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How to show that if $9n^2=a^2+b^2$, $a$ and $b$ are multiples of $3$ To be honest, I don't know where to start with this problem:
Let $n\in \mathbb{N}$. Prove that if $9n^2$ is the sum of two perfect squares $(a^2,b^2)$, then $a$ and $b$ are multiples of $3$.
| Taking $9n^2=a^2+b^2$ modulo $3$, we get $$0\equiv a^2+b^2\pmod{3}$$
Now, mod $3$, we have $0^2\equiv 0, 1^2\equiv 1, 2^2\equiv 1$. The only way two of these sum to $0$ is $0+0$, so $a^2,b^2$ are both equivalent to $0$ mod $3$.
Hence, $3|a^2$ and $3|b^2$. However, $3$ is prime, so in fact $3|a$ and $3|b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\theta} + \frac{\sin 3\theta}{\cos 3\theta}= \frac {\sin \theta}{\cos \theta} \frac {\sin 2\theta} {\cos 2\theta} \frac{\sin 3\theta}{\cos 3\theta}$$
$$ \frac {\sin \theta \cos 2\theta \cos 3\theta + \cos \theta \sin 2\theta \cos 3\theta + \cos \theta \cos 2\theta \sin 3\theta - \sin \theta \sin 2\theta \sin 3\theta }{\cos\theta \cos 2\theta \cos 3\theta} = 0 $$
$$\cos 2\theta \{\sin\theta \cos 3\theta + cos \theta \sin 3\theta \} + \sin 2\theta \{\cos \theta \cos 3\theta - \sin \theta \sin 3\theta \} = 0 $$
$$\cos 2\theta \sin(3\theta + \theta) +\sin2\theta \cos(3\theta + \theta) = 0 $$
$$ \cos 2\theta \sin 4\theta + sin 2\theta cos 4\theta = 0$$
$$ \sin (2\theta + 4\theta) = 0$$
$$\sin 6\theta = 0 $$
$$ \theta = \frac {n\pi}{6}, n \in Z$$
I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos 2\theta \cos 3\theta $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.
I'd like :
*
*To understand the intuition behind why removing $\cos \theta \cos 2\theta cos 3\theta $ is a mistake.
*To know how to solve this question correctly
*How do I avoid making these types of mistakes when solving trigonometric equations
| your intuition should be correct here's why
$\left(\frac{\sin\theta}{\cos\theta}+\frac{\sin2\theta}{\cos2\theta }+\frac{\sin3\theta}{\cos3\theta}=\frac{\sin\theta\sin2\theta\sin3\theta}{\cos\theta\cos2\theta\cos3\theta}\right)*\cos\theta\cos2\theta\cos3\theta$
$\Rightarrow \sin\theta\cos2\theta\cos3\theta+\cos\theta\sin2\theta\cos3\theta+\cos\theta\cos2\theta\sin3\theta= \sin\theta\sin2\theta\sin3\theta$
$\Rightarrow \cos\theta(\sin2\theta\cos3\theta+\cos2\theta\sin3\theta )+\sin\theta(\cos2\theta\cos3\theta-\sin2\theta\sin3\theta)=0 $
$\Rightarrow \cos\theta\sin(2\theta+3\theta)+\sin\theta\cos(2\theta+3\theta)=0 $
$\Rightarrow sin([2\theta+3\theta]+\theta)=0 $
$\Rightarrow sin(6\theta)=0 $
So, I'm uncertain that your answer is wrong...
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \frac32.$
Prove that $$\dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{9a^2+b^2}}+\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \dfrac{3}{2}.$$
When is equality attained ?
My Attempt :
I could not think of anything suitable for the entire LHS. The last term $\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}} $ suggested C-S
. So I applied C-S on the values $\left(\dfrac{2a}{\sqrt{a^2+b^2}}\right)$ and $\left(\dfrac{b}{\sqrt{9a^2+b^2}}\right)$ to get :
$$\left(\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2}}\right)^2\leq \left(\dfrac{4a^2}{a^2+b^2}\right)\left(\dfrac{b^2}{9a^2+b^2}\right)\leq\left(\dfrac{4a^2}{b^2}\right)\times \left(\dfrac{b^2}{9a^2}\right)=\dfrac{4}{9}.\,\,\,\,(♦)$$
For the first term $\dfrac{a}{\sqrt{a^2+b^2}}$, I applied C-S on the terms $\left(a\right)$ and $\left(\dfrac{1}{\sqrt{a^2+b^2}}\right)$ to get :
$$\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)\leq \left(\dfrac{a^2}{a^2+b^2}\right)^{1/2}\leq 1.\,\,\,\,(♣)$$
Using the same logic for the second term, I get :
$$\dfrac{b}{\sqrt{9a^2+b^2}}\leq 1 \,\,\,\,(♠)$$
Adding all the inequalities, I get a very "weak" inequality when compared to the problem.
What is the best way to prove this inequality ?
| Hint: Apply the Cauchy-Schwarz inequality to the vectors $(2a, \sqrt{a^2+b^2}, 2a)$ and $(\sqrt{9a^2+b^2}, 2b, 2b)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\frac{\sum_\limits{k=0}^{6}\csc^2\left(x+\frac{k\pi}{7}\right)}{7\csc^2(7x)}$ Find the value of
$\dfrac{\sum_\limits{k=0}^{6}\csc^2\left(x+\dfrac{k\pi}{7}\right)}{7\csc^2(7x)}$
when $x=\dfrac{\pi}{8}$.
The Hint given is: $n\cot nx=\sum_\limits{k=0}^{n-1}\cot\left(x+\dfrac{k\pi}{n}\right)$
I dont know how it comes nor how to use it
| with the help of hint $\displaystyle \sum^{n-1}_{k=0}\cot\left(x+\frac{k\pi}{n}\right) = n\cot(nx)$
differentiate with respect to $x$
$\displaystyle \displaystyle -\sum^{n-1}_{k=0}\csc^2\left(x+\frac{k\pi}{n}\right) = -n^2\csc^2(nx)$
$\displaystyle \sum^{n-1}_{k=0}\csc^2\left(x+\frac{k\pi}{n}\right) = n^2\csc^2(nx)$
put $n=7$ and $\displaystyle x = \frac{\pi}{8}$
| {
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"url": "https://math.stackexchange.com/questions/2154639",
"timestamp": "2023-03-29T00:00:00",
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Considering $0Problem: Considering $0<s<1$ for the series $\sum_{i=1}^{\infty} \frac {1}{i^s}$, I want to show that $a_{2^{n+1} -1}$ $<\sum_{j=0}^{n}$ $({\frac {1} {2^{s-1}})}^j$ where $a_n$ is the partial sums of the aforementioned series..
So for $n=1$ this would yield the inequality:
$\frac {1}{1^s} +\frac {1}{2^s} + \frac {1}{3^s}+ \frac {1}{4^s} < (\frac {1} {2^{s-1}})^{0} + (\frac {1} {2^{s-1}})^{1}=1+\frac {1} {2^{s-1}}$ which would imply that
$\frac {1} {2^{s-1}} > \frac {1}{2^s} + \frac {1}{3^s} + \frac {1}{4^s} $
which I think I can show since $ \frac {1}{2^s} + \frac {1}{3^s} + \frac {1}{4^s} < \frac {3} {2^{s-1}} < \frac {2} {2^{s-1}}= \frac {1} {2^s} < \frac {1} {2^{s-1}}$.
I just need a way to generalize this notion of grouping the terms in groups of $log_2$ to complete the desired statement. Any hints/help appreciated.
Edit: It seems as if I have mis-written the question several times. I will post it from the notes I have and if it is deemed an invalid question, well then I guess we have our answer. We are answering part (b)
| $$ \begin{align}
{\large\sum_{\normalsize i=1}^{\normalsize 2^{n+1}}}\frac{1}{i^s} &= \left[\frac{1}{1^s}\right]+\left[\frac{1}{2^s}+\frac{1}{3^s}\right]+\left[\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}+\frac{1}{7^s}\right]+\cdots+\left[\frac{}{}\cdots\frac{}{}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&\lt \left[\frac{1}{1^s}\right]+\left[\frac{1}{2^s}+\frac{1}{2^s}\right]+\left[\frac{1}{4^s}+\frac{1}{4^s}+\frac{1}{4^s}+\frac{1}{4^s}\right]+\cdots+\left[\frac{}{}\cdots\frac{}{}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&= \left[\frac{1}{1^s}\right]+\quad\left[\frac{2}{2^s}\right]\quad+\qquad\qquad\left[\frac{4}{4^s}\right]\qquad\qquad+\cdots+\left[\frac{2^n}{\left(2^n\right)^s}\right]\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&= \frac{1}{\left(2^0\right)^{s-1}}+\frac{1}{\left(2^1\right)^{s-1}}+\frac{1}{\left(2^2\right)^{s-1}}+\cdots+\frac{1}{\left(2^n\right)^{s-1}}\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}} \\[3mm]
&={\large\sum_{\normalsize j=0}^{\normalsize n}}\left(\frac{1}{2^{s-1}}\right)^j\color{red}{+\frac{1}{\left(2^{n+1}\right)^s}}
\end{align} $$
Hence,
$$ \boxed{ \,\\ \quad {\Huge s_{{\normalsize 2^{n+1}\color{red}{-1}}}}={\large\sum_{\normalsize i=1}^{\normalsize 2^{n+1}\color{red}{-1}}}\frac{1}{i^s} \quad{\Large\lt}\quad {\large\sum_{\normalsize j=0}^{\normalsize n}}\left(\frac{1}{2^{s-1}}\right)^j \qquad\colon\quad s\,\gt\,0 \quad \\\, } $$
NB: Part of this answer is a replication of an existing answer by @B.Mehta with corrections.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving uniform continuity for: $f(x) = \frac{1}{1+x^2}$ on $(-\infty, \infty)$
Show $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $(-\infty, \infty)$
I have recently shown that: $\frac{|x|}{1+x^2} \leq \frac{1}{2}$ for all real $x$ and will use this in the proof. I will prove the result using $\varepsilon$-$\delta$ argument. I've just learned these arguments and was wondering if I could get some verification/advice.
Note that,
$\left| \frac{1}{1+x^2} - \frac{1}{1+y^2} \right| = \left| \frac{y^2 - x^2}{(1+y^2)(1+x^2)} \right| = \frac{|x-y||x+y|}{(1+y^2)(1+x^2)} \leq |x-y|\left(\frac{|x|}{(1+y^2)(1+x^2)} + \frac{|y|}{(1+x^2)(1+y^2)} \right) \leq \\ |x-y|\left(\frac{|x|}{1+x^2} + \frac{|y|}{1+y^2} \right)$
Using the result I mentioned above.
$|x-y|\left(\frac{|x|}{1+x^2} + \frac{|y|}{1+y^2} \right) \leq |x-y|(\frac{1}{2} + \frac{1}{2}) = |x-y|$.
Here is the start of proof:
For $\varepsilon > 0$ and $x,y \in (-\infty, \infty)$ then let $\delta = \varepsilon > 0$. If $|x-y| < \delta$, then:
$|f(x) - f(y)| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2} \right| \leq |x-y| <
\delta = \varepsilon$
| Your proof is correct.
| {
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write the following as a polynomial in $\alpha$ with rational coefficient Let $\alpha = \sqrt[3]{2}$
write $\frac{\alpha^2 -1}{\alpha + 2}$ as a polynomial in $\alpha$ with rational coefficient.
First I write the following:
$\frac{\alpha^2 -1}{\alpha + 2} = \frac{2^{2/3} - 1}{2^{1/3} + 2}$
Then I proceed to manipulate the R.H.S hoping to achieve all rational terms. But it's been a mess so far. Have interpret the question wrongly or is there some algebra trick here I need to do.
Any help or insight is deeply appreciated.
| We have
$$
\alpha^3-2=0 \implies \alpha^2+8=10 \implies \frac{1}{\alpha+2}=\frac{\alpha^2-2\alpha+4}{10}
$$
Thus
$$
\frac{\alpha^2 -1}{\alpha + 2}=\frac{(\alpha^2-1)(\alpha^2-2\alpha+4)}{10}=\frac {1}{10}\,{\alpha}^{4}-\frac{1}{5}\,{\alpha}^{3}+\frac{3}{10}\,{\alpha}^{2}+\frac{1}{ 5}\,\alpha-\frac{2}{5}
$$
| {
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"url": "https://math.stackexchange.com/questions/2157476",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| Okay you have $A = \frac {7^{55}}{5^{72}}=(\frac 75)^{55}\times(\frac 15)^{17}$
well, since you choose to go that way:
$= (\frac {49}{25})^{27}(\frac 1{5})^7\times [\frac 7{5}]$
$= (2*\frac{49}{50})^{27}(\frac 14\times \frac45)^{17}[\frac 7{5}]$
$=2^{27}\times2^{-34}\times[(\frac{49}{50})^{27}\times (\frac 45)^{17}\times \frac 75]$
As $\frac {49}{50} < 1$ and $\frac 45 < 1$ then
$< 2^{27}\times2^{-34}\times \frac 75$
$= \frac 7{2^7*5} < 1$.
So $7^{55} < 5^{72}$. By quite a lot actually.
====
Another way of doing it would be
$\log 7^{55} = 55 \log 7$ and $\log 5^{72} = 72\log 5$
And $\log 7 = \frac 12 \log 49 \approx^- \frac 12 \log \frac{100}2 = 1 - \frac {\log 2}2$. So $\log 7^{55} \approx^- 55 - 22\frac 12 \log 2$
$\log 5 = \log \frac {10}2 = 1 - \log2$. So $\log^{72} = 72 - 72\log 2$
So $7^{55} ??? 5^{72}$
if $55 - 22\frac 12 \log 2 ???^- 72 - 72\log 2$
$49\frac 12 \log 2 ???^- 17$
And $\log 2 = \frac 1{10} \log 2^{10} = \frac 1{10} \log 1024 \approx^+ \frac 3{10}$
So $49\frac 12 \log 2 \approx^+ 4.95\times 3 < 17$.
There is a bit of margin of error as $\log 2 > 3/10$ and $\log 7 < 1 - \frac {\log 2} 2$ but the margin is not significant.
| {
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"url": "https://math.stackexchange.com/questions/2158143",
"timestamp": "2023-03-29T00:00:00",
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value of $abc$ in algebraic expression If $a+b+c=0$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 3$. where $a,b,c$ are non zero real numbers
then $ab(a+b)+bc(b+c)+ca(c+a)$ is
Attempt: from $a+b+c=0$ we have to find value of $-3abc$
and from $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 3$ we have $\displaystyle \frac{ab+bc+ca}{abc} = 3$
could some help how to get value of $abc,$ thanks
| We have $$(a+b+c)^2 =a^2+b^2+c^2+2 (ab+bc+ca) =0$$ $$\Rightarrow ab+bc+ca = -0.5 (a^2+b^2+c^2) $$
We already know $ab + bc + ca =3abc $, so hope you can take it from here.
On a side note, observe that $$ (a+b+c)^3 =a^3+b^3+c^3+ (a+b+c)(ab+bc+ca) \Rightarrow a^3+b^3+c^3=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:
What will happen to the roots of the quadratic equation
$$ax^2 + bx + c = 0$$
if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)
Request:
Is there indeed another root to find? If so, how to I find it?
| If $a = 0$ then $ax^2 + bx + c = bx + c$ and the equation $bx +c=0$ is a linear equation with a single root;$x = -\frac cb$.
The set of equations $\alpha x^2 + bx^2 + c = 0$ will each have two roots of $\frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}$
We can solve $\lim_{\alpha \rightarrow 0}\frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}$ with L'Hopital.
$\lim \frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}= \frac{\mp 2c\frac 1{\sqrt{b^2 - 4\alpha c}}}{2}=\mp \frac c{|b|} = \mp \frac c{\pm b} = - \frac cb$
Okay.... so what happens to the "other" root?
For sake of argument let's assume $b > 0$. Then one of th the two roots is$\frac {-b + \sqrt{b^2 - 4ac}}{2a}$. As $a$ "gets small" $\sqrt{b^2 - 4ac}$ gets close to $b$ and this expression gets close to $0/0$ so we can use L'Hopital to get that the limit. So that was the answer we got above.
But the other root is $\frac {-b + \sqrt{b^2 - 4ac}}{2a}$. As $s$ "gets small" $\sqrt{b^2 - 4ac}$ gets close to $b$ and this expression gets close to $-2b/0$ which diverges to negative infinity.
