Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\dfrac{\sin{5x}}{\sin{x}}\in\left({-\dfrac54,5}\right)$ Prove that $\dfrac{\sin{5x}}{\sin{x}}\in\left({-\dfrac54,5}\right)$ for any $x\in\mathbb{R}\setminus{k\pi}$ where $k\in\mathbb{Z}$.
I wrote $\sin5x$ as $5\cos^4x\sin{x}-10\cos^2 x\sin^3x+\sin^5x$ and now I need to prove that $5\cos^4x-10\cos^2 x\sin^2x+\sin^4x\in\left({-\dfrac54,5}\right)$, but it seems to be very difficult. Is there any easy way?
| For $\sin x\ne0,$
\begin{align}
\frac{\sin5x}{\sin x} &= 5\cos^4x-10\cos^2x(1-\cos^2x)+(1-\cos^2x)^2\\
&= 16\cos^4x-12\cos^2x+1\\
&= \left(4\cos^2x-\frac32\right)^2+1-\frac94
\end{align}
Since anything squared is $\ge0$, we have:
\begin{align}
0 &\le \left(4\cos^2x-\frac32\right)^2\\
1-\frac94 &\le \left(4\cos^2x-\frac32\right)^2+1-\frac94\\
-\frac54 &\le 16\cos^4x-12\cos^2x+1
\end{align}
Also as $\cos x\le1,$ $$16\cos^4x-12\cos^2x+1\le16-12+1=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding the generating function of a series with a binomial coefficient and a exponential coefficient So I am given this series
$$2^8, 2^7 \binom{8}{1}, 2^6 \binom{8}{2}, 2^5 \binom{8}{3}, 2^4 \binom{8}{4}, 2^3 \binom{8}{5}, 2^2 \binom{8}{6}, 2^1 \binom{8}{7}, \binom{8}{8}, 0, 0, 0, 0, ...$$
which I converted to the summation
$$\sum_{n=0}^\infty \frac{(2^8 \cdot \binom{8}{n} \cdot x^n)}{2^n} $$
The problem is to find the closed form generating function for this series
I know that the closed form generating function for
$$\sum_{n=0}^\infty \binom{8}{n}x^n = (1+x)^8$$ and
$$\sum_{n=0}^\infty \frac{(2^8 \cdot x^n)}{2^n} = \frac{2^8}{1-x/2}$$
I just can't find anything in my book or online on how to deal with the combination of the two
| Hint:
$$ \sum \frac{2^8 \binom{8}{n} x^n}{2^n} = \sum 2^8 \binom{8}{n} \left(\frac x2\right)^n = \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Eisenstein integers and applications to Diophantine equations Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?
I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.
| Note that $N(a+b\omega)=a^2-ab+b^2$ is the sum of squares, because
$$
a^2-ab+b^2=\frac{1}{4}((2a-b)^2+3b^2).
$$
Hence we have to solve the equation $(2a-b)^2+3b^2=4\cdot 7\cdot 13\cdot 19=6916$, which is
straightforward, since we only have to test a few integers $a,b \in \mathbb{N}$. In particular, $3b^2\le 6916$, so that $b<49$. Similarly, $(2a-b)^2\le 6916$ then gives $a< 66$.
We find, that the integer solutions with $b>a>0$ are given by
$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$ It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed up. And I found that the rational solutions of $w^2+x^2+y^2=1$ is
$$w = \frac{2s}{1+s^2+t^2}$$
$$x = \frac{2t}{1+s^2+t^2}$$
$$y = \frac{1-s^2 - t^2}{1+s^2+t^2}$$
where $s$ and $t$ are both rational numbers. From this I found out that
$$(2mnq^2)^2 + (2pqn^2)^2 + (n^2q^2 - m^2q^2-n^2p^2)^2 = (n^2q^2 + m^2q^2+n^2p^2)^2.$$
But it helps little because I cannot prove that the solutions have to be in such form.
Can any one help?
| Aspects of your question have been covered here about Lebesgue's Identity,
lebesgue's identity
and a general approach to $x_1^2+x_2^2+\dots+x_n^2 = z^2$,
Diophatine equation $x^2+y^2+z^2=t^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$.
Is there an easy way to obtain this closed-form?
Also, if there are any good references for binomial coefficient identities like these I'd appreciate it. I searched some but did not find any similar to this.
| Generating functions to the rescue!
\begin{align*}
\sum_{\substack{n\ge 0 \\ n \text{ even}}} x^n \sum_{i=0}^{n/2} \binom{n-i}{i}2^i &= \sum_{i=0}^\infty 2^i \sum_{\substack{n\ge 2i \\ n \text{ even}}} \binom{n-i}{i} x^n \\
&= \sum_{i=0}^\infty 2^i \sum_{\substack{m\ge 0 \\ m \text{ even}}} \binom{m+i}{i} x^{m+2i} \\
&= \sum_{i=0}^\infty 2^i x^{2i} \sum_{m\ge0} \frac{1+(-1)^m}2 \binom{m+i}{i} x^m \\
&= \frac12 \sum_{i=0}^\infty (2x^2)^i \sum_{m\ge0} \binom{m+i}{i} \big( x^m + (-x)^m \big).
\end{align*}
Using the known series $\sum_{m\ge0} \binom{m+i}{i} x^m = (1-x)^{-(i+1)}$, we get
\begin{align*}
\sum_{\substack{n\ge 0 \\ n \text{ even}}} x^n \sum_{i=0}^{n/2} \binom{n-i}{i}2^i &= \frac12 \sum_{i=0}^\infty (2x^2)^i \big( (1-x)^{-(i+1)} + (1+x)^{-(i+1)} \big) \\
&= \frac1{2(1-x)} \sum_{i=0}^\infty \bigg( \frac{2x^2}{1-x} \bigg)^i + \frac1{2(1+x)} \sum_{i=0}^\infty \bigg( \frac{2x^2}{1+x} \bigg)^i \\
&= \frac1{2(1-x)} \frac1{1-2x^2/(1-x)} + \frac1{2(1+x)} \frac1{1-2x^2/(1+x)} \\
&= \frac{1-2x^2}{1-5x^2+4x^4} \\
&= \frac{2}{3 (1-4x^2)}+\frac{1}{3(1-x^2)} \\
&= \sum_{m=0}^\infty \bigg( \frac23 (4x^2)^m + \frac13 (x^2)^m \bigg) \\
&= \sum_{\substack{n\ge0 \\ n\text{ even}}} \frac{2^{n+1} + 1}3 x^n.
\end{align*}
Comparing coefficients of $x^n$ on both sides establishes the result (without knowing what the answer was in advance, for that matter). Note that $(-1)^n=1$ for $n$ even, so we really got the right answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to show distributivity in a ring, and what is wrong with my algebra? I am trying to show the following is a commutative ring with unity, however I am encountering a problem.
First, addition and multiplication are defined as:
$$a \oplus b=a+b-1$$$$a \odot b=ab-(a+b)+2$$
I have shown this "addition" is an abelian group. The identity is $1$ and the negative is $2-a$. Now, I am stuck on distributivity. I found that:
$$a \odot(b \oplus c)$$
$$= a \odot (b+c-1)$$
$$=(a\odot b) + (a \odot c) - (a \odot 1)$$
$$=ab-(a+b)+2+ac-(a+c)+2-(a1-(a+1)+2)$$
$$=ab-a-b+2+ac-a-c+2-a+a+1-2$$
$$=ab+ac-2a-b-c+3$$
Why is this not equal to:
$$= a \odot (b+c+(-1))$$
$$=(a\odot b) + (a \odot c) + (a \odot (-1))$$
$$=ab-(a+b)+2+ac-(a+c)+2+a(-1)-(a+(-1))+2)$$
$$=ab-a-b+2+ac-a-c+2-a-a+1+2$$
$$=ab+ac-4a-b-c+7$$
However, the first equation is equal to
$$(a \odot b) \oplus (a \odot c)$$
$$=(ab-(a+b)+2) \oplus (ac-(a+c)+2)$$
$$=ab-(a+b)+2+ac-(a+c)+2-1$$
$$=ab-a-b+2+ac-a-c+2-1$$
$$=ab+ac-2a-b-c+3$$
So, why is the first equation not equal to the second, but equal to the third? Which one is correct?
| We want to show that $\odot$ distributes over $\oplus$. However, we can NOT assume that $\odot$ distributes over $+$.
For the first part, we have:
\begin{align*}
a \odot(b \oplus c)
&= a \odot (b + c - 1) \\
&= (a)(b + c - 1) - ((a) + (b + c - 1)) + 2 \\
&= (ab + ac - a) - (a + b + c - 1) + 2 \\
&= ab + ac - 2a - b - c + 3 \\
\end{align*}
For the last part, we have:
\begin{align*}
(a \odot b) \oplus (a \odot c)
&= (ab - (a + b) + 2) \oplus (ac - (a + c) + 2) \\
&= (ab - (a + b) + 2) + (ac - (a + c) + 2) - 1 \\
&= ab + ac - 2a - b - c + 3 \\
\end{align*}
which matches.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
| using the lemma $$t+\frac{1}{t}\geq 2$$ for $t>0$ we get the inequality $$\tan(x)+\frac{1}{\tan(x)}\geq 2$$
your way is also ok.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
changing $r=1+2r\cos \theta$ to its cartesian equivalent My textbook says the polar equation, $r=1+2r\cos \theta$, its cartesian equivalent is $y^2-3x^2-4x-1=0.$ I understand that I get this if I square $r$; $r^2=x^2+y^2=(1+2x)^2.$ But don't I need to square root this back or something?? I thought the answer is supposed to be $ \sqrt{y^2-3x-4x-1}=0 $... I'm confused..
| You are right that the polar equation $r = (1+2r \cos \theta)$
and the cartesian equation $y^2 - 3x^2 - 4x -1 = 0$ are not equivalent.
The solutions of the first one are a subset of the solutions of the second.
$y^2 - 3x^2 - 4x -1 = 0$ is equivalent to
$r^2=(1+2r \cos \theta)^2$ and that is equivalent to
$$
r = (1+2r \cos \theta) \quad\textbf{ or }\quad r = -(1+2r \cos \theta) \quad .
$$
So your polar equation $r = (1+2r \cos \theta)$ describes one branch of the hyperbola
$y^2 - 3x^2 - 4x -1 = 0$ and $r = -(1+2r \cos \theta) $ is the other branch.
Here is an image from Wolfram Alpha:
Addendum: I did assume that $r$ must be positive (see below comments), and it
is easy to see that then $x \ge -1/3$ for the first equation and $x \le -1$ for
the second, so each polar equation produces one branch of the hyperbola.
Without this restriction, either equation produces the entire hyperbola,
because the transformation $r \to -r, \theta \to \theta + \pi$ transforms
one equation into the other.
In other words: If $x = r \cos \theta, y = r \sin \theta $ is a solution of
the cartesian equation $y^2 - 3x^2 - 4x -1 = 0$ then either
$(r, \theta)$ or $(-r, \theta + \pi)$ is a solution of the polar equation
$r = 1 + 2r \cos \theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$ How can I prove this equality?
$$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$
| Assuming convergence of both expressions, you have $A^2=x+A$ and $B^2=x-B$, so
$$
(A-B)(A+B)=A^2-B^2=(A+B).
$$
We conclude that $A-B=1$. Then $$A^2+B^2=2x+(A-B)=2x+1,$$ and therefore
$$
AB=\frac{1}{2}(A^2+B^2-(A-B)^2)=\frac{1}{2}(2x+1-1)=x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Intersection multiplicity of the curves I want to find the intersection multiplicity of the curves $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$ at the point $P=(0,0)$.
That`s what I have tried:
$f$ and $g$ have a common tangent, the $y=0$.
So $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$
$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg f(x, 0)=5$$
$$g(x, 0)=x^6-x^5 \Rightarrow r =\deg g(x, 0)=6$$
$$s \leq r$$
So we consider $h(x, y)=g(x, y)-x f(x, y)$
$$h(x, y)=x^6-x^5+y^2-x(x^5+x^4+y^2)=x^6-x^5+y^2-x^6-x^5-xy^2 \\ \Rightarrow h(x, y)=-2x^5+y^2-xy^2$$
$\deg h(x, 0)=5<r$
So $I(P, f\cap g)=I(P,f\cap h)$
$$f(x,0)=x^5+x^4\Rightarrow \deg f(x,0)=5=s$$
$$h(x,0)=-2x^5\Rightarrow \deg h(x,0)=5=p$$
They have a common tangent, $x=0$, so they don`t intersect traverrsally.
We consider the polynomial $$h_1(x,y)=h(x,y)+2f(x,y)=3y^2-xy^2+2x^4$$
$deg h_1(x,0)=4<s,p$
So, $I(P, f\cap h)=I(P,f\cap h_1)$
$f(x,0)=x^5+x^4 \Rightarrow \deg f(x,0)=5=s$
$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=t$
They have a common tangent $x=0$,so they don`t intersect traversally.
We consider the polynomial $h_2(x,y)=2f(x,y)-xh_1(x,y)=2x^4+2y^2-3xy^2+x^2y^2$
$\deg h_2(x,0)=4<s$
So $I(P, f\cap h_1)=I(P, h_1\cap h_2)$
$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=s$
$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$
They have a common tangent $x=0$, so they don`t intersect traversally.
We consider the polynomial $h_3(x,y)=h_1(x,y)-h_2(x,y)=y^2(1+2x-x^2)$
$\deg h_3(x,0)=0<s,m$
So $I(P,h_1\cap h_2)=I(P,h_2\cap h_3)$
$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$
$h_3(x,0)=0\Rightarrow \deg h_3(x,9)=9=n$
So $I(P,h_2\cap h_3)=I(P,h_2\cap y^2)+I(P,h_2\cap (1+2x-x^2))$
$I(P,h_2\cap y^2)=8$
$I(h_2\cap (1+2x-x^2))=0$
Therefore, $I(P, f \cap g)=8$.
Is it right? Do we find that $f$ and $h$ have a common tangent from $f(x,0)$ and $h(x,0)$ ?
| See here for the definining properties of the intersection property. These allow us to compute:
$~~~I_P(x^5+x^4+y^2,x^6-x^5+y^2)\\
=I_P(x^5+x^4+y^2,(x^6-x^5+y^2)-(x^5+x^4+y^2))\\
=I_P(x^5+x^4+y^2,x^4 (x^2-2x-1))\\
=I_P(x^5+x^4+y^2,x^4)+I_P(x^5+x^4+y^2,x^2-2x-1)\\
=I_P(x^5+x^4+y^2,x^4)\\
=I_P(y^2,x^4)\\
=2 \cdot 4 \cdot I_P(y,x)\\
=8$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$ My Work
$x = \tan\theta$
$dx = \sec^2\theta d\theta$
$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$
$\int \sec\theta \tan\theta ln(tan\theta)$
$u = \ln(\tan\theta)$
$du = \frac{\sec^2 \theta}{\tan \theta}$
$dv = \sec \theta \tan \theta d\theta$
$v = \sec \theta$
$\sec\theta\ln\tan\theta - \int \frac{\sec^3\theta}{\tan \theta}$
$\sec\theta\ln\tan\theta - \int \sec^2\theta \csc\theta d\theta$
I'm stuck after here. Parts doesn't look particularly appealing. I don't see an easy substitution. Brain is pretty tired at this point. Anyone know what to do?
| $$
\begin{aligned}
\int \frac{x \ln x}{\sqrt{x^{2}-1}} d x &=\int \ln x d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{\sqrt{x^{2}-1}}{x} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\
&=\sqrt{x^{2}-1} \ln x-\int \frac{x^{2}-1}{x^{2}} d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\int\left(1-\frac{1}{x^{2}}\right) d \sqrt{x^{2}-1} \\
&=\sqrt{x^{2}-1} \ln x-\sqrt{x^{2}-1}+\int \frac{d \sqrt{x^{2}-1}}{\left(\sqrt{x^{2}-1}\right)^{2}+1}\\
&=\sqrt{x^{2}-1} \ln x-\sqrt{x^{2}-1}+\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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If 8 divides $a^2$ + $3b^2$, prove that both a and b are even. Question: Prove that if $a$ and $b$ are integers and 8 divides $a^2$ + $3b^2$, then both a and b are even.
I'm able to prove that 8 divides $a^2$ + $3b^2$ if both $a$ and $b$ are even using the fundamental theorem of arithmetic and expressing 8 as $2^3$, but I'm not sure how to prove that 8 does not divide $a^2$ + $3b^2$ if either $a$ or $b$ is odd or both $a$ and $b$ are odd.
| $a^2+3b^2=(a+b)^2+2b^2-2ab.$ Clearly $2b^2-2ab$ is even. If one of $a$ or $b$ is odd, then $(a+b)^2$ is odd. But then $$ (a+b)^2+2b^2-2ab$$ is the sum of an odd and even number and hence odd. Thus $8$ cannot divide it.
If they are both odd then we have
\begin{align*}
a^2+3b^2&=(2m+1)^2+3(2n+1)^2 \\
&= 4m^2+4m+1+12n^2+12n+3\\
&=4(m^2+3n^2+m+3n+1).
\end{align*}
Let $k= m^2+3n^2+m+3n+1$
If $m,n$ are both even, then $k$ is odd.
If one of $m,n$ is odd, $k$ is again seen to be odd.
Finally, when $m,n$ are both odd, then $k$ is odd again.
$k$ being odd in all cases, we see that $8$ cannot divide the sum, which is a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Disk/Washer method proof checking This is a homework question, but i am just checking if what i am saying is correct,
The question in the book states that
A sphere of radius $r$ is cut by a plane of $h$ ($h < r$) units above the equator. Find the volume of the solid (spherical segment) above the plane.
Now i know that the upper hemisphere of any circle can be mapped on a 2 dimensional graph as
$$
f(x) = \sqrt{r^2-x^2}
$$
If this plane would be represented as a line on this graph that intersects with this function at 2 points, calling those points $a$ and $b$
$$
h = \sqrt{r^2-x^2}
$$
The positive point of intersection would be
$$
\sqrt{r^2-h^2} = a
$$
If i were to revolve this around the x-axis the volume would be
$$
V = 4\pi \int_{0}^{a} (\sqrt{r^2-x^2}-h^2)^2dx
$$
Due to symmetry, then i would continue with the integration
$$
V = 4\pi\int_0^{\sqrt{r^2-h^2}}(r^2-x^2 - 2h^2 \sqrt{r^2-x^2} - h^4)dx
$$
Therefore
$$
V = 4\pi [r^2\sqrt{r^2 - h^2} - \frac{(r^2-h^2)^{3/2}}{3} - h^4 \sqrt{r^2-h^2} + \int_{0}^{\sqrt{r^2-h^2}} 2h^2 \sqrt{r^2-x^2} dx]
$$
$$
\int_0^\sqrt{r^2-h^2} 2h^2 \sqrt{r^2-x^2} dx = h^3 \sqrt{r^2-h^2} + r^2 \tan^{-1}(\frac{\sqrt{r^2-h^2}}{h})
$$
Therefore the volume of this spherical segment would be
$$
V = 4\pi [r^2\sqrt{r^2 - h^2} - \frac{(r^2-h^2)^{3/2}}{3} - h^4 \sqrt{r^2-h^2} + h^3 \sqrt{r^2-h^2} + h^2 r^2 \tan^{-1}(\frac{\sqrt{r^2-h^2}}{h})]
$$
This was a problem for my calcus 2 class, and it seems like i over did this problem since i have never done a problem of this magnitude in the class thus far. Did i do this correctly? Was there an easier way to go about doing this?
EDIT: I know what i did wrong, i made a mistake with the washer method and did
$$
V = 2 \pi \int_a^b (R-r^2)^2 dx
$$
Instead of
$$
V = 2 \pi \int_{a}^b (R^2-r^2)dx
$$
I answered my own question down below.
| I found my mistake,
$$
V = 2 \pi \int_a^b (R^2 - r^2)dx
$$
Therefore
$$
V = 4 \pi \int_0^\sqrt{r^2-h^2} (r^2-x^2-h^2) dx
$$
$$
V = 4 \pi [r^2 \sqrt{r^2-h^2} - \frac{(r^2-h^2)^{3/2}}{3}
- h^2 \sqrt{r^2-h^2}]$$
So much easier
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluating a product of tangents
(1) Evaluation of $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right) = $
(2) Evaluation of $\displaystyle \tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right) = $
$\bf{My\; Try::(1):}$ Let $\displaystyle \frac{\pi}{7} = \theta\Rightarrow \pi = 7\theta\Rightarrow (\pi-3\theta) = 4\theta\Rightarrow \tan(\pi-3\theta) = \tan(4\theta)$
So $\displaystyle \tan (3\theta) = -\tan (4\theta)\Rightarrow \frac{3\tan \theta - \tan^3\theta}{1-3\tan^2 \theta} = -\left(\frac{2\tan 2\theta}{1-\tan^2 2\theta}\right) = \frac{4\tan^3 \theta-4\tan \theta}{1+\tan^4 \theta-6\tan^2 \theta}$
So $\displaystyle \left(\frac{3-\tan^2 \theta}{1-3\tan^2 \theta}\right) = \left(\frac{4\tan^2 \theta-4}{1+\tan^4 \theta-6\tan^2 \theta}\right)\Rightarrow \tan^6\theta-21\tan^4 \theta+21\tan^2 \theta-7=0$
Now Let $\displaystyle \tan \theta = y\;,$ then eqn. convert into $\displaystyle y^6-21y^4+21y^2-7=0$
and equation has a roots
$\displaystyle y = \tan \left(\frac{\pi}{7}\right)\;,\tan \left(\frac{2\pi}{7}\right)\;,\tan \left(\frac{3\pi}{7}\right)\;,\tan \left(\frac{4\pi}{7}\right)\;,\tan \left(\frac{5\pi}{7}\right)\;,\tan \left(\frac{6\pi}{7}\right)\;$.
