end{align}
Let us test the cases $\theta = \pi, 3\pi, 5\pi$ (all the other cases are similar to these three) since all three are arguments of $-1$. Observe
\begin{align}
\exp\left(i\frac{\pi}{3} \right) =&\ \cos\frac{\pi}{3} + i\sin\frac{\pi}{3}= \frac{1}{2}+i\frac{\sqrt{3}}{2}\\
\exp\left(i\frac{3\pi}{3} \right) =&\ \cos \pi + i \sin \pi = -1\\
\exp\left(i\frac{5\pi}{3} \right) =&\ \cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3} = \frac{1}{2}-i\frac{\sqrt{3}}{2}
\end{align}
which means $(-1)^{1/3}$ have three interpretations.
In my interpretation of $z^{1/3}$, I have used the branch where $-\frac{\pi}{2}<\theta\le \frac{3\pi}{2}$, which contains $\pi$.Whereas, robjohn chose a branch that contains $3\pi$ ( or something similar) say $\frac{3\pi}{2}<\theta\le\frac{7\pi}{2}$ which will lead to the conclusion
\begin{align}
\int^\infty_{-\infty} \frac{x^{1/3}}{1+x^2}\ dx = 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x \leq y \leq z$, and both sides of the equation are defined, then $\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.$ Show that if $x$, $y$, and $z$ are consecutive terms of an arithmetic sequence, with $x \leq y \leq z$, and both sides of the equation are defined, then
$$\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.$$
I have no idea how to even start this problem, I'm stuck. Solutions are greatly appreciated.
| Let $x=y-d$ and $z=y+d$ with $d>0$. Then
\begin{align*}
\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z}& = \frac{\sin (y-d) + \sin y + \sin (y+d) }{\cos (y-d) + \cos y + \cos (y+d)}\\
& = \frac{2\sin y \cos d + \sin y}{2 \cos y \cos d + \cos y}\\
& = \tan y.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inductive proof of $1^2-2^2+3^2-\dots+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}2$ $$P(n):1^2-2^2+3^2-\dots+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}2$$
I'm having trouble proving $P(k+1)$ through mathematical induction for this problem.
For $P(k+1)$, I have
$$1^2-2^2+3^2-\dots+(-1)^{(k+1)-1}(k+1)^2=\frac{(-1)^{(k+1)-1}(k+1)(k+2)}2$$
But I am unsure how to simplify/prove $P(k+1)$ from here. Any suggestions?
Thank you!
| Here it is better to derive $P(n+1)$ from $P(n)$ rather than the other way round. I'll use sigma notation for the sum:
$$1^2-2^2+3^2-\dots+(-1)^{n-1}n^2=\sum_{k=1}^n(-1)^{k-1}k^2$$
Having shown $P(n)$:
$$\sum_{k=1}^n(-1)^{k-1}k^2=\frac{(-1)^{n-1}n(n+1)}2$$
we now add the next term of the summation, $(-1)^n(n+1)^2$:
$$\sum_{k=1}^{n+1}(-1)^{k-1}k^2=\frac{(-1)^{n-1}n(n+1)}2+(-1)^n(n+1)^2$$
$$=\frac{(-1)^{n-1}n(n+1)}2-\frac{2(-1)^{n-1}(n+1)^2}2$$
$$=\frac{(-1)^{n-1}(n+1)(n-2(n+1))}2$$
$$=\frac{(-1)^{n-1}(n+1)(-n-2)}2$$
$$=\frac{(-1)^n(n+1)(n+2)}2$$
This is $P(n+1)$ and completes the inductive step. Of course, you still need the base case, which is $P(1):1^2=\frac{(-1)^0×1×2}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}$ I want to evaluate following integral
\begin{align}
\int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}
\end{align}
Simple try on this integral is using branch cut and apply residue theorem.
Usual procedure gives for $0 < \alpha < 1$, with $Q(x)$ deg $n$ and $P(x)$ deg $m$, for $x>0$, $Q(x) \neq 0$
\begin{align}
\int_0^{\infty} \frac{x^\alpha P(x)}{Q(x)} dx = \frac{2\pi i}{1- e^{i\alpha 2 \pi}} \sum_j Res[\frac{z^\alpha P(z)}{Q(z)} , z_j]
\end{align}
where $z_j$ are poles which does not make $\frac{P}{Q}$ be zero.
This formula comes from Mathews and Howell's complex analysis textbook.
And this is nothing but applying branch cut to make $x^{\frac{1}{4}}$ singled valued function. I think this formula works for above improper integral but results seems different.
Apply $\alpha=\frac{1}{4}$ and take poles $z_0=-1$, $z_1 = e^{\frac{i \pi}{3}}$
, $z_2 = e^{\frac{i5 \pi}{3}}$, i got different things.
Am i doing right?
\begin{align}
\frac{2\pi i}{1-i}\frac{1}{3} \left( e^{\frac{1}{4} \pi i} + e^{-\frac{7}{12}\pi i} + e^{-\frac{35}{12} \pi i}\right)
\end{align}
| Letting $y=\frac{1}{1+x^3}$ yields
$$
\begin{aligned}
I &=\int_{1}^{0} y\left(\frac{1}{y}-1\right)^{\frac{1}{12}} \frac{1}{3}\left(\frac{y}{1-y}\right)^{\frac{2}{3}} \frac{d y}{-y^{2}} \\
&=\frac{1}{3} \int_{0}^{1} y^{\frac{1}{12}-1}(1-y)^{\frac{5}{12}-1} d y\\&= \frac{1}{3} B\left(\frac{7}{12}, \frac{5}{12}\right)
\end{aligned}
$$
Using the property of Beta function$$
B(x, 1-x)=\pi \csc (\pi x), \quad \textrm{ where } x \notin \mathbb{Z},
$$
we can now conclude that
$$
\boxed{I=\frac{\pi}{3} \csc \frac{5 \pi}{12}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that $\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$? How to show that the following is true? $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} \in \mathbb{Z}$$
I have tried to set $$\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} = r,$$ $$a=\sqrt[3]{26+15\sqrt{3}},$$ $$b=\sqrt[3]{26-15\sqrt{3}},$$ and used the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3})$$ but I got nowhere. I am stuck at $$\left(a^{1/3}+b^{1/3}\right)^3=52+ 3 \cdot \left(a^{1/3}+b^{1/3}\right)$$
I'd be glad at any help.
| Find $p$ and $q$ such that
$$
-\frac{q}{2}=26
\qquad
\frac{q^2}{4}+\frac{p^3}{27}=675
$$
We get $q=-52$ and $p=-3$. So your number is a root of
$$
x^3-3x-52=0
$$
and you can see that $4$ is the only real solution. Cardan's formula tells you that
$$
\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=4
$$
Alternatively,
$$
\frac{1}{26-15\sqrt{3}}=\frac{26+15\sqrt{3}}{26^2-675}=26+15\sqrt{3}
$$
so $b=a^{-1}$. Then
$$
r^3=(a+a^{-1})^3=a^3+3a+3a^{-1}+a^{-3}=3r+52
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can't solve lagrange multiplier problem I've been working on this particular problem for some time and can't seem to see what I'm doing wrong. The solution provided says the extrema are $\pm 7^{1/2}/8$. If someone could be so kind as to show me where my error lies, it would be much appreciated.
The questions posed is as follows:
Use lagrange multipliers to find the extrema of $f(x,y,z) = x + y + z$ subject to the constraint $x^2 + 2y^2 + 4z^2 = 1/16$.
I've attached a photo of my work
Thanks so much!!
| So you have
$\lambda2x = \lambda 4y = \lambda 8z = 1\\
y = \frac x2\\
z = \frac x4$
$x^2 + 2 y^2 + 4z^2 = \frac 1{16}\\
x^2 + 2 (\frac x2)^2 + 4(\frac x4) = \frac {1}{16}$
And this is where you made your mistake. Carrying it through.
$x^2 + \frac {x^2}{2} + \frac {x^2}{4} = \frac {1}{16}\\
16x^2 + 8x^2 + 4x^2 = 1\\
28 x^2 = 1\\
x = \frac {1}{2\sqrt{7}}\\
y = \frac {1}{4\sqrt{7}}\\
z = \frac {1}{8\sqrt{7}}$
$x+y+z = \frac {7}{8\sqrt {7}}\\
x+y+z = \frac {\sqrt 7}{8}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
Calculating the Discriminant for The Following Quadratic: Consider $$ p_{k+1} \zeta_{k+1}^2 + q_{k+1}\zeta_{k+1} + r_{k+1} = 0 $$
where
$$p_{k+1} = p_0P_k^2 + q_0P_kQ_k + r_0Q_k^2$$
$$q_{k+1} = 2p_0P_kP_{k-1} + q_0 ( P_kQ_{k-1} + Q_kP_{k-1} ) + 2r_0Q_kQ_{k-1}$$
$$r_{k+1} = p_0P_{k-1}^2 + q_0 P_{k-1}Q_{k-1} + r_0Q_{k-1}^2 = p_k $$
Given that $p_k \neq 0$ for any given $k$
Now the discriminant of this equation turns out to be:
$$\Delta = (q_0^2 - 4p_0r_0 )( P_kQ_{k-1} - P_{k-1}Q_k)^2 $$
Specifically, $$C_k = \frac{ P_k } { Q_k } $$ denote the $k-th$ convergent of a simple continued fraction, so the identity $$P_kQ_{k-1}+Q_kP_{k-1} = \pm 1 $$ may be useful for our calculation.
I noticed that the first bracketed term of $\Delta$ was simply when $k=-1$, so is there some sort of easy way of achieving that expression without any intense algebra?
Kind Regards,
| Let's see, for you, $\alpha = P_k,\gamma = Q_k, $ $\beta = P_{k-1},\delta = Q_{k-1}, $
$$
p =
\left(
\begin{array}{cc}
P_k & P_{k-1} \\
Q_k & Q_{k-1}
\end{array}
\right)
$$
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
We have a binary form $\langle A,B,C \rangle$ meaning $f(x,y) = A x^2 + B xy + C y^2.$ We create the Hessian matrix
$$
h =
\left(
\begin{array}{cc}
2A & B \\
B & 2 C
\end{array}
\right)
$$
with discriminant
$$ \Delta = B^2 - 4 AC. $$
To get back to the triple of coefficients we halve the diagonal entries but keep one of the off diagonal entries as is, for $B.$
Given a matrix
$$
p =
\left(
\begin{array}{cc}
\alpha & \beta \\
\gamma & \delta
\end{array}
\right)
$$
we calculate the symmetric matrix $p^T h p$ to give a new one,
$$ \langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2 \rangle $$
Note that, about the discriminant $\Delta,$ we automatically have the new discriminant being $\Delta \det^2 p.$
Given
$$
Q =
\left(
\begin{array}{ccc}
\alpha^2 & 2 \alpha \beta & \beta^2 \\
\alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\
\gamma^2 & 2 \gamma \delta & \delta^2
\end{array}
\right),
$$
there is a typographical error on page 23 of Magnus, actually $\det Q = (\alpha \delta - \beta \gamma)^3.$
If we now write the triple $(A,B,C)$ as a row vector, we find
$$
(A,B,C)
\left(
\begin{array}{ccc}
\alpha^2 & 2 \alpha \beta & \beta^2 \\
\alpha \gamma & \alpha \delta + \beta \gamma & \beta \delta \\
\gamma^2 & 2 \gamma \delta & \delta^2
\end{array}
\right) =
$$
$$
( A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2)
$$
Compare our earlier $ p^T h p =$
$$ \langle A \alpha^2 + B \alpha \gamma + C \gamma^2, 2A \alpha \beta + B(\alpha \delta + \beta \gamma) + 2 C \gamma \delta, A \beta^2 + B \beta \delta + C \delta^2 \rangle $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Limit of sum with cube roots Do you have any hints for the following limit?
$$\lim\limits_{n \to \infty}{1 \over n^2}\left( {1 \over \sqrt[3]{1}} + {1 \over \sqrt[3]{2}} + {1 \over \sqrt[3]{3}} + \dots + {1 \over \sqrt[3]{n^2-1}} + {1 \over \sqrt[3]{n^2}} \right)$$
| One may use a Riemann sum, as $n \to \infty$,
$$
\frac{1}{n^2} \sum_{k=1}^{n^2} \sqrt[3]{\frac{n^2}{k}}\to \int_0^1 \frac{1}{\sqrt[3]{x}}\:dx=\frac32
$$ giving, as $n \to \infty$,
$$
\frac{1}{n^2} \sum_{k=1}^{n^2} \frac{1}{\sqrt[3]{k}}=\frac{1}{n^{2/3}} \cdot \frac{1}{n^2} \sum_{k=1}^{n^2} \sqrt[3]{\frac{n^2}{k}} \sim \frac{1}{n^{2/3}} \cdot \frac32 \to 0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $f(1)+\cdots+f(60)$
Let $f(x) = \dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}}$. Find $f(1)+\cdots+f(60)$.
I considered rationalizing the denominator, but that seems to make the fraction more complicated. We get $$\dfrac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} = \dfrac{1}{2}\left(4x+\sqrt{4x^2-1}\right)\left(\sqrt{2x+1}-\sqrt{2x-1}\right).$$ is there an easier way?
| Using the fact that $(\sqrt{2x+1}+\sqrt{2x-1})^2=4x+2\sqrt{4x^2-1}$, you can produce :
\begin{align}
\frac{4x+\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} &=
\frac{(\sqrt{2x+1}+\sqrt{2x-1})^2-\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\
&= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\ &= \sqrt{2x+1}+\sqrt{2x-1} - \frac{\sqrt{4x^2-1}(\sqrt{2x+1}-\sqrt{2x-1})}{2} \\
&= \sqrt{2x+1}+\sqrt{2x-1} - \frac{(2x+1)}{2}\sqrt{2x-1} + \frac{(2x-1)}{2}\sqrt{2x+1} \\
&= \frac{(2x+1)^{3/2}}{2} - \frac{(2x-1)^{3/2}}{2}
\end{align}
So your sum is telescopic, and you find :
$$\sum_{k=1}^{60} f(k) = \frac{(2\times 60+1)^{3/2}}{2} - \frac{(2\times 1-1)^{3/2}}{2} = 665$$
Note : I'm sure there's a much cleaner way to obtain this, but I don't see how :-)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find a generating function for $3r^3 - 5r^2 + 4r$ a. Show how $r^2$ and $r^3$ can be written as linear combinations of $P(r,3)$, $P(r,2)$, and $P(r,1)$.
b. Use part (a) to find a generating function for $3r^3 - 5r^2 + 4r$
I believe that I have part a, simply by calculating each of the permutations.
\begin{align*}
P(r,3) &= (r-2)(r-1)r \\
&= r^3 - 3r^2 + 2r
\end{align*}
\begin{align*}
P(r,2) &= (r-1)r \\
&= r^2 - r
\end{align*}
\begin{align*}
P(r,1) &= r
\end{align*}
Then we can see that
\begin{align*}
P(r,2) + P(r,1) &= (r^2 - r) + r\\
&= r^2
\end{align*}
and
\begin{align*}
P(r,3) + 3P(r,2) + P(r,1) &= r^3 - 3r^2 + 2r + 3(r^2 - r) + r \\
&= r^3 - 3r^2 + 2r + 3r^2 - 3r + r \\
&= r^3
\end{align*}
As for part b, I have found that it can be written with the linear combination $3P(r,3) + 4P(r,2) + 2P(r,1)$. However, I am not sure how to use this to calculate a generating function.
| Hint: Take the geometric series i.e., $$\sum _{n=0}^{\infty}x^n=\frac{1}{1-x},$$
differentiate both sides, you will get $$\sum _{n=0}^{\infty}nx^{n-1}=(\frac{1}{1-x})'=\frac{1}{(1-x)^2},$$ multiply both sides by $x,$
$$\sum _{n=0}^{\infty}nx^n=\sum _{n=0}^{\infty}P(n,1)x^n=\frac{x}{(1-x)^2}.$$
Do it for the other ones and use the combination you got.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
First and second derivative of $|x|^3$ I need to prove that $|x|^3$ is twice differentiable, by showing that the first and second derivatives exist using the definition. I've tried several ways, this is what I've got:
$$\lim_{h\to 0} \frac{|x+h|^3 - |x|^3}{h} = \lim_{h\to 0}\frac{\sqrt{(x+h)^2}^3 - \sqrt{x^2}^3}{h} = \lim_{h\to 0} \frac{\sqrt{(x+h)^6} - \sqrt{x^6}}{h}$$
then I rationalized the numerator:
$$= \lim_{h\to 0} \frac{(x+h)^6 - x^6}{h\left(\sqrt{(x+h)^6} + \sqrt{x^6}\right)}$$
and I'm stuck on what to do next, I'm skeptic that this is the right way, but I was not able to reach any answer using other ways either.
Any help would be great.