So.. one root converges to $-\frac cb$ and the other diverges to negative infinity.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $2^a|(x-1)$ iff $2^a|(y-1)$. Suppose that $2^n + 1 =xy$, where $x$ and $y$ are integers > $1$ and $n>0$.
Show that $2^a|(x-1)$ iff $2^a|(y-1)$.
Here $a|b$ implies $a$ divides $b$.
| $$(x-1)(y-1) = 1 + xy - x - y = 2^n - (x - 1) - (y - 1)$$
Then if $2^a$ divides $(x-1)$ then it definetile divides $(y-1)$, because
$$(y - 1) = 2^n - (x - 1) - (x-1)(y-1).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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equation $x^4+ax^3-6x^2+ax+1 = 0$ has two distinct positive roots Finding parameter of $a$ for which equation $x^4+ax^3-6x^2+ax+1 = 0$
has two distinct positive roots
Attempt: writing equation $\displaystyle \bigg(x^2+\frac{1}{x^2}\bigg)+a\bigg(x+\frac{1}{x}\bigg)-6=0\;,$ where $x\neq 0$
So $\displaystyle \bigg(x+\frac{1}{x}\bigg)^2+a\bigg(x+\frac{1}{x}\bigg)-8=0$
put $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t\;,$ where $|t|\geq 2$
so $t^2+at-8=0.$ So for real roots $D\geq 0.$ So $a^2+32\geq 0$
So $\displaystyle t = \frac{-a\pm \sqrt{a^2+32}}{2}.$
could some help me how to solve it, thanks
| Your strategy is good. The equation $t^2+at-8=0$ has two real roots, namely
$$
r_1=\frac{-a+\sqrt{a^2+32}}{2}
\qquad
r_2=\frac{-a-\sqrt{a^2+32}}{2}
$$
Note that $r_1>0$ and $r_2<0$. Consider now
$$
x+\frac{1}{x}=r_i
$$
which becomes
$$
x^2-r_ix+1=0
$$
If we take $i=2$, we see that the equation either has no real root or it has two negative real roots. So we need to take $i=1$. The equation has (distinct) positive roots if and only if its discriminant is positive:
$$
r_1^2-4>0
$$
This becomes$$(\sqrt{a^2+32}-a)^2-16>0$$and, doing some simplifications,$$a^2+8>a\sqrt{a^2+32}$$This is true whenever $a<0$. If $a\ge0$ we can square and get$$a^4+16a^2+64>a^4+32a^2$$that is, $a^2<4$, which gives $0\le a<2$. Thus the final solution is $a<2$.
| {
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Find the $A$ :$A=\frac{\tan 30^{\circ}+\tan 40^{\circ}+\tan 50^{\circ}+\tan 60^{\circ}}{\cos20^{\circ}}=?$ Find the $A$ :
$$A=\frac{\tan 30^{\circ}+\tan 40^{\circ}+\tan 50^{\circ}+\tan 60^{\circ}}{\cos20^{\circ}}=?$$
My Try :
$$\tan q+\tan q=\frac{\sin (q+p)}{\cos q\cos p}\\\cos q \cos p=\frac{\cos (q+p)+\cos(q-p)}{2}\\A=\frac{\frac{\sin(90^{\circ})}{\cos 40^{\circ} \cos 50}+\frac{\sin(90^{\circ})}{\cos 30^{\circ} \cos 40^{\circ}}}{\cos 20^{\circ}}=\frac{\cos 30^{\circ} \cos 40^{\circ} +\cos 40^{\circ} \cos 50^{\circ}}{\cos 30^{\circ}\cos 60^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 20^{\circ} }$$
now ?
| Generalization:
$$\tan(2a-3d)+\tan(2a-d)+\tan(2a+d)+\tan(2a+3d)$$
$$=\dfrac{\sin(2a-3d+2a+3d)}{\cos(2a-3d)\cos(2a+3d)}+\dfrac{\sin(2a-d+2a+d)}{\cos(2a-d)\cos(2a+d)}$$
$$=\dfrac{\sin4a\{2\cos(2a-3d)\cos(2a+3d)+2\cos(2a-d)\cos(2a+d)\}}{2\cos(2a-3d)\cos(2a-d)\cos(2a+d)\cos(2a+3d)}$$
$$=\dfrac{\sin4a(2\cos4a+\cos6d+\cos2d)}{2\cos(2a-3d)\cos(2a-d)\cos(2a+d)\cos(2a+3d)}$$
If $4a=90^\circ,2a+3d=90^\circ-(2a-3d);2a+d=90^\circ-(2a-d)$
$$\tan(45^\circ-3d)+\tan(45^\circ-d)+\tan(45^\circ+d)+\tan(45^\circ+3d)$$
$$=\dfrac{\cos6d+\cos2d}{2\cos(2a-3d)\cos(2a-d)\cos(2a+d)\cos(2a+3d)}$$
$$=\dfrac{4\cos4d\cos2d}{2\cos(2a-3d)\sin(2a-3d)\cdot2\sin(2a-d)\cos(2a-d)}$$
$$=\dfrac{4\cos4d\cos2d}{\sin(4a-6d)\sin(4a-2d)}$$
$$=\dfrac{4\cos4d\cos2d}{\cos6d\cos2d}\text{ as }4a=90^\circ$$
$$=\dfrac{4\cos4d}{\cos6d}\text{ for }\cos2d\ne0$$
Here $d=5^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\arcsin x- \arccos x= \pi/6$, difficult conclusion I do not understand a conclusion in the following equation. To understand my problem, please read through the example steps and my questions below.
Solve:
$\arcsin x - \arccos x = \frac{\pi}{6}$
$\arcsin x = \arccos x + \frac{\pi}{6}$
Substitute $u$ for $\arccos x$
$\arcsin x = u + \frac{\pi}{6}$
$\sin(u + \frac{\pi}{6})=x$
Use sum identity for $\sin (A + B)$
$\sin(u + \frac{\pi}{6}) = \sin u \cos\frac{\pi}{6}+\cos u\sin\frac{\pi}{6}$
$\sin u \cos \frac{\pi}{6} + \cos u\sin\frac{\pi}{6} =x$
I understand the problem up to this point. The following conclusion I do not understand:
$\sin u=\pm\sqrt{(1 - x^2)}$
How does this example come to this conclusion? What identities may have been used? I have pondered this equation for a while and I'm still flummoxed. All advice is greatly appreciated.
Note: This was a textbook example problem
| The easy way to do this is to start with:
$\arccos x = \frac {\pi}{2} - \arcsin x$
Then your problem becomes
$2\arcsin x - \frac {\pi}{2} = \frac {\pi}{6}\\
x = \sin \frac {\pi}{3} = \frac {\sqrt3}2$
Now what have you done?
$\sin (u + \frac \pi6) = x\\
\frac 12 \cos u + \frac {\sqrt{3}}{2}\sin u = x\\
\frac 12 x + \frac {\sqrt{3}}{2}\sqrt{1-x^2} = x$
Why does $\sin (\arccos x) =\sqrt {1-x^2}?$
Draw a right triangle with base $= x$ and hypotenuse $= 1.$
For that angle $u, \cos u = x.$ What is the length of the opposite leg? $\sqrt {1-x^2}$
The range of $\arccos x$ is $[0, \pi]$ so $\sin (\arccos x) \ge 0$
$\sqrt{3}\sqrt{1-x^2} = x\\
\sqrt{3}(1-x^2) = x^2\\
4x^2 = 3\\
x = \sqrt{\frac 34}
$
We can reject the negative root as $x$ must be greater than $0$ as $\arcsin x - \arccos x < 0$ when $x<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ring of integers of $\mathbb{Q}(a)$ I am stuck with this exercise in my home work, could use a hand:
Let $a$ be a root of the irreducible polynomial $f(x)=x^3+x+1$.
Show that the ring of integers of $F:=\mathbb{Q}(a)$ is $\mathbb{Z} + \mathbb{Z}a + \mathbb{Z}a^2$.
Thanks in advance!
| Here is just an ad hoc solution, there is probably a better one using algebraic number theory. Let
$$x^3+x+1=(x-a)(x-b)(x-c)$$
and let $$x+ya+za^2$$ be an algebraic integer.
Multiplying by $a$ and then by $a^2$ and simpifying we get that
$$-z+(x-z)a+ya^2$$
and
$$-y-(y+z)a+(x-z)a^2$$
are also integers.
Now consider
$$x+ya+za^2$$
$$x+yb+zb^2$$
$$x+yc+zc^2$$
adding then together we have
$3x-2z$ is an integer.
Where we use that $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=-2$
Applying the same trick to the other integers we get that $-3z-2y$ and $-3y-2x+2z$ are also integers.
This means that
$$\begin{pmatrix}3&0&-2\\
0&-2&-3\\
-2&-3&2\\
\end{pmatrix}\begin{pmatrix}x\\y\\z\\
\end{pmatrix}$$ is an integer vector.
Using (integer!) row reduction we have that
$$\begin{pmatrix}1&0&-11\\
0&1&-14\\
0&0&31\\
\end{pmatrix}\begin{pmatrix}x\\y\\z\\
\end{pmatrix}$$ is an integer vector.
Thus we see that $31z$ is an integer and $x=n+11z$, and $y=m+14z$.
Thus the whole problem reduces to the question of whether
$$\alpha=\frac{a^2+14a+11}{31}$$ is an integer or not.
It is not an integer, as one sees by calculating $\alpha^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the minimum value of $f=x^2+y^2+x(1-y)+y(1-x)$, holds on $x,y$ are integers. Let $x$ and $y$ are integers such that $-2\le x\le3$ and $-8\le y\le4$
Find the minimum value of $f=x^2+y^2+x(1-y)+y(1-x)\tag{1}$
From $(1)$, I get $f=(x-y)^2+x+y$ and I don't know what I should do next.
I had tried to use differential of f then I get no critical point (the minimum is on the bound $x=-2,x=3,y=-8$ or $y=4$). I found the minimum on $x=-2$ but $y=-2.5$ that isn't integer, the equation f that holds on $x=-2$ is $(y+2.5)^2-4.25,$ So no conclusion for $x,y$ are integers or Can it conclude that $y=2,3$ minimize f ? (I think this solution is suitable for real numbers).
I had tried to substitute all the possible cases then I got the minimum is $-4$, when $(x,y)$=$(-2,-2)$, $(-2,-3)$.
All help would be appreciated.
| Because $x$ and $y$ are integers, we have $$f(x,y)=(y-x)^2+(y-x)+2x=\left(y-x+\frac{1}{2}\right)^2+2x-\frac{1}{4}\geq \frac{1}{4}+2x-\frac{1}{4}=2x\geq-4\,.$$
Therefore, $f(x,y)\geq-4$. The equality holds iff $y-x+\frac{1}{2}=\pm\frac{1}{2}$ and $x=-2$, which leads to $(x,y)=(-2,-2)$ and $(x,y)=(-2,-3)$.
If $x$ and $y$ may take non-integral values (within the required ranges), then
$$f(x,y)\geq 2x-\frac{1}{4}\geq -\frac{17}{4}\,.$$
The equality holds iff $y-x+\frac{1}{2}=0$ and $x=-2$, i.e., $(x,y)=\left(-2,-\frac{5}{2}\right)$.
For the maximum, we note that $-\frac{21}{2}\leq y-x+\frac{1}{2}\leq \frac{13}{2}$, so $$f(x,y)\leq \frac{441}{4}+2x-\frac{1}{4}= 110+2x\leq 116\,.$$
The equality holds if and only if $y-x+\frac{1}{2}=-\frac{21}{2}$ and $x=3$, which means $(x,y)=(3,-8)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Derivative of $ \sin^x(x) $ I was trying to find the derivative for $\sin^x(x)$
I followed two methods, to get to different answers and after comparing the answer with Wolfram Alpha, I found the one which was correct and which was wrong, however I am unable to reconcile why the one which was wrong is incorrect.
The method which leads to the answer from Wolfram Alpha is as follows:
$ y = \sin^x(x) $
$ \implies \ln(y) = x\ln(\sin(x)) $
Taking derivative with respect to x on both sides, applying chain rule for the first term on the LHS:
$ \implies \frac{d}{dx}ln(y) = x.\frac{1}{\sin(x)}.\cos(x) + \ln(\sin(x)) $
$ \implies \frac{d}{dy}ln(y).\frac{dy}{dx} = x.\cot(x) +\ln(\sin(x)) $
$ \implies \frac{1}{y}.\frac{dy}{dx} = x.\cot(x) + \ln(\sin(x)) $
$ \implies \frac{dy}{dx} = y.[x.\cot(x) + \ln(\sin(x))] $
$ \implies \frac{dy}{dx} = \sin^x(x).[x.\cot(x) + \ln(\sin(x))]$
Which is the same as I get from Wolfram: https://www.wolframalpha.com/input/?i=derivative+of+sinx%5Ex
Now the second method I tried and it leads to a partial answer:
$ y = \sin^x(x) $
$ \frac{dy}{dx} = \frac{d}{dx}\sin^x(x) $
$ \frac{dy}{dx} = \frac{d}{dx}\sin(x).\sin^{x-1}(x) $
$ \frac{dy}{dx} =\sin(x).\frac{d}{dx}\sin^{x-1}(x) + \sin^{x-1}(x)\cos(x) $
$ \frac{dy}{dx} =\sin(x).\frac{d}{dx}\sin(x).\sin^{x-2}(x) + \sin^{x-1}(x)\cos(x)$
$ \frac{dy}{dx} =\sin(x).[\sin(x).\frac{d}{dx}\sin^{x-2} + \cos(x).\sin^{x-2}(x)] + \sin^{x-1}(x)\cos(x)$
$ \frac{dy}{dx} = \sin^2(x).\frac{d}{dx}\sin^{x-2} + 2.\sin^{x-1}(x).cos(x) $
Repeating this process
$ \frac{dy}{dx} = \sin^3(x).\frac{d}{dx}\sin^{x-3} + 3.\sin^{x-1}(x).cos(x) $
Again
$ \frac{dy}{dx} = \sin^4(x).\frac{d}{dx}\sin^{x-4} + 4.\sin^{x-1}(x).cos(x) $
until
$ \frac{dy}{dx} = \sin^{x-1}(x).\frac{d}{dx}\sin^{x-(x-1)} + (x-1).\sin^{x-1}(x).cos(x) $
$ \frac{dy}{dx} = \sin^{x-1}(x).\frac{d}{dx}\sin(x) + (x-1).\sin^{x-1}(x).cos(x) $
$ \frac{dy}{dx} = \sin^{x-1}(x).\cos(x) + (x-1).\sin^{x-1}(x).cos(x) $
$ \frac{dy}{dx} = x.\sin^{x-1}(x).cos(x) $
When I compare this with the previous result:
$ \frac{dy}{dx} = \sin^x(x).[x.\cot(x) + \ln(\sin(x))] $
$ \frac{dy}{dx} = x.\sin^{x-1}(x).\cos(x) + \sin^x(x)\ln(\sin(x)) $
There is this extra term
$ \sin^x(x)\ln(\sin(x)) $
Now looking at both terms:
$ x.\sin^{x-1}(x).\cos(x) $
and
$ \sin^x(x)\ln(\sin(x)) $
I see the first one is something I would get from considering y to a be a polynomial in x and finding the derivative, while the second one is something I would get if I considered y to be an exponential in x and solved that.
Perhaps I am forgetting some basic calculus here, I hope someone can help me reconcile the reasoning here.
|
Quote:
[...] Repeating this process
$$ \frac{dy}{dx} = \sin^3(x)\cdot\frac{d}{dx}\sin^{x-3} + 3\cdot\sin^{x-1}(x).cos(x) $$
Again
$$ \frac{dy}{dx} = \sin^4(x)\cdot\frac{d}{dx}\sin^{x-4} + 4\cdot\sin^{x-1}(x)\cdot\cos(x) $$
until
$$ \frac{dy}{dx} = \sin^{x-1}(x)\cdot\frac{d}{dx}\sin^{x-(x-1)} + (x-1)\cdot\sin^{x-1}(x).cos(x) $$
[...]
... "until" you decide arbitrarily that $x$ is a natural number which does not depend on $x$, thus making a major mistake.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$ Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$
Wolfram Alpha gave this result.