So Product of Roots is
$\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right)\cdot \tan \left(\frac{4\pi}{7}\right)\cdot \tan \left(\frac{5\pi}{7}\right)\cdot \tan \left(\frac{6\pi}{7}\right) = -7$
Now $\displaystyle \tan\left(\frac{4\pi}{7}\right) = \tan\left(\pi-\frac{3\pi}{7}\right)=-\tan\left(\frac{3\pi}{7}\right)$
Similarly $\displaystyle \tan\left(\frac{5\pi}{7}\right) = -\tan\left(\frac{2\pi}{7}\right)$ and $\displaystyle \tan\left(\frac{6\pi}{7}\right) = -\tan\left(\frac{\pi}{7}\right)$
So $\displaystyle \tan\left(\frac{\pi}{7}\right)\cdot \tan\left(\frac{2\pi}{7}\right)\cdot \tan\left(\frac{3\pi}{7}\right)=\sqrt{7}$
I did not understand how can i calculate $(2)$ one using the same method:, bcz it is very lengthy
can we solve the $(1)$ and $(2)$ question any other method.
If yes then plz explain me, Thanks
::
Thanks
| Like Prove the trigonometric identity $(35)$,
$$\prod_{k=1}^m\tan \left(\frac{k \pi}{2m+1}\right)=\sqrt {2m+1}$$ as all the angles lies in $\left(0,\frac\pi2\right)$
| {
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"url": "https://math.stackexchange.com/questions/1055379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Understanding expanding trig identities I already have the answer, but I am trying to figure out why because it's not making much sense. Or maybe my professor misspoke I am not sure.
I have $\sin^5x $ and I am trying to express it in terms containing only first powers of sine and cosine.
Expanding into:
$\sin x\cdot \sin^2x\cdot \sin^2x $
and further:
$\sin x(\frac{1-\cos2x}{2})(\frac{1+\cos2x}{2}) $
And this is the point where I get a little confused. I don't recall any identity saying that $\sin^2x+ \sin^2x $ = $\ (\frac{1-\cos2x}{2})(\frac{1+\cos2x}{2}) $
Anyways the answer is: $\ \frac{\sin x-\sin^2 x\cos 2x + \sin x\cos^2 (2x)}{4} $ But I'm not quite sure how to get there. I figured I would post the answer so people wouldn't think that this is homework, this is for my own understanding.
| Since $\sin^2 x = \frac{1 -\cos(2x)}{2}$, $$\sin^5 x = \sin x \cdot \left(\frac{1 - \cos(2x)}{2}\right)^2 = \sin x \cdot \frac{1 - 2\cos(2x) +\cos^2(2x)}{4} = \frac{\sin x - 2\sin x \cos(2x) + \sin x \cos^2(2x)}{4}.$$
Since $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$,
\begin{align}\sin x - 2\sin x \cos(2x) + \sin x \cos^2(2x) &= \sin x - 2\sin x \cos(2x) + \frac{1}{2}\sin x + \frac{1}{2}\sin x \cos(4x)\\
&=\frac{3}{2}\sin x - 2\sin x \cos(2x) + \frac{1}{2}\sin x \cos(4x).
\end{align}
Therefore
$$\sin^5 x = \frac{3}{8}\sin x - \frac{1}{2}\sin x \cos(2x) + \frac{1}{8}\sin x \cos(4x).$$
Note: The answer you had written does not equal to $\sin^5 x$; at $x = \frac{\pi}{2}$, $\sin^5 x = 1$, but the expression $\frac{\sin x - \sin^2 x \cos(2x) + \sin x \cos^2(2x)}{4}$ equals $\frac{1 - 1(-1) + 1(-1)^2}{4} = \frac{3}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$ Solve the initial value problem: $\frac{dx}{dt}=2t \sin x$; $x(0)=\frac{\pi}{2}$
I am almost done. But I cannot seem to solve the equation for x(t) in terms of t.
This is what I got:
$$
\frac{dx}{\sin x}=2tdt\\
\csc x =2tdt\\
-\ln|\csc x+\cot x|=t^2+C\\
\ln \frac{1}{|\csc x+\cot x|}=t^2+C\\
\frac{1}{\csc x+\cot x}=\pm e^Ce^{t^2}
$$
So now I need to know what x(t) is. How do get the expression for x(t)?
| \begin{align*}
\frac{dx}{dt}&=2t \sin x\\
\frac{dx}{\sin x}&=2t \ dt\\
\csc x \ dx&= 2t \ dt\\
-\log(\csc x + \cot x)&=t^2+C_1\\
\log((\frac{1+\cos x}{\sin x})^{-1})&=t^2+C_1\\
\log (\tan\frac{x}{2})&=t^2+C_1\\
\tan{\frac{x}{2}}&=Ce^{t^2}\\
x(t)&=2\arctan(Ce^{t^2})
\end{align*}
For the initial condition:
\begin{align*}
\frac{\pi}{2}&= 2\arctan(Ce^{0})\\
\tan( \frac{\pi}{4}) &=C\\
C&=1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| You're essentially asking to prove $\sqrt{1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k}=2\sqrt{1+3\cdot10^n+8\cdot\sum\limits_{k=0}^{n-1}10^k}$
So it's sufficient to prove:
*
*$1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k=4\cdot(1+3\cdot10^n+8\cdot\sum\limits_{k=0}^{n-1}10^k)$
*$1+10^{n+1}+5\cdot\sum\limits_{k=0}^{n}10^k=4+12\cdot10^n+32\cdot\sum\limits_{k=0}^{n-1}10^k$
*$5\cdot\sum\limits_{k=0}^{n}10^k=3+2\cdot10^n+32\cdot\sum\limits_{k=0}^{n-1}10^k$
*$5\cdot10^n=3+2\cdot10^n+27\cdot\sum\limits_{k=0}^{n-1}10^k$
*$3\cdot10^n=3+27\cdot\sum\limits_{k=0}^{n-1}10^k$
*$10^n=1+9\cdot\sum\limits_{k=0}^{n-1}10^k$
*$10^n=10^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of
$$x^4-2(a+b)x^2+(a-b)^2=0$$
are......
$$x=\sqrt{a}+\sqrt{b}$$
$$x=\sqrt{a}-\sqrt{b}$$
$$x=-\sqrt{a}+\sqrt{b}$$
$$x=-\sqrt{a}-\sqrt{b}$$
When $a$ and $b$ are real numbers(negative or positive)
I proved this by substituting $x=\sqrt{a}+\sqrt{b}$ in the main equation , but I think this way not effective because It needs to be repeated four times to complete the proving.
Is there another effective proving.
| Calculate the determinant:$$4(a+b)^2-4(a-b)^2 = 16ab$$
Then find the roots as following:
$$x^2 = \frac{2(a+b)+\sqrt{16ab}}{2},\frac{2(a+b)-\sqrt{16ab}}{2}$$
$$x^2 = a+b+2\sqrt {ab}, a+b-2\sqrt {ab}$$
$$x = \pm(\sqrt a + \sqrt b), \pm(\sqrt a - \sqrt b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Computing the trigonometric sum $ \sum_{j=1}^{n} \cos(j) $ I have a task to compute such a sum:
$$ \sum_{j=1}^{n} \cos(j) $$
Of course I know that the answer is $$ \frac{1}{2} (\cos(n)+\cot(\frac{1}{2}) \sin(n)-1) = \frac{\cos(n)}{2}+\frac{1}{2} \cot(\frac{1}{2}) \sin(n)-\frac{1}{2} $$
but I don't have any idea how to start proving it.
| You could also use $\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2})=2\cos n\sin\frac{1}{2}$; $\;\;$ so if we let $a=2\sin\frac{1}{2}$,
$\cos n =\frac{1}{a}\left[\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2})\right]\implies$
$\cos 1+\cos2+\cos 3+\cdots+\cos(n-1)+\cos n=$
$\frac{1}{a}\left[(\sin\frac{3}{2}-\sin\frac{1}{2})+(\sin\frac{5}{2}-\sin\frac{3}{2})+(\sin\frac{7}{2}-\sin\frac{5}{2})+\cdots+(\sin(n+\frac{1}{2})-\sin(n-\frac{1}{2}))\right]=$
$\frac{1}{a}\left[\sin(n+\frac{1}{2})-\sin\frac{1}{2}\right]=\frac{1}{2\sin\frac{1}{2}}\left[\sin n\cos\frac{1}{2}+\cos n\sin\frac{1}{2}-\sin\frac{1}{2}\right]$
$=\frac{1}{2}\left[\sin n\cot\frac{1}{2}+\cos n-1\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Ways to write "50" A really good friend of mine is an elementary school math teacher. He is turning 50, and we want to put a mathematical expression that equals 50 on his birthday cake but goes beyond the typical "order of operations" problems. Some simple examples are
$$e^{\ln{50}}$$
$$100\sin{\frac{\pi}{6}}$$
$$25\sum_{k=0}^\infty \frac{1}{2^k}$$
$$\frac{300}{\pi^2}\sum_{k\in \mathbb{N}}\frac{1}{k^2}$$
What are some other creative ways I can top his cake?
I should note that he is an elementary school teacher. Now he LOVES math, and I can certainly show him a lot of expressions. I don't want them so difficult that it takes a masters degree to solve, but they should certainly be interesting enough to cause him to be wowed. Elementary functions are good, summations are also good, integrals can be explained, so this is the type of expression I'm looking for...
EDIT:: I would make a note that we are talking about a cake here, so use your judgement from here on out. Think of a normal rectangular cake and how big it is. Hence, long strings of numbers, complex integrals, and long summations are not going to work. I appreciate the answers but I need more compact expressions.
| We also have
\begin{align*}
50 &= 11+12+13+14 \\
&= (8+4)+(8-4)+(8\cdot 4)+(8/4) \\
&= 4^2 + 4^2 + 3^2 + 3^2\\
&= 6^2 + 3^2 + 2^2 + 1^2\\
&= (7+i)(7-i) \\
&= (10-\color{red}{5})(10-\color{red}{0})\\
&= 10(\color{red}{5}+\color{red}{0})\\
&= \sqrt{30^2+40^2}\\
&= \sqrt[3]{170^2+310^2}\\
&= \sqrt[3]{146^2+322^2}\\
&= \sqrt[3]{50^2+350^2}\\
\end{align*}
Finally $50= 2 + 4 + 8 + 12 + 24$ and $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 20,
"answer_id": 15
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Evaluate $\int\frac{\cot x}{\cos^2 x-\cos x+1}\,\,dx$ $$\int\frac{\cot x}{\cos^2 x-\cos x+1}\,\,dx$$
Please guide me by which term it should be substituted to get the result of this integration. I have tried it by using $\cos x =t$, but it went so long and more problematic.
| Hint :
\begin{align}
\int\frac{\cot x}{\cos^2 x-\cos x+1}\;\mathrm dx&=\int\frac{\frac{\cos x}{\sin x}}{\cos^2 x-\cos x+1}\cdot\frac{\sin^2 x}{\sin^2 x}\;\mathrm dx\\[10pt]
&=\int\frac{\cos x}{\cos^2 x-\cos x+1}\cdot\frac{\sin x}{1-\cos^2 x}\;\mathrm dx\\[10pt]
&=\int\frac{y}{y^2 -y+1}\cdot\frac{\mathrm dy}{y^2 - 1}\qquad\Rightarrow\qquad\text{set}\;y=\cos x\\[10pt]
&=-\frac{1}{3}\underbrace{\int\frac{2y-1}{y^2 -y+1}\mathrm dy}_{\large\color{red}{\text{set}\;t\,=\,y^2 -y+1}}+\frac{1}{2}\underbrace{\int\frac{\mathrm dy}{y-1}}_{\large\color{red}{\text{set}\;u\,=\,y-1}}+\frac{1}{6}\underbrace{\int\frac{\mathrm dy}{y+1}}_{\large\color{red}{\text{set}\;v\,=\,y+1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Splitting field of $X^6-7X^4+3X^2+3$ over $\mathbb Q$ and $\mathbb F_{13}$ I want to find the splitting field and the degree of the splitting field over $\mathbb Q$ and $\mathbb F_{13}$ for the polynomial $X^6-7X^4+3X^2+3$.
Over $\mathbb Q$ the polynomial factors as $(X-1)(X+1)(X^4-6X^2-3)$, but how can I find the splitting field of $(X^4-6X^2-3)$? For $\mathbb F_{13}$ I have no idea.
| Observe that $Y^2-6Y-3$ splits over $\mathbb{F}_{13}$ because its roots are
$$\frac{6\pm\sqrt{36+12}}{2}=3\pm4\sqrt{3}=3\pm4(4)=7\text{ and }12$$
The roots of $X^4-6X^2-3$ are obviously the square roots of the roots of $Y^2-6Y-3$. The element $12$ is already a square in $\mathbb{F}_{13}$ (its square roots are $5$ and $-5=8$) so the splitting field of $X^4-6X^2-3$ over $\mathbb{F}_{13}$, and hence the splitting field of $X^6-7X^4+3X^2+3$ over $\mathbb{F}_{13}$, is
$$\mathbb{F}_{13}(1,-1,5,-5,\sqrt{7},-\sqrt{7})=\mathbb{F}_{13}(\sqrt{7})\cong\mathbb{F}_{13^2}$$
The idea of analyzing $Y^2-6Y-3$ should help you figure out the case of $\mathbb{Q}$ on your own.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the limit of function using Taylor series Good evening,
I'm somehow stuck on solving some easy exercises :
$$\lim_{x\to\infty} x^{3/2}\bigl(\sqrt{x+1}+\sqrt{x-1}-2\,\sqrt{x}\bigr)$$
| $$\begin{align}&x\sqrt x\left(\sqrt{x+1}+\sqrt{x-1}-2\sqrt x\right)=x\left(\sqrt{x^2+x}+\sqrt{x^2-x}-2x\right)=\\{}\\
&=x\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{x^2+x}+\sqrt{x^2-x}+2x}=\frac{-2x^2+2\sqrt{x^4-x^2}}{\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2}=\\{}\\
&=\frac{4(x^4-x^2)-4x^4}{\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\right)\left(2\sqrt{x^4-x^2}+2x^2\right)}\\{}\\
&=\frac{-4x^2}{2x^2\left(\sqrt{1+\frac1x}+\sqrt{1-\frac1x}+2\right)\left(\sqrt{1-\frac1{x^2}}+1\right)}\xrightarrow[x\to\infty]{}\frac{-2}{(1+1+2)(1+1)}=-\frac14\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve the recurrence relation: $2a_n = 7a_{n-1} - 3a_{n-2}; a_0 = a_1 = 1$ $2a_n = 7a_{n-1} - 3a_{n-2}\\
a_0 = a_1 = 1$
My attempt:
$2t^2 - 7t + 3 = 0\\
t = -\frac{1}{2}, -3\\
\\
U_n = b(-\frac{1}{2})^n + d(-3)^n\\
b+d = 1 = -\frac{1}{2}b-3d\\
b = \frac{8}{5}, d = -\frac{3}{5}\\
a_n = \frac{8}{5}(-\frac{1}{2})^n - \frac{3}{5}(-3)^n\\
a_2 = -5 \neq 2 =\frac{1}{2}(7-3)$
Where did I go wrong?
Update: changing the t values to positive, new solution:
$a_n = \frac{4}{7}(\frac{1}{2})^n+\frac{3}{7}(3)^n\\
a_2 = 4$
But following the initial equation given, $2a_2 = 4$, so shouldn't $a_2$ be 2?
$a_2$ according to Wolfram Alpha is indeed 2.
| The answer given by Wolfram Alpha does match your textbook. (Try multiplying the $2^{-n}$ through.)
Your mistake was when you changed the roots of the characteristic to positive; if
$$a_n = A\left(\frac{1}{2}\right)^n + B\left(3\right)^n$$
then the correct method of solving for $A$ and $B$ would give $A=\frac{4}{5}$ and $B=\frac 1 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Mclaurins with $e^{\sin(x)}$ To evaluate $e^{\sin(x)}$ I use the standard series $e^t$ and $\sin(t)$, combining them gives me:
$e^t = 1+t+\dfrac{t^2}{2!}+\dfrac{t^3}{3!}+\dfrac{t^4}{4!}+O(t^5)$
$\sin(t) = t-\dfrac{t^3}{3!}+O(t^5)$
$\therefore e^{\sin(x)} = 1+\sin(x)+\dfrac{\sin^2(x)}{2!}+\cdots+O(x^5)$
$\iff e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{x^3}{6}+O(x^5) = 1+x +\dfrac{x^2}{2}-\dfrac{x^4}{6} + O(x^5)$
In the last step, I only evaluate up to $\sin^3x$ term, everything above has a grade equal to or greater than $x^5$
However, I'm wrong. According to Wolfram, the series expansion for $e^{\sin x} = 1+x+\dfrac{x^2}{2}-\dfrac{x^4}{8}+O(x^5)$.
What did I do wrong? I can't seem to find what or where.
| $$=e^{\sin(x)}=1+\left(x-\dfrac{x^3}{6}\right)+\dfrac{\left(x-\frac{x^3}{6}\right)^2}{2}+\dfrac{(x-x^3/6)^3}{6}+\dfrac{(x-x^3/6)^4}{24}+O(x^5)$$
and this term
$$\dfrac{(x-x^3/6)^4}{24}=\dfrac{x^4}{24}+o(x^5)$$
so
$$e^{\sin{x}}=1+x+\dfrac{x^2}{2}+\left(-\dfrac{x^4}{6}+\dfrac{x^4}{24}\right)+o(x^5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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To evaluate limit of sequence $\left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$ How do I evaluate the limit of the following sequence
$$a_n = \left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}$$
I tried to take log and then using cauchy $1^{st}$ theorem of limits but I coudn't do it.
Could someone kindly help?
| Rewrite $\displaystyle a_n = \prod\limits_{k=1}^{n} \left(1+\frac{k}{n}\right)^{1/n} $
Since, $\displaystyle \lim\limits_{n\to \infty} b_n^{1/n} = \lim\limits_{n\to \infty} \frac{b_{n+1}}{b_n}$ (that is if the limit exists)
We may write:
$\displaystyle \begin{align} \lim\limits_{n\to \infty} a_n =\lim\limits_{n\to \infty} \frac{\prod\limits_{k=1}^{n+1} \left(1+\frac{k}{n+1}\right)}{\prod\limits_{k=1}^{n} \left(1+\frac{k}{n}\right)} &= \lim\limits_{n \to \infty} \frac{\prod\limits_{k=1}^{n+1} (n+k+1)}{\prod\limits_{k=1}^{n} (n+k)}.\frac{(n+1)^{-(n+1)}}{n^{-n}} \tag{1}\\ &=\lim\limits_{n\to \infty} \frac{(n+2)\cdots(2n+2)}{(n+1)\cdots(2n)}.\frac{\left(1+\frac{1}{n}\right)^{-n}}{n+1} \tag{2}\\&= \lim\limits_{n\to \infty} 2(2n+1).\frac{\left(1+\frac{1}{n}\right)^{-n}}{n+1} \tag{3} \\&= \frac{4}{e}\end{align}$
2nd Approach:
Write $\displaystyle a_n = \left(\binom{2n}{n}\frac{n!}{n^n}\right)^{1/n}$
Using the identity $\displaystyle \sum\limits_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$ we make an application of Cauchy-Schwarz Ineqality
$$\displaystyle (n+1)\binom{2n}{n} = \left(\sum\limits_{k=0}^{n} 1^2\right)\left(\sum\limits_{k=0}^{n} \binom{n}{k}^2\right) \ge \left(\sum\limits_{k=0}^{n} \binom{n}{k}\right)^2 = 4^n$$
Thus, $\displaystyle \frac{4^n}{n+1} \le \binom{2n}{n} \le 4^n \implies \lim\limits_{n \to \infty} \frac{4}{\sqrt[n]{n+1}} = \lim\limits_{n \to \infty} \binom{2n}{n}^{1/n} = 4$ (form, squeeze theorem).
Since, $\displaystyle \lim\limits_{n \to \infty} \frac{(n!)^{1/n}}{n} = e^{-1}$ we get $\displaystyle \lim\limits_{n \to \infty} a_n = \frac{4}{e}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
$(1-a)(1-b)(1-c)(1-d)\geq abcd$ for $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=1$. Prove that $$(1-a)(1-b)(1-c)(1-d)\geq abcd.$$
I thought about substituting $a=\sqrt{w},b=\sqrt{x}$, etc. (assuming first that $a,b,c,d$ are positive), and then looking at the convexity of the function $f(r)=(1-\sqrt{r})/\sqrt{r}$ and applying some Jensen-type inequality. But such an inequality applies to the sum of functions, not the product.
| Observe that as long as $a,b,c,d \ge 0$: $$\begin{align}(1-a)(1-b) - cd &\ge 1 - a - b + ab - \left(\frac{c^2+d^2}{2}\right) \\ &= 1 - a - b + ab - \left(\frac{1-a^2-b^2}{2}\right) \\ &= \frac{1}{2}(1-a-b)^2 \ge 0\end{align}$$
Similarly, $$(1-c)(1-d) -ab \ge \frac{1}{2}(1-c-d)^2\ge 0$$
Hence, $$(1-a)(1-b)(1-c)(1-d) \ge abcd$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1072899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| 1073177
Use Sylvester's dialytic eliminant:
$\begin{vmatrix}
-2&-7x&3x^2&2x^3-20&0\\
0&-2&-7x&3x^2&2x^3-20\\
1&x&x^2-7&0&0\\
0&1&x&x^2-7&0\\
0&0&1&x&x^2-7\\
\end{vmatrix}$
$(=79x^6-1022x^4-80x^3+2891x^2+1120x-972)$
to get rid of the $y$s. Equate it to $0$, solve it for $x$, then substitute each value into one of the original equations (preferably the simpler one), then solve for the corresponding $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| A key to this kind of question is a formed by Newton's identities, which relate the power sums $p_k=\sum_{i=1}^nx^i$ for different values of $k$ to the elementary symmetric polynomials $e_k$; one can define $e_k$ as the coefficient of $X^k$ in the expansion of $\prod_{i=1}^n(1+x_iX)$. The relations are most easily stated recursively as
$$
ke_k=\sum_{i=1}^k(-1)^{i-1}p_ie_{k-i}
$$
(note that the final term involves $e_0=1$). This makes converting between values $p_1,p_2,\ldots,p_k$ and values $e_1,e_2,\ldots,e_k$ easy.