Thanks.
| Just separate the last limit, i.e.,
\begin{aligned}
L&=\lim_{h\to 0}\frac{(x+h)^6 - x^6}{h\left(\sqrt{(x+h)^6} + \sqrt{x^6}\right)}\\ &=
\lim_{h\to 0}\frac{(x+h)^6 - x^6}{h}\cdot \lim_{h\to 0}{\frac1{\sqrt{(x+h)^6} + \sqrt{x^6}}} \\&= \frac{\mathrm{d}x^6}{\mathrm{d}x}\cdot\frac1{2\sqrt{x^6}}
\\&= 6x^5\cdot\frac1{2|x^3|} \\ &=6x^5\cdot\frac{1}{2x^2|x|}
\\ &=3\frac{x^3}{|x|}
\end{aligned}
and you're done.
If you need to prove $\frac{\mathrm{d}x^6}{\mathrm{d}x}=6x^5$, you can use this \begin{aligned}
\frac{\mathrm{d}x^6}{\mathrm{d}x}&=\lim_{h\to0}\frac{(x+h)^6-x^6}h\\
&=\lim_{h\to0}\frac{(x+h-x)\left((x+h)^5+x(x+h)^4+x^2(x+h)^3+x^3(x+h)^2+x^4(x+h)+x^5\right)}h\\
&=\lim_{h\to0}\frac{h\left((x+h)^5+x(x+h)^4+x^2(x+h)^3+x^3(x+h)^2+x^4(x+h)+x^5\right)}h\\
&=\lim_{h\to0}\left((x+h)^5+x(x+h)^4+x^2(x+h)^3+x^3(x+h)^2+x^4(x+h)+x^5\right)\\
&=6x^5
\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
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find the volume in the octant bounded by x+y+z=9,2x+3y=18 and x+3y=9 Find the volume in the first octant bounded by x+y+z=9,2x+3y=18 and x+3y=9.
it's obvious that z=9-x-y, but for range of x and y i am confused. from 2x+3y=18 and x+3y=9, i find that x=9-(3/2)y and x=9-3y, so 9-(3/2)y≤x≤9-3y. is that correct? also for y, 3≤y≤6( assume x=0).
so volume=∫(3to6)∫(9-(3/2)y to 9-2y)(9-x-y)dx dy. answer is not correct(should be 81/2)i don't know which part i got wrong.
| Looking at the two constraints that don't involve $z$, we have the following:
Take the limits of $x$ as $0 \le x \le 9$
Then the lower bound of $y$ is given by $x+3y=9 \Rightarrow y=\frac {9-x}{3}$
and the lower bound of $y$ is given by $2x+3y=18 \Rightarrow y=\frac {18-2x}{3}$
The integral you want is $V=\int_{x=0}^{x=9} \int_{y=\frac {9-x}{3}}^{y=\frac {18-2x}{3}}(9-x-y) dy dx $
$V=\int_{x=0}^{x=9} [9y-xy-\frac 12y^2]_{y=\frac {9-x}{3}}^{y=\frac {18-2x}{3}} dx $
$V=\int_{x=0}^{x=9} [9\frac {18-2x}{3}-x\frac {18-2x}{3}-\frac 12(\frac {18-2x}{3})^2-9\frac {9-x}{3}+x\frac {9-x}{3}+\frac 12(\frac {9-x}{3})^2] dx $
$V=\int_{x=0}^{x=9} \frac 1{18}[54(18-2x)-6x(18-2x)-(18-2x)^2-54(9-x)+6x(9-x)+(9-x)^2] dx $
$V=\int_{x=0}^{x=9} \frac 1{18}[972-108x-108x+12x^2-324+72x-4x^2-486+54x+54x-6x^2+81-18x+x^2] dx $
$V=\int_{x=0}^{x=9} \frac 1{18}[243-54x+3x^2]dx $
$V=[\frac 1{18}(243x-27x^2+x^3)]_{x=0}^{x=9} $
$V=[\frac 1{18}(243.9-27.81+729)]$
$V=\frac {729}{18}$
$V=\frac {81}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018970",
"timestamp": "2023-03-29T00:00:00",
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How to prove using induction that $ \sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n} = \frac{2n}{n+1}$ Using Mathematical Induction I need to prove that
$ \sum_{i = 1}^n \frac{1}{\sum_{n = 0}^i n} = \frac{2n}{n+1}$
As A first step I verified it for numbers 1 and 2 which worked
Secondly I simplified the LHS as:
$ \frac{1}{\sum n} = \frac{1}{\frac {n(n+1)}{2}} = \frac{2}{n(n+1)}$
which converts my problem into
$ \frac{2}{1 \cdot 2} + \frac{2}{2 \cdot 3} + \frac{2}{3 \cdot 4} ......... + \frac{2}{n \cdot (n+1)} = \frac{2n}{n+1} $
After which I cancel out 2 as the common factor from both sides. But at this stage I am stuck and am unable to go forward. What should I do next ?
| Straightforward. After simplifying your expression, proceed with your induction.
$$\sum_{i=1}^{n+1} \dfrac{2}{i(i+1)} = \sum_{i=1}^{n} \dfrac{2}{i(i+1)} + \frac{2}{(n+1)(n+2)} = \frac{2n}{n+1} + \frac{2}{(n+1)(n+2)} = \frac{2n^2+4n+2}{(n+1)(n+2)} = \frac{2(n+1)}{n+2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the range of $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$, for $a$, $b$, $c$ the sides of a triangle
If $a$, $b$, and $c$ are the three sides of a triangle, then
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$$
lies in what interval?
| For a $\triangle ABC,$ we have $a+b>c$ and $b+c>a$ and $c+a>b$
For $\bf{Upper\; bound}$
$$a+b>c\Rightarrow 2(a+b)>a+b+c\Rightarrow \frac{1}{2(a+b)}<\frac{1}{a+b+c}\Rightarrow \frac{c}{a+b}<\frac{2c}{a+b+c}$$
In a Similar way for other two expression, And then adding , We get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c} = 2$$
Similarly For $\bf{Lower\; bound}$
$$a>0\Rightarrow a+b+c>b+c\Rightarrow \frac{1}{a+b+c}<\frac{1}{b+c}\Rightarrow \frac{a}{a+b+c}<\frac{a}{b+c}$$
Similar way for other two expression, And then Adding, We get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c} = 1$$
So we get $$1<\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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find n, if the coefficient of $x^2$ in $(3 + 2x)^n$ is 20412. Find $n$, if the coefficient of $x^2$ in $(3+2x)^n$ is 20412
I solved it using binomial theorem and got
$n(n-1)3^{n-2} = 10206$
The term like $n(n-1)3^{n-2}$, I am not sure how to solve it.
| The coefficient should be $2n(n-1)3^{n-2}$.
To obtain $n$ use prime factorisation: $\;20412=2^2\cdot3^6\cdot 7$. Maybe $n$ or $n-1$ is divisible by $3$, so the equality
$$2n(n-1)3^{n-2}=2^2\cdot3^6\cdot 7$$
implies $\;n\le 8$.
By Euclid's lemma, it also implies $7$ divides $n$ or $n-1$. Combining with $n\le 8$, we have that $n=7$ or $n=8$. However $n=8$ is impossible, as the expression is divisible only by $2^2$. Thus there is only one solution: $\;n=\color{red}7$.
| {
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"url": "https://math.stackexchange.com/questions/2027024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving limits of two sequences I've trouble proving that:
a) $\lim(\frac{1}{\sqrt[n]{2}-1} - \frac{2}{\sqrt[n]{4}-1})=\frac{1}{2}$
b) $\lim \frac{2n-1}{2n+1}\cdot \frac{2n-2}{2n+2}\cdot ... \cdot \frac{n}{3n}=0$.
In a) I've tried using formula $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ which gives me $\lim(\frac{1}{\sqrt[n]{2}-1} - \frac{2}{\sqrt[n]{4}-1})=\lim \sum_{k=1}^{n}\frac{2}{\sqrt[n]{2^k}}(1- \frac{4}{3\sqrt[n]{2^k}})$ and I can't go any further.
In b) I've tried using Squeeze Theorem $\lim \frac{2n-1}{2n+1}\cdot \frac{2n-2}{2n+2}\cdot ... \cdot \frac{n}{3n}>(\frac{1}{3})^n$ but i cant find second inequality.
| For part a), we have
$${1\over2^{1/n}-1}-{2\over4^{1/n}-1}={1\over2^{1/n}-1}-{2\over2^{2/n}-1}={1\over2^{1/n}-1}-{2\over(2^{1/n}-1)(2^{1/n}+1)}={(2^{1/n}+1)-2\over(2^{1/n}-1)(2^{1/n}+1)}={1\over2^{1/n}+1}\to{1\over1+1}={1\over2}$$
For part b), we have
$$\left(2n-1\over2n+1\right)\left(2n-2\over2n+2\right)\cdots\left(n+1\over3n-1\right)\left(n\over3n\right)=\left(n\over2n+1\right)\left(n+1\over2n+2\right)\cdots\left(2n-2\over3n-1\right)\left(2n-1\over3n\right)$$
and, for $0\le k\le n-1$, we have
$${n+k\over2n+1+k}\lt{2\over3}$$
since $3n+3k\lt4n+2+2k$ is equivalent to $k\lt n+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cover-up rule for transfert function:$\frac{z^2+2z-1}{z^2(z-1)}$ $$\frac{Y(z)}{z}=\frac{z^2+2z-1}{z^2(z-1)}=\frac{A}{z-1}+\frac{B}{z}+\frac{C}{z^2}$$
I find $A=2$ but what are the values for $B$ and $C$ ?
From the book Introduction to Digital Signal Processing
written by Bob Meddins
Regards
| You have obtained
$$
\frac{z^2+2z-1}{z^2(z-1)}=\frac{2}{z-1}+\frac{B}{z}+\frac{C}{z^2} \tag1
$$ then multiplying out by $z^2$ one gets
$$
\frac{z^2+2z-1}{(z-1)}=\frac{2z^2}{z-1}+z\: B+C
$$ letting $z:=0$ gives $C=1$ then one has
$$
\frac{z^2+2z-1}{z^2(z-1)}=\frac{2}{z-1}+\frac{B}{z}+\frac{1}{z^2} \tag2
$$letting $z:=-1$ gives
$$
1=\frac{2}{-2}+\frac{B}{-1}+\frac{1}{1} \tag2
$$ and $B=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove using Taylor's Theorem: $1-\cos(\frac{x}{n})=\frac{x^2}{n^2}-\cos\theta(x_n,n)\frac{x^3}{6n^3}$ let $F_n(x)=n^2(1-\cos(\frac{x}{n}))$. I am asked to prove that this converges pointwise to limit function $F$ on a set $S$. In this case using geogebra, I see that it converges to $F=x^2$ where $x\in \mathbb{R}$.
From the solutions manual, I found they got this using Taylor's theorem,
$1-\cos(\frac{x}{n})=\frac{x^2}{n^2}-\cos\theta(x_n,n)\frac{x^3}{6n^3}$
where $\theta(x_n,n)$ is between $0$ and $\frac{x}{n}$
Furthermore, they state that $|F_n(x)-x^2|\leq \frac{x^4}{24n^2}$.
How can I show,
$1-\cos(\frac{x}{n})=\frac{x^2}{n^2}-\cos\theta(x_n,n)\frac{x^3}{6n^3}$
and $|F_n(x)-x^2|\leq \frac{x^4}{24n^2}$.
What I have so far:
$1-\cos(\frac{x}{n})=1-\cos(\frac{a}{n})+\sin(\frac{a}{n})*(\frac{1}{n})*(\frac{x}{n}-\frac{a}{n})^2+\cos(\frac{a}{n})*(\frac{1}{n})*(\frac{x}{n}-\frac{a}{n})^3$....
How would I proceed.
| Hint. Assume $h>0$. Integrating by parts three times, one has
$$
\begin{align}
\cos h-1&=\int_0^h-\sin t\:dt
\\&=\left[(h-t) \frac{}{}\sin t\right]_0^h-\int_0^h (h-t)\cos t\:dt
\\&=\color{red}{0}+\left[\frac{(t-h)^2}{2}\cdot\cos t\right]_0^h+\int_0^h \frac{(t-h)^2}{2}\cdot\sin t\:dt
\\&=-\frac{h^2}{2}+\frac12\int_0^h (t-h)^2\cdot\sin t\:dt
\\&=-\frac{h^2}{2}+\left[\frac{(t-h)^3}{6}\cdot\sin t\right]_0^h-\int_0^h \frac{(t-h)^3}{6}\cdot\cos t\:dt
\\&=-\frac{h^2}{2}+\color{red}{0}-\int_0^h \frac{(t-h)^3}{6}\cdot\cos t\:dt
\end{align}
$$ or
$$
\cos h=1-\frac{h^2}{2}+\int_0^h \frac{(h-t)^3}{6}\cdot\cos t\:dt
$$
then one may use the mean value theorem for definite integrals to conclude.
| {
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"timestamp": "2023-03-29T00:00:00",
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What points on surface is the tangent plane parallel to $xy$- plane? At what pts. on the surface $z = x^{2}y + y^{2}x + 3x$ is the tangent plane parallel to the $xy$-plane?
So first I define a function $F(x, y, z) = x^{2}y + y^{2}x + 3x - z$ which has gradient $grad F = (2xy + y^2 + 3, x^{2} + 2yx, -1)$. So the equation of our tangent plane is: $(2xy + y^{2} + 3)(x - x_{0}) + (x^{2} + 2yx)(y - y_{0}) - (z - z_{0}) = 0$.
So we get as a normal line:
$r(t) = (x_{0} + (2x_{0}y_{0} + y_{0}^{2} + 3)t, y_{0} + (x_{0}^{2} +2y_{0}x_{0})t, z_{0} - t)$.
Now I know two planes are parallel if their normal lines are parallel, but I'm not quite sure how to complete the problem.
| If the tangent plane to the surface defined by $f(x,y,z) = 0$ with $f(x,y,z) = x^2y+y^2x+3x-z$ at the point $P = (x,y,z)$ is parallel to the $xy$ plane, then its normal line must be parallel to the vector $\vec{a} = <0,0,1>$. But $\vec{n} = \nabla{f}= <2xy+y^2+3, x^2+2xy, -1>$. Thus we must have: $2xy+y^2+3 = 0 = x^2+2xy$. Observe that $x \neq 0$ since $y^2+3 > 0$. Thus: $x(x+2y) = 0 \implies x = -2y\implies -4y^2+y^2+3 = 0\implies y^2 = 1 \implies y = \pm 1\implies x = \mp 2$. Thus the points are $(2,-1,4), (-2,1,-4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why are the roots of $x^3+2=0 $: $-2^{1/3}, (-2)^{1/3}$, and $-(-1)^{2/3} 2^{1/3}$? For some reason I can't understand why the roots of $x^3+2=0$ are:
$$x = -\sqrt[3]{2}$$
$$x = \sqrt[3]{-2}$$
$$x = -(-1)^\frac{2}{3} \sqrt[3]{2}$$
I thought it was easier to solve this equation by doing this (I'm probably wrong):
$$x^3+2=0
\iff x^3=-2 \iff x= \pm \sqrt[3]{-2} $$
I put the function on a graph and saw that $x = -\sqrt[3]{2}$ was indeed the real root of it. However, I can't understand why.
(I started learning imaginary numbers and complex roots a few days ago so that might explain my reasoning.)
Thank you!
Kenny
| Note the three solutions of $x^3+2=0$ are
\begin{align*}
&\{-\sqrt[3]{2},+\sqrt[3]{-1}\cdot\sqrt[3]{2},-\sqrt[3]{-1}\cdot\sqrt[3]{2}\}=\left\{\left.-\sqrt[3]{2}\cdot \exp\left(\frac{2k\pi i}{3}\right)\right|k=0,1,2\right\}
\end{align*}
with
$\exp\left(\frac{2k\pi i}{3}\right), k=0,1,2$ the three roots of unity.
*
*The key to the three different solutions are the three roots of unity, the vertices of a regular $3$-gon at $\left\{1,\pm \exp\left(\frac{2\pi i}{3}\right)\right\}$ scaled by $\sqrt[3]{2}$ and rotated by $\pi$.
\begin{align*}
z^3=1\quad\rightarrow\quad\left(\frac{z}{\sqrt[3]{2}}\right)^3=1\quad\rightarrow\quad\left(-\frac{z}{\sqrt[3]{2}}\right)^3=1\\
\end{align*}
which is
\begin{align*}
z^3=1\quad\rightarrow\quad z^3=2\quad\rightarrow\quad z^3=-2\\
\end{align*}
*
*In general we consider $x^n=1$ and $n$ roots of unity forming a regular $n$-gon.
