Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the limit of $\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}$
Find the following limit: \begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}
I have tried to divide the numerator and denominator by $\sqrt{x}$, but it did not work.
I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work.
I have tried to multiply by the conjugates of the numerator only but it did not work.
So what shall I do?
| Rationalise the numerator.
$\lim_{x\rightarrow4}\left(\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}\right)=\lim_{x\rightarrow4}\left(\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}\times\dfrac{\sqrt{1+2x}+3}{\sqrt{1+2x}+3}\right)=\lim_{x\rightarrow4}\left(\dfrac{2(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{1+2x}+3)}\right)=\dfrac{2(\sqrt{4}+2)}{\sqrt{1+2(4)}+3}=\dfrac{8}{9}=\dfrac43$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$ Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$
I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$
A tangent on $f(x)=\sqrt{1-x}$ at $x={1\over 3}$ is $$y=-{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$
So we have, for all $x\in[0,1]$:
$$\sqrt{1-x} \leq -{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$ and we are done...
I wonder if there is elegant method avoiding calculus?
| How about this:
$$
(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2=2(x+y+z)+2\left[\sqrt{(x+y)(y+z)}+\sqrt{(y+z)(z+x)}+\sqrt{(z+x)(x+y)}\right]\leq
2(x+y+z)+(x+2y+z)+(x+y+2z)+(2x+y+z)=6(x+y+z)
$$
where one uses the GM-AM inequality.
Finally, use $0\leq x\leq y\implies \sqrt{x}\leq\sqrt{y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Help with closed form of $\int_0^\infty\frac{\tanh(ax)}{e^x-1}dx$ I have been trying to find the value of:$$\int_0^\infty\frac{\tanh(ax)}{e^x-1}dx=\int_0^\infty\frac{e^{2ax}-1}{(e^{2ax}+1)(e^x-1)}dx$$
Under u-substitution:
Let $u=e^{ax}$, $x=\frac{\log(u)}{a}$, $dx=\frac{du}{ua}$
$$\int_1^\infty \frac{u^2-1}{(u^2+1)(u^{\frac{1}{a}}-1)}\frac{du}{ua}$$
I don't really know where to proceed further from here or if I should use a series expansion of $\tanh(ax)$.
| Not an answer.
Since marty cohen started with some values, I shall add a few other (including those already given)
$$\left(
\begin{array}{cc}
a & \text{result} \\
\frac{1}{2} & \log (2) \\
1 & \frac{1}{4} (\pi +2\log (2)) \\
\frac{3}{2} & \frac{1}{9} \left(2 \sqrt{3} \pi +3\log (2)\right) \\
2 & \frac{1}{8} \left(\pi +2 \sqrt{2} \pi +2\log (2)\right) \\
3 & \frac{1}{36} \left(\left(15+4 \sqrt{3}\right) \pi +6\log (2)\right) \\
4 & \frac{1}{16} \left(\pi +2 \sqrt{2} \pi +8 \pi \cos \left(\frac{\pi
}{8}\right)+2\log (2)\right)
\end{array}
\right)$$ Using numerical integration (let $a=10^k$), we have as results
$$\left(
\begin{array}{cc}
k & \text{integral} \\
0 & 1.13197 \\
1 & 3.15568 \\
2 & 5.42741 \\
3 & 7.72688 \\
4 & 10.0292 \\
5 & 12.3317 \\
6 & 14.6343 \\
7 & 16.9369 \\
8 & 19.2395 \\
9 & 21.5420 \\
10 & 23.8446
\end{array}
\right)$$ which seems to be almost a straight line when plotted as a function of $k$ $(R^2=0.99997)$ and the slope seems to be very close to $\log(10)$.
$$\begin{array}{clclclclc}
\text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
a & 0.928645 & 0.049392 & \{0.814746,1.042544\} \\
b & 2.287000 & 0.008349 & \{2.267748,2.306253\} \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$ If $a.b,c \in \mathbb{R^+}$ and $ab+bc+ca=1$
Then Prove $$S=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$$
My try we have $$S=\sum \frac{a}{\sqrt{a^2+ab+bc+ca}}=\sum \frac{a}{\sqrt{a+b}\sqrt{a+c})}$$
any hint here?
| Not in the spirit you want to g, but here is alternative solution:
Let $x = \arctan {1\over a}$, $y = \arctan {1\over b}$ and $z = \arctan {1\over c}$
Then $a= \cot x$ and so $\frac{a}{\sqrt{a^2+1}} = \cos x\sin x$ ....
Now we have to prove $$\sin 2x+\sin 2y+\sin 2z\leq 3$$ which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Prove $\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$ If $a,b,c \in \mathbb{R+, }$ Then Prove that $$\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{6^3}$$
My try:
Consider $$P=\frac{a}{3\sqrt{3}}+\frac{b}{4\sqrt{4}}+\frac{c}{5\sqrt{5}}$$
BY Cauchy Scwartz Inequality we have
$$P \le \sqrt{3} \times \sqrt {\left(\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \right)}$$
any way proceed here?
| Using Titu's lemma(special case of CS inequiality): $$\frac{a^2}{3^3}+\frac{b^2}{4^3}+\frac{c^2}{5^3} \ge \frac{(a+b+c)^2}{(3^3+4^3+5^3)}=\frac{(a+b+c)^2}{6^3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How check if the sequence $x_n=(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing or not? I want to check if the sequence $(1+\frac{1}{n})^{\frac{1}{n}}$ is monotonically increasing. I tried computing $\frac{x_{n+1}}{x_n}$ to check if the ratio is less than 1 or greater than 1, but I am unable to simplify:
$$\frac{x_{n+1}}{x_n} = \frac{n^{\frac{1}{n}}\left(n+2\right)^{\frac{1}{n+1}}}{\left(n+1\right)^{\frac{2n+1}{n\left(n+1\right)}}}$$
I also tried $x_{n+1} - x_n = \left(\frac{n+2}{n+1}\right)^{\frac{1}{n+1}}-\left(\frac{n+1}{n}\right)^{\frac{1}{n}}$
Is there any other way to check the monotonic behaviour of the sequence $x_n$?
Also I would like to know if I can check $y_n=(1-\frac{1}{n})^{\frac{1}{n}}$ using similar arguments?
| Here is a simpler solution (which does not use any advanced calculus).
Note that
\begin{alignat*}{2}
&&\frac{-1}{n} & < \frac{-1}{n+1}\\
\Rightarrow &\qquad&1-\frac{1}{n} &< 1-\frac{1}{n+1}\\
\Rightarrow &\qquad&\left(1-\frac{1}{n}\right)^{\frac{1}{n}} & < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}}\label{1}\tag{1}
\end{alignat*}
Now,
\begin{alignat*}{2}
\text{Since,}&\qquad& \left(1-\frac{1}{n+1}\right) & < 1\\
\Rightarrow &\qquad& \left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} & < 1\\
\Rightarrow &\qquad&
\left(1-\frac{1}{n+1}\right)^{\frac{1}{n}}\left(1-\frac{1}{n+1}\right) & <
\left(1-\frac{1}{n+1}\right)\\
\Rightarrow &&\left(1-\frac{1}{n+1}\right)^{\frac{n+1}{n}} & <
\left(1-\frac{1}{n+1}\right)
\end{alignat*}
taking $n+1$-th root,
$$\left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} < \left(1-\frac{1}{n+1}\right)^{\frac{1}{n+1}}\label{2}\tag{2}$$
Thus from (\ref{1}) and (\ref{2}), we get,
$$a_n=\left(1-\frac{1}{n}\right)^{\frac{1}{n}} <
\left(1-\frac{1}{n+1}\right)^{\frac{1}{n}} <
\left(1-\frac{1}{n+1}\right)^{\frac{1}{n+1}}=a_{n+1}$$
Thus we get $a_{n}<a_{n+1}$. Hence the sequence $(a_n)$ is monotonically increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum values of $a^3$ and $b^3$ from the following given cubic function. Today after teaching AM,GM,HM relations my teacher proposed this question.
It isn't a HW though, since we don't have to tell that we solved the question and submit a solution.
Question
Given that $a,b$ are two positive real numbers.
If $$f(x)=2x^3+ax^2+bx+4=0$$
Find the minimum values of $a^3$ and $b^3$.
Well I have progressed a bit.
Applying $AM\ge GM${taking all the terms in $f(x)$ positive}
We get $8x^6ab\le0$ which is impossible since $x$ can't be $0$ and $a,b$ are positive.
So we have a conclusion that roots of $f(x)$ are negative.
Now putting $x=-X$ where $X$ is positive real number, we get
$2X^3+bX=aX^2+4$
This is all useful I got. Using AM,GM on the both the sides one by one yields two inequalities of no use.
Thanks for any solution or hint in advance.
| Here's a sketch.
Assume $a, b \geq 0$. Let the roots be denoted $x_1, x_2, x_3$.
Using Vieta's formula we find
\begin{align}
x_1+x_2+x_3 &= -\frac{a}{2} \\
x_1 x_2 + x_2 x_3 +x_3 x_1&= \frac{b}{2} \\
x1 x_2 x_3 &= -2
\end{align}
The last inequality implies that
$$2 = (-x_1)(-x_2)(-x_3) = |x_1||x_2||x_3|$$
Applying AM-GM, then
\begin{align}
\frac{a}{6} &= -\frac{1}{3}(x_1+x_2+x_3) \\
&\geq ((-x_1)(-x_2)(-x_3))^{1/3} \\
&= \ldots
\end{align}
Also,
\begin{align}
\frac{b}{6} &= \frac{1}{3}(x_1 x_2 + x_2 x_3 +x_3 x_1)\\
&\geq (x_1 x_2 x_1 x_3 x_2 x_3)^{1/3}\\
&\vdots
\end{align}
Then, once you have found $a, b$ above the minimums will follow.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration by substitution vs. polynomial expansion first....different results?? I came across a textbook problem that showed an integral solved with the substitution method:
$$\int_a^b(b-x)^2dx = \left(-\frac{(b-x)^3}{3}\right)\Biggl\vert_a^b$$
I then attempted to solve the same definite integral by expanding out the polynomial first and then integrating. Step 1 I got this:
$$\int_a^b(b^2 - 2bx +x^2)dx$$
Next, I integrated each part of the polynomial and got this:
$$=(b^2x - bx^2 + \frac{x^3}{3})\Biggl\vert_a^b$$
I must have done something wrong here since these results are not equivalent. Where am I missing something? Thanks for your help!
| Not sure why you think the results are not equivalent.
If you evaluate the first, you get
\begin{align*}
\left.\left(-\frac{(b-x)^3}{3}\right)\right|_a^b &= -\frac{(b-b)^3}{3} + \frac{(b-a)^3}{3}\\
&= \frac{1}{3}b^3 - b^2a + ba^2 - \frac{1}{3}a^3.
\end{align*}
On the other hand, if you evaluate your second expression, you get
\begin{align*}
\left.\left(b^2x-bx^2 +\frac{x^3}{3}\right)\right|_a^b &= b^3 - b^3 + \frac{b^3}{3} - b^2a + ba^2 - \frac{a^3}{3}\\
&= \frac{1}{3}b^3 - b^2a + ba^2 -\frac{1}{3}a^3,
\end{align*}
Exactly the same answer. Just because they look different at first glance does not mean they are different.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can $pk+1$ divide $(p-k)^2$? Let $p>3$ be a prime, $0<k<p$. Then is it possible that $pk+1\mid(p-k)^2$?
For $k=1$, since $(p-1)^2\equiv(-2)^2\equiv4\pmod{p+1}$, and $p>3$, this is not possible.
And if $pk+1\mid(p-k)^2$ then $k^2-3pk+p^2-1\geq0$, so $k\leq\frac{3p-\sqrt{5p^2+4}}2$.
This arises from another question. I am running out of ideas, so any help is sincerely appreciated, thanks in advance.
| Suppose that $p$ and $k$ are integers such that $p>k>0$ and $$n=\frac{(p-k)^2}{pk+1}$$
for some integer $n>0$. Note that $(x,y)=(p,k)$ is a solution to the quadratic equation
$$x^2+y^2-(n+2)xy=n.\tag{1}$$
Let $(x,y)=(a,b)$ be a solution to (1) such that $a>b>0$ and $a+b$ takes the smallest possible value. We claim that $n=b^2$.
Observe that $$(x,y)=\big(b,(n+2)b-a\big)=\left(b,\frac{b^2-n}{a}\right)\tag{2}$$
is an integer solution to (1), but not necessarily positive. If $b^2>n$, then we see that (2) gives an integer solution in which
$$b>\frac{b^2-n}{a}>0\text{ and }b+\frac{b^2-n}{a}<a+b.$$
This contradicts the assumption that $a+b$ is minimum. Hence, $b^2\leq n$.
If $b^2<n$, then observe that
$$(x,y)=\left(\frac{b^2-n}{a},\frac{b^2-n}{a}(n+2)-b\right)=\left(\frac{b^2-n}{a},\frac{\left(\frac{b^2-n}{a}\right)^2-n}{b}\right)$$
is another integer solution to (1), with both numbers being negative. Since (1) is invariant under the assignment $(x,y)\mapsto (-y,-x)$, we see that
$$(x,y)=\left(\frac{n-b^2}{a}(n+2)+b,\frac{n-b^2}{a}\right)$$
is a positive integer solution with
$$\frac{n-b^2}{a}(n+2)+b>\frac{n-b^2}{a}>0.$$
By minimality of $a+b$, we must have
$$\left(\frac{n-b^2}{a}(n+2)+b\right)+\frac{n-b^2}{a}\geq a+b.$$
Hence,
$$n(n+3)\geq a^2+(n+3)b^2.$$
Recall from (2) that
$$(n+2)b-a=\frac{b^2-a}{n}<0.$$
That is, $a>(n+2)b$ and we get
$$\begin{align}n(n+3)&\geq a^2+(n+3)b^2>(n+2)^2b^2+(n+3)b^2\\&\geq (n+2)^2+(n+3)=n(n+3)+(2n+7)>n(n+3).\end{align}$$
This is a contradiction, so $n=b^2$ is the only possibility.
With $n=b^2$, (2) becomes $(x,y)=\big(b,(b^2+2)b-a\big)=(b,0)$, so
$$a=b(b^2+2).$$
Note also that
$$(x,y)=\big((b^2+2)p-k,p\big)=\left(\frac{p^2-n}{k},p\right)$$
is also an integer solution to (1) satisfying $x>y>0$, provided that $(x,y)=(p,k)$ is a solution to (1) with $p>k>0$. Indeed, we can see that all integer solutions $(x,y)=(p,k)$ with $n=b^2$ satisfying $p>k>0$ are given by
$$(p,k)=(t_{j+1},t_j)$$
for some positive integer $t_j$. Here, we define $t_0=0$, $t_1=b$, and
$$t_j=(b^2+2)t_{j-1}-t_{j-2}$$
for $j\geq 2$. Note that $t_j$ is divisible by $b$ at all $j$. Since $t_j>b$ for $j>1$, we conclude that $p$ is not prime when $b>1$.
For $b=1$, however, $t_j=F_{2j}$ (where $\left(F_m\right)$ is the Fibonacci sequence). Since $F_j$ divides $F_{2j}$ for all $j$ and $F_j>1$ for $j>2$, we conclude that $p$ is prime only when $p=F_4=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why the $\lim_{n\to\infty} (\frac{n}{n^2+1}+\frac{n}{n^2+4}+.....+\frac{n}{n^2+n^2})= \frac{\pi}{4}$? Why the $\lim_{n\to\infty} (\frac{n}{n^2+1}+\frac{n}{n^2+4}+.....+\frac{n}{n^2+n^2})= \frac{\pi}{4}$?
I read somewhere that it is related to $f(x)=\frac{1}{1+x^2}$ but dont know why...
| Use the def of Riemann integration. Write
$$ \lim_{n \to \infty} \sum_{i=1}^n \frac{ n}{n^2+ i^2} = \lim_{n \to \infty} \sum_{1}^n \frac{1/n}{1+(i/n)^2} = \int\limits_0^1 \frac{1}{1+x^2} dx = \frac{\pi}{4} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $f(x)=x^4$ is concave up by definition? I know $f(x)=x^4$ is concave up, by calculating its second derivative. However, how to prove that $f(x)=x^4$ is concave up by definition, say $f(\frac{x_1+x_2}{2})<(1/2)f(x_1)+(1/2)f(x_2)$ for all $x_1,~x_2$? I'd tried binomial theorem, but can't get anything.
| As an intermediate step, use the inequality
$$ \frac{x^2 + y^2}{2} \ge \left(\frac{x + y}{2}\right)^2 $$
which is relatively easy to prove.
This gives
$$ \frac{x^4 + y^4}{2} = \frac{(x^2)^2 + (y^2)^2}{2} \ge \left(\frac{x^2 + y^2}{2}\right)^2 \ge \left(\left(\frac{x + y}{2}\right)^2\right)^2 = \left(\frac{x + y}{2}\right)^4. $$
Alternatively, using the "usual" definition of convexity, we have
$$ [(1 - \lambda)x^2 + \lambda y^2] - [(1 - \lambda)x + \lambda y]^2 = (1 - \lambda)\lambda (x - y)^2 \ge 0. $$
Being a quadratic form, this is an easy enough sum of squares to compute. One can do it by hand even. Then we do what we did above:
$$ (1 - \lambda)x^4 + \lambda y^4 \ge [(1 - \lambda)x^2 + \lambda y^2]^2 \ge [(1 - \lambda)x + \lambda y]^4 . $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Commutative Semigroup Let $S$ be a Semigroup with the two following properties,
$(1):$ for all $x$ in $S$ we have $x^3=x$
$(2):$ for any $x,y$ in $S$ we have $xy^2x=yx^2y$.
Then prove that this Semigroup $S$ is commutative.
I have found the following identities for any $x,y$ in $S$
*
*$(xy)^3=xy=x^3y^3$
*$xy^2x=y^2(xy^2x)$
*$(xy)^2=y(xy)^2y$
*$xy^2x^2x=yx^2yx^2$
| Suppose $S$ is a semigroup such that
\begin{align*}
x^3&=x,\;\text{for all}\;x\in S\tag{1}\\[4pt]
xy^2x&=yx^2y,\;\text{for all}\;x,y\in S\tag{2}\\[4pt]
\end{align*}
Our goal is to show $xy=yx$, for all $x,y\in S$.
We can recast $(2)$ as
$$(xy)(yx)=(yx)(xy),\;\text{for all}\;x,y\in S\tag{3}$$
so $xy$ commutes with $yx$, for all $x,y\in S$.
Next, working on $y^2x^2y^2$, we get
\begin{align*}
y^2x^2y^2&=(y^2x)(xy^2)\\[4pt]
&=(xy^2)(y^2x)&&\text{[by $(3)$]}\\[4pt]
&=xy^4x\\[4pt]
&=xy^2x&&\text{[by $(1)$]}\\[4pt]
\end{align*}
Thus we have
$$y^2x^2y^2=xy^2x,\;\text{for all}\;x,y\in S\tag{4}$$
Next, working on $x^2y^2$, we get
\begin{align*}
x^2y^2&=(x^2y^2)^3\\[4pt]
&=(x^2y^2)(x^2y^2)(x^2y^2)\\[4pt]
&=x^2(y^2x^2y^2)x^2y^2\\[4pt]
&=x^2(xy^2x)x^2y^2&&\text{[by $(4)$]}\\[4pt]
&=x^3y^2x^3y^2\\[4pt]
&=xy^2xy^2&&\text{[by $(1)$]}\\[4pt]
&=(xy^2x)y^2\\[4pt]
&=(yx^2y)y^2&&\text{[by $(2)$]}\\[4pt]
&=yx^2y^3\\[4pt]
&=yx^2y&&\text{[by $(1)$]}\\[4pt]
&=xy^2x&&\text{[by $(2)$]}\\[4pt]
\end{align*}
Thus we have
$$x^2y^2=yx^2y=xy^2x,\;\text{for all}\;x,y\in S$$
hence, by symmetry, we get
$$x^2y^2=y^2x^2,\;\text{for all}\;x,y\in S\tag{5}$$
so $x^2$ commutes with $y^2$, for all $x,y\in S$.
Next, working on $x^2y$, we get
\begin{align*}
x^2y&=(x^2y)^3&&\text{[by $(1)$]}\\[4pt]
&=x^2(yx^2y)x^2y\\[4pt]
&=x^2(xy^2x)x^2y&&\text{[by $(2)$]}\\[4pt]
&=x^3y^2x^3y\\[4pt]
&=xy^2xy&&\text{[by $(1)$]}\\[4pt]
&=(xy^2x)y\\[4pt]
&=(yx^2y)y&&\text{[by $(2)$]}\\[4pt]
&=y(x^2y^2)\\[4pt]
&=y(y^2x^2)&&\text{[by $(5)$]}\\[4pt]
&=y^3x^2\\[4pt]
&=yx^2&&\text{[by $(1)$]}\\[4pt]
\end{align*}
Thus we have
$$x^2y=yx^2,\;\text{for all}\;x,y\in S\tag{6}$$
so squares commute with everything.
Finally, working on $xy$, we get
\begin{align*}
xy&=(xy)^3&&\text{[by $(1)$]}\\[4pt]
&=x(yx)^2y\\[4pt]
&=(yx)^2xy&&\text{[by $(6)$]}\\[4pt]
&=yx(yx^2y)\\[4pt]
&=yx(xy^2x)&&\text{[by $(2)$]}\\[4pt]
&=yx^2y^2x\\[4pt]
&=y(x^2y^2)x\\[4pt]
&=y(y^2x^2)x&&\text{[by $(5)$]}\\[4pt]
&=y^3x^3\\[4pt]
&=yx&&\text{[by $(1)$]}\\[4pt]
\end{align*}
Thus we have
$$xy=yx,\;\text{for all}\;x,y\in S$$
as was to be shown.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Condition on $k$ for $x^2+y^2-12x-6y-4=0$ and $x^2+y^2-4x-12y-k=0$ to have simultaneous solutions $(x,y)$ The two equations: $$x^2+y^2-12x-6y-4=0$$ and $$x^2+y^2-4x-12y-k=0$$
have simultaneous real solutions $(x,y) \iff a \le k\le b$.
Then, what is the value of $a+b$?
| The two equations represent the circles:
$$\begin{cases}x^2+y^2-12x-6y-4=0 \\ x^2+y^2-4x-12y-k=0 \end{cases} \Rightarrow \\
\begin{cases}(x-6)^2+(y-3)^2=7^2 \\ (x-2)^2+(y-6)^2=40+k \end{cases} \Rightarrow \\
\begin{cases}\text{center: A(6,3), radius: AC=AD=7} \\ \text{center: B(2,6), radius: min BC, max BD} \end{cases} $$
Refer to the graph:
$\hspace{1cm}$
Note that: $AB=\sqrt{(6-2)^2+(3-6)^2}=5$. Hence: $BC=2, BD=12$. It implies:
$$2\le \sqrt{40+k}\le 12 \Rightarrow -36\le k\le 104.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry problem $\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$
What is the value of:
$$\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ$$
I've done trigonometry in my earlier years of high school but I forgot a lot of rules. This is where I'm stuck on this problem:
$\large{{\sin100^\circ+\cos70^\circ\over\cos80^\circ-\cos20^\circ}=\\{\sin(90^\circ+10^\circ)+\cos(60^\circ+10^\circ)\over\cos(90^\circ-10^\circ)-\cos(30^\circ+10^\circ)}=\\{\sin90^\circ\cos10^\circ+\cos90^\circ\sin10^\circ+\cos60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ\over\cos90^\circ\cos10^\circ+\sin90^\circ\sin10^\circ-\cos30^\circ\cos10^\circ+\sin30^\circ\sin10^\circ}=\\{\cos10^\circ+{1\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over\sin10^\circ-{\sqrt3\over2}\cos10^\circ+{1\over2}\sin10^\circ}=\\{{3\over2}\cos10^\circ-{\sqrt3\over2}\sin10^\circ\over{3\over2}\sin10^\circ-{\sqrt3\over2}\cos10^\circ}}$
Not sure what I should do further with this.
| Given $$\dfrac{\sin100+\cos100}{\cos80-\cos20}$$
Now to solve the denominator use the formula $\cos A-\cos B=-2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)$
$\cos80-\cos20=-2\sin\left(\dfrac{80+20}{20}\right)\sin\left(\dfrac{80-20}{2}\right)$
$$=\dfrac{\sin100+\cos70}{-2\sin\left(\dfrac{80+20}{20}\right)\sin\left(\dfrac{80-20}{2}\right)}$$
$$=-\dfrac{\sin100+\cos70}{\sin50}$$
Now to solve the numerator use the identity $\sin A+\sin B=2\cos\left(\dfrac{A-B}{2}\right)\sin\left(\dfrac{A+B}{2}\right)$
$\sin100+\sin20=2\cos\left(\dfrac{100-20}{2}\right)\sin\left(\dfrac{100+20}{2}\right)$
$$=-\dfrac{2\cos\left(\dfrac{100-20}{2}\right)\sin\left(\dfrac{100+20}{2}\right)}{\sin50}$$
$$=-\dfrac{2\cos40\sin60}{\sin50}$$
$$=-\dfrac{2\cos40\cdot\dfrac{\sqrt{3}}{2}}{\sin50}$$
$$-\dfrac{\sqrt{3}\sin(90-40)}{\sin50}=-\sqrt{3}$$
Therefore, $$\dfrac{\sin100+\cos100}{\cos80-\cos20}=-\sqrt{3}$$
| {
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$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational?