The problem actually gives $p_k=k$ for $k=1,2,3$, and asks for $p_4$, which normally Newton's identities cannot do. However the additional fact that makes this doable is that one has $n=3$ here: with only three values $x_i$, we know that $e_k=0$ for all $k>3$, in particular for $k=4$.
So first solve $e_1,e_2,e_3$ from
$$ \begin{align} e_1&=p_1e_0=1 \\ 2e_2&=p_1e_1-p_2e_0=e_1-2 \\
3e_3&=p_1e_2-p_2e_1+p_3e_0=e_2+2e_1+3
\end{align}
$$
giving $(e_1,e_2,e_3)=(1,-\frac12,\frac16)$, and then use $e_4=0$ in the reversed identity
$$
p_4=e_1p_3-e_2p_2+e_3p_4-4e_4=1\times3+\frac12\times2+\frac16\times1-0=\frac{25}6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ Show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea:
Let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+(a^2+1)(b^2+1)\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-\dfrac{4ab(a+b+a^2b+b^2a)^2}{4ab}\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-(a+b+a^2b+b^2a)^2\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a+b)^2+(ab-1)^2+(a+b)^2(1+ab)^2
\end{align*}
so I think this idea is not good, maybe this have good methods,because this result is nice
|
$$$$
Let us start to calculate it.
\begin{eqnarray}
I&=&\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}\\
&=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{\frac{a^2+1}{2a}-x}\sqrt{\frac{b^2+1}{2b}-x}}\\
&=&\frac{1}{2\sqrt{ab}}\int_{-1}^{1}\dfrac{dx}{\sqrt{m-x}\sqrt{n-x}}\\
&=&\frac{1}{\sqrt{ab}}\int_{-1}^{1}\dfrac{-d(\sqrt{n-x})}{\sqrt{(\sqrt{n-x})^2+m-n}}\\
&=&-\frac{1}{\sqrt{ab}}\ln(\sqrt{n-x}+\sqrt{m-x}\sqrt{n-x})|_{-1}^{1}\\
&=&\frac{1}{\sqrt{ab}}\frac{ln(\sqrt{n+1}+\sqrt{m+1}\sqrt{n+1})}{ln(\sqrt{n-1}+\sqrt{m-1}\sqrt{n-1})}\\
&=&\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}
\end{eqnarray}
Where $m=\frac{a^2+1}{2a} \\
n=\frac{b^2+1}{2b}$
So we can get it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
What property of summation is used while solving this problem? Saw this problem on a website.
Can someone explain how the summation is split into summation of summation?
What property of summation was used here?
$$ T(n) = \sum_{k=1}^n \lfloor \sqrt{k} \rfloor. $$
Therefore
$$
\begin{align*}
T(m^2-1) &= \sum_{k=1}^{m^2-1} \lfloor \sqrt{k} \rfloor \\ &=
\sum_{r=1}^{m-1} \sum_{\ell=r^2}^{(r+1)^2-1} \lfloor \sqrt{\ell} \rfloor \\ &=
\sum_{r=1}^{m-1} \sum_{\ell=r^2}^{(r+1)^2-1} r
\end{align*}
$$
| Both the single sum $\displaystyle \sum_{\ell=1}^{m^2-1}$ and the nested sum $\displaystyle \sum_{r=1}^{m-1} \sum_{\ell=r^2}^{(r+1)^2-1}$ run over the same set of values:
*
*for $r=1$, $\ell$ runs from $1^2 = 1$ to $2^2-1 = 3$;
*for $r=2$, $\ell$ runs from $2^2 = 4$ to $3^2-1 = 8$;
*for $r=3$, $\ell$ runs from $3^2 = 9$ to $4^2-1 = 15$;
*...
*for $r=m-1$, $\ell$ runs from $(m-1)^2$ to $m^2-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a_n$ is bounded and decreasing my second question from
[An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$
Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$
Show that $a_{n}$ is decreasing and bounded
note that $ \forall x\in [0,\dfrac{\pi}{2}],\quad \cos(x) \geq 0$
since $\pi/2^n \in [0,\dfrac{\pi}{2}]$ then $cos(\dfrac{\pi}{2})\geq 0$ then $a_n\geq 0 $
let $a_n=a_{n−1}\cos(\dfrac{\pi}{2^n})\quad \forall n\geq 3$
then $a_{n+1}−a_n=a_n(\cos(\dfrac{\pi}{2^{n+1}})-1)$ which is negative since $(\cos(\dfrac{\pi}{2^{n+1}})-1)$ its.
but for bounded :
we've already that $a_n\geq 0,\quad \forall n \geq 2$
note that $|(cos(\pi/2^k)|\leq 1\quad \forall k\geq 2$ then $\prod_{k=3}^{n}|(cos(\pi/2^k)|\leq 1 $ thus $0\leq a_n \leq 1$
am i right ?
any help would be appreciated
| The reasoning about $a_n$ decreasing is fine; that is, for $n\ge2$, $0\lt\cos\left(\frac\pi{2^n}\right)\lt1$. This also shows that $a_n\gt0$. However, we can do a bit better.
Trigonometric Approach
Since
$$
2\sin\left(\frac\pi{2^k}\right)\cos\left(\frac\pi{2^k}\right)
=\sin\left(\frac\pi{2^{k-1}}\right)
$$
We have
$$
\prod_{k=2}^n\cos\left(\frac\pi{2^k}\right)=\frac2{2^n\sin\left(\frac\pi{2^n}\right)}
$$
and
$$
\lim_{n\to\infty}\frac2{2^n\sin\left(\frac\pi{2^n}\right)}=\frac2\pi
$$
Therefore, $a_n\gt\frac2\pi$.
Another Approach
Since $\sin(x)\le x$, $\cos(x)\ge\left(1-x^2\right)^{1/2}\ge\left(1-x^2\right)$. Therefore,
$$
\begin{align}
\prod_{k=2}^\infty\cos\left(\frac\pi{2^k}\right)
&\ge\prod_{k=2}^\infty\left(1-\frac{\pi^2}{4^k}\right)\\
&=\prod_{k=2}^\infty\left(1+\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{4^k}}\right)^{-1}\\
&\ge\exp\left(\sum_{k=2}^\infty\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{4^k}}\right)^{-1}\\
&\ge\exp\left(\sum_{k=2}^\infty\frac{\frac{\pi^2}{4^k}}{1-\frac{\pi^2}{16}}\right)^{-1}\\
&=\exp\left(\frac{-4\pi^2}{48-3\pi^2}\right)\\[9pt]
&\doteq0.116881538733185
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding solutions to $y^2 = x^3 - 27$ I am trying to find integer solutions to this equation:
$$ y^2 = x^3 - 27 $$
With the other problem I tried I was able to use unique factorization in $\mathbb Z [\sqrt{n}]$. I don't know how to get started on this one. I tried looking online and there's some stuff about elliptic curves but I don't understand them.
Here's what I tried so far:
$x$ is odd since otherwise $y^2 = -27 = 5 \mod 8$ but $5$ is not a square mod $8$ also $y$ is even.
$ x^3 = y^2 + 27 $
$ x^3 = (y + 3\sqrt{-3})(y-3\sqrt{-3})$ in $\mathbb Z[\sqrt{-3}]$ which is unique prime factorization.
Any common divisor must divide $6\sqrt{-3}$ and so norm of common divisor must divide $108$?
???
Thanks for any help, sorry English is not my first language.
edit: Here's some new work:
Let $d$ be greatest common divisor of $y+3\sqrt{-3}$ and $y-3\sqrt{-3}$. then $N(d) | 108$ and $N(d) | y^2 + 27 $, this is odd number. $N(d)$ is odd number so $N(d) = 1, 3, 9, 27$. (I know really I need $N(d) = 27$ because that gives $y=0$.)
| It seems the following.
This equation is a special case of the Mordell equation, $y^2 =x^3 +A$. A tremendous amount of work has been done on the Mordell equation, and solutions have been tabulated for large ranges of values of $A $, for example, here.
Also it is new for me that you consider a non-natural factorization. We have $(x-3)(x^2+3x+9)=y^2$. Put $a=GCD(x-3,x^2+3x+9)$. It is easy to show that $a|27,$ so it rests to solve the systems
$$\cases{
x-3=du^2\\
x^2+3x+9=dv^2},$$ where $u$ and $v$ are coprime and $d=3^k$, $0\le k\le 3$. For instance, for $d=1,9$ we obtain $(x+1)^2<(\sqrt{d}v)^2<(x+2)^2$ provided $x>5$, so we have $x=3$. I'll think about the cases $d=3,27$.
Assume $d=3$. Then $x=3(u^2+1)$. So
$$3^2(u^2+1)^2+3^2(u^2+1)+3^2=3v^2$$
$$3(u^2+1)^2+3(u^2+1)+3=v^2$$
So $v=3w$ and
$$(u^2+1)^2+(u^2+1)+1=3w^2$$
$$u^4+2u^2+1+u^2+1+1=3w^2$$
$$u^4+3u^2+3=3w^2$$
Then $u$ is divisible by $3$ too, a contradiction, because $u$ and $v$ are coprime.
Assume $d=27$. Then $x=3(3^2u^2+1)$. So
$$3^2(3^2u^2+1)^2+3^2(3^2u^2+1)+3^2=3^3v^2$$
$$(3^2u^2+1)^2+(3^2u^2+1)+1=3v^2,$$
$$3^4u^4+2\cdot 3^2u^2+1+3^2u^2+1+1=3v^2,$$
$$3^3u^4+3^2u^2+1=v^2$$
My program checked $1\le u\le 2\cdot 10^4$ and found no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$
For any triangle with sides $a$, $b$, $c$, prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$
This is IMO 1983 problem 6.
I tried substituting $a=x+y$, $b=y+z$, $c=z+x$ but well it doesn't help in any sense except wasting 3 pages that lead to nothing (please don't mind the joke). Using $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ also didn't lead to anything for me. Could you give me a hint for finding the proper substitution?
| $$ \it{a^2 b(a-b)+b^2c(b-c)+c^2 a(c-a)}$$
$$ \it{={\frac {bc\, \left( a+b-c \right) ^{2} \left( a-b \right) ^{2}+a \left( c+a-b \right) \left\{ b \left( a+b-c \right) \left( a-c \right) ^{2} +a \left( b+c-a \right) \left( b-c \right) ^{2} \right\} }{a \left( c +a-b \right) +b \left( a+b-c \right) }}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
How can I evaluate this limit with or without applying derivatives? $$ \lim_{n \to \infty} \frac{1^{1/3} + 2^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} $$
High school student here! This was a question from our Mathematics exam (prior to learning derivatives). Now there was some sort of a bounty here in our school, but nobody could solve it even after weeks passed.
WolframAlpha gives $3/4$ but no other explanation. I'm curious how one could tackle the expression in the numerator.
| Consider that, since $a-b=\frac{a^3-b^3}{a^2+ab+b^2},$
$$ (n+1)^{\frac{4}{3}}-n^{\frac{4}{3}} = \frac{(n+1)^4-n^4}{3n^{\frac{8}{3}}+O\left(n^{\frac{5}{3}}\right)}=\frac{4}{3}n^{\frac{1}{3}}+O\left(n^{-\frac{2}{3}}\right).\tag{1}$$
By multiplying both sides of the previous equation by $\frac{3}{4}$, then summing over $n$ we get that the limit is $\color{red}{\frac{3}{4}}$, as expected. This is just an application of creative telescoping and trivial inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| Observing that
$$
f(x)=\frac{1}{x}\sqrt{\frac{1-x}{1+x}}=\frac{1}{x}\sqrt{\frac{1-x}{1+x}\cdot\frac{1-x}{1-x}}=\frac{1}{x}\frac{1-x}{\sqrt{1-x^2}}
$$
so we have
$$
I=\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}\mathrm d x=\underbrace{\int \frac{1}{x\sqrt{1-x^2}}\mathrm d x}_J-\underbrace{\int \frac{1}{\sqrt{1-x^2}}\mathrm d x}_K=J-K
$$
Putting $x=\sin u$ we have
$$
K=\int \mathrm d u=u+C_1
$$
that is $K=\arcsin x+ C_1$; and for $J$
$$
J=\int \frac{1}{\sin u}\mathrm d u
$$
Putting $t=\tan(\frac{u}{2})$, so that $\sin u=\frac{2t}{1+t^2}$ and $\mathrm{d}u=\frac{2}{1+t^2}\mathrm{d}t$ we find
$$
J=\int \frac{1}{t}\mathrm{d}t=\log t+C_2
$$
Substituting back $t=\tan(\frac{u}{2})=\tan\left(\frac{\arcsin(x)}{2}\right)$
we have
$$
J=\log \left(\tan\left(\frac{\arcsin(x)}{2}\right)\right)+C_2
$$
Finally
$$
I=\arcsin x+\log \left(\tan\left(\frac{\arcsin(x)}{2}\right)\right)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Evaluating an alternating sum using contour integrals Evaluate:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{3n-1}$$
Using contour integration.
Normally I would use $\pi\csc(\pi z)f(z)$ and evaluate the residue multiply by (-1) and divide by $2$ if the function $f(n)$ were even, in this case it is not even.
I was wondering if I could use contour integration here.
I converted it into an integral.
Without showing the working, (well it was mostly integration and division)
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{3n-1} = \int_{0}^{1} \frac{1}{1+x^3}dx$$
I considered a contour $C$ quarter circle, radius $1$ the y-axis is the imaginary axis.
into section $A,B,D$
With the quad formula, I found three roots.
$x = ${$ \displaystyle \frac{1 + \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2}, -1 $}
Let the poles be called $a,b,c$ respectively, we will only consider $a$ since its the only one in the contour, the radius $R = 1$.
$$\oint_{C} f(z) dz = \int_{0}^{1} f(x) dx + \int_{B} f(z) dz - \int_{0}^{1} f(iy) d(iy)$$
By the residue theorem I got:
$$\oint_{C} f(z) dz = \frac{4\pi i}{3 + \sqrt{3}i}$$
But I am not sure what to do next.
| $$\begin{align}\int_{0}^{1} \frac{1}{1+x^3}dx&=\frac13\int_0^1\left(\frac1{x+1}-\frac{x-2}{x^2-x+1}\right)dx\\&=\frac13\int_0^1\left(\frac1{x+1}-\frac12\frac{2(x-1/2)}{(x-1/2)^2+3/4}-\frac{3/2}{(x-1/2)^2+3/4}\right)dx\\
&=\frac13\left(\ln|x+1|-\frac12\ln|(x-1/2)^2+3/4|-\frac32\frac1{\sqrt{3/4}}\arctan\frac{x-1/2}{\sqrt{3/4}}\right)_0^1\\
&=\frac13\left(\ln2-\frac12\ln\frac11-{\sqrt3}\left(\frac{\pi}6-\left(-\frac{\pi}6\right)\right)\right)\\
&=\frac13\left(\ln2-\frac{\pi}{\sqrt3}\right)=\frac19\left(3\ln2-3\frac{\pi}{\sqrt3}\right)=\frac19\left(\ln8-\sqrt3\pi\right)
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 \ \text{mod} \ 7= & &6 \\
555 \ \text{mod} \ 7= & &2 \\
5555 \ \text{mod} \ 7= & &4 \\
55555 \ \text{mod} \ 7= & &3 \\
555555 \ \text{mod} \ 7= & &0 \\
5555555 \ \text{mod} \ 7= & &5 \\
55555555 \ \text{mod} \ 7= & &6 \\
555555555 \ \text{mod} \ 7= & &2 \\
5555555555 \ \text{mod} \ 7= & &4 \\
\end{array}$$
It can be seen that the cycle is: $\{5,6,2,4,3,0\}$.
$$\begin{array}
& 1 \ \text{number =} &5 \\
7 \ \text{numbers =} &5 \\
13 \ \text{numbers =} &5 \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots & \\
985 \ \text{numbers =} &5 \\
991 \ \text{numbers =} &5 \\
997 \ \text{numbers =} &5 \\
998 \ \text{numbers =} &6 \\
999 \ \text{numbers =} &2 \\
\color{red}{1000} \ \color{red}{\text{numbers =}} &\color{red}{4} \\
\end{array}$$
From here, we can conclude that $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7 = 4$.
However, I wasn't allowed to use a calculator and solved this in about 12 minutes. Another problem was that there was a time limit of about 5 minutes. My question is: Is there an easier and faster way to solve this?
Thanks a lot in advance!
| ${\rm mod}\ 7\!:\,\ \overbrace{55\cdots 55}^{1+3n\rm\,\ fives}\, =\, \dfrac{5(10^{1+3n}\!-1)}9\, \equiv\, \dfrac{-2\,(3^{1+3n}-1)}2 \,\equiv\, -3(\color{#c00}{3^3})^{n}\!+1 \equiv 4\ $ by $\ \color{#c00}{3^3\equiv -1},\ n$ odd
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$ Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):
$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$$
Transforming it into $2 \sin \frac{3\alpha-180}4 \cos \frac{\beta - \gamma}4 + \sin \frac{3\alpha}2 =\frac{3}2$
and
$\cos \frac{\beta - \gamma}4 =\cos \frac{180- \alpha}4 + 2 \sin \frac{\beta}4 \sin \frac{\gamma}4$ is as far as I came.
| When we exchange $\beta$ and $\gamma$ in the equation, we get the same equation back. To me, that means $\beta = \gamma+4k\pi$, where $k$ is an integer, is a third equation that we can use. Of course, only $k=0$ satisfies the constraints that $\alpha+\beta+\gamma=\pi$ and $\alpha, \gamma, \beta>0$, since they are angles in a triangle. I haven't been able to derive it, but $\alpha=5\pi/9$, $\beta=2\pi/9$, and $\gamma=2\pi/9$ does appear to be the solution.
| {
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if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.
| Note that $x$ is positive, so we can write $x=y^2-2$, for some real number $y>\sqrt{2}$ and the equation becomes
$$y^2-2=\sqrt{y^2-1}+y+\sqrt{y^2+1}$$
we obtain thus
$\sqrt{y^2+1}=y^2-y-2-\sqrt{y^2-1}$, hence
$$y^2+1=(y^2-y-2)^2+(y^2-1)-2(y^2-y-2)\sqrt{y^2-1}.$$
This yields
$$2(y^2-y-2)\sqrt{y^2-1}=(y^2-y-2)^2-2$$
so
$4(y^2-y-2)^2(y^2-1)=((y^2-y-2)^2-2)^2$. Hence $y$ is a positive real root of the polynomial
$$\begin{array}{rcl}
P(y)&=&((y^2-y-2)^2-2)^2-4(y^2-y-2)^2(y^2-1)\\
&=&y^8-4y^7-6y^6+28y^5+13y^4-56y^3-24y^2+32y+20\end{array}$$
There is only one real root $y>\sqrt{2}$ of this polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Slight problem with solving a trigonometric equation. I had to prove the identity $4\cos^3x-3\cos x=\cos 3x$ and then use it to solve the equation $(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)=1$.
After proving the identity I proceeded to simplify the equation, obtaining
\begin{align*}
(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)&=1\\
(4\cos^3x-3\cos x)(4\cos^23x-3)(4\cos^29x-3)&=\cos x\\
\cos3x(4\cos^23x-3)(4\cos^29x-3)&=\cos x\\
(4\cos^33x-3\cos3x)(4\cos^29x-3)&=\cos x\\
\cos9x(4\cos^29x-3)&=\cos x\\
4\cos^39x-3\cos9x&=\cos x\\
\cos27x&=\cos x.
\end{align*}
The only thing is that when I plotted $y=(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)-1$ (the original equation which appears in red) and $y=\cos27x-\cos x$ (this is in black) the following graph emerged
Notice that after $1.5$ and $-1.5$ there is a root that is not shared by both curves. What do I have wrong?
| By multiplying by $\cos(x)$ you created additional solutions precisely at the points where $\cos(x)=0$. This is because at these points you just multiplied your original equation by zero which makes any equation true. Consider for example the equation $x = 1$ which has precisely one solution. Contrast this with the equation we get after multiplying with $\cos(x)$:
$$x\cos(x) = \cos(x)$$
which has infinitely many solutions, namely, for all $x$ such that $\cos(x)$ the equation is solved.