But may be there is simple way to get all a?
| $$\text{Given}\quad a^4-6a^2b-8ac-3b^2=0\quad\implies 3b^2+6a^2b -(a^4-8ac)=0$$
The solution for $b$ using the quadratic equation is
$$b = \pm\frac{2 \sqrt{a^4 - 2 a c}}{\sqrt{3}} - a^2\quad\text{for}\quad a\in\mathbb{R}\land c\in\mathbb{R}\land b\in\mathbb{R}\iff a^4-2ac\ge0$$
Technically, this defines all values of $a$ but perhaps you seek integers.
Regular algebras solves for $c$ as $$c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$$
The second equation looks easier to work with (though you can try the first if you want) and experimentation in a spreadsheet offers these sample integer triples.
\begin{equation}
(1,-11,-37)\quad
(1,-7,-13)\quad
(1,-3,-1)\quad
(1,1,-1)\quad
(1,5,-13)\quad
(1,9,-37)\quad\\
(2,-8,1)\quad
(2,-4,4)\quad
(2,0,1)\quad
(2,4,-8)\quad
(2,8,-23)\quad
(2,12,-44)\quad\\
(3,-11,13)\quad
(3,-7,13)\quad
(3,-3,9)\quad
(3,1,1)\quad
(3,5,-11)\quad
(3,9,-27)\quad
(3,13,-47)\quad\\
(4,-8,26)\quad
(4,0,8)\quad
(4,8,-22)\quad\\
(5,5,-5)\quad\\
(6,-8,59)\quad
(6,-4,44)\quad
(6,0,27)\quad
(6,4,8)\quad
(6,8,-13)\quad
(6,12,-36)\quad\\
(7,-7,77)\quad\\
(8,-8,109)\quad
(8,0,64)\quad
(8,8,13)\quad
\end{equation}
WolframAlpha offer only the roots $a=0,b=0$ or
$a\ne0\land c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$ as shown here. This solution for $c$ (as I pointed out above) appears to have infinite integer solutions and, when $a=1$, it appears that these have a symmetry for values of $b$ around zero. Let me know if you have other questions about this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $. Proof explanation Question: Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
Solution: Let's suppose that $3^{n-1}+5^{n-1}\mid 3^n+5^n$, so there is some positive integer $k$ such that $3^n+5^n=k(3^{n-1}+5^{n-1})$. Now, if $k\ge 5$ we have $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$$
This means that $k\le 4$. On the other hand, $3^n+5^n=3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$, then $k\ge 4$ and thus we deduce that $k=4$. In this case we have $3^n+5^n=4(3^{n-1}+5^{n-1})$, which gives us the equation $5^{n-1}=3^{n-1}$, but if $n>1$ this equation is impossible.
Hence $n=1$
The above solution is from this answer. Could someone please explain a few things:
*
*why was the value of $k\le4$ and $k\ge4$ picked? where did they get 4?
*why does $3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$ mean that $k\ge4$ ?
*Why does $5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$ mean that $k\le4$ ?
| The line $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n$$
makes a contradiction, since $k(3^{n-1}+5^{n-1})=3^n+5^n$, and $3^n+5^n>3^n+5^n$ is obviously false. Since they assumed $k\geq 5$, that must be false; thus, we conclude $k\leq 4$.
Then they say $$3^n+5^n>3(3^{n-1}+5^{n-1})$$ which you could rewrite to
$$3^n+5^n=k(3^{n-1}+5^{n-1}>3(3^{n-1}+5^{n-1})$$
thus $k>3$, and since $k\leq 4$, we know $k=4$.
They're basically just first eliminating all numbers greater than four, and then all integers smaller than four, to conclude $k=4$ (which they eliminate too).
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x^d \equiv a$ mod $p$ if and only if $a^\frac{p-1}{d} \equiv 1$ mod $p$ Let $p$ be an odd prime and $d$ be a natural number such that $d \mid (p-1)$. For $a \in Z$ coprime with p, show that $x^d \equiv a$ mod $p$ if and only if $a^\frac{p-1}{d} \equiv 1$ mod $p$. I have no idea about this $x^d \equiv a$ mod p.
| Assumptions:
*
*$p$ is prime.
*$d$ is a positive integer such that $d \mid (p-1)$.
*$a$ is an integer, $a$ not a multiple of $p$.
To be shown:
*
*$x^d \equiv a \pmod{p}$, for some integer $x$, if and only if $a^\frac{p-1}{d} \equiv 1 \pmod{p}$.
First we show $x^d \equiv a \pmod{p}$, for some integer $x$, implies $a^\frac{p-1}{d} \equiv 1 \pmod{p}$.
Thus, suppose there is an integer $x$ such that $x^d \equiv a \pmod{p}$.
Since $a,p$ are relatively prime, $x^d \equiv a \pmod{p}$ implies $x,p$ are relatively prime.
Then
\begin{align*}
a^\frac{p-1}{d} &\equiv (x^d)^\frac{p-1}{d} \pmod{p}\\[6pt]
&\equiv x^{p-1} \pmod{p}\\[6pt]
&\equiv 1 \pmod{p}\\[6pt]
\end{align*}
as was to be shown.
Next we show $a^\frac{p-1}{d} \equiv 1 \pmod{p}$ implies $x^d \equiv a \pmod{p}$, for some integer $x$.
Thus, suppose $a^\frac{p-1}{d} \equiv 1 \pmod{p}$.
Since $p$ is prime, there exists a primitive element, mod $p$. Let $y$ be such an element.
Since $y$ is a primitive element, mod $p$, the multiplicative order, mod $p$, of $y$ is $p-1$.
It follows that the $p-1$ elements $1,y,y^2,...y^{p-2}$ represent, in some order, all nonzero residues, mod $p$. In particular, there is an integer $k \in \{0,1,...,p-2\}$ such that $y^k \equiv a \pmod{p}$.
Then
\begin{align*}
&a^\frac{p-1}{d} \equiv 1 \pmod{p}\\[6pt]
\implies\; &(y^k)^\frac{p-1}{d} \equiv 1 \pmod{p}\\[6pt]
\implies\; &(p-1)\; \text{ divides} \left(k{\small{\left(\frac{p-1}{d}\right)}}\right)\\[6pt]
\implies\;
&
{\frac
{k{\large{\left(\frac{p-1}{d}\right)}}}
{p-1}} \in \mathbb{Z}^+
\\[6pt]
\implies\; &{\frac{k}{d}} \in \mathbb{Z}^+\\[6pt]
\implies\; &k = jd,\text{ for some }j \in \mathbb{Z}^+\\[6pt]
\end{align*}
Let $x = y^j$. Then
\begin{align*}
x^d &\equiv (y^j)^d \pmod{p}\\[6pt]
&\equiv y^{jd} \pmod{p}\\[6pt]
&\equiv y^k \pmod{p}\\[6pt]
&\equiv a \pmod{p}\\[6pt]
\end{align*}
thus, we have an integer $x$ such that $x^d \equiv a \pmod{p}$, as was to be shown.
This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inverse hyperbolic functions are equal to each other but how I want to prove the following statement:
$${arctanh}(\frac{x}{\sqrt{1+x^2}})= {arccosech}(\frac{1}{x})$$
I consulted Schaum's Outlines of Mathematical Handbook of Formulas and Tables. It says:
$${arctanh}(x)=\frac{1}{2}ln(\sqrt{\frac{1+x}{1-x}})$$
$${arccosech}(x)=ln(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1})$$
So I wrote my question in terms of those statements above:
$${arctanh}(\frac{x}{\sqrt{1+x^2}})= \frac{1}{2} ln(\frac{1+\frac{x}{\sqrt{1+x^2}}}{1-\frac{x}{\sqrt{1-x^2}}}) $$
$${arccosech}(\frac{1}{x})= ln(\frac{1}{\frac{1}{x}}+\sqrt{ \frac{1}{(\frac{1}{x})^2}+1}) $$
Now, the two equations those I must prove whether they are equal or not are:
$$ (\frac{1+\frac{x}{\sqrt{1+x^2}}}{1-\frac{x}{\sqrt{1-x^2}}})^\frac{1}{2} = (\frac{1}{\frac{1}{x}}+\sqrt{\frac{1}{(\frac{1}{x})^2}+1}) $$
Then I squared them both:
$$ \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}-x} = x^2 + 2x\sqrt{x^2+1}+x^2+1$$
But it seems to me it is going to anywhere. Any help or solving method is appreciated.
Regards.
| $\DeclareMathOperator{\atanh}{arctanh}\DeclareMathOperator{\acsch}{arccosech}$Two key identities to note are
$$
1 \pm \frac{x}{\sqrt{1 + x^{2}}} = \frac{\sqrt{1 + x^{2}} \pm x}{\sqrt{1 + x^{2}}},
$$
so that
$$
\frac{1 + \dfrac{x}{\sqrt{1 + x^{2}}}}{1 - \dfrac{x}{\sqrt{1 + x^{2}}}}
= \frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x},
\tag{1}
$$
and the difference of squares factorization
$$
(\sqrt{1 + x^{2}} + x)(\sqrt{1 + x^{2}} - x) = (1 + x^{2}) - x^{2} = 1,
$$
which gives
$$
\sqrt{1 + x^{2}} + x = \frac{1}{\sqrt{1 + x^{2}} - x}.
\tag{2}
$$
Since $\atanh x = \ln \sqrt{\dfrac{1 + x}{1 - x}} = \dfrac{1}{2} \ln \dfrac{1 + x}{1 - x}$ (cf. your formula),
\begin{align*}
\atanh \frac{x}{\sqrt{1 + x^{2}}}
&= \frac{1}{2} \ln\frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x} && (1) \\
&= \frac{1}{2} \ln (\sqrt{1 + x^{2}} + x)^{2} && (2) \\
&= \ln (\sqrt{1 + x^{2}} + x) && \\
&= \acsch \frac{1}{x}.
\end{align*}
| {
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Solving $x + \sqrt{x(x-a)} = b + a/2$ This algebraic equation came up in some work I was doing, and I haven't been able to solve it. I know what the solution is, but I'm quite bothered by not knowing the techniques to get there.
$$x + \sqrt{x(x-a)} = b + \frac{a}{2} $$
My attempts:
$$x + \sqrt{x(x-a)} = \frac{x^2 - x(x-a)}{x - \sqrt{x(x-a)}}$$
This led to the same issues in the denominator that I was having previously.
$$(x + \sqrt{x(x-a)})^2= x^2 + x(x-a) + 2x \sqrt{x(x-a)} = \frac{(a + 4b^2)^2}{2}$$
This RHS looks fairly close to the actual solution for $x $, but I don't see how to work with this unruly equation
| Well, we have that:
$$x+\sqrt{x\left(x-\text{a}\right)}=\text{b}+\frac{\text{a}}{2}\tag1$$
Subtract $x$ from both sides:
$$\sqrt{x\left(x-\text{a}\right)}=\text{b}+\frac{\text{a}}{2}-x\tag2$$
Square both sides:
$$\left(\sqrt{x\left(x-\text{a}\right)}\right)^2=\left(\text{b}+\frac{\text{a}}{2}-x\right)^2=x\left(x-\text{a}\right)=\frac{\text{a}^2}{4}+\text{a}\text{b}+\text{b}^2-x\left(\text{a}+2\text{b}\right)+x^2\tag3$$
So:
$$x\left(x-\text{a}\right)-\frac{\text{a}^2}{4}-\text{a}\text{b}-\text{b}^2+x\left(\text{a}+2\text{b}\right)-x^2=$$
$$\text{a}^2-4\text{a}\text{b}-4\text{b}^2+8\text{b}x=0\tag4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do we find the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}?$ We would like to find out the closed form for integral $(1)$
$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}\tag1$$
An attempt:
We may write
$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$
Let $x=\sin x$ and $y=\cos x$
$x^6+y^6=(x^4-x^2y^2+y^4)$
Simplified down to
$x^6+y^6=\sin^4 x+\cos2x\cos^2x$
$x^6+y^6=\sin^2 x-{1\over 4}\sin^2 2x+\cos2x\cos^2x$
$(1)$ becomes
$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^2 x-{1\over 4}\sin^2 2x+\cos2x\cos^2x}}\tag2$$
The power has reduced but looked more messier
How else can we evaluate $(1)$?
| Using the substitution $x=\arctan t$ the given integral boils down to:
$$ I=\int_{0}^{+\infty}\sqrt{\frac{1+t^2}{1+t^6}}\,dt=\int_{0}^{+\infty}\frac{dt}{\sqrt{t^4-t^2+1}}=\int_{0}^{+\infty}\frac{dt}{\sqrt{(t^2+\omega^2)(t^2+\bar{\omega}^2)}} $$
(with $\omega=\exp\left(\frac{\pi i}{3}\right)$ ) that is a complete elliptic integral of the first kind. Such integral can be computed through the AGM mean $\text{AGM}(a,b)=\text{AGM}\left(\frac{a+b}{2},\sqrt{ab}\right)$:
$$ I = \frac{\pi}{2\,\text{AGM}(\omega,\bar{\omega})} = \color{red}{\frac{\pi}{\text{AGM}(1,2)}}.$$
In particular, the integral is bounded between $\pi(12-8\sqrt{2})$ and $\pi\sqrt[4]{\frac{2}{9}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Finding a basis for a set of $2\times2$ matrices
Find a basis for $M_{2\times2}$ containing the matrices $$\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix}$$
I know that every $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ can be written as:
$$a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$
so those matrices are a basis for the vector space of $2\times2$ matrices, but how do I apply this to specific matrices? I know how to find a basis for a set of vectors, but matrices confuse me.
| You can go with more trial & error-like approaches, or go for a more systematic approach. If you can easily add two linearly independent matrices "by inspection", then you are done. This is doable in your case (see Emilio Novati's answer) but can get hard(er) in a more general case.
Knowing the standard basis, you could look at:
$$\left\{
\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix},
\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix},
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},
\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
\right\}$$
Because the last 4 matrices form a basis for $M_{2 \times 2}$, this set clearly spans $M_{2 \times 2}$. But since $M_{2 \times 2}$ is 4-dimensional, this set cannot be linearly independent. Work from left to right and omit any matrix which can be written as a linear combination of the matrices before; i.e. deleting the linearly dependent matrices to end up with a basis for $M_{2 \times 2}$ containing the given two matrices.
This can be done in a more systematic way by entering the matrices in columns, identifying the matrix $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ with the column vector $(a,b,c,d)^T$ and then row reducing this matrix; the linearly independent columns (and thus corresponding matrices) will be the pivot columns. This approach is explained in Omnomnomnom's answer as well.