The number $x$ defined below is rational or irrational?
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$
From: IMO 1973 - Longlist
My attempt (my real question is at the end):
the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-ac-bc)$ when $a+b+c=0$, leads to $$a^3+b^3+c^3=3abc \tag{1}$$
Now considering
$$a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2},c=-x$$
from (1) it is true that
$$x^3-3x-2\sqrt{5}=0 \tag{2}$$
That is the number $x$ is a root from (2).
Note: By trial and error I've found that answer is $x=\sqrt{5}$ (the other 2 roots are complex), that is irrational. But my question is more subtle.
Question: Can I conclude just inspecting (2), judging by the coefficient $2\sqrt{5}$, that $x$ is irrational, without actually solving the equation? In a math contest that might be helpful, if possible, as it would avoid extra steps.
| If $a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2}$
Clearly, $a-b>0$
$ab=1$ and $a^3-b^3=4$
$$(a-b)^3+3ab(a-b)=4$$
$$\iff(a-b)^3+3(a-b)-4=0$$
Observe that only positive real root of $t^3+3t-4=0$ is $1$
$\implies a-b=1$
$\implies a+b=+\sqrt{(a-b)^2+4ab}=?$
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditioning and Independence of probability Three marksmen hit the target with probabilities $ \frac{1}{2}$,$ \frac{2}{3}$ and $ {\frac{3}{4}}$ respectively.
They shoot simultaneously and there are two hits.
Who missed? Find the probabilities.
I have tried as follows:-
$P(A~\text{hits the target})=\frac{1}{2}= P(A)$. Similarly, $P(B)=\frac{2}{3}$ and $P(C)= \frac{3}{4}$.
Now, $P(A~\text{does not hit}) = P(A')= 1- \frac{1}{2}= \frac{1}{2}$.
Similarly, $P(B')= \frac{1}{3}$ and $P(C')= \frac{1}{4}$.
Since all of them shoot but only two hit so we have three cases as follows:
$$P(A~\text{and}~B~\text{hit but}~C~\text{does not})=P(ABC')= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} = \frac{1}{12}$$
Similarly,
$$P(AB'C)= \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{8}$$
and
$$P(A'BC)= \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$$
So,
$$P(\text{they all shoot simultaneously but only two hit})= \frac{1}{12} + \frac{1}{8} + \frac{1}{4} = \frac{11}{24}$$
Therefore,
\begin{align*}
P(A~\text{missed the target}) & = \frac{P(A'BC)}{P(\text{they all shoot simultaneously but only two hit})}\\
& = \frac{\frac{1}{4}}{\frac{11}{24}}\\
& = \frac{1}{4} \cdot \frac{24}{11}\\
& = \frac{6}{11}
\end{align*}
Similarly,
$$P(B~\text{missed})= \frac{\frac{1}{8}}{\frac{11}{24}} = \frac{3}{11}$$
$$P(C~\text{missed})= \frac{\frac{1}{12}}{\frac{11}{24}} = \frac{2}{11}$$
Please correct me wherever I am wrong, whether explanation or calculation.
| Let $A,B,C$ be the events that person A,B,C hit the target respectively.
Let $X$ be the event that exactly one person missed.
The probability that exactly one misses is:
$$Pr(X)=Pr((A^c\cap B\cap C)\cup (A\cap B^c\cap C)\cup (A\cap B\cap C^c))$$
Using properties of mutually exclusive events and using properties of independent events, this expands out as:
$$=Pr(A^c)Pr(B)Pr(C)+Pr(A)Pr(B^c)Pr(C)+Pr(A)Pr(B)Pr(C^c)$$
Plugging in values:
$$\frac{1}{2}\cdot\frac{2}{3}\cdot\frac34+\frac12\cdot\frac13\cdot\frac34+\frac12\cdot\frac23\cdot\frac14 = \frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{11}{24}$$
We are tasked with calculating the values: $Pr(A^c\mid X)$, $Pr(B^c\mid X), Pr(C^c\mid X)$.
Remembering the definition of conditional probability: $Pr(E\mid F) = \dfrac{Pr(E\cap F)}{Pr(F)}$ we have:
$$Pr(A^c\mid X) = \frac{\frac12\cdot\frac23\cdot\frac34}{\frac{11}{24}}=\frac{6}{11}$$
Similarly calculated, we have $\frac{3}{11}$ and $\frac{2}{11}$ for the probabilities that $B$ and $C$ were the ones who missed respectively. Your attempt was correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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For any primitive pythagorean triple $(a,b,c)$ either $a$ or $b$ must be a multiple of $3$ I'm reading "Friendly Introduction to Number Theory". Now I'm working on Primitive Pythagorean Triples Exercises 2.1 (a) on P18.
We showed that in any primitive Pythagorean triple $(a, b, c)$, either $a$ or $b$ is even. Use the same sort of argument to show that either $a$ or $b$ must be a multiple of 3.
(1) $a^2 + b^2 = c^2$ with a odd, b even, a,b,c having no common factors
(2) $a^2 = c^2 - b^2 = (c-b)(c+b)$
(3) $c + b = s^2$ and $c - b = t^2$
(4) $c = \frac{(s^2 + t^2)}{2}$ and $b = \frac{(s^2 - t^2)}{2}$
(5) $a = \sqrt{(c-b)(c+b)} = st$
(6) $a = st$, $b = \frac{(s^2 - t^2)}{2}$, $c = \frac{(s^2 + t^2)}{2}$
https://www.math.brown.edu/~jhs/frintch1ch6.pdf
I have no idea how I start doing this. Can you give me a hint? I think I need to show that both the following (1) and (2) are satisfied.
$X \neq 0$
(1) $a\equiv 0\pmod 3$ and $b\equiv X\pmod 3$
(2) $b\equiv 0\pmod 3$ and $a\equiv X\pmod 3$
| The numbers $a$ and $b$ cannot both be multiples of $3$, because then $c$ would be a multiple of $3$ and the triple would not be primitive.
And if none of them is a multiple of $3$, then both of them are of the form $3k\pm1$, for some integer $k$, from which it follows that both squares $a^2$ and $b^2$ are of the form $3k+1$. But then $c^2(=a^2+b^2)$ is of the form $3k+2$. You should be able to show that this is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to use the Law of Sines to Find an Angle I am trying to figure out how to find an angle with the law of sines.
I have a triangle where:
A = $120^\circ$
B = unmarked
C = $\theta$
a = 45
b = unmarked
c = 36
How can I find the angle for C?
I have tried:
$$\frac{sin120^\circ}{45} = \frac{sinB}{b} =\frac{sin\theta}{36} $$
$$36(\frac{sin\theta}{36}) = 36(\frac{sin120^\circ}{45})$$
$$sin\theta = \frac{36sin120^\circ}{45}$$
$$ \theta = \frac{36(120^\circ)}{45} = 95^\circ$$
But, the answer is supposed to be $44^\circ$.
| The following step is completely wrong
$$\sin\theta = \frac{36 \sin120^\circ}{45} \iff \color{red}{\theta = \frac{36(120^\circ)}{45} = 95^\circ}$$
we have that $\sin120^\circ=\sqrt 3/2$ and then
$$\sin\theta = \frac{36\sin120^\circ}{45} \iff \sin \theta = \frac{36\sqrt 3}{2\cdot 45} \iff \theta = \arcsin \left(\frac{2\sqrt 3}{9}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding range of function $$f(x) = \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}$$
Prove that for all positive real number $a$, $1<f(x)<2$
According to me i think question is not correct.
as at $a= 16$, we have case when function reaches
infinite value in left of $-1/2$.
| You are right, it seems to be true for
$$f(x) = \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}$$
only for $x>0$, indeed
$$f'(x) = -\frac12\frac{1}{\sqrt[2]{(1+x)^3}}+\frac{a}{ax+8}\frac{1-\frac{ax}{ax+8}}{2\sqrt{\frac{ax}{ax+8}}}<0$$
and $$f(0)=1+\frac{1}{\sqrt{1+a}}<2, \quad f(x)\to 1 \quad x\to \infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Where is the mistake in finding $c$ and $n$ of $\sqrt{x^3+4x} - \sqrt{x^3+x} \sim cx^n$ for $x \to 0$ and $x \to +\infty$ First we simplify in the following way:
$$\frac{3}{\sqrt{x}\left( (1+\frac{4}{x^2})^{1/2}+(1+\frac{1}{x^2})^{1/2} \right)}$$
For $x \to 0$:
We then postulate that since the above must be equivalent to some constant times $x$ to some power, then the inverted fraction is equivalent to some $\frac{1}{cx^n}$, I actually think that this step is flawed, because for $x \to 0$, $\frac{1}{cx^n}$ is indeterminate for $x \neq 0$. We then have
$$\lim_{x \to 0} \frac{\sqrt{x}}{3}\left(\left(1+\frac{4}{x^2}\right)^{1/2}-1 \right) + \lim_{x \to 0} \left( \left(1+\frac{1}{x^2} \right)^{1/2}-1 \right) = \\ \frac{\sqrt{x}}{6} \left( \frac{4}{x^2} + \frac{1}{x^2} \right)$$
And so $c=6/5$ and $n=1.5$, because we have to invert back.
For $x \to +\infty$:
We invert again, but this time also divide by $cx^n$. So that we end up with:
$$\lim_{x \to +\infty} \frac{c}{3} x^n \sqrt{x} \left(\left( 1+\frac{4}{x^2} \right)^{1/2} + \left( 1+\frac{1}{x^2}\right)^{1/2} \right) = \\
\frac{2c}{3}(x^n \sqrt{x})$$
that means that $c=2/3$ and $n=0.5$.
But one or both of these solutions are, in fact, incorrect.
| Your first line is useful when $x\to +\infty$:
$$\frac{3}{\sqrt{x}\left( (1+\frac{4}{x^2})^{1/2}+(1+\frac{1}{x^2})^{1/2} \right)}\sim \frac{3}{\sqrt{x}(1+1)}=\frac{3}{2}\cdot x^{-1/2}.$$
For $x\to 0$, you may simply note that
$$\sqrt{x^3+4x} - \sqrt{x^3+x}=\sqrt{x}(\sqrt{x^2+4} - \sqrt{x^2+1})\sim \sqrt{x} (\sqrt{4}-1)=x^{1/2}.$$
Alternative method.
In both cases, it suffices to use the following fact: as $t\to 0$,
$$\sqrt{1+t}=1+\frac{t}{2}+o(t).$$
1) As $x\to 0$,
$$\begin{align}
\sqrt{x^3+4x} - \sqrt{x^3+x}&=x^{1/2}\left(2\sqrt{1+(x/2)^2} - \sqrt{1+x^2}\right)
\\&=x^{1/2}\left(2(1+\frac{(x/2)^2}{2}+o(x^2)) - (1+\frac{x^2}{2}+o(x^2)\right)\\&=x^{1/2}+o(x^{1/2}).
\end{align}$$
2) As $x\to +\infty$,
$$\begin{align}\sqrt{x^3+4x} - \sqrt{x^3+x}&=x^{3/2}\left(\sqrt{1+4/x^2} - \sqrt{1+1/x^2}\right)
\\&=x^{3/2}\left((1+\frac{4/x^2}{2}+o(1/x^2)) - (1+\frac{1/x^2}{2}+o(1/x^2)\right)
\\&=x^{3/2}\left(\frac{(4-1)/x^2}{2}+o(1/x^2)\right)\\&=\frac{3}{2}\cdot x^{-1/2}+o(x^{-1/2}).\end{align}$$
| {
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How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$ So, I was watching this video by blackpenredpen where he mentions that $$\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$$ so I wanted to try and prove it myself.
Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$
But$\sqrt {a-\sqrt {a+\underbrace{\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}_x}}=x$
$\therefore \sqrt {a-\sqrt {a+x}}=x$
$a-\sqrt {a+x}=x^2$
$x^2-a=-\sqrt {a+x}$
$x^4-2ax^2+a^2=a+x$
$x^4-2ax^2-x+a^2-a=0$
Note that this is of the form $y^4+py^2+qy+r=0$ so we can use Ferrari-Cardano.
We need to find a $z$ such that $(2z-p)y^2-qy+(z^2-r)$ has a discriminant of $0$. The discriminant of $(2z-p)y^2-qy+(z^2-r)$ is equal to $q^2 - 4(2z - p)(z^2 - r),$ which simplifies to $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$
Substituting values from $x^4-2ax^2-x+a^2-a=0$ into $8z^3 - 4pz^2 - 8rz + (4pr - q^2) = 0$ gives us
$8z^3-4\cdot(-2a)\cdot z^2-8\cdot(-a)\cdot z+\left(4\cdot (-2a) \cdot (-a) - (-1)^2 \right)=0 \implies 8z^3+8az^2+8az+(8a^2-1)=0$
Using Cardano's formula, or in my case Wolfram Alpha, we get that $$z_1 = \frac {\sqrt [3]{-16 a^3 - 144 a^2 + 3 \sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}}{6\sqrt[3]2} - \frac {192 a - 64 a^2}{48\cdot 2^{\frac 23} \sqrt [3]{-16 a^3 - 144 a^2 + 3\sqrt 3 \sqrt {256 a^5 + 512 a^4 + 224 a^3 - 288 a^2 + 27} + 27}} - \frac a3$$
I simply can not solve that quintic and continue as it is already too cluttered. Was there a mistake in my problem or is there any other way to do it? Also, I am sincerely sorry but I am not sure how to tag this question.
Edit $1:$
After Ross Millikan's answer, I snooped around in the comment section of the video and found someone who found that it is true using alternating root series. Was his proof correct as $\frac {\sqrt {4a-3}-1}2$ does not seem to have real values for $a \lt \frac 34$? Thank you!
| It's wrong! You can try $a=0$.
Also, try $a=1$.
If your sequence converges then we need to solve the following equation $$\sqrt{a-\sqrt{a+x}}=x.$$
Let $a+x=y^2,$ where $y\geq0$.
Hence, $$y^2-x=a$$ and $$x^2+y=a,$$ where $x\geq0$ and $a\geq0.$
Thus, $$y^2-x^2-x-y=0$$ or
$$(x+y)(y-x-1)=0.$$
If $x+y=0$ so $x=y=a=0$ and your formula is still wrong.
If $y=x+1$ then $x^2+x+1-a=0,$ which gives
$$x=\frac{-1+\sqrt{4a-3}}{2}.$$
Now, since $x\geq0$, we have
$$\frac{-1+\sqrt{4a-3}}{2}\geq0,$$ which gives $a\geq1$.
But for $a=1$ our sequence divergences and it should be $a>1$.
| {
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Your evil probability professor has an urn with 9 balls. Your evil probability professor has an urn with 9 balls: 2 red, 3 white and 4 blue. He draws two balls from the urn without replacement. Let X be the number of red balls drawn and Y the number of white balls.
a) Determine the joint probability mass function of X and Y.
b) Are X and Y independent random variables?
c) Compute the covariance between X and Y.
For point A:
$P(0,0)=\frac{4}{9} \cdot \frac{3}{8} = \frac{1}{6}$ That is correct for the solution.
$P(0,1)=\frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{4}{8} = \frac{1}{3}$ That is correct for the solution.
$P(1,0)=\frac{2}{9} \cdot \frac{4}{8} + \frac{4}{9} \cdot \frac{2}{8}= \frac{2}{9}$ That is correct for the solution.
$P(1,1)=\frac{2}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{2}{8}= \frac{1}{6}$ That is correct for the solution.
$P(2,0)=\frac{2}{9} \cdot \frac{1}{8} = \frac{1}{36}$ That is correct for the solution.
$P(0,2)=\frac{3}{9} \cdot \frac{2}{8} = \frac{1}{12}$ That is correct for the solution.
For point B to check the independancy I have just to check if for example $P(X=0,Y=0) = P(X=0) \cdot P(Y=0)$.
$\frac{1}{6} \neq (\frac{7}{9} \cdot \frac{6}{8}) \cdot (\frac{6}{9} \cdot \frac{5}{8})$.
So X and Y are not independent.
For point C I know that the covariance $Cov(X,Y)=E[X \cdot Y]-E[X]\cdot E[Y]$, but how can compute the expectations, do I have to figure put with distribution is? How can I do it?
Can someone help me?
Thanks in advance, Fabio!
|
but how can compute the expectations, do I have to figure put with distribution is?
You have the distribution, $P(x,y)$. Use it.
The expectation of function $g$ of $X,Y$ is:
$$\begin{align}\mathsf E(g(X,Y)) &=\sum_{x}\sum_{y} g(x,y)~P(x,y)\\[1ex]\textsf{so...}\\[2ex]\mathsf E(X) &=\sum_{x}\sum_{y} x~P(x,y) \\ &=0+P(1,0)+P(1,1)+2P(2,0)\end{align}$$
And so on.
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Is $\frac{x^2y^2}{\sqrt{x^2+y^2}}$ continuous? Let $f: \mathbb{R}^2 $ -> $\mathbb{R}$
I was wondering if this function is a continuous function.
Can I just say that $\frac{x^2y^2}{\sqrt{x^2+y^2}}$ is continuous everywhere except perhaps at $x=0$, because $\lim_{x\rightarrow \infty}$ $\frac{x^2y^2}{\sqrt{x^2+y^2}}$ = $\infty$, and the definition is that if $\lim_{x\rightarrow c}$ $f(x)$ = $f(c)$ then $f(x)$ is continuous at $c$.
| Let $|x| <1$, $|y|<1$.
$|f(x,y)| =|\dfrac{x^2y^2}{\sqrt{x^2+y^2}} | \lt$
$\dfrac{|x||y|}{\sqrt{x^2+y^2}} =$
$\dfrac{\sqrt{x^2}\sqrt{y^2}}{\sqrt{x^2+y^2}}\le$
$\dfrac{\sqrt{x^2+y^2}\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}=$
$\sqrt{x^2+y^2} .$
Choose $\delta = \epsilon.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Can a divisor of $n^2 +n+1$ be $2\operatorname{mod} 3$? Let $n$ be a natural number. It seems that a divisor of $n^2+n+1$ cannot be $2 \operatorname{mod} 3$.
Couldn't $n^2+n+1$ have an even number of divisors which are $2 \operatorname{mod} 3$?
| Let $p$ be a prime $\equiv 2\bmod 3$. Suppose $p|(n^2+n+1)$. Then also $p|(n^3-1)$ since $n^3-1=(n-1)(n^2+n+1)$.
So $n^{p-1}\equiv 1\bmod p$ from Fermat's Little Theorem and also $n^3\equiv 1\bmod p$ from above. Since $p$ is taken to be $\equiv 2\bmod 3$, the latter implies $n^{p+1}\equiv 1\bmod p$. Then
$n^{3+(p-1)-(p+1)}=n\equiv 1\bmod p$
which is contradictory because $1^2+1+1\equiv 3\not\equiv 0\bmod p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solve for the coefficient of an even generating function Using a generating function, find the number of ways to select 10 candies from a huge pile of red, blue, and green lollipops if the selection has an even number of blue lollipops.
I started, but I don't understand how to continue
$(1 + x + x^2 + x^3 ...)^2 \cdot (1 + x^2 + x^4 + x^6 ...)$
$=(\frac{1}{1-x})^2 \cdot \frac{1}{1-x^2}$
$=\big(1 + \binom{1 + 2 - 1}{1}x + \binom{2 + 2 - 1}{2}x^2 ...\big) \cdot ???$
How do you find coefficient to $x^{10}$?
Also, our textbook tells us $(1 + x^2 + x^4 + x^6 ...)$ = $\frac{1}{1-x^2}$ how?
| Notice that $\left(\frac{1}{1-x}\right)^2=\frac{d}{dx}\left(\frac{1}{1-x}\right)$, and hence $\left(\frac{1}{1-x}\right)^2=1+2x+3x^2+\dots$ You can then find the $x^{10}$ coefficient just by multiplying out.
The equality stated in the textbooks holds for the same reason that $1+x+x^2+\dots=\frac{1}{1-x}$.
| {
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"url": "https://math.stackexchange.com/questions/2979775",
"timestamp": "2023-03-29T00:00:00",
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Find all functions such that $f(x^2+y^2f(x))=xf(y)^2-f(x)^2$ I am dealing with the test of the OBM (Brasilian Math Olympiad), University level, 2016, phase 2.
I hope someone can help me to discuss this test. Thanks for any help.
The question 2 says:
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that
$f(x^2+y^2f(x))=xf(y)^2-f(x)^2$
for all $x,y\in\mathbb{R}$.
My attempt:
Note that $f(0)\in\{0,-1\}$. In fact, by taking $x=y=0$, we have $f(0)=-f(0)^2$.
Case 1 $f(0)=0$
By taking $y=0$, we have
$f(x^2)=-f(x)^2\forall x\in\mathbb{R}$
Particularly, $f(1)=-f(1)^2$, so $f(1)\in\{0,-1\}$.
(a) f(1)=0
By taking $x=1$, we have $f(1)=f(y)^2\forall y\in\mathbb{R}$.
So, $f\equiv 0$. Is trivial that it respects the statement.
(b) f(1)=-1
By taking $x=1$, we have $f(1-y^2)=f(y)^2-1=-f(y^2)-1\forall y\in\mathbb{R}$. So, to $t\leq 0$, we have $f(1-x)=-f(x)-1$.
By taking $y=1$, we have $f(x^2+f(x))=x-f(x)^2=x+f(x^2) \forall x\in\mathbb{R}$.
I could not finish this subcase
Case 2 $f(0)=-1$
By taking $x=0$,
$f(-y^2)=-1\forall y\in\mathbb{R}$
So, $f(t)=-1\forall t\leq0$.
By taking $y=0$,
$f(x^2)=-x-f(x)^2 \forall x\in\mathbb{R}$
So, $f(t)=-\sqrt{t}-1\forall t\geq0$.
But this function does not is correct. For instance, to $x=y=1$, $f(x^2+y^2f(x))=f(1+1(-2))=f(-1)=-1$, but $xf(y)^2-f(x)^2=1(-2)^2-(-2)=6\not=-1$.
| Partial progress, but not a complete answer, I'm afraid.
$$f(x^2+y^2f(x)) = xf(y)^2-f(x)^2$$
$f$ has a root
Let $y=x$; then $f(x^2(1+f(x)) = (x-1)f(x)^2$. In particular, letting $x=1$ we obtain $f(1+f(1)) = 0$, so $f$ does have a root.
$f$ is $0$ or has exactly the root $0$
Suppose $f(x) = 0$. Then $f(x^2) = x f(y)^2$ for all $y$, and so either $x = 0$ or $f(y)^2$ is constant as $y$ varies.
Suppose $f(x) = 0$ but $x \not = 0$. Then $f(y)^2$ is constant as $y$ varies; but substituting $y = x$ we obtain that $f(y)^2 = 0$ and hence $f$ is the constant $0$.
So the only possible nonzero case is that $f$ has exactly one root, and it is the root $x = 0$.
$f$ is very nearly symmetric
Substitute $y \to -y$ to obtain the following:
$$x f(y)^2-f(x)^2 = f(x^2+y^2f(x)) = x f(-y)^2-f(x)^2$$
from which $$x f(y)^2 = x f(-y)^2$$ for all $x$ and $y$; in particular, $$f(y) = \pm f(-y)$$ for all $y$.
$f$ is odd or $0$
Suppose $f(x) = f(-x)$. Then $$x f(y) - f(x)^2 = f(x^2 + y^2 f(x)) = -x f(y) - f(-x)^2 = -x f(y) - f(x)^2$$
and so $-x f(y) = x f(y)$ for all $y$; so (since wlog $f$ is not the constant zero function) $-x = x$ and hence $x=0$.
So if $f(x) = f(-x)$ then $x = 0$; hence $f(-x) = -f(x)$ for all $x$.
$f$ is sign-reversing or $0$
Note also that since $f(x^2) = -f(x)^2$ (by letting $y=0$), for every $x > 0$ we have $f(x) < 0$.
$f(n) = -n$ or $f=0$
Substituting $x=-1$ gives $f(1+y^2) = -f(y)^2-1$
and in particular $$f(x^2+1) = f(x^2)-1$$
Therefore $f(x+1) = f(x)-1$ whenever $x>0$.
This fixes the value of $f$ on the natural numbers: we have $f(n) = -n$.
We already know that the root occurs at $x=1+f(1)$, so $f(1) = -1$ (as you noted). Moreover, by letting $x=y$ and supposing $f(x)=-1$, we get $f(0) = x-1$ at any such $x$, and so $x=1$ is the only time $f$ hits $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence.
Let $n\in \mathbb N$ and:
$$
x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n
$$
Show that $\{x_n\}$ is a bounded sequence.
This sequence appears a bit tricky because it involves harmonic series. Below are steps I take.
Lower bound:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \ge \sqrt{\sum_{k=1}^n\left(\frac{k}{k}\right)^2} = \sqrt{\sum_{k=1}^n1}=\sqrt{n}\implies\\
\implies x_n \ge \sqrt n - \sqrt n \ge 0
$$
Lower bound is simple.