*The special case with only two solutions ($\pm$) is valid for quadratic equations as in
\begin{align*}
x^2+2=0\quad\iff \quad x^2=-2 \quad\iff \quad x= \pm \sqrt{-2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $x + y + z = 0$ then $x\left(\frac1y+\frac1z\right)+y\left(\frac1z+\frac1x\right)+z\left(\frac1x+\frac1y\right) = -3$ Im asked to prove given $x + y + z = 0$, that:
$$x\left(\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{z}+\frac{1}{x}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right) = -3$$
Im stuck, I've tried to use that: $$x=-y-z$$ $$y=-x-z$$ $$z = -y-x$$
and trying to expand after that.
But I don't know if that is a good way of proving something like this?
| $
\begin{align*}
x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{z}+\frac{1}{x})+z(\frac{1}{x}+\frac{1}{y})
&= \frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{y}
\\ &= \frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}
\\ &= \frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}
\\ &= -\frac{y}{y}-\frac{z}{z}-\frac{x}{x}
\\ &= -3
\end{align*}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Multivariate-Multi-objective Optimization Problem: $x+y+z = 1$ and $x^2+y^2+z^2 = 1$ $x, y, z $ are three distinct positive real numbers such that
$$\begin{cases}&x + y + z = 1\\
&{x^2} + {y^2} + {z^2} = 1\\
&x \ne y \ne z\\
&0 < x,y,z < 1
\end{cases}$$
Is there any solution for $x, y, z $ ? If yes, how we can find the solution. Thank you.
| Since $0<x,y,z<1$, you have that $x^2<x, y^2<y$ and $z^2<z$. Plugging in the first two equations you get $$1=x^2+y^2+z^2<x+y+z=1$$ which is a contradiction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Just giving another way.
$$x+\frac 1x=\frac{1+\sqrt2}{2}\\ x^2+\frac{1}{x^2}=\frac{-1+\sqrt5}{2}\\ x^4+\frac{1}{x^4}=-\frac{1+\sqrt5}{2}\\ x^8+\frac{1}{x^8}=\frac{-1+\sqrt5}{2}$$ Hence $$\left(x^8+\frac{1}{x^8}\right)\left(x^2+\frac{1}{x^2}\right)=\left(\frac{-1+\sqrt5}{2}\right)^2\\ x^{10}+\frac{1}{x^{10}}=\left(\frac{-1+\sqrt5}{2}\right)^2-\left(-\frac{1+\sqrt5}{2}\right)=2\iff(x^{10}-1)^2=0$$
Thus $x^{10}=1$ from which the result $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 6
} |
A class consists of 15 boys of whom 5 are prefects. How many committees of 8 can be formed if each consists of at least 2 prefects? A class consists of $15$ boys of whom $5$ are prefects. How many committees of 8 can be formed if each consists of at least $2$ prefects?
I had the solution following solution in mind but that seems to be incorrect. can anyone please explain to me why?
First, I select $2$ students from the perfect category using $\binom{5}{2}$ and then I select remaining $6$ students combined from both the perfect and unperfect category i.e. $\binom{13}{6}$. So the answer becomes $\binom{5}{2}\times \binom{13}{6} = 17160$.
But, the correct answer is $5790$. Can you please help me?
| The total number of committees of $8$ is $\binom{15}{8}$. The number of committees of $8$ with no prefects is equal to $\binom{10}{8}$. The number of committees with exactly one prefect is equal to $\binom{5}{1} \cdot \binom{10}{7}$.
Then the answer to your question is $\binom{15}{8} - \binom{10}{8} - \binom{5}{1} \cdot \binom{10}{7}=6435 - 45 - 5\cdot 120 = 5790.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$ Show that $\sum ^{\infty}_0 \frac{1}{(n-\frac{1}{2})^2}$=$\frac{\pi^2}{6}$
i know that $ \sum _{k=1} 1/k^2 = \pi^2/6$ how to do prove this
| Simplify the term inside the summation.
$$
\sum \frac{1}{(n-\frac 12)^2} = \sum \frac 4{(2n-1)^2} = 4 \sum \frac 1{(2n-1)^2}
$$
This is just the summation of squares of all odd numbers.
Now, use the fact that:
$$
\frac {\pi^2}6 = \sum \frac 1{k^2} = \left(1 + \frac 1{3^2} + \frac 1{5^2} + \ldots\right) + \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots\right)
$$
And finally, note that:
$$
\left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \ldots\right) = \frac 14 \left(\sum \frac 1{k^2}\right) = \frac{\pi^2}{24}
$$
See if you can finish it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to prove induction when n is part of the sum How to Prove
$$\sum_{k=1}^{2n} (-1)^{k-1}\frac{1}{k} = \sum_{k=1}^{n}\frac{1}{k+n}$$
by induction
we did a few of induction but the n was never a part of the sum so im kinda lost on where to start because if I would try to use n+1 i have no idea how the second sum changes
| For $n=1$, LHS(=lefthand side) is
$1 - \frac 12= \frac 12$, and
RHS is $\frac{1}{1+1} = \frac 12$.
Take this as the base case.
Now do the induction step. The induction hypothesis is that formula is true for $n$. We will show that this implies that it also holds for $n+1$. Let's write the LHS for $n+1$. This will be
$$ \sum_{k=1}^{2(n+1)} (-1)^{k-1} \frac1k = \underset{(I)}{\underbrace{\sum_{k=1}^{2n} (-1)^{k-1} \frac 1k}} +\frac{1}{2n+1}- \frac{1}{2n+2}=(*)$$
By the induction hypothesis,
$$ (I) = \sum_{k=1}^n \frac{1}{k+n}\underset{j=k-1}{=} \sum_{j=0}^{n-1} \frac{1}{j+n+1}.$$
Plugging this back into $(*)$, and isolating the $j=0$ summand, we have:
\begin{align*} (*) &= \frac{1}{n+1} + \sum_{j=1}^{n-1} \frac{1}{j+(n+1)} + \frac{1}{n+(n+1)}-\frac{1}{(n+1)+n+1}.\\
& =\sum_{j=1}^n \frac{1}{j+(n+1)}+ \frac{1}{n+1} - \frac 12 \times\frac{1}{n+1}\\
&= \sum_{j=1}^n \frac{1}{j+(n+1)}+ \frac{1}{2(n+1)}\\
& = \sum_{j=1}^{n+1} \frac{1}{j+(n+1)},
\end{align*}
which is what we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I solve: $\frac {3x-2}2 - \frac {4x-5}3=2$? I am a a student and I am having difficulty with answering this question. I keep getting $6$ as my value for x whereas it is wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Question: Solve $$\frac {3x-2}2 - \frac {4x-5}3=2$$
This is what I did to get the answer:
$$2(3x-2) - 3(4x-5) = 12$$
$$6x-4 - 12x+15 = 12$$
$$6x - 12x = 12+4-15$$
$$-6x = 1$$
$$x = -6 $$
Kind Regards help would be appreciated
| There are different ways you could solve for $x;$ here is one way:
$\frac{3x-2}{2} - \frac{4x-5}{3} = 2 \rightarrow$
$\frac{3}{2}x - 1 - (\frac{4}{3}x - \frac{5}{3}) = 2 \rightarrow$
$\frac{1}{6}x + \frac{2}{3} = 2 \rightarrow$
$\frac{1}{6}x = \frac{4}{3} \rightarrow$
$x = 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to calculate $\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}e^{-x^2+ik_0x}e^{-ikx}dx$? $$A(k)=\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}u(x,0)e^{-ikx}dx$$
When $$u(x,0)=e^{-x^2+ik_0x}$$
We get $$A(k)=\frac{1}{\sqrt2}e^{-\frac{(k-k_0)^2}{4}}$$
How to get A(k)?
I am stuck at here,
$$
A(k)
=\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}e^{-x^2+ik_0x}e^{-ikx}dx
=\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}e^{-x^2}\int^∞_{-∞}e^{ik_0x-ikx}dx=\frac{1}{\sqrt{2\pi}}\sqrt{\pi}[\frac{e^{ik_0x-ikx}}{ik_0-ik}]^∞_{-∞}dx=\frac{1}{\sqrt{2}}[\frac{e^{ik_0x-ikx}}{ik_0-ik}]^∞_{-∞}dx$$
| Let $a=k_{0}-k$ and
\begin{align}
I &= \int\limits_{-\infty}^{\infty} \mathrm{e}^{-x^{2}+iax} dx \\
&= \mathrm{e}^{-a^{2}/4} \int\limits_{-\infty}^{\infty} \mathrm{e}^{-(x-ia/2)^{2}} dx
\end{align}
we completed the square in the exponent.
\begin{align}
I_{1} &= \int \mathrm{e}^{-(x-ia/2)^{2}} dx \\
&= \int \mathrm{e}^{-y^{2}} dy \\
&= \frac{\sqrt{\pi}}{2} \mathrm{erf}(y) \\
&= \frac{\sqrt{\pi}}{2} \mathrm{erf}\left(x-i\frac{a}{2} \right)
\end{align}
Applying limits to $I_{1}$, we obtain
\begin{align}
\int\limits_{-\infty}^{\infty} \mathrm{e}^{-(x-ia/2)^{2}} dx
&= \frac{\sqrt{\pi}}{2} \mathrm{erf}\left(x-i\frac{a}{2} \right) \Big|_{-\infty}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} \Big[\lim_{x \to \infty} \mathrm{erf}\left(x-i\frac{a}{2} \right)
- \lim_{x \to -\infty} \mathrm{erf}\left(x-i\frac{a}{2} \right) \Big] \\
&= \sqrt{\pi}
\end{align}
Thus
\begin{equation}
I = \sqrt{\pi} \mathrm{e}^{-a^{2}/4}
\end{equation}
and
\begin{equation}
A(k) = \frac{1}{\sqrt{2}} \mathrm{e}^{-(k_{0}-k)^{2} /4}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this nice $10$th power identity for $x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6$? Ramanujan's 6-10-8 Identity turns out to depend on a special case of,
$$u_1^k+u_2^k+u_3^k =v_1^k+v_2^k+v_3^k$$
simultaneously valid for $k=2,4$. I was investigating if the next system $k=2,6$,
$$x_1^2+x_2^2+x_3^2 =y_1^2+y_2^2+y_3^2\\x_1^6+x_2^6+x_3^6 =y_1^6+y_2^6+y_3^6\tag1$$
would have something similar. I observed empirically that,
$$\left(\sum_{i=1}^3\big(x_i^{10}-y_i^{10}\big)\right)\left(\sum_{i=1}^3\big(x_i^{4}-y_i^{4}\big)\right)^2=20\prod_{i=1}^3\prod_{j=1}^3\big(x_i^2-y_j^2\big)\tag2$$
Example:
$$10^k+15^k+23^k = 3^k+19^k+22^k$$
yields,
$$\small \text{LHS}= \big(10^{10} + 15^{10} + 23^{10} - 3^{10} - 19^{10} - 22^{10}\big)\big(10^4 + 15^4 + 23^4 - 3^4 - 19^4 - 22^4\big)^2$$
$$\small \text{RHS}=20(10^2 - 3^2)(10^2 - 19^2)(10^2 - 22^2)(15^2 - 3^2)(15^2 - 19^2)(15^2 - 22^2)(23^2 - 3^2)(23^2 - 19^2)(23^2 - 22^2)$$
$$\small\text{LHS}=\text{RHS}=37739520^2\times3830610$$
I've also tested it with more general parametric solutions and it works just fine.
Q: But how do we prove $(2)$ rigorously?
| Define $$R_n=(a_1^n+a_2^n+a_3^n -b_1^n-b_2^n-b_3^n)/n$$
I found and proved that
$$R_1S+R_3^3-2R_2R_3R_4+R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
where $S$ is a polynomial containing 9 items.
If $R_1=R_2=0$, then $$R_3^3=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $2^n+13^n+21^n=6^n+7^n+23^n$, $(n=1,2)$
If $R_1=R_3=0$, then $$R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $2^n+10^n+12^n=3^n+8^n+13^n$, $(n=1,3)$
If $R_1=R_4=0$, then $$R_3^3+R_2^2R_5=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $3^n+25^n+38^n=7^n+20^n+39^n$, $(n=1,4)$
If $R_1=R_5=0$, then $$R_3^3-2R_2R_3R_4=\prod_{i=1}^3\prod_{j=1}^3\big(a_i-b_j\big)$$
Example: $3^n+54^n+62^n=24^n+28^n+67^n$, $(n=1,5)$
Let $a_i=x_i^2,b_i=y_i^2$, $(i=1,2,3)$, the system $(1)$ satisfies $R_1=R_3=0$, then we can prove equation $(2)$.
For more identities of the similar form, please refer to my website on Algebraic Identities
For more examples of numerical solutions, please refer to my website on Equal Sums of Like Powers
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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strategies to find explicit formulae for series I have been manipulating a certain series for several hours without finding any pattern. Hence I am wondering what some of the better strategies are to find patterns and thus an explicit formula for a series. Among the things I have tried so far are:
*
*looking for a common difference between terms
*looking for a common ratio between terms
*reversing the order of the terms and summing them up, to check whether the result will be the same for all terms
*bringing the terms to a common denominator and looking for a obvious pattern in the numerator
I had no luck with any of these and others. The series btw. is $\sum_{k = 1}^n\frac{k - 1}{k(k + 1)(k + 2)}$
This is oen of the things I tried:
$\begin{align*}
S_n & = \frac{0}{6} + \frac{1}{24} + \frac{2}{60} + \frac{3}{120} + \frac{4}{210} + \frac{6}{336}\\
& = \frac{0}{6} + \frac{1}{24} + \frac{1}{30} + \frac{1}{40} + \frac{1}{52,5} + \frac{1}{56}\\
& = \frac{0}{1680} + \frac{70}{1680} + \frac{56}{1680} + \frac{42}{1680} + \frac{32}{1680} + \frac{30}{1680}\\
a_n - a_{n+1} : & -\frac{70}{1680}; \frac{14}{1680}; \frac{14}{1680}; \frac{10}{1680}; \frac{8}{1680};
\end{align*}$
The differences between the terms get ever smaller and the sum approaches $.25$, but any internal pattern remains hidden after the things I tried. So, are there a number of useful methods to uncover patterns in series?
| Hints:
*
*$\require{cancel}\;\sum_{k = 1}^n\frac{k - 1}{k(k + 1)(k + 2)} = \sum_{k = 1}^n\frac{\cancel{k}}{\cancel{k}(k + 1)(k + 2)} - \sum_{k = 1}^n\frac{1}{k(k + 1)(k + 2)}$
*$\;\sum_{k = 1}^n\frac{1}{(k + 1)(k + 2)} = \sum_{k = 1}^n\left(\frac{1}{k + 1} - \frac{1}{k + 2}\right)$
*$\;\sum_{k = 1}^n\frac{1}{k(k + 1)(k + 2)} = \frac{1}{2}\sum_{k = 1}^n\left(\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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partial fraction decomposition: product in denominator A fraction $\dfrac{a}{bc}$ can be split into $\dfrac{x}{b} + \dfrac{y}{c}$ by solving for $x$ and $y$ from $a = by + cx$.
Now, then a term $\dfrac{a}{cd}$ should be able to be split like this
$$\frac{x}{bc} + \frac{y}{d}$$
But apparently that is not the case, as I have redone the decomposition for the expression $\dfrac{1}{k(k + 1)(k + 2)}$ several times now:
\begin{align*}
\frac{1}{k(k + 1)(k + 2)} &=
\frac{\overbrace{1}^a}{\underbrace{(k^2 + k)}_b\underbrace{(k + 2)}_c}\\
&= \frac{x}{(k^2 + k)} + \frac{y}{(k + 2)}&\\
&= \frac{x(k + 2)}{(k + 2)(k^2 + k)} + \frac{y(k^2 + k)}{(k + 2)(k^2 + k)}
\end{align*}
Solving for $x$ and $y$:
\begin{align*}
& 1 = x(k + 2) + y(k^2 + k)\\
& k := -2 \implies y = \frac{1}{2}\\
& 1 = x(k + 2) + \frac{1}{2}(k^2 + k)\\
& k := -1 \implies x = 1
\end{align*}
Thus:
$$\frac{1}{k(k + 1)(k + 2)} = \frac{1}{k(k + 1)} + \frac{\frac{1}{2}}{(k + 2)}\\$$
But that is not correct, as this example shows:
$$\frac{1}{3(3 + 1)}+ \frac{\frac{1}{2}}{3 + 2} = \frac{11}{60}$$
while
$$\frac{1}{3(3 + 1)(3 + 2)} = \frac{1}{60}$$
I hope I have not made some silly obvious arithmetic or algebraic mistake, but I have redone this several times, leading me to assume I am making a wrong assumption about how this works. So, is my assumption that there are $x$ and $y$ such that $\dfrac{a}{bcd} = \dfrac{x}{bc} + \dfrac{y}{d}$ wrong?
| You have three factors without multiplicity, so try
$$\frac{1}{k(k+1)(k+2)} \equiv \frac{A}{k}+\frac{B}{k+1}+\frac{C}{k+2}$$
Spoiler below
$$\frac{1}{k(k+1)(k+2)} \equiv \frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of selecting two numbers with a sum of squares divisible by 10
Two natural numbers $x$ and $y$ are chosen at random. Find the probability that $x^2 + y^2$ is divisible by 10.