Alternatively, you can check this answer for another systematic approach. I'll leave out the details here, but identifying the matrix $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ with the vector $(a,b,c,d) \in \mathbb{R}^4$, this comes down to finding a basis for the null space of the matrix (WolframAlpha):
$$\begin{pmatrix} 1 & 1 & 2 & 3 \\ 1 & 1 & 3 & 2 \end{pmatrix}$$
So a basis would also be:
$$\left\{
\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix},
\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix},
\begin{pmatrix} -5 & 0 \\ 1 & 1 \end{pmatrix},
\begin{pmatrix} -1 & 1 \\ 0 & 0 \end{pmatrix}
\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proving $\int_0^\infty\frac{\sin(x)}x\ dx=\frac{\pi}2$. Why is this step correct? I came across a different approach on the proof: $$\int_0^\infty \frac{\sin(x)}x\ dx=\frac{\pi}2$$
First, recall the identity: $$\sin(A)-\sin(B)=2\sin\left(\frac{A}2-\frac{B}2\right)\cos\left(\frac{A}2+\frac{B}2\right)$$
Applying the identity for: $$A=kx+\frac{x}2\ \land\ B=kx-\frac{x}2$$
We obtain:$$\sin\left(kx+\frac{x}2\right)-\sin\left(kx-\frac{x}2\right)=2\sin\left(\frac{x}2\right)\cos\left(kx\right)\Rightarrow \\\cos\left(kx\right)=\frac{\sin\left(kx+\frac{x}2\right)-\sin\left(kx-\frac{x}2\right)}{2\sin\left(\frac{x}2\right)}$$
Using the previous result, we can easily show that:
$$\frac12+\cos(x)+\cos(2x)+\cdots+\cos(\lambda x)=\frac{\sin\left(\lambda x+\frac{x}2\right)}{2\sin\left(\frac{x}2\right)} \quad \text{where $\lambda \in \mathbb{N}$}$$
Integrating the last expression: $$\int_0^\pi\frac{\sin\left(\lambda x+\frac{x}2\right)}{\sin\left(\frac{x}2\right)}\ dx=\int_0^\pi\left(1+2\cos(x)+2\cos(2x)+\cdots+2\cos(\lambda x)\right)\ dx=\pi$$
We can also prove (since $f(x)$ is continuous on $[0,\pi]$), using Riemann-Lebesgue Lemma, that: $$\lim_{\lambda\to\infty}\int_0^\pi\underbrace{\left(\frac2t-\frac1{\sin\left(\frac{t}2\right)}\right)}_{f(x)}\sin\left(\lambda t+\frac{t}2\right)dt=\lim_{\lambda\to\infty}\int_0^\pi\left(\frac{2\sin\left(\lambda t+\frac{t}2\right)}t-\frac{\sin\left(\lambda t+\frac{t}2\right)}{\sin\left(\frac{t}2\right)}\right)=0$$
Therefore: $$\left(1\right)\ \lim_{\lambda\to\infty}\int_0^\pi\frac{2\sin\left(\lambda t+\frac{t}2\right)}t=\lim_{\lambda\to\infty}\int_0^\pi\frac{\sin\left(\lambda t+\frac{t}2\right)}{\sin\left(\frac{t}2\right)}=\pi$$
$$$$Returning to the initial problem:
$$\\$$
Let: $$x=\lambda t+\frac{t}2$$
Thus:
$$\int_0^\infty \frac{\sin(x)}x\ dx \stackrel{\eqref{*}}=\frac12\lim_{\lambda\to\infty}\int_0^{\color{teal}{\pi}}\frac{2\sin\left(\lambda t+\frac{t}2\right)}{t}\ dt$$
Using the result obtained from $(1)$:$$\int_0^\infty \frac{\sin(x)}x\ dx=\boxed{\frac{\pi}2}$$
$$$$ My question comes from $\color{teal}{(???)}$, Why is it correct to have $\pi$ instead of $\infty$ when changing the limits of integration?
| Let $x = (\lambda + \frac{1}{2}) t$. Then
$$\int_0^T \frac{\sin((\lambda + \frac{1}{2}) t)}{t} \mathrm{d}t =
\int_0^{(\lambda + \frac{1}{2}) T} \frac{\sin(x)}{\frac{x}{\lambda + \frac{1}{2}}}\frac{\mathrm{d}x}{(\lambda + \frac{1}{2})}=\int_0^{(\lambda + \frac{1}{2}) T} \frac{\sin(x)}{x} \mathrm{d}x $$
Take the limit:
$$\lim_{\lambda\to\infty} \int_0^{(\lambda + \frac{1}{2}) T} \frac{\sin(x)}{x} \mathrm{d}x = \int_0^{\infty} \frac{\sin(x)}{x} \mathrm{d}x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 0
} |
searching for $f(x)$ when knowing $f(2^{2x})$ I have a function that works for powers of 2
This is only for INTEGERS
is there a way to calculate any integer x?
$$f(2^{2x})=\frac{4^x+2}3.$$
$f(x)$=?
here are the first 100 values of f(x)
1, 2, 2, 2, 2, 4, 4, 6, 6, 8, 8, 6, 6, 8, 8, 6, 6, 8, 8, 6, 6, 8, 8, \ 14, 14, 16, 16, 14, 14, 16, 16, 22, 22, 24, 24, 22, 22, 24, 24, 30, \ 30, 32, 32, 30, 30, 32, 32, 22, 22, 24, 24, 22, 22, 24, 24, 30, 30, \ 32, 32, 30, 30, 32, 32, 22, 22, 24, 24, 22, 22, 24, 24, 30, 30, 32, \ 32, 30, 30, 32, 32, 22, 22, 24, 24, 22, 22, 24, 24, 30, 30, 32, 32, \ 30, 30, 32, 32, 54, 54, 56, 56, 54 is there a pattern to calculate any f(x)?
thanx
| $$f(2^{2x})=\frac{4^x+2}{3}=\frac{(2^2)^x+2}{3}=\frac{2^{2x}+2}{3}$$ so $$f(x)=\frac{x+2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume between a sphere and a plane in Cartesian system How can I calculate the volume between the sphere $r=R$ and the plane $z=R/2$
(above the plane and below the sphere) in Cartesian system?
edit - using triple integral $dxdydz$
| Using a triple integral is not a good idea..... but can be a good exercise!
So, as a first step note that the projection on the plane $x-y$ of the itersection of the sphere of radius $R$, centered at the origin, with the plane of equation $z=R/2$ is a circle of equation $x^2+y^2=\frac{3}{4}R^2$.
This means that the limits of integration for $y$ are:
$$
-\sqrt{\frac{3}{4}R^2-x^2}<y<\sqrt{\frac{3}{4}R^2-x^2}
$$
and $y$ is a real number if $x$ has limits:
$$
-\frac{\sqrt{3}}{2}R<x<\frac{\sqrt{3}}{2}R
$$
The equation of the sphere in cartesian coordinates is $x^2+y^2+z^2=R^2$, so we have the limits of integration for $z$:
$$
\frac{R}{2}<z<\sqrt{R^2-x^2-y^2}
$$
Putting all toghether we have that the volume is:
$$
\int_{-\frac{\sqrt{3}}{2}R}^{\frac{\sqrt{3}}{2}R} \int_{-\sqrt{\frac{3}{4}R^2-x^2}}^{\sqrt{\frac{3}{4}R^2-x^2}} \int _{\frac{R}{2}}^{\sqrt{R^2-x^2-y^2}} dz dy dx
$$
This integral is difficult and requires a trigonometric substitution. A simpler solution is to use cylindrical coordinates. Can you find what the integral becomes in this case?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Simplification of $\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$ $$\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$$
The above sequence amounts to one of the following options:
*
*$\binom{100}{50}$
*$\binom{100}{51}$
*$\binom{50}{25}$
*$\binom{50}{25}^2$
| $$\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$$
Using $\binom{n}{r}=\binom{n}{n-r}$ ,
$$= \binom{50}{0}\binom{50}{49}+\binom{50}{1}\binom{50}{48}+\cdots+\binom{50}{49}\binom{50}{0}$$
Using Vandermonde's identity:
$$=\binom{50+50}{49}=\binom{100}{49}=\binom{100}{51}$$
Option b is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution Show that for $p$ to be odd prime and $p \equiv 3$ mod $4$, then $x^2+y^2 = p$ has no integer solution. I have no idea how can i apply quadratic reciprocity to the equation $x^2+y^2 = p$ or should use other method.
| Generalization,
If $k$ be a positive integer satisfying $k \equiv 3 \pmod 4$ then $x^2+y^2=k$ has no integer solutions.
Proof
$k \equiv 3 \pmod 4$ is given and if $x,y$ be integers satisfying $x^2+y^2=k$ then we must have,
$x^2+y^2\equiv 3 \pmod 4$
But,
Given any integer $j$ we know that $j^2 \equiv 0 \pmod 4$ or $j^2 \equiv 1 \pmod 4$
And hence $x \in \mathbb{Z} , y \in \mathbb{Z}$ implies, $x^2+y^2 \equiv 0+0 = 0 \pmod 4$ or $\equiv 1+0 = 1 \pmod 4$ or $\equiv 1+1 = 2 \pmod 4$or $\equiv 0+1 = 1 \pmod 4$
And again, None of the numbers $0,1,2$ is $\equiv 3 \pmod 4$ .
Hence contradiction !!
Thus our assumption was false.
Hence Proved
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
For $\{a,b\}\subset(0,1]$ prove that $a^{b-a}+b^{a-b}+(a-b)^2\leq2$
Let $\{a,b\}\subset(0,1]$. Prove that:
$$a^{b-a}+b^{a-b}+(a-b)^2\leq2$$
I don't see even how to begin the proof.
We can rewrite our inequality in the following form
$$\frac{a^b}{a^a}+\frac{b^a}{b^b}+(a-b)^2\leq2$$ or
$$(ab)^b+(ab)^a\leq(2-(a-b)^2)a^ab^b.$$
By Jensen easy to show that $a^ab^b\geq\left(\frac{a+b}{2}\right)^{a+b}$.
Thus, it remains to prove that
$$\left(\frac{a+b}{2}\right)^{a+b}(2-(a-b)^2)\geq(ab)^b+(ab)^a$$
and I not sure that the last inequality is true.
Thank you!
| Nice inequality. We need some lemmas first.
Let $\varphi\colon [0,1] \rightarrow \mathbb{R}$ be the function given by
$$\varphi(x) = \begin{cases} (x-1) \log \left(\frac{1-x}{2}\right)+(x+1) \log \left(\frac{x+1}{2}\right) & \text{if $x \neq 1$,} \\[1ex] 0 & \text{if $x = 1.$} \end{cases}
$$
Lemma 1. $\varphi \geq 0$.
Proof.
We have $$\lim\limits_{x \to 1} \varphi(x) = 0,$$ thus $\varphi$ is continuous and attains its minimum at a point $x \in [0,1]$. If $x \in \{0,1\}$, then $\varphi(x) = 0$ and we are done.
If $x \in (0,1)$, we have
$$\varphi''(x)=\frac{2 x}{x^2-1} < 0.$$
Since $\varphi$ is differentiable at $x$ and $x$ is a minimum point, we also have $\varphi'(x) = 0$. By the second derivative test, $x$ is a strict local maximum point. This is inconsistent with $x$ being a global minimum point. Thus $x \in \{0,1\}$ and we are done.
$$\tag*{$\Box$}$$
For all $x \in [0,1]$, we let $\psi_x\colon (0,1] \rightarrow \mathbb{R}$ be the function given by
$$\psi_x(z) = (x-1) \log (z)+(x+1) \log (x+z).$$
Lemma 2. If $x \in [0,1]$, then $\psi_x \geq 0$.
Proof.
Let $x \in [0,1]$.
If $x=1$, we have $\psi_x(z)=2 \log (z+1)\geq 0$ for all $z \in (0,1]$ and we are done.
If $x \in [0,1)$, then for all $z \in (0,1]$, we have
$$\psi_x'(z) = \frac{2 x}{z (x+z)} \left(z-\frac{1-x}{2}\right).$$
For all $z \in (0,\frac{1-x}{2}]$, we have $\psi_x'(z) \leq 0$, thus $\psi_x$ is decreasing on $(0,\frac{1-x}{2}]$.
For all $z \in [\frac{1-x}{2},1]$, we have $\psi_x'(z) \geq 0$, thus $\psi_x$ is increasing on $[\frac{1-x}{2},1]$.
Since $\psi_x\left(\frac{1-x}{2}\right) = \varphi(x) \geq 0$ by Lemma 1, we are done.
$$\tag*{$\Box$}$$
For all $x \in [0,1]$, we let $\gamma_x\colon (0,1] \rightarrow \mathbb{R}$ be the function given by
$$\gamma_x(z) = x^2+z^x+(x+z)^{-x}.$$
Lemma 3. If $x \in [0,1]$, then $\gamma_x$ is increasing.
Proof.
Let $x \in [0,1]$.
For all $z \in (0,1]$, we have
\begin{align}
\gamma_x'(z) = x \left(z^{x-1}-(x+z)^{-x-1}\right) &\geq 0
\\ \impliedby z^{x-1} &\geq (x+z)^{-x-1}
\\ \iff (x-1) \log z &\geq (-x-1) \log(x+z)
\\ \iff \psi_x(z) &\geq 0
\\ \impliedby &\text{Lemma 2}.
\end{align}
$$\tag*{$\Box$}$$
Claim. If $a,b \in (0,1]$, then $a^{b-a}+b^{a-b}+(a-b)^2 \leq 2$.
Proof.
Let $a,b \in (0,1]$. Since the inequality is symmetric, we can assume $b \geq a$. Let $x = b-a \in [0,1)$.
We have
\begin{align}
a^{b-a}+b^{a-b}+(a-b)^2 &= a^{x}+(x+a)^{-x}+x^2
\\ &= \gamma_{x}(a)
\\ \text{By Lemma 3:}
\\ &\leq \gamma_{x}(1-x)
\\[5pt] &= 1+x^2+(1-x)^x
\\ \text{By Bernoulli's inequality:}
\\ &\leq 2.
\end{align}
$$\tag*{$\Box$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding a holomorphic complex function We have that $\varphi:R->R$ can be derived twice, we want to find a function f=u+iv which is holomorphic. $u(x,y)=\varphi ( \frac{x^2+y^2}{x} )$
Solution:
We calculate that:
$u_{x}=\frac{x^2-y^2}{x^2} \varphi' (\frac{x^2+y^2}{x})$ and $u_{xx}= \frac{2y^{2}}{x}\varphi' (\frac{x^2+y^2}{x})+(\frac{x^2-y^{2}}{x^2})^2\varphi'' (\frac{x^2+y^2}{x})$
$u_{y}=\frac{2y}{x}\varphi'(\frac{x^2+y^2}{x})$ and $u_{yy}=\frac{2}{x}\varphi'(\frac{x^2+y^2}{x})+\frac{4y^2}{x^2}\varphi''(\frac{x^2+y^2}{x})$
I have to use this information and:
$$u_{xx}+u_{yy}=0$$
but I end up having this massive equaion which I can't solve:
$$\frac{2(y^2+1)}{x}\varphi'(\frac{x^{2}+y^2}{x})+((\frac{x^2-y^2}{x^2})^2+\frac{4y^2}{x^2})\varphi''(\frac{x^{2}+y^2}{x})=0$$
So I either misscalculated something or there is a magic trick which I am missing. Can you help me find my mistake and fix it ?
| Note first that your differential equation (which contains a small typo, because the derivative of $-y^{2}/x^{2}$ with respect to $x$ is $2y^{2}/x^{3}$) can be written
\begin{align*}
0 &= \frac{2(y^{2} + x^{2})}{x^{3}} \varphi'\left(\frac{x^{2} + y^{2}}{x}\right)
+ \left[\left(\frac{x^{2}-y^{2}}{x^{2}}\right)^{2} + \frac{4y^{2}}{x^{2}}\right] \varphi''\left(\frac{x^{2}+y^{2}}{x}\right) \\
&= \frac{2(y^{2} + x^{2})}{x^{3}} \varphi'\left(\frac{x^{2} + y^{2}}{x}\right)
+ \left(\frac{x^{2} + y^{2}}{x^{2}}\right)^{2} \varphi''\left(\frac{x^{2}+y^{2}}{x}\right).
\tag{*}
\end{align*}
In particular, (*) holds where $y = 0$, and here it reduces to
$$
\frac{2}{x} \varphi'(x) + \varphi''(x) = 0.
$$
The solutions are easily checked to have the form
$$
\varphi(x) = c_{1} + \frac{c_{2}}{x},
$$
for which
$$
u(x, y) = c_{1} + c_{2}\, \frac{x}{x^{2} + y^{2}}
$$
is the real part of $f(z) = c_{1} + \dfrac{c_{2}}{z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $1 + \frac{n}{2} \leq 1+ \frac{1}{2} +\frac{1}{3} +\cdots + \frac{1}{2^n}$ for all natural numbers $n$ Definitions
*
*$H_n = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ for all $n \in \mathbb{N}$
The Question
*
*Prove $1 + \frac{n}{2} \leq H_{2^n}$ for all $n \in \mathbb{N}$
My Work
*
*
Base Case:
$1+\frac{1}{2} \leq 1+\frac{1}{2} = H_1$
*
Inductive Hypothesis: $1 + \frac{k}{2} \leq H_{2^k}$ for all $k \in \mathbb{N}$
*
Induction Step: $1+\frac{k+1}{2} = 1+\frac{k}{2} + \frac{1}{2} \leq H_{2^k}+\frac{1}{2} \leq H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}} = H_{2^{k+1}} $
My Problem
*
*My problem is actually understanding the $H_{2^k}+\frac{1}{2} \leq H_{2^k} + \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}$ step. I think that's how the proof should finish, but I don't know why.
My Question
*
*Can someone explain why the inequality under the "My Problem" header is true? Or if it even is true, am I going about this proof the wrong way?
| you only require:
$$
\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}} = \frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^k+2^k} \\ =\sum_{j=1}^{2^k} \frac1{2^k+j} \\ \ge \sum_{j=1}^{2^k} \frac1{2^{k+1}} \\ = \frac12
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Expansion of this expression Let $x$ be a real number in $\left[0,\frac{1}{2}\right].$ It is well known that
$$\frac{1}{1-x}=\sum_{n=0}^{+\infty} x^n.$$
What is the expansion or the series of the expression $(\frac{1}{1-x})^2$?