Upper bound:
To get rid of radical lets use Cauchy-Schwarz (note the below is incorrect as shown in Zvi's answer):
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \le \sqrt{\left(\sum_{k=1}^n\left(\frac{k+1}{k}\right)\right)^2} =\sum_{k=1}^n\left(\frac{k+1}{k}\right) = n+\sum_{k=1}^n{1\over k} = n + H_n
$$
So this doesn't show $x_n$ is bounded above. I've tried another approach:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}
$$
Consider nominator:
$$
\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=n+\sum_{k=1}^n{2\over k}+\sum_{k=1}^n{1\over k^2}-n = \sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}
$$
For denominator:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n = \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n
$$
So $x_n$ is:
$$
x_n = \frac{\sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}}{ \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n }
$$
But i don't see how to proceed from this point. What else could i try? How to show $x_n$ is bounded above?
Please note the precalculus tag.
| You have, using that $a^2-b^2=(a+b)(a-b)$,
\begin{align}
\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=\sum_{k=1}^n \left[ \left(1+\frac1k\right)^2-1\right]
=\sum_{k=1}^n \frac1k\left(2+\frac1k\right)\leq3\sum_{k=1}^n\frac1k.
\end{align}
Using the comparison with the integral we get
$$
\sum_{k=1}^n\frac1k\leq1+\int_1^n\frac1t\,dt=1+\log n.
$$
Then
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}\leq\frac{3(1+\log n)}{\sqrt n}.
$$
Since $\lim_{x\to\infty}\frac{3(1+\log x)}{\sqrt x}=0$, the continous function $f(x)=\frac{3(1+\log x)}{\sqrt x}$ is bounded on $[1,\infty)$. Thus there exists $C>0$ with
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n \leq\frac{3(1+\log n)}{\sqrt n}\leq C.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Can we proof that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?; given $0 < a < b < 1$ I have found in some test problem:
Given $0 < a < b < 1$, can we conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?
I divide the combination of a and b into a few cases. Then, see what happens.
case 1 : $a \rightarrow 0$ and $b \rightarrow 0$
When $a$ approaches to zero and $b$ approaches to $a$ (which is zero), the expression will be evaluated as $\sqrt{0 + 0}$ and $\sqrt{0} + \sqrt{0}$. Since $0 = 0$, this case cannot conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 2 : $a \rightarrow 1$ and $b \rightarrow 1$
When $b$ approaches to one and $a$ approches to $b$ (which is one), the expression will be evaluated as $\sqrt{1 + 1}$ and $\sqrt{1} + \sqrt{1}$. Since
$\sqrt{2} < 2$, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 3 : $a \rightarrow 0$ and $b \rightarrow 1$
When $a$ approaches to zero and $b$ approches to one, the expression will be evaluated as $\sqrt{0 + 1}$ and $\sqrt{0} + \sqrt{1}$. Since
$1 = 1$, then this case cannot concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 4 : $a \rightarrow b$
When $a$ approaches to $b$ the expression will be evaluated as $\sqrt{b + b}$ and $\sqrt{b} + \sqrt{b}$. Since $\sqrt{2b} < 2\sqrt{b} $, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
From above cases, am I still missing some point? If not, How should I write the conclusion from that pieces of thinking mathematically? Because I instinctively believe that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$ should be true (and not $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$).
| A bit of geometry.
Set $x^2 = a$; $y^2 =b$, where $x,y >0.$
Choose a.point $C(x,y)$ in the first quadrant.
Points : $O(0,0)$, $A(x,0)$ and $C(x,y)$.
$\triangle OAC$ is a right triangle with side lengths $x,y$ and $\sqrt{x^2+y^2.}$ (Pythagoras)
The sum of the lengths of two sides in a triangle is greater than the length of the third side:
$\sqrt{x^2+y^2} < x+y$, or in original
notation $\sqrt{a+b} < √a+√b.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $y=\frac{x^2}{x^4+25}$ ,prove: $0 \leq y \leq \frac{1}{10}$ We know that:
$$y=\frac{x^2}{x^4+25}$$
Then we have to prove that:
$$0 \leq y \leq \frac{1}{10}$$
How to use what we obtain from first fraction to prove what it wanted? Do you have an easy idea?
| $x^2, x^4, x^4 + 25 \ge 0$ so $\frac {x^2}{x^4 + 25} \ge 0$.
If $x = 0$ then $\frac {x^2}{x^4 + 25} = 0 < \frac 1{10}$.
If $x \ne 0$ then $\frac {x^2}{x^4 + 25} \le \frac 1{10}\iff \frac {x^4 + 25}{x^2} \ge 10$.
And by AM-GM... $\frac {x^4 + 25}{x^2} = x^2 + \frac {25}{x^2} \ge 2\sqrt{x^2*\frac {25}{x^2}}= 2\sqrt{25} =10$.
.... But perhaps the most direct way as all terms are positive.
$\frac {x^2}{x^4 + 25} \le \frac 1{10} \iff 10x^2 \le x^4 + 25 \iff x^4 -10x^2 + 25=(x^2 -5)^2 \ge 0$. which must be true as $(x^2 - 5)^2$ is a perfect square.
[Note: the second argument is essentially replicating a prove of the AM-GM: Namely for positive $a,b$ we know $a + b \ge 2\sqrt{ab}\iff a-2\sqrt{ab} + b = (\sqrt a - \sqrt b) \ge 0$ which.... must be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Implicit functions and related differential equations I'm seeking guidance in derivation of implicit equation solutions to second degree differential equations.In the example below, differentiating twice just produced a tangle of terms which did not obviously lead to the required result.
Example: If
$$y^3 +3yx +2x^3 = 0, $$
prove that $$x^2(1+x^3)y'' - (3/2)xy' +y =0$$
| Borrow ideas from the numerical variant of Cardano's method for cubic equations.
Set $y=2\sqrt{x}\sinh(u)$, then
\begin{align}
0&=y^3+3xy+2x^3\\
&=2x\sqrt x(4\sinh(u)^3+3\sinh(u))+2x^3\\
&=2x\sqrt x\sinh(3u)+2x^3
\end{align}
$$\\~\\
\implies u=-\frac13\text{asinh}(x^{3/2}),
~~
u'=-\frac12\frac{\sqrt{x}}{\sqrt{1+x^3}}=-\frac12(x^{-1}+x^2)^{-1/2},
~~
u''=\frac14\frac{-x^{-2}+2x}{(x^{-1}+x^2)^{3/2}}
$$
and on the other side
$$
\left(\frac{y}{2\sqrt x}\right)'=\cosh(u)u',~~\left(\frac{y}{2\sqrt x}\right)''=\sinh(u)u'^2+\cosh(u)u''
\\~\\
\implies
u'\left(\frac{y}{2\sqrt x}\right)''=u''\left(\frac{y}{2\sqrt x}\right)'+u'^3\left(\frac{y}{2\sqrt x}\right).
$$
This is a linear ODE in $y$, as $u$ and its derivatives are solved as functions of $x$.
Now insert $u', u''$, apply the quotient rule, and simplify.
Observing the coefficient structure, one could first collect the terms as
$$
\left(\frac1{u'}\left(\frac{y}{\sqrt x}\right)'\right)'=u'\left(\frac{y}{\sqrt x}\right).
$$
As $(x^{-1/2}y)'=x^{-1/2}y'-1/2x^{-3/2}y=x^{-3/2}(xy'-\frac12y)$, the last formula results in
$$
\left(2\sqrt{x^{-1}+x^{-4}}\left(xy'-\frac12y\right)\right)'
=\frac1{2x\sqrt{1+x^{-3}}}\left(\frac{y}{\sqrt x}\right)
\\~\\
-\frac{x^{-2}+4x^{-5}}{\sqrt{x^{-1}+x^{-4}}}\left(xy'-\frac12y\right)
+2\sqrt{x^{-1}+x^{-4}}\left(xy''+\frac12y'\right)
-\frac1{2x\sqrt{1+x^{-3}}}\left(\frac{y}{\sqrt x}\right)=0
\\~\\
-(x^3+4)\left(xy'-\frac12y\right)
+2x(1+x^{3})\left(xy''+\frac12y'\right)
-\frac{x^3}{2}y=0
\\~\\
2x^2(1+x^3)y''+[x(1+x^3)-x(x^3+4)]y'+2y=0
\\~\\
x^2(1+x^3)y''-\frac32xy'+y=0
$$
which is indeed the claimed differential equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Data transformation: new min, max and mean I have a dataset of 31 values, of which the $min = -0,8,\ max = 11, 1$ and $mean = 5,0$.
Is there a way to transform these data to a $\text{new} \ minimum (0.3), \ \text{new} \ maximum (11.4)$ and $\text{new} \ mean (5.3)$?
Thanks!
6,7
4,9
4,5
3,0
4,6
1,9
2,7
4,0
2,3
6,4
7,6
6,2
6,7
6,3
11,1
9,1
8,3
8,3
4,5
5,6
5,6
5,9
4,4
5,9
2,4
0,7
1,0
-0,8
| One way would be to use a quadratic equation to change every value in your dataset from $x_i$ to $y_i= a \cdot x_i^2+b \cdot x_i+ c$. The equation for the new minimum would then be $$0.3 = a \cdot (-0.8)^2+b \cdot (-0.8) + c\tag{1}$$
The equation for the new maximum would be $$11.4 = a \cdot (11.1)^2+b \cdot (11.1) + c \tag{2}$$
And the equation for the new mean would be $$5.3 = \frac{1}{31}\sum_{i=1}^{31} (a \cdot x_i^2+b \cdot x_i + c)$$
or
$$5.3 = a \cdot \frac{1}{31}\sum_{i=1}^{31} x_i^2+b \cdot (5.0) + c \tag{3}$$
Assuming equations $1$, $2$ and $3$ are linearly independent you could work out the values of $a$, $b$ and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that x + $\frac{9}{x}$ $\ge$ 6 for all real numbers x $>$ 0 I have:
$x^2+9\ge6x$
$x^2-6x+9\ge0$
$(x-3)^2\ge0$
Is this a sufficient proof for all real numbers? Or do I need to prove that it works from $1<x<3$?
| Option:
AM-GM:
$x>0.$
$x+9/x \ge 2\sqrt{x(9/x)}= 2\sqrt{9} =6.$
Equality for $x=9/x$.
Your answer, as other users have pointed out already:
For $x>0$.
Start from:
1)$(x-3)^2 \ge 0$, this is true for any real $x$ (why?)
Expand :
2) $x^2-6x +9\ge 0.$
Divide inequality by $x>0:$
3) $x -6 +9/x \ge0.$
4) $x+9/x \ge 6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve recurring sequence using a generating function I have the sequence $a_n=3a_{n-1}-3a_{n-2}+a_{n-3}$, $\forall\ n \ge 3$, with $a_0=2$, $a_1=2$, $a_2=4$ being the known terms, and I want to find a non-recursive equation for $a_n$ using a generating function.
What I have done:
$$
\begin{align}
A(x)
& = \sum_{n\ge0}{a_nx^n}
= a_0+a_1x+a_2x^2+\sum_{n\ge 3}\left({3a_{n-1}-3a_{n-2}+a_{n-3}}\right)x^n\\
& = 2+2x+4x^2+3x\sum_{n\ge 3}{a_{n-1}x^{n-1}}
-3x^2\sum_{n\ge 3}{a_{n-2}x^{n-2}}
+x^3\sum_{n\ge 3}{a_{n-3}x^{n-3}}\\
& = 2+2x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)\\
& = \frac{2+2x+4x^2}{1-3x+3x^2-x^3}\\
& = \frac{4x^2+2x+2}{(1-x)^3}
\end{align}
$$
As shown above, I have reached a solution for $A(x)$, but I'm not sure how to use it to find a solution for $a_n$. Any tips pointing me in the right direction would be greatly appreciated.
| $A(x)=2+2x+4x^2+3x\sum_{n\geq3}a_{n-1}x^{n-1}-3x^2\sum_{n\geq3}a_{n-2}x^{n-2}+x^3\sum_{n\geq3}a_{n-3}x^{n-3}$
$\Rightarrow A(x)=2-4x+4x^2+3x\sum_{n\geq1}a_{n-1}x^{n-1}-3x^2\sum_{n\geq2}a_{n-2}x^{n-2}+x^3A(x)$
$\Rightarrow A(x)=2-4x+4x^2+3xA(x)-3x^2A(x)+x^3A(x)$
$\Rightarrow (-x^3+3x^2-3x+1)A(x)=2-4x+4x^2$
$\Rightarrow A(x)=\frac{2-4x+4x^2}{-x^3+3x^2-3x+1}=\frac{2(2x^2-2x+1)}{(1-x)^3}=\frac{4}{1-x}-\frac{4}{(1-x)^2}+\frac{2}{(1-x)^3}$
I will use $\sum_{n\geq 0}x^n=\frac{1}{1-x}$
claim: $\frac{1}{(1-x)^k}=\sum_{n\geq 0}\binom{n+k-1}{n}x^n$
So, $\frac{1}{(1-x)^k}=(\sum_{n\geq 0}x^n)^k=\sum_{n\geq0}x^n(\sum_{{i_1}+{i_2}+...+{i_k}=n}1)=\sum_{n\geq0}x^n\binom{n+k-1}{n}$
So, now you need to find $a_n$ which is the coefficient of $x^n$ in the generating function
For your problem, it is $4\cdot\binom{n+1-1}{n}-4\cdot\binom{n+2-1}{n}+2\cdot\binom{n+3-1}{n}=4-4(n+1)+\frac{2(n+1)(n+2)}{2}=n^2+3n+2-4n\\=n^2-n+2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Tangential planes of $f(x,y) := (y^2-x)(y^2-2x) $ in $(-1,1)$ and $(-1,-1)$ Let $f:\mathbb{R^2} \to \mathbb{R}$ with $f(x,y) := (y^2-x)(y^2-2x) $
How can I find the function rules $\tau_{(-1,1)}(x,y)$ and $\tau_{(-1,-1)}(x,y)$ of the tangential planes on the graph of the function $f$ in the points $(-1,1)$ and $(-1,-1)$?
I know that the general equation of a plane $E$ is given by
$$E: z = z_0 + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0)$$
So first I have to find out the partial derivatives:
$$f_x = 4x-3y^2$$
$$f_y = 4y^3-6xy$$
For the point $(-1,1)$ I get:
$$f_x = 4\cdot (-1) -3\cdot 1^2 = -7$$
$$f_y = 4\cdot1^3 - 6\cdot(-1)\cdot1 = 2$$
For the point $(-1,-1)$ I get:
$$f_x = 4\cdot (-1) - 3\cdot (-1)^2 = -1$$
$$f_y = 4\cdot (-1)^3 - 6\cdot (-1)\cdot (-1)= -10$$
To get $z_0$ we also evaluate $f$ on the point $(-1,1)$:
$$z_0 = f(-1,1) = (1^2 - (-1)) \cdot ((1^2) - 2\cdot (-1)) = 4$$ and on the point $(-1,-1)$:
$$z_0 = f(-1,-1) = ((-1)^2 - (-1)) \cdot ((-1)^2 - 2\cdot (-1)) = 6$$
Now I can use the formula for the tangential plane for the point $(-1,1)$:
$$E: z = z_0 + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0) \\
= 4 + (-7)\cdot(x-(-1)) + 2 \cdot (y-1) \\
= 4-7x + 7 + 2y - 2 \\
= -7x + 2y + 9$$
So the first tangential plane is $E: z = -7x + 2y + 9$
I continue with the formula for the tangential plane for the point $(-1,-1)$:
$$E: z = z_0 + f_x(x_0,y_0) (x-x_0) + f_y(x_0,y_0) (y-y_0) \\
= 6 + (-1) \cdot (x- (-1)) + (-10) \cdot (y- (-1)) \\
= 6 -x - 1 - 10 y - 10 \\
= -x - 10y - 5 $$
So the second tangential plane is $E: z = -x - 10y - 5$
Two questions:
- Is this correct?
- How can I find the intersecting line of those two tangential planes (or how can I show that those areas are parallel/matching?)
| Given $f(x,y) := (y^2-x)(y^2-2x)$ find the equations of the tangent planes at $(-1,1)$ and $(-1.-1)$.
Note to OP: You made errors in computing some of the constants.
\begin{eqnarray}
f_x(x,y)&=&-3y^2+4x\\
f_x(-1,1)&=&-7\\
f_x(-1,-1)&=&-7\\
f_y(x,y)&=&2y(2y^2-3x)\\
f_y(-1,1)&=&10\\
f_y(-1,-1)&=&-10\\
f(-1,1)&=&6\\
f(-1,-1)&=&6
\end{eqnarray}
\begin{eqnarray}
\tau_{(-1,1)}(x,y)&=&f(-1,1)+f_x(-1,1)(x+1)+f_y(-1,1)(y-1)\\
&=&6-7(x+1)+10(y-1)\\
&=&-7x+10y-11
\end{eqnarray}
\begin{eqnarray}
\tau_{(-1,-1)}(x,y)&=&f(-1,-1)+f_x(-1,-1)(x+1)+f_y(-1,-1)(y+1)\\
&=&6-7(x+1)-10(y+1)\\
&=&-7x-10y-11
\end{eqnarray}
Find the equation of the line of intersection of the two tangent planes.
First, rewrite the equations of the tangent planes in the following form:
\begin{eqnarray}
7x-10y+z&=&-11\\
7x+10y+z&=&-11
\end{eqnarray}
In this form we see that the normal vectors for the two planes are
\begin{eqnarray}
\mathbf{N}_{(-1,1)}&=&(7,-10,1)\\
\mathbf{N}_{(-1,-1)}&=&(7,10,1)
\end{eqnarray}
The cross-product of these two vectors (or any non-zero multiple of the cross-product) will be a direction vector of the line of their intersection.
$$ (7,-10,1)\times(7,10,1)=(-20,0,140) $$
So let the normal vector of the line of intersection be
$$ \mathbf{N}=(1,0,-7) $$
But we also need a point common to both planes. But clearly, $(0,0,-11)$ lies on both planes. Thus the equation of the line of intersection is
\begin{eqnarray}
(x,y,z)=(0,0,-11)+(1,0,-7)t
\end{eqnarray}
In symmetric form this is
$$ x=\frac{z+11}{-7},\quad y=0 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$
My Work
I was able to prove identity using half angle formula and $\cos3\theta $ expansion.
Since
$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$
$$\therefore \cos4\theta=\cos3\theta$$
I cannot prove the final part.
Please help me. Thanks in advance.
| see $ \cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are the 4 roots and go by sum of roots formula i.e. **sum of roots = -b/a **
hence,
$ \cos\frac{0\pi}{7}+ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} = \frac{4}{8}$
$ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} =\frac{1}{2} -1 $
=$ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} =\frac{-1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the number of arrangements in the word “EDUCATION” where vowels are never together? I know the answer is
$$ 4! P(5,5) $$
Because we can arrange the consonants amongst themselves in 4! ways and then independently insert the five vowels into the five spaces available.
My question is; why the inclusion-exclusion principle is not working?
Since there are nine distinct letters, there are $9!$ total arrangements. The arrangements in which two vowels come together is
$$\binom{5}{2}2!8!$$
three vowels come together is
$$\binom{5}{3}3!7!$$
four vowels come together is
$$\binom{5}{4}4!6!$$
and five vowels come together is
$$\binom{5}{5}5!5!$$
But
$$
9!- \binom{5}{2}2!8!+ \binom{5}{3}3!7!- \binom{5}{4}4!6!+ \binom{5}{5}5!5!
$$
is not giving the correct answer which is $5!4!$.
Edited: This question has an answer using inclusion exclusion principle where the vowels are three. My question is with five vowels.
| You have not accounted for those arrangements such as EAUDCTOIN in which there are disjoint pairs of adjacent vowels.
What we need to exclude are adjacent pairs of vowels. Notice that we can partition $5$ in the following ways:
\begin{align*}
5 & = 4 + 1\\
& = 3 + 2\\
& = 3 + 1 + 1\\
& = 2 + 2 + 1\\
& = 2 + 1 + 1 + 1\\
& = 1 + 1 + 1 + 1 + 1
\end{align*}
Think of these numbers as the number of consecutive vowels.
The case $1 + 1 + 1 + 1 + 1$ in which no two vowels are adjacent is what we want to count.
The case $2 + 1 + 1 + 1$ has one pair of adjacent vowels.
The case $2 + 2 + 1$ has two disjoint pairs of adjacent vowels.
The case $3 + 1 + 1 + 1$ has two overlapping pairs of adjacent vowels (such as the string AEI, which has the pairs AE and EI).
The case $3 + 2$ has three pairs of adjacent vowels, two of which are overlapping.
The case $4 + 1$ has three pairs of adjacent vowels.
The case $5$ has four pairs of adjacent vowels.
We use the Incluson-Exclusion Principle to count the number of admissible choices for the positions of the vowels and consonants first, then arrange the vowels and consonants in the those positions.
There are $\binom{9}{5}$ ways to choose the positions of the vowels. From these, we must exclude those arrangements in which there are one or more pairs of adjacent vowels.
A pair of adjacent vowels: We have eight positions, one for a block of two vowels, three for single vowels, and four for consonants. Choose one position for the block and three of the remaining seven positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{8}{1}\binom{7}{3}$$
such choices.
Two pairs of adjacent vowels: This can occur in two ways. Either there are two overlapping pairs (a block of three consecutive vowels) or two disjoint pairs (two blocks of two vowels each).
Two overlapping pairs: We have seven positions to fill with a block of three vowels, two single vowels, and four consonants. Choose one position for the block and two of the remaining six positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{7}{1}\binom{6}{2}$$
such choices.
Two disjoint pairs: We have seven positions to fill with two blocks of two vowels, one single vowel, and four consonants. Choose two positions for the blocks and one of the remaining five positions for the single vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{7}{2}\binom{5}{1}$$
such choices.
Three pairs of adjacent vowels: There are again two cases. Either there are three overlapping pairs of adjacent vowels (a block of four consecutive vowels) or two overlapping pairs of vowels (a block of three consecutive vowels) and a disjoint pair of adjacent vowels (a block of two consecutive vowels).
Three overlapping pairs of adjacent vowels: There are six positions to fill with a block of four consecutive vowels, a single vowel, and four consonants. There are six ways to choose the position of the block and five ways to choose the position of the single vowel. The remaining four positions must be reserved for the four consonants. There are
$$\binom{6}{1}\binom{5}{1}$$
such choices.
Two overlapping pairs of adjacent vowels and a disjoint pair of adjacent vowels: There are six positions to fill with a block of three consecutive vowels, a block of two consecutive vowels, and four consonants. There are six ways to choose the position of the block of three vowels and five ways to choose the position of the block of two vowels. The remaining four positions must be reserved for the four consonants. There are
$$\binom{6}{1}\binom{5}{1}$$
such choices.
Four pairs of adjacent vowels: There are five positions to fill with a block of five consecutive vowels and four consonants. Choose the position of the block. The remaining four positions must be reserved for the four consonants. There are
$$\binom{5}{1}$$
such choices.
By the Inclusion-Exclusion Principle, the number of ways of positioning the vowels and consonants so that no two vowels are consecutive is
$$\binom{9}{5} - \binom{8}{1}\binom{7}{3} + \binom{7}{1}\binom{6}{2} + \binom{7}{2}\binom{5}{1} - \binom{6}{1}\binom{5}{1} - \binom{6}{1}\binom{5}{1} + \binom{5}{1} = 1$$
namely, VCVCVCVCV.
There are $5!$ ways to arrange the vowels in their five positions and $4!$ ways to arrange the consonants in their four positions. Hence, the number of admissible arrangements is
$$\left[\binom{9}{5} - \binom{8}{1}\binom{7}{3} + \binom{7}{1}\binom{6}{2} + \binom{7}{2}\binom{5}{1} - \binom{6}{1}\binom{5}{1} - \binom{6}{1}\binom{5}{1} + \binom{5}{1}\right]5!4! = 5!4!$$
which agrees with your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $BP+BQ=2PQ $ Let consider a circle of diameter $CA $ and $B\in CA $ such that $A\in [CB] $ and $AB=\frac {CA}{2} $.
If $M \in [CA] $ such that $AM=\frac {CA}{3} $ and $P, Q $ on circle such that $P, M, Q $ collinear.
Show that $BP+BQ=2PQ $
My idea. I notice that $M $ is the middle of $[CB] $. I take the middle of $[CP] $ and $[CQ] $. Now I am stuck.
| The problem (with $AM=\frac{CA}{3}$) cannot be correct. Choose $P$ and $Q$ so that $PQ\perp CA$ at $M$. If $R$ is the radius of the circle, then we have $$OM=\frac{R}{3}\ \Rightarrow\ PM=\sqrt{OP^2-OM^2}=\sqrt{R^2-\left(\frac{R}{3}\right)^2}=\frac{2\sqrt{2}}{3}R.$$
This gives $PQ=2\ PM=\frac{4\sqrt{2}}{3}R$. On the other hand,
$$BM=AB+AM=\frac{CA}{2}+\frac{CA}{3}=\frac{2R}{2}+\frac{2R}{3}=\frac{5}{3}R,$$
so
$$BP=\sqrt{BM^2+PM^2}=\sqrt{\left(\frac{5}{3}R\right)^2+\left(\frac{2\sqrt{2}}{3}R\right)^2}=\frac{\sqrt{33}}{3}R.$$
By symmetry, $BQ=BP=\frac{\sqrt{33}}{3}R$. Clearly,
$$BP+BQ=\frac{2\sqrt{33}}{3}R\neq \frac{8\sqrt{2}}{3}R=2\ PQ.$$
What may be the correct condition is that $AM=\frac{CA}{4}$. In this case, $A$ and $C$ harmonically divide $B$ and $M$. Thus, if $\Gamma$ is the circle with diameter $CA$ and $X\in \Gamma$, then $AX$ and $CX$ are the internal and external angular bisectors of $\angle BXM$, respectively. This proves that $A$ is the incenter of the triangle $BPQ$. This means $AB$ internally bisects $\angle BPQ$.