I could not understand how select two numbers from any natural number (infinite).
| Its simple logic. Count numbers on the basis of last digits whose sum of square makes 0.
So Favourable cases = {(0,0),(1,3),(1,7),(2,4),(2,6),(3,1),(3,9),(4,2),(4,8),(5,5),(6,2),(6,8),(7,1),(7,9),(8,4)(8,6),(9,3),(9,7)} = 18
Total cases = Value of x can be any digit from 0 to 9.
Similarly for y.
$= 10 * 10 = 100$
$Probability = \frac{\text{No. of favourable cases}}{\text{Total no. of cases}}$
$= \frac{18}{100} = \frac{9}{50} $
| {
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"url": "https://math.stackexchange.com/questions/2054883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Computing large modulus by hand I am having trouble computing $12^{15}$ mod $2016$ due to the large size of the modulus. I need to do this and list the steps out by hand
| Remember $a \equiv b \mod n$ means $\frac a{\gcd(a,n)} \equiv \frac b{\gcd(a,n)} \mod \frac n{\gcd(a,n)}$.
Proof: $a \equiv b \mod n \implies a + nk = b \implies \gcd(a,n)(\frac a{\gcd(a,n)} + \frac n{\gcd(a,n)}k \implies \frac a{\gcd(a,n)} \equiv \frac b{\gcd(a,n)} \mod \frac n{\gcd(a,n)}$
$2016 = 7*2^53^2$ so $\gcd(2016, 12^{15}) = 2^53^2$
So $12^{15} \equiv x \mod 2016 \implies 2^{25}3^{13} \equiv x/288 \mod 7$
$2^6 \equiv 3^6 \equiv 1 \mod 7$ by Fermat's Little Theorem so
$2^{25}3^{13}\equiv 2*3 \equiv 6 \equiv -1 \mod 7$.
So $x = -288$.
So $12^{15} \equiv -288 \mod 2016$
| {
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"url": "https://math.stackexchange.com/questions/2057859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove an equality The matrices $A,B,C$ all $2\times 2$ dimensions so: $$A^2 +B^2 +C^2 =AB+BC+CA.$$
Prove that $$(A^2 +B^2 +C^2 -BA-CB-AC)^2=O_2$$
Can someone help me with this? Thank you.
| Set $X=A-B,Y=B-C$, the condition is equivalent to $X^2+YX+Y^2=0$. Hence $A^2+B^2+C^2-BA-CB-AC=X^2+XY+Y^2=XY-YX$, whose trace is obviously $0$. We will frequently use the following special case of Hamilton-Cayley theorem:
Let $M$ be a $2\times 2$ matrix, then
$$M^2=(\operatorname{tr}M)M-(\det M)I$$
For example, $(XY-YX)^2=-(\det(XY-YX))I$, so we only need to prove that $XY-YX$ is not invertible. Note that $X^2+YX+Y^2=0$ implies
$$0=(X^2+YX+Y^2)X-Y(X^2+YX+Y^2)=X^3-Y^3$$
So $X^3=Y^3$. Then one may obtain
$$X(YX+Y^2)=-X^3=-Y^3=(X^2+YX)Y=Y(X^2+YX)$$
or equivalently
$$(XY-YX)X=Y^2X-XY^2,\hspace{1em}Y(XY-YX)=YX^2-X^2Y$$
Now Hamilton-Cayley implies $Y^2X-XY^2=(\operatorname{tr}Y)(YX-XY)$, so
$$(\det X)\det(XY-YX)=(\operatorname{tr}Y)^2\det(YX-XY)=(\operatorname{tr}Y)^2\det(XY-YX)$$
Similarly,
$$(\det Y)\det(XY-YX)=(\operatorname{tr}X)^2\det(YX-XY)=(\operatorname{tr}X)^2\det(XY-YX)$$
Now we suppose on the contrary that $XY-YX$ is invertible, then $\det(XY-YX)\not=0$, hence $(\operatorname{tr}X)^2=\det Y,(\operatorname{tr}Y)^2=\det X$. But $X^3=Y^3$ implies $\det X=\det Y$, thus
$$(\operatorname{tr}X)^2=(\operatorname{tr}Y)^2=\det X=\det Y$$
Applying Hamilton-Cayley again, we see that $\det X=0$ would result in $X^2=Y^2=0$, hence $XY=0$ and $(XY-YX)^2=(YX)(YX)=Y(XY)X=0$. So we may assume $\det X\not=0$. In that case we have
\begin{align}
X^3=X\cdot X^2 & =X((\operatorname{tr}X)X-(\det X)I) \\
& =(\operatorname{tr}X)X^2-(\det X)X \\
& =(\operatorname{tr}X)((\operatorname{tr}X)X-(\det X)I)-(\det X)X\\
& =-(\operatorname{tr}X)^3I
\end{align}
For the same reason, $Y^3=-(\operatorname{tr}Y)^3I$. Now $X^3=Y^3$ yields $\operatorname{tr} X=\operatorname{tr} Y$. Let $\lambda=\operatorname{tr} X\not=0$. Then $0=X^2+YX+Y^2=\lambda(X+Y)+YX-\lambda^2 I$. On the other hand, $$0=(X^2+YX+Y^2)X=-\lambda^3 I+Y(\lambda X-\lambda^2 I)+(\lambda Y-\lambda^2 I)X=\lambda(2YX-\lambda(X+Y)-\lambda^2I)$$
"Solving" these two equations we get
$$X+Y=\frac 1 3\lambda I,\hspace{1em} YX=\frac 2 3\lambda^2 I$$
But $\det(YX)=(\det Y)(\det X)=\lambda^4$, while $\det(2\lambda^2I/3)=4\lambda^4/9$, contradiction. Now we're done.
| {
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"source": "stackexchange",
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Need help with RHS of Induction Problem about $1^{3} + 2^{3} + ... + n^{3}$ = $(1 + 2 + ... + n)^{2}$ Prove $1^{3} + 2^{3} + ... + n^{3}$ = $(1 + 2 + ... + n)^{2}$ for all positive integers n.
I've tried to work with this problem using mathematical induction. However, I really don't understand how to manipulate the right side of the equation. Any help would be greatly appreciated.
Thank you in advance.
| Assume the result holds for $n$. Then
$$1^3+2^3+\dotsb+n^3+(n+1)^3 = (1+2+\dotsb+n)^2+(n+1)^3$$
$$(1+2+\dotsb+n+(n+1))^2 = (1+2+\dotsb+n)^2 + 2(1+2+\dotsb+n)(n+1)+(n+1)^2
\\=1^3+2^3+\dotsb + n^3+2\left(\frac{n(n+1)}{2}\right)(n+1)+(n+1)^2
\\=1^3+2^3+\dotsb + n^3+n(n+1)(n+1)+(n+1)^2=1^3+2^3+\dotsb + n^3+(n+1)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^2+3x+24$ is a perfect square then find the value of $x$ Find all integer values of $x$ where $x^2+3x+24$ is a perfect square. By guessing I found one solution that $x=5$.I found the problem in the exercise of a book in the chapter of polynomials. So please help me.
| $x^2+3x+24 = n^2$
so
$x^2+3x+9/4 = n^2-24+9/4$
or
$(x+3/2)^2 =n^2-87/4$
or
$(2x+3)^2 =4n^2-87$.
Therefore
$87
=4n^2-(2x+3)^2
=(2n-2x-3)(2n+2x+3)
$.
$87=3*29$,
so it can be factored as
$87
=1*87,3*29
$.
Therefore
$(2n-2x-3,2n+2x+3)
=(-87, -1), (-29, -3), (3, 29), (1, 87)
$.
If
$(2n-2x-3,2n+2x+3)
=(a, b)
$,
then
$4x+6 = b-a$
and
$4n=b+a$
so
$x = (b-a-6)/4$
and
$n = (b+a)/4$.
For these 4 sets,
$x = (-87+1-6)/4
=-92/4 = -23
$,
$x = (-29+3-6)/4
=-32/4 = -8
$,
$x = (29-3-6)/4
=20/4 = 5
$,
$x = (87-1-6)/4
=80/4 = 20
$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The complex numbers $z_1, z_2,z_3$ are satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$..... Problem :
The complex numbers $z_1, z_2,z_3$ are satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ $z_1,z_2,z_3$ are vertices of a triangle , which type of triangle is it :
My approach :
$|\frac{z_1-z_3}{z_2-z_3}|=|\frac{1-i\sqrt{3}}{2}| = 1 $
argument $\frac{z_1-z_3}{z_2-z_3} = \cos\frac{\pi}{3}-\sin\frac{\pi}{3}$ =$\frac{\pi}{3}$
So, I am unable to work out, whether it a isosceles or equilateral triangle. Please suggest will be of great help, thanks.
| Given $$\frac{z_1-z_3}{z_2-z_3} = \frac{1-i\sqrt{3}}{2} =\frac{1}{2}-i\frac{\sqrt{3}}{2} = \cos(-\frac{\pi}{3}) +i\sin (-\frac{\pi}{3})$$ $$\Rightarrow \frac{z_2-z_3}{z_1-z_3} =\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$$ $$\Rightarrow |\frac{z_2-z_3}{z_1-z_3}| = 1 \text{and arg}(\frac{z_2-z_3}{z_1-z_3}) =\frac{\pi}{3}$$ If $z_1, z_2, z_3$ represent the vertices $A, B, C$ of a triangle ABC, then, $$|\frac{z_2-z_3}{z_1-z_3}| =1 \Rightarrow |z_2-z_3| = |z_1-z_3|$$ This gives us $BC=AC$. Also, $\text{arg}(\frac{z_2-z_3}{z_1-z_3}) =\frac{\pi}{3}$ gives $\angle BCA =\frac{\pi}{3}$. So, ABC is equilateral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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real values of $k$ in quadratic equation in trigonometric form real values of $k$ for which the equation $\cos^2 x-(k^2+k+5)|\cos x|+(k^3+3k^2+2k+6)=0$ has real solutions
equation is $|\cos x|^2-(k^2+k+5)|\cos x|+(k^3+3k^2+2k+6)=0$
$\displaystyle |\cos x| = \frac{k^2+k+5\pm \sqrt{(k^2+k+5)^2-4(k^3+3k^2+2k+5)}}{2}$
equation has real solution for $0 \leq|\cos x|\leq 1$
$(k^2+k+5)^2-4(k^3+3k^2+2k+5)\geq0$
i wan,t go after that, could some help me with this
| Let $t=|\cos x|$ and $f(t)=t^2-(k^2+k+5)t+k^3+3k^2+2k+6$.
Then,
$$f(t)=\left(t-\frac{k^2+k+5}{2}\right)^2-\left(\frac{k^2+k+5}{2}\right)^2+k^3+3k^2+2k+6\tag1$$
Now
$$\begin{align}&-\left(\frac{k^2+k+5}{2}\right)^2+k^3+3k^2+2k+6\\\\&=\frac 14(-(k^2+k+5)^2+4(k^3+3k^2+2k+6))\\\\&=-\frac 14(k^4-2k^3-k^2+2k+1)\\\\&=-\frac 14((k^2)^2+(-k)^2+(-1)^2+2\cdot k^2\cdot (-k)+2\cdot k^2\cdot (-1)+2\cdot (-k)\cdot (-1))\\\\&=-\left(\frac{k^2-k-1}{2}\right)^2\end{align}$$
So, from $(1)$, we get
$$f(t)=\left(t-\frac{k^2+k+5}{2}\right)^2-\left(\frac{k^2-k-1}{2}\right)^2$$
Since
$$\frac{k^2+k+5}{2}=\frac 12\left(k+\frac 12\right)^2+\frac{19}{8}\gt 1\quad\text{and}\quad -\left(\frac{k^2-k-1}{2}\right)^2\le 0$$
for any $k\in\mathbb R$, considering the graph of a parabola $y=f(t)$ gives that we want to find $k$ such that
$$\frac{k^2-k-1}{2}\not=0\quad\text{and}\quad f(0)=(k+3)(k^2+2)\ge 0\quad\text{and}\quad f(1)=(k+2)(k^2+1)\le 0$$
Hence, the answer is $\ \color{red}{-3\le k\le -2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $\sum_{k=0}^\infty \frac{2^k}{2^{2^k}+1}=1$? How can I prove this result?
$$\sum_{k=0}^\infty \frac{2^k}{2^{2^k}+1}=1$$
The sum converges very quickly: The term at $k=4$ is already smaller than $2^{-12}$ and each further term is much smaller than its predecessor.
I've tried replacing the $2$'s by $x$'s and looking for power series, as well as seeing if the sum telescopes, but nothing so far has worked.
| Make it into a telescopic sum as follows
\begin{align*}
\sum_{k=0}^n \frac{2^k}{2^{2^k}+1}&=\sum_{k=0}^n \frac{2^k(2^{2^k}-1)}{(2^{2^{k+1}}-1)}\\
&=\sum_{k=0}^n \frac{2^k(2^{2^k}+1-2)}{(2^{2^{k+1}}-1)}\\
&=\sum_{k=0}^n \frac{2^k(2^{2^k}+1)}{(2^{2^{k+1}}-1)}-\frac{2^{k+1}}{(2^{2^{k+1}}-1)}\\
&=\sum_{k=0}^n \frac{2^k}{(2^{2^{k}}-1)}-\frac{2^{k+1}}{(2^{2^{k+1}}-1)}\\
&=1-\frac{2^{n+1}}{(2^{2^{n+1}}-1)}\rightarrow 1
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a+b+c+d=1$ so $\sum\limits_{cyc}\sqrt{a+b+c^2}\geq3$ Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=1$. Prove that:
$$\sqrt{a+b+c^2}+\sqrt{b+c+d^2}+\sqrt{c+d+a^2}+\sqrt{d+a+b^2}\geq3.$$
I tried C-S, Holder and more, but without success.
| A computer solution using the Buffalo Way
After homogenization, it suffices to prove that, for all $a, b, c, d \ge 0$,
$$\sum_{\mathrm{cyc}} \sqrt{(a + b)(a+b+c+d) + c^2} \ge 3(a+b+c+d).$$
Denote
\begin{align}
&X = (a + b)(a+b+c+d) + c^2,\ Y = (b+c)(a+b+c+d) + d^2, \\
&Z = (c+d)(a+b+c+d) + a^2,\ W = (d+a)(a+b+c+d) + b^2.
\end{align}
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} + \sqrt{W} \ge 3(a+b+c+d)$.
WLOG, assume $d = \min(a, b, c, d)$. We split into four cases:
*
*$a\ge b \ge c \ge d$ or $a\ge c \ge b \ge d$ or $c\ge a\ge b \ge d$: By GM-HM, we have $\sqrt{X} + \sqrt{W} = \sqrt{X + W + 2\sqrt{XW}}\ge \sqrt{X + W + \frac{4XW}{X+W}}$
and $\sqrt{Y} + \sqrt{Z} \ge \sqrt{Y + Z + \frac{4YZ}{Y+Z}}$.
It suffices to prove that
$$\sqrt{X + W + \frac{4XW}{X+W}} + \sqrt{Y + Z + \frac{4YZ}{Y+Z}} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + W + \frac{4XW}{X+W}} \sqrt{Y + Z + \frac{4YZ}{Y+Z}}\\
\ge\ & 9(a+b+c+d)^2 - X - W - \frac{4XW}{X+W} - Y - Z - \frac{4YZ}{Y+Z}.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + W + \frac{4XW}{X+W}\right)\left(Y + Z + \frac{4YZ}{Y+Z}\right)\\
\ge \ & \left(9(a+b+c+d)^2 - X - W - \frac{4XW}{X+W} - Y - Z - \frac{4YZ}{Y+Z}\right)^2.
\end{align}
The Buffalo Way works.
*$b\ge c \ge a \ge d$: By GM-HM, we have
$\sqrt{X} + \sqrt{Y} \ge \sqrt{X + Y + \frac{4XY}{X+Y}}$ and $\sqrt{Z} + \sqrt{W} \ge \sqrt{Z + W + \frac{4ZW}{Z+W}}$.
It suffices to prove that
$$\sqrt{X + Y + \frac{4XY}{X+Y}} + \sqrt{Z + W + \frac{4ZW}{Z+W}} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + Y + \frac{4XY}{X+Y}} \sqrt{Z + W + \frac{4ZW}{Z+W}}\\
\ge\ & 9(a+b+c+d)^2 - X - Y - \frac{4XY}{X+Y} - Z - W - \frac{4ZW}{Z+W}.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + Y + \frac{4XY}{X+Y}\right)\left(Z + W + \frac{4ZW}{Z+W}\right)\\
\ge\ & \left(9(a+b+c+d)^2 - X - Y - \frac{4XY}{X+Y} - Z - W - \frac{4ZW}{Z+W}\right)^2.