UPDATE:
Your method of finding solution is correct except that all solutions such that $\cos(x)=0$ are not necessarily solutions at all. This is because of what I said before. These solutions are $x = \frac{\pi}{2}+ n \pi$. You can manually check whether these $x$ values give solutions though by just working out the expression
$$(4\cos^2x-3)(4\cos^23x-3)(4\cos^29x-3)$$
and check if it equals $1$. You will find that this is not the case, as all cosine terms disappear and you are left with $(-3)(-3)(-3) \neq 1$. Thus you must remove from your set of solutions all $x$ such that $x = \frac{\pi}{2}+ n \pi$.
| {
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Prove by induction that $\forall n \in \mathbb{N} \cup \{0\}: \sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n + 2}{2^{n}}$ Prove by induction $\forall n \in \mathbb{N} \cup \{0\}: \sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n + 2}{2^{n}}$
Step 1: Show true for n = 0:
LHS: $\frac{0}{2^{0}}$ = 0
RHS = $2 - \frac{0+2}{2^{0}}$ = 0
Step 2: Show that it is true for $n = p$, it is true for $n = p + 1$
Starting with the LHS of the $n = p + 1$ expression, breaking out the largest term and substituting in the $n = p$ equality gives:
$\sum_{k=0}^{p+1} \frac{k}{2^{k}} = \frac{p+1}{2^{p+1}} + \sum_{k=0}^{p} \frac{k}{2^{k}} = \frac{p+1}{2^{p+1}} + 2 - \frac{p + 2}{2^{p}}$
Making everything besides the 2 have the same denominator $2^{p+1}$
$\frac{p+1}{2^{p+1}} + 2 - \frac{p + 2}{2^{p}} = 2 + \frac{p + 1 - 2(p+2)}{2^{p+1}} = 2 - \frac{p - 3}{2^{p+1}} $
...I get lost here.
Question: Where have I gone wrong in the above attempt?
| $\dfrac{p+1}{2^{p+1}} + 2 - \dfrac{p + 2}{2^{p}} = 2 + \dfrac{p + 1 - 2(p+2)}{2^{p+1}} = 2 + \dfrac{(-p - 3)}{2^{p+1}} = 2 - \dfrac{p+3}{2^{p+1}} = 2 - \dfrac{(p + 1)+2}{2^{p+1}}.$
| {
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Find the volume of the region of a sphere bounded by two planes Calculate the volume of a sphere $x^2+y^2+z^2=R^2$ which is bounded by $z=a$ and $z=b$, where $0\leq a<b<R$ using double integral. I can imagine the picture but I don't know how to set it up.
| I show below how you do this calculation as a difference of two volumes:
The volume is given by $I(a)-I(b)$ where (for $0<s<R$), $I(s)$ is defined to be the volume of the ball $x^2+y^2+z^2\leq R^2$ such that $z>s$. Let us calculate $I(s)$ (you complete the details):
The domain is bounded from below by the plane $z=s$ and from above by the surface $z=\sqrt{R^2-x^2-y^2}$. The domain projected onto the $xy$-plane is a disc $D:x^2+y^2\leq R^2-s^2$. Thus,
$$
I(s)=\iint_D\Bigl(\sqrt{R^2-x^2-y^2}-s\Bigr)\, dx\,dy
$$
Changing to polar coordinates $(r,\phi)$, this becomes
\begin{align}
I(s)&=\int_0^{2\pi}\, d\phi\int_0^{\sqrt{R^2-s^2}}\Bigl(\sqrt{R^2-r^2}-s\Bigr)r\,dr\\
&=2\pi\Bigl[-\frac{1}{3}(R^2-r^2)^{3/2}-\frac{1}{2}sr^2\Bigr]_0^{\sqrt{R^2-s^2}}\\
&=2\pi\Bigl(-\frac{1}{3}s^3-\frac{1}{2}s(R^2-s^2)+\frac{1}{3}R^3\Bigr).
\end{align}
Thus (collect and simplify), the volume you look for is:
$$
I(a)-I(b) = \pi R^2(b-a) -\frac{\pi}{3}(b^3-a^3).
$$
Just as a side-note we see that the expression above is non-negative:
Since $a$ and $b$ are bounded by $R$,
$$
\frac{\pi}{3}(b^3-a^3)=\frac{\pi}{3}(b-a)(b^2+ab+a^2)\leq \frac{\pi}{3}(b-a)(R^2+R^2+R^2)=\pi R^2(b-a).
$$
OK!
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$ Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$
Thanks for any help
| Let we sketch an alternative technique. Since:
$$\sum_{n\geq 0} x^{2n} = \frac{1}{1-x^2} = \frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right),\tag{1}$$
we have:
$$\sum_{n\geq 0} \frac{x^{2n+1}}{2n+1} = \frac{1}{2}\left(\log(1+x)-\log(1-x)\right)=\operatorname{arctanh} x,\tag{2} $$
and integrating again:
$$\sum_{n\geq 0} \frac{x^{2n+2}}{(2n+1)(2n+2)} = x\operatorname{arctanh} x+\frac{1}{2}\log(1-x^2),\tag{3} $$
$$\sum_{n\geq 0} \frac{x^{2n+3}}{(2n+1)(2n+2)(2n+3)} =\frac{1}{2}\left( -x+(x^2+1)\operatorname{arctanh} x+x\log(1-x^2)\right),\tag{4} $$
and by setting $g(x)=\sum_{n\geq 0} \frac{x^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)} $ we have: $$g(x)=\frac{1}{12}\left( -5x^2+(2x^4+6x^2)\frac{\operatorname{arctanh} x}{x}+(3x^2+1)\log(1-x^2)\right),\tag{5} $$
so:
$$\sum_{n\geq 0}\frac{x^{n+2}}{(2n+1)(2n+2)(2n+3)(2n+4)}=\frac{1}{12}\left(-5x+(2x^2+6x)\frac{\operatorname{arctanh}\sqrt{x}}{\sqrt{x}}+(3x+1)\log(1-x)\right)\tag{6}$$
and finally:
$$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}=\color{red}{\frac{5-\pi-2\log 2}{12}}.\tag{7}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum $\sum_{n=1}^\infty \frac{n^2}{(n+2)!}$ Problem is to find sum
$$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$
What I knew doesn't apply on this problem
Some series are telescoping, some types are solvable using binomial , both look useless here
Binomial gives
$$n (n-1) (n-2)x^3=1$$
$$n (n-1)( n-2)(n-3)x^4=4$$
What approach to use here?
| $$\sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}} = - 5 + 2\sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 1} \right)^2 }}{{n!}}} = - 5 + 2e
$$
$$
\begin{array}{l}
s = \sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{\left( {n + 2} \right)!}}} = \sum\limits_{n = 2}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}} = - \frac{{\left( {0 - 2} \right)^2 }}{{0!}} - \frac{{\left( {1 - 2} \right)^2 }}{{1!}} + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}} \\
s = - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}} = - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{\left( {n - 2} \right)^2 }}{{n!}}} \\
s = - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 - 4n + 4}}{{n!}}} = - 5 + \sum\limits_{n = 0}^{ + \infty } {\frac{{n^2 }}{{n!}}} - \sum\limits_{n = 0}^{ + \infty } {\frac{{4n}}{{n!}}} + \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}} \\
s = - 5 + \frac{{0^2 }}{{0!}} + \sum\limits_{n = 1}^{ + \infty } {\frac{{n^2 }}{{n!}}} - \frac{{4\left( 0 \right)}}{{0!}} - \sum\limits_{n = 1}^{ + \infty } {\frac{{4n}}{{n!}}} + \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}} \\
s = - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n - 1} \right)!}}} - \sum\limits_{n = 1}^{ + \infty } {\frac{4}{{\left( {n - 1} \right)!}}} + \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}} \quad ;\quad \left( {\sum\limits_{n = 1}^{ + \infty } {\frac{4}{{\left( {n - 1} \right)!}}} = \sum\limits_{n = 0}^{ + \infty } {\frac{4}{{n!}}} } \right) \\
s = - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n - 1} \right)!}}} = - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{{n - 1 + 1}}{{\left( {n - 1} \right)!}}} = - 5 + \sum\limits_{n = 1}^{ + \infty } {\frac{{n - 1}}{{\left( {n - 1} \right)!}}} + \sum\limits_{n = 1}^{ + \infty } {\frac{1}{{\left( {n - 1} \right)!}}} \\
s = - 5 + 2e \\
\end{array}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $2^{2x+1} - 2^{x+4} = 2^3 - 2^x$ $$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$
How can I solve an exponential equation that has many terms as the one above. Include more than one method if available.
| The key here is to transform this into a quadratic equation.
$$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$ can be rewritten as $$2 \cdot 2^{2x} - 2^4 \cdot 2^x = 2^3 -2^x.$$
Temporarily we will express $2^x=u$ and we find that the equation turns into $$2u^2 - 16u = 8 - u.$$ Finally writing this as $$2u^2 - 15u - 8 = 0$$ we can factor or use the quadratic equation to solve for $u$. Then remembering that $u=2^x$ we can find the solution for $x$ by writing $\log_2(u)=\ln(u)/\ln(2)=x$. Just be sure to try plugging the answers back into the original equation at the end.
| {
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Prove inequality $\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2 $ I cannot prove the following inequality.
Let $a,b,c$ be positive numbers.Prove that:
$\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2. $
I tried to use Cauchy inequality. I think the functions $f(a,b,c)=\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}$ and $g(a,b,c)=(a-b)^2+(b-c)^2+(c-a)^2$ are symmetric with respect to $a,b,c.$
| Using Cauchy inequality:$$\frac{a^3}{b+c}+a(b+c)\geq2a^2(1)$$
$$\frac{b^3}{a+c}+b(a+c)\geq2b^2(2)$$
$$\frac{c^3}{a+b}+c(a+b)\geq2c^2(3)$$
$$(1)+(2)+(3):\frac{a^3}{b+c}+a(b+c)+\frac{b^3}{a+c}+b(a+c)+\frac{c^3}{a+b}+c(a+b)\geq2(a^2+b^2+c^2)$$
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}+2(ab+bc+ac)\geq2(a^2+b^2+c^2)$$
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq2(a^2+b^2+c^2-(ab+bc+ac))$$
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)$$
$$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2$$
| {
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Summation to Equation I have a summation and I want to be able to find the sum for given $n$ without having to go through $1,\dots,n$.
$$\sum_{x=1}^{n - 1}x+300\cdot2^{x/7}$$
It's been awhile since I've done summations and I can't figure this one out. Also, this isn't for homework if that makes a difference.
Thanks
| Might as well give an answer. Note that we can split this sum into the following two sums: $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \sum_{x=1}^{n-1} x + \sum_{x=1}^{n-1} 300 \cdot 2^{x/7}.$$
Given that $$\sum_{x=1}^n x = \frac{n(n+1)}{2}$$ we know that $$\sum_{x=1}^{n-1} x = \frac{n(n+1)}{2} - n = \frac{n(n-1)}{2}.$$ We're halfway done! Turning our attention to the second sum, we have that
\begin{align*}
\sum_{x=1}^{n-1} 300 \cdot 2^{x/7} & = 300 \sum_{x=1}^{n-1} 2^{x/7} \\
& = 300 \left(\frac{2^{n/7} - 2^{1/7}}{2^{1/7} - 1}\right)
\end{align*}
by the formula for a finite geometric series.
Therefore, the total sum is $$\sum_{x=1}^{n-1} x + 300 \cdot 2^{x/7} = \frac{n(n-1)}{2} + 300\left(\frac{2^{n/7} - \sqrt[7]{2}}{\sqrt[7]{2} - 1}\right).$$ Plug in $n$ and you will have your solution.
| {
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Proof that $a^b>b^a$ if $aI would like a proof that if $a<b$ are integers with $2\leq a,b$ we have $a^b>b^a$ unless $a=2,b=3$ or $a=2,b=4$ . I would like to use as little calculus as possible. Here is my current solution:
Case 1: $a>2$
Fix $a$ and start with $b=a$. notice $a^a=a^a$. Now increase $b$ by $1$ to now get $a^{a+1}>(a+1)^a$. When $a$ is at least three this is true because the left side was multiplied by $a$ and the right side by $(1+\frac{1}{a})^a$.
Since $(1+\frac{1}{a})^a=\sum\limits_{i=0}^a \binom{a}{i} \frac{1}{a^i}$ is the sum of $a+1$ elements all of them smaller than $1$ and the last two terms add $\binom{a}{a-1}\frac{1}{a^{a-1}}+\binom{a}{a}\frac{1}{a^a}=\frac{1}{a^{a-2}}+\frac{1}{a^a}<\frac{1}{3}+\frac{1}{27}<1$ we conclude the sum is less than $a$. Therefore the inequality holds. Notice that each time we add $1$ to $b$ the left side is multiplied by $a$ and the right by something smaller than $(1+\frac{1}{a})^a$, hence the inequality holds for all $b$ larger than $a$.
Case $a=2$ is the same thing only we check $b=3,4$ by hand and then do something similar.
I am not really happy with the current solution, can we find something simpler? If it can be more combinatorial it would be better. Notice I want as little calculus or inequalities as possible.
| Consider $\mathbb{N}^a$ as a Cartesian lattice. Consider the collection of paths of length $b$ that
*
*start at the origin
*grow in legs of length $1$, moving parallel to one of the $a$ axes
*only grow outward from the origin
There are $a^b$ such paths, since for each leg you have $a$ directional options. We can identify a subset of these paths whose cardinality is greater than $b^a$, proving the claim.
Let $S$ be the subset of such paths where there is one direction which precisely $b-a$ of the legs run parallel to, another direction with precisely two legs running parallel, and each of the remaining directions has precisely one parallel leg.
When $b>a+2$, combinatorics tells us that there are $$|S|=\binom{b!}{b-a;2;\overbrace{1;\cdots;1}^{a-2}}\cdot a(a-1)$$ such paths. The multinomial coefficient counts how many paths have $b-a$ legs specifically parallel to the first direction, and two legs specifically parallel to the second. The $a(a-1)$ factor then accounts for all the permutations. Since $b>a+2$, there is no over counting. (We can handle $b=a+1$ and $b=a+2$ similarly, later below).
So $$\begin{align}
a^b
&>\binom{b!}{b-a;2;\overbrace{1;\cdots;1}^{a-2}}\cdot a(a-1)\\
&=b(b-1)\cdots(b-a+1)\cdot\binom{a}{2}\\
\end{align}$$
$b(b-1)\cdots(b-a+1)$ is shy of $b^a$, but the $\binom{a}{2}$ factor should compensate if $a$ is at least $3$. The limit of the ratio $\frac{b(b-1)\cdots(b-a+1)}{b^a}$ is $1$, so this certainly happens for large enough $b$. (It would be nice if we could establish it holds for all $b>a+2$.) Alternatively we could include more paths in $S$ that contribute like $\binom{b!}{b-a-1;2;2;\overbrace{1;\cdots;1}^{a-3}}$ which would cancel the subleading term in $b(b-1)\cdots(b-a+1)$.
When $b=a+2$, $|S|=\binom{b!}{2;2;\overbrace{1;\cdots;1}^{a-2}}\cdot \binom{a}{2}$, and when $b=a+1$, $|S|=\binom{b!}{1;2;\overbrace{1;\cdots;1}^{a-2}}\cdot \binom{a}{1}$. The above parts needs slight modification to compensate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find and specify the type of all singularities
Find and specify the type of all singularities of the function $$f(z)=\frac{(z-\frac{\pi}{2})z\sin(\frac{1}{z})}{\cos z}$$
I have some difficulties doing this
When it comes to singularities, there are $w_0:=0$ and $z_n:=\frac{\pi}{2}+n\pi$, $n\in\mathbb{Z}$
Do you have any hint for me? Should I use d'Hospital?
| First $z\sin(\frac{1}{z})$ has an essential singularity at $z=0$ since $z^n\cdot (z\sin(\frac{1}{z}))$ is not differentiable for any $n>0$ at that point. Regarding the rest of the expression. Notice that for $z=\frac{\pi}{2}+k\pi$ where $k\in\mathbb{Z}$, we have $\cos(z)=0$. On the other hand for $k=0$ we have both the numerator and the denominator equal to zero. Evaluating the value of the function as $z\to\frac{\pi}{2}$ we get$$\lim_{z\to\frac{\pi}{2}}f(z)=\lim_{z\to\frac{\pi}{2}}\Big(\frac{z-\frac{\pi}{2}}{\cos(z)}\cdot z\sin(\frac{1}{z})\Big)=\frac{\pi}{2}\sin(\frac{2}{\pi})\lim_{z\to\frac{\pi}{2}}\Big(\frac{z-\frac{\pi}{2}}{\cos(z)}\Big)$$$$=\frac{\pi}{2}\sin(\frac{2}{\pi})\frac{1}{\lim_{z\to\frac{\pi}{2}}\Big(\frac{\cos(z)-\cos(\frac{\pi}{2})}{z-\frac{\pi}{2}}\Big)}=\frac{\pi}{2}\sin(\frac{2}{\pi})\cdot\frac{1}{-\sin(\frac{\pi}{2})}=-\frac{\pi}{2}\sin(\frac{2}{\pi})<\infty$$
So for $z=\frac{\pi}{2}$ the function is finite. For all other $k\neq0$ we have the denominator equal to zero while the numerator a finite number different from zero and hence the function has singularities for all $z=\frac{\pi}{2}+k\pi$ where $k\in\mathbb{Z}-\{0\}$. These singularities are simple poles with residue
$$\lim_{z\to\frac{\pi}{2}+k\pi}(z-\frac{\pi}{2}-k\pi)f(z)=k\pi\Big(\frac{\pi}{2}+k\pi\Big)\sin(\frac{1}{\frac{\pi}{2}+k\pi})\cdot\lim_{z\to\frac{\pi}{2}+k\pi}\frac{z-\frac{\pi}{2}-k\pi}{\cos(z)}$$
$$=k\pi\Big(\frac{\pi}{2}+k\pi\Big)\sin(\frac{1}{\frac{\pi}{2}+k\pi})\cdot\lim_{z\to\frac{\pi}{2}+k\pi}\frac{z-\frac{\pi}{2}-k\pi}{\cos(z)-\cos(\frac{\pi}{2}+k\pi)}$$
$$=k\pi\Big(\frac{\pi}{2}+k\pi\Big)\sin(\frac{1}{\frac{\pi}{2}+k\pi})\cdot\frac{1}{-\sin(\frac{\pi}{2}+k\pi)}=k\pi\Big(\frac{\pi}{2}+k\pi\Big)\sin(\frac{1}{\frac{\pi}{2}+k\pi})(-1)^{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
An $n \times n$ matrix with rational entries such that $A^{n+1}=I$ I'm working on finding $A \in M_n(\mathbb{Q})$ such that $A^{n+1}=I$. If $n$ is odd, $A=-I$ satisfies the condition. When $n$ is even, clearly it should have eigenvalues $e^{2 \pi ik/(n+1)}(k=1,\cdots ,n)$ as a complex matrix. Now I proved this matrix should be similar to $\begin{pmatrix} R_\theta & 0 & \cdots & 0\\0&R_{2\theta} & \cdots &0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & R_{n\theta/2} \end{pmatrix} $, where $R_\theta = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ and $\theta = \frac{2\pi}{n+1}$. Unfortunately, however, $R_\theta$ is not rational if $n \ge 4$. Is there any other method?
| First step: decompose $x^{n+1}-1$ in irreducible polynomials of $\mathbb{Q}[x]$.
*
*About your example, we can generalize as follows: let $p$ be a prime number and let $A\in M_{p-1}(\mathbb{Q})$ s.t. $A^p=I_{p-1}$. One has $x^p-1=(x-1)cyclo(p,x)$. Thus the minimal polynomial of $A$ is $x-1$ and $A=I_4$ or $r(x)=cyclo(p,x)$ and $A$ is similar, over $\mathbb{Q}$, to $C_r$, that is the companion matrix of $r$.
*$A\in M_8(\mathbb{Q})$ s.t. $A^9=I_8$. One has $x^9-1=(x-1)(x^2+x+1)(x^6+x^3+1)$. The minimal polynomial of $A$ is $x-1$ and $A=I$ or $q(x)=x^2+x+1$ and $A\sim diag(C_q,C_q,C_q,C_q)$ or $q(x)r(x)=(x^2+x+1)(x^6+x^3+1)$ and $A\sim diag(C_q,C_r)$ or $(x-1)r(x)$ and $A\sim diag(1,1,C_r)$ or $(x-1)q(x)$ and [$A\sim diag(1,1,1,1,1,1,C_q)$ or $A\sim diag(1,1,1,1,C_q,C_q)$ or $A\sim diag(1,1,C_q,C_q,C_q)$].
*$A\in M_3(\mathbb{Q})$ s.t. $A^{4}=I_3$. One has $x^{4}-1=(x-1)(x+1)(x^2+1)$. The minimal polynomial of $A$ is $x-1$ and $A=I$ or $x+1$ and $A=-I$ or $x^2-1$ and $A$ is a symmetry or $(x-1)q(x)=(x-1)(x^2+1)$ and $A\sim diag(1,C_q)$ or $(x+1)q(x)$ and $A\sim diag(-1,C_q)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Simplify $\cos 1^\circ + \cos 3^\circ + \cdots+ \cos 43^\circ$? I am currently working on a problem and reduced part of the equations down to
$$\cos(1^\circ)+\cos(3^\circ)+\cdots+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ)$$
How can I calculate this easily using the product-to-sum formula for $\cos(x)+\cos(y)$?
| The trick is to multiply the whole sum by $2\sin 1^\circ$. Since:
$$ 2\sin 1^\circ \cos 1^{\circ} = \sin 2^\circ - \sin 0^\circ,$$
$$ 2\sin 1^\circ \cos 3^{\circ} = \sin 4^\circ - \sin 2^\circ,$$
$$\ldots $$
$$ 2\sin 1^\circ \cos 43^{\circ} = \sin 44^\circ - \sin 42^\circ,$$
by adding these identites we get that the original sum $S$, multiplied by $2\sin 1^\circ$, equals $\sin 44^\circ$.