Let $Q'$ be the reflection of $Q$ about the line $CA$. Then, $Q'$ is on the line $BP$ and also on the circle $\Gamma$. Therefore,
$$BP\cdot BQ=BP\cdot BQ'=BA\cdot BC= R\cdot 3R=3R^2,$$
if $R$ denotes the radius of $\Gamma$. This shows that
$$\frac{BM^2}{BP\cdot BQ}=\frac{\left(\frac{3}{2}R\right)^2}{3R^2}=\frac{3}{4}.$$
It is well-known (via Stewart's theorem and the angle bisector theorem, for example) that
$$BM^2=BP\cdot BQ\left(1-\frac{PQ^2}{(BP+BQ)^2}\right).$$
That is
$$1-\frac{PQ^2}{(BP+BQ)^2}=\frac{3}{4},$$
so $BP+BQ=2\ PQ$, as desired.
Or as Lozenges remarked, by the angle bisector theorem, $$\frac{BP}{PM}=\frac{BA}{AM}=2$$ and
$$\frac{BQ}{QM}=\frac{BA}{AM}=2.$$
So, $BP=2\ PM$ and $BQ=2\ QM$, making $$BP+BQ=2\ PM+2\ QM=2\ PQ.$$
If instead let $B$ be the point on the line $CA$ such that $A$ lies between $B$ and $C$, and that $$BA=\frac{\alpha\ CA}{2}$$ for some $\alpha>0$, then we can take $M$ to be the point between $A$ and $C$ such that $$AM=\frac{\alpha\ CA}{2(\alpha+1)}.$$ If $P$ and $Q$ are two distinct points on the circle with diameter $CA$ such that $P$, $M$, and $Q$ are collinear, then we have $$BP+BQ=(\alpha+1)\ PQ.$$
The proof is the same as the case $\alpha=1$.
| {
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"url": "https://math.stackexchange.com/questions/2993773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\cos^8x.\sec^6y,\frac12,\sin^8x.\csc^6y$ in AP if $\cos^4x.\sec^2y,\frac12,\sin^4x.\csc^2y$ in A.P If $\cos^4x.\sec^2y,\dfrac{1}{2},\sin^4x.\csc^2y$ are in A.P, then prove that $\cos^8x.\sec^6y,\dfrac{1}{2},\sin^8x.\csc^6y$ in AP.
My Attempt
$$
\cos^4x.\sec^2y+\sin^4.x\csc^2y=\frac{\cos^4x}{\cos^2y}+\frac{\sin^4x}{\sin^2y}=1\\
\implies\sin^2y.\cos^4x+\cos^2y.\sin^4x=\sin^2y.\cos^2y
$$
$$
\cos^8x.\sec^6y+\sin^8x.\csc^6y=\frac{\cos^8x}{\cos^6y}+\frac{\sin^8x}{\sin^6y}
$$
How do I know that the given terms are in A.P, G.P or H.P ?
| Hint:
$$\dfrac12-\cos^4x=2\left(\sin^4x-\dfrac12\right)$$
$$\iff2-(1+\cos2x)^2=2(1-\cos2x)^2-4$$
Solve for $\cos2x$
$$\csc^2y-\dfrac12=2\left(\dfrac12-\sec^2y\right)$$
$$\iff2-4(1+\tan^2y)=2(1+\cot^2y)-1$$
$$4\tan^2y+2\cdot\dfrac1{\tan^2y}-5=0$$
Solve for $\tan^2y\ge0$
| {
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Evaluating Derivative of $\sqrt{1+\sqrt{1+\sqrt{1-2x}}}$ Did I evaluate the following derivative correctly? I know I have not simplified to the utmost extent, but I want to know if this method is correct.
Consider, $\sqrt{1+\sqrt{1+\sqrt{1-2x}}}$
Let A = $1+\sqrt{1+\sqrt{1-2x}}$
Let B = $1+\sqrt{1-2x}$
Let C = $1-2x$
$$\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' = \frac{1}{2}A^{-\frac{1}{2}} \cdot \frac{1}{2}B^{-\frac{1}{2}} \cdot \frac{1}{2}C^{-\frac{1}{2}} \cdot -2 = \frac{-1}{4\sqrt{ABC}}$$
| Yes, this is valid. Maybe some explanatory steps in between would help to someone reading it though. It strikes me as not being obvious according to standard well-known derivative rules without a few intermediate steps.
$$\begin{align}
\left(\sqrt{1+\sqrt{1+\sqrt{1-2x}}}\right)' &= \left(\sqrt{A}\right)'\\
&=\frac{1}{2}A^{-\frac{1}{2}}A'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\left(1+\sqrt{B}\right))'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}B'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\left(1+\sqrt{C}\right)'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\frac{1}{2}C^{-\frac{1}{2}}C'\\
&=\frac{1}{2}A^{-\frac{1}{2}}\frac{1}{2}B^{-\frac{1}{2}}\frac{1}{2}C^{-\frac{1}{2}}(-2)\\
&= \frac{-1}{4\sqrt{ABC}}
\end{align}$$
| {
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Square coloring problem only using 2 colors
"We need to color $4×4$ square using $4$ black color and $12$ white color. Then, how many cases it may be? Flip is prohibited but rotating is ok"
I tried case by case (inner square and rest) anf obtained answer 389. But I don't know if it's correct. Help me please.
| We apply PET (Polya Enumeration Theorem) here and this needs the cycle
index. There are four rotations. The first is the identity which
contributes
$$a_1^{16}.$$
There are the rotations by $90$ degrees and by $270$ degrees, which
contribute
$$2 a_4^4.$$
The rotation by $180$ degrees contributes
$$a_2^8.$$
This yields the cycle index
$$Z(G) = \frac{1}{4} (a_1^{16} + 2 a_4^4 + a_2^8).$$
We thus have for the desired quantity
$$[B^4 W^{12}] Z(G; B + W)
\\ = [B^4 W^{12}]
\frac{1}{4} ((B+W)^{16} + 2 (B^4+W^4)^4 + (B^2+W^2)^8)
\\ = \frac{1}{4} {16\choose 4} +
\frac{1}{2} [B W^3] (B+W)^4 + \frac{1}{4} [B^2 W^6] (B+W)^8
\\ = \frac{1}{4} {16\choose 4} + \frac{1}{2} {4\choose 1}
+ \frac{1}{4} {8\choose 2}.$$
This yields
$$\bbox[5px,border:2px solid #00A000]{464}$$
confirming the data from the comment.
| {
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Prove that $\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}=4\sqrt{3}$ I'm trying to calculate the expression: $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ and show that it is equal $4\sqrt{3}$.
I was trying to group the summands and calculate sums of $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}} \hspace{0.5cm}\text{and} \hspace{0.5cm} -\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}$$ where we get $$\frac{2\cos\frac{2\pi}{15}+1}{\sin\frac{2\pi}{15}}-\frac{2\cos\frac{4\pi}{15}-1}{\sin\frac{8\pi}{15}}$$ but unfortunately this sum is not simplified.
How to prove this equality?
| $$\frac{1}{\sin\frac{\pi}{15}}+\frac{1}{\sin\frac{2\pi}{15}}-\frac{1}{\sin\frac{4\pi}{15}}+\frac{1}{\sin\frac{8\pi}{15}}\\ =\frac{1}{\sin12°}+\frac{1}{\sin24°}-\frac{1}{\sin48°}+\frac{1}{\sin96°} $$
We split them into two groups as shown:
$$\frac{1}{\sin12°}-\frac{1}{\sin48°} \hspace{0.5cm}\text{and}\hspace{0.5cm} \frac{1}{\sin24°}+\frac{1}{\sin96°}$$
And we see that (from sum-to-product and product-to-sum)
$\Large{\frac{1}{\sin12°}-\frac{1}{\sin48°}\\ = \frac{\sin48°-\sin12°}{\sin12°\sin48°}\\ = \frac{2\cos30°\sin18°}{{\frac12}(\cos36°-\cos60°)}\\=\frac{2(\frac{\sqrt3}{2})\sin18°}{{\frac12}(\cos36°-{\frac12})}\\=\frac{4\sqrt3\sin18°}{2\cos36°-1}}$
and
$\Large{\frac{1}{\sin24°}+\frac{1}{\sin96°}\\=\frac{\sin24°+\sin96°}{\sin24°\sin96°}\\=\frac{2\sin60°\cos36°}{{\frac12}(\cos72°-\cos120°)}\\=\frac{2({\frac{\sqrt3} 2})\cos36°}{{\frac12}(\sin18°+{\frac12})}\\=\frac{4\sqrt3\cos36°}{2\sin18°+1}}$
Hence it remains to find
$$\frac{4\sqrt3\sin18°}{2\cos36°-1} + \frac{4\sqrt3\cos36°}{2\sin18°+1}$$
From here we can determine $2\cos36°-1=2\sin18°\ $ and $\ 2\sin18°+1=2\cos36°$. By plugging in these into the denominators and simplifying we get $4\sqrt3$, which is what we want. $\ _\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of $ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+ \cdots$
Finding series sum of $$ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+\frac{1}{1\cdot 3\cdot 5\cdot 7\cdot 9}+ \cdots$$
Try: Let $\displaystyle a_{k}=\frac{1}{1\cdot 3\cdot 5\cdot 7\cdots (2k+1)}=\frac{2\cdot 4\cdot 6\cdots 2k}{(2k+1)!}$
So we have $\displaystyle a_{k}=\frac{2^k\cdot k!}{(2k+1)!}$
So our desired sum is $$\sum^{\infty}_{k=1}\frac{2^k\cdot k!}{(2k+1)!}$$
Now i am struck at that point
I did not understand how can i solve further,
Could some help me plaese , thanks
| That term in the denominator is also called the double factorial, namely: $$(2n+1)!!=\frac{(2n+1)!}{2^n n!}$$
And our series can be rewritten as: $$S=\sum_{n=0}^\infty \frac{1}{(2n+1)!!}$$
Now from this link see (21), is given that: $$\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!!}=\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt 2} \right) e^{\frac{x^2}{2}}$$
$$\Rightarrow S=\sqrt{\frac{e\pi}{2}}\text{erf}\left(\frac{1}{\sqrt 2}\right)$$ Where $\text{erf(z)}$ is the error function, defined as: $\displaystyle{\text{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}dt}$
| {
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Compute $\lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{x} {1+k(k+1)x^2}$, $x>0$ I tried using the squeeze theorem, but I had no luck. Then I thought that maybe the sum is telescoping, but also I couldn't see any pattern.
| You can use the following formula
\begin{equation}
\arctan{u}+\arctan{v}=\arctan{\frac{u+v}{1-u v}}
\end{equation}
which is valid for $u v <1$. Now consider the sum. For $k=1$ you have
\begin{equation}
\arctan{\frac{x}{1+2 x^2}}
\end{equation}
and for $k=2$
\begin{equation}
\arctan{\frac{x}{1+6 x^2}}
\end{equation}
If you apply the formula above to the first two term you get:
\begin{equation}
\arctan{\frac{x}{1+2 x^2}}+\arctan{\frac{x}{1+6 x^2}}=\arctan{\frac{2x}{1+3 x²}}
\end{equation}
You can repeat the same computation for the term obtained using $k=3$ and $k=4$ and so on and find the expression:
\begin{equation}
\arctan{\frac{2x}{1+x²(2 m+1)(2m-1)}}
\end{equation}
where the index $m$ runs from 1 to N/2. This procedure can be repeated again and again. For instance the $\arctan{\frac{2x}{1+3 x²}}$ can be added to the sum of the terms obtained by using $k=3$ and $k=4$ and the sum can be written as
\begin{equation}
\arctan{\frac{4x}{1+x²(4w-31)(4w+1)}}
\end{equation}
where the index $w$ runs from 1 to N/4. By applying the formula of the sum of the two arctan it is possible to compute:
\begin{equation}
\sum_{k=1}^{n} \arctan \frac{x} {1+k(k+1)x^2}=\frac{2 x n}{1+x²(2 n +1)}
\end{equation}
So that
\begin{equation}
\lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{x} {1+k(k+1)x^2}=arccot(x)
\end{equation}
please note that $arccot(x)=\frac{\pi}{2}-\arctan{x}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve $\lim_{x\to\infty}{\sqrt{x^4+x^2}+\sqrt{x^2+5x}-x^2-x}$ I tried solve this limit.
$\lim_{x\to\infty}{\sqrt{x^4+x^2}+\sqrt{x^2+5x}-x^2-x}$
I tried multiply by $\sqrt{x^4+x^2}-\sqrt{x^2+5x}$, and apply L'Hospital. but this led to alot of work.. and this question seems to had a very easy and fast way to do. I know the answer is 3.
thanks any help.
| Continue with ajotatxe's hint:
$$\lim_{x \to \infty} \sqrt{x^4 + x^2} - x^2$$
$$\lim_{x \to \infty} x^2 (\sqrt{1 + \frac{1}{x^2}} - 1)$$
$$\lim_{x \to \infty} \frac{x^2 (\sqrt{1 + \frac{1}{x^2}} - 1)(\sqrt{1 + \frac{1}{x^2}} + 1)}{\sqrt{1 + \frac{1}{x^2}} + 1} $$
$$\lim_{x \to \infty} \frac{x^2 \frac{1}{x^2}}{\sqrt{1 + \frac{1}{x^2}} + 1} $$
$$\lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x^2}} + 1} $$
| {
"language": "en",
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Link between polynomial and derivative of polynomial I can't seem to solve this problem, can anyone help me please? The problem is:
Let real numbers $a$,$b$ and $c$, with $a ≤ b ≤ c$ be the 3 roots of the polynomial $p(x)=x^3 + qx^2 + rx + s$. Show that if we divide the interval $[b, c]$ into six equal parts, then one of the root of $p'(x)$ (the derivative of the polynomial $p(x)$) will be in the 4th part.
What I did was:
Because we know the roots of $p(x)$ are $a$,$b$ and $c$, we can write the polynomial $p(x)$ like this: $p(x) = (x-a)(x-b)(x-c)$
So we have $p(x) = x^3 + qx^2 + rx + s = (x-a)(x-b)(x-c)$
We find the value of $q$,$r$ and $s$:
$q = -(a+b+c)$
$r = (ab + ac + bc)$
$s = -abc$
We have that $p'(x) = 3x^2 + 2qx + r$
We want to find the roots of $p'(x)$, so if we apply the quadratic formula, we get:
$(-2q ± 2*\sqrt{q^2 - 3r})/6$
Because we know the value of q,r and s, we can rewrite this expression like this:
$(2(a+b+c) ± 2\sqrt{a^2 + b^2 + c^2 - ab - bc -ca})/6$
But I am stuck here, any help would be great. Thank you in advance.
| Since $p(x) = (x-a)(x-b)(x-c)$,
$$\begin{align}p'(x) &= (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)\\
&= (x-b)(x-c) + (x-a)(2x - (b+c))\end{align}$$
At $x = \frac{b+c}{2}$, we have
$$p'(x) = \frac{c-b}{2}\cdot\frac{b-c}{2} + (x-a)\cdot 0 = -\frac{(c-b)^2}{4} \le 0$$
At $x = \frac{b+2c}{3}$, we have
$$\begin{align}p'(x) &= \frac{2(c-b)}{3}\cdot\frac{b-c}{3}
+ \left(\frac{b+2c}{3} - a\right)\cdot\frac{c-b}{3}\\
&= -\frac{2(c-b)^2}{9} + \left(\frac{2(c-b)}{3} + (b-a)\right)\cdot\frac{c-b}{3}\\
&= \frac{(b-a)(c-b)}{3}\\&\ge 0\end{align}$$
This implies $p'(x)$ has a root in $\left[\frac{b+c}{2},\frac{b+2c}{3}\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt:
we have $x+2y=100-5z$
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$
we need to collect coefficient of $100-5z$ in the above given by
$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$
Total number of solutions is
$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$
$$S(z)=540.5$$
what went wrong in my analysis?
| Given: $x+2y=100-5z$, tabulate:
$$\begin{array}{c|c|c}
z&x&\text{count}\\
\hline
0&100,98,\cdots, 0&\color{red}{51}\\
1&\ \ 95,93,\cdots, 1&\color{blue}{48}\\
2&\ \ 90,88,\cdots, 0&\color{red}{46}\\
3&\ \ 85,83,\cdots, 1&\color{blue}{43}\\
4&\ \ 80,78,\cdots, 0&\color{red}{41}\\
\vdots&\vdots&\vdots\\
17&15,13,\cdots,1&\color{blue}{8}\\
18&10,8,\cdots,0&\color{red}{6}\\
19&5,3,1&\color{blue}{3}\\
20&0&\color{red}{1}\\
\hline
&&\color{red}{286}+\color{blue}{255}=541
\end{array}$$
| {
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Prove that $5$ is irreducible in $\mathbb Z[\sqrt{2}]$ Consider the UFD $\mathbb Z[\sqrt{2}]$. Prime and irreducible elements there are the same. How do I show that $5$ is irreducible?
I tried to write $5=(a+b\sqrt 2)(c+d\sqrt 2)$ or $(2bd+ac-5)+(bc+ad)\sqrt 2=0$. We have $bc+ad=0$. If $c=0$, then either $d=0$ (in which case $5=0$) or $a=0$ (in which case $2bd=5$). Either case gives a contradiction. Suppose $c\ne 0$. Then $b=-(da)/c$. So $-2d^2a/c+ac-5=0$. But I don't see how to proceed. Maybe it's the wrong path?
| Your equation
$$5=(a+b\sqrt 2)(c+d\sqrt 2) = (ac + 2bd) + (ad+bc)\sqrt2$$
gives the system
$$\begin{cases}
ac+2bd = 5\\
ad+bc = 0
\end{cases}$$
Multiplying the first equation by $d$, the second by $-c$ and adding them yields $$2bd^2 - bc^2 = 5d \implies b(2d^2-c^2) = 5d$$
Hence $5 \mid b$ or $5 \mid (2d^2 - c^2)$. However, the second possibility is impossible because $x^2 \equiv \pm 1 \pmod 5$.
Therefore $5 \mid b$.
Similarly, multiplying the first equation by $c$, the second by $-2d$ and adding them gives
$$ac^2-2ad^2 = 5c \implies a(c^2-2d^2) = 5c$$
As above we conclude $5 \mid a$.
Therefore $\exists \hat{a}, \hat{b} \in \mathbb{Z}$ such that $a = 5\hat{a}$ and $b = 5\hat{b}$.
We have
$$5 = (a+b\sqrt 2)(c+d\sqrt 2) = (5\hat{a}+5\hat{b}\sqrt 2)(c+d\sqrt 2) = 5(\hat{a}+\hat{b}\sqrt 2)(c+d\sqrt 2)$$
Dividing be $5$ gives
$$1 = (\hat{a}+\hat{b}\sqrt 2)(c+d\sqrt 2)$$
so $c + d\sqrt{2}$ is invertible in $\mathbb{Z}[\sqrt{2}]$ with $(c + d\sqrt{2})^{-1} = \hat{a}+\hat{b}\sqrt 2$.
| {
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Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$ I need some help with this exercise.
Given that $$\tan\alpha=2$$
calculate the value of:
$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$$
I've tried everything but I always end up having something irreducible in the end. Maybe this is an easy question, but I'm new at trigonometry. If anyone can help me by providing a step-by-step solution to this, I would be really thankful! :)
| Note that
$$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$
Thus if
$$\tan \alpha = 2$$
then
$$\sin \alpha = 2 \cos \alpha$$
Now just plug for sine
$$\frac{\sin^3 \alpha - 2 \cos^3 \alpha + 3 \cos \alpha}{3 \sin \alpha + 2 \cos \alpha} = \frac{8 \cos^3 \alpha - 2 \cos^3 \alpha + 3 \cos \alpha}{6 \cos \alpha + 2 \cos \alpha}$$
which then simplifies to
$$\frac{6 \cos^3 \alpha + 3 \cos \alpha}{8 \cos \alpha} = \frac{1}{8} [6 \cos^2 \alpha + 3]$$
Now note that
$$\frac{1}{\cos \alpha} = \sec \alpha$$
and we have the trigonometric identity
$$1 + \tan^2 \alpha = \sec^2 \alpha$$
thus $$\sec^2 \alpha = 1 + (2)^2 = 1 + 4 = 5$$ and thus $\cos^2 \alpha = \frac{1}{5}$. Thus we can plug that into the prior expression to get
$$\frac{\sin^3 \alpha - 2 \cos^3 \alpha + 3 \cos \alpha}{3 \sin \alpha + 2 \cos \alpha} = \frac{1}{8} [6 \cos^2 \alpha + 3] = \frac{1}{8} [6 \frac{1}{5} + 3] = \frac{21}{40}$$.
| {
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"answer_count": 6,
"answer_id": 1
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Solving $x^2 \equiv 140 \pmod{221}$ I'm stuck with the last part of this question: solve $x^2 \equiv 140 \pmod{221}$.
We know that $140 = 7 \times2^2\times5$ and $221 = 13 \times 17$.
We split the original congruence in two, so we have:
$x^2 \equiv 140 \pmod{13}$
$x^2 \equiv 140 \pmod{17}$
Applying the properties of moduli we have:
$x^2 \equiv 10\pmod{13} \rightarrow x=\pm6$
$x^2 \equiv 4\pmod{17} \rightarrow x=\pm2$
After this point, it's not clear for me how I can arrive to the complete solution. Any advice?
| Note that
$$x^2 -140\equiv x^2 -140-221=x^2−19^2=(x-19)(x+19)\pmod{221}.$$
It follows that
$$x\equiv \pm 19\equiv \pm 6 \pmod{13}\quad\text{and}\quad x\equiv \pm 19\equiv \pm 2 \pmod{17}.$$ Hence we have four solutions (actually we already have two of them, i.e. $19$ and $-19$) that can be obtained by using the Chinese Remainder Theorem:
$$\begin{cases}x\equiv 6 \pmod{13}\\
x\equiv 2 \pmod{17}
\end{cases}\quad
\begin{cases}x\equiv 6 \pmod{13}\\
x\equiv -2 \pmod{17}
\end{cases}\\
\begin{cases}x\equiv -6 \pmod{13}\\
x\equiv 2 \pmod{17}
\end{cases}\quad
\begin{cases}x\equiv -6 \pmod{13}\\
x\equiv -2 \pmod{17}
\end{cases}$$
Can you take it from here?
P.S. By the first remark, $\pm 19$ are two solutions and all you need is to solve just ONE system:
$$\begin{cases}x\equiv 6 \pmod{13}\\
x\equiv -2 \pmod{17}
\end{cases}$$
if $m$ is the solution then the fourth one is $-m$.
| {
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"source": "stackexchange",
"question_score": "2",
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Finding $\lim\limits_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)$ What is$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$So it is$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $\dfrac{1}{4\sqrt{2}}$.
| Formally substitute $n=1/t$; if the function you get has a limit for $t\to0^+$, then it is the same as the limit you are looking for. So consider
$$
\lim_{t\to0^+}\frac{1}{t^3}\left(
\sqrt{\frac{1}{t^2}+\sqrt{\frac{1}{t^4}+1}}-\frac{\sqrt{2}}{t}
\right)=
\lim_{t\to0^+}\frac{\sqrt{1+\sqrt{1+t^4}}-\sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
\lim_{u\to0^+}\frac{\sqrt{1+\sqrt{1+u}}-\sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=\sqrt{1+\sqrt{1+u}}$.
Since $$f'(u)=\frac{1}{2\sqrt{1+\sqrt{1+u}}}\frac{1}{2\sqrt{1+u}}$$ we have $$f'(0)=\frac{1}{4\sqrt{2}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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What's wrong with this Penrose pattern? I programmed the Penrose tiling by projecting a portion of 5D lattice to 2D space, by the "cut and project" method described in
*
*Quasicrystals: projections of 5-D lattice into 2 and 3 dimensions, H. Au-Yang and J. Perk.