\end{align}
The Buffalo Way works.
*$b \ge a \ge c \ge d$: By GM-HM, we have
\begin{align}
\sqrt{X} + \sqrt{Y} + \sqrt{W} &= \sqrt{X + Y + W + 2\sqrt{XY} + 2\sqrt{YW} + 2\sqrt{WX}}\\
&\ge \sqrt{X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}}.
\end{align}
It suffices to prove that
$$\sqrt{X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}} + \sqrt{Z} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}} \sqrt{Z}\\
\ge\ & 9(a+b+c+d)^2 - X - Y - W - \frac{4XY}{X+Y} - \frac{4YW}{Y+W} - \frac{4WX}{W+X} - Z.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + Y + W + \frac{4XY}{X+Y} + \frac{4YW}{Y+W} + \frac{4WX}{W+X}\right)Z\\
\ge\ & \left(9(a+b+c+d)^2 - X - Y - W - \frac{4XY}{X+Y} - \frac{4YW}{Y+W} - \frac{4WX}{W+X} - Z\right)^2.
\end{align}
The Buffalo Way works.
*$c \ge b \ge a \ge d$: By GM-HM, we have
\begin{align}
\sqrt{X} + \sqrt{Y} + \sqrt{Z} &= \sqrt{X + Y + Z + 2\sqrt{XY} + 2\sqrt{YZ} + 2\sqrt{ZX}}\\
&\ge \sqrt{X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}}.
\end{align}
It suffices to prove that
$$\sqrt{X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}} + \sqrt{W} \ge 3(a+b+c+d).$$
Squaring both sides, it suffices to prove that
\begin{align}
&2\sqrt{X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}}\sqrt{W}\\
\ge\ & 9(a+b+c+d)^2 - X - Y - Z - \frac{4XY}{X+Y} - \frac{4YZ}{Y+Z} - \frac{4ZX}{Z+X} - W.
\end{align}
Squaring both sides, it suffices to prove that
\begin{align}
&4\left(X + Y + Z + \frac{4XY}{X+Y} + \frac{4YZ}{Y+Z} + \frac{4ZX}{Z+X}\right)W\\
\ge\ & \left(9(a+b+c+d)^2 - X - Y - Z - \frac{4XY}{X+Y} - \frac{4YZ}{Y+Z} - \frac{4ZX}{Z+X} - W\right)^2.
\end{align}
The Buffalo Way works.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of $p^2+q$
Let $p$ and $q$ be real numbers such that $p^2+q>q^2+p$. Then the minimum value of $p^2+q$ is what?
My argument is as follows:
Since $p,q$ are real numbers, it should not matter if we take $p$ as $x$ and $q$ as $y$.
\begin{align*} p^2+q &> q^2+p \\ x^2-y^2 &> x-y \\ (x-y)(x+y-1)&>0\end{align*}
So, we have to maximize $x^2+y$ subject to the conditions that:
$$x-y>0,\quad x+y-1>0$$ or $$x-y<0,\quad x+y-1<0$$
From the graph of thse, it's clear that the minimum value occurs at $(1/2,1/2)$ in region $1$ (labeled on the graph). So, $p^2+q = (1/2)^2+(1/2) = 3/4$.
However this is not the correct answer.
Can anyone suggest where I'm making a mistake?
| I'll solve the following problem.
Let $p^2+q\geq q^2+p$. Find a minimal value of $p^2+q$.
Indeed, $p^2+q+\frac{1}{4}=\frac{1}{2}\left(\left(p+\frac{1}{2}\right)^2+\left(q+\frac{1}{2}\right)^2\right)+\frac{1}{2}(p^2+q-q^2-p)\geq0$.
The equality occurs for $p=q=-\frac{1}{2}$.
Id est, the answer is $-\frac{1}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it possible to have three real numbers that have both their sum and product equal to $1$? I have to solve $ x+y+z=1$ and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers?
Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$?
| Given $x$ and the requirements $x + y + z = xyz = 1$, we can derive possible values for $y$ which in turn fully determines $z$.
Substituting $z = 1 - x - y$ in $xyz = 1$:
$$xy (1 - x - y) = 1$$
$$xy - x^2y - xy^2 = 1$$
$$(-x)y^2 + (x - x^2)y - 1 = 0$$
The quadratic formula now tells us that y is given by:
$$y = {-(x - x^2) \pm \sqrt{(x - x^2)^2 - 4x} \over -2x}$$
$$y = \frac{1-x}{2} \pm {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$
Before tackling the domain, consider this:
$$y = \frac{1-x}{2} \pm R(x)$$
$$z = 1 - x - (\frac{1-x}{2} \pm R(x))$$
$$z = \frac{1 - x}{2} - (\pm R(x))$$
Thus the sign of the square root is irrelevant, as it merely swaps $y$ and $z$ around. Now all that remains is finding the domain for $x$ such that the solution is real. This means $x^4 - 2x^3 + x^2 - 4x > 0$. You can take the time to solve this by hand (it's actually a cubic), or plug it straight into your favourite CAS and find out it's:
$$x < 0 \vee x > \frac{2+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6\sqrt{78}}}{3}$$
$$X < 0 \vee x > 2.3146...$$
So the final parametric solution is:
$$R(x) = {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$
$$\{x, y, z\} = \left\{x, \frac{1 - x}{2} - R(x), \frac{1 - x}{2} + R(x)\right\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating a definite integral: $\int_0^{\pi}\frac{x}{1-\sin{x}\cos{x}}\,\mathrm dx$ Ok so I'm supposed to evaluate:
$$\int_0^{\pi}\frac{x}{1-\sin{x}\cos{x}}\,\mathrm dx$$
I tried using definite integral properties, but this didn't yield any good follow up, because I couldn't evaluate the resulting integrals.
How exactly can one go about solving this? (preferably without contour integration)
| $$I=\int_0^\pi\frac{x}{1-\sin(x)\cos(x)}\ dx=\int_0^{\pi}\frac{2x}{2-\sin(2x)}\ dx$$
$$=\frac12\int_0^{2\pi}\frac{x}{2-\sin(x)}\ dx\overset{x+\pi/2=t}{=}\frac12\int_{\pi/2}^{5\pi/2}\frac{t-\pi/2}{2+\cos(t)}\ dt$$
Using the identity
$$\frac{1}{a+b\cos(t)}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
We have
$$\frac{1}{2+\cos(x)}=\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}\sum_{n=1}^{\infty}\left(\sqrt{3}-2\right)^n\cos{(nx)}$$
$$\Longrightarrow I=\frac{1}{2\sqrt{3}}\underbrace{\int_{\pi/2}^{5\pi/2}\left(t-\frac{\pi}{2}\right)\ dt}_{2\pi^2}+\frac{1}{\sqrt{3}}\sum_{n=1}^\infty (\sqrt{3}-2)^n\underbrace{\int_{\pi/2}^{5\pi/2}\left(t-\frac{\pi}2\right)\cos(nt)\ dt}_{A}$$
$$A=\int_{\pi/2}^{5\pi/2}t\cos(nt)\ dt-\frac{\pi}{2}\underbrace{\int_{\pi/2}^{5\pi/2}\cos(nt)\ dt}_{0}=\frac{2\pi}{n}\sin\left(\frac{\pi}{2}n\right), \quad n=1,2,3,...$$
$$\Longrightarrow I=\frac{\pi^2}{\sqrt{3}}+\frac{2\pi}{\sqrt{3}}\underbrace{\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^n}{n}\sin\left(\frac{\pi}{2}n\right)}_{-\pi^2/12}=\frac{5\pi^2}{6\sqrt{3}}$$
Details for the last result;
$$\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^n}{n}\sin\left(\frac{\pi}{2}n\right)=\Im\sum_{n=1}^\infty \frac{(\sqrt{3}-2)^ne^{in\pi/2}}{n}=\Im\sum_{n=1}^\infty \frac{[i(\sqrt{3}-2)]^n}{n}$$
$$=-\Im\ln\left(1-i(\sqrt{3}-2)\right)=-\tan^{-1}(-\sqrt{3}+2)=-\frac{\pi^2}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Need help with proof for Dedekind cuts on $\mathbb{Q}^+$ I am working on a proof about Dedekind cuts on the positive rational numbers. I have been stuck for a while on the following point and would appreciate any help.
Given $x\in \mathbb{Q}^+$ such that $x^2<2$, how to find $y\in \mathbb{Q}^+$ such $y>x$ and $y^2<2$?
| Here is an exercise (which is not too far from what @HanDeBruijn is doing) from Hardy's "A Course of Pure Mathematics", 3rd edition, page 12
The book is old and should be freely available online.
Basically
*
*$\frac{m}{n}<\sqrt{2} \Rightarrow \sqrt{2} < \frac{m+2n}{m+n}$.
$$\frac{m}{n}<\sqrt{2}=\frac{2-\sqrt{2}}{\sqrt{2}-1}\Rightarrow \\ m(\sqrt{2}-1)<n(2-\sqrt{2})\Rightarrow \\ \sqrt{2}(m+n)<m+2n$$
*Similarly if $\frac{m}{n}>\sqrt{2} \Rightarrow \sqrt{2} > \frac{m+2n}{m+n}$.
*$\left|\frac{m+2n}{m+n} - \sqrt{2}\right|< \left|\frac{m}{n} - \sqrt{2}\right|$. $$\left|\frac{\frac{m+2n}{m+n} - \sqrt{2}}{\frac{m}{n} - \sqrt{2}}\right|=\left|\frac{\frac{m+2n}{m+n} - \sqrt{2}}{\frac{m^2}{n^2} - 2}\cdot \left(\frac{m}{n} + \sqrt{2}\right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left(\frac{m+2n}{m+n} - \sqrt{2}\right)\cdot \left(\frac{m}{n} + \sqrt{2}\right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left( \frac{m^2+2mn}{nm+n^2} +\sqrt{2}\frac{m+2n}{m+n} -\sqrt{2}\frac{m}{n} -2 \right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left( \frac{m^2+2mn}{nm+n^2} - 2 +\sqrt{2}\left(\frac{m+2n}{m+n} -\frac{m}{n}\right) \right)\right|\\=\left|\frac{n^2}{m^2 - 2n^2}\cdot \left( \frac{m^2-2n^2}{nm+n^2} +\sqrt{2}\frac{2n^2-m^2}{nm+n^2} \right)\right|\\=\left|\frac{n^2}{nm+n^2}\cdot \left( 1-\sqrt{2} \right)\right|=\left|\frac{n^2}{nm+n^2}\right| \cdot \left| 1-\sqrt{2} \right|<1$$
And this process can be repeated, by taking $$m_1=m+2n \\n_1=m+n \\...\\m_{k+1}=m_k+2n_k \\n_{k+1}=m_k+n_k$$ considering alterations of course, jumping around $\sqrt{2}$ and becoming closer to $\sqrt{2}$ each step.
This will lead to $\frac{m}{n} < \frac{m_2}{n_2} < \sqrt{2} < \frac{m_1}{n_1}$ or $\frac{m^2}{n^2} < \frac{m_2^2}{n_2^2} < 2< \frac{m_1^2}{n_1^2}$. So if $x=\frac{m}{n}$ then the first suitable $y$ is $y=\frac{m_2}{n_2}$.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
In a determinant prove $xyz = -1$ If we are given the following determinant
$$\begin{vmatrix}
x^3+1 & x^2 & x \\
y^3+1 & y^2 & y \\
z^3+1 & z^2 & z \\
\end{vmatrix}=0
$$
and $x, y, z$ are all different, then we have to prove that $xyz = -1$.
I tried to expand the determinant, but using that, it is getting too complicated.
| Hint:
Apply $R'_2=R_2-R_1$ and $R'_3=R_3-R_1$ to find
$$\begin{vmatrix}
x^3+1 & x^2 & x \\
y^3+1 & y^2 & y \\
z^3+1 & z^2 & z \\
\end{vmatrix}
=\begin{vmatrix}
x^3+1 & x^2 & x \\
y^3-x^3 & y^2-x^2 & y-x \\
z^3-x^3 & z^2-x^2 & z-x \\
\end{vmatrix}$$ $$=(y-x)(z-x)\begin{vmatrix}
x^3+1 & x^2 & x \\
y^2+xy+x^2 & y+x &1 \\
z^2+zx+x^2 & z+x &1 \\
\end{vmatrix}$$
Now apply $R_3'=R_3-R_2$ $$\begin{vmatrix}
x^3+1 & x^2 & x \\
y^2+xy+x^2 & y+x &1 \\
z^2+zx+x^2 & z+x &1 \\
\end{vmatrix}=\begin{vmatrix}
x^3+1 & x^2 & x \\
y^2+xy+x^2 & y+x &1 \\
z^2-y^2+zx-xy & z-y &0 \\
\end{vmatrix}=(z-y)\begin{vmatrix}
x^3+1 & x^2 & x \\
y^2+xy+x^2 & y+x &1 \\
z+y+x & 1 &0 \\
\end{vmatrix}$$
Finally $R_1'=R_1-x\cdot R_2$ $$\begin{vmatrix}
x^3+1 & x^2 & x \\
y^2+xy+x^2 & y+x &1 \\
z+y+x & 1 &0 \\
\end{vmatrix}=\begin{vmatrix}
1-xy(x+y) & -xy &0 \\
y^2+xy+x^2 & y+x &1 \\
z+y+x & 1 &0 \\
\end{vmatrix}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum _{i=0}^{p} \left( \binom {p+1}{i+1} \times \sum _{j=1}^{n} j^{p-i} \right)=(n+1)^{p+1}-1$ Problem as follows:
Let $S^p_n := 1^p+2^p+3^p+...n^p$
Prove that $$\binom {p+1}{1}S^{p}_n+\binom {p+1}{2}S^{p-1}_n+\binom {p+1}{3}S^{p-2}_n+...+\binom {p+1}{p+1}S^{0}_n=(n+1)^{p+1}-1$$
Or, condensed in the capital sigma notation, prove that
$$\sum _{i=0}^{p} \left( \binom {p+1}{i+1} \times \sum _{j=1}^{n} j^{p-i} \right)=(n+1)^{p+1}-1$$
I tried induction on $p$ and got stuck on the inductive step:
Assume that
$\binom {p+1}{1}S^{p}_n+\binom {p+1}{2}S^{p-1}_n+\binom {p+1}{3}S^{p-2}_n+...+\binom {p+1}{p+1}S^{0}_n=(n+1)^{p+1}-1$, then show that
$\binom {p+2}{1}S^{p+1}_n+\binom {p+2}{2}S^{p}_n+\binom {p+2}{3}S^{p-1}_n+...+\binom {p+2}{p+2}S^{0}_n=(n+1)^{p+2}-1$.
I tried subtracting the first sum from the second by subtracting alike $S^x_n$'s and using Pascal's identity, and got
$$\binom {p+2}{1}S^{p+1}_n+\binom {p+1}{2}S^{p}_n+\binom {p+1}{3}S^{p-1}_n+...+\binom {p+1}{p+1}S^{1}_n=n(n+1)^{p+1}$$
I have no idea where to go from here. I also tried subtracting the two sums by subtracting alike coefficients, but didn't get anything sensible either.
| Firstly it should be a bit easier to use induction in $n$ instead of $p$, since you have $S_{n+1}^p = S_{n}^p + (n+1)^p$ in the induction step.
Secondly, this can be proven directly without using induction if you consider binomial expansions of $2^p$, $3^p$, ..., $(n+1)^p$ as shown in this table:
\begin{array}{ccc}
2^{p+1}&=&\binom{p+1}{0}1^{p+1}&+&\binom{p+1}{1}1^{p}&+&\ldots&+&\binom{p+1}{p}1^{1}&+&\binom{p+1}{p+1}1^{0}\\
3^{p+1}&=&\binom{p+1}{0}2^{p+1}&+&\binom{p+1}{1}2^{p}&+&\ldots&+&\binom{p+1}{p}2^{1}&+&\binom{p+1}{p+1}2^{0}\\
\vdots&&&&&&&&\\
(n+1)^{p+1}&=&\binom{p+1}{0}n^{p+1}&+&\binom{p+1}{1}n^{p}&+&\ldots&+&\binom{p+1}{p}n^{1}&+&\binom{p+1}{p+1}n^{0}\\\hline
S_{n+1}^{p+1}-1^{p+1}&=&\binom{p+1}{0}S_{n}^{p+1}&+&\binom{p+1}{1}S_n^{p}&+&\ldots&+&\binom{p+1}{p}S_{n}^{1}&+&\binom{p+1}{p+1}S_{n}^{0}
\end{array}
Now just subtract $S_{n}^{p+1}$ from last line and notice that $S_{n+1}^{p+1} - S_{n}^{p+1} = (n+1)^{p+1}$, and the identity follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\sqrt[3]x-\sqrt[3]{x-14}=2$, find $x - 1/x$
$$ \sqrt[3]x-\sqrt[3]{x-14}=2$$
Then find $x - \frac{1}{x} = ?$
I've tried to take third power of both sides but I couldn't find anything.
| Hint: $(a-b)^3=a^3-b^3-3ab(a-b)\,$ so: $$8 = \left(\sqrt[3]{x}-\sqrt[3]{x-14}\right)^3=x - (x-14) - 3\cdot\sqrt[3]{x\,(x-14)} \cdot 2 = 14 - 6 \sqrt[3]{x\,(x-14)}$$ It follows that: $$x(x-14) = 1 \iff x^2-14x - 1 = 0 \iff x - \frac{1}{x} - 14=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$. I have found a new problem which asks:
Find the remainder when $10^{10}+10^{10^2}+.........+10^{10^{10}}$ is divided by $7$.