Many thanks.
| Method 1 : Binomial Theorem for any index
$$(1-x)^{-n}= 1+nx+ \frac{n(n+1)}{2!}x^2+ \frac{n(n+1)(n+2)}{3!}x^3 \dots $$
Putting $n=1$ , we get :
$$\frac{1}{1-x} = 1+x+x^2+x^3+ \dots $$
Putting $n=2$ , we get :
$$\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$
Method 2 : Derivatives
Let: $$f(x)=\frac{1}{1-x} = 1+x+x^2+x^3+ \dots $$
Take derivative w.r.t. $x$ both the sides;
$$f'(x)=\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$
$$ \implies \frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$
NOTE : These expansions are valid not only for $ x \in \left[0,\dfrac{1}{2}\right]$ but $\forall ~x \in (-1,1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How have I incorrectly computed $\int \frac{x^{2}+4}{x(x-1)^{2}} dx$? My work:
$\displaystyle \int \frac{x^{2}+4}{x(x-1)^{2}}dx$$ =\int \frac{A}{x} dx+\int \frac{B}{x-1}dx +\int \frac{C}{(x-1)^{2}}dx$
$x^{2} + 4 = A(x-1)^{2} + Bx(x-1) + C(x) $
$x^2 + 4 = x^{2}(A+B) + x(-2A+C) + (A-B)$
equating coefficients:
$1 = A + B $
$0 = - 2A + C $
$4 = A - B$
then
$1 - B = A$
so
$4 = 1 - B - B$
$\frac{-3}{2} = B$
and
$4 = A - \frac{-3}{2}$
$\frac{5}{2} = A$
then
$0 = - 2(\frac{5}{2}) + C$
$0 = - 5 + C$
$5 = C$
plugging A, B, and C back into the integral I got:
$= \frac{5}{2}ln|x| - \frac{3}{2}ln|x - 1| - 5ln|x - 1| + C$
wolfram state this is wrong. What have I done wrong here?
Thank you
| $$\int \frac{1}{(x-1)^2} dx \neq \ln|x-1|+C$$
$$\int \frac{1}{(x-1)^2} dx= \int (x-1)^{-2} dx =\frac{(x-1)^{-1}}{-1}+C=-\frac{1}{x-1}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof by induction, dont know how to represent range The question asks for me to prove the following through induction:
$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n} \geq 1 + \frac{n}{2}$
This is my proof thus far:
Proving true for $n = 1$
\begin{align*}
1 + \frac{1}{2} &\geq 1 + \frac{1}{2}\\
\end{align*}
Assuming true for $n = k$
\begin{align*}
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} \geq 1 + \frac{k}{2}
\end{align*}
Proving true for $n = k + 1$
\begin{align*}
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} &\geq 1 + \frac{k}{2} + \frac{1}{2}\\
(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}) + \frac{1}{2^{k+1}} &\geq (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}) + \frac{1}{2}\\
\frac{1}{2^{k+1}} &\geq \frac{1}{2}\\
\end{align*}
I realized at the last statement that something was off, because the last statement contradicts the thing I'm trying to prove. I then realized that it was because the difference between $\frac{1}{2^k}$ and $\frac{1}{2^{k + 1}}$ sets was not simply the addition of $\frac{1}{2^{k + 1}}$, but rather, all numbers between $\frac{1}{2^k}$ and $\frac{1}{2^{k + 1}}$.
For example n = 1 is $1 + \frac{1}{2}$, but $n = 2$ is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$. (note the addition of $\frac{1}{3}$.)
So how can I represent this range?
I believe my proof works except for the fact that $\frac{1}{2^{k + 1}}$ needs to be replaced with something else, I just don't know what that is.
| Start with $$1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} \ge 1+\frac{k}{2}.$$
On the left-hand side, you want to add $\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}$ (all the terms from $2^k+1$ to $2^{k+1}$), as you noted. On the right-hand side you want to add $\frac{1}{2}$, as you have done.
It remains to show
$$\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}} \ge \frac{1}{2}.$$
Can you do this?
Hint: each term on the left-hand side is larger than $\frac{1}{2^{k+1}}$, and there are $2^k$ terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solving $\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }} = 2$ with 100 nested radicals I have seen a book that offers to solve the following equation:
$$\underbrace {\sqrt[3]{{x\sqrt {x\sqrt[3]{{x \ldots \sqrt x }}} }}}_{{\text{100 radicals}}} = 2$$
The book also contains the answer:
$$x = {2^{\left( {\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}} \right)}}$$
How did they get the answer for such equation? I tried to obtain the recurrence relation, but could not find the way to get the above answer.
EDIT
$${u_{100}} = 2,$$
$$\sqrt[3]{{x{u_{99}}}} = 2,$$
$$x{u_{99}} = {2^3},$$
$${u_{99}} = \sqrt {x{u_{98}}} ,$$
$$x\sqrt {x{u_{98}}} = {2^3},$$
$${x^2}x{u_{98}} = {({2^3})^2},$$
$${x^3}{u_{98}} = {2^6},$$
$${u_{98}} = \sqrt[3]{{x{u_{97}}}},$$
$${x^3} \times \sqrt[3]{{x{u_{97}}}} = {2^6},$$
$${x^9}x{u_{97}} = {({2^6})^3},$$
$${x^{10}}{u_{97}} = {2^{18}},$$
$${u_{97}} = \sqrt {x{u_{96}}} ,$$
$${x^{10}}\sqrt {x{u_{96}}} = {2^{18}},$$
$${x^{20}}x{u_{96}} = {({2^{18}})^2},$$
$${x^{21}}{u_{96}} = {2^{36}},$$
$${u_{96}} = \sqrt[3]{{x{u_{95}}}},$$
$${x^{21}} \times \sqrt[3]{{x{u_{95}}}} = {2^{36}},$$
$${x^{63}}x{u_{95}} = {2^{108}}$$
$${x^{64}}{u_{95}} = {2^{108}},$$
$${u_{95}} = \sqrt {x{u_{94}}} ,$$
$${x^{64}}\sqrt {x{u_{94}}} = {2^{108}},$$
$${x^{128}}x{u_{94}} = {2^{216}},$$
$${x^{129}}{u_{94}} = {2^{216}},$$
$$ \ldots $$
but I still have no idea how to find a generalized formula which allows to obtain the answer.
| Here is an unannotated version of one possible solution:
$$\begin{array}{l}
{E_1} = \sqrt[3]{{x\sqrt x }} = {x^{\frac{{{A_1}}}{6}}} = {x^{\frac{3}{6}}}, \\
{E_2} = \sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt x }}} }} = {x^{\frac{{{A_2}}}{{36}}}} = {x^{\frac{{21}}{{{6^2}}}}} = {x^{\frac{{6{A_1} + 3}}{{{6^2}}}}}, \\
{E_3} = \sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt {x\sqrt[3]{{x\sqrt x }}} }}} }} = {x^{\frac{{{A_3}}}{{216}}}} = {x^{\frac{{129}}{{{6^3}}}}} = {x^{\frac{{6{A_2} + 3}}{{{6^3}}}}}, \\
\ldots, \\
{E_{50}} = {x^{\frac{{{A_{50}}}}{{{6^{50}}}}}} = {x^{\frac{{6{A_{49}} + 3}}{{{6^50}}}}}, \\
B = 3, \\
q = 6, \\
d = 3, \\
{A_1} = B, \\
{A_{n + 1}} = q \times {A_n} + d, \\
{A_n} = {u_n} + C, \\
C = q \times C + d = d/(1 - q) = - \frac{3}{5}, \\
{u_1} = B - C, \\
{u_{n + 1}} = q \times {u_n}, \\
{u_n} = (B - C) \times {q^{n - 1}}, \\
{A_n} = (B - C) \times {q^{n - 1}} + C, \\
{A_{50}} = (3 + \frac{3}{5}) \times {6^{49}} - \frac{3}{5} = (6 \times \frac{3}{5}) \times {6^{49}} - \frac{3}{5} = \frac{3}{5} \times {6^{50}} - \frac{3}{5} = \frac{3}{5} \times ({6^{50}} - 1), \\
{x^{\frac{{\frac{3}{5} \times ({6^{50}} - 1)}}{{{6^{50}}}}}} = 2 \Rightarrow {x^{\frac{{3 \times ({6^{50}} - 1)}}{{5 \times {6^{50}}}}}} = 2 \Rightarrow x = {2^{\frac{{5 \times {6^{50}}}}{{3 \times ({6^{50}} - 1)}}}}.
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
If $\gcd(a,b)=9,$ then what is $\gcd(a^2,b^3)\,?$ I know that by the euclidean algorithm, I can obtain the following equations.
I tried some algebraic manipulation but I can't seem to determine, if $$\text{if }\;\gcd(a,b)=9,\text{ then what is }\gcd(a^2,b^3)\;?$$
| $\gcd(a,b) = 9$. So $a = 9k$ and $b=9j$ and $k$ and $j$ have no factors in common.
$\gcd(a^2,b^3) = \gcd(9^2k^2, 9^3j^3)$. $9^2k^2$ and $9j^3$ have $9^2$ in common.
So $\gcd(a^2,b^3) = \gcd(9^2k^2, 9^3j^3) = 9^2*\gcd(k^2, 9j^3)$. $k, j$ have no factors in common with each other but $9$ and $k^2$ may have factors is common. If they do then that prime factor in common is $3$ and $9|k^2$.
So
CASE 1: $a = 9k$ and $b=9j$ and $3|k$. so $k= 3l$ for some $l$ and $a = 27l$. If so then $3 \not \mid j$ (as $j,k$ have no factors in common).
So $\gcd(a^2, b^3) = \gcd(27^2l^2, 9^3j^3)$. These have $9^3$ in common but $l$ and $j$ have no factors in common and no power of $3$ divides $j^3$.
So $\gcd (a^2,b^3) = 9^3$.
CASE 2: $a = 9k$ and $b = 9j$ and $3 \not \mid k$. Then $\gcd(a^2,b^3) = \gcd(9^2k^2, 9^3j^3)$. These have $9^2$ in common. $j$ and $k$ have nothing in common and no power of $3$ divides $k^2$.
So $\gcd(a^2,b^3) = 9^2$.
So the $\gcd(a^2,b^3) = 9^3$ if $27|a$ but $\gcd(a^2,b^3) = 9^2$ if $27 \not \mid a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty$$
$$e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty$$
$$(e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}$$
Now I need help.
| You have almost done it,note that $$\sum _{ n=1 }^{ \infty } \frac { n^{ 2 } }{ (n+1)! } =\sum _{ n=1 }^{ \infty } \frac { n^{ 2 }-1+1 }{ (n+1)! } =\sum _{ n=1 }^{ \infty } \left( \frac { n-1 }{ n! } +\frac { 1 }{ (n+1)! } \right) =\sum _{ n=1 }^{ \infty } \left( \frac { 1 }{ \left( n-1 \right) ! } -\frac { 1 }{ n! } +\frac { 1 }{ (n+1)! } \right) =$$
here $$\sum _{ n=1 }^{ \infty } \left( \frac { 1 }{ \left( n-1 \right) ! } -\frac { 1 }{ n! } \right) =1$$ is telescoping series so
$$\sum _{ n=1 }^{ \infty } \left( \frac { 1 }{ \left( n-1 \right) ! } -\frac { 1 }{ n! } +\frac { 1 }{ (n+1)! } \right) =1+\left( e-2 \right) = \color {blue}{e-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Can anything interesting be said about this fake proof? The Facebook account called BestTheorems has posted the following. Can anything of interest be said about it that a casual reader might miss?
Note that \begin{align}
\small 2 & = \frac 2{3-2} = \cfrac 2 {3-\cfrac2 {3-2}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-2}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\[10pt]
\text{and } \\
\small 1 & = \frac 2 {3-1} = \cfrac 2 {3 - \cfrac 2 {3-1}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3-1}}}} = \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {3 - \cfrac 2 {\ddots}}}}} \\
\text{So } & 2=1.
\end{align}
| Suppose
$$
x = 2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 {\ddots}}}}}}
$$
Then
$$
x = 2 + \cfrac 1 {3 + \cfrac 1 x}.
$$
Solving this by the usual method, one gets
$$
x = \frac {3\pm\sqrt{15}} 3.
$$
At this point I'd be inclined to say we obviously want the positive solution. And that that's what the fraction converges to. However, the answer from Doug M. suggests this point of view: The function
$$
x \mapsto 2 + \cfrac 1 {3 + \cfrac 1 x}
$$
has one attractive fixed point and one repulsive fixed point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 3
} |
multiplying a linear system by an invertible diagonal matrix let $Ax = b$ be a linear n by n system
if we multiply this system by a non-singular diagonal $D$ I can say that the new system still has the same solution as the previous one , right ?
I ran some tests on matlab and noticed that the condition number stays all the time the same but our teacher claims that by this method we can make the condition number smaller of the system and he didn't explain why
can someone please clarify or give me an example where it is true ? thanks !
| Original linear system
$$
\mathbf{A}x = b
$$
with the matrix $\mathbf{A}\in\mathbb{C}^{n\times n}$, and the data vector $b\in\mathbb{C}^{n}$ is not in the null space $\mathcal{N}\left(\mathbf{A}\right)$. The least squares solution defined as
$$
x_{LS} =\left\{
x \in \mathbb{C}^{n} \colon \lVert \mathbf{A}x - b \rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Introduce a set of weighting factors $w_{k}$ in the form of an invertible diagonal matrix $\mathbf{D}\in\mathbb{C}^{n\times n}$.
$$
\mathbf{D} = \left[ \begin{array}{ccccc}
\sqrt{w_{1}} \\
& \sqrt{w_{2}} \\
& & \ddots &\\
& & & \sqrt{w_{n}}
\end{array} \right]
$$
The weighted linear system is expressed as
$$
\mathbf{D}\, \mathbf{A}x = \mathbf{D} \, b
$$
The least squares solution is defined as
$$
\tilde{x}_{LS} =\left\{
\tilde{x} \in \mathbb{C}^{n} \colon \lVert \mathbf{D}\, \mathbf{A}\,x - \mathbf{D} \, b \rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Establish the normal equations:
$$
\left( \mathbf{D} \, \mathbf{A}\right)^{T} \, \mathbf{D} \, \mathbf{A} \, x = \left( \mathbf{D} \, \mathbf{A}\right)^{T} \, \mathbf{D} \, b
\qquad \Rightarrow \qquad
\mathbf{A}^{T} \, \mathbf{W} \, \mathbf{A} \, x = \mathbf{A}^{T} \, \mathbf{W} \,b
$$
Notice
$$
\text{rank } \mathbf{A} = \text{rank } \left( \mathbf{D} \, \mathbf{A} \right)
$$
The weighting does not affect the existence or uniqueness of solutions. The solution is
$$
\tilde{x}_{LS} = \left( \mathbf{A}^{T} \, \mathbf{W} \, \mathbf{A} \right)^{-1} \mathbf{A}^{T} \, \mathbf{W} \, b.
$$
Example
From Data Reduction and Error Analysis for the Physical Sciences, 1e, by Philip Bevington, table 6.1:
$$
\begin{align}
\mathbf{A} x &= b \\
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
1 & 4 \\
1 & 5 \\
1 & 6 \\
1 & 7 \\
1 & 8 \\
1 & 9 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
a_{0} \\
a_{1}
\end{array}
\right]
%
&=
%
\frac{1}{10}
\left[
\begin{array}{cc}
156 \\
175 \\
366 \\
438 \\
582 \\
616 \\
642 \\
704 \\
988
\end{array}
\right]
%
\end{align}
$$
The least squares solution is
$$
\left[
\begin{array}{cc}
a_{0} \\
a_{1}
\end{array}
\right]_{LS}
=
\frac{1}{360}
\left[
\begin{array}{cc}
1733 \\
3387
\end{array}
\right]
\approx
\left[
\begin{array}{cc}
4.81389 \\
9.40833
\end{array}
\right]
$$
Introduce the weighting factors
$$
\mathbf{W} =
\left[
\begin{array}{ccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \sqrt{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \sqrt{3} & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & \sqrt{5} & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & \sqrt{7} & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \sqrt{2} & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 \\
\end{array}
\right]
$$
and the least squares solution becomes
$$
\left[
\begin{array}{cc}
\tilde{a}_{0} \\
\tilde{a}_{1}
\end{array}
\right]_{LS}
\approx
\left[
\begin{array}{c}
4.27364 \\
9.49364
\end{array}
\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Probability that the number $(a^3 + b^3)$ is divisible by $3$
Two numbers $a$ and $b$ are chosen from the set $\{1, 2, 3, \ldots 3n\}$. Find the probability that $(a^3 + b^3)$ is divisible by $3$.