*Generalised 2D Penrose tilings, A. Pavlovitch and M. Kléman
The orthonormal basis is chosen as
$$
M=\sqrt{\frac{2}{5}} \begin{bmatrix}
\cos 0 & \cos \frac{2\pi}{5} & \cos \frac{4\pi}{5}& \cos \frac{6\pi}{5}& \cos \frac{8\pi}{5} \\
\sin 0 & \sin \frac{2\pi}{5} & \sin \frac{4\pi}{5}& \sin \frac{6\pi}{5}& \sin \frac{8\pi}{5} \\
\cos 0 & \cos \frac{4\pi}{5} & \cos \frac{8\pi}{5}& \cos \frac{12\pi}{5}& \cos \frac{16\pi}{5} \\
\sin 0 & \sin \frac{4\pi}{5} & \sin \frac{8\pi}{5}& \sin \frac{12\pi}{5}& \sin \frac{16\pi}{5} \\
\frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}}\\
\end{bmatrix}
$$
Each row presents a basis vector, i.e.
$$
M_i\cdot M_j=0, \;\;\textrm{for } i<j.$$
and
$$||M_i||=1, \;\;\textrm{for } 1\leq i \leq 5.
$$
$M$ consists of the parallel operator (representing the physical space)
$$
A=\begin{bmatrix}
M_1\\
M_2 \\
\end{bmatrix}=
\begin{bmatrix}
\cos 0 & \cos \frac{2\pi}{5} & \cos \frac{4\pi}{5}& \cos \frac{6\pi}{5}& \cos \frac{8\pi}{5} \\
\sin 0 & \sin \frac{2\pi}{5} & \sin \frac{4\pi}{5}& \sin \frac{6\pi}{5}& \sin \frac{8\pi}{5} \\
\end{bmatrix}
$$
and the perpendicular operator
$$
B=\begin{bmatrix}
M_3\\
M_4 \\
M_5 \\
\end{bmatrix}=\begin{bmatrix}
\cos 0 & \cos \frac{4\pi}{5} & \cos \frac{8\pi}{5}& \cos \frac{12\pi}{5}& \cos \frac{16\pi}{5} \\
\sin 0 & \sin \frac{4\pi}{5} & \sin \frac{8\pi}{5}& \sin \frac{12\pi}{5}& \sin \frac{16\pi}{5} \\
\frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}} & \frac{1}{\sqrt {2}}\\
\end{bmatrix}
$$
The 5D lattice points are integer combinations of basis such as
$$
p=i \begin{bmatrix}
1\\
0\\
0\\
0\\
0\\
\end{bmatrix}
+ j\begin{bmatrix}
0\\
1\\
0\\
0\\
0\\
\end{bmatrix}
+\dots, \;\; i,j,\dots \in \mathbb{Z}
$$
A 5D cube (centered at origin) is projected into 3D as polytope
$$
v'= B v, \;\; v\in hypercube
$$
so that I can check whether a $p$ is inside this polytope (20 faces). This is called "cutting" the 5D lattice points.
The resultant 2d projection $Ap$ is
Everything works fine, however, my result differs from the "standard" one (e.g. in wiki page) as follows
Is this a mistake or an alternative view of the same tiling?
Finally, I find this image (from Vertex Frequencies in Generalized Penrose Patterns, by E. Zobetz and A. Preisinger)
where the center of standard tiling exhibits the "S" pattern, while the center of my version has the "ST" pattern. But what does it mean exactly?
| I believe your constant $\gamma$ to be essentially the same as the constant $y$ in de Bruijn 1981 cited below. I don't use your exact projection matrices though like you I do use matrices involving sines and cosines of multiples of $2\pi/5$. I ran into similar issues, and for me, setting $y_i=\epsilon$ for $\epsilon>0$ as small as I could represent it got rid of the forbidden intersections like the ST intersection. Somewhere, distantly, in my tiling is an ST intersection but I regard that as approximation error and simply work to keep it distant.
So I recommend $\gamma_i=\epsilon$. It may not work because your matrices are in a different basis from mine, but I bet some straightforward combination of $\pm\epsilon$ will do the trick for you.
Additionally if it would be helpful I can dig through my notes to get my derivation.
de Bruijn, N. G. "Algebraic theory of Penrose's non-periodic tilings of the plane. I, II: dedicated to G. Pólya." Indagationes Mathematicae 43.1 (1981): 39-66.
| {
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Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots(x^{12}-12)$.
Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots (x^{12}-12)$.
I tried to solve this problem using theory of equation the coefficient of $x^{70}$ will be the sum of products taking two at a time. But this very very exhaustive I want to know some another method as it will be proficient in higher powers.
| Since $1+2+3+\ldots+12=78$, the term $x^{70}$ must arise from taking $x^k$ from $(x^k-k)$ for almost every $k\in\{1,2,\ldots,12\}$, except for some $j_1,j_2,\ldots,j_r\in\{1,2,\ldots,k\}$ such that $j_1<j_2<\ldots<j_r$ and $j_1+j_2+\ldots+j_r=8$. There are very few such tuples $(j_1,j_2,\ldots,j_r)$:
*
*for $r=1$, $j_1=8$;
*for $r=2$, $(j_1,j_2)=(1,7),(2,6),(3,5)$;
*for $r=3$, $(j_1,j_2,j_3)=(1,2,5),(1,3,4)$.
Therefore, the coefficient of $x^{70}$ is
$$(-1)^1\cdot 8+(-1)^2\cdot (1\cdot 7+2\cdot 6+3\cdot 5)+(-1)^3\cdot(1\cdot 2\cdot 5+1\cdot 3\cdot 4)=4\,.$$
| {
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"question_score": "5",
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} |
Finding the area of an orthic triangle (DEF) when given vertices of triangle ABC. Given the triangle ABC whose vertices are endpoints of the altitudes from $A$,$B$ and $C$ is called the orthic triangle. The triangle $ABC$ has vertices at $A=(2,4)$; $B=(8,5)$ and $C=(3,9)$. The altitude from $B$ to $AC$, meets AC at point $D=(2.42,6.12)$. Find the area if the orthic triangle.
To attempt this problem I decided to use the formula
$$area = \frac{abc|cosAcosBcosC|}{2R}$$ where $R$ is the circumradius of the triangle $ABC$. I calculated the length of each sides by using the length equation and got $a=\sqrt41$, $b=\sqrt26$, and $c=\sqrt37$.
Next I used the cosine rule to get the angle at vertix $A$. Then used the sine rule to get angle at $B$ and subtracted these 2 angles from 180 to get the third angle. $A=69.23$, $B=48.12$ and $C=62.65$.
To get the circumradius of the triangle $ABC$ I used $$R=\frac{abc}{4(Area)}$$ I found the area using heron's formula to be 14.5, putting all these values into the original equation I got the area of the orthic triangle to be 3.154. Is this correct? Is there an easier method? Thanks
| I suggest to use
$$ \Delta_{orthic} = 2\Delta\left|\cos A\cos B\cos C\right|$$
and to compute $\Delta$ from the shoelace formula and the remaining part from the cosine theorem, such that we do not need to extract any square root. By the shoelace formula
$$ 2\Delta = |2\cdot 5+8\cdot 9+3\cdot 4-4\cdot 8-5\cdot 3-9\cdot 2|=29 $$
and by the cosine theorem
$$\left|\cos A\cos B\cos C\right|=\frac{(a^2+b^2-c^2)(a^2-b^2+c^2)(-a^2+b^2+c^2)}{8a^2b^2 c^2} $$
where $a^2=41,b^2=26,c^2=37$ are given by the Pythagorean theorem. It follows that
$$\Delta_{orthic} = 29\cdot\frac{30\cdot 52\cdot 22 }{8\cdot 41\cdot 26\cdot 37}=\frac{4785}{1517}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding numbers by given XOR values. Given XOR values of 3 indices how can we find the numbers?
Like say if I have indices from 1 to 7, how can I find the numbers by given XOR values?
I have:
*
*$X_{1} \oplus X_{3} \oplus X_{5}=V_1$
*$X_{1} \oplus X_{3} \oplus X_{6}=V_2$
*$X_{1} \oplus X_{4} \oplus X_{6}=V_3$
*$X_{2} \oplus X_{4} \oplus X_{6}=V_4$
*$X_{2} \oplus X_{4} \oplus X_{7}=V_5$
*$X_{2} \oplus X_{5} \oplus X_{7}=V_6$
*$X_{3} \oplus X_{5} \oplus X_{7}=V_7$
How can I find any $X_{i}$ from the above data? Is there any pattern?
| Build the matrix $A$ in which the entry $A_{i,j}$ is $1$ if in the $i$th equation there is $X_j$ appearing on the left hand side and $0$ otherwise.
The XOR system is equivalent to the following system of linear equations over the field $\mathbb F_2$:
$$
\left(
\begin{array}{ccccccc}
1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 1 \\
\end{array}
\right) \cdot
\begin{pmatrix} X_1\\X_2\\X_3\\X_4\\X_5\\X_6\\X_7 \end{pmatrix}
=
\begin{pmatrix} V_1\\V_2\\V_3\\V_4\\V_5\\V_6\\V_7 \end{pmatrix}
$$
The matrix on the left is the matrix $A$ that I made you build at the beginning.
The system is certainly solvable if the matrix $A$ is invertible in $\mathbb F_2$.
In our case, $\det(A)=-3$, so it is possible to solve the system.
This is the inverse matrix:
$$
\left(
\begin{array}{ccccccc}
1 & 1 & 1 & 0 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 & 1 & 0 & 1 \\
1 & 0 & 1 & 1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1 & 0 & 1 & 1 \\
0 & 1 & 1 & 0 & 1 & 1 & 1 \\
\end{array}
\right) .
$$
To compute $X_i$, read the $i$th row and look for the columns where $1$ is present. Those are the indices of your original equations that you have to XOR together. For instance, for $X_1$ you have to XOR together
$$
X_1 = V_1 \oplus V_2 \oplus V_3 \oplus V_5 \oplus V_6 .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Taylor series of $\ln\frac{1+x}{1-x}$
Let $f(x)=\ln\frac{1+x}{1-x}$ for $x$ in $(-1,1)$. Calculate the Taylor series of $f$ at $x_0=0$
I determined some derivatives:
$f'(x)=\frac{2}{1-x^2}$; $f''(x)=\frac{4x}{(1-x^2)^2}$; $f^{(3)}(x)=\frac{4(3x^2+1)}{(1-x^2)^3}$; $f^{(4)}(x)=\frac{48x(x^2+1)}{(1-x^2)^4}$; $f^{(5)}(x)=\frac{48(5x^2+10x^2+1)}{(1-x^2)^5}$
and their values at $x_0=0$:
$f(0)=0$; $f'(0)=2$; $f''(0)=0$; $f^{(3)}(0)=4=2^2$
$f^{(4)}(0)=0$; $f^{(5)}(0)=48=2^4.3$; $f^{(7)}(0)=1440=2^5.3^2.5$
I can just see that for $n$ even, $f^{(n)}(0)=0$, but how can I generalize the entire series?
| The derivative is
$$\biggl(\ln\frac{1+x}{1-x}\biggr)'=\frac{1-x}{1+x}\,\frac 2{(1-x)^2}=\frac 2{1-x^2}$$
Now
$$\frac 2{1-x^2}=2\sum_{n=0}^\infty x^{2n},\enspace\text{so}\quad\ln\frac{1+x}{1-x}=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}, $$
taking into account that both sides are $0$ for $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3035412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Absolute extrema of $f(x,y,z)=xy-yz+xz$ on the paraboloid $x^2+y^2 \leq z\leq1$ I'm asked to find the global (=absolute) extremas of the function $$f(x,y,z)=xy-yz+xz$$ on the domain $$x^2+y^2 \leq z\leq1$$ which is obviously a paraboloid.
First, I find the extremas in the inside of the domain $$\nabla f=(y,x-z,x-y)=(0,0,0)$$ and I get the trivial solution $$x=y=z=0$$. Then I thought about using Lagrange multiplier for the border with constraint $$g(x,y,z)=x^2+y^2-z$$ where where $z \leq 1$. So :
$$\nabla f=(y,x-z,x-y)=\lambda \nabla g=\lambda(2x,2y,-1)$$ and $$x^2+y^2=z$$ After a bit of solving, I get $y=\frac{x}{2x-1}$ and $z=x^2(1+\frac{1}{(2x-1)^2})$.
At that point, the problem is that expecially the $x$ equivalence of $z$ seems a bit tricky, so I'm not sure if a made any mistake, took the wrong path or if there is an easier way to proceed in this situation. So my question is if my approach is correct, and if so, how should I proceed next, and also if there is an easier way to proceed on that conic domain than with Lagrange multiplier ?
Thanks for your help !
| We first find any critical point in the interior of the domain, which is $x^2 + y^2 < 1.$ So we have $f'(x, y, z) = (y + z, x - z, x - y)$ and $f'(x, y, z) = 0$ can only happen when $x = y = z$ and $y = -z,$ so $x = y = z = 0,$ but this is not in the interior of the domain. Hence, there are no critical points.
Consider now one part of the frontrier, the one defined by $x^2 + y^2 = z = 1.$ Hence we have to maximise $u(x, y) = f(x, y, 1) = x + xy - y$ with $x^2 + y^2 = 1.$ Set $g(x, y) = x^2 + y^2 - 1.$ Since $g'(x, y) = (2x, 2y),$ the derivative of $g$ is never zero and the Lagrange multipliers method apply. We have to solve $u'(x, y) = \dfrac{\lambda}{2} g'(x, y),$ which signifies $(1 + y, x - 1) = \lambda (x, y).$ Observe that $x = 1$ implies $y = 0$ and so we can consider the point $(1, 0)$ as one of the critical points in the restriction. Putting it aside, we have (by division), $\dfrac{1 + y}{x - 1} = \dfrac{x}{y}$ or else $y + y^2 = x^2 - x = 1 -y^2 - x$ or $x = 1 - 2y^2 - y.$ From the relation $x^2 + y^2 = 1$ we get $1 + 4y^4 + y^2 - 4y^2-2y+y^2 = 1,$ that is, $4y^4-2y^2-2y=0$ which leads to $y = 0$ (and hence $x = 1$) or else $2y^3-y-1=0.$ By inspection, we can factor $(y - 1)(2y^2+2y+1) = 0$ and the only solution is $y = 1$ with $x = -3,$ which is not in the restriction. Hence, we got only one point so far $(1, 0, 1).$
We finally tackle the other part of the restriction, which is $x^2 + y^2 - z = 0.$ Here we solve $f'(x, y, z) = \lambda (2x, 2y, -1)$ and this leads to (upon summing) $y + x = 2\lambda(x + y)$ and $y - x = \lambda.$ If $x = -y,$ then $z = 2x^2 \leq 1$ and we are optimising $h(x) = f(x, -x, 2x^2) = 4x^3 - x^2$ for $|x| \leq \dfrac{1}{2}$ and the optimisers for $h$ are $x = -\dfrac{1}{\sqrt{2}}$ and $x = \sqrt{2} - \dfrac{1}{2},$ this allow deducing the critical points for $f$ to be $\left(-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}, 1\right)$ and $\left(\dfrac{1}{\sqrt{2}}, -\dfrac{1}{\sqrt{2}}, 1\right).$ If $x \neq -y,$ then $\lambda = \dfrac{1}{2}$ and $y = x + \dfrac{1}{2}.$ Again, substituting $x^2 + y^2 = z \leq 1$ leads to the point (details ommited) $\left(-\dfrac{1+\sqrt{7}}{4}, \dfrac{1 - \sqrt{7}}{4}, 1\right).$
To terminate the problem, recall the region is compact and evaluate in each of the four points.
Take my calculations with a grain of salt, I did them and checked them twice, but with calculus of several variables, I always commit arithmetic errors.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Every odd divisor is $\equiv 1\bmod 3$ If $n$ is a positive integer with $n\equiv 2\pmod 3$ then I want to show that each odd divisor of $n^2+n+1$ is congruent to $1\pmod 3$.
$$$$
I have done the following:
Let $d$ be an odd divisor of $n^2+n+1$.
Then $$n^2+n+1\equiv 0\bmod d \Rightarrow (n-1)(n^2+n+1)\equiv 0\bmod d \Rightarrow n^3-1\equiv 0 \bmod d \Rightarrow n^3\equiv 1\bmod d$$
Since $n\equiv 2\pmod 3$ we have that $n=2+3k$. Then $$n^3=(2+3k)^3=27k^3+54k^2+36k+8$$
We have that $n^3\equiv 1\bmod d$ therefore we get $$27k^3+54k^2+36k+8\equiv 1\bmod d \Rightarrow 27k^3+54k^2+36k+7\equiv 0\bmod d$$
Is everything correct so far? How could we continue?
| Hint: Start with odd prime divisors. As you have correctly shown, for any $p\mid n^2+n+1$ we have that
$$n^3\equiv 1\pmod p.$$
Hence $\text{ord}_{p}(n)\mid 3$. Since $n\equiv 2\not\equiv 1\pmod p$ we know that $\text{ord}_{p}(n)=3$. However, we also know that for every $m$ and $x$ coprime to $m$ that $\text{ord}_m(x)\mid \phi(m)$. How can we apply this in this scenario? Once we know the statement holds for every odd prime divisor, what can we say about every odd divisor?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3041946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find all the solutions of $z^2-(1+3i)z-8-i=0$ I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.
I want to find all the roots of:
$$z^2-(1+3i)z-8-i=0$$
There are two ways I tried to approach this.
*
*Quadratic formula:
$$\begin{align} z_1,z_2 &=-\frac{p}{2}\pm\sqrt{\left( \frac{p}{2}\right)^2-q} \\ & \implies z_1,z_2=\frac{1+3i}{2}\pm\sqrt{\left( \frac{(-1-3i)}{2}\right)^2-(-8-i)}\\ &= \frac{1+3i}{2} \pm\sqrt{\frac{24+10i}{4} }=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{align}$$
*Completing the square:
$$\begin{aligned} z^2-(1+3i)z-8-i &=0 \\ & \iff \left(z-\left( \frac{1+3i}{2}\right) \right)^2-8-i=\left( \frac{1+3i}{2}\right)^2 \\ & \iff u^2=\frac{24+10i}{4} \\ & \iff u=\pm\frac{\sqrt{24+10i}}{2} \\ & \iff z_{1,2}=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{aligned} $$
Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:
$$z_1=-2+i \\ z_2=3+2i$$
I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?
| That's just because the square roots of $24+10i$ are $\pm(5+i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
range of $3x^2-2xy$ subjected to $x^2+y^2=1$ If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done try here is use arithmetic geometric inequality
$\displaystyle x^2+y^2\geq 2xy$
$\displaystyle -2xy\geq -(x^2+y^2)$
$\displaystyle 3x^2-2xy\geq 2x^2-y^2$
this will not help more
how do i solve it help me please
| For a calculs-free approach, note that
$$
\frac{3+\sqrt{13}}{2}\left[\frac{2y-(3+\sqrt{13})x}{\sqrt{26+6\sqrt{13}}}\right]^2+\frac{3-\sqrt{13}}{2}\left[\frac{2y-(3-\sqrt{13})x}{\sqrt{26-6\sqrt{13}}}\right]^2=3x^2-2xy
$$
and
$$
\left[\frac{2y-(3+\sqrt{13})x}{\sqrt{26+6\sqrt{13}}}\right]^2+\left[\frac{2y-(3-\sqrt{13})x}{\sqrt{26-6\sqrt{13}}}\right]^2=x^2+y^2
$$
So the range of $3x^2-2xy$, as $(x,y)$ varies on the unit circle, is the closed interval with endpoints $\dfrac{3\pm\sqrt{13}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3044271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Computing $\sqrt[4]{28+16 \sqrt 3}$ I want to compute following radical
$$\sqrt[4]{28+16 \sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
We know that $ 28 = 2 \cdot 7^{\frac{1}{2}}$
$$(2 \cdot 7^{\frac{1}{2}}+16\cdot 3^{\frac{1}{2}})^{\frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
| To find the fourth root, we need to use Bill Dubuque's denesting algorithm for square roots twice.
First time round, we can find that $\sqrt{\text{norm}} = \sqrt{28^2 - 3 \cdot 16^2} = \sqrt{16} = 4$. Subtracting out the norm gives $24 + 16 \sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{2 \cdot 24} = 4 \sqrt{3}$, we get $\frac{8 \sqrt{3}\sqrt 3}{4 \sqrt 3} + \frac{16 \sqrt3}{4 \sqrt3} = 2\sqrt{3}+4 = 4 + 2\sqrt{3}$.
Second time round, we can find that $\sqrt{\text{norm}} = \sqrt{4^2 - 3 \cdot 2^2 -8} = \sqrt{4} = 2$. Subtracting out the norm gives $2 + 2\sqrt{3}$, and when you divide this by $\sqrt{\text{trace}} = \sqrt{4} = 2$, you get $1 + \sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Peetre's Inequality - not strict?
(Peetre's inequality)
Let $x,y \in \Bbb R^n$ and $s \in \Bbb R$. Then
$$ \frac{(1+|x|^2)^s}{(1+|y|^2)^s} \le 2^{|s|} (1+|x-y|^2)^{|s|}.$$
Proof: By switching roles of $x,y$ we may suppose $s \ge 0$, and taking $s$th root may assume $s=1$. Then the argument i found online:
\begin{align*}
(1+|x|^2) &= 1 + |x-y|^2 + |y|^2 +2(x-y)y \\
& \le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \\
& \le 2(1+|y|^2 + |x-y|^2 +|y|^2|x-y|^2 ) \\
& = 2(1+|y|^2)(1+|x-y|^2).
\end{align*}
What I don't understand is that on the third inequality, isn't this clearly a strict inequality when $s\not= 0$? (at least by 1)
| Yes, you are correct. We have that for $x,y\in \mathbb{R}^n$,
\begin{align*}
(1+|x|^2) &=1+|(x-y)+y|^2\\
&= 1 + |x-y|^2 + |y|^2 +2\langle (x-y), y \rangle \\
&\leq 1 + |x-y|^2 + |y|^2 +2|x-y| |y| \\
& \le 1 + |x-y|^2 + |y|^2 + (|x-y|^2+|y|^2) \\
& = 1+ 2|y|^2 + 2|x-y|^2 \\
& < 2+ 2|y|^2 +2|x-y|^2 +2|y|^2|x-y|^2 \\
& = 2(1+|y|^2)(1+|x-y|^2).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the volume of intersection between cylinders
Find the volume of intersection of the cylinder
{$ x^2 +
y^2 \leq 1 $} , {$ x^2 + z^2 \leq 1$}, {$ y^2 + z^2 \leq 1$}.
i am having tough time finding the volume how do i solve this kind of questions ? .
my trial :
i will move to the cylinder coordinates of the xy cylinder let :
$x^2 + y^2 = r^2 $
$ z = z $
$0\leq\theta \leq 2\pi$
solving the inequalties i get :
$ 0 \leq r^2 \leq 1$
$ -\sqrt{1-\frac{r^2}{2}}\leq z \leq \sqrt{1-\frac{r^2}{2}}$
$0\leq\theta \leq 2\pi$
the integral is :
$ \int_{z=-\sqrt{1-\frac{r^2}{2}}}^{z=\sqrt{1-\frac{r^2}{2}}}\int_{r=0}^{r=1}\int_0^{2\pi} dz \ dr \ d{\theta}$ = $ \frac{4\pi}{\sqrt{2}}\frac{(2-r^2)^{\frac{3}{2}}}{-3} |_{r=0}^{r=1}$
|
$$V = 16 \int\limits_0^{\pi/4} \int\limits_0^1 s \sqrt{1 - s^2 \cos^2 \theta}\ ds\ d\theta = 8(2 - \sqrt{2}) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof verification: $d$ and $\tilde{d}$ are topologically equivalent Given $X = (0,1]$ a metric space with $\tilde{d}$ defined as $$\tilde{d}(x,y) = \bigl|\frac{1}{x} - \frac{1}{y}\bigr| \ \ \text{for} \ \ x,y \in X$$
I'm trying to prove that $\tilde{d}$ and $d$ the standard metric on $(0,1]$ are topologically equivalent, using the open sets criteria and would like to know if it's correct.
Proof:
$U$ is open in $(X,\tilde{d}) \implies U$ is open in $(X,d)$
Let $x \in U$ and $\epsilon > 0$ such that $B_{\epsilon}(x,\tilde{d}) \subseteq U$
If $y \in B_{\epsilon}(x,\tilde{d})$ which means $ \tilde{d}(x,y) =\bigl|\frac{1}{x} - \frac{1}{y}\bigr| = \bigl|\frac{y-x}{xy}\bigr| = \frac{|x-y|}{|xy|} < \epsilon$ and that's equivalent to $d(x,y) = |x-y| < |xy|\epsilon$, then $y \in U$
Therefore, for any $x \in U$ we can find an epsilon $\tilde{\epsilon} = |xy|\epsilon$ such that: $y \in B_{|xy|\epsilon}(x, d) \implies y \in U$
$\Longleftarrow$
Similarly, If $y \in B_{\epsilon}(x,d)$, it follows that $d(x,y) = |x-y| < \epsilon$ which is equivalent to $\tilde{d}(x,y) =\bigl|\frac{1}{x} - \frac{1}{y}\bigr| < \frac{\epsilon}{|xy|}$
Therefore, for any $x \in U$ we can find an epsilon $\tilde{\epsilon} = \frac{\epsilon}{|xy|}$ such that: $y \in B_{\frac{\epsilon}{|xy|}}(x, \tilde{d}) \implies y \in U$
| There is a fault in your proof. Kindly see the proof below for the other side
\begin{align}\bar{d}(x,y)<r_2 &\iff \left| \dfrac{1}{x}-\dfrac{1}{y} \right| <r_2 \iff \dfrac{1}{x}-r_2<\dfrac{1}{y}<\dfrac{1}{x}+r_2\\&
\iff \dfrac{1-x r_2 }{x}<\dfrac{1}{y}<\dfrac{1+xr_2 }{x}
\end{align}
Choose $r_2<\dfrac{1}{x},$ then
\begin{align}
\dfrac{1-x r_2 }{x}<\dfrac{1}{y}<\dfrac{1+xr_2 }{x}&\iff \dfrac{x }{1+x r_2}<y<\dfrac{x }{1-xr_2}\\& \iff - \dfrac{x^2 r_2 }{1+x r_2}<y-x<\dfrac{x^2 r_2 }{1-xr_2}.