I am thinking to find the remainder using Fermet's theorem, but I think I am unable to do it.
Please help.
| Yes, You can use Fermat's theorem here.
By Fermat's theorem, $10^6\equiv 1 \pmod 7$.
Hence, $10^{6m}\equiv 1 \pmod 7$ for all $m$.
Now, $$10\equiv 4 \pmod 6, 10^2\equiv 40 \equiv 4 \pmod 6$$
By induction, $10^n \equiv 4 \pmod 6$ for all $n$
Thus, $10^n=6m+4$
and $$10^{10n}=10^{6m}\times10^4\equiv 10^4 \pmod 7 \equiv 4 \pmod 7$$
Consequently,
$$10^{10}+10^{10^2}+..........+10^{10^{10}} \equiv 4 \times 10 \pmod 7\equiv 5 \pmod 7$$
So, remainder is $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How do you find the integral of $\int_a^bdx\int_a^b \frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy$ $$\int_a^bdx\int_a^b \frac{y^2}{(x^2+y^2)^\frac{3}{2}}dy$$
I've tried using the trigonometric substitution $x=\cos t, y=\sin t$ but it didn't work out
| Interchanging the order of integration, we have
\begin{align}
I&=\int_a^b dyy^2\int_a^b\frac{dx}{(x^2+y^2)^{\frac{3}{2}}}
\end{align}
Now let $x=y\tan\phi$, then $dx=y\sec^2\phi d\phi$. Plugging this in, we have
\begin{align}
I &= \int_a^b dyy^2\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\frac{y\sec^2\phi}{(y^2\tan^2\phi + y^2)^{\frac{3}{2}}}\\
&=\int_a^b dyy^2\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\frac{y\sec^2\phi}{y^3(\tan^2\phi + 1)^{\frac{3}{2}}}\\
&=\int_a^b dy\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\frac{\sec^2\phi}{\sec^3\phi}\\
&=\int_a^b dy\int_{\arctan\left(\frac{a}{y}\right)}^{\arctan\left(\frac{b}{y}\right)}d\phi\cos\phi\\
&=\int_a^b dy\left[\sin\left(\arctan\left(\frac{b}{y}\right)\right)-\sin\left(\arctan\left(\frac{a}{y}\right)\right)\right]\\
&=\int_a^b dy\left[\frac{b}{\sqrt{b^2+y^2}}-\frac{a}{\sqrt{a^2+y^2}}\right]\\
&=\left[b\log(\sqrt{a^2+y^2}+y)-a\log(\sqrt{b^2+y^2}+y)\right]_{y=a}^{y=b}\\
&=b\log(\sqrt{a^2+b^2}+b) - a\log((\sqrt{2}+1)b)-b\log((\sqrt{2}+1)a) + a\log(\sqrt{a^2+b^2}+a)\\
&=a\log\left(\frac{\sqrt{a^2+b^2}+a}{(\sqrt{2}+1)b}\right) + b\log\left(\frac{\sqrt{a^2+b^2}+b^2}{(\sqrt{2}+1)a}\right)
\end{align}
This presumes that the substitution $x=y\tan\phi$ is invertible over the interval $(a,b)$ - otherwise, you'll have to split up the integral, which changes things a little. The gist of the calculation should stay the same, though.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $x^3+y^3+z^3-3xyz=1$ defines a surface of revolution
Prove that the equation $x^3+y^3+z^3-3xyz=1$ defines a surface of
revolution and find the analytical equation of its axis of revolution.
I think that I need to apply Euler's formula, so that I get rid of the third-grade polynomial there: $x^3+y^3+z^3-3xyz=1 \Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-xz-yz)=1$ but then I stuck on how to prove it defines a surface of revolution.
The textbook notes that: $f(x,y,z)=0$ defines a surface of revolution around the axis with equation $\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$ if and only if it can be written as a polynomial of $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$ and $ax+by+cz$. Any hint for that one?
| $x^2+y^2+z^2−xy−xz−yz = \mathbf x^T A \mathbf x$
Diagonlize $A$
and find and ortho-normal basis.
As it turns out $(1,1,1)^T$ is an eigenvector of A. The other eigenvalue is a duplicated eigenvalue. And between those two bits of information, that suggests a surface of revolution.
Along similar lines:
$u = \frac 1{\sqrt 3} (x + y + z)\\
v = \frac 1{\sqrt 2} (x+y)\\
w = \frac 1{\sqrt 6} (x+y+2z)$
make this substitution and
$(x+y+z)(x^2+y^2 + z^2 - xy-xz - yz) = 1$ becomes
$\frac {3\sqrt 3}{2} u (v^2 + w^2) = 1$
$v = r \cos \theta\\
w = r \sin \theta$
$u= \frac 2{3r\sqrt3}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why must you have a data point per term when exactly fitting an equation to data points? If you have a linear function $y=Ax+B$, you can uniquely identify it by two points on that line.
If you have a quadratic function $y=Ax^2+Bx+C$ you can uniquely identify it with three points.
The pattern continues for higher functions as well as lower (0th degree constant functions can be uniquely identified by a single point).
This extends to higher dimensions as well.
If you have a function $z=Axy+Bx+Cy+D$ aka a bilinear surface, you can uniquely identify it by 4 points.
Taking surfaces, volumes, or higher and extending them to higher degrees, the pattern continues as far as I know and can tell.
Why is this? While I see the pattern, I can't understand why it's true.
| There is considerable nuance in this equation, because there is no mathematical requirement to match the number of data points to the number of fit parameters.
Start with the polynomial approximation through order $d$ of $f(x,y)$:
$$
\begin{align}
f(x,y)
&= a_{0,0} + a_{1,0}x + a_{0,1}x + a_{2,0}x^{2} + a_{1,1}xy + a_{0,2}y^{2} + \dots + a_{0,d}y^{d} \\
&= a_{0,0} + \sum_{k=1}^{d} \sum_{j=0}^{k} a_{}x^{k-j} y^{j}
\end{align}
$$
The number of terms $n = \frac{1} {2}(d+1)(d+2)$.
Now assume a series of $m$ independent measurements $\left\{ x_{i}, y_{i}, f(x_{i}, y_{i}) \right\}_{i=1}^{m}$. Use the method of least squares to find the amplitudes $a$.
The linear system is
$$
\begin{align}
\mathbf{A} a &= f \\
\left[
\begin{array}{ccccc}
1 & x_{1} & y_{1} & x_{1}^{2} & x_{1}y_{1} & y_{1}^{2} & \dots & y_{1}^{d} \\
1 & x_{2} & y_{2} & x_{2}^{2} & x_{2}y_{2} & y_{2}^{2} & \dots & y_{2}^{d} \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \dots & \vdots\\
1 & x_{m} & y_{m} & x_{m}^{2} & x_{m}y_{m} & y_{m}^{2} & \dots & y_{m}^{d}
\end{array}
\right]
%
\left[
\begin{array}{c}
a_{0,0} \\
a_{1,0} \\
a_{0,1} \\
a_{2,0} \\
a_{1,1} \\
a_{0,2} \\
\vdots \\
a_{0,d}
\end{array}
\right]
&=
\left[
\begin{array}{c}
f(x_{1},y_{1}) \\
f(x_{2},y_{2}) \\
\vdots \\
f(x_{m},y_{m})
\end{array}
\right]
\end{align}
$$
Mathematically the linear system has three classifications. Overdetermined: $m>n$, A is tall, more rows than columns.
Underdetermined: $m<n$, A is wide, more columns than rows.
Full rank: $m=n$, A is square. Provided that the data vector is not in the null space $\mathcal{N}(\mathbf{A}^{*})$, there is a solution in each case:
$$
a = \mathbf{A}^{\dagger} f + \left(\mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)z, \qquad z \in \mathbb{C}^{n}
$$
The matrix operator $\left(\mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right)$ is $\mathbf{P}_{\mathcal{N}(\mathbf{A})}$, a projector onto $\mathcal{N}(\mathbf{A})$, the null space of $\mathbf{A}$.
The singular value decomposition is
$$
\mathbf{A} = \mathbf{U} \mathbf{\Sigma} \mathbf{V}^{*},
$$
where the domain matrices $U\in\mathbb{C}^{m\times m}$, $V\in\mathbb{C}^{n\times n}$ are unitary. That is,
$$
\mathbf{U} \mathbf{U}^{*} = \mathbf{U}^{*}\mathbf{U} = \mathbf{I}_{m}, \qquad \mathbf{V} \mathbf{V}^{*} = \mathbf{V}^{*} \mathbf{V} = \mathbf{I}_{n}.
$$
The matrix $\Sigma$ contains the ordered singular values $\sigma_{k}\in\mathbb{R}$, where $k$ runs from 1 to the matrix rank $\rho$:
$$
\sigma_{1} \ge \sigma_{2} \ge \dots \ge \sigma_{\rho}.
$$
The diagonal matrix
$$
\mathbf{S} =
\left[
\begin{array}{cccc}
\sigma_{1} & 0 & \dots & 0 \\
0 & \sigma_{2} & \dots & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & 0 & 0 & \sigma{\rho}
\end{array}
\right]
$$
is the primary block in the matrix
$$
\Sigma =
\left[
\begin{array}{cc}
\mathbf{S} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array}
\right] .
$$
The pseudo inverse matrix is constructed via
$$
\mathbf{A}^{\dagger} = \mathbf{V} \mathbf{\Sigma}^{\dagger} \mathbf{U}^{*}
$$
where
$$
\mathbf{\Sigma}^{\dagger} =
\left[
\begin{array}{lc}
\mathbf{S}^{-1} & \mathbf{0} \\
\mathbf{0} & \mathbf{0}
\end{array}
\right].
$$
Example Pencil and paper exercise
Part I: Full rank The linear system has $m=n=\rho=2$ and is
$$
\begin{align}
\mathbf{A} x &= b\\
\left[
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right]
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
&=
\left[
\begin{array}{c}
b_{1} \\
b_{2}
\end{array}
\right]
\end{align}.
$$
The matrix inverse exists and the solution is
$$
\begin{align}
x &= \mathbf{A}^{-1} b, \\
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
&=
\left[
\begin{array}{c}
b_{1} \\
b_{2}
\end{array}
\right].
\end{align}
$$
Part II: Underdetermined The linear system has $m=n=2$ with $\rho=1$ and is
$$
\begin{align}
\mathbf{A} x &= b\\
\left[
\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}
\right]
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
&=
\left[
\begin{array}{c}
b_{1} \\
b_{2}
\end{array}
\right]
\end{align}.
$$
The singular value decomposition of $\mathbf{A}$ is painless to compute:
$$
\mathbf{A} = \mathbf{U} \Sigma \mathbf{V}^{*} =
\left[
\begin{array}{cc}
\color{blue}1 & \color{red}0 \\
\color{blue}0 & \color{red}1
\end{array}
\right]
\left[
\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}
\right]
\left[
\begin{array}{cc}
\color{blue}1 & \color{blue}0 \\
\color{red}0 & \color{red}1
\end{array}
\right]
$$
Blue vectors are in a range space, red vectors a null space.
The pseudoinverse follows immediately:
$$
\mathbf{A}^{\dagger} = \mathbf{V} \Sigma^{\dagger} \mathbf{U}^{*} =
\left[
\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}
\right]
$$
The projector matrix is
$$
\mathbf{P}_{\mathcal{N}(\mathcal{A})} =
\mathbf{I}_{2} - \mathbf{A}^{\dagger}\mathbf{A} =
\left[
\begin{array}{cc}
0 & 0 \\
0 & 1
\end{array}
\right].
$$
The least squares minimizers are
$$
\begin{align}
x_{LS}
& = \mathbf{A}^{\dagger} b +
\left( \mathbf{I}_{2} - \mathbf{A}^{\dagger}\mathbf{A} \right) \\
%
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
& =
\left[
\begin{array}{c}
b_{1} \\
0
\end{array}
\right]
+
\alpha
\left[
\begin{array}{c}
0 \\
1
\end{array}
\right]
\end{align}
$$
where the arbitrary parameter $\alpha \in \mathbb{R}$.
The range space of $\mathbf{A}$ is the $x-$axis; the null space is the $y-$axis. The data is in the plane, be we can only see along the $x-$axis. The affine space of the underdetermined solution contains the full rank solution with the proper selection of the parameter $\alpha$, that is, $\alpha = b_{1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the remainder when $7^{7^{7^{7..........Infinity }}}$ is divided by $5$? What is the remainder when $7^{7^{7^{7.........Infinity }}}$ is divided by $5$ ?
My try :
$7^7$ when divided by $5$ gives the remainder $3$,and
similarly, $7^{7^7}$ when divided by $5$ again gives the remainder $3$,and
so, i know that upto infinity it will give remainder $3$.
But, Above approach does not shows any real Maths. How to approach for such questions ?
Edit :
Can I stop the power tower to something like $7^{4k + R}$ ?
| $7\equiv 2 \bmod 5$
$2^x \equiv 2^{x \bmod \phi(5)} \bmod 5$
$2^x \equiv 2^{x \bmod 4} \bmod 5$
$7 \equiv 3\bmod 4$
$3^x \equiv -1^{x} \equiv -1^{x\bmod 2} \bmod 4$
$7^{\text{anything}} \equiv 1 \bmod 2$
$7^{7^{\text{anything}}} \equiv 3^{7^{\text{anything}}} \equiv 3 \bmod 4$
$7^{7^{7^{\text{anything}}}} \equiv 2^{3^{7^{\text{anything}}}} \equiv 2^{3} \equiv 3 \bmod 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How do you calculate $\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{t^2}{\sqrt{3t^4-4}}dt$? How do you calculate $$\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{t^2}{\sqrt{3t^4-4}}dt?$$ Mathematica fails to do it.
| Let $I$ represent the original integral and
$$ I(a)=\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{1}{2t}\arctan\frac{at^2}{\sqrt{3t^4-4}}dt.$$
Clearly $I(0)=0$ and $I(1)=I$.
Note
\begin{eqnarray}
I'(a)&=&\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{\partial}{\partial a}\frac{1}{2t}\arctan\frac{at^2}{\sqrt{3t^4-4}}dt\\
&=&\int_{(4/3)^{1/4}}^{2^{1/4}}\frac{t \sqrt{3 t^4-4}}{\left(a^2+3\right) t^4-4}dt\\
&=&\frac14\int_{(4/3)^{1/2}}^{2^{1/2}}\frac{\sqrt{3 t^2-4}}{\left(a^2+3\right) t^2-4}dt\\
&=&\frac{a \arctan\left(\frac{a t}{\sqrt{3 t^2-4}}\right)+\sqrt{3} \log
\left(\sqrt{9 t^2-12}+3 t\right)}{4 \left(a^2+3\right)}\bigg|_{(4/3)^{1/2}}^{2^{1/2}}\\
&=&-\frac{\pi a-2 a \arctan(a)+\sqrt{3} \left(\log (6)-2 \log
\left(3+\sqrt{3}\right)\right)}{8 \left(a^2+3\right)}.
\end{eqnarray}
So
\begin{eqnarray}
I(a)
&=&-\int_0^1\frac{\pi a-2 a \arctan(a)+\sqrt{3} \left(\log (6)-2 \log
\left(3+\sqrt{3}\right)\right)}{8 \left(a^2+3\right)}da\\
&=&\frac{1}{48} \pi \log \left(\frac{27}{64} \left(2+\sqrt{3}\right)\right)+\frac14\int_0^1\frac{a\arctan(a)}{a^2+3}da.
\end{eqnarray}
Now we solve
\begin{eqnarray}
\int_0^1\frac{a\arctan(a)}{a^2+3}da.