This just gives $S=\color{red}{\frac{\sin 44^\circ}{2\sin 1^{\circ}}}.$
Footnote. Such sum must be greater than $\frac{44\sqrt{2}}{\pi}$, but not greater than $\frac{44\sqrt{2}}{\pi}+\frac{\pi}{22\sqrt{2}}$, by a Riemann sum $+$ concavity argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$? I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
| Here are the steps
$$
\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right] \left[{\frac{\sqrt{1 + x + x^2} + 1}{\sqrt{1 + x + x^2} + 1}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{1 + x + x^2 - 1}{ x\left(\sqrt{1 + x + x^2} + 1\right)}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{x\left(1+ x\right)}{x\left(\sqrt{1 + x + x^2} + 1\right)}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{1+ x}{\sqrt{1 + x + x^2} + 1}}\right]
$$
$$
={\frac{1+ 0}{\sqrt{1 + 0+ 0} + 1}}
$$
$$
={\frac{1}{\sqrt{1} + 1}}
$$
$$
=\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Integrate rational function with multiple complex roots I want to integrate
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx.
$$
By WolframAlpha I know the solution is
$$
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx = \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}.
$$
Do derive it on my own, I did a partial fraction decomposition and found
$$
\frac{x^2}{(1+n^2x^2)^2} = \frac{\frac{1}{2n^2}}{1+n^2x^2} + \frac{\frac{1}{4n^2}}{(x-i/n)^2} + \frac{\frac{1}{4n^2}}{(x+i/n)^2}.
$$
I checked this multiple times, but it must be wrong, because when I proceed with it I get
\begin{align*}
\int_{-1}^1 \frac{x^2}{(1+n^2x^2)^2} dx
& = \frac{1}{n^2} \left( \frac{1}{2} \int_{-1}^1 \frac{1}{1+n^2x^2} d x + \frac{1}{4} \int_{-1}^1 \frac{1}{(x-i/n)^2} d x + \frac{1}{4} \int_{-1}^1 \frac{1}{(x+i/n)^2} d x \right) \\
& = \frac{1}{n^2}\left( \frac{1}{2} \frac{\arctan(xn)}{n}\bigg\vert_{-1}^1 + \frac{1}{4} \left( \frac{-1}{x-i/n}\bigg\vert_{-1}^1 + \frac{-1}{x+i/n}\bigg\vert_{-1}^1 \right)\right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} + \frac{1}{4} \left( \frac{-2}{1+1/n^2} + \frac{-2}{1+1/n^2}\right)\right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} -\frac{1}{1+1/n^2} \right) \\
& = \frac{1}{n^2} \left( \frac{\arctan(n)}{n} - \frac{n^2}{n^2+1} \right) \\
& = \frac{\arctan(n)}{n^3} - \frac{1}{n^2+1}
\end{align*}
This result is consistent which what I found with WolframAlpha, but surely
$$
\frac{\arctan(n)}{n^3} - \frac{1}{n^2+1} \ne \frac{\arctan(n) - \frac{n}{n^2+1}}{n^3}
$$
so what went wrong, I am sitting here since hours and do not see any fault...
| To avoid multiple complex conjugate roots, you can use the formula in Application to symbolic integration
\begin{align}
\frac{x^2}{(1+n^2x^2)^2}
&=\frac{d}{dx}\left[\frac{Ax+B}{1+n^2x^2}\right]+\frac{Cx+D}{1+n^2x^2}\\
&=\frac{Cn^2x^3+(D-A)n^2x^2+(C-2Bn^2)x+(D+A)}{(1+n^2x^2)^2}
\end{align}
so that
$$\left\{
\begin{align}
A&=-\frac{1}{2n^2}\\
B&=0\\
C&=0\\
D&=+\frac{1}{2n^2}
\end{align}
\right.
$$
and the integral becomes
\begin{align}
\int\frac{x^2}{(1+n^2x^2)^2}dx
&=\frac{Ax+B}{1+n^2x^2}+\int\frac{Cx+D}{1+n^2x^2}dx\\
&=-\frac{x}{2n^2(1+n^2x^2)}+\frac{1}{2n^2}\int\frac{1}{1+n^2x^2}dx\\
&=-\frac{x}{2n^2(1+n^2x^2)}+\frac{1}{2n^3}\arctan(nx)+c
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why can't I prove this statement by simple induction? Sum of $1/2^1 + \cdots+ n/2^n = x$ I have to prove the following:
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I am trying to prove this by simple induction. First, I proved that $P(1)$ holds. It clearly does.
I then assume that $n$ is a positive number $> 1$ and that $P(n)$ holds. Hence,
$$
\frac{1}{2}+\ldots+\frac{n}{2^n}=2-\frac{2 + n}{2^n}.
$$
I now add ${\displaystyle \frac{n + 1}{2\cdot 2^n}}$ to both sides to get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2\cdot 2^n}=2-\frac{2+n}{2^n}+\frac{n+1}{2\cdot 2^n}.
$$
But after manipulating the right side I get the following:
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{3n + 5}{2^{n+1}}.
$$
This is definitely not right since I should be getting
$$
\frac{1}{2}+\cdots+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}=2-\frac{n + 3}{2^{n+1}}=2-\frac{2+(n+1)}{2^{n+1}}
$$
to prove that $P(n)$ implies $P(n+1)$.
What am I doing wrong?
| $$\begin{array}{l}
\left| x \right| < 1:\frac{1}{{1 - x}} = \sum\limits_{k = 0}^\infty {x^k } = 1 + \sum\limits_{k = 1}^\infty {x^k } \\
\Rightarrow \sum\limits_{k = 1}^\infty {kx^k } + \sum\limits_{k = 1}^\infty {x^k } = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{x}{{1 - x}} \\
\Rightarrow \sum\limits_{k = 1}^\infty {\left( {k + 1} \right)x^k } = \frac{x}{{\left( {1 - x} \right)^2 }} + \frac{x}{{1 - x}} \\
\left| x \right| < 1;x = \frac{1}{2} \Rightarrow \sum\limits_{k = 1}^\infty {\left( {k + 1} \right)\left( {\frac{1}{2}} \right)^k } = \frac{{\frac{1}{2}}}{{\left( {1 - \frac{1}{2}} \right)^2 }} + \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = 3 \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$ Let $a,b,c\in{\mathbb{R^+}}$. Prove that
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$
I tried to expand both, but I did not get anything useful.
| As Macavity says, this is the same as
$$
\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +\left(\frac{b}a+\frac{a}c+\frac{c}b\right)\ge \left(\frac{a}b+\frac{b}c+\frac{c}a\right)+3
$$
which is the same after rearrangement as
$$
\underbrace{\left(\frac ab-1\right)^2+\left(\frac bc-1\right)^2+\left(\frac ca-1\right)^2}_{\ge0}
\ge\underbrace{6-\left(\frac ab+\frac bc+\frac ca\right)-\left(\frac ba+\frac ac+\frac cb\right)}_{\le0}
$$
since $x+\frac1x\ge2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Calculate $ S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. $ Calculate $S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. This sequence is neither arithmetic nor geometric. How can you solve this. Thanks!
| Since $$\frac{1}{k(k+2)} = \frac{1}{2} \frac{2}{k(k+2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k+2}\right)$$
for all $k$, we have
$$\sum_{k = 1}^n \frac{1}{k(k+1)(k+2)} = \frac{1}{2}\sum_{k = 1}^n \left(\frac{1}{k(k+1)} - \frac{1}{(k+1)(k+2)}\right),$$
which telescopes to $$\frac{1}{2}\left(\frac{1}{2} - \frac{1}{(n+1)(n+2)}\right) = \frac{1}{4} - \frac{1}{2(n+1)(n+2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Sum $\sum_{n=1}^{\infty}\ln(1+2^{-2^n})$ How can I solve this: Find the sum of:
$$\sum_{n=0}^{\infty}\ln(1+2^{-2^n})=?$$ and $$\sum_{n=1}^{\infty}{(-1)^{n+1}}\cdot{1\over n}=$$
Can you please give me not the solution of the problem, but just how I can begin. Thanks.
| $$I=\left(1+\dfrac{1}{2^{2^0}}\right)\left(1+\dfrac{1}{2^{2^1}}\right)\left((1+\dfrac{1}{2^{2^2}}\right)\cdots\left(1+\dfrac{1}{2^{2^n}}\right)$$
use this
$$(x+y)(x-y)=x^2-y^2$$
then
$$\dfrac{(1-\dfrac{1}{2^{2^0}})I}{1-\dfrac{1}{2^{2^0}}}=\dfrac{1-\dfrac{1}{2^{2^{n+1}}}}{1-\dfrac{1}{2}}\to 2,n\to \infty$$
second 2
Note
\begin{align*}
&\sum_{k=1}^{2n}(-1)^{k-1}\dfrac{1}{k}=1-\dfrac{1}{2}+\dfrac{1}{3}-\cdots+\dfrac{1}{2n-1}-\dfrac{1}{2n}\\
&=H_{2n}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\cdots+\dfrac{1}{2n}\right)\\
&=H_{2n}-H_{n}\\
&=\ln{(2n)}+C+o(2n)-\ln{n}-C-o(n)=\ln{2},n\to \infty
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Street Fighter: is the game balanced? Suppose that $A$ is a matrix that describes the matchup information of any pair of Street Fighter characters e.g., considering $3$ characters, assume that the first row/collumn is associated with a character like Ryu, the second row/collumn is associated with Chun-li, the third row/collumn with Dhalsim, etc. Then for a $3\times3$ matrix we have:
$$A=\begin{pmatrix} 5 & 3 & 9 \\ 7 & 5 & 4 \\ 1 & 6 & 5 \end{pmatrix},$$
which tell us e.g., if Ryu fights Chun-li he wins $3$ out of $10$ matches.
Which kind of processing can I do to $A$, so that I know more information about the game?
| Let's start by subtracting $5$ from each entry to turn it into a payoff matrix for a zero-sum game: $$A'=\begin{pmatrix} 0 & -2 & 4 \\ 2 & 0 & -1 \\ -4 & 1 & 0 \end{pmatrix}$$
Now we can look for a mixed strategy (a probability with which we select each character) which is a Nash equilibrium. That boils down to
$$\begin{pmatrix} 0 & -2 & 4 \\ 2 & 0 & -1 \\ -4 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} k \\ k \\ k \\ 1\end{pmatrix}$$ where $k$ is unknown. But we can subtract the first row from each of the other rows of $A'$ to get $$\begin{pmatrix} 0 & -2 & 4 \\ 2 & 2 & -5 \\ -4 & 3 & -4 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} k \\ 0 \\ 0 \\ 1\end{pmatrix}$$ and then discard the first row to get $$\begin{pmatrix} 2 & 2 & -5 \\ -4 & 3 & -4 \\ 1 & 1 & 1 \end{pmatrix}\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$$ which gives $$\begin{pmatrix} p_0 \\ p_1 \\ p_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{7} \\ \frac{4}{7} \\ \frac{2}{7}\end{pmatrix}$$
So the optimal strategy is to roll a seven-sided die and play Ryu on a 1, Chun-Li on a 2 to 5, and Dhalsim on a 6 or 7. As a tool to assess balance, you could consider using a metric on how far the mixed strategy is from $(\frac13, \frac13, \frac13)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.
| Taking the contrapositive, we suppose $x+y+z<\sqrt 3$ and want to prove
$$xy+yz+zx<\frac 1{\sqrt{x^2+y^2+z^2}}\iff (xy+yz+zx)^2(x^2+y^2+z^2)<1$$
Scale $(x,y,z)$ by dividing all variables by $\sqrt3$. Then $x+y+z<3$ and we want to prove
$$(xy+yz+zx)^2(x^2+y^2+z^2)<27$$
And that's a simple AM-GM:
$$(xy+yz+zx)^2(x^2+y^2+z^2)\le\left(\frac{2(xy+yz+zx)+x^2+y^2+z^2}3\right)^3=\left(\frac{(x+y+z)^2}3\right)^3<27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Trigonometric functions - finding solutions Question:
Find the general solution for the equation:
$$\sin x + 2\sin2x - \sin3x = 3$$
Approach:
Well using the identity of $\sin a - \sin b $, I merged together $\sin x - \sin3x$
And as $\sin 2x = 2\sin x\cos x$, I got up till here:
$$2\sin x(2\cos x - \cos 2x) = 3$$
Edit:
Considering $\sin x - \sin 3x = -2\cos 2x\sin x$
The equation becomes:
$$-2\cos 2x\sin x+2\sin 2x = 3$$
$$(-2\sin x)\cos 2x +2\sin 2x = 3$$
The maximum value of the equation $asinx + bcosx$ is $\sqrt{a^2 + b^2}$
Using this in here: (considering $b = -2\sin x$ and $a = 2$)
$$3 \leq \sqrt{4\sin^2 x + 4}$$
$$3/2 \leq \sqrt{1 + sin^2x}$$
Obviously cannot happen as maximum value of $\sqrt{1 + sin^2x}$ is $\sqrt2$
| There are no real solutions. $s = \sin(x)$ must satisfy
$16s^6-24s^3-12s^2+12s+9=0$, which has no real root.
| {
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"url": "https://math.stackexchange.com/questions/1113679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why's the derivative of $f(x) = x^3-5x-2 $ not $3x^2-7$? I wanted to resolve this problem :
$$
f(x) = 3 x^2 - 5 x - 2
$$
to a derivative, and I did it like this :
$
\begin{align}
f(x) &= x^3-5x-2 \\
f'(x) &= 3x^2-5-2 \\
&= 3x^2-7
\end{align}
$
but once I checked the correction, I found this:
$$
f'(x) = 3 x^2 - 5
$$
And I really don't know how they manage to get $3x^2 -5$ instead of $3x^2-7$.
Thanks for you time
| The derivative of a constant is zero, so if $$f(x) = x^3 - 5x - 2 = x^3 - 5 x^1 - 2,$$ the correct derivative is $$f'(x) = 3x^{3-1} - 5 \cdot 1 x^{1-1} - \color{red}{0} = 3x^2 - 5x^0 - 0 = 3x^2 - 5.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $x^2 + y^2 + 1 \le \sqrt{(x^3 + y + 1)(y^3 + x + 1)}$ For $x, y \ge 0$ prove that:
$$x^2 + y^2 + 1 \le \sqrt{(x^3 + y + 1)(y^3 + x + 1)}$$
What I think would apply is the AM-GM Inequality, so first,
$$(x^2 + y^2 + 1)^2 \le (x^3 + y + 1)(y^3 + x + 1)$$
$$\implies (x^3 + y + 1)(y^3 + x + 1) - (x^2 + y^2 + 1)^2 \ge 0$$
$$\implies x^3y^3 + x^3 - 2x^2y^2 - 2x^2 + xy + x + y^3 - 2y^2 + y \ge 0$$
From AM-GM,
$$\frac{ x^3y^3 + x^3 - 2x^2y^2 - 2x^2 + xy + x + y^3 - 2y^2 + y}{9} \ge \sqrt[9]{x^3y^3(x^3)(-2x^2y^2) (-2x^2)(xy) (x) (y^3) (- 2y^2) (y)}$$
The negative is the issue we have: $(-2x^2)(-2x^2y^2)(-2y^2)$ which is a problem.
How can this be done? Please, HINTS only.
| Solution 1:
Only use Cauchy-Schwarz inequality :
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$
let $$a_{1}=x^{\frac{3}{2}},a_{2}=y^{\frac{1}{2}},a_{3}=1$$
$$b_{1}=x^{\frac{1}{2}},b_{2}=y^{\frac{3}{2}},b_{3}=1$$
so we have
$$(x^3+y+1)(x+y^3+1)\ge (x^2+y^2+1)^2$$
so
$$\sqrt{(x^3+y+1)(y^3+x+1)}\ge x^2+y^2+1$$
Solution 2:
you can also
$$\Longleftrightarrow (x^3y^3-2x^2y^2+xy)+(x^3-2x^2+x)+(y^3-2y^2+y)\ge 0$$
since use AM-GM inequality
$$ x^3y^3+xy\ge 2\sqrt{x^3y^3\cdot xy}=2x^2y^2$$
$$x^3+x\ge 2\sqrt{x^3\cdot x}=2x^2$$
$$y^3+y\ge 2\sqrt{y^3\cdot y}=2y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
| Let $p=a+b$, and suppose that $p$ divides at least one of $a^a+b^b$ or $a^b+b^a$.
Then $p$ divides the product
$$(a^a+b^b)(a^b+b^a)=a^p+b^p+(ab)^a+(ab)^b $$
By Fermat's little theorem
*
*$a^p= a\mod p$
*$b^p= b\mod p$
*$a^p+b^p= a+b = 0 \mod p$
So (remember we suppose that $p$ divides the product) we have :
$$ (ab)^a+(ab)^b=0\mod p$$
But $b=-a\mod p$.
$$ (-a^2)^a+(-a^2)^b=0 \mod p$$
As $a+b$ is odd, one of $a$ and $b$ is odd and the other is even. So
$$ a^{2a}=a^{2b} \mod p $$
it means that $2b-2a$ is multiple of the order $r$ of $a$, and that order $r$ divides $p-1=a+b-1$. But if $r$ divides $(2b-1)-(2a-1)$ and $2(a+b-1)=(2a-1)+(2b-1)$, then either $r=2$ or it divides $2a-1$ and $2b-1$ (they are both different prime numbers) so $r=1$.
So $a=1\mod p$ or $a=-1\mod p$. But $a+b=p$. So $a<p$, so $a=1$ (but then $2a-1$ is not prime) or $a=p-1$ (but then $2b-1$ is not prime), this is not possible.
A contradiction. So, our first hypothesis that $p$ divides the product is false.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the general term of $a_{n+1}=\frac{2a_n-1}{5a_n-1} \ , \ \ a_1=1$? I've struggled to solve this exercise
$$a_{n+1}=\frac{2a_n-1}{5a_n-1}\ , \ \ a_1=1$$
$$b_{n+1}=(5a_n-1)b_n \ , \ \ b_1=1$$
Find $b_{\ 40}$ .
$$$$
I thought 'taking inverse' will be useful, but not yet... :-(
$\color{red}{01.}$ How can I find $b_{\ 40}$ ?
$\color{red}{02.}$ How can I find the general term of $a_n$, $b_n$ ?
$$$$
Thank you for your attention to this matter.
$$$$
| since
$$b_{n+1}=(5a_{n}-1)b_{n}\Longrightarrow a_{n}=\dfrac{1}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)$$
take in
$$a_{n+1}=\dfrac{2a_{n}-1}{5a_{n}-1}\Longrightarrow \dfrac{1}{5}\left(\dfrac{b_{n+2}}{b_{n+1}}+1\right)=\dfrac{\frac{2}{5}\left(\dfrac{b_{n+1}}{b_{n}}+1\right)-1}{\frac{b_{n+1}}{b_{n}}}$$
$$\Longrightarrow \dfrac{b_{n+2}+b_{n+1}}{5b_{n+1}}=\dfrac{\frac{2}{5}b_{n+1}+\frac{2}{5}b_{n}-b_{n}}{b_{n+1}}$$
$$\Longrightarrow b_{n+2}+b_{n+1}=2b_{n+1}-3b_{n}$$
$$\Longrightarrow b_{n+2}=b_{n+1}-3b_{n},b_{1}=1,b_{2}=(5a_{1}-1)b_{1}=4$$
The corresponding characteristic equation is
$$r^2=r-3\Longrightarrow r=\dfrac{1\pm\sqrt{11}i}{2}$$
so
$$b_{n}=A\left(\dfrac{1+\sqrt{11}i}{2}\right)^n+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^n$$
since $b_{1}=1,b_{2}=4$
so we have
$$\begin{cases}
A\left(\dfrac{1+\sqrt{11}i}{2}\right)+B\left(\dfrac{1-\sqrt{11}i}{2}\right)=1\\
A\left(\dfrac{1+\sqrt{11}i}{2}\right)^2+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^2=4
\end{cases}$$
$$\Longrightarrow \begin{cases}
3A+B\left(\dfrac{1-\sqrt{11}i}{2}\right)^2=\dfrac{1-\sqrt{11}i}{2}\\
A\left(\dfrac{1+\sqrt{11}i}{2}\right)^2+3B=\dfrac{1+\sqrt{11}i}{2}
\end{cases}\Longrightarrow \begin{cases}
A=\dfrac{7+\sqrt{11}i}{\sqrt{11}i-11}\\
B=\dfrac{\sqrt{11}i-7}{11+\sqrt{11}i}
\end{cases}$$
so
$$b_{n}=\dfrac{7+\sqrt{11}i}{\sqrt{11}i-11}\cdot\left(\dfrac{1+\sqrt{11}i}{2}\right)^n+\dfrac{\sqrt{11}i-7}{11+\sqrt{11}i}\cdot\left(\dfrac{1-\sqrt{11}i}{2}\right)^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Where does the sum of $\sin(n)$ formula come from? I have seen Lagrange's formula for the sum of $\sin(n)$ from $1$ to $n$ during one of my classes last week, but I never saw how it came to be. I tried googling it to find a proof but couldn't seem to find any as it kept bringing up his other work instead and just statements of the formula rather than derivations/proofs.
I'm really interested to see where it comes from so if anyone has any nice proofs that would be appreciated.