\end{align}
As $r_2\to 0,$ then $\dfrac{x^2 r_2 }{1+x r_2}\to 0$ and $\dfrac{x^2 r_2 }{1-xr_2}\to 0.$ Hence, $\forall \, r>0,$ and $\forall \,x\in (0,1],$ there exists $r_2\in (0,1]$ such that $r_2<(r/2x^2).$ So,
\begin{align}
y-x<\dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <\dfrac{r }{2}<r.
\end{align}
Also, \begin{align}
r_2<\dfrac{r}{2x^2}\iff -r_2>-\dfrac{r}{2x^2}\iff -x^2 r_2>-\dfrac{r}{2}.
\end{align}
But, $r_2\in (0,1]\implies r_2>-\dfrac{1}{2x},$ implies $x r_2>-\dfrac{1}{2} \iff 1+x r_2>\dfrac{1}{2}\iff \dfrac{1}{1+x r_2}<2.$ Hence,
\begin{align}
y-x> - \dfrac{x^2 r_2 }{1+x r_2}>-\dfrac{r}{2(1+x r_2)} > -\dfrac{2r}{2}=-r
\end{align}
Hence, we have that
\begin{align}
-r=-\dfrac{2r}{2}<-\dfrac{r}{2(1+x r_2)}< - \dfrac{x^2 r_2 }{1+x r_2}<y-x<\dfrac{x^2 r_2 }{1-xr_2}<x^2 r_2 <\dfrac{r }{2}<r.
\end{align}
Therefore, \begin{align}
|x-y|<r \iff d(x,y)<r
\end{align}
And you are done. Kindly get back if you have some questions. Remember that they don't necessarily need to have the same radius.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Mistake in solving an equation involving a square root I want to solve $2x = \sqrt{x+3}$, which I have tried as below:
$$\begin{equation}
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\left(x - \frac12 \right)^2 = 1 \\
x = \frac32 , -\frac12
\end{equation}$$
This, however, is incorrect.
What is wrong with my solution?
| $\require{cancel}$Additional details in $\color{blue}{blue}$. Important detail in $\color{green}{green}$. Mistakes in $\color{red}{\cancel{\text{canceled red}}}$. Corrections in $\color{purple}{purple}$.
$\begin{equation}
\color{blue}{2x = \sqrt{x+3}}\color{green}{\ge 0}\\
\color{blue}{4x^2 = x+3}\color{green}{\text{!AND!} x\ge 0}\\
4x^2 - x -3 = 0 \\
x^2 - \frac14 x - \frac34 = 0 \\
x^2 - \frac14x = \frac34 \\
\color{blue}{x^2 -2\cdot\frac 18=\frac 34}\\
\color{blue}{x^2 -2\cdot\frac 18 +(\frac 18)^2=\frac 34+\frac 1{64}}\\
\color{red}{\cancel{(x - \frac12 )^2 = 1}}\color{purple}{(x - \frac18)^2 = \frac {49}{64}} \\
\color{blue}{x-\frac 18=\pm \frac 78}\\
\color{red}{\cancel{x = \frac32 , -\frac12}}\color{purple}{x = 1 , -\frac34}\color{green}{\text{!AND!} x\ge 0}\\
\color{purple}{x = 1}\\
\end{equation}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 4
} |
Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$. I have managed to solve it in one way, but I became very interested in this failed attempt.
$$
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x
= \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x
- \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x
$$
We only have to show that those two on the right are equal. And numerical evaluations seem to suggest that they both are in fact $-\frac{\pi}{4}$ but I don't know how to break these down.
I am currently really interested in proving this
$$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$
Anyway, here's my trivial solution using $u = \frac1x$:
$$
\begin{align}
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x & = \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = \int_\infty^1 \frac{\frac{1}{u^2} \ln(\frac1u)}{(1+\frac{1}{u^2})^3} \frac{-1}{u^2} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = -\int_1^\infty \frac{\ln(u)}{u(u+\frac{1}{u})^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = - \int_1^\infty \frac{u^2 \ln(u)}{(1+u^2)^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = 0
\end{align}
$$
I'm sure there are many more interesting methods for cracking this integral, since it's so closely related to the popular $\int_0^\infty \frac{\ln(x)}{1+x^2} {\rm d}x = 0$. Please share them if you do come up with any.
| I found a way to directly evaluate both of those integrals!
We first use the well known result from $\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral
$$
\int_0^\infty \frac{\ln(x)}{x^2 +\alpha^2} {\rm d}x = \frac{\pi}{2\alpha} \ln(\alpha)
$$
and use differentiation under the integral sign.
$$
\int_0^\infty \frac{-2\alpha\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x
= \frac{\pi}{2} \frac{1-\ln(\alpha)}{\alpha^2} \\
\implies \int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x
= \frac{\pi}{4\alpha^3} (\ln(\alpha)-1)
$$
And again, to get
$$
\int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^3} {\rm d}x
= \frac{\pi}{16\alpha^5}(3\ln(\alpha) - 4)
$$
And so setting $\alpha = 1$ gives us the immediate result
$$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$
Another method to evaluate this is to use the integral from the 2015 MIT Integration Bee, and is also how I first came across this integral.
From this result (solved by using the substitution $u=\frac1x$)
$$ \int_0^\infty \frac{1}{(1+x^2)(1+x^\alpha)} {\rm d}x = \frac\pi4 $$
we will get, by differentiating with respect to $\alpha$,
$$ \int_0^\infty \frac{x^\alpha \ln(x)}{(1+x^2)(1+x^\alpha)^2} {\rm d}x = 0 $$
And finally setting $\alpha = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Different ways of proving that $|\sum^{\infty}_{k=1}(1-\cos(1/k))|\leq 2 $ I've found two ways of proving that
\begin{align} \left|\sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]\right|&\leq 2 \end{align}
Are there any other ways out there, for proving this?
METHOD 1
Let $k\in \Bbb{N}$, then
\begin{align} f:[ 0&,1]\longrightarrow \Bbb{R}\\&x \mapsto
\cos\left(\dfrac{x}{k}\right) \end{align}
is continuous. Then, by Mean Value Theorem, there exists $c\in [ 0,x]$ such that
\begin{align} \cos\left(\dfrac{x}{k}\right)-\cos\left(0\right) =-\dfrac{1}{k}\sin\left(\dfrac{c}{k}\right)\,(x-0), \end{align}
which implies \begin{align} \left|\sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]\right| &=\left|\sum^{\infty}_{k=1}\dfrac{x}{k}\sin\left(\dfrac{c}{k}\right)\right| \leq \sum^{\infty}_{k=1}\dfrac{\left|x\right|}{k}\left|\sin\left(\dfrac{c}{k}\right)\right|\leq \sum^{\infty}_{k=1}\dfrac{\left|x\right|}{k}\dfrac{\left|c\right|}{k}\\&\leq \sum^{\infty}_{k=1}\left(\dfrac{\left|x\right|}{k}\right)^2\leq \sum^{\infty}_{k=1}\dfrac{1}{k^2}=1+ \sum^{\infty}_{k=2}\dfrac{1}{k^2}\\&\leq 1+ \sum^{\infty}_{k=2}\dfrac{1}{k(k-1)}\\&= 1+ \lim\limits_{n\to\infty}\sum^{n}_{k=2}\left(\dfrac{1}{k-1}-\dfrac{1}{k}\right)\\&=1+ \lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)\\&=2, \end{align}
METHOD 2
Let $x\in [0,1]$ be fixed, then
\begin{align} \sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]&=\sum^{\infty}_{k=1}\dfrac{1}{k}\left[-k\cos\left(\dfrac{x}{k}\right)\right]^{1}_{0}=\sum^{\infty}_{k=1}\dfrac{1}{k}\int^{1}_{0}\sin\left(\dfrac{x}{k}\right)dx \\&=\sum^{\infty}_{k=1}\int^{1}_{0}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx \end{align}
The series $\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)$ converges uniformly on $[0,1]$, by Weierstrass-M test, since
\begin{align} \left|\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right) \right|\leq \sum^{\infty}_{k=1}\dfrac{1}{k}\left|\sin\left(\dfrac{x}{k}\right) \right|\leq \sum^{\infty}_{k=1}\dfrac{1}{k^2}. \end{align}
Hence,
\begin{align} \sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]&=\sum^{\infty}_{k=1}\int^{1}_{0}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx=\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx, \end{align}
and
\begin{align} \left|\sum^{\infty}_{k=1}\left[1-\cos\left(\dfrac{1}{k}\right)\right]\right|&=\left|\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k}\sin\left(\dfrac{x}{k}\right)dx\right|\leq\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k}\left|\sin\left(\dfrac{x}{k}\right)\right|dx \\&\leq\int^{1}_{0}\sum^{\infty}_{k=1}\dfrac{1}{k^2}dx \\&\leq 2 \end{align}
| Note that
$$1 - \cos(x) = \cos(0) - \cos(x) = 2 \sin^2\left( \frac{x}2 \right) $$
by the sum to product identities for all $x$. Therefore,
$$ \sum_{k=1}^{\infty} \left(1- \cos\left( \frac{1}k \right) \right) = 2 \sum_{k=1}^{\infty} \sin^2\left( \frac{1}{2k} \right). $$
Now using the inequality $\sin(x) \le x$ for all positive $x$, we get
$$ 2\sum_{k=1}^{\infty} \sin^2\left( \frac{1}{2k} \right) \le \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k^2} \approx 0.822.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Rationalizing denominator of $\frac{7}{2+\sqrt{3}}$. Cannot match textbook solution I am given this expression and asked to simplify by rationalizing the denominator:
$$\frac{7}{2+\sqrt{3}}$$
The solution is provided:
$14 - 7\sqrt{3}$
I was able to get to this in the numerator but am left with a 4 in the denominator. Here are my steps:
$$\frac{7}{2+\sqrt{3}} * \frac{2-\sqrt{3}}{2-\sqrt{3}}$$
=
$$\frac{14 - 7\sqrt{3}}{4}$$
Presumably I should not have a denominator here since the solution I'm given is just whats in the numerator. Presumably my numerator calculation is correct, where did I go wrong on the denominator?
| note that:
$$(a+b)(c+d)=ac+ad+bc+bd$$
so in some cases it simplifies to:
$$(a+b)(a-b)=a^2-b^2$$
for you, $a=2$ and $b=\sqrt{3}$ so $a^2-b^2=4-3=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that base is twice the height if base angles of a triangle are $22.5^\circ$ and $112.5^\circ$
The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Show that the base is twice the height.
My Attempt
$$
h=c.\sin22.5^\circ=c.\cos 67.5^\circ\\
=b\sin 67.5^\circ=b\cos 22.5^\circ
$$
$$
a=c\cos22.5^\circ- b\sin22.5\circ=\frac{h}{b}-\frac{h}{c}=h\cdot\frac{c-b}{bc}
$$
I have no clue of how to prove this.
| The two right angled triangles in your picture are similar and both have smaller angle $22.5$. Let the shortest unmarked side be $x$
Then $$\frac{h}{a+x}=\frac xh=\tan22.5=\sqrt{2}-1$$
Eliminating $x$ gives $$h^2=ah(\sqrt{2}-1)+h^2(\sqrt{2}-1)^2$$
Rearranging gives $$\frac ah=\frac{2\sqrt{2}-2}{\sqrt{2}-1}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove $ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $ Show that if $a,b,c > 0$, such that $ab + bc + ca = 1$, then the following inequality holds:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq 20 $$
What I have tried is firstly using the inequality:
$\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, for any $x, y$ and any $a,b > 0$.
Using this inequality we obtain $\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} \geq \frac{a + b + c}{2}$, and then we have:
$$ \frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b} + \frac{36}{a + b + c} \geq \frac{a + b + c}{2} + \frac{36}{a + b + c} = \frac{(a + b + c)^2 + 72)}{2(a + b + c)} $$.
Using $ab + bc + ca = 1$, we would then have to prove that:
$$a^2 + b^2 + c^2 + 74 - 40(a + b + c) \geq 0 $$ and then I tried replacing in this inequality $c = \frac{1 - ab}{a + b}$, but I didn't get anything nice.
I also tried rewriting the lhs:
$$\frac{a^2}{b + c} = \frac{a^2(ab + bc + ca)}{b + c} = a^3 + \frac{a^2bc}{b + c}$$
And this would result in: $a^3 + b^3 + c^3 + abc(\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b}) + \frac{36}{a + b + c} \geq 20$, but I didn't know how to continue from here.
Do you have any suggestions for this inequality?
| I can easy get an SOS (Sum of Squares)!
We have: $$\text{LHS-RHS} =\frac{A}{(a+b)(b+c)(c+a)(a+b+c)}\geqq 0$$
Where: $$A=\Big[4\, \left( a+b+c+5+\sqrt {3} \right) \left( a+b+c-\sqrt {3} \right)
+20\,\sqrt {3}-24\Big]abc$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b+c \right) \Big[ \left( a+b+c \right) ^{2
}+4\,a+4\,b+4\,c+9\Big] \left( a+b+c-2 \right) ^{2} \geqq 0$$
Equality holds when $a=b,\,c=0$ and its permutations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Least Squares solution for a symmetric singular matrix I want to solve this system by Least Squares method:$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\5\\-2\end{pmatrix} $$ This symmetric matrix is singular with one eigenvalue $\lambda1 = 0$, so $\ A^t\cdot A$ is also singular and for this reason I cannot use the normal equation: $\hat x = (A^t\cdot A)^{-1}\cdot A^t\cdot b $.
So I performed Gauss-Jordan to the extended matrix to come with $$\begin{pmatrix}1 & 2 & 3\\\ 0 & 1 & 2 \\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\3\\-1\end{pmatrix} $$
Finally I solved the $\ 2x2$ system: $$\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix} $$ taking into account that the best $\ \hat b\ $ is $\begin{pmatrix}1\\3\\0\end{pmatrix}$
The solution is then $\ \hat x = \begin{pmatrix}-5\\3\\0\end{pmatrix}$
Is this approach correct ?
EDIT
Based on the book 'Lianear Algebra and its applications' from David Lay, I also include the Least Squares method he proposes: $(A^tA)\hat x=A^t b $
$$A^t b =\begin{pmatrix}5\\9\\13\end{pmatrix}, A^tA = \begin{pmatrix}14 & 20 & 26 \\
20 & 29 & 38 \\
26 & 38 & 50\end{pmatrix}$$
The reduced echelon from the augmented is:
$$ \begin{pmatrix}14 & 20 & 26 & 5 \\
20 & 29 & 38 & 9 \\
26 & 38 & 50 & 13 \end{pmatrix} \sim \begin{pmatrix}1 & 0 & -1 & -\frac{35}{6} \\
0 & 1 & 2 & \frac{13}{3} \\
0 & 0 & 0 & 0
\end{pmatrix} \Rightarrow \hat x = \begin{pmatrix}-\frac{35}{6} \\ \frac{13}{3} \\ 0 \end{pmatrix}$$ for the independent variable case that $z=\alpha , \alpha=0 $
| Your solution is incorrect for the following reason. When you perform the Gauss-Jordan elimination, you transform the original system
$$\tag{1}
Ax=b
$$ to another
$$\tag{2}
SAx=Sb.
$$
But the least squares solutions of (1) and (2) do not coincide in general.
Indeed, the least squares solution of (1) is $A^{+}b$, the least squares solution of (2) is $(SA)^{+}Sb$. If $A$ is invertible, then
$$(SA)^{+}S=(SA)^{-1}S=A^{-1}=A^{+}$$
and everything is OK, but in general case $(SA)^{+}S\ne A^{+}$.
In your case, in particular,
$$
S=\left(\begin{array}{rrr}
1 & 0 & 0 \\
2 & -1 & 0 \\
1 & -2 & 1 \\
\end{array}\right),\quad
(SA)^{+}S=\left(\begin{array}{rrr}
-11/6 & 4/3 & 0 \\
-1/3 & 1/3 & 0 \\
7/6 & -2/3 & 0 \\
\end{array}\right),
$$
$$
A^{+}=\left(\begin{array}{rrr}
-13/12 & -1/6 & 3/4 \\
-1/6 & 0 & 1/6 \\
3/4 & 1/6 & -5/12\\
\end{array}\right).$$
You can calculate the pseudoinverse matrix by using the rank factorization:
$$
A=BC,\quad B=\left(\begin{array}{rr}
1 & 3\\
2 & 4\\
3 & 5\\
\end{array}\right),\quad
C=\left(\begin{array}{rrr}
1&1/2&0\\
0&1/2&1
\end{array}\right)
$$
(this decomposition comes from the fact the second column of $A$ is the arithmetic mean of the remaining columns).
It remains only to calculate the pseudoinverse matrix
$$
A^{+}=C^{+}B^{+}=C^T(CC^T)^{-1}(B^TB)^{-1}B^T
$$
and the least squares solution is $A^{+}b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Trying to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$ to be $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ I am asked to simplify $\frac{\sqrt{8}}{1-\sqrt{3x}}$. The solution is provided as $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$ and I am unable to arrive at this. I was able to arrive at $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Here is my working:
$\frac{\sqrt{8}}{1-\sqrt{3x}}$ = $\frac{\sqrt{8}}{1-\sqrt{3x}}$ * $\frac{1+\sqrt{3x}}{1+\sqrt{3x}}$ = $\frac{1+\sqrt{8}\sqrt{3x}}{1-3x}$ = $\frac{1+\sqrt{2}\sqrt{2}\sqrt{2}\sqrt{3x}}{1-3x}$ = $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$
Is $\frac{1+2\sqrt{2}\sqrt{3x}}{1-3x}$ correct and part of the way? How can I arrive at the provided solution $\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$?
| Hints:$\sqrt8=\sqrt{4×2}=2\sqrt{2}.$
$\sqrt24=2\sqrt6$. (Why?)
$(1+\sqrt{3x})×(1-\sqrt{3x})=1-3x $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Solve the integral $\int_0^1\int^1_xy^4e^{xy^2}dydx$.
Solve the integral $\int_0^1\int^1_xy^4e^{xy^2}dydx$.
I think that variables substituation is neede here. I've substitute
$$
\\ \left\{\begin{matrix}
u=xy^2\\
v=y
\end{matrix}\right. \
$$
and calculated
$$\\J=\begin{vmatrix}
y^2 & 2xy\\
0 & 1
\end{vmatrix}=y^2\
$$
Then, the new integrand is $v^2e^u$. But what is the new domain? Thanks.
| Well, solving a much more general problem:
$$\mathcal{I}_\text{n}\left(\alpha\right):=\int_0^\alpha\int_x^\alpha\text{y}^\text{n}\cdot\exp\left(x\cdot\text{y}^{\text{n}-2}\right)\space\text{d}\text{y}\space\text{d}x\tag1$$
Using that (for all $x$):
$$\exp\left(x\right)=\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\tag2$$
We can write:
$$\mathcal{I}_\text{n}\left(\alpha\right)=\int_0^\alpha\int_x^\alpha\text{y}^\text{n}\cdot\sum_{\text{k}=0}^\infty\frac{\left(x\cdot\text{y}^{\text{n}-2}\right)^\text{k}}{\text{k}!}\space\text{d}\text{y}\space\text{d}x=$$
$$\int_0^\alpha\int_x^\alpha\text{y}^\text{n}\cdot\sum_{\text{k}=0}^\infty\frac{x^\text{k}\cdot\text{y}^{\text{k}\left(\text{n}-2\right)}}{\text{k}!}\space\text{d}\text{y}\space\text{d}x=$$
$$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\int_x^\alpha\text{y}^\text{n}\cdot\text{y}^{\text{k}\left(\text{n}-2\right)}\space\text{d}\text{y}\right\}\space\text{d}x=$$
$$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\int_x^\alpha\text{y}^{\text{n}+\text{k}\left(\text{n}-2\right)}\space\text{d}\text{y}\right\}\space\text{d}x=$$
$$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\cdot\left[\frac{\text{y}^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\right]_x^\alpha\right\}\space\text{d}x=$$
$$\int_0^\alpha\left\{\sum_{\text{k}=0}^\infty\frac{x^\text{k}}{\text{k}!}\cdot\left(\frac{\alpha^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}-\frac{x^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\right)\right\}\space\text{d}x=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left\{\frac{\alpha^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\int_0^\alpha x^\text{k}\space\text{d}x-\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\int_0^\alpha x^{1+\text{n}+\text{k}\left(\text{n}-1\right)}\space\text{d}x\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left\{\frac{\alpha^{1+\text{n}+\text{k}\left(\text{n}-2\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{\alpha^{1+\text{k}}}{1+\text{k}}-\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{\alpha^{1+1+\text{n}+\text{k}\left(\text{n}-1\right)}}{1+1+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left\{\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{1}{1+\text{k}}-\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\left\{\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{1+\text{k}}-\frac{\alpha^{2+\text{n}+\text{k}\left(\text{n}-1\right)}}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}\tag3$$
When $\alpha=1$, we get:
$$\mathcal{I}_\text{n}\left(1\right):=\int_0^1\int_x^1\text{y}^\text{n}\cdot\exp\left(x\cdot\text{y}^{\text{n}-2}\right)\space\text{d}\text{y}\space\text{d}x=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{n}+\text{k}\left(\text{n}-2\right)}\cdot\left\{\frac{1}{1+\text{k}}-\frac{1}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\right\}=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+\text{k}}\cdot\frac{1}{2+\text{n}+\text{k}\left(\text{n}-1\right)}\tag4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ is $\frac{3\sqrt[4]{2x^2}}{2}$ (I've been posting a lot today and yesterday, not sure if too many posts are frowned upon or not. I am studying and making sincere efforts to solve on my own and only post here as a last resort)
I'm asked to simplify $\sqrt[4]{\frac{162x^6}{16x^4}}$ and am provided the text book solution $\frac{3\sqrt[4]{2x^2}}{2}$.
I arrived at $\frac{3\sqrt[4]{2x^6}}{2x^4}$. I cannot tell if this is right and that the provided solution is just a further simplification of where I've gotten to, or if I'm off track entirely.
Here is my working:
$\sqrt[4]{\frac{162x^6}{16x^4}}$ = $\frac{\sqrt[4]{162x^6}}{\sqrt[4]{16x^4}}$
Denominator: $\sqrt[4]{16x^4}$ I think can be simplified to $2x^4$ since $2^4$ = 16
Numerator: $\sqrt[4]{162x^6}$ I was able to simplify (or over complicate) to $3\sqrt[4]{2}\sqrt[4]{x^6}$ since:
$\sqrt[4]{162x^6}$ = $\sqrt[4]{81}$ * $\sqrt[4]{2}$ * $\sqrt[4]{x^6}$ = $3 * \sqrt[4]{2} * \sqrt[4]{x^6}$
Thus I got:
$\frac{3\sqrt[4]{2}\sqrt[4]{x^6}}{2x^4}$ which I think is equal to $\frac{3\sqrt[4]{2x^6}}{2x^4}$ (product of the radicals in the numerator).
How ca I arrive at the provided solution $\frac{3\sqrt[4]{2x^2}}{2}$?
| $$\sqrt[4]{\frac{162x^6}{16x^4}}=\sqrt[4]{\frac{81\cdot2x^2}{16}}=\frac{3\sqrt[4]{2x^2}}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving for $x$ in $\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$
Given an equation: $$\sin^{-1}(2x) + \sin^{-1}(3x) = \frac \pi 4$$
How do I find $x$?
I tried solving by differentiating both sides, but I get $x=0$.