I am not able to check whether the cube of a number is divisible by $3$.
| Hint:
$a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 - 3ab)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit of a 2-variable function Given problem:
$$
\lim_{x\to0,y\to0}(1+x^2y^2)^{-1/(x^2+y^2)}
.$$
I tried to do it with assigning y to $y = kx$, but that didn't help me at all. Also one point, I can't use L'Hospital's rule.
| I assume here that you mean $\lim_{(x,y) \to (0,0)}$ then
$$
\begin{align}
\lim_{(x,y) \to (0,0)} (1+x^2y^2)^{-\frac{1}{x^2+y^2}} =
\lim_{(x,y) \to (0,0)} \exp\left( \ln \left( (1+x^2y^2)^{-\frac{1}{x^2 + y^2}} \right) \right) &= \\ \\\lim_{(x,y) \to (0,0)} \exp\left( -\frac{\ln (1+x^2y^2)}{x^2 + y^2} \right) &= \\ \exp\left( -\lim_{(x,y) \to (0,0)} \frac{\ln (1+x^2y^2)}{x^2 + y^2} \right) \tag{1}
\end{align}
$$
Assuming the limit exists and by continuity of the exponential.
Now if we look at the function which is clearly non-negative and using the fact that $\ln(1+x) \leq x$ we have
$$
0 \leq \frac{\ln (1+x^2y^2)}{x^2 + y^2} \leq \frac{x^2y^2}{x^2+y^2} \\
\lim_{(x,y) \to (0,0)} 0 \leq \lim_{(x,y) \to (0,0)}\frac{\ln (1+x^2y^2)}{x^2 + y^2} \leq \lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^2+y^2} = \\
\lim_{r\to 0}\frac{r^4 \cos(t)^2\sin(t)^2}{r^2} = 0
$$
Hence
$$
\lim_{(x,y) \to (0,0)}\frac{\ln (1+x^2y^2)}{x^2 + y^2} = 0 \tag{2}
$$
$(1)$ and $(2)$ allow you to conclude.
| {
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"url": "https://math.stackexchange.com/questions/2223838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Determine the limit of a function based on a given function I received this question on my past midterm and the professor did not offer a solution and I am perplexed as to what the answer is.
Suppose f is a function such that
$${\lim_{x\rightarrow \infty }} [f(x) + 2x^{2}] = 1$$
Determine, if possible,
$$\lim_{x\rightarrow \infty} {\frac{f(x)}{x^{2}+1}}$$
| Write: $\frac{f(x)}{x^2+1} = \frac{f(x)+2x^2}{x^2+1} - \frac{2x^2}{x^2+1}$.
We know that $\lim_{x \to \infty} \frac 1{x^2+1} = 0$. Hence, by product of limits formula:
$$
\lim_{x \to \infty} \frac{f(x)+2x^2}{x^2+1} = \lim_{x \to \infty} [f(x)+2x^2] \lim_{x \to \infty} \frac 1{x^2+1} = 1 \times 0 = 0
$$
Note that $\frac{2x^2}{x^2+1} = 2 - \frac{2}{x^2+1}$, so the limit of this quantity as $x \to \infty$ is $2$.
Hence, the limit of $\frac{f(x)}{x^2+1}$ exists, and equals $0 - 2 = -2$.
EXAMPLE : $f(x) = -2x^2 + 1$, then $f$ satisfies the conditions given, and $\frac{f(x)}{x^2+1}= \frac{-2x^2 - 2 + 1}{x^2+1} = -2 +\frac{1}{x^2+1}$, whence the limit is $-2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ (A^2+B^2) <<1$ I need to evaluate the definite integral $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A<<1, B<<1, (A^2+B^2) <<1,$
For unresricted (but real) A&B, Wolfram Alpha provides the following indefinite general solution:-
$$\int \frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \left( \frac{1}{A^2+B^2}\right) \left[ \frac{2B}{K} \tanh^{-1} \left( \frac{A-(B-1)\tan\left(\frac{\theta}{2}\right)}{K}\right) + F(\theta)\right]$$
where $F(\theta) = A \ln(1 + A\sin\theta+B\cos\theta)+B\theta$,
and $K = \sqrt{A^2 + B^2 -1}$, therefore K is complex for the range of A,B I am interested in.
In a previous question seeking a solution for the similar, but slightly simpler, definite integral (with numerator $1$ rather than $\cos\theta$) user Dr. MV found a solution given by:
$$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{2\pi}{\sqrt{1-A^2-B^2}}\text{ } (\text { for} \sqrt{A^2+B^2}<1) $$.
My question: Can similar solutions be found for these two definite integrals $$\int_0^{2\pi}\frac{\cos\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \tag 1$$
and
$$\int_0^{2\pi}\frac{\sin\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \tag 2$$?
EDIT
I have taken the solution proposed by user Chappers.
By simultaneous equations in A,B,I,J it turns out that
$$I=\int_0^{2\pi}\frac{\cos\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{B}{A^2+B^2} 2\pi (1-\frac{1}{\sqrt{1-A^2-B^2}})$$ and
$$J=\int_0^{2\pi}\frac{\sin\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{A}{A^2+B^2} 2\pi (1-\frac{1}{\sqrt{1-A^2-B^2}})$$.
These were confirmed in a numerical model.
| There's a trick to this: let the cosine one be $I$, the sine one $J$. We add to get something easy to integrate:
$$ BI+AJ = \int_0^{2\pi} \frac{A\sin{\theta}+B\cos{\theta}}{1+A\sin{\theta}+B\cos{\theta}} \, d\theta = \int_0^{2\pi} \left( 1 - \frac{1}{1+A\sin{\theta}+B\cos{\theta}} \right) d\theta, $$
which you can do using the previous answer to get
$$ BI+AJ = 2\pi - \frac{2\pi}{\sqrt{1-A^2-B^2}}. $$
The harder bit is to come up with a second equation. Remembering our derivatives, we try
$$ AI-BJ = \int_0^{2\pi} \frac{A\cos{\theta}-B\sin{\theta}}{1+A\sin{\theta}+B\cos{\theta}} \, d\theta, $$
which has antiderivative $ \log{(1+A\sin{\theta}+b\cos{\theta})} $, continuous if $A^2+B^2<1$. But this has the same value at both of the endpoints, so the integral is zero, and $AI=BJ$. Now you can solve this simultaneously with the first equation to find the values of $I$ and $J$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating the volume of rotation between $y=4x-16$, $x^2-y^2=9$ around $x$ axis Having the following equations:
$y=4x-16$,
$x^2-y^2=9$
I calculated the mutual points - $(53/15,-28/15),(5,4)$
How can I find the volume of rotation of the area in between them around the $x$ axis? Should it be split into two different volumes?
I've tried calculating the volume above axis x and the volume below it separately.
Above axis x I'm getting $2\pi$ :
$$V_1 = \pi\cdot \int^5_4(4x-16)^2dx - \pi \int^4_3(\sqrt{x^2-9})^2dx = 2\pi$$
and below I got $1.446\pi$:
$$V_2 =\pi\cdot \int ^4_{53/15}(4x-16)^2dx + \pi\int^{53/15}_{3}(\sqrt{x^2-9}^2dx = 1.446\pi$$
So In total I get a final result of $V= V_1+V_2 = 3.446\pi$. Is there anything I did wrong in my calculation?
| You can perform the calculation via the method of cylindrical shells or with washers.
First, a sketch of the region in question:
This shows that you have an issue if this region is to be rotated about the $x$-axis, since the relevant cross-section is the portion enclosed by the two curves above the $x$-axis (else you will double-count the volume from the portion of the region below the $x$-axis when you revolve it). Hence, by the method of cylindrical shells, the radius of the shell ranges from $y = 0$ to $y = 4$, and the height of the shell is given by $$h(y) = (y+16)/4 - \sqrt{y^2+9}.$$ Thus a representative shell has differential volume $$dV = 2 \pi y h(y) \, dy = 2 \pi y \left(\frac{y}{4} + 4 - \sqrt{y^2+9}\right) \, dy,$$ and the total volume is $$V = 2\pi \int_{y=0}^4 \frac{y^2}{4} + 4y - y\sqrt{y^2+9} \, dy = 2\pi \left[\frac{y^3}{12} + 2y^2 - \frac{(y^2+9)^{3/2}}{3}\right]_{y=0}^4 = \frac{28}{3}\pi.$$
If we perform the calculation using washers, we have from $x = 3$ to $x = 4$ the volume $$V_1 = \int_{x=3}^4 \pi( x^2-9) \, dx = \pi \left[\frac{x^3}{3} - 9x\right]_{x=3}^4 = \frac{10}{3}\pi,$$ and from $x = 4$ to $x = 5$, $$V_2 = \int_{x=4}^5 \pi \left(x^2 - 9 - (4x-16)^2\right) \, dx = \pi \left[-5x^3 + 64x^2 - 265x\right]_{x=4}^5 = 6 \pi.$$
From this, the total volume is again $$\frac{28}{3}\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding value of limit involving cosine I know we can use Maclaurin expansion and l'hopital's rule for solving it but I want another way . Find value of $\lim_{x \to 0} \frac{\cos^2x - \sqrt{\cos x}}{x^2}$.
My try : I multiplied and then divided by conjugate of numerator but it didn't help .
| $$\frac{\cos^2x-\sqrt{\cos x}}{x^2}=\frac{\cos^4x-\cos x}{x^2(\cos^2x+\sqrt{\cos x})}=\frac{\cos x}{\cos^2x+\sqrt{\cos x}}\frac{\cos^3x-1}{x^2}=\frac{\cos x}{\cos^2x+\sqrt{\cos x}}\frac{(\cos x-1)(\cos^2 x+\cos x+1)}{x^2}=\frac{\cos x(\cos^2 x+\cos x+1)}{\cos^2x+\sqrt{\cos x}}\frac{(\cos x-1)}{x^2}$$
Now, use $\cos 2x=1-2\sin^2 x \implies 2\sin^2 x=1-\cos 2x$ or $2\sin^2 (x/2)=1-\cos(x).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving with AM-GM Inequality $$\frac{4}{abcd}\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}$$
Given: $a+b+c+d=4$ and $a$, $b$, $c$ abd $d$ are positives.
How to prove the above inequality using Arithmetic Geometric Mean Inequality?
I tried the following but, I am getting stuck after last step.
| Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.
Hence, by Rearrengement and AM-GM we obtain:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\frac{1}{abcd}(a^2cd+b^2da+c^2ab+d^2bc)=$$
$$=\frac{1}{abcd}(a\cdot acd+b\cdot bda+c\cdot cab+d\cdot dbc)\leq$$
$$\leq\frac{1}{abcd}(x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt)=$$
$$=\frac{1}{abcd}(x^2yz+y^2xt+z^2xt+t^2yz)=\frac{1}{abcd}(xy+zt)(xz+yt)\leq$$
$$\leq\frac{1}{abcd}\left(\frac{xy+zt+xz+yt}{2}\right)^2=$$
$$=\frac{1}{4abcd}\left((x+t)(y+z)\right)^2\leq\frac{1}{4abcd}\left(\frac{x+y+z+t}{2}\right)^4=\frac{4}{abcd}.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$ for all $n \in N$ using mathematical induction I'm having a horrible time completing this since this is the first time I'm learning mathematical induction.
So far I have this:
Let $P(n)$ be:
$P(n):\sum_{i=1}^{n} {\frac{1}{4i^2-1}} = \frac{n}{2n +1}$
For the base case, $n=1$
$LHS = \sum_{i=1}^{0} {\frac{1}{4i^2-1}} = {\frac{1}{4(1^2)-1}} = {\frac{1}{4-1}} = {\frac{1}{3}}$
$RHS = \frac{n}{2n +1} = \frac{1}{2(1) +1} = \frac{1}{2 +1} = \frac{1}{3}$
Therefore $P(1)$ holds.
Assuming $P(k)$, $P(k)$ holds for $n=k$:
$P(k): \sum_{i=1}^{k} {\frac{1}{4i^2-1}} = \frac{k}{2k +1}$
Proving for $P(k+1)$:
$P(k+1):\\ \\\ \\ \sum_{i=1}^{k+1} {\frac{1}{4i^2-1}}$
$= \sum_{i=1}^{k} {\frac{1}{4i^2-1}} + \frac{1}{4(k+1)^2-1}$
$=\frac{k}{2k +1} + \frac{1}{4(k+1)^2-1}$
Can anyone help me from this point onward?
| Note that the difference between the $(n+1)$th and $n$th putative partial sums is $$\begin{align*}\frac{n+1}{2(n+1)+ 1} - \frac{n}{2n+1} &= \frac{(n+1)(2n+1) - n (2n+3)}{(2n+3)(2n+1)} \\
&= \frac{1}{4n^2 + 8n + 3} \\
&= \frac{1}{4 (n+1)^2 - 1}\end{align*}$$
and this is the $(n+1)$th term in the series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding solutions to a polynomial by using De Moivre's theorem Use de Moivre’s theorem to show that
$$\ \cos 6θ = 32\cos^6θ − 48\cos^4θ + 18\cos^2θ − 1 $$.
Hence solve the equation
$$\ 64x^6 − 96x^4 + 36x^2 − 1 = 0$$
giving each root in the form$\ \cos kπ$.
Attempt
I have completed almost the whole question and obtained an equation
$$\ \cos\theta = \cos\frac{k\pi}{9} = -\frac{1}{2}$$
The problem is the equation requires six roots. Whereas the above equation is satisfied by $$\ k= \pm1,\pm2,\pm4,\pm5,\pm7,\pm8$$
My question is
*
*Which six roots I am supposed to choose? (I am aware that both plus-minus values will give the same value for x, but it is also possible for the same value of x to be the root of the equation twice. Like $\ x=1$ is a root of the equation $\ x^2-2x+1$ twice.)
*Why is such a complication arising?
| Assuming you have proved
$$
\cos 6\theta = 32\cos^6\theta − 48\cos^4\theta + 18\cos^2\theta − 1
$$
you get
$$
32\cos^6\theta − 48\cos^4\theta + 18\cos^2\theta=1+\cos6\theta
$$
Now set $x=\cos\theta$, so your equation becomes
$$
2(1+\cos6\theta)-1=0
$$
that is
$$
\cos6\theta=-\frac{1}{2}
$$
which has the solutions
$$
6\theta=\frac{2\pi}{3}+2k\pi
\qquad\text{or}\qquad
6\theta=-\frac{2\pi}{3}+2k\pi
$$
You can choose $0\le k\le 5$, because every other solution will repeat values for $x$. On the other hand, you can discard the second set of solutions, for the same reason, because $\cos(-\alpha)=\cos\alpha$.
So you get
$$
\frac{\pi}{9},\quad
\frac{\pi}{9}+\frac{\pi}{3},\quad
\frac{\pi}{9}+\frac{2\pi}{3},\quad
\frac{\pi}{9}+\frac{3\pi}{3},\quad
\frac{\pi}{9}+\frac{4\pi}{3},\quad
\frac{\pi}{9}+\frac{5\pi}{3}
$$
and the solutions for $x$ are
$$
\cos\frac{\pi}{9},\quad
\cos\frac{4\pi}{9},\quad
\cos\frac{7\pi}{9},\quad
\cos\frac{10\pi}{9},\quad
\cos\frac{13\pi}{9},\quad
\cos\frac{16\pi}{9}
$$
which are pairwise distinct.
To summarize: the equation $\cos6\theta=-1/2$ has twelve solutions in the interval $[0,2\pi)$, but your equation in $x$ has only six, because the solutions in $\theta$ can be grouped in pairs that yield the same value for $\cos\theta$.
| {
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"url": "https://math.stackexchange.com/questions/2234530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$
My attempt:
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies that $\frac{dx}{d\theta}=\frac{\cos \theta}{2\sqrt{\sin \theta}}.$
So we have $\int \frac{1}{1-x^4}dx=\int\frac{1}{\cos^2\theta} \cdot \frac{\cos \theta}{2\sqrt{\sin \theta}} d\theta = \int\frac{1}{2\cos\theta \sqrt{\sin\theta}}d\theta.$ Then I stuck here.