Thank you!
| A comment on a follow-up question suggests using the following identity:
$$
2 \sin \left(\tfrac12\right) \sin (k)
= \cos\left(k - \tfrac12\right) - \cos\left(k + \tfrac12\right).
$$
Then
\begin{align}
2 \sin \left(\tfrac12\right)&\left(\sin(1) + \sin(2) + \sin(3) + \cdots + \sin(n-1) + \sin(n)\right) \\
&= \left(\cos\left(\tfrac12\right) - \cos\left(\tfrac32\right)\right)
+\left(\cos\left(\tfrac32\right) - \cos\left(\tfrac52\right)\right)
+\left(\cos\left(\tfrac52\right) - \cos\left(\tfrac72\right)\right)\\
& \qquad + \cdots
+\left(\cos\left(n - \tfrac32\right) - \cos\left(n - \tfrac12\right)\right)
+\left(\cos\left(n - \tfrac12\right) - \cos\left(n + \tfrac12\right)\right) \\
&= \cos\left(\tfrac12\right) - \cos\left(n + \tfrac12\right) \\
&= \sin\left(\tfrac{n+1}2\right) \sin\left(\tfrac n2\right).
\end{align}
Therefore
$$
\sin(1) + \sin(2) + \cdots + \sin(n)
= \frac{\sin\left(\frac{n+1}2\right) \sin\left(\frac n2\right)}
{\sin\left(\frac12\right)}.
$$
There is another derivation of this formula
(using $2 \sin(1) \sin(k) = \cos(k-1) -\cos(k + 1)$)
in another answer.
| {
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"url": "https://math.stackexchange.com/questions/1119043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Trying to understand a step in Ramanujan's Proof of Bertrand's Postulate regarding the gamma function My question relates to this step in the proof here:
But it is easy to see that
$$\log \Gamma(x)-2\log\Gamma(\frac12x+\frac12) \le \log\left\lfloor x\right\rfloor!-2\log\left\lfloor\frac12x\right\rfloor!$$
if I use $\left\{\dfrac{x}{2}\right\} = \dfrac{x}{2} - \left\lfloor\dfrac{x}{2}\right\rfloor$ to represent the fractional part.
It is clear to me that the statement is true for when $\left\{\dfrac{x}{2}\right\} \le \dfrac{1}{2}$. I am having trouble understanding why it is necessarily true when $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$
When $\left\{\dfrac{x}{2}\right\} \le \dfrac{1}{2}$, I can use the answer from here.
where
$x_1 = x$, $\Delta{t_1} = 1-\left\{x\right\}$, $x_1+\Delta t_1 = \lfloor{x}\rfloor+1$
$x_2 = \dfrac{x}{2}+\dfrac{1}{2}$, $\Delta{t_2} = \dfrac{1}{2}-\left\{\dfrac{x}{2}\right\}$, $x_2+\Delta t_2 = \lfloor\dfrac{x}{2}\rfloor+1$
$x_3 = \dfrac{x}{2}+\dfrac{1}{2}$, $\Delta{t_3} = \dfrac{1}{2}-\left\{\dfrac{x}{2}\right\}$, $x_3+\Delta t_3 = \lfloor\dfrac{x}{2}\rfloor+1$
to get:
$$\dfrac{\Gamma(\left\lfloor{x}\right\rfloor+1)}{\Gamma({x})} \ge \frac{\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1)}{\Gamma(\dfrac{x}{2}+\dfrac{1}{2})}\dfrac{\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1)}{\Gamma(\dfrac{x}{2}+\dfrac{1}{2})}$$
This approach fails for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$ since $\Delta{t_2}, \Delta{t_3} < 0$
Can someone provide me the argument for why this inequality is true for the condition where $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$?
Thanks very much.
Edit: I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$
I posted it as the answer below.
| I figured out an argument that works for $\left\{\dfrac{x}{2}\right\} > \dfrac{1}{2}$
$$\log\Gamma(x) \le \log\Gamma(x + 1 - \left\{ x\right\}) = \log\Gamma(\left\lfloor x\right\rfloor + 1) = \log\left\lfloor x\right\rfloor!$$
$$2\log\Gamma(\dfrac{x}{2} + \dfrac{1}{2}) \ge 2\log\Gamma(\dfrac{x}{2}+1 - \left\{\dfrac{x}{2}\right\}) = 2\log\Gamma(\left\lfloor\dfrac{x}{2}\right\rfloor+1) = 2\log\left\lfloor\dfrac{x}{2}\right\rfloor! $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is ... A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}$, where $a$, $b$, and $c$ are integers. Find $a+b+c$.
| Manuel Lafond’s suggestion that you try some small values of $n$ is always worth pursuing if you don’t see a cleverer approach. By actual calculation you can fairly readily discover that $$\begin{align*}
S_0&=\frac12=1-\frac12\;,\\
S_1&=\frac9{10}=1-\frac1{10}\;,\\
S_2&=\frac{409}{410}=1-\frac1{410}\;,\text{ and}\\
S_3&=\frac{1345209}{1345210}=1-\frac1{1345210}\;.
\end{align*}$$ We’d like to find $a,b$, and $c$ such that $$\begin{align*}
\frac12&=\frac{a^b}{3^{2^c}-1}\;,\\
\frac1{10}&=\frac{a^{1+b}}{3^{2^{1+c}}-1}\;,\\
\frac1{410}&=\frac{a^{2+b}}{3^{2^{2+c}}-1}\;,\text{ and}\\
\frac1{1345210}&=\frac{a^{3+b}}{3^{2^{3+c}}-1}\;.
\end{align*}$$
The third one looks like a good place to start: the fourth involves uncomfortably large numbers, and the first and second look as if they might allow too many reasonable possibilities. For the third we need $2^{2+c}$ to be a number $m$ such that $3^m-1$ is a multiple of $410$. The first possibility is $m=8$, and indeed $3^8-1=6560=16\cdot410$. That suggests that we try the hypothesis that $$S_n=1-\frac{2^{n+2}}{3^{2^{n+1}}-1}\;.$$
This yields the correct values for $n=0,1$, and $3$, so it’s almost certainly correct, and all that remains is to prove it by induction, as you suggested. The induction step requires showing that $$1-\frac{2^{n+2}}{3^{2^{n+1}}-1}+\frac{2^{n+2}}{3^{2^{n+1}}+1}=1-\frac{2^{n+3}}{3^{2^{n+2}}-1}\;,$$ which is a straightforward bit of algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability question please guide There are $5$ red and $3$ blue chips in a bowl. The red ones are numbered $1,2,3,4,5$ and the blue ones as $1,2,3$ respectively. if $2$ chips are drawn without repacement, find the probability that these chips have either the same number or the same colour.
What i think ::
For same colour, for the blue ones it will be $\binom{5}{1}*\binom{4}{1}$ and for the red ones $\binom{3}{1}*\binom{2}{1}$. So total would be, $\binom{5}{1}*\binom{4}{1}$ + $\binom{3}{1}*\binom{2}{1}$.
For the same colour, it would be, $3(\binom{5}{1}*\binom{3}{1})$.
So answer would be :: $$\frac{\binom{5}{1}*\binom{4}{1} + \binom{3}{1}*\binom{2}{1}+3(\binom{5}{1}*\binom{3}{1})}{\binom{8}{2}}$$
Is this correct? if not, then please help me.
help appreciated. This is my first post so please go easy on me if i did somthing wrong :)
| The total number of ways to choose $2$ out of $8$ chips is $\binom82=28$.
The number of ways to choose $2$ chips with the same color is $\binom52+\binom32=13$.
The number of ways to choose $2$ chips with the same number is $\min(5,3)=3$.
So the probability to choose $2$ chips with the same color or number is $\frac{3+13}{28}=\frac{4}{7}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$
I was able to notice that: $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx $$
Where should I take it from here? (assuming I'm not the right way)
EDIT
Following the anwer:
$$\int \sum_{n=0}^\infty x^{2n-1} dx = \int \frac{\sum_{n=0}^\infty (x^2)^n}{x} dx = \int \frac{\frac{1}{1-x^2}}{x} dx= \int \frac{1}{x(1-x^2)}dx = \int \frac{1-x^2+x^2}{x(1-x^2)} = \int \frac{1}{x} dx + \int \frac{x}{1-x^2}dx = \ln(x) + \int \frac{1}{2}\frac{1}{1-t}dt + C = \ln(x) - \frac{\ln(x^2)}{2} + C$$
*
*Is that right?
*How to evaluate $C$?
| $$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \\
\log \frac{1}{1-x} = x + \frac{x^2}{2} + \frac{x^3}{3} +\frac{x^4}{4} + \cdots $$
and so
$$\frac{1}{2}(\, \log(1+x) + \log \frac{1}{1-x}\,) = x + \frac{x^3}{3}+\frac{x^5}{5}+ \cdots$$
or
$$ x + \frac{x^3}{3}+\frac{x^5}{5}+ \cdots= \log \sqrt{ \frac{1+x}{1-x}}$$
for $|x| \le 1$, $x \ne \pm 1$.
$\tiny{\log \sqrt{ \frac{1+x}{1-x}}= \text{arctanh} x= \frac{\arctan(ix)}{i}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Need help understanding Fibonacci Fast Doubling Proof From this website, http://www.nayuki.io/page/fast-fibonacci-algorithms (fast doubling proof close to the bottom of the page).
I have understood the proof for the most part but I am struggling to see how this part of the proof works especially when the the F(n) function is squared.
\begin{align}
F(2n+1) &= F(n+1)^2 + F(n)^2. \\
F(2n) &= F(n) \left[ F(n+1) + F(n-1) \right] \\
&= F(n) \left[ F(n+1) + (F(n+1) - F(n)) \right] \\
&= F(n) \left[ 2F(n+1) - F(n) \right]. \\
F(2n-1) &= F(n)^2 + F(n-1)^2.
\end{align}
| Start with:
$$F(n+1) = F(n) + F(n-1)$$
Rewrite as:
$$\begin{align} F(n + 1) &= F(n) + F(n-1) \\
F(n) &= F(n) \end{align}$$
which is:
$$\begin{align}
\begin{bmatrix} F(n+1) \\ F(n)\end{bmatrix}
&=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix} F(n) \\ F(n-1)\end{bmatrix} \\ &=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^2
\begin{bmatrix} F(n-1) \\ F(n-2)\end{bmatrix} \\ &=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^3
\begin{bmatrix} F(n-1) \\ F(n-3)\end{bmatrix} \\ & \dots \\ &=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n
\begin{bmatrix} F(1) \\ F(0)\end{bmatrix}
\end{align}$$
Taking $F(0) = 0$ and $F(1) = 1$, you get:
$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix}$$
So to use "doubling", we just plug $2n$ into the formula:
$$\begin{align}
\begin{bmatrix} F(2n+1) \\ F(2n)\end{bmatrix}
&=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{2n}
\begin{bmatrix} F(1) \\ F(0)\end{bmatrix} \\ \\ &=
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n}
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n}
\begin{bmatrix} F(1) \\ F(0)\end{bmatrix} \\ \\ &=
\begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix}
\begin{bmatrix} F(n+1) & F(n) \\ F(n) & F(n-1) \end{bmatrix}
\begin{bmatrix} 1 \\ 0 \end{bmatrix} \\ \\ &=
\begin{bmatrix} F(n+1)^2 + F(n)^2 \\ F(n)F(n+1) + F(n-1)F(n) \end{bmatrix} \\
\\ ~
\\ & \text{And if you want...}
\\ ~
\\ &=
\begin{bmatrix} F(n+1)^2 + F(n)^2 \\ F(n)F(n+1)+ \bigg(F(n+1) - F(n)\bigg)F(n) \end{bmatrix}\\ \\&=
\begin{bmatrix} F(n+1)^2 + F(n)^2 \\ 2F(n+1)F(n) - F(n)^2 \end{bmatrix}
\end{align}$$
Which isn't actually better than matrix exponentiation asymptotically. And regardless, since the Fibonacci sequence grows exponentially, it will always require exponential time to compute just due to the size of the output. The matrix or "doubling" approach takes you from $O({\rm exp}~x^2)$ to $O({\rm exp}~x)$ asymptotic calculation time, which isn't nothing, but it still isn't exactly tractable either.
| {
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"url": "https://math.stackexchange.com/questions/1124590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Cylindrical limits of integration for a particular triple integral In cylindrical coordinates, what would be the limits of integration for the triple integral serving to find the volume of the region in $\mathbb R^3$ bounded by:
$x^2 + y^2 = y$ and the sphere of center $O$ and radius $1$, in the first octant?
$\theta$ moves between $0$ and $\pi/2$, and $z$ moves between $0$ and $\sqrt{1 - r^2}$; but what is the behaviour of $r$?
Thanks a lot.
| $x^2+y^2 = y$ describes an infinite cylinder $$x^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^2$$
The interior of intersection of two regions is thus given by
$$
\mathcal{R} = \left\{ x^2 + y^2 + z^2 \leqslant 1, x^2 + \left(y - \frac{1}{2}\right)^2 \leqslant \left(\frac{1}{2}\right)^2, x \geqslant 0, y \geqslant 0, z \geqslant 0 \right\}
$$
which is
$$
\mathcal{R} = \left\{ x^2 + y^2 + z^2 \leqslant 1, x^2 + y^2 \leqslant y, x \geqslant 0, y \geqslant 0, z \geqslant 0\right\}
$$
Switching to polar coordinates $x = r \cos(\theta)$, $x = r \sin(\theta)$ we have
$$\begin{eqnarray}
\mathcal{R} &=& \left\{ r^2 + z^2 \leqslant 1, r \leqslant \sin(\theta), 0 \leqslant \theta \leqslant \frac{\pi}{2}, r \geqslant 0, z \geqslant 0\right\} \\
&=& \left\{0 \leqslant \theta \leqslant \frac{\pi}{2}, 0 \leqslant r \leqslant \sin(\theta), 0 \leqslant z \leqslant \sqrt{1-r^2} \right\}
\end{eqnarray}
$$
Therefore the volume
$$\begin{eqnarray}
\operatorname{Vol}\left(\mathcal{R}\right) &=& \int_0^{\pi/2} \mathrm{d} \theta \int_0^{\sin \theta} r \mathrm{d} r \int_0^{\sqrt{1-r^2}} \mathrm{d} z \\
&=& \int_0^{\pi/2} \mathrm{d} \theta \int_0^{\sin \theta} r \sqrt{1-r^2} \mathrm{d} r \\
&=& \int_0^{\pi/2} \frac{1}{3} \left(1-\cos^3(\theta)\right) \mathrm{d} \theta = \frac{\pi}{6} - \frac{2}{9} \approx 0.301377
\end{eqnarray}
$$
Confirming in Mathematica:
In[1]:= NIntegrate[1,
Element[xvec,
RegionIntersection[Cuboid[{0, 0, 0}, {1, 1, 1}], Ball[{0, 0, 0}],
Cylinder[{{0, 1/2, 0}, {0, 1/2, 1}}, 1/2]]]]
Out[1]= 0.301377
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving an underdetermined system of equations I've had a little break in solving systems of equations and so I wanted to verify here that my own answer to this problem is 100% correct and done :) So I have the following problem:
Solve the system of equations:
$$3x_1+x_2-4x_3+5x_4=2$$ $$2x_1-3x_2-2x_3+3x_4=5$$
My attempt:
Because this is an underdetermined system we will have infinite amount of solutions. I will solve $x_1$ and $x_2$ explicitly in terms of $x_3$ and $x_4$. By multiplying the first equation by $3$ and adding it to the second we get:
$$11x_1-14x_3+18x_4=11$$
$$x_1=1+\frac{14}{11}x_3-\frac{18}{11}x_4$$
By substituting $x_2$ into the first equation we get:
$$3\left(1+\frac{14}{11}x_3-\frac{18}{11}x_4\right)+x_2-4x_3+5x_4=2$$
From which we get
$$x_2=\frac{2}{11}x_3-\frac{1}{11}x_4-1$$
So the solution is:
$$x_1=1+\frac{14}{11}x_3-\frac{18}{11}x_4$$
$$x_2=\frac{2}{11}x_3-\frac{1}{11}x_4-1,$$
where $x_3,x_4$ are free variables.
| Note that
$$
\DeclareMathOperator{rref}{rref}
\rref
\begin{bmatrix}
3 & 1 & -4 & 5 & 2 \\
2 & -3 & -2 & 3 & 5
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & -\frac{14}{11} & \frac{18}{11} & 1 \\
0 & 1 & -\frac{2}{11} & \frac{1}{11} & -1
\end{bmatrix}
$$
This implies
\begin{align*}
x_1 &= \frac{14}{11}\,x_3-\frac{18}{11}\,x_4+1 \\
x_2 &= \frac{2}{11}\,x_3-\frac{1}{11}\,x_4-1
\end{align*}
which confirms your answer.
Do you know how to row-reduce a matrix into its reduced row-echelon form? This is a much more efficient method than substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $\cos x + 3^{1/2} \sin x = 1$ for $0\leq x \leq 360^{\circ}$ $\cos x + \sqrt3 \sin x = 1$
Not sure what the step would be to get an answer.
| Note that our expression is equal to $2\left(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\right)$. Find an angle $\phi$ such that $\sin\phi=\frac{1}{2}$ and $\cos \phi=\frac{\sqrt{3}}{2}$. Then our expression is equal to $2\sin(\phi+x)$.
Remark: The start you mentioned in a comment will work, though it is a bit messier. Divide by $\cos x$. We get $1+\sqrt{3}\tan x=\sec x$. Square both sides. We get
$$1+2\sqrt{3}\tan x+3\tan^2 x=\sec^2 x=\tan^2 x+1.$$ Solve for $\tan x$, and then for $x$.
Since we squared, we may have introduced extraneous solutions, so we need to check that the candidate solutions we find are solutions of the original equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given any $a,b$, find $A,B$ such that $a\sin(x) + b\cos(x) = A\sin(x+B)$ I understand that if $f''+f = 0, f(0) = a$, $f'(0) = b$ we can have $f = b\sin +a\cos$. so if $f(x) = A\sin(x+B)$, we choose $a = A\cos(B)$ and $b = A\sin(B)$
How do we find $a$ and $b$ conversely?
| Let $\theta$ be such that $\cos \theta = a/\sqrt{a^2 + b^2}$ and $\sin \theta = b/\sqrt{a^2 + b^2}$. Then $$a\sin(x) + b\cos(x) = \sqrt{a^2 + b^2}(\cos(\theta)\sin(x) + \sin(\theta)\cos(x)) = \sqrt{a^2 + b^2}\sin(x + \theta).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\frac{a-bc}{a+bc}+\frac{b-ac}{b+ac}+\frac{c-ab}{c+ab}\le \frac32$ Let $a,b,c>0$ such that $a+b+c=1$. Prove that $\dfrac{a-bc}{a+bc}+\dfrac{b-ac}{b+ac}+\dfrac{c-ab}{c+ab}\le \dfrac32$.
My attempt:
First I thought if each term could be less than $\dfrac12$. That did not help. So, a little manipulation yielded
$\dfrac{bc}{a+bc}+\dfrac{ca}{b+ca}+\dfrac{ab}{c+ab}\ge \dfrac34$.
Again I thought that each term is greater than $\dfrac14$.
Proceeding that way yielded $3(ab+bc+ca)\ge 1$
Now, $a^2+b^2+c^2+2ab+2bc+2ca\ge 3(ab+bc+ca)\ge 1$.
Here, we get $1\ge 1$, here, I am confused whether the solution is done or not. I don't think it is. So, please help. Thank you.
| Using Cauchy-Schwarz Inequality:$$\left(\dfrac{bc}{a+bc}+\dfrac{ca}{b+ca}+\dfrac{ab}{c+ab}\right)\left(\sum\limits_{cyc} bc(a+bc)\right)\ge \left(\sum\limits_{cyc} bc\right)^2$$
So it suffices to prove: $\displaystyle 4\left(\sum\limits_{cyc} bc\right)^2 \ge 3\left(\sum\limits_{cyc} bc(a+bc)\right) = 9abc + 3\sum\limits_{cyc} b^2c^2$
i.e, $\displaystyle \sum\limits_{cyc} b^2c^2 + 8abc(a+b+c) \ge 9abc \iff \sum\limits_{cyc} b^2c^2 \ge abc(a+b+c)$ which is just Cauchy-Schwarz Inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1136050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$ The limit is:
$$
\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}
$$
When I use calculator and substitute $h$ with $0.000001$ and $-0.000001$, the result is:
$$
\frac{1}{5x^{\frac45}}
$$
My question is:
*
*How to do it without calculator.
*Show me the steps on how it's being done.
| $$\lim_{h \to 0} \frac{(x(1+\frac{h}{x}))^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \frac{x^{1/5}(1+\frac{h}{x})^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \bigl( x^{1/5} \cdot \frac{(1+\frac{h}{x})^{1/5} -1}{h} \bigr) = x^{1/5}\lim_{h \to 0} \frac{(1+\frac{h}{x})^{1/5} -1}{h} =x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} $$
Since $\lim\limits_{h \to 0} \frac{1}{5} \cdot \ln{(1+\frac{h}{x})}=0$,
$$x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} = x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{h} = \\ x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \lim_{h \to 0} \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{\frac{h}{x}} \cdot \lim_{h \to 0} \frac{1}{x} = x^{1/5} \cdot 1 \cdot 1/5 \cdot \frac{1}{x} = \frac{x^{1/5-1}}{5}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the orthogonal decomposition of a symmetric matrix unique? Suppose I have a real symmetric matrix $M$, and it can be decomposed in two ways: $M = AUA^T$ and $M = BVB^T$, where
*
*$U$ and $V$ are diagonal matrices
*The columns of $A$ are orthonormal to each other
*The columns of $B$ are also orthonormal to each other.