How do you solve it, purely using trigonometric techniques?
| Start with $$2x\sqrt{1-9x^2}+3x\sqrt{1-4x^2} = \frac{1}{\sqrt{2}}$$ and square both sides to get
$$-72 x^4+13x^2+12 \sqrt{1-9 x^2} \sqrt{1-4 x^2} x^2=\frac 12$$ that is to say
$$72 x^4-13x^2+\frac 12=12 \sqrt{1-9 x^2} \sqrt{1-4 x^2} x^2$$ Let $y=x^2$ to make
$$72 y^2-13y+\frac 12=12 y\sqrt{1-9 y} \sqrt{1-4 y} $$
Square again, expand and simplify to get
$$97 y^2-13 y+\frac{1}{4}=0$$ which is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Some weird results in complex number computing The question I met is to show that if $z=\cos (\theta)+i\sin(\theta)$ with $i=\sqrt{-1}$,then $ Re(\frac{z-1}{z+1})=0$
In the normal way, we found that: $$\frac{z-1}{z+1}=\frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}\\
=\frac{\bigl(\cos(\theta)-1+i\sin(\theta)\bigl)\bigl(\cos(\theta)+1-i\sin(\theta)\bigl)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\
=\frac{\cos^2(\theta)+\cos(\theta)-\cos(\theta)-1+\sin^2(\theta)+2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}\\
=\frac{2i\sin(\theta)}{\bigl(\cos(\theta)+1\bigl)^2+\sin^2(\theta)}$$
So $Re(\frac{z-1}{z+1})=0$
If we do it in another way:
$$
\frac{\cos(\theta)-1+i\sin(\theta)}{\cos(\theta)+1+i\sin(\theta)}=\frac{-2\sin^2(\frac{\theta}{2})+2i\cos(\frac{\theta}{2})\sin(\frac{\theta}{2})}{2cos^2(\frac{\theta}{2})+2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})}\\
=-\tan(\frac{\theta}{2})\frac{\sin(\frac{\theta}{2})-i\cos(\frac{\theta}{2})}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\
=-\tan(\frac{\theta}{2})\frac{\cos\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)+i\sin\bigl(-(\frac{\theta}{2}+\frac{\pi}{2})\big)}{\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})}\\
=-\tan(\frac{\theta}{2})\bigl(\cos(-\theta-\frac{\pi}{2})+i\sin(-\theta-\frac{\pi}{2})\bigl)
$$
So the real part of it will be $-\tan(\frac{\theta}{2})\cos(-\theta-\frac{\pi}{2})$ which is not $0$.
Which step I made mistake or they are equivalent?
| We have
$$
\sin\left(\frac\theta2\right) - i\cos\left(\frac\theta2\right) = -\sin\left(-\frac\theta2\right) - i\cos\left(\frac\theta2\right)\\
= -\cos\left(-\frac\theta2-\frac\pi2\right) - i\sin\left(\frac\theta2 + \frac\pi2\right)\\
= -\left(\cos\left(\frac\theta2 + \frac\pi2\right) + i\sin\left(\frac\theta2 + \frac\pi2\right)\right)
$$
so you have a sign error in the numerator between line 2 and line 3 of your alternate approach.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Calculating $\int \frac{dx}{(x^2+1)^3}$ integral Would anyone help me calculate the following integral? $\int \frac{dx}{(x^2+1)^3}$
During our lecutre we've done very similiar one, $\int \frac{dx}{(x^2+1)^2}$ like that:
$\int \frac{dx}{(x^2+1)^2} = \int \frac{x^2+1-x^2}{(x^2+1)^2}dx = \int \frac{1}{x^2+1}dx - \int \frac{x^2}{(x^2+1)^2}dx = $
$= \Biggr\rvert \begin{equation}
\begin{split}
& u = x \quad v' =\frac{x}{(x^2+1)^2} =\frac{1(x^2+1)'}{2(x^2+1)^2}\\
& u' = 1 \quad v = -\frac{1}{2} \frac{1}{x^2+1}
\end{split}
\end{equation} \Biggr\rvert$
$= \arctan x - (-x\frac{1}{2}\frac{1}{x^2+1} + \frac{1}{2} \int \frac{dx}{x^2+1})$
$= \arctan x + \frac{x}{2(x^2+1)} - \frac{1}{2}\arctan x + C = \frac{1}{2}\arctan x + \frac{x}{2(x^2+1)} + C$
Thank you.
| Use $$\int \frac{dx}{(x^2+1)^3}=\frac{x}{4(x^2+1)^2}+\frac{3}{4}\int \frac{dx}{(x^2+1)^2}$$ and then use the value for the integral you already calculated.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $N = 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$ Please show how to solve this step by step, because I don't even have an idea to begin with.
$$N = 1 + \frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}$$
| Hint: $$N = 1 + \cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{\ddots}}}}}$$
$$N -1= \cfrac{1}{2+\color{blue}{\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{\ddots}}}}}}$$
Notice that the blue part is also $N-1$.
$$N-1 = \frac{1}{2+\color{blue}{N-1}}$$
$$N-1 = \frac{1}{N+1}$$
It should be simple enough from here. Also, note that by inspection, $N > 1$, so it should be easy to discard an extraneous solution.
| {
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"url": "https://math.stackexchange.com/questions/3072227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find closed form of sum of fraction of binomial coefficients can somebody give me a hint for this exercise, where I have to find the specific closed form?
$\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}}, m,n\in\mathbb{N}$ and $m\leq n$
What I have done so far:
$\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}} = \sum_{k=0}^m \frac{\frac{m!}{(m-k)!\cdot k!}}{\frac{n!}{(n-k)!\cdot k!}} = \sum_{k=0}^m \frac{m!}{n!}\cdot \frac{(n-k)!}{(m-k)!} = \frac{m!}{n!}\cdot \sum_{k=0}^m \frac{(n-k)!}{(m-k)!}$
Look at example 1: $n=8, m=5$
$\frac{5!}{8!}\cdot(\frac{8!}{5!} + \frac{7!}{4!} +\frac{6!}{3!} +\frac{5!}{2!} + \frac{4!}{1!} +\frac{3!}{0!}) = \\\frac{5!}{8!} \cdot (8\cdot7\cdot6+7\cdot6\cdot5+6\cdot5\cdot4+5\cdot4\cdot3+4\cdot3\cdot2+3\cdot2\cdot1) = \\
\frac{5!}{8!} \cdot (336+210+120+60+24+6) = \frac{5!}{8!}\cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution.
$\sum_{k=0}^m \frac{m!}{n!}\cdot \sum_{k=0}^m \frac{(n-k)!}{(m-k)!}= \frac{m!}{n!}\cdot\frac{1}{4}\cdot t\cdot(t+1)\cdot(t+2)\cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$\frac{3!}{9!}\cdot(\frac{9!}{3!} + \frac{8!}{2!} +\frac{7!}{1!} +\frac{6!}{0!}) = \\\frac{3!}{9!} \cdot (9\cdot8\cdot7\cdot6\cdot5\cdot4+8\cdot7\cdot6\cdot5\cdot4\cdot3+7\cdot6\cdot5\cdot4\cdot3\cdot2+6\cdot5\cdot4\cdot3\cdot2\cdot1) = \\
\frac{3!}{9!} \cdot (60480+20160+5040+720) = \frac{3!}{9!}\cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
| With $m\le n$ we have the sum
$$\sum_{k=0}^m {m\choose k} {n\choose k}^{-1}
= \sum_{k=0}^m \frac{m!}{(m-k)! k!} \frac{k! (n-k)!}{n!}
\\ = \frac{m!}{n!} \sum_{k=0}^m \frac{(n-k)!}{(m-k)!}
= {n\choose m}^{-1} \sum_{k=0}^m {n-k\choose m-k}
\\ = {n\choose m}^{-1} \sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {n\choose m}^{-1} [z^m] (1+z)^n \sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${n\choose m}^{-1} [z^m] (1+z)^n \sum_{k\ge 0} z^k (1+z)^{-k}
= {n\choose m}^{-1} [z^m] (1+z)^n \frac{1}{1-z/(1+z)}
\\ = {n\choose m}^{-1} [z^m] (1+z)^{n+1}
= {n\choose m}^{-1} {n+1\choose m}
= \frac{n+1}{n+1-m}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\tan 2\alpha \cdot \tan \alpha = 1$, then what is $\alpha$? Different methods give different answers.
If $\tan 2\alpha\cdot\tan \alpha = 1$, then what is $\alpha$?
I tried two methods but got two different answers.
Method 1:
$$\begin{align}
\tan 2\alpha\cdot\tan \alpha = 1
&\implies
\frac{2\tan \alpha}{1 - \tan ^2 \alpha}\;\tan \alpha = 1 \tag{1a}\\[6pt]
&\implies 2\tan ^2\alpha = 1 - \tan ^2\alpha \tag{1b}\\[6pt]
&\implies \tan ^2 \alpha = \frac{1}{3} \tag{1c}\\[6pt]
&\implies \tan \alpha = \pm\frac{1}{\sqrt 3} \tag{1d}\\[6pt]
&\implies \tan \alpha = \tan\left(\pm\frac{\pi}{6}\right) \tag{1e}\\[6pt]
&\implies \alpha = n\pi \pm \frac{\pi}{6} \;\text{where}\; n \in \mathbb{Z} \tag{1f}
\end{align}$$
Method 2:
$$\begin{align}
\tan 2\alpha \cdot \tan \alpha = 1
&\implies \tan 2\alpha = \frac{1}{\tan \alpha} \tag{2a}\\[6pt]
&\implies \tan 2\alpha = \cot \alpha \tag{2b}\\[6pt]
&\implies \tan 2\alpha = \tan\left(\frac{\pi}{2} - \alpha\right) \tag{2c}\\[6pt]
&\implies 2\alpha = n\pi + \frac{\pi}{2} - \alpha \text{?} \tag{2d}\\[6pt]
&\implies \alpha = \frac{1}{3}\left(n\pi + \frac{\pi}{2}\right)\;\text{where}\; n \in \mathbb{Z} \tag{2e}
\end{align}$$
Which one is correct? Is there any mistake in the above solutions?
| @Toky,
There is nothing wrong with your methods. Both are correct. You are getting two different representations for the same set of solutions. According to the second solution
$$
\dots -\frac{\pi}{6}, \frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6},\dots
$$
Your first solutions gives you half of these if you choose the $+$ sign, and gives you the other half if you choose the $-$ sign.
Hope this helps.
| {
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"url": "https://math.stackexchange.com/questions/3072765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
} |
Find a ratio of triangle's height segment Given a right angle triangle ABC (C = 90) and a median AM. CD is the height of the triangle and it intersects median in point K. What is the CK / KD ratio?
| Let $AC=b$, $BC=a$ and $N$ be placed on $DB$ such that $MN\perp AB.$
Thus, $$MN=\frac{1}{2}CD,$$
$$AD=\frac{b^2}{\sqrt{a^2+b^2}}$$ and
$$BD=\frac{a^2}{\sqrt{a^2+b^2}}.$$
Id est, $$\frac{KD}{MN}=\frac{AD}{AN}=\frac{\frac{b^2}{\sqrt{a^2+b^2}}}{\frac{b^2}{\sqrt{a^2+b^2}}+\frac{a^2}{2\sqrt{a^2+b^2}}}=\frac{2b^2}{a^2+2b^2}.$$
Thus, $$\frac{KD}{CD}=\frac{b^2}{a^2+2b^2}$$ or
$$\frac{CD}{KD}=\frac{a^2+2b^2}{b^2}$$ or
$$1+\frac{CK}{KD}=\frac{a^2+b^2}{b^2}+1,$$ which gives
$$CK:KD=(a^2+b^2):b^2.$$
We see that the needed ratio depends on the ratio $a:b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $z=\cos\theta + i\sin \theta$ prove $\frac{z^2-1}{z^2+1}=i\tan\theta$
If $z=\cos\theta + i\sin \theta$ prove $$\frac{z^2-1}{z^2+1}=i\tan\theta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$
$$\frac{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)-1}{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)+1}$$
$$\frac{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta) -1}{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta)+1}$$
$$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}$$
I understand how I can do it with using $z=e^{i \theta}$, however I want to solve it using double angle identities.
| $z=e^{i\theta}$, $\frac{z^2-1}{z^2+1}=\frac{e^{2 i\theta}-1}{e^{2i\theta}+1}=\frac{e^{ i\theta}-e^{ -i\theta}}{e^{i\theta}+e^{ i\theta}}=\frac{2 i\, sin(\theta)}{2\,cos(\theta)}=itan(\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073776",
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"source": "stackexchange",
"question_score": "3",
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Precalc Trig Identity, verify: $1 + \cos(x) + \cos(2x) = \frac 12 + \frac{\sin(5x/2)}{2\sin(x/2)}$ Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2\cos(3x/2)\cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
| Start with the RHS. Notice that
$$ \sin(5x/2) = \sin(2x + x/2) = \sin 2x \cos (x/2) + \cos 2x \sin (x/2) $$
Also, notice that
$$ \sin 2x = 2 \sin x \cos x = 4 \sin (x/2) \cos (x/2) \cos x $$
Therefore,
$$ \frac{ \sin (5x/2) }{2 \sin(x/2) } = \frac{4 \sin (x/2) \cos^2 (x/2) \cos x + \cos 2x \sin (x/2)}{2 \sin(x/2)} $$
$$ = 2 \cos^2 (x/2) \cos x + \frac{ \cos 2x }{2} $$
Also, we have that $\cos^2 (x/2) = \frac{ \cos x + 1 }{2}$ and so
$$ = \cos^2 x + \cos x + \frac{ \cos 2x }{2} $$
$$ = \frac{1+\cos 2x }{2} + \cos x + \frac{ \cos 2x }{2} $$
$$ \cos x + \cos 2x + \frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x$ and $y$ are acute, and $\sin y = 3 \cos (x+y) \sin x$, then find the maximum value of $\tan y$
Given $x,y$ are acute angles such that
$$\sin y = 3 \cos(x+y)\sin x$$
Find the maximum value of $\tan y$.
Attempt:
We have
$$\begin{aligned} 3(\cos x \cos y - \sin x \sin y) \sin x & = \sin y \\ 3 \cos x \sin x - 3 \sin^2 x \tan y & = \tan y \\ 3 \cos x \sin x & = \tan y(1 +3 \sin^2 x) \\ \tan y & = \dfrac{3 \sin x \cos x} {1+3 \sin^2 x} \end{aligned}$$
Now, how about the next step? Or maybe I did some mistakes?
| Put $t = \tan x$, then $\tan y = \dfrac{3t}{1+4t^2}\le \dfrac{3t}{4t} = \dfrac{3}{4}$, which is the max of $\tan y$ with equality occurs when $t = \dfrac{1}{2}$ or $2\sin x = \cos x$...the rest is simple...
| {
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"url": "https://math.stackexchange.com/questions/3076662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the convergence of $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ I want to find what the series $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ converges to exactly or show that it diverges.
By taking the partial sum of the series $S_N$ = $ \sum_{n=1}^{N} (n+1)^\frac{1}{3} - n^\frac{1}{3}$ then $S_N = 2^\frac{1}{3} - 1 + 3^\frac{1}{3} - 2^\frac{1}{3} +4^\frac{1}{3}-3^\frac{1}{3} + ... + (N+1)^\frac{1}{3} - N^\frac{1}{3}$
And at the end I'm left with $S_N = -1 + (N+1)^\frac{1}{3}$ and $\lim_{N \to \infty} S_N = -1 + \infty= \infty $
So $ \sum_{n=1}^{\infty} (n+1)^\frac{1}{3} - n^\frac{1}{3} = \infty$
Is this correct is my first question and my second question is does there exist any other method of finding what series to converge exactly?
Thank you in prior.
|
While summation of the telescoping series is trivial and immediately shows divergence of the series, the OP has asked if there are alternative approaches. Herein, we give two straightforward ways forward.
METHODOLOGY $1$:
Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ reveals
$$(n+1)^{1/3}-n^{1/3}=\frac{1}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}> \frac{1}{3(n+1)^{2/3}}>\frac{1}{3(n+1)}$$
Hence, we have
$$\sum_{n=1}^N \left((n+1)^{1/3}-n^{1/3} \right)>\frac13 \sum_{n=2}^{N+1}\frac1n$$
Inasmuch as the harmonic series diverges, the series of interest diverges also.
METHODOLOGY $2$:
Using $(n+1)^{1/3}-n^{1/3}=n^{1/3}\left(1+\frac1{3n}+O\left(\frac1{n^2}\right)\right)-n^{1/3}=\frac1{3n^{2/3}}+O\left(\frac1{n^{5/3}}\right)$
Inasmuch as the series $\sum_{n=1}^\infty \frac{1}{n^p}$ diverges for $p\le 1$, the series of interest diverges also.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I have been able to prove that this holds for $n=1$ , $n=2$, and $n=3$ using the following lemma:
Lemma 1: Let $a,b \in \Bbb{R}^+$. If $ab =1$ then $a+b \ge 2$
For example, the case for $n=3$ can be proven like this:
Let $a,b,c \in \Bbb{R}^+$. Then we have:
$(a+b+c)\big(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\big) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 $
$= 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) $
By lemma 1, $\big(\frac{a}{b} + \frac{b}{a}\big) \ge 2$, $ \big(\frac{a}{c} + \frac{c}{a}\big) \ge 2$ and $\big(\frac{b}{c} + \frac{c}{b}\big) \ge 2$ , therefore:
$3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare $
However I'm not sure the generalized version for all natural $n$ is true. I can't come up with a counterexample and when I try to prove it by induction I get stuck.
Here is my attempt:
Let $P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$
Base case: $\big(\sum_{i=1}^{1}a_{i}\big) \big(\sum_{i=1}^{1} \frac{1}{a_{i}}\big) = a_{1} \frac{1}{a_{1}} = 1 = 1^2$ , so $P(1)$ is true.
Inductive hypothesis: I assume $P(n)$ is true.
Inductive step:
$$\left(\sum_{i=1}^{n+1}a_{i}\right) \left(\sum_{i=1}^{n+1} \frac{1}{a_{i}}\right) = \left[\left(\sum_{i=1}^{n}a_{i}\right) + a_{n+1}\right] \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right) \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right] + a_{n+1} \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +a_{n+1} \frac{1}{a_{n+1}}$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +1 $$
$$\underbrace{\ge}_{IH} n^2 + \left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + 1$$
And here I don't know what to do with the $\big( \sum_{i=1}^{n}a_{i} \big) \frac{1}{a_{n+1}} + a_{n+1} \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big)$ term.
Is this inequality true? If it is, how can I prove it? If it isn't, can anyone show me a counterexample?
| Just for completeness:
You may also show the inequality by direct calculation using
*
*$x + \frac{1}{x} \geq 2$ for $x > 0$
\begin{eqnarray*} \left(\sum_{i=1}^{n} a_{i} \right) \left(\sum_{j=1}^{n} \frac{1}{a_{j}} \right)
& = & \sum_{i,j =1}^n \frac{a_i}{a_j} \\
& = & \sum_{\stackrel{i,j =1}{\color{blue}{i=j}}}^n \frac{a_i}{a_j} + \sum_{\stackrel{i,j =1}{\color{blue}{i\neq j}}}^n \frac{a_i}{a_j} \\
& = & n + \sum_{\stackrel{i,j =1}{\color{blue}{i< j}}}^n \left(\frac{a_i}{a_j} + \frac{a_j}{a_i} \right)\\
& \color{blue}{\geq} & n + \sum_{\stackrel{i,j =1}{\color{blue}{i< j}}}^n 2\\
& = & n + 2\binom{n}{2} = n^2\\
\end{eqnarray*}
| {
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} |
Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$. Find the smallest prime number $p$ such that $p\, | \,n^2-n-2023$ for some integer $n$.
since $n^2-n =n(n-1)$ is the product of two consecutive integers they must be even so the difference between an even and odd number is always odd so $n^2-n-2023$ is always odd which implies $p$ is not even and the only even prime is $2$ so $p\neq 2$ but after this I do not know what to do please help.
| $n^2-n-2023$ is odd, so for prime $p$ we have, modulo $p,$ that $n^2-n-2023\equiv 0\iff 4n^2-4n-8092\equiv 0\iff (2n-1)^2\equiv 8093.$
Now $8093\equiv 2 \mod 3,$ and $2$ is not the residue$\mod 3$ of a square, so $p\ne 3.$
And $8093\equiv 3 \mod 5,$ and $3$ is not the residue$\mod 5$ of a square, so $p\ne 5.$
Since $8093\equiv 1\equiv 1^2 \mod 7,$ we have $n=1\implies (2n-1)^2= 1\equiv 8093 \mod 7\implies n^2-n-2023\equiv 0\mod 7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Is my proof of $x^n-y^n = \ldots$ correct? I am solving problems from Spivak's calculus book's 1st chapter. Basically Spivak wants me to prove this:
$$x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$$
Question 1: Is my proof correct?
Question 2: How to make it perfect? Any advise?
My proof:
I guess Spivak wants to say:
$$
x^n-y^n = (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
$$
Let:
$$
f(x,y,n) = \left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
$$
Then we can say (is this sub-proof perfect?):
$$\begin{split}
x^{n+1}-y^{n+1} &= (x-y)\left(\sum_{i=1}^{i=n+1} x^{(n+1)-i}y^{i-1}\right)\\
&= (x-y)\left(
\begin{split}
&x^{(n+1)-(n+1)} y^{(n+1)-1}\\
&+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
\end{split}
\right)\\
&= (x-y)\left(
x^{0} y^{n}
+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
\right)\\
&= (x-y)\left(
y^{n}
+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1}
\right)\\
&= (x-y)\left(
y^{n}
+ x\sum_{i=1}^{i=n} x^{n-i}y^{i-1}
\right)\\
&= (x-y)\left(
y^{n}
+ xf(x,y,n)
\right)\\
\end{split}$$
We can also rewrite what Spivak wants into:
$$
x^n-y^n = (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)
$$
We've already proven in [spivak_calc_probs.1.1.2]:
$$\begin{split}
x^2-y^2 &= (x-y)(x+y)\\
&= (x-y)\left(\sum_{i=1}^{i=2} x^{2-i}y^{i-1}\right)\\
&= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\
\end{split}$$
Then, by induction, we prove that:
$$\begin{split}
x^2-y^2 &= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\
x^3-y^3 &= (x-y)\Big(y^{3-1} + xf(x,y,3-1)\Big)\\
\vdots\\
x^n-y^n &= (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)\\
\end{split}$$
And since $(x-y)(y^{n-1} + xf(x,y,n-1))$ is only a rewrite of what Spivak
wants, i.e. $(x-y)(x^{n-1}$ $+$ $x^{n-2}y$ $+$ $\ldots$ $+$ $xy^{n-2})$,
therefore Q.E.D already.
alternative proof:
Using the distributive axiom:
$$\begin{split}
& (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(x^{n-n}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(x^{0}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-y\left(y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\
&= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x\left(x^{n-1}y^{1-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x\left(x^{n-1}y^{0} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x\left(x^{n-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=2}^{i=n} x^{n-i+1}y^{i-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-(i+1)+1}y^{(i+1)-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i-1+1}y^{i+1-1}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)
-\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\
\end{split}$$
Then using additive associative axiom:
$$\begin{split}
& x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\
&= x^{n}
- y^{n}
+ \left(\sum_{i=1}^{i=n-1} x^{n-i}y^{i}
- \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\
&= x^{n}
- y^{n}
+ \left(0\right)\\
&= x^{n}
- y^{n}
\end{split}$$
| A proof by induction (although in this case it's not the simplest approach, as pointed out in comments) would proceed as follows:
1) Base case - show the the expansion is correct for $n=1$. In this case the sum $\sum_{i=1}^{i=n} x^{n-i}y^{i-1}$ has only one term which is $x^0y^0=1$ so this is straightforward.
2) Assume the expansion is correct for a specifc value of $n$, say $n=k$.
3) Find an expression that relates the $n=k+1$ case to the $n=k$ case. You could try:
$\quad x^{k+1}-y^{k+1} = x(x^k-y^k) + (x-y)y^k$
4) Use (2) to expand your expression from (3):
$\quad x^{k+1}-y^{k+1} = x(x-y)\sum_{i=1}^{i=k} x^{k-i}y^{i-1} + (x-y)y^k\\ \quad=(x-y)\left( x\sum_{i=1}^{i=k} x^{k-i}y^{i-1} + y^k\right) \\\quad =(x-y)\left(\sum_{i=1}^{i=k+1} x^{k+1-i}y^{i-1}\right)$
So now you have shown that if the expansion is true for $n=k$ then it is also true for $n=k+1$. Together with your base case $n=1$, this proves by induction that the expansion is true for all $n \in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find mistake in Fourier series calculation.
The function $f$ is $3$-periodic and $$f(t)=\left\{
\begin{array}{ll}
t,\quad &0\leq t\leq1\\
1,\quad &1<t<2\\
3-t,\quad &2\leq t \leq 3
\end{array}
\right. $$
Expand $f$ as a (real) Fourier series.
Attempt: I simply use the formulae for $a_0,a_n$ and $b_n$. The value of $L$ is $2L=3\Leftrightarrow L=3/2.$ this gives
$$a_0=\frac1{2L}\int\limits_{-L}^Lf(t)\,\mathrm dt=\frac13\cdot2=\frac23,$$
which also is easy to see by drawing $f(t)$. Now
\begin{align}
a_n &=\frac23\left[\,\int\limits_{-3/2}^{-1}\cos\left(\frac{2\pi nt}3\right)\,\mathrm dt-\int\limits_{-1}^0t\cos\left(\frac{2\pi nt}3\right)\,\mathrm dt+\int\limits_0^1t\cos\left(\frac{2\pi nt}3\right)\,\mathrm dt+\int\limits_1^{3/2}\cos\left(\frac{2\pi nt}3\right)\,\mathrm dt\,\right]\\
&=\frac{\sin(\pi n)-\sin\left(\frac{2\pi n}{3}\right)}{\pi n},
\end{align}
since the two middle integrals cancel and the two outer are the same. For $b_n$ I have, with identical integrals as above but replacing $\cos$ by $\sin$, that $b_n=0$. This means that
\begin{align}
f(t)&=\frac{2}{3}-\sum_{n=1}^{\infty}\frac{\sin(\pi n)-\sin\left(\frac{2\pi n}{3}\right)}{\pi n}\cos\left(\frac{2\pi nt}{3}\right)\\
&=\frac{2}{3}-\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\sin{\left(\frac{2\pi n}{3}\right)}}{n}\cos\left(\frac{2\pi nt}{3}\right),
\end{align}
since $\sin(\pi n)=0 \ \forall \ n\in\mathbb{Z}.$ However the book wants the answer to be
$$f(t)=\frac{2}{3}-\frac{3}{\pi^2}\sum_{n=1}^{\infty}\frac{1-\cos\left(\frac{2\pi n}{3}\right)}{n^2}\cos\left(\frac{2\pi nt}{3}\right).$$
Can someone help me find the mistake?
|
since the two middle integrals cancel...