Any hint would be appreciated.
| Hint: $$\frac{1}{1-x^4}=\frac{1}{2}\left(\frac{1-x^2+1+x^2}{1-x^4}\right)=\frac{1}{2(1+x^2)}+\frac{1}{2(1-x^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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inequation of sinuses of triangle's angles Let $\alpha, \beta, \gamma$ be inner angels of a triangle. Prove that the follwoing inequation is fulfilled:
$\sin\alpha+\sin\beta+\sin\gamma-\sin\alpha\sin\beta\sin\gamma \ge \sin^3\alpha+\sin^3\beta+\sin^3\gamma$
| we will prove that
$$\sin(\alpha)+\sin(\beta)+\sin(\gamma)-\sin(\alpha)\sin(\beta)\sin(\gamma)\le \sin(\alpha)^3+\sin(\beta)^3+\sin(\gamma)^3$$
we use that
$$\sin(\alpha)=\frac{a}{2R}$$ etc and $$A=\frac{abc}{4R}$$ and $$A=\sqrt{s(s-a)(s-b)(s-c)}$$
plugging this in the given inequality we have to prove that
$$a^3+b^3+c^3+abc-\frac{(abc)^2(a+b+c)}{4s(s-a)(s-b)(s-c)}\geq 0$$ with $$s=\frac{a+b+c}{2}$$ this is equivalent to
$${\frac {{a}^{6}-{a}^{5}b-{a}^{5}c-{a}^{4}{b}^{2}+3\,{a}^{4}bc-{a}^{4}{
c}^{2}+2\,{a}^{3}{b}^{3}-2\,{a}^{3}{b}^{2}c-2\,{a}^{3}b{c}^{2}+2\,{a}^
{3}{c}^{3}-{a}^{2}{b}^{4}-2\,{a}^{2}{b}^{3}c+6\,{a}^{2}{b}^{2}{c}^{2}-
2\,{a}^{2}b{c}^{3}-{a}^{2}{c}^{4}-a{b}^{5}+3\,a{b}^{4}c-2\,a{b}^{3}{c}
^{2}-2\,a{b}^{2}{c}^{3}+3\,ab{c}^{4}-a{c}^{5}+{b}^{6}-{b}^{5}c-{b}^{4}
{c}^{2}+2\,{b}^{3}{c}^{3}-{b}^{2}{c}^{4}-b{c}^{5}+{c}^{6}}{ \left( a+b
-c \right) \left( a-b+c \right) \left( a-b-c \right) }}
\le 0$$
now using the Ravi-substitution:
$$a=y+z,b=x+z,c=x+y$$
and we get after simplification:
$${\frac { \left( x-z \right) ^{2} \left( x+y \right) ^{2}}{-2\,z+x-y}}\le0$$
which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple method to determine if a graph exists only in the positive quadrant.
How can one tell by inspection or by some simple tests, preferably without using calculus, that a graph exists only in the positive quadrant only (or for that matter in any one quadrant only)?
An example is the graph
$$(bx-ay)^2=2ab\left(bx+ay-\frac {ab}2\right)$$
from the question/answer here.
It is not immediately obvious. Setting RHS$>0$ does not help either.
| (Revised solution)
The equation of the curve (assuming $a, b>0$) is
$$\begin{align}
(bx-ay)^2&=2ab\left(bx+ay-\frac {ab}2\right)\\
b^2x^2-2abxy+a^2y^2&=2ab(bx+ay)-a^2b^2\end{align}$$
$$\small\begin{array}
{l|l}
\hline
\text{Form quadratic in $x$:}
&\text{Form quadratic in $y$:}\\
\;\;b^2x^2-2ab(b+y)x+a^2(y^2+b^2)-2a^2by=0
&\;\;a^2y^2-2ab(ax+x)y+b^2(x^2+a^2)-2ab^2x=0\\
\;\;b^2x^2-2ab(b+y)x+a^2(y-b)^2=0
&\;\;a^2y^2-2b(a+x)y+b^2(x-a)^2=0\\
\text{For real $x$},
&\text{For real $y$}, \\
\;\;[-2ab(b+y)]^2\geq 4\cdot b^2\cdot a^2(y-b)^2
&\;\;[-2ab(a+x)]^2\geq 4\cdot a^2\cdot b^2(x-a)^2\\
\;\;(y+b)^2-(y-b)^2\geq 0
&\;\;(x+a)^2-(x-a)^2\geq 0\\
\;\;4by\geq 0
&\;\;4ax\geq 0\\
\;\;y\geq 0
&\;\;x\geq 0\\
\hline
\end{array}
$$
i.e., for real $x,y$, the curve exists for
$$\begin{cases}
x\geq 0\\
y\geq 0\end{cases}
\qquad
\text{i.e. positive quadrant only}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Integration of $\frac{e^{-x^2/4}}{(2+x^2)^2}$ I was doing a question where I needed to do this integration
$$\int\frac{e^{-x^2/4}}{(2+x^2)^2}dx$$
This answer on Wolfram alpha agrees with the answer I expect to get. However I can't see how to do this integration (and I do not have Wolfram Pro, so cannot see any step-by-step solution).
How can I do this integral?
| Write $$f(x) = \frac{e^{-x^2/4}}{(2+x^2)^2} = \frac{2x}{(2+x^2)^2} \cdot \frac{e^{-x^2/4}}{2x}$$ and integrate by parts with the choice $$u = \frac{e^{-x^2/4}}{2x}, \quad du = -e^{-x^2/4}\frac{(2+x^2)}{4x^2} \, dx, \\ dv = \frac{2x}{(2+x^2)^2} \, dx, \quad v = -\frac{1}{2+x^2}.$$ Hence $$\int f(x) \, dx = -\frac{e^{-x^2/4}}{2x(2+x^2)} - \int \frac{e^{-x^2/4}}{4x^2} \, dx.$$ We again perform another integration by parts with the choice $$u = e^{-x^2/4}, \quad du = -\frac{x}{2} e^{-x^2/4} \, dx, \\ dv = \frac{1}{4x^2} \, dx, \quad v = -\frac{1}{4x},$$ to obtain $$\int f(x) \, dx = -\frac{e^{-x^2/4}}{2x(2+x^2)} + \frac{e^{-x^2/4}}{4x} + \frac{1}{8} \int e^{-x^2/4} \, dx.$$ This last integral does not have an elementary closed-form antiderivative but it is related to the familiar "error function", thus $$f(x) = \frac{x e^{-x^2/4}}{4(2+x^2)} + \frac{\sqrt{\pi}}{8}\operatorname{erf}(x/2) + C.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Which value of $a$ maximizes $\int_{a-1}^{a+1}\frac{1}{1+x^{8}}dx$? I am not being able to understand the graphical method of solving this, any simple explanation will be appreciated.
A non-graphical calculation will be very helpful too.
Thank you so much in advance!
| We can apply the fundamental theorem of calculus. Let $f(x)=\frac{1}{1+x^8}$. Then if $F(t)=\int_0^t f(x)dx$, we have $F'(t)=f(t)$. We also have $\int_a^b f(x)dx=F(b)-F(a)$. Putting this together, we can write $$\int_{a-1}^{a+1}f(x)dx=F(a+1)-F(a-1).$$ Taking the derivative with respect to $a$, we get $$F'(a+1)-F'(a-1)=f(a+1)-f(a-1)=\frac{1}{1+(a+1)^8}-\frac{1}{1+(a-1)^8}.$$ We find the critical points by setting this equal to zero, so we end up with $1+(a+1)^8=1+(a-1)^8$, or $(a+1)^8-(a-1)^8=0$. We can apply the binomial theorem to rewrite this as $$\sum_{k=0}^8\binom{8}{k}(1+(-1)^k)a^k=16a^7+112a^5+112a^3+16a=0.$$ We simplify slightly to obtain $a(a^6+7a^4+7a^2+1)=0$. We can check that two more roots are $\pm i$. Factoring leaves us with $$a(a^2+1)(a^4+6a^2+1)=a(a^2+1)\left(a^2-\dfrac{-6+\sqrt{32}}{2}\right)\left(a^2-\dfrac{-6-\sqrt{32}}{2}\right).$$ We can check that all of the factors except $a$ have only imaginary roots, so the only real root is $a=0$. Now, we should check that this critical point is a maximum. We compute the second derivative of the original expression: $$\frac{d}{da}\left(\frac{1}{1+(a+1)^8}-\frac{1}{1+(a-1)^8}\right)=-\frac{8(a+1)^7}{(1+(a+1)^8)^2}+\frac{8(a-1)^7}{(1+(a-1)^8)^2}.$$ Plugging in $a=0$, we obtain $-4$, so $0$ is indeed a local maximum (and since it is the only local maximum on an unbounded domain, it is indeed the global maximum).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Intuitive meaning of $E(X^2)$ and $E(X+a)$? I understand conceptually that $E(X)$ is the average expected value of random variable $X$ over multiple trials over a long period of time i.e. the mean. Similarly, I understand for conditionals that $E(X|Y)$ is the average value of $X$ for all cases where $Y$ has already happened.
However, what is the intuitive meaning behind expressions like $E(X^2)$, and other expectations besides $E(X)$? For instance, the meaning of $E(X+a)$, where $a$ is a constant? I can't seem to grasp the concept of those expressions.
| $X^2$ or $X + a$ are both random variables, pretty much the same as $X$. In the first case, you are to compute the mean value of $X^2$ and in the second one, you should compute the mean of $X + a$.
A little example:
Let $X$ be a random variable that takes value $-2$ with probability $\dfrac{1}{2}$, $1$ with probability $\dfrac{1}{3}$ and $3$ with probability $\dfrac{1}{6}$. Hence, $$E(X) = -2\cdot \dfrac{1}{2} + 1 \cdot \dfrac{1}{3} + 3 \cdot \dfrac{1}{6} = -\dfrac{1}{6}$$
Then $X^2$ is $(-2)^2$ with probability $\dfrac{1}{2}$, $1^2$ with probability $\dfrac{1}{3}$ and $3^2$ with probability $\dfrac{1}{6}$.
$$E(X^2) = 4\cdot \dfrac{1}{2} + 1 \cdot \dfrac{1}{3} + 9 \cdot \dfrac{1}{6} = \dfrac{23}{6}$$
As regards to, say, $X+2$, we can merely write down
$$E(x+2) = E(x) + 2 =\dfrac{11}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Is the equation $x^3-y^3\equiv k \mod n$ always solveable, when neither $7$ nor $9$ divides $n$? Here
A Diophantine equation involving factorial
solutions of $x^3-y^3=z!$ are wanted. To rule out some $z$, I wanted to find modules $n$, such that $$x^3-y^3\equiv k\mod n$$ is not solveable for some $k$.
For $n=7$, we have no solution , if $k$ is $3$ or $4$. And if $n=9$, we have no solution if $k$ is $3,4,5$ or $6$.
But the verification of the numbers $n=2$ to $500$ showed that always a solution exists if neither $7$ nor $9$ divides $n$.
Is the following conjecture true ?
If $n>1$ and neither $7$ nor $9$ divides $n$, then the equation $$x^3-y^3\equiv k\mod n$$ is solveable for all $k$ ?
If the above conjecture is actually true does this imply that $x^3-y^3\equiv z!\mod n$ is always solveable for $z\ge 7$ and $n>1$ ?
| For $n=3$ the equation is clearly solvable for all $k$. To show that there is a solution for every $k$ whenever $7$ and $9$ do not divide $n$, by the Chinese remainder theorem it suffices to show that solutions exists for all prime powers $p^m$ with $p\neq3,7$. By Hensel's lemma it suffces to show that solutions exist for all primes $p\neq3,7$.
If $p\equiv2\pmod{3}$ then $3$ is coprime to $\varphi(p^m)=p^{m-1}(p-1)$, the order of the group $(\Bbb{Z}/p^m\Bbb{Z})^{\times}$. Hence every element of $\Bbb{Z}/p^m\Bbb{Z}$ that is not a multiple of $p$ is a cube, so the equation has solutions for every $k$.
If $p\equiv1\pmod{3}$ then the set of cubes $C$ is a subgroup of index $3$ in $(\Bbb{Z}/p\Bbb{Z})^{\times}$.
Fact: If $p>7$ then there exists $c\in C$ with $c\neq\pm1$ and $c-1\notin C$.
The cosets of $C$ in $(\Bbb{Z}/p\Bbb{Z})^{\times}$ are then $C$, $(c-1)C$ and $(c-1)^2C$. For every $k\in C$ there is clearly a solution, and the fact that $c-1\in C+C$ shows that $(c-1)C\subset C+C$ so for every $k\in(c-1)C$ there is also a solution. It remains to show that $(c-1)^2C\subset C+C$.
Suppose toward a contradiction that $(c-1)^2C\not\subset C+C$. Then $(c-1)^2C\cap(C+C)=\varnothing$ and so
$$1+c\notin(c-1)^2C\qquad\text{ and }\qquad(c+1)(c-1)=c^2-1\notin(c-1)^2C.$$
By cancelling $c-1$ we get $c+1\notin(c-1)C$, so $c+1\in C$.
Similarly $2=1+1\notin(c-1)^2C$, and $2\notin(c-1)C$ because otherwise
$$c+1=(c-1)+2\in(c-1)C+(c-1)C,$$
contradicting $C\cap((c-1)C+(c-1)C)=\varnothing$, hence also $2\in C$ and so $-2c\in C$.
Now $c^2+1\in C+C$ and hence $c^2+1\notin(c-1)^2C$ by assumption. If $c^2+1\in(c-1)C$ then
$$c^4-1=(c^2+1)(c+1)(c-1)\in(c-1)^2C,$$
but $c^4-1\in C+C$, a contradiction. Hence $c^2+1\in C$, but then
$$(c-1)^2=(c^2+1)+(-2c)\in C+C,$$
again contradicting $(c-1)^2C\cap(C+C)=\varnothing$. This shows that also $(c-1)^2C\subset C+C$ and therefore $C+C=\Bbb{Z}/p\Bbb{Z}$. So if $p\equiv1\pmod{3}$ and $p\neq7$ then there is also a solution for every $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
| Since you have $\sin A + 1 = 4 \cos A$, you can square both sides to yield
$$
\sin^2 A + 2 \sin A + 1 = 16 \cos^2 A = 16 (1 - \sin^2 A).
$$
Rearranging, we obtain
$$
17 \sin^2 A + 2 \sin A - 15 = 0
$$
which is a quadratic equation in $\sin A$ and can be solved to obtain
$$
\sin A = \frac{ -2 \pm 32}{34} = \left\{ \frac{15}{17}, -1 \right\}.
$$
and then we must have
$$
\cos A = \pm \sqrt{ 1 - \sin^2 A} = \left\{ \pm \frac{8}{17}, 0 \right\}.
$$
We now have three candidate solutions for this equation. However, since we squared the equation and at one point multiplied by $\cos A$, we may have introduced spurious solutions; so we should check to ensure that these results actually satisfy the original equation. It turns out that only one of these solutions actually satisfies the original equation; which one is left as an exercise to the reader.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
} |
Finding the foci and vertices of an ellipse. How would you find the foci and vertices of the following ellipse:
$$\frac{2x^2}{15} +\frac{8y^2}{45} -\frac{2\sqrt3}{45}xy=1?$$
| Consider a rotation:
$$
x=X\cos\alpha-Y\sin\alpha,
\qquad
y=X\sin\alpha+Y\cos\alpha
$$
Then your ellipse becomes
$$
6(X\cos\alpha-Y\sin\alpha)^2+8(X\sin\alpha+Y\cos\alpha)^2
-2\sqrt{3}(X\cos\alpha-Y\sin\alpha)(X\sin\alpha+Y\cos\alpha)=45
$$
The term in $XY$ has coefficient
$$
4\cos\alpha\sin\alpha
-2\sqrt{3}\cos^2\alpha+2\sqrt{3}\sin^2\alpha
$$
which you want to be vanishing; dividing by $\cos^2\alpha$ we get
$$
2\sqrt{3}\tan^2\alpha+4\tan\alpha-2\sqrt{3}=0
$$
so
$$
\tan\alpha=\frac{-2+4}{2\sqrt{3}}=\frac{1}{\sqrt{3}}
\quad\text{or}\quad
\tan\alpha=\frac{-2-4}{2\sqrt{3}}=-\sqrt{3}
$$
So we can take $\alpha=\pi/6$ and the equation of the ellipse becomes
$$
X^2(6\cos^2\alpha+8\sin^2\alpha-2\sqrt{3}\cos\alpha\sin\alpha)
+Y^2(6\sin^2\alpha+8\cos^2\alpha+2\sqrt{3}\cos\alpha\sin\alpha)=45
$$
that is
$$
5X^2+9Y^2=45
$$
or
$$
\frac{X^2}{9}+\frac{Y^2}{5}=1
$$
Find foci and vertices, then use the inverse rotation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating $\int_{0}^{\frac{\pi}{6}} \cos{(x)} \sqrt{2\sin (x)+1} dx$ I am trying to calculate the value of the following
$$\int_{0}^{\frac{\pi}{6}} cosx \sqrt{2sin x+1} dx$$
I used a substitution method.
$$u = 2 \sin (x) + 1$$
$$\frac{du}{dx} = 2\cos (x)$$
$$\frac{u}{2 \cos (x)}du = dx$$
hence
$$\int_{0}^{\frac{\pi}{6}} \cos (x) \sqrt{2 \sin (x)+1} dx$$ =
$$\int_{1}^{2} \cos (x) \sqrt{u} \times\frac{u}{2 \cos (x)}du$$
= $$\int_{1}^{2} \frac{1}{2}u^\frac{3}{2}du$$
= $$\left[\frac{1}{5}u^\frac{5}{2}\right]_1^2$$
but I can't seem to move any further.
Can anyone help please?
Many thanks.
UPDATE
The third line is incorrect. It should be
$$\frac{du}{2 \cos (x)} = dx$$
hence
= $$\int_{1}^{2} \frac{1}{2}u^\frac{1}{2}du$$
= $$\left[\frac{1}{3}u^\frac{3}{2}\right]_1^2$$
= $$\frac{1}{3}\times (2\sqrt3 - 1)$$
= $$\frac{2\sqrt3-1}{3}$$
| $$\int \cos(x) \sqrt{2\sin(x)+1} dx = \frac{1}{2}\int \sqrt{2\sin(x)+1}d(2\sin(x)+1) = \frac{1}{2}(\frac{2}{3}(2\sin(x)+1)\sqrt{2\sin(x)+1}+C)=F(x)$$
So the result is $F(\frac{\pi}{6}) - F(0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding side-length proof in double-angle triangle. In triangle $ABC$, $|AC| = b$ and $|AB| = c$. Angle $A$ is twice angle $B$. Prove that $$|BC| = \sqrt{b\cdot \left(b+c\right)}$$
I understand how to apply laws such as the cosine and sine law to this triangle, but I have no idea how to end up at the desired expression.
Here's a diagram of the triangle:
| Applying the cosine rule in two ways ... we have
\begin{eqnarray*}
a^2=b^2+c^2-2bc \cos(B) \\
c^2=a^2+b^2-2ab \cos(2B)
\end{eqnarray*}
Now reacll the double angle formula $\cos(2B)=2 \cos^2(B)-1$ & after a little algebra
\begin{eqnarray*}
(a^2-bc-c^2)(a+b-c)(a-b+c)(b+c)b=0
\end{eqnarray*}
The latter four factors can be rejected as either the triangle inequality or quantities that are clearly positive ... so we have $a^2-bc-c^2=0$ & the required result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{\cos x}=2\cos x$$
$$\sin x + 1=2\cos^2 x$$
$$\sin x +1=2-2\sin^2 x$$
$$2\sin^2 x +\sin x - 1=0$$
$$2\sin^2 x +2\sin x -\sin x - 1=0$$
$$2\sin x (\sin x +1) -1(\sin x+1)=0$$
$$(\sin x +1)(2\sin x-1)=0$$
So, what's the next?
| Obviously $x$ has possible values of $\frac{3 \pi}{2}$ (by $\sin x=-1$) and $\frac{ \pi}{6}$ , $\frac{5 \pi}{6}$ by $\sin x=\frac{1}{2}$
Now put These roots in intial conditions to verify if they satisfy it or not.
Note: $(x-a)(x-b)=0$ means $x=a$ or $x=b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
| Ok, you want an intuitive proof, not a better proof. That's fine.
First note that for $x=y=z$, we have $x^2+y^2+z^2 = 3\left(\frac13\right)^2= \frac13$. Next let us show that $x^2+y+2+z^2$ is not minimal if $x\ne y$ or $y\ne z$. Indeed if $a\ne b$ then
$$
a^2 + b^2 - \left(\left(\frac{a+b}2\right)^2+\left(\frac{a+b}2\right)^2\right) = \frac12\left(
a^2 + b^2 -2ab
\right)=\frac12(a-b)^2>0,
$$
so we can replace two of the numbers be their averages and lower the sum of squares.
Unfortunately we not yet quite done. We are done if we can show that there is a global minimum of $x^2+y^2+z^2$, given that $x+y+z=1$. Because then that global minimum must also be a local minimum, but for that we cannot have $x\ne y$ or $y\ne z$, so $x=y=z$ must be the global minimum, and its value is $\frac13$.
For that we can first argue that we can restrict ourselves to $x,y,z\ge0$ (e.g. if $x<0$ replace $(x,y,z)$ by $\frac1{-x+y+z}(-x,y,z)$, which yields a smaller value since $-x+y+z = 1 - 2x > 1$) and then appeal to the compactness of $\{(x,y,z)\colon \text{$x+y+z=1$, $x,y,z\ge0$}\}$.
I did not say that this would be pretty, but it is pretty intuitive to me :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct answer is $2 (-1 + 2^{1 + n})$
How do I make this transition?
| Your expression for $a_n$ is correct. So $$a_n=2+\cdots+2^n+2^{n+1}.$$ Now note that this is a sum of $n+1$ terms in geometric progression with first term $a=2$ and the common ratio $r=2$. The sum $S$ can be found by $$S=\frac{a(r^{n+1}-1)}{r-1}$$ So substituting $a=2$ and $r=2$ we get $$a_n=\frac{2(2^{n+1}-1)}{2-1}=2(2^{n+1}-1)$$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$?
I feel this method does not work here.
| Suppose $x\equiv 0\bmod 5$
Then $2x^2+3x+4\equiv4\not\equiv0\bmod 5$
Suppose $x\equiv 1\bmod 5$
Then $2x^2+3x+4\equiv 2+3+4\equiv4\not\equiv0\bmod 5$
Suppose $x\equiv 2\bmod 5$
Then $2x^2+3x+4\equiv 8+6+4\equiv 3\not\equiv0\bmod 5$
Suppose $x\equiv 3\bmod 5$
Then $2x^2+3x+4\equiv3+4+4\equiv1\not\equiv0\bmod 5$
Suppose $x\equiv 4\bmod 5$
Then $2x^2+3x+4\equiv 2+2+4\equiv 3\not\equiv0\bmod 5$
Then by exhaustion of the possible cases modulo $5$, we can conclude that $5\not |2x^2+3x+4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Construct a differentiable and periodic function I am looking for a differentiable and periodic function $f$ such that the following conditions are satisfied:
$1)$ $f(x_i) = y_i,~ $ for given $x_i,y_i$, $ \forall i \in \{0,1,...,n-1\}$;
$2)$ $f(x_{i+kn}) = f(x_i), ~ \forall k \in \mathbb{Z}$;
$3)$ $ x_{j} - x_{j-1} = \Delta x~ (constant),~\forall j \in \mathbb{Z}.$
Is there some method to construct a such function?
I appreciate some hints/references to solve this problem. Thanks in advance!
| Suppose that the number of points $n$ is odd. For example, consider $x_0 = -2$, $x_1 = -1$, $x_2 = 0$, $x_3 = 1$ and $x_4 = 2$, and $y_0 = 4$, $y_1 = 2, y_2 = 1, y_3 = -1$ and $y_4 = -2$ ($n=5$).
In this case $\Delta x = 1$, while the period is $T = n\Delta x = 5.$
You can try to find your function using a truncated Fourier series of order $m = \frac{n-1}{2}:$
$$f(x) = \frac{a_0}{2} + \sum_{k=1}^{m}\left(a_k \cos\left(\frac{2\pi xk}{T}\right) + b_k \sin\left(\frac{2\pi xk}{T}\right)\right). $$
In your case, $m=2$, and you have to find the parameters $a_0$, $a_1$,$a_2$, $b_1$ and $b_2$. They are $5$, which is the same number of available points $n$. This is very important.
Now, you can substitute the available points to the function $f$:
$$\begin{cases}
f(x_0) = y_0 \\
f(x_1) = y_1 \\
f(x_2) = y_2 \\
f(x_3) = y_3 \\
f(x_4) = y_4
\end{cases} \Rightarrow\\
\begin{cases}
\frac{a_0}{2} + \sum_{k=1}^{2}\left(a_k \cos\left(\frac{-4\pi k}{5}\right) + b_k \sin\left(-\frac{4\pi k}{5}\right)\right) = 4\\
\frac{a_0}{2} + \sum_{k=1}^{2}\left(a_k \cos\left(\frac{-2\pi k}{5}\right) + b_k \sin\left(-\frac{2\pi k}{5}\right)\right) = 2\\
\frac{a_0}{2} + \sum_{k=1}^{2}\left(a_k \cos\left(0\right) + b_k \sin\left(0\right)\right) = 1\\
\frac{a_0}{2} + \sum_{k=1}^{2}\left(a_k \cos\left(\frac{2\pi k}{5}\right) + b_k \sin\left(\frac{2\pi k}{5}\right)\right) = -1\\
\frac{a_0}{2} + \sum_{k=1}^{2}\left(a_k \cos\left(\frac{4\pi k}{5}\right) + b_k \sin\left(\frac{4\pi k}{5}\right)\right) = -2\\
\end{cases}.$$
This system can be rewritten in matricial form:
$$\begin{bmatrix}
\frac{1}{2} & \cos\left(-\frac{4\pi}{5}\right) & \cos\left(-\frac{8\pi}{5}\right) & \sin\left(-\frac{4\pi}{5}\right) & \sin\left(-\frac{8\pi}{5}\right)\\
\frac{1}{2} & \cos\left(-\frac{2\pi}{5}\right) & \cos\left(-\frac{4\pi}{5}\right) & \sin\left(-\frac{2\pi}{5}\right) & \sin\left(-\frac{4\pi}{5}\right)\\
\frac{1}{2} & \cos\left(0\right) & \cos\left(0\right) & \sin\left(0\right) & \sin\left(0\right)\\
\frac{1}{2} & \cos\left(\frac{2\pi}{5}\right) & \cos\left(\frac{4\pi}{5}\right) & \sin\left(\frac{2\pi}{5}\right) & \sin\left(\frac{4\pi}{5}\right)\\
\frac{1}{2} & \cos\left(\frac{4\pi}{5}\right) & \cos\left(\frac{8\pi}{5}\right) & \sin\left(\frac{4\pi}{5}\right) & \sin\left(\frac{8\pi}{5}\right)\\
\end{bmatrix} \cdot \begin{bmatrix}
a_0 \\
a_1 \\
a_2\\
b_1\\
b_2
\end{bmatrix} = \begin{bmatrix}
4 \\
2 \\
1\\
-1\\
-2
\end{bmatrix}.$$
If the matrix is invertible, then you are done!
In this case, I get the following:
$$ a_0 \simeq 1.6, a_1 \simeq -0.1236,
a_2 \simeq 0.32,
b_1 \simeq -2.5520,
b_2 \simeq 1.5772.$$
Here is how this function looks like:
Addition The number of points must be odd, since the number of parameters of the truncated Fourier series is odd too. In order to have a well posed system of equation, you need to have as many unknowns (the parameters) as the available data, as well as the same number of equations.
For more details on Fourier series, look here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using the bound $\cosh(x) \le e^{x^2/2}$
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2} \le \lim_{x \to 0} \frac{\left(x^2-xc\right)^2}{2x^2}=\frac{c^2}{2}
\end{align}
My question: How finish this argument.
| PRIMER:
In This Answer , I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential and logarithm functions satisfies the inequalities
$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$
for $x<1$, and
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 2$$
for $x>0$,respectively.
It can also be shown that
$$|\sinh(x)|\ge |x| \tag 3$$
for all $x$.
METHODOLOGY $1$: DEVELOPING A USEFUL LOWER BOUND
The OP has expressed a preference to finding a useful lower bound for the function of interest. To that end, we proceed.
If we wish to find a lower bound then we can apply $(2)$ and $(3)$ and proceed to write
$$\begin{align}
\log(\cosh(x^2-cx)) &=\log\left(1-(1-\cosh(x^2-cx))\right)\\\\
&=\log\left(1+2\sinh^2\left(\frac{x^2-cx}{2}\right)\right)\\\\
&\ge \frac{2\sinh^2\left(\frac{x^2-cx}{2}\right)}{1+2\sinh^2\left(\frac{x^2-cx}{2}\right)}\\\\
&\ge \frac{2\left(\frac{x^2-cx}{2}\right)^2}{1+2\sinh^2\left(\frac{x^2-cx}{2}\right)}\tag 4
\end{align}$$
Then, dividing $(4)$ by $x^2$ we find that
$$\frac{\log(\cosh(x^2-cx))}{x^2}\ge \frac12 \frac{(x-c)^2}{1+2\sinh^2\left(\frac{x^2-cx}{2}\right)} \tag 5$$
The limit of the right-hand side of $(5)$ approaches $\frac{c^2}2$ as $x\to 0$
METHODOLOGY $2$: APPLYING TAYLOR'S THEOREM
The OP has stated that application of Taylor's is an acceptable way forward. To that end, we proceed.
Note from Taylor's Theorem,
$$\cosh(s)=1+\frac12s^2+O(s^4)\tag6$$
and
$$\log(1+t)=t-\frac12t^2+O(t^4) \tag7$$
Using $(6)$ and $(7)$ yields
$$\begin{align}
\log(\cosh(x^2-cx))&=\log\left(1+\frac12(x^2-cx)^2+O((x^2-cx)^4)\right)\\\\
&=\frac12(x^2-cx)^2+O((x^2-cx)^4)\tag 8
\end{align}$$
Dividing $(8)$ by $x^2$ as taking the limit as $x\to 0$, we obtain the coveted limit
$$\lim_{x\to 0}\frac{\log(\cosh(x^2-cx))}{x^2}=\lim_{x\to 0}\frac{\frac12(x^2-cx)^2+O((x^2-cx)^4)}{x^2}=\frac{c^2}{2}$$
as was to be shown!
| {
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"source": "stackexchange",
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Minimum value of an expression What would be the minimum value of
$$
(1+a)(1+b)(1+c)(1+d)
\left({\frac{1}{a}}+{\frac{1}{b}}+{\frac{1}{c}}+{\frac{1}{d}}\right)
$$
Given that $a,b,c,d$ are all positive real non zero number
THE ANSWER IS ${\frac{4^5}{27}}$
| By Symmetry an optimal value will be attained when $a=b=c=d$ so it suffice to consider
\begin{eqnarray*}
S=(1+a)^4\frac{4}{a}.
\end{eqnarray*}
Differentiation gives
\begin{eqnarray*}
\frac{dS}{da}=\frac{-4(1+a)^4}{a^2}+\frac{16(1+a)^3}{a}=\frac{4(1+a)^3(3a-1)}{a^2}
\end{eqnarray*}
Setting $\frac{dS}{da}=0$ gives $a=\frac{1}{3}$ and the minimum value is $\color{red}{\frac{4^5}{27}}$. (as stated in the question).
EDIT ...
In order to show that an optimal value is attained when $a=b=c=d$. Calculate the partial derivatives
\begin{eqnarray*}
S&=&(1+a)(1+b)(1+c)(1+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\\
\frac{\partial S}{\partial a}&=&(1+b)(1+c)(1+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)+(1+a)(1+b)(1+c)(1+d)\frac{-1}{a^2}
\end{eqnarray*}
So we have
\begin{eqnarray*}
\frac{1+a}{a^2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1+b}{b^2}=\frac{1+c}{c^2}=\frac{1+d}{d^2}
\end{eqnarray*}
Now note that the function $f(x)=\frac{1+x}{x^2}$ is monotonic ... $f(x)=f(y)$ implies $x=y$ so $\color{blue}{a=b=c=d}$.
| {
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"source": "stackexchange",
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If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \sqrt{ab}$.
Because $\gcd(a,b) = 1$, there exists integers $k$ and $l$ such that $ak + bl = 1$. This means that $\sqrt{a}(k\sqrt{}) + \sqrt{b}(l\sqrt{b}) = 1$.
Hence $\sqrt{a}$ and $\sqrt{b