Can we say that $U = V$ and $A = B$?
| No. Unless all of the eigenvalues are the same, $U$ is not necessarily equal to $V$ because we can permute the diagonal entries to get a different diagonal matrix.
Also, $A$ is not necessarily equal to $B.$ Following is an example in which even if we fix $U,$ we can find an infinite number of orthogonal matrices $A$ such that $M = AUA^{\mathrm T}.$ The key to the example is that an eigenvalue of $M$ is repeated.
Consider
$$M = \begin{pmatrix}
10 & 2 & 2 \\
2 & 13 & 4 \\
2 & 4 & 13
\end{pmatrix}.$$
The eigenvalues are $18, 9, 9,$ and the corresponding eigenvectors are
$$\begin{pmatrix}
1 \\
2 \\
2
\end{pmatrix},
\begin{pmatrix}
-2 \\
0 \\
1
\end{pmatrix},
\begin{pmatrix}
-2 \\
1 \\
0
\end{pmatrix}.$$
Let us fix $U$ as
$$U = \begin{pmatrix}
18 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{pmatrix}.$$
If all of the eigenvalues were different, we would use the corresponding eigenvectors, normalized, for the columns of $A.$ Because, however, $9$ is a repeated eigenvalue, its two eigenvectors are not necessarily orthogonal, which is the case here. But we can use, for example, the Gram-Schmidt process, to get an orthogonal basis and then normalize those vectors to get $A:$
$$A = \begin{pmatrix}
\frac13 & \frac{-2}{\sqrt{5}} & \frac{-2}{\sqrt{45}} \\
\frac23 & 0 & \frac5{\sqrt{45}} \\
\frac23 & \frac1{\sqrt{5}} & \frac{-4}{\sqrt{45}}
\end{pmatrix}.$$
Now the eigenspace associated with eigenvalue $9$ is the plane through the origin and perpendicular to the other eigenvector $(1, 2, 2)^{\mathrm T},$ so we can use any pair of orthonormal vectors in that plane to form the second two columns of $A.$ One way to get other such pairs of vectors is to rotate the two that we already have (the second and third columns of $A$) about the axis in line with $(1, 2, 2)^{\mathrm T}.$ I used the formula after Lemma 98 in the article "Rotation About an Arbitrary Axis" to get a matrix that performs that rotation:
$$\begin{pmatrix}
\frac{1- \cos\theta}9 + \cos\theta & \frac{2(1- \cos\theta)}9 - \frac{2\sin\theta}3 & \frac{2(1 - \cos\theta)}9 + \frac{2\sin\theta}3 \\
\frac{2(1- \cos\theta)}9 + \frac{2\sin\theta}3 & \frac{4(1- \cos\theta)}9 + \cos\theta & \frac{4(1- \cos\theta)}9 - \frac{\sin\theta}3 \\
\frac{2(1 - \cos\theta)}9 - \frac{2\sin\theta}3 & \frac{4(1- \cos\theta)}9 + \frac{\sin\theta}3 & \frac{4(1- \cos\theta)}9 + \cos\theta
\end{pmatrix} = R_\theta$$
where $\theta$ is the angle of rotation.
Then $M = (R_\theta A)U(R_\theta A)^{\mathrm T}$ where the columns of $R_\theta A$ are orthonormal to each other. Thus, there are an infinite number of ways (by choosing any real number for $\theta$) to decompose $M$ as you requested.
| {
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"timestamp": "2023-03-29T00:00:00",
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How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$? When integrating $\int \frac{1-x}{x^2-1} dx$ Maple rewrote it as $-\int\frac{1}{x+1}dx$
How is $\frac{1-x}{x^2-1}=\frac{1}{x+1}$?
| $\dfrac{1-x}{x^2-1} = \dfrac{-(x-1)}{(x+1)(x-1)}$ and if $ x \ne 1 $, then $\dfrac{-1}{x+1}$.
With this, $\int\dfrac{1-x}{x^2-1}dx = -\int\dfrac{1}{x+1}=- \ln|x+1|+C$.
| {
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$\mathbb Z[\sqrt{-5}]$ is not a UFD Prove that the ring of integers of $\mathbb Q (\sqrt{-5})$ does not have unique factorisation.
Since $-5\equiv 3\pmod 4$, I know that the ring of integers of $\mathbb Q (\sqrt{-5})$ is $\mathbb Z [\sqrt{-5}]$.
I assume the way to prove that this does not have a unique factorisation is to give two different factorisations, but what exactly do I use to factorise? Do I consider the polynomial $x+\sqrt{-5}$?
| In a UFD, an element is irreducible if and only if it is prime.
Observe that $2$ is irreducible in $\mathbb Z + \mathbb Z\sqrt{-5}$: Suppose $$2 = (a + b \sqrt{-5})(c + b \sqrt{-5}),$$ taking the norm of both sides gives us $$4 = (a^2 + 5b^2)(c^2 + 5d^2)$$ which means $a^2 + 5b^2 = 1, 2$ or $4$. If $a^2 + 5b^2 = 1$, then $a = 1$ and $b = 0$ which means $a + b \sqrt{-5} = 1$ which is a unit and we're done. If $a^2 + 5b^2 = 4$, then $a = 2$ and $b = 0$ which means $$c + d \sqrt{-5} = \frac{2}{a + b\sqrt{-5}} = \frac{2}{2} = 1,$$ a unit, which means we're done. Notice that $a^2 + 5b^2 = 2$ can never happen: $b$ will have to be zero because if it's not, then the sum is greater than 5, which means it's greater than $2$, which means $a^2 = 2$ which only holds when $a = \sqrt 2 \notin \mathbb Z$, so this case can't occur. Conclude by definition that $2$ is irreducible in $\mathbb Z + \mathbb Z \sqrt{-5}$.
Observe that $2 \mid 6 = (1 + \sqrt{-5})(1 - \sqrt{-5})$ but $2 \nmid (1 + \sqrt{-5}), (1 - \sqrt{-5})$.
Say, by way of contradiction, that $2 \mid 1 + \sqrt{-5}$, then there exist $a, b \in \mathbb Z$ such that $1 + \sqrt{-5} = 2(a + b \sqrt{-5})$ which means $2a = 1$ and $2b = 1$ which can only happen if $a = b = 1/2 \notin \mathbb Z$, a contradiction. Similar reasoning works for $1 - \sqrt{-5}$.
Conclude that $2$ is not prime, but it is irreducible. Hence we're not in a unique factorization domain.
| {
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If $abc=1$ then $\sum\limits_{cyc}^{}{\frac{1}{b(a+b)}}\ge \frac{3}{2}$
If $abc=1$ for positive $a,b,c$, then $\sum\limits_{cyc}^{}{\dfrac{1}{b(a+b)}}\ge \dfrac{3}{2}$
I have tried the following,in decreasing order of success:
1)AM-GM:$a+b+c\ge 3$ and $ab+bc+ca\ge 3$
2)Substituting $1=abc$ yields nothing
3)Substituting $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ yields something weird
4)Rearrangement inequality on the sequences $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a}$
5)Titu's lemma doesn't help
A little nudge in the right direction would help.
| Hint:
Indeed, as @Macavity pointed out, the sum $s(a,b,c)$ only has circular symmetry. In fact it is not hard to show ( the same method as below) that if $a\le b\le c$ then
$s(a,b,c)\le s(c,b,a)$.
Here is a method amenable to a computer algebra system. As suggested, substitute $a=u/v$, $b=v/w$, $c=w/u$. Get an equivalent inequality homogeneous of degree $6$ in $u$,$v$, $w\ge 0$
$$2 u^4 v^2-3 u^3 v^3+2 u v^5+2 u^5 w-3 u^4 v w+4 u^2 v^3 w-
3 u v^4 w+4 u^3 v w^2-\\-6 u^2 v^2 w^2+2 v^4 w^2-3 u^3 w^3+4 u v^2 w^3-3 v^3 w^3+2 u^2 w^4-3 u v w^4+2 v w^5\ge 0$$
with circular symmetry. It is enough to consider the cases $u\le v \le w$ and $u \ge v \ge w$.
In the first case, make the substitution $u=p$, $v=p+q$, $w=p+q+r$, with $p$, $q$, $r\ge 0$. One gets
$$8 p^4 q^2+18 p^3 q^3+15 p^2 q^4+6 p q^5+q^6+8 p^4 q r+31 p^3 q^2 r
+38 p^2 q^3 r+22 p q^4 r+\\+5 q^5 r+8 p^4 r^2+41 p^3 q r^2
+66 p^2 q^2 r^2+46 p q^3 r^2+13 q^4 r^2+14 p^3 r^3+43 p^2 q r^3+\\
+43 p q^2 r^3+17 q^3 r^3+9 p^2 r^4+17 p q r^4+10 q^2 r^4+2 p r^5+2 q r^5$$
clearly positive, with equality when $q=r=0$, that is, the $a$,$b$,$c$ are equal.
The other case is treated similarly, giving the positive expression
$$8 p^4 q^2+18 p^3 q^3+15 p^2 q^4+6 p q^5+q^6+8 p^4 q r+23 p^3 q^2 r+22 p^2 q^3 r+8 p q^4 r+q^5 r+8 p^4 r^2+33 p^3 q r^2+42 p^2 q^2 r^2+18 p q^3 r^2+3 q^4 r^2+14 p^3 r^3+35 p^2 q r^3+23 p q^2 r^3+5 q^3 r^3+9 p^2 r^4+11 p q r^4+2 q^2 r^4+2 p r^5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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First order partial differential equation - how to finish the solution? I have problems solving this simple pde. I don't know what I'm doing wrong:
$$yu_x - u_y = y, \ \ \ u(x,0 ) = \frac{1}{x}$$
Here is how I do it. The equation is equivalent to this one:
$$\frac{dx}{y} = \frac{dy}{-1} = \frac{du}{y}$$
So $\frac{dx}{y} = - dy, \ \ dx=du$
This means that $dx = -ydy$ and $u = x+ C$
So $x + C' = - \frac{y^2}{2} $ and $u = x+C$
This above gives us an integral curve $F(x,y) = - x - \frac{y^2}{2}$ of $yu_x - u_y=0$, but not of the main pde.
If we plug $F(x,0) = -x$, we get $u(x,0) = - \frac{1}{F(x,0)}$, so $u(x,y) = - \frac{1}{F(x,y)} = \frac{1}{x+ \frac{y^2}{2}}$, which is not the solution.
Could you help me fix it?
| Use method of characteristics.
First, reparameterise your curve letting
$$x \to x(s) \\
y \to y(s)$$
Hence you have
$$u = u(x(s), y(s))$$
Taking the total derivative wrt to $s$
$$\begin{align}
\implies \frac{d}{ds} u &= \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial y} \frac{dy}{ds} \\
&= \frac{\partial u}{\partial x} \cdot y + \frac{\partial u}{\partial y} \cdot (-1) \\
&= y \\
\end{align}$$
Equating, we find
$$\begin{align}
\frac{dy}{ds} &= -1 \\
\implies dy &= -ds \ \ \ \ \ (1) \\
\frac{dx}{ds} &= y \\
\implies \frac{dx}{-dy} &= y \ \ \ \ \ \ \ \ \ \ (2) \\
\frac{du}{ds} &= y \\
\implies \frac{du}{-dy} &= y \ \ \ \ \ \ \ \ \ \ (3) \\
\end{align}$$
Solving
$$\begin{align}
(2) \implies x(s) &= \frac{-y^{2}}{2} + x_{0} \\
(3) \implies u(x, y) &= \frac{-y^{2}}{2} + f(x_0) \\
&= \frac{-y^{2}}{2} + f \bigg(x + \frac{y^{2}}{2} \bigg) \ \ \ \ \ \ (4) \\
\end{align}$$
Using our initial condition
$$\begin{align}
u(x,0) &= f(x) \\
&= \frac{1}{x} \\
\implies f \bigg(x + \frac{y^{2}}{2} \bigg) &= \frac{1}{x + \frac{y^{2}}{2}} \\
\end{align}$$
Substituting into $(4)$, we find
$$u(x, y) = \frac{-y^{2}}{2} + \frac{1}{x + \frac{y^{2}}{2}}$$
You can check by differentiation that this satisfies the PDE and initial condition.
| {
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What's the method to finding the scale factor of enlargement and rotation of a 2D matrix? The matrix M is defined by:
\begin{bmatrix}
-1 & -1 \\
1 & -1 \\
\end{bmatrix}
Assuming the matrix represents an enlargement followed by a rotation
My idea here was to make an equation so you're left with simultaneous equations to solve.
$\begin{bmatrix}
\cos \left(θ\right) & \sin \left(θ\right) \\
-\sin \left(θ\right) & \cos \left(θ\right) \\
\end{bmatrix}
\begin{bmatrix}
x & 0 \\
0 & x \\
\end{bmatrix}
=
\begin{bmatrix}
-1 & -1 \\
1 & -1 \\
\end{bmatrix}$
$\begin{bmatrix}
x\cos \left(θ\right) & x\sin \left(θ\right) \\
-x\sin \left(θ\right) & x\cos \left(θ\right) \\
\end{bmatrix}
=
\begin{bmatrix}
-1 & -1 \\
1 & -1 \\
\end{bmatrix}$
This is where I get stuck. I don't think you can solve this problem like this but if you can, please answer. Regards
A couple more questions,
Does the type of enlargement and type of rotation alter this method? e.g. a scale factor more or less than 1 and a clockwise or counter clockwise rotation.
Also if there is an easier method to finding the matrices could someone please answer with working?
Regards Tom
| If you suspect that this matrix is a scaling followed by a rotation, you can apply it to some basis vectors to get a clue.
For instance multiplying your matrix on $[1,0]^T$ yields $[-1, 1]$. And applying it to $[0,1]^T$ yields $[-1, -1]$.
The scale factor for $[1,0]^T$ is $\sqrt{2}$. The rotation can be figured out from the dot product $[1,0]^T \cdot [-1, 1]^T = -1 = \sqrt{2} \cos(\theta)$. Thus $\cos(\theta) = -1/\sqrt{2}$.
The scale factor for $[0,1]^T$ is again $\sqrt{2}$. The dot product gives us $[0,1]^T \cdot [-1,-1]^T = -1 = \sqrt{2} \cos(\theta)$. So again $\cos(\theta) = -1/\sqrt{2}$.
This tells us that $\theta = 3\pi/4$ or $\theta=-3\pi/4$.
Now we just test some matrices:
For $\theta = 3\pi/4$:
$$\left[\begin{array}{cc} -1/\sqrt{2} & 1/\sqrt{2}\\ -1/\sqrt{2} & -1/\sqrt{2} \end{array}\right] \left[ \begin{array}{cc} \sqrt{2} & 0\\ 0 & \sqrt{2} \end{array}\right] = \left[ \begin{array}{cc} -1 & 1\\ -1& -1\end{array}\right]$$
For $\theta = -3\pi/4$:
$$\left[\begin{array}{cc} -1/\sqrt{2} & -1/\sqrt{2}\\ 1/\sqrt{2} & -1/\sqrt{2} \end{array}\right] \left[ \begin{array}{cc} \sqrt{2} & 0\\ 0 & \sqrt{2} \end{array}\right] = \left[ \begin{array}{cc} -1 & -1\\ 1& -1\end{array}\right].$$
And this gives the decomposition you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of natural solutions to the equation $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000$ I'm trying to tackle some question and I think I solved it, but I'm not sure and would like someone to check my work.
The question goes like this:
Find the number of non-negative solutions of the equation $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000$.
My try:
First I have written the equation as $x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=10^6=2^6\cdot{5^6}$, hence $x_i$ is of the form $\displaystyle 2^{a_i}\cdot{5^{b_i}}$ where $0\le a_i,b_i \le 6$.
Now we can write $$2^{a_1}5^{b_1}\cdot{2^{a_2}5^{b_2}}\cdot{2^{a_3}5^{b_3}}\cdot{2^{a_4}5^{b_4}}=2^{a_1+a_2+a_3+a_4}\cdot{5^{b_1+b_2+b_3+b_4}}=2^6\cdot{5^6}$$
Thus, we are looking to find the number of non-negative solutions to the equations $$\begin{cases}a_1+a_2+a_3+a_4=6\\b_1+b_2+b_3+b_4=6\end{cases}$$
We know that $0 \le a_i,b_i \le 6$, hence the generating function would be $$g(x)=(1+x+...+x^6)^4=\left(\frac{1-x^7}{1-x}\right)^4=(1-x^7)^4\cdot{\frac{1}{(1-x)^4}}\\g(x)=(1-x^7)^4\cdot{\sum_{n=0}^{\infty}{{n+3}\choose 3}x^n}$$
Now, $\displaystyle (1-x^7)^4=x^0-4x^7+6x^{14}-4x^{21}+x^{28}$, hence the only relavant term is $x^0$ (we want to find the coefficient of $x^6$). We find that the required coefficient is $\displaystyle {9\choose 3}=84$, so we have 84 optional solutions to any of the equations.
Edit: There is no dependence between the equations, hence the number of solutions is $84^2$.
Is my reasoning correct? Please help me fix my errors if there any. Thanks!
| Let's try it for $100=x_1\cdot x_2$. Then the numbers of solutions to $a_1+a_2=2$ is $3$ and the number of solutions to $b_1+b_2=2$ is $3$. Your answer would give us $3+3=6$ answers.
Let's now enumerate:
$$100\cdot 1\\50\cdot 2\\25\cdot 4\\20\cdot 5\\10\cdot 10\\5\cdot 20\\4\cdot 25\\2\cdot 50\\1\cdot 100$$
That's nine solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding number of integer solutions using Generating Functions This is a problem for a practice test my professor gave me.
$$\text{How many integer solutions are there to } x_1+x_2+x_3+x_4 \leq 50 \\ \text{with } x_i \geq 2 \text{ for all } i = 1,2,3,4 \text{ and } x_1,x_2 \leq 7 \text{?}$$
This is how I approached the problem, using generating functions:
$\text{Same as}$
$$x_1+x_2+x_3+x_4+x_5 = 50, \space x_5 \geq 0 $$
$\text{Find the coefficient of } x^{50} \text{ in}$
$$(x^2 + x^3 + x^4 + \dotso)^2 (x^2 + x^3 + x^4 + \dotso + x^7)^2 (1 + x + x^2 + x^3 + \dotso)$$
After some factoring, we'll have:
$$x^8(1+x^2+x^3+ \dotso + x^5)^2 (1+x+x^2+\dotso)^3$$
This is the same as:
$\text{Find the coefficient of } x^{42} \text{ in}$
$$(1+x^2+x^3+ \dotso + x^5)^2 (1+x+x^2+\dotso)^3$$
To simplify further:
$$(1-x^6)^2 \frac{1}{(1-x)^3}$$
So, this is where I'm confused. I was using the formula which is based off this answer on Math.SE, but I don't get the correct answer. According to my professor, the correct answer is:
$$\dbinom{30+5-1}{30} - 2\dbinom{36+5-1}{36} + \dbinom{42+5-1}{42} = 26,781$$
What I end up doing mirrors that of the linked question on Math.SE:
$$(1-x^6)^2 = 1-2x^6+x^{12} \\
\\
(1-2x^6+x^{12}) \frac{1}{(1-x)^3}$$
Using the formula from the linked question:
$\text{We do this three times, for } k=0, k=6, \text{ and } k=12$. The result is
$$(1-2x^6+x^{12})\frac{1}{(1-x)^3}={m-0+2 \choose 2}- 2{m-6+2 \choose 2} + {m-12+2 \choose 2}
\\
\
\\m = 42 \\
\
\\={42-0+2 \choose 2}- 2{42-6+2 \choose 2} + {42-12+2 \choose 2} = 36$$
As you can see, my answer differs greatly from what my professor said was correct. I don't understand why this formula I used doesn't work; I've used it for lots of other problems of this same type, and I calculated the correct number; for this one though, it doesn't seem to be working.
What am I doing wrong?
| The actual generating function should be $$\frac{(1-x^6)^2}{(1-x)^5}$$ When you said, "To further simplify...," you forgot that $$1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x},$$ not simply $1-x^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Rank of a matrix summation Assume $M\in\{0,1\}^{n\times n}$ of rank $r$. Is there an example of $M$ such that $rk(M+M')/rk(M)> 2$? Is there an example of $M$ such that $rk(M)/rk(M+M')> 2$?
| Due to the rank inequalities, the answer to the first question is no:
$$
rank(M+M') \le rank(M) + rank(M') = 2 rank(M).
$$
Here is an example of an $8\times 8$ matrix $M$ such that $rank(M)=8$, $rank(M+M^T)=2$:
$$
M:=
\left(
\begin{array}{cc|cc|cc|cc}
0& 0& 1& 0& 0& 1& 1& 0\\
1& 0& 0& 0& 0& 0& 0& 1\\
\hline
0& 0& 0& 1& 0& 0& 0& 0\\
0& 1& 0& 0& 0& 0& 1& 0\\
\hline
0& 1& 1& 0& 0& 0& 1& 0\\
0& 0& 0& 1& 1& 0& 0& 0\\
\hline
0& 0& 0& 0& 0& 0& 0& 1\\
0& 0& 1& 0& 0& 1& 0& 0\\
\end{array}
\right),
M+M^T=
\left(
\begin{array}{cc|cc|cc|cc}
0& 1& 1& 0& 0& 1& 1& 0\\
1& 0& 0& 1& 1& 0& 0& 1\\
\hline
1& 0& 0& 1& 1& 0& 0& 1\\
0& 1& 1& 0& 0& 1& 1& 0\\
\hline
0& 1& 1& 0& 0& 1& 1& 0\\
1& 0& 0& 1& 1& 0& 0& 1\\
\hline
1& 0& 0& 1& 1& 0& 0& 1\\
0& 1& 1& 0& 0& 1& 1& 0\\
\end{array}
\right).
$$
(Matlab did the rank computation). I could not find a smaller example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$ My problem is to find the sum of the series
$$
S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots
$$
where the terms are the reciprocals of the positive integers whose only prime factors are $2$s and $3$s.