They don't. Our $f(t)=\begin{cases}1&-\frac32 \le t\le -1\\-t&-1\le t\le 0\\t&0\le t\le 1\\1&1\le t\le \frac32\end{cases}$ is even. Integrate it against an odd function like $\sin\left(\frac{2\pi nt}{3}\right)$ and we'll get zero by odd symmetry. Integrate it against something even like $\cos\left(\frac{2\pi nt}{3}\right)$, and the parts for negative $t$ will mirror the parts for positive $t$ and reinforce each other, doubling the total value. By even symmetry,
\begin{align*}a_n &= \frac23\int_{-\frac32}^{\frac32}f(t)\cos\left(\frac{2\pi nt}{3}\right)\,dt = \frac43\int_{0}^{\frac32}f(t)\cos\left(\frac{2\pi nt}{3}\right)\,dt\\
a_n &= \frac43\int_0^1 t\cos\left(\frac{2\pi nt}{3}\right)\,dt+\frac43\int_1^{\frac32}\cos\left(\frac{2\pi nt}{3}\right)\,dt\\
a_n &= \frac43\left[\frac{3t}{2\pi n}\sin\left(\frac{2\pi nt}{3}\right)+\frac{9}{4\pi^2 n^2}\cos\left(\frac{2\pi nt}{3}\right)\right]_0^1 +\frac43\left[\frac{3}{2\pi n}\sin\left(\frac{2\pi nt}{3}\right)\right]_1^{\frac32}\\
a_n &= \frac2{\pi n}\sin\left(\frac{2\pi n}{3}\right) +\frac{3}{\pi^2 n^2}\left(1-\cos\left(\frac{2\pi n}{3}\right)\right) + \frac2{\pi n}\left(\sin\pi n - \sin\left(\frac{2\pi n}{3}\right)\right)\\
a_n &= \frac{3}{\pi^2 n^2}+\frac2{\pi n}(\sin(\pi n))-\frac{3}{\pi n^2}\cos\left(\frac{2\pi n}{3}\right)\\
a_n &= \begin{cases}\frac23& n=0\\ 0& 3\mid n,n>0\\ \frac{9}{2\pi^2 n^2}& 3\nmid n\end{cases}\end{align*}
In that last step, we use that $\sin(\pi n) = 0$ and $\cos\left(\frac{2\pi n}{3}\right)$ has a cycle of values $1,-\frac12,-\frac12$; subtract that from $1$ and it becomes $0,\frac32,\frac32$. Also, we incorporate the previously calculated (by a different method) $a_0$. The book's version of the answer applies the $\sin(\pi n)=0$ fact, but doesn't do anything with the $\cos\left(\frac{2\pi n}{3}\right)$ term.
So there it is. Incorporate the middle part of the integral as it should be for this even function, and we get the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Expected value. Basic problem We have pawn on a board and we roll dice maximally two times. When we roll six we can roll dice again. We move pawn as many times as we threw dots. What's the expected value of move on the board.
Am i thinking correctly ?
$$E(X)= 1 \cdot \frac16 + 2 \cdot \frac16 + 3 \cdot \frac16 + 4 \cdot \frac16 + 5 \cdot \frac16 + 7 \cdot \frac1{36} + 8 \cdot \frac1{36} \\
\qquad + 9 \cdot \frac1{36} + 10 \cdot \frac1{36} + 11 \cdot \frac1{36} + 12 \cdot \frac1{36} $$
| Yes
Alternatively the expectation for the first die is $3.5$ and for the second (which may not be thrown) is $\frac16 \times {3.5}$ and you can add these together
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that there is at least $1$ solution and finitely many solutions.
Fix an integer $n$ such that $n \geq 2$. Consider the system of equations
$$\begin{cases}
a+b-c=n \\
a^2+b^2-c^2=n
\end{cases}$$
in three variables $a,b,c\in\mathbb{Z}$.
Prove that it has at least $1$ solution and finitely many solutions.
For example, for $n = 2$, it seems that the only solutions are $\left(1,1,0\right)$ and $\left(3,3,4\right)$, as found by the following python code:
def find(n, N):
for a in range(-N, N+1):
for b in range(-N, N+1):
c = a + b - n
if a**2 + b**2 - c**2 == n:
print a, b, c
find(2, 25)
This question is probably one of the hardest I have attempted yet. I have tried to solve it but couldn't. Is there any way I could get help here?
| For even $n$,
$$a=2n-1, b=\frac{3n}2, c=\frac{5n}2-1$$
is a solution, while for odd $n$,
$$a=2n, \space b=\frac{3n-1}2, \space c=\frac{5n-1}2$$
is one.
To prove that there are only finitely many solutions for a given $n\ge2$, we observe that
$$b-c=n-a, \space b^2-c^2=n-a^2$$
and hence $n-a|n-a^2$. Since $n-a|n^2-a^2$ it follows that $n-a|n^2-n$. Since $n \ge 2$, we have $n^2-n > 0$ and thus it has only finitely many integer devisors and thus only finitely many $a$ are possible. It remains to be shown that for each such $a$, only finitely many $b,c$ can exist.
$a=n$ is impossible, as that would imply $0|n^2-n$, which is impossible. So we get
$$b+c = \frac{b^2-c^2}{b-c} = \frac{n-a^2}{n-a}$$
Together with $b-c=n-a$ this gives exactly one solution $(b,c)$ in rationals, so at most one for integers, which concludes the proof. $\blacksquare$
These equations for $b-c$ and $b+c$ are also the way I found the special solutions given above. $n-a|n^2-n$ implies looking for divisors of $n^2-n$, where $n$ and $n-1$ obviously stand out. A little experimentation with signs lead to considering $a-n=n$ and $a-n=n-1$, and the equations for $b-c$ and $b+c$ then lead straightforward to the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089667",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Showing that $f(4x) + 4f(2x) =8\cos^4(x)-3$, where $f(x) = \cos x$
It is given that
$$f(x)= \sin(x+30^\circ) + \cos(x+60^\circ)$$
A) Show that $$f(x)= \cos(x)$$
B) Hence, show that
$$f(4x) + 4f(2x) =8\cos^4(x)-3$$
I managed to prove $f(x)$ equals $\cos(x)$, but after that I'm stumped.
| $$f(x)=\sin(30^{\circ}+x)+\sin(30^{\circ}-x)=2\sin30^{\circ}\cos{x}=\cos{x}.$$
Thus, $$f(4x)+4f(2x)=\cos4x+4\cos2x=8\cos^4x-8\cos^2x+1+8\cos^2x-4=8\cos^4x-3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x} \qquad $$
Thank all of you for your answers.
| If you already know the concept of the first derivative you can do the following:
$$\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}= \frac{2\sqrt{1-x} - 2}{x} - \frac{\sqrt[3]{8-x}-2}{x} \stackrel{x\to 0}{\longrightarrow} \left.\frac{d}{dx}\left(2\sqrt{1-x}- \sqrt[3]{8-x}\right) \right|_{x=0}$$
Differentiating and setting $x= 0$ gives $-\frac{11}{12}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Multivariable critical points I have the following function:
$\ f(x,y) = (x^3-y^2)(x-1) $
Expanding the brackets gives:
$\ f(x,y) = x^4 - x^3 - xy^2 + y^2 $
I understand to find the first derivative of both variables and set to zero:
$\ f_x ==> 4x^3 - 3x^2 - y^2 = 0.....[1] $
$\ f_y ==> y - 2xy = 0.....[2] $
$\ y = 2xy.....[3] $
Substituting [3] into [1]:
$\ 4x^3 - 3x^2 - 4x^2y^2 = 0 $
$\ x^2(4x - 3 - y^2) = 0$
I am stuck at this point and can not factorise further to find the critical points. Any hints or clues would be helpful. Thank you.
| The second partial derivative is $-2xy+2y$ which gives $-xy+y=y(-x+1)=0$, $y=0$ or $x=1$. By replacing $y=0$ you have $4x^3-3x^2=0$ or $x=1$, $4-3-y^2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The definite integrals of $e^{-x} \sin^n{(x)}$ and $e^{-x} \cos^n{(x)}$ between $0$ and $\infty$ For the integral
$$\begin{align}S_n =\int_{0}^{\infty} e^{-x} \sin^n{(x)}dx\end{align}$$
I have found a formula for $S_n$:
$$\begin{align} S_n=2^{\frac{(-1)^n-1}{2}} n!! (n-1)!! \prod_{2\le k\le n}^{{k\equiv n \mod{2}}} \frac{1}{k^2+1} \end{align}$$
By solving the following recurrence relation:
$$\begin{align} S_n=\frac{n(n-1)}{n^2+1}S_{n-2} \end{align}$$
But I cannot find a similar general formula for the integral
$$C_n=\int_{0}^{\infty} e^{-x} \cos^n{(x)}dx$$
Given that I have found the following recurrence relation for $C_n$:
$$\begin{align} C_n = \frac{1}{n^2+1}+ \frac{n(n-1)}{n^2+1}C_{n-2}\end{align}$$
Is it possible to find a general formula for $C_n$ in a similar format to that of $S_n$?
| I'm not sure this will help, but using the fact that $\cos^n(x) = \displaystyle \frac{(e^{ix}+e^{-ix})^n}{2^n}$, we have
$$\begin{align*}
\int_0^\infty e^{-x}\cos^n(x)\mathrm d x &= \frac1 {2^n}\int_0^\infty e^{-x}\sum_{k=0}^n \binom n k e^{i(2k-n)x}\mathrm d x\\
&=\frac1 {2^n}\sum_{k=0}^n \binom n k\frac{1}{1+i(n-2k)}\\&=
\frac1 {2^{n+1}}\sum_{k=0}^n \binom n k\left(\frac{1}{1+i(n-2k)}+\frac{1}{1-i(n-2k)}\right)\\
&=\frac 1 {2^{n}}\sum_{k=0}^n \binom n k\frac{1}{1^2+(n-2k)^2}.
\end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$.
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
| Visual proof of
$$\sum_{k=1}^{n}k^2=\sum_{k=1}^{n} (2k-1)(n+1-k).$$
Take a look at this picture for $n=5$.
$$1+2^2+3^2+4^2+5^2=\underbrace{1\cdot 5}_{k=1}+\underbrace{3\cdot 4}_{k=2} +\underbrace{5\cdot 3}_{k=3}+ \underbrace{7\cdot 2}_{k=4}+ \underbrace{9\cdot 1}_{k=5}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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find the coefficient of the term when the expression is expanded. $a^2x^3$: $(a + x + c)^2(a + x + d)^3$
I am considering to list all cases of $a^nx^m$ from both expansion that sum of n=2, sum of m=3.
Like in the first case: $a^0x^0$ from first expansion and $a^2x^3$ from the second expansion.
Then calculate the coefficient of $a^2x^3$ in each case by trinomial expansion and sum them up. But this way is way to complex. I am wondering if there is any idea to make it easier?
Thank you!
| It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can focus at the terms which contribute to $[a^2x^3]$ and can skip other terms.
We obtain
\begin{align*}
\color{blue}{[a^2x^3]}&\color{blue}{(a+x+c)^2(a+x+d)^3}\\
&=[a^2x^3](a+(x+c))^2(a+(x+d))^3\\
&=[a^2x^3]\left(a^2+2a(x+c)+(x+c)^2\right)\\
&\qquad\qquad\quad\cdot\left(a^3+3a^2(x+d)+3a(x+d)^2+(x+d)^3\right)\\
&=[a^2x^3]\left(a^2(x+d)^3+2a(x+c)3a(x+d)^2+(x+c)^23a^2(x+d)\right)\tag{1}\\
&=[a^2x^3]\left(a^2x^3+6a^2x^3+3a^2x^3\right)\tag{2}\\
&\,\,\color{blue}{=10}
\end{align*}
Comment:
*
*In (1) we multiply out leaving only terms which contribute to $[a^2x^3]$.
*In (2) we multiply out once more again leaving only terms which contribute to $[a^2x^3]$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^1 \frac{\mathrm dx}{(x^2+ax+1)^{n+1}}$ with real methods I would like to know if there are any (preferably easier) methods of evaluating
$$Q_n(a)=\int_0^1 \frac{\mathrm dx}{(x^2+ax+1)^{n+1}}$$
With real methods. Here's the way I did it.
Complete the square:
$$Q_n(a)=4^{n+1}\int_0^1\frac{\mathrm dx}{((2x+a)^2+4-a^2)^{n+1}}$$
Then $u=2x+a$ gives
$$Q_n(a)=2^{2n+1}\int_a^{a+2}\frac{\mathrm du}{(u^2+4-a^2)^{n+1}}$$
Then consider the indefinite integral
$$F_n^{w}(x)=\int\frac{\mathrm dx}{(x^2+w)^{n+1}}$$
Integration by parts yields the recurrence relation (for $n\in\Bbb Z\geq 1$)
$$F_n^{w}(x)=\frac{x}{2wn(x^2+w)^n}+\frac{2n-1}{2wn}F_{n-1}^{w}(x)$$
With the base case
$$F_0^{w}(x)=\frac1{\sqrt{w}}\arctan\frac{x}{\sqrt{w}}$$
And since this recurrence is in the form $$f_n=\alpha_n+\beta_nf_{n-1}$$
the solution to which is $$f_n=f_0\prod_{k=1}^{n}\beta_k+\sum_{k=0}^{n-1}\alpha_{n-k}\prod_{j=1}^{k}\beta_{n-j+1}$$
We have (I omit the simplification steps)
$$F_n^{w}(x)=\frac{{2n\choose n}}{2^{2n}w^{n+1/2}}\arctan\frac{x}{\sqrt{w}}+S_n^w(x)$$
With $$S_n^w(x)=\frac{x}{2w}\sum_{r=0}^{n-1}\frac{(x^2+w)^{r-n}}{(2w)^r}R_r^{(n)}$$
and $$R_r^{(n)}=\frac1{n-r}\prod_{j=1}^{r}\frac{2n-2j+1}{n-j+1}$$
So we have that
$$Q_n(a)=2^{2n+1}\left[F_n^{4-a^2}(a+2)-F_n^{4-a^2}(a)\right]$$
Which is, after some simplification,
$$\begin{align}
Q_n(a)=&\frac{2{2n\choose n}}{(4-a^2)^{n+1/2}}\left[\arctan\sqrt{\frac{2+a}{2-a}}-\arctan\frac{a}{\sqrt{4-a^2}}\right]\\
&+\frac1{4-a^2}\sum_{r=0}^{n-1}\frac{2^rR_r^{(n)}}{(4-a^2)^r}\left[(2+a)^{r-n+1}-a\right]
\end{align}$$
| $$Q_n(a)=\int_0^1\frac{dx}{(x^2+ax+1)^{n+1}}$$
and since $x^2+ax+1=\left(x+\frac a2\right)^2+\left(1-\frac{a^2}{4}\right)$. If we let $u=x+\frac a2$ and $\alpha^2=1-\frac{a^2}{4}$ we get similar to what you got:
$$Q_n(a)=\int_{a/2}^{1+a/2}\frac{du}{(u^2+\alpha^2)^{n+1}}$$
now by observing that: $u^2+\alpha^2=\alpha^2\left(\left[\frac u\alpha\right]^2+1\right)$ and letting $v=\frac u\alpha$ we obtain:
$$Q_n(a)=\int_{a/2}^{1+a/2}\frac{\alpha^{-(2n+1)}dv}{(v^2+1)^{n+1}}$$
now by letting $v=\tan\omega$ and $\beta=\arctan(1/2),\gamma=\arctan(1+a/2)$ we get:
$$Q_n(a)=\alpha^{-(2n+1)}\int_\beta^\gamma\frac{\sec^2\omega}{(\sec^2\omega)^{n+1}}d\omega=\alpha^{-(2n+1)}\int_\beta^\gamma\cos^{2n}(\omega)d\omega$$
Now by making the substitution $\sigma=\cos\omega$ we get:
$$Q_n(a)=-\alpha^{-(2n+1)}\int_{\cos\beta}^{\cos\gamma}\sigma^{2n}(1-\sigma^2)^{-1}d\sigma$$
Where we can now let $\epsilon=\sigma^2$, giving:
$$Q_n(a)=-\frac{\alpha^{-(2n+1)}}{2}\int_{\cos^2\beta}^{\cos^2\gamma}\epsilon^{\frac{2n-1}{2}}(1-\epsilon)^{-1}d\epsilon$$
Now we can use the standard function for the incomplete beta function which is:
$$B(x;a,b)=\int_0^xt^{a-1}(1-t)^{b-1}dt$$
And so we can now say that:
$$Q_n(a)=-\frac{\alpha^{-(2n+1)}}{2}\left[B\left(\cos^2\gamma;\frac{2n+1}{2},0\right)-B\left(\cos^2\beta;\frac{2n+1}{2},0\right)\right]$$
We can now go back and calculate that:
$$\cos^2\gamma=\frac{4}{(2+a)^2+4}$$
$$\cos^2\beta=\frac{4}{a^2+4}$$
$$-\frac{\alpha^{-(2n+1)}}{2}=-\frac{2^{4n+1}}{(4-a^2)^{-(2n+1)}}$$
Overall if we combine this together we get:
$$Q_n(a)=-\frac{2^{4n+1}}{(4-a^2)^{-(2n+1)}}\left[B\left(\frac{4}{(2+a)^2+4};\frac{2n+1}{2},0\right)-B\left(\frac{4}{a^2+4};\frac{2n+1}{2},0\right)\right]$$
I believe this is right but correct me if there are any mistakes
| {
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"url": "https://math.stackexchange.com/questions/3109642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Determine the range of a function Find the range of the following function :
$$y= (3\cos{2x} -\sin{2x} -1)^2 -8$$
First solution:
$$ (3\cos{2x} -\sin{2x} -1)^2= (\sqrt{10} \cos{2x+ \theta})^2$$
$$-\sqrt{10} =<(\sqrt{10} \cos{(2x+ \theta}))=< \sqrt{10}$$
$$-1-\sqrt{10} =<(\sqrt{10} \cos{(2x+ \theta}))-1 =< \sqrt{10} -1$$
$$-8=< \left((\sqrt{10} \cos{(2x+ \theta}))-1\right)^2 -8=< 3-2\sqrt{10}$$
Second solution:
$$ (3\cos{2x} -\sin{2x} -1)^2=(2\sin{2x}-3\cos{2x} +1)^2 = (\sqrt{10} \sin\left(2x -\theta\right))^2 $$
Similarly as the first we get
$$-8 =< y =< 3+2\sqrt{10}$$
Which solution is correct ? and whats the mistake ?
| $$3\cos2x-2\sin2x=\sqrt{3^2+1^2}\cos\left(2x+\arccos\dfrac3{\sqrt{10}}\right)$$
$\implies-\sqrt{10}-1\le3\cos2x-\sin2x-1\le\sqrt{10}-1$
$$3\cos2x-\sin2x-1\le\sqrt{10}-1\implies(3\cos2x-\sin2x-1)^2-8\le(\sqrt{10}-1)^2-8$$
$$3\cos2x-\sin2x-1\ge-\sqrt{10}-1\implies(3\cos2x-\sin2x-1)^2-8\le(-\sqrt{10}-1)^2-8$$
Take the union of the two ranges
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find if terms are terms of the same arithmetic progression
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Hint: Yes. Try $\frac{1}{30}$ as a difference.
I was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find:
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$.
| Let the terms be $a_1=\frac12, a_m=\frac13,a_n=\frac15$. Then:
$$d=\frac{a_m-a_1}{m-1}=\frac{a_n-a_1}{n-1} \Rightarrow \\
d=\frac{\frac13-\frac12}{m-1}=\frac{\frac15-\frac12}{n-1} \Rightarrow \\
-\frac16(n-1)=-\frac3{10}(m-1) \Rightarrow \\
n=\frac{9}{5}(m-1)$$
As $m,n\in \mathbb N_+$ and $n>m$, we get $m=5k+1,n=9k,d=-\frac1{30k},k\in\mathbb N_+$:
$$\begin{array}{c|c|c|c|c}
m&6&11&16&\cdots\\
\hline
n&9&18&27&\cdots\\
\hline
d&-\frac1{30}&-\frac1{60}&-\frac1{90}&\cdots
\end{array}$$
Note: It was considered a decreasing AP. It can also be considered an increasing AP.
| {
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"source": "stackexchange",
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Evaluate $\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$ Evaluate $$\int_{y=-\infty}^1\int_{x=0}^\infty 4xy\sqrt{x^2+y^2}\mathrm dx\mathrm dy$$
I tried to solve this for hours without any success. I tried substitution method of $u = x^2+y^2$ but doesn't seem to work.
Here is my attempt:
$$\int_{-\infty}^1\int_0^\infty (16x^2y^2)^{\frac{1}{2}}(x^2 + y^2)^\frac{1}{2} \; \mathrm{d}x \mathrm{d}{y} = \int_{-\infty}^1 \int_0^\infty (16x^3y^2 + 16x^2 y^3)^{\frac{1}{2}} \; \mathrm{d}x\mathrm{d}{y}.$$
Let
\begin{align}
u &= 16x^3 y^2 + 16x^2 y^3 \\
\mathrm{d}u &= (48x^2y^2 + 32xy^3) \; \mathrm{d}{x} \\
\mathrm{d}u &= 16xy^2(3x + 2y) \; \mathrm{d}{x} \\
\mathrm{d}u &= 4xy(12xy + 8y^2) \; \mathrm{d}{x} \\
4xy \; \mathrm{d}{x} &= \frac{\mathrm{d}{u}}{12xy + 8y^2}.
\end{align}
$$\int_{-\infty}^1 \int_0^\infty 4xy \sqrt{x^2 + y^2} \; \mathrm{d}x \mathrm{d}y = \int_{-\infty}^1 \int_0^\infty \frac{\mathrm{d}u}{(12xy + 8y^2)}.$$
Attaching image for reference
EDITS
Original Post asked integrate: 4xy sqrt(x^2+y^2) where 0<x, y<1, and added this image. .
Later edits misread 0<x,y<1 as if 0<x<1and 0<y<1
| You first integrate with respect to $x$ while treating $y$ as a constant. Then, you integrate the result of that with respect to $y$:
$$
\int_{0}^{1}\int_{0}^{1} 4xy\sqrt{x^2+y^2}\,dx\,dy=
\int_{0}^{1}\left(\frac{4}{2}y\int_{0}^{1}\sqrt{x^2+y^2}\frac{d}{dx}(x^2+y^2)\,dx\right)\,dy=
\int_{0}^{1}\left(2y\int_{0}^{1}\sqrt{x^2+y^2}\,d(x^2+y^2)\right)\,dy=\\
\int_{0}^{1}\left(2y\frac{2\sqrt{(x^2+y^2)^3}}{3}\bigg|_{0}^{1}\right)\,dy=
\int_{0}^{1}\left[2y\left(\frac{2\sqrt{(1^2+y^2)^3}}{3}-\frac{2\sqrt{(0^2+y^2)^3}}{3}\right)\right]\,dy=\\
\frac{4}{3}\int_{0}^{1}\left(y\sqrt{(1+y^2)^3}-y^4\right)\,dy=\\
\frac{4}{3}\int_{0}^{1}y\sqrt{(1+y^2)^3}\,dy-\frac{4}{3}\int_{0}^{1}y^4\,dy=\\
\frac{4/3}{2}\int_{0}^{1}\sqrt{(1+y^2)^3}\frac{d}{dy}(1+y^2)\,dy-\frac{4}{3}\frac{y^5}{5}\bigg|_{0}^{1}=\\
\frac{2}{3}\int_{0}^{1}(1+y^2)^{3/2}\,d(1+y^2)-\frac{4}{3}\left(\frac{1^5}{5}-\frac{0^5}{5}\right)=\\
\frac{2}{3}\frac{2(1+y^2)^{5/2}}{5}\bigg|_{0}^{1}-\frac{4}{15}=\\
\frac{2}{3}\left(\frac{2(1+1^2)^{5/2}}{5}-\frac{2(1+0^2)^{5/2}}{5}\right)-\frac{4}{15}=\\
\frac{2}{3}\left(\frac{2\sqrt{2^5}}{5}-\frac{2}{5}\right)-\frac{4}{15}=\\
\frac{4(4\sqrt{2}-1)}{15}-\frac{4}{15}=\frac{16\sqrt{2}-4-4}{15}=\frac{16(\sqrt{2}-8)}{15}=\frac{8(2\sqrt{2}-1)}{15}.
$$
Wolfram Alpha check.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.