\end{eqnarray}
Let
$$ J(b)=\int_0^1\frac{a\arctan(ab)}{a^2+3}da.$$
Then
$$ J'(b)=\int_0^1\frac{a^2}{(a^2+3)(1+a^2b^2)}da=\frac{6 \arctan (b)-\sqrt{3} \pi b}{6 \left(b-3 b^3\right)}.$$
So
\begin{eqnarray}
\int_0^1\frac{a\arctan(a)}{a^2+3}da&=&\int_0^1\frac{6 \arctan (b)-\sqrt{3} \pi b}{6 \left(b-3 b^3\right)}db\\
&=&\frac{1}{48} \left(8 C+\pi \log \left(64 \left(97-56
\sqrt{3}\right)\right)\right)
\end{eqnarray}
which is obtained from Mathematica. Here $C$ is Catalan constant. Thus
$$ I=\frac{1}{192} \left(8 C+\pi \left(\log \left(64 \left(97-56
\sqrt{3}\right)\right)+4 \log \left(\frac{27}{64}
\left(2+\sqrt{3}\right)\right)\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Writing the equation of a perpendicular bisector Write the equation of the perpendicular bisector of the line segment between the points $(1,-2)$ and $( -1,-2)$.
What I have worked out so far:
The first part is $m = \dfrac{y_2-y_1}{x_2-x_1}$
$$m = \frac{-2-(-2)}{-1-1} = \frac{0}{-2}$$
$$m_{perpendicular} = \frac{2}{0}$$
Is $\frac{2}{0}$ correct? (But I can't divide by zero??) Where do I go from here?
| I'll approach this as a more general problem,
using techniques that are probably beyond what this question was
expecting the solution to use. Consider this an answer for future reference.
Problem:
Given $A=(x_A,y_A)$ and $B=(x_B,y_B),$ find an equation of the
perpendicular bisector of the segment $AB.$
Solution:
Define vectors
$\newcommand{\a}{\mathbf a}\a$ and $\newcommand{\b}{\mathbf b}\b$
equal to the displacements of $A$ and $B$ from the origin:
$$
\a = \begin{pmatrix} x_A \\ y_A \end{pmatrix}, \qquad
\b = \begin{pmatrix} x_B \\ y_B \end{pmatrix}.
$$
Now let $\newcommand{\v}{\mathbf v}\v = \b - \a.$
Then the equation
$$
\v \cdot \begin{pmatrix} x \\ y \end{pmatrix} = c, \tag1
$$
where $\v\cdot\mathbf u$ is the inner product (vector "dot" product)
of $\v$ and $\mathbf u$,
is the equation of a line perpendicular to $\v,$
and therefore perpendicular to the segment $AB.$
The constant $c$ determines which member of that family of lines
the equation describes.
We want a line perpendicular to $AB$ that passes through the midpoint
of the segment $AB,$ that is, the point
$\newcommand{\xC}{\frac{x_A+x_B}{2}}\newcommand{\yC}{\frac{y_A+y_B}{2}}
\left(\xC, \yC\right).$
In order for this point to be on the line described by Equation $1,$
it must be true that
$$
\v \cdot \begin{pmatrix}\xC \\ \yC\end{pmatrix} = c.
$$
We can use this fact to substitute for $c$ in Equation $1,$ with the result
$$
\v \cdot \begin{pmatrix} x \\ y \end{pmatrix} =
\v \cdot \begin{pmatrix}\xC \\ \yC\end{pmatrix}.
$$
This is a perfectly valid equation of the perpendicular bisector of segment $AB.$ In a question such as posed here, however, no doubt a "simpler"
form of the equation is desired.
Such a form can be obtained using the fact that
$$
\v = \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}.
$$
Making this substitution for $\v,$ multiplying term-by-term to evaluate
the inner product, and simplifying algebraically, we have
\begin{align}
\begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}
\cdot \begin{pmatrix} x \\ y \end{pmatrix}
&= \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}
\cdot \begin{pmatrix}\xC \\ \yC\end{pmatrix},\\
(x_B - x_A)x + (y_B - y_A)y
&= (x_B - x_A)\left(\xC\right) + (y_B - y_A)\left(\yC\right),\\
(x_B - x_A)x + (y_B - y_A)y &= \frac12(x_B^2 - x_A^2 + y_B^2 - y_A^2).
\end{align}
For any given points $A$ and $B$ we can then substitute the given values
of $x_A,$ $y_A,$ $x_B,$ and $y_B$ into the equation to obtain the
equation of a line in the simple format $px + qy = k$ for constants
$p,$ $q,$ and $k.$
The advantage of this approach over most others is that it has no special
cases to watch out for (such as when the segment $AB$ is horizontal
or vertical); it works exactly the same for every pair of points
$A$ and $B,$ requiring only that they be distinct points.
But if you must have the equation in a different format
(such as, "$y=mx+b$ unless the line is vertical, in which case write $x=k$"),
it is easy to convert the equation above into the desired format.
In the particular instance given in the question,
$x_A=1,$ $y_A=-2,$ $x_B=-1,$ and $y_B=-2,$
and the equation of the line simplifies to
$$
-2x + 0y = 0,
$$
or even more simply,
$$
x = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 1
} |
problem with joint probability function The joint probability density function of $X$ and $Y$
is given by
$$f(x,y) = \begin{cases}c(y^2-x^2) e^{-y}, & -y \le x < +y, \ 0 \le y < \infty,\\
0, & \text{otherwise}. \end{cases}$$
(a) Find $c$.
(b) Find the marginal densities of $X$ and $Y$.
(c) Find $\operatorname{E}[X]$.
I have some problems with the point (b)
(a) $$
\int_0^\infty\ \int_{-y}^y c(y^2-x^2)e^{-y}\,dy\,dx = 1 \Leftrightarrow c= \frac 1 8.
$$
(b) I calculate the marginal density of $Y$ as
$$
\int_{x=-y}^{x=+y} \frac 1 8 (y^2-x^2)e^{-y}\,dx = \frac 1 6 y^3 e^{-y},
$$
and the density of $X$ as
$$
\int_{y=0}^\infty\ c(y^2-x^2)e^{-y}\,dy,
$$
but there something wrong because the solution is different.
Can someone help me to understand my mistake?
| The marginal density of $Y$ is
$$\int_{-y}^y c(y^2-x^2) e^{-y} \mathop{dx}=ce^{-y}(2y^3 - (2/3)y^3 ) = \frac{4c}{3} y^3 e^{-y}.$$
Integrating this over $y$ gives
$$1=\int_0^\infty \frac{4c}{3} y^3 e^{-y} \mathop{dy} = \frac{4c}{3} \cdot 3! = 8c \implies c=1/8.$$
So, the marginal density of $Y$ is $\frac{1}{6} y^3 e^{-y}$ as you obtained.
The marginal density of $X$ is
$$\int_{|x|}^\infty c(y^2-x^2) e^{-y} \mathop{dy}
.$$
The tricky part is the limits of integration, which comes from the condition $|x| \le y$.
First, integration by parts twice gives
\begin{align}
\int_{|x|}^\infty y^2 e^{-y} \mathop{dy}
&=[-y^2 e^{-y}]_{y=|x|}^\infty + 2\int_{|x|}^\infty ye^{-y} \mathop{dy}\\
&= x^2 e^{-|x|} + 2[-ye^{-y}]_{y=|x|}^\infty + 2\int_{|x|}^\infty e^{-y}\mathop{dy}\\
&= x^2 e^{-|x|} + 2|x|e^{-|x|} + 2e^{-|x|}\\
&= e^{-|x|}(|x|^2+2|x|+2).
\end{align}
So, the marginal density of $X$ is
$$\int_{|x|}^\infty c(y^2-x^2) e^{-y} \mathop{dy}
= c(e^{-|x|}(|x|^2+2|x|+2) - x^2 e^{-|x|}) = \frac{1}{4} e^{-|x|}(|x|+1).$$
Edit (explanation for why the lower limit of integration is $|x|$):
In general, the marginal density of $X$ is
$$f_X(x) = \int_{-\infty}^\infty f(x,y) \mathop{dy}.$$
For a specific $x$, note that by definition $f(x,y)$ is zero if $|x|>y$; otherwise $f(x,y) = c(y^2-x^2)e^{-y}$. So
$$f_X(x) = \int_{-\infty}^\infty f(x,y) \mathop{dy} = \int_{-\infty}^{|x|} f(x,y) \mathop{dy} + \int_{|x|}^\infty f(x,y) \mathop{dy}
= 0 + \int_{|x|}^\infty c(y^2-x^2)e^{-y} \mathop{dy}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show that $f(z)=\frac{z^{5}}{|z|^4}$ satisfies the Cauchy Riemann equations at $z=0$ but not differentable at $z=0$? $$
f(z) = \left\{
\begin{array}{ll}
\frac{z^{5}}{|z|^4} & \quad z \neq 0 \\
0 & \quad z= 0
\end{array}
\right.
$$
I am trying to solve this past exam question. Similar question was asked in Show that $f(z)=\frac{z^5}{|z|^4}$ but has not been answered.
My attempt:
Let $z=x+iy$ then we can separate the real and imaginary parts of $f(z)$ as
$$
f(z)=u+iv=\left(\frac{x^5-10 x^3 y^2+5 x y^4}{\left(x^2+y^2\right)^2}\right)+i\left(\frac{5 x^4 y-10 x^2
y^3+y^5}{\left(x^2+y^2\right)^2}\right)
$$
$$
\frac{\partial u}{\partial x}=\frac{5 x^4-30 x^2 y^2+5 y^4}{\left(x^2+y^2\right)^2}-\frac{4 x
\left(x^5-10 x^3 y^2+5 x y^4\right)}{\left(x^2+y^2\right)^3}
$$
$$
\frac{\partial v}{\partial y}=\frac{5 x^4-30
x^2 y^2+5 y^4}{\left(x^2+y^2\right)^2}-\frac{4 y \left(5 x^4 y-10 x^2
y^3+y^5\right)}{\left(x^2+y^2\right)^3}
$$
$$
\frac{\partial u}{\partial y}=\frac{20 x y^3-20 x^3 y}{\left(x^2+y^2\right)^2}-\frac{4 y \left(x^5-10
x^3 y^2+5 x y^4\right)}{\left(x^2+y^2\right)^3}
$$
$$
\frac{\partial v}{\partial x}=\frac{20 x^3 y-20 x
y^3}{\left(x^2+y^2\right)^2}-\frac{4 x \left(5 x^4 y-10 x^2
y^3+y^5\right)}{\left(x^2+y^2\right)^3}
$$
I am not sure what to do next. If I substitute $z=0\rightarrow x=0,y=0$ in the equations above then they become infinite because the bottom term $(x^{2}+y^{2})$ becomes zero. So how can I show that $f(z)$ satisfies the Cauchy Riemann equations at $z=0$. Also, how can I show that $f(z)$ is not differentiable at $z=0$?
Thanks.
| It must satisfy the CR condition in $0$, because they only involve partial derivatives in the $x-$ and $y-$ directions, and the function coincides with $f(z)=z$ both on $x$ and $y$-axis. (Note that $|ix|^4=x^4=(ix)^4$ for real $x$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all values of $a$ for which the system of equations it has exactly two solutions
Find all values $a$ for which the system of equations it has exactly two solutions
$$\begin{cases}
x^2-8x+y^2-8(|x|+|y|-2)=4;
\\
y=ax-4a-6
\end{cases}$$
My attempt:
1) $$x^2-8x+y^2-8(|x|+|y|-2)=4$$
2) $\forall a \in \mathbb R$ $$(4;-6) \in y=ax-4a-6$$
| As you wrote, $L:y=ax-4a-6$ is a line passing through $P(4,-6)$.
For $x\lt 0$, we have
$$x^2-8x+y^2-8(-x+|y|-2)=4\iff x^2+(|y|-4)^2=4$$
For $x\ge 0$, we have
$$x^2-8x+y^2-8(x+|y|-2)=4\iff (x-8)^2+(|y|-4)^2=68$$
These are the parts of four circles
$$C_1 : x^2+(y-4)^2=4\ \ (x\lt 0,y\ge 0),\quad C_2 : x^2+(y+4)^2=4\ \ (x\lt 0,y\lt 0)$$
$$C_3 : (x-8)^2+(y-4)^2=68\ \ (x\ge 0,y\ge 0),\quad C_4 : (x-8)^2+(y+4)^2=68\ \ (x\ge 0,y\lt 0)$$
$\qquad\qquad$
*
*$C_1$ and $C_3$ intersect at $A(0,2)$ and $B(0,6)$.
*$C_3$ and $C_4$ intersect at $C(8-2\sqrt{13},0)$ and $D(8+2\sqrt{13},0)$.
*$C_2$ and $C_4$ intersect at $E(0,-2)$ and $F(0,-6)$.
*The line $L:y=ax-4a-6$ is tangent to $C_1$ at $G$ when $a=\frac{-10+2\sqrt 7}{3}$.
*The slope of the line $l$ passing through $P,F$ is $0$.
*The slope of the line $g$ passing through $P,D$ is $\frac{\sqrt{13}-2}{3}$.
*The slope of the line $h$ passing through $P,B$ is $-3$.
*The slope of the line $i$ passing through $P,A$ is $-2$.
*The slope of the line $j$ passing through $P,C$ is $\frac{-2-\sqrt{13}}{3}$.
*The slope of the line $n$ passing $P,G$ is $\frac{-10+2\sqrt 7}{3}$.
*The slope of the line $k$ passing through $P,E$ is $-1$.
We see that for $-2\le a\lt\frac{-10+2\sqrt 7}{3}$, the line $L$ and $C_1$ has two intersection points, so these $a$ have to be eliminated.
Also, for $a=\frac{-10+2\sqrt 7}{3}$, there are three intersection points, so this $a$ has to be eliminated.
Therefore, the answer is
$$\color{red}{a\lt -2\quad\text{or}\quad a\gt \frac{-10+2\sqrt 7}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving $x^4+y^4=z^2$ has no integer solutions I need solution check to see if I overlooked something:
If $x^4+y^4=z^2$ has an integer solution then $(\frac{x}{y})^4+1=(\frac{z}{y^2})^2$ has a solution in rationals.
Second equation is equivalent to $X^4+1=Y^2$ which can be written as cubic Weierstrass form $$v^2=u^3-4u$$ where $u=\frac{2(Y+1)}{X^2}$ and $v=\frac{4(Y+1)}{X^3}$.
By using SAGE we can see that Mordell-Weil group of this elliptic curve is empty (SAGE returns empty set, but I'm not sure if that is enough to conclude this), so we can conclude that there are no rational solutions to the second equation and no integer solution to the first one.
edit: No integer solution in positive integers.
| There are no positive integer solutions to $x^4+y^4=z^2.$ By contradiction suppose $(x,y,z)$ is a positive solution with the least possible $z.$
If a prime $p$ divides $x$ and $y$ then $p^4$ divides $z^2,$ so $p^2$ divides $z.$ But then $(x',y',z')=(x/p,y/p,z/p^2)$ is a positive solution with $z'<z,$ contradicting the minimality of $z.$
Therefore $x,y$ are co-prime. So $(x^2,y^2,z)$ is a Fundamental Pythagorean triplet. So there exist co-prime positive integers $m,n$, not both odd, with $\{x^2,y^2\}=\{m^2-n^2, 2mn\}$ and $z=m^2+n^2.$ WLOG $x^2=m^2-n^2$ and $y^2=2mn.$ Note that $x$ is odd and $y$ is even.
Now $x^2+n^2=m^2.$ A prime divisor of both $x$ and $n$ must divide $m,$ but $\gcd(m,n)=1.$ So $x,n$ are co-prime and $(x,n,m)$ is a Fundamental Pythagorean triplet. So there are co-prime positive integers $a,b$ with $\{a^2-b^2, 2ab\}=\{x,n\}$ and $a^2+b^2=m.$ Since $x$ is odd we have $x=a^2-b^2$ and $n=2ab.$
Now $y^2=2mn=2(a^2+b^2)(2ab)=4(a^2+b^2)ab.$ But $\gcd (a,b)=1$ so the members of $\{a,b, a^2+b^2\}$ are pair-wise co-prime and their product $(y/2)^2$ is a square. Therefore each of $a,b,a^2+b^2$ is a square: There are positive integers $d,e,f$ with $a=d^2,b=e^2,a^2+b^2=f^2.$ Therefore $$ f^2=a^2+b^2=d^4+e^4.$$ But $f^2=a^2+b^2=m<m^2+n^2=z,$ so $f\leq f^2<z.$ This contradicts the minimality of $z.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
For $a+b+c=2$ prove that $2a^ab^bc^c\geq a^2b+b^2c+c^2a$ Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove that:
$$2a^ab^bc^c\geq a^2b+b^2c+c^2a$$
I tried convexity, but without success:
We need to prove that
$$\ln2+\sum_{cyc}a\ln{a}\geq\ln\sum_{cyc}a^2b$$ and since $f(x)=x\ln{x}$ is a convex function, by Jensen we obtain:
$$\ln2+\sum_{cyc}a\ln{a}\geq\ln2+3\cdot\frac{2}{3}\ln\frac{2}{3}=\ln\frac{8}{9}.$$
Thus, we need to prove that
$$\frac{8}{9}\geq\sum_{cyc}a^2b$$ or
$$(a+b+c)^3\geq9(a^2b+b^2c+c^2a),$$ which is wrong for $c\rightarrow0^+$.