I can see that every term in $S$ is of the form $\dfrac{1}{2^m3^n}$ where $m,n\geq 0$. I can also see that each term only occurs once, but I'm not really sure how to use this information effectively. Any ideas (apparently geometric series was a hint)?
| Given what you have observed, note the following:
\begin{align}
S &= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{2^m}{3^n}\\[1em]
&= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{2^m}\frac{1}{3^n}\\[1em]
&= \sum_{m=0}^\infty\frac{1}{2^m}\sum_{n=0}^\infty\frac{1}{3^n}\\[1em]
&= \frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{2}{3}}\\[1em]
&= 2\cdot \frac{3}{2}\\[1em]
&= 3.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to prove this floor function equation? How can I prove the following equation?
$$
\lfloor nx \rfloor =
\lfloor x \rfloor +
\Big\lfloor x + \frac{1}{n} \Big\rfloor +
\Big\lfloor x + \frac{2}{n} \Big\rfloor +
\Big\lfloor x + \frac{3}{n} \Big\rfloor +
\Big\lfloor x + \frac{4}{n} \Big\rfloor +
\Big\lfloor x + \frac{5}{n} \Big\rfloor+ \dotsb +
\Big\lfloor x + \frac{n-1}{n} \Big\rfloor
$$
$n∈N$ and $x∈R$
| Replacing $x$ by $x':=x+{1\over n}$ adds $1$ on the left hand side. On the right hand side the first term $\lfloor x\rfloor$ disappears, and on the far right the term $\lfloor x+1\rfloor$ appears. It follows that the right hand side increases by $1$ as well.
Therefore it suffices to prove the formula for $0\leq x<{1\over n}$. Inspection immediately reveils that both sides are $=0$ in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve this quadratic congruence modulo a non-prime number
$(1)$ Find all $x$, that solve $7x^2 + x + 22 \equiv 0 \pmod{60}$.
I tried to solve this by first considering the prime factorization $60 = 2^2\cdot 3\cdot 5$ and then using the Chinese Remainder Theorem, that is for $n_1, ..., n_k$ coprime and $a_1, ..., a_k$ there exists a $x \in \mathbb{Z}$ that solves the system of simultaneous congruences
\begin{cases} x \equiv a_1 & \pmod{n_1} \\ \quad \cdots \\ x \equiv a_k &\pmod{n_k} \end{cases}
In this case i just tried to find a solution to $(1)$ mod $3$, mod $4$, mod $5$ one at a time by testing $0,1,2$, $0,1,2,3$ and $0,1,2,3,4$ respectively. While all of the congruences have solutions I couldn't find a single $x \in \mathbb{Z}$ that solves all of them. However, Wolfram-Alpha says there are two solutions $x_1 = 31$ and $x_2 = 46$. So how do I apply the CRT to this in the correct way?
| By CRT $7x^2+x\equiv -22\equiv 38 \bmod 60$ if and only if:
$x(7x+1)\equiv x(3x+1)\equiv 2 \bmod 4$
$x(7x+1)\equiv x(x+1) \equiv 2 \bmod 3$
$x(7x+1)\equiv x(2x+1) \equiv 3 \bmod 5$
Finding the solutions to each of these can be done by hand.
You want numbers such that:
$x\equiv 2$ or $3\bmod 4$
$x\equiv 1 \bmod 3$
$x\equiv 1 \bmod 5$.
By CRT there is one congruence when $x\equiv 2\bmod 4$ and another when $x\equiv 3\bmod 4$
I will solve the first of these, the other one is analogous. We use the algebraic approach explained in wikipedia:
$x\equiv2\bmod 4$
$x\equiv 1\bmod 3$
$x\equiv 1 \bmod 5$
We have $x=2+4t$ so $2+4t\equiv 1 \bmod 3$
so $4t\equiv 2 \bmod 3\implies t\equiv 2 \bmod 3\implies t=2+3k$. Therefore $x=2+4(2+3k)=10+12k$.
We now have $x=10+12k\equiv 1 \bmod 5\implies 12k\equiv 1 \bmod 5\implies 2k\equiv 1\bmod 5\implies k\equiv 3 \bmod 5$. therefore $k=5s+3$
So $x=10+12(5s+3)=60s+46$. So $x\equiv 46 \bmod 60$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$dx$ being a desginator (with respect to $x$) or being a term? I am confused as to what $dx$ truly is. I am doing some u-substitution problems and this is what I came across:
$$\int 2x(x-1)^{1/2}\,dx$$
$u=x-1$ and therefore $du=1$
when we substitute we get:
$$2 \int (u^{3/2}+u^{1/2}) \, du$$
(here the du simply replaces the dx because our variable changed)
In another example:
$$\int 4x^5(x^2+1)^{1/3} \, dx$$
$$u=x^3+1$$
$$du=3x^2$$
therefore it becomes:
$$\int 4(u-1)(du/3)(u)^{1/3}\,$$
-here my teacher didn't put $d$x at the end, she just left it off
So my question is this: why is it that sometimes $dx$ and $du$ are treated as values that can be multiplied to other terms in the integrand and sometimes they are simply treated as a command (do "blank" with respect to $x$, or $u$ or whatever is used)?
| Consider
$$\begin{array}{lll}
\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx&=&\displaystyle\int 4x^5(x^2+1)^\frac{1}{3}dx\\
&=&\displaystyle2\int (x^2)^2(2x)(x^2+1)^\frac{1}{3}dx\\
&=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\\
\end{array}$$
Next, let's make the substitution
$u=x^2+1$, implying that $du/dx=2x$.
We do this by multiplying by $1$.
$$\begin{array}{lll}
&=&\displaystyle2\int ((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}dx\cdot\frac{\frac{du}{dx}}{\frac{du}{dx}}\\
&=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}\frac{dx\cdot du}{dx}}{\frac{du}{dx}}\\
&=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}du}{2x}\\
\end{array}$$
Notice how I interpreted $du/dx$ as a ratio of differential in the numerator, but as a derivative in the numerator. This "inconsistency" isn't a bad thing, for I think that everyone can remember in middle/high school how useful it was to interpret a division as a multiplication (e.g. $5\div2=5\times\frac{1}{2}$) while manipulating fractions. Extending this technique of "multiple interpretations" to calculus, can only make your mathematical skills stronger.
Continuing
$$\begin{array}{lll}
&=&\displaystyle2\int \frac{((x^2+1)-1)^2(2x)(x^2+1)^\frac{1}{3}du}{2x}\\
&=&\displaystyle2\int ((x^2+1)-1)^2(x^2+1)^\frac{1}{3}du\\
&=&\displaystyle2\int (u-1)^2u^\frac{1}{3}du\\
&=&\dots\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Partial Integration $ \int \frac{x\cos x}{\sin^3x}dx $ The problem:
$$ \int \frac{x\cos x}{\sin^3x}dx $$
Can someone give me a hint on how to solve this without using cosecant?
The solution provided is:
$$ -\frac{1}{2}\left(\frac{x}{\sin^2x}+\cot x\right)\:+\:C $$
| $$\begin{gathered}
I = \int {\frac{{x\cos x}}
{{{{\sin }^3}x}}dx} = \int {\frac{{x\cos x}}
{{\sin x}}.\frac{1}
{{{{\sin }^2}x}}dx} = - \int {\frac{{x\cos x}}
{{\sin x}}.d\left( {\cot x} \right)} \hfill \\
= - \frac{{x\cos x}}
{{\sin x}}.\cot x + \int {\cot xd\left( {\frac{{x\cos x}}
{{\sin x}}} \right)} \hfill \\
= - \frac{{x{{\cos }^2}x}}
{{{{\sin }^2}x}} + \int {\cot x.\frac{{\left( {\cos x - x\sin x} \right)\sin x - \left( {x\cos x} \right)\cos x}}
{{{{\sin }^2}x}}dx} \hfill \\
= - x{\cot ^2}x + \int {\cot x.\frac{{\sin x\cos x - x}}
{{{{\sin }^2}x}}dx} \hfill \\
= - x{\cot ^2}x + \int {\frac{{\cos x}}
{{\sin x}}.\left( {\frac{{\cos x}}
{{\sin x}} - \frac{x}
{{{{\sin }^2}x}}} \right)dx} \hfill \\
= - x{\cot ^2}x + \int {{{\cot }^2}xdx} - I \hfill \\
\end{gathered} $$
So,
$$2I = - x{\cot ^2}x + \int {{{\cot }^2}xdx} \Rightarrow I = \frac{1}
{2}\left( { - x{{\cot }^2}x + \int {{{\cot }^2}xdx} } \right).$$
But
$$\int {{{\cot }^2}xdx} = \int {\left( {\frac{1}
{{{{\sin }^2}x}} - 1} \right)dx} = - \cot x - x + C.$$
So,
$$I = \frac{1}{2}\left( { - x{{\cot }^2}x - \cot x - x + C} \right).$$
Finally,
$$I = \frac{1}
{2}\left[ { - x\left( {1 + {{\cot }^2}x} \right) - \cot x + C} \right] = - \frac{1}
{2}\left[ {x\left( {1 + {{\cot }^2}x} \right) + \cot x - C} \right] = - \frac{1}
{2}\left( {\frac{x}
{{{{\sin }^2}x}} + \cot x - C} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solve for x in the following Quadratic equations I am not able to solve this problem for my son. Is ther any error in the question itself.
$\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$.
Thanks in advance..
| $\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$
$\implies \dfrac{1}{(x+5)}-\dfrac{1}{(x+4)}=\dfrac{1}{(x+2)}-\dfrac{1}{(x+1)} $
$\implies \dfrac{x+4-x-5}{(x+4)(x+5)}=\dfrac{x+1-x-2}{(x+1)(x+2)}$
$\implies (x+4)(x+5)=(x+1)(x+2)$
$\implies x^2+9x+20=x^2+3x+2$
$\implies 9x-3x+20-2=0$
Or,$$x=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Problem with understanding theorem on Riccati Equation. `The matrices $A,B,C,D,X$ are real, square, $n \times n$.
I have trouble understanding theorem 7.1.2 from Lancaster & Rodman "Algebraic Riccati Equations".
The part that I understand is as follows.
A matrix $X$ solves equation $XBX + XA -DX - C = 0$ iff we can find a matrix $Z$ such that we have $ \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} I \\ X \end{array} \right] = \left[ \begin{array}{c} I \\ X \end{array} \right] Z$.
The part I don't understand is this.
Denote $T = \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] $.
Consider the Jordan decomposition of $T$: $T= W J W^{-1}$.
Denote a matrix containing a subset of Jordan chains of $T$ (a subset of columns of $W$ such that each chain is either completely included or completely excluded) as as $V = \left[ \begin{array}{c} Y \\ Z \end{array} \right]$. Denote the corresponding sub-matrix of $J$ as $J_s$. Assume furthermore that $Y$ is invertible.
We have $TV = VJ_s$, i.e. $ \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} Y \\ Z \end{array} \right] = \left[ \begin{array}{c} Y \\ Z \end{array} \right] J_s $.
Now the bit where I get lost is Theorem 7.1.2, which seems to imply (maybe I understand it wrong) the following.
$ \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} Y \\ Z \end{array} \right] = \left[ \begin{array}{c} Y \\ Z \end{array} \right] J_s \quad \Longrightarrow \quad \exists Z'.\; \left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} I \\ ZY^{-1} \end{array} \right] = \left[ \begin{array}{c} I \\ ZY^{-1} \end{array} \right] Z'$
Edit: Thanks to the answer, I now see that a good choice of $Z'$ is $Y J_s Y^{-1} $.
| $J$ is not the Jordan block in this context. If it was, there would be dimension mismatch. Since $T$ is $2n \times 2n$, so is $V$. Therefore, $Y$ or $Z$ could not be square.
I believe the authors tried to prove this: Suppose the Riccati equation has a solution. We need to show that there exists matrices $Y$, $Z$ and $J$ with $Y$ is invertible, such that
$$\left[ \begin{array}{cc} A & B \\ C & D\end{array} \right] \left[ \begin{array}{c} Y \\ Z \end{array} \right] = \left[ \begin{array}{c} Y \\ Z \end{array} \right] J$$
Rewriting the equations
$$\begin{align}
AX + BY &= YJ \\
Y^{-1}AX + Y^{-1}BY &= J \\
\\
CY + DZ &= ZJ \\
&= ZY^{-1}AY + ZY^{-1}BZ \\
C + DZY^{-1} &= ZY^{-1}A + ZY^{-1}BZY^{-1}
\end{align}$$
Hence, $ZY^{-1}$ is a solution to the Riccati equation, so these matrices exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show by induction on $n$ that: $$1^4 + 2^4 +\cdots+n^4=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}$$
I proved true for case $n=1$ , assumed true for $n=k$ , but cannot get things to work out. I tried putting the right side over a common denominator and going from there but I'm sure how much it helps.
| First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^4=\frac{6\cdot1^5+15\cdot1^4+10\cdot1^3-1}{30}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^4=$
$\color{red}{\sum\limits_{k=1}^{n}k^4}+(n+1)^4=$
$\color{red}{\frac{6n^5+15n^4+10n^3-n}{30}}+(n+1)^4=$
$\frac{6n^5+15n^4+10n^3-n}{30}+n^4+4n^3+6n^2+4n+1=$
$\frac{6n^5+15n^4+10n^3-n}{30}+\frac{30n^4+120n^3+180n^2+120n+30}{30}=$
$\frac{6n^5+45n^4+130n^3+180n^2+119n+30}{30}=$
$\frac{6(n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)}{30}$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Deriving an expression for $\cos^4 x + \sin^4 x$
Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$
I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.
So now I need to show that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac{1}{4} \cos (4x) +\frac{3}{4}$, which has stumped me.
| $$\cos^4x+\sin^4x\\=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x\\=1-\frac12\sin^22x=1-\frac12\left(\frac12(1-\cos4x)\right)=\frac34+\frac14\cos4x$$
As:
$$\sin^2x+\cos^2x=1;\\(a+b)^2=a^2+b^2+2ab;\\\sin2x=2\sin x\cos x;\\\cos2x=\cos^2x-\sin^2x=1-2\sin^2x$$
Also:
$$\cos^4x+\sin^4x=\left(\frac{e^{ix}+e^{-ix}}2\right)^4+\left(\frac{e^{ix}-e^{-ix}}2\right)^4\\=\frac1{16}({\small e^{-4 ix}+4 e^{-2 ix}+4 e^{2 ix}+e^{4 ix}+6+e^{-4 ix}-4 e^{-2 ix}-4 e^{2 ix}+e^{4i x}+6})\\=\frac34+\frac14\left(\frac{e^{4ix}+e^{-4ix}}2\right)=\frac34+\frac14\cos4x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Euler phi Function and Floor Function Let $\phi(x)$ be the Euler phi function and $\lfloor x\rfloor $ the floor function
Count the sum of $$ \phi(1)\cdot\lfloor \frac{2015}{1} \rfloor + \phi(2)\cdot\lfloor \frac{2015}{2} \rfloor + \phi(3)\cdot\lfloor \frac{2015}{3} \rfloor + \phi(4)\cdot\lfloor \frac{2015}{4} \rfloor + \cdots + \phi(2014)\cdot\lfloor \frac{2015}{2014} \rfloor + \phi(2015)\cdot\lfloor \frac{2015}{2015} \rfloor $$
I have do for $1,2, \ldots, 10$ but I can see no pattern there.
| Let
$$f(n)=\sum_{k=1}^n\varphi(k)\left\lfloor\frac{n}k\right\rfloor\;,$$
so that you’re interested in $f(2015)$. Calculate the first few values:
$$\begin{align*}
f(1)&=1\\
f(2)&=2+1=3\\
f(3)&=3+1+2=6\\
f(4)&=4+2+2+2=10\\
f(5)&=5+2+2+2+4=15
\end{align*}$$
There is a very clear pattern here: at least for $n=1,2,3,4,5$ we have the very familiar triangular numbers, suggesting that
$$f(n)=\sum_{k=1}^nk=\frac{n(n+1)}2=\binom{n+1}2\;.\tag{1}$$
Another way to put it is that it appears that we might have $f(1)=1$ and $f(n)=f(n-1)+n$ for $n>1$. We know that $f(1)=1$, so we’d prove $(1)$ if we could show that the recurrence $f(n)=f(n-1)+n$ really does hold for $n>1$. That would mean proving that
$$\sum_{k=1}^n\varphi(k)\left\lfloor\frac{n}k\right\rfloor=n+\sum_{k=1}^{n-1}\varphi(k)\left\lfloor\frac{n-1}k\right\rfloor\tag{2}$$
or, equivalently, that
$$\sum_{k=1}^{n-1}\varphi(k)\left(\left\lfloor\frac{n}k\right\rfloor-\left\lfloor\frac{n-1}k\right\rfloor\right)=n-\varphi(n)\;.\tag{3}$$
*
*Justify the assertion that $(3)$ is equivalent to $(2)$.
*$n-\varphi(n)$ is the number of integers in $\{1,2,\ldots,n\}$ with what property?
Note that $$\left\lfloor\frac{n}k\right\rfloor-\left\lfloor\frac{n-1}k\right\rfloor\tag{4}$$
is always $0$ or $1$.
*
*For which values of $k$ is the difference in $(4)$ equal to $1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find the equation of the plane knowing that it passes through 3 points I have to find the equation of the plane that passes through $(0, 0, 0), (4, 0, -2), (0, 8, -6)$.
I have done the following:
The equation of the plane is of the form $$ax+by+cz+d=0$$
Since the points $(0, 0, 0), (4, 0, -2), (0, 8, -6)$ are points o fthe plane, we have that $$d=0 \\ 4a-2c=0 \Rightarrow a=\frac{c}{2} \\ 8b-6c=0 \Rightarrow b=\frac{3c}{4}$$
So $$\frac{c}{2}x+\frac{3c}{4}y+cz=0 \Rightarrow \frac{1}{2}x+\frac{3}{4}y+z=0$$
Is it correct??
Is there also an other way to find the equation of the plane??
Maybe using the cross-product??
| Hint: first you find the normal to the plane: $N = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 8 \\-6 \end{pmatrix} = (a,b,c)$, then your plane is:
$a(x-0)+b(y-0)+c(z-0) = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How do I calculate: $\int \frac{dx}{3\sin^2 x+5\cos^2x}?$ How can I calculate this integral ?
$$\int \frac{dx}{3\sin^2 x+5\cos^2x}=\text{?}$$
Thank you! I've tried using universal substitution but the result was too complicated to be somehow integrated. Can you please give me a useful hint ?
| $$\begin{gathered}
\int {\frac{1}
{{3{{\sin }^2}x + 5{{\cos }^2}x}}dx} = \int {\frac{1}
{{\left( {3\frac{{{{\sin }^2}x}}
{{{{\cos }^2}x}} + 5} \right){{\cos }^2}x}}dx} = \int {\frac{1}
{{3{{\tan }^2}x + 5}}d\left( {\tan x} \right)} \hfill \\
= \frac{1}
{3}\int {\frac{1}
{{{{\tan }^2}x + \frac{5}
{3}}}d\left( {\tan x} \right)} = \frac{1}
{3}.\frac{1}
{{\sqrt {5/3} }}\arctan \left( {\frac{{\tan x}}
{{\sqrt {5/3} }}} \right) + C = \frac{1}
{{\sqrt {15} }}\arctan \left( {\sqrt {\frac{3}
{5}} \tan x} \right) + C \hfill \\
\end{gathered} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Implicit differentiation of $(x^2 + y^2)^2 = (x-y)^2$ $$
(x^2 + y^2)^2 = (x-y)^2$$
Wolfram alpha yields this answer:
$$
y'(x) = \frac{(-2 x^3-2 x y^2+x-y)}{((2 x^2-1) y+x+2 y^3)}$$
But it's impossible to get $-y$ in the denominator
Actually, my answer is pretty much the same, except it's $+ y$ in the denominator
When differentiating the left side, you'll get:
$$(2x^2 + 2y^2) * (2x + 2yy') = 4x^3 + 4x^2 y y' + 4y^2x+4y^3 y'$$
dividing by 2 both sides and taking y' out, the left side would look:
$$y'(2x^2 y + 2y^3 + x + y)$$
as you can see, it's $+y$
so why does wolfram alpha says otherwise?
| $$(x^2+y^2)^2=(x-y)^2$$
$$x^4+2x^2y^2+y^4=x^2-2xy+y^2$$
$$4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}=2x-2y-2x\frac{dy}{dx}+2y\frac{dy}{dx}$$
$$y'(4x^2y+4y^3)+4x^3+4xy^2=y'(-2x+2y)+2x-2y$$
$$y'(4x^2y+4y^3+2x-2y)=-4x^3-4xy^2+2x-2y$$
$$y'=\frac{-4x^3-4xy^2+2x-2y}{4x^2y+4y^3+2x-2y}$$
Take out factor of $2$
$$y'=\frac{-2x^3-2xy^2+x-y}{2x^2y+2y^3+x-y}$$
$\large\color{green}{\checkmark}$Verified by Wolframalpha
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq (b^2 + c^2)/2$? If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq \frac{b^2 + c^2}{2}$? I can only see how to get $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.
| With question as it is now, it holds because $\frac{b^2+c^2}{2}\geq(\frac{b+c}{2})^2$. Can you show why the last inequality holds?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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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.