The equality occurs for $a=b=c=\frac{2}{3}$.
| Proof: By taking logarithm on both sides, it suffices to prove that
$$a\ln a + b\ln b + c\ln c \ge \ln \tfrac{a^2b + b^2c + c^2a}{2}.$$
We will use the following bounds (their proof is not hard and thus omitted):
$$x\ln x \ge f(x) = \tfrac{2}{3}\ln \tfrac{2}{3} + (1 + \ln \tfrac{2}{3})(x- \tfrac{2}{3})
+ \tfrac{9}{20}(x-\tfrac{2}{3})^2, \quad \forall x\in (0, 2]$$
and
$$\ln\tfrac{4}{9} + \tfrac{9}{4}(y - \tfrac49) \ge \ln y, \quad \forall y > 0.$$
With the bounds above, it suffices to prove that
$$f(a) + f(b) + f(c) \ge \ln\tfrac{4}{9} + \tfrac{9}{4}(\tfrac{a^2b + b^2c + c^2a}{2} - \tfrac49)$$
which is simplified to (by using $a+b+c=2$)
$$18a^2+18b^2+18c^2+16 - 45a^2b - 45b^2c - 45c^2a \ge 0.$$
After homogenization, it suffices to prove that
$$(18a^2+18b^2+18c^2)\tfrac{a+b+c}{2}+16(\tfrac{a+b+c}{2})^3- 45a^2b - 45b^2c - 45c^2a \ge 0.$$
The Buffalo Way works. WLOG, assume that $c = \min(a, b, c)$.
Let $b = c + s, \ a = c + t$ for $s, t\ge 0$.
It suffices to prove that
$$(18s^2-18st+18t^2)c +11s^3+15s^2t-30st^2+11t^3\ge 0.$$
It is not hard to prove that $11s^3+15s^2t-30st^2+11t^3\ge 0$ for $s, t\ge 0$. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Integral $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ We have to evaluate the following integral:
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I tried this:
I multiplied both the numerator and denominator by $\sec x$
And substituted $\tan x = t$.
But after that I got stuck.
The book where this is taken from gives the following as the answer: $$\ln(1+t)-\frac14\ln(1+t^4)+\frac1{2\sqrt2}\ln\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}-\frac12\tan^{-1}t^2+c$$ where $t=\sqrt{\cot x}$
| $\displaystyle \mathcal{I} = \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx = \int \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$
substitute $\tan x= t^2$ and $\displaystyle dx = \frac{1}{1+t^4}dt$
$\displaystyle \mathcal{I}= \int\frac{t}{(1+t)(1+t^4)}dt = \frac{1}{2}\int\frac{\bigg((1+t^4)+(1-t^4)\bigg)t}{(1+t)(1+t^4)}dt$
$\displaystyle = \frac{1}{2}\int\frac{t}{1+t}dt+\frac{1}{2}\int\frac{(t-t^2)(1+t^2)}{1+t^4}dt$
$\displaystyle = \frac{1}{2}\int \frac{(1+t)-1}{1+t}dt+\frac{1}{2}\int \frac{t+t^3-(t^2-1)-t^4-1}{1+t^4}dt$
$\displaystyle =-\frac{t}{2}+\frac{1}{2}\ln|t+1|+\frac{1}{4}\int\frac{2t}{1+t^4}+\frac{1}{2}\int\frac{t^3}{1+t^4}dt-\frac{1}{2}\int \frac{t^2-1}{1+t^4}dt-\frac{1}{2}t+\mathcal{C}$
all integrals are easy except $\displaystyle \mathcal{J} = \int\frac{t^2-1}{1+t^4}dt = \int\frac{1-t^{-2}}{\left(t+t^{-1}\right)^2-2}dt = \int\frac{(t-t^{-1})'}{(t-t^{-1})^2-2}dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Help to prove that $\int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$ I am trying to prove that
$$\displaystyle \int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$$
$u=(x^2+1)^{1/2}$ then $du=(x^2+1)^{-1/2}dx$
$$\int_{1}^{\infty}{u^2+(u^2-1)\sqrt{u^2+1}\over u^2\sqrt{u^2+1}}du$$
$v=(u^2+1)^{1/2}$ then $dv=(u^2+1)^{-1/2}du$
$$\int_{1}^{\infty}{v^3+v^2-2v-1\over v^2-1}dv$$
$\int_{1}^{\infty}{v^2\over v^2-1}-{1\over v^2-1}dv$ -$\ln{(v^2-1)}|_{1}^{\infty}$
I am sure I when wrong somewhere, but I can figured it out.
Any help?
| on primitive function should be $${\frac {\sqrt {{x}^{2}+1}\sqrt {{x}^{2}+2}}{\sqrt { \left( {x}^{2}+1
\right) \left( {x}^{2}+2 \right) }}\arctan \left( {\frac {x}{\sqrt {
{x}^{2}+2}}} \right) }+{\frac {\sqrt {{x}^{2}+2} \left( {\rm arcsinh}
\left(x\right)\sqrt {{x}^{2}+1}-x \right) }{\sqrt { \left( {x}^{2}+1
\right) \left( {x}^{2}+2 \right) }}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to simplify this fraction with different powers? I happen to be stuck trying to simplify this:
$$\left[\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}\right]$$
here's the simplified solution that I'm trying to figure out how it was reached
| $$\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}=$$
take the common factor $(x+1)^\frac{1}{2}$
$$=\frac{[(3x+2)(x+1)-(\frac{3}{2}x^2+2x)(\frac{3}{2})]\color{red}{(x+1)^\frac{1}{2}}}{(x+1)^3}=$$
$$=\frac{(3x+2)(x+1)-(\frac{3}{2}x^2+2x)(\frac{3}{2})}{(x+1)^3\color{red}{(x+1)^{-\frac{1}{2}}}}=\frac{(3/4)x^2+2x+2}{(x+1)^{\frac{5}{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Probability: One red ball and $N$ green balls (AMC 10 question) Let $N$ be a positive multiple of $5$. One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\frac35$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\frac45$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N) < \frac{321}{400}$?
This is question from AMC 10, however, I am lost with their explanations. Thanks.
| Since $N$ is a multiple of $5$, let $N = 5n$ for some positive integer $n$. Then we have a total of $5n + 1$ balls. Consider the complementary event that less than $3/5$ of the green balls are on each side of the red ball, which suggests that the red ball is "near the center" of the line.
In particular, we can see that if there exists a side for which at least $3n$ (green) balls are on that side of the red ball, then the red ball is "too far away" from the center. Specifically, if we number the balls $1, 2, \ldots, 5n+1$, the largest number we can pick for the red ball is $3n$, since if we pick $3n+1$ or greater, then the balls $1, 2, \ldots, 3n$ are all green. Similarly, the smallest number we can pick for the red ball is $(5n+1) - (3n-1) = 2n+2$, since if we pick $2n+1$ or less, the balls $2n+2, \ldots, 5n+1$ are all green. It follows that we can only pick the numbers $$\{2n+2, \ldots, 3n\}.$$ If $n = 1$, this is the empty set, since $4 > 3$. For $n > 1$, there are $$3n - (2n+2) + 1 = n-1$$ permissible choices of the red ball out of $5n+1$ choices, therefore the complementary probability is $$1 - \Pr[N] = \frac{n-1}{5n+1},$$ and the desired probability is $$\Pr[N] = \frac{4n+2}{5n+1}.$$ Solving the inequality $$\frac{4n+2}{5n+1} < \frac{321}{400},$$ we get $400(4n+2) < 321(5n+1)$ or $5n > 479$, or $n > 95.8$. The smallest such integer satisfying this inequality is $96$, hence $N = 5(96) = 480$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary operations on matrices
$$A \cdot
\begin{bmatrix}
1 & 3 & 4\\
3 & -1 & 5\\
-2 & 4 & -3\\
\end{bmatrix}
=
\begin{bmatrix}
3 & -1 & 5\\
1 & 3 & 4\\
4 & -8 & 6\\
\end{bmatrix}
$$ Find the $3 \times 3$ matrix $A$.
According to my textbook, the question requires elementary row operations on the given matrices.
I read somewhere that for an equation of the form $AB=X$ ,we can apply elementary row operation on $A$ and $X$ only. I don't know why do these contradict. Where am I wrong?
| Since
$$\det \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\end{bmatrix} = 20 \neq 0$$
we can right-multiply both sides of the linear matrix equation by elementary matrices until we obtain
$$\mathrm A = \begin{bmatrix} 3 & -1 & 5\\ 1 & 3 & 4\\ 4 & -8 & 6\end{bmatrix} \begin{bmatrix} 1 & 3 & 4\\ 3 & -1 & 5\\ -2 & 4 & -3\end{bmatrix}^{-1}$$
We would be doing elementary column operations. If you must do elementary row operations, then do transpose both sides of the linear matrix equation, then do left-multiply both sides by elementary matrices, obtain $\mathrm A^{\top}$ and then transpose to obtain $\mathrm A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ax=b with parametric b vector. A=$\begin{pmatrix}
1 & 3 & 1\\
0 & 1 & 1\\
2 & -2 & -6
\end{pmatrix}$
B=$\begin{pmatrix}
8\\
k\\
8
\end{pmatrix}$
Discuss the solutions of the system S: Ax=B.
I have used the Rouche Capelli theorem wich implicate that S have solutions only if:
$rkA\leq$rk$A'$
where A'= A|B.
Well, $rkA$=2 still. But $rkA'=2$ only if $\begin{vmatrix}
1 & 3 & 8\\
0 & 1 & k\\
2 & -2 & 8
\end{vmatrix}$=$0$
This happens for $k=1$ wich is ok with my textbook but... ...if $k=1$ I can also calculate the other possible determinant:
$\begin{vmatrix}
1 & 1 & 8\\
0 & 1 & k\\
2 & -6 & 8
\end{vmatrix}$ wich for $k=1$ is not equal to $0$ $\Rightarrow $ for $k=1$, $rkA'$=$0$
My textbook still saying that for $k=1$, S have one solution. Why i'm wrong?
My solution will be to create a system made of the possible determinant equalized to 0 (in this case 2 deterimnants).
| Use row reduction for the augmented matrix:
\begin{align}
\begin{bmatrix}1&3&1&8\\0&1&1&k\\2&-2&-6&8\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&3&1&8\\0&1&1&k\\0&-8&-8&-8\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&3&1&8\\0&1&1&k\\0&0&0&(k-1)8\end{bmatrix}\end{align}
Thus $\operatorname{rank}A'=2$ if and only if $k=1$.
Moreover, the solutions are an affine subspace of dimension $1$: if we continue row reduction to obtain the reduced row echelon form, dropping the last (zero) row,we have
$$\begin{bmatrix}1&3&1&8\\0&1&1&1\end{bmatrix}\rightsquigarrow\begin{bmatrix}1&0&-2&5\\0&1&1&1\end{bmatrix}$$
whence the solutions:
\begin{align}
\begin{cases}x&=2z-5,\\
y&=-z-1,\end{cases}\quad\text{or, in vector form:}\quad \begin{bmatrix}x\\y\\z\end{bmatrix}=z\begin{bmatrix}2\\-1\\1\end{bmatrix}-\begin{bmatrix}5\\1\\0\end{bmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$.
Find the maximum value of the expression :
$${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
| Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$\sum_{cyc}\frac{x}{x^2+(x+y+z)^2}\leq\frac{9}{10(x+y+z)}$$ or
$$9\prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)\sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)\geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)\geq0$, which says that
$$9\prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)\sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)\geq$$
$$\geq\frac{1}{81}\left(\sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)\right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)\geq0,$$
which is obviously true.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Finding the derivative of $7^{x^2-x}$ using the chain rule $7^{x^2-x}$
I know there is a formula ($(a^u)' = u' \cdot a^u \cdot \ln a$) for this but I wanted to understand the logic behind that formula so I tried using the chain rule to solve this:
$$(7^{x^2-x})' = (x^2-x)'\cdot7^{x^2-x-1}\cdot 7' = (2x-1)\cdot7^{x^2-x-1}\cdot0 = 0$$
Then I tried replacing $7^{x^2-x}$ with $(e^x)^{\ln7\cdot(x-1)}$:
$$((e^x)^{\ln 7 (x-1)})' = (\ln 7 (x-1))' \cdot (e^x)^{\ln 7 (x-1)-1} \cdot (e^x)' = (\ln7) \cdot (e^x)^{\ln 7 (x-1)-1} \cdot e^x = (\ln7) \cdot (e^x)^{\ln 7 (x-1)} = (\ln7) \cdot 7^{x(x-1)} =(\ln7) \cdot 7^{x(x-1)}$$
But the right answer is $(\ln7) \cdot 7^{x(x-1)} \cdot (2x-1)$. What did I do wrong?
| As you have written $$(a^u)' = u' \cdot a^u \cdot \ln a$$ so :$$(a^{ u })'=u'\cdot a^{ u }\cdot \ln { a } ={ \left( 7^{ x^{ 2 }-x } \right) }^{ \prime }={ \left( { x }^{ 2 }-x \right) }^{ \prime }7^{ x^{ 2 }-x }\ln { 7 } =\left( 2x-1 \right) { 7 }^{ { x }^{ 2 }-x }\ln { 7 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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In a triangle $ABC$ let $D$ be the midpoint of $BC$ . If $\angle ADB=45^\circ$ and $\angle ACD=30^\circ$ then find $\angle BAD$ In a triangle $\Delta ABC$ let $D$ be the midpoint of $BC$. If angle $\angle ADB=45^{\circ}$ and angle $\angle ACD=30^{\circ}$ then find angle $\angle BAD$.
NOW this is a special case do we need a construction.
| Drop perpendicular from $A$ on extended $BC$ intersecting at $E$ to form right triangle $\triangle AEC$.
Let $AC = b$
Then by basic trigonometry we get:
$EA=\frac{b}{2}$
$DC=EC-ED=\frac{\sqrt{3}-1}{2}b$
$EB=EC-BC=\frac{2-\sqrt{3}}{2}b$
$\angle EAB=\tan^{-1}\frac{EB}{EA}=\frac{\pi}{12}$
Therefore,
$$\angle BAD=\angle EAD-\angle EAB=\frac{\pi}{4}-\frac{\pi}{12}=\frac{\pi}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Another way to express $\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$ The number
$$\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$$ is expressed in the form $r \, \text{cis } \theta$, where $0 \le \theta < 360^\circ$. Find $\theta$ in degrees.
Hint: $\text{cis}\ \theta=\cos \theta+i \sin \theta$
Edit:
I simplified the expression down to$$\frac{\text{cis} \ 75^\circ \sin 40^\circ \ \text{cis} \ 40^\circ }{\sin 4^\circ \ \text{cis} \ 4^\circ}.$$
What should I do now?
| \begin{align}
&\text{cis} 75 ^\circ+\text{cis} 83 ^{\circ}+\ldots + \text{cis}147^\circ \\
&= \text{cis} 75^{\circ}(1+\text{cis}8 ^\circ+\ldots+\text{cis}72^{\circ}) \\
\end{align}
Using the geometric series formula, the answer is the same as:
The expression $\frac{(\text{cis}\ 75^\circ)(\text{cis}\ 80^\circ−1)}{\text{cis}\ 8^\circ−1}$ can be written as $r\ \text{cis}\ \theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove the inequality....
Let $x,y$ be positive reals such that $x+y=2$. Prove that :
$x^3y^3(x^3+y^3) \leq 2$
Source : INMO 2002
My attempt :
I started with the left side of the inequality to be proved.
$x^3y^3(x^3+y^3) = x^3y^3(x+y)(x^2+y^2-xy) = 2 x^3y^3(x^2+y^2+2xy-3xy)$
$=2x^3y^3(4-3xy)$
How to proceed ?
Do I have to some AM-GM or Cauchy-Schwarz on particular set of values ?
| You are nearly done.
Applying AM-GM gives us that $$ xy \times xy \times xy \times (4-3xy) \le \frac {(4-3xy+xy+xy+xy)^4}{4^4}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$ How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?
| $2^{n-1} = \frac12 2^n \equiv 10 \cdot 2^n \pmod {19}$. Hence:
$5^{2n+1} +3^{n+2} \cdot 2^{n-1} \equiv 5 \cdot 6^n + 90 \cdot 6^n \pmod {19} \equiv 95 \cdot 6^n \pmod {19} \equiv 0 \pmod {19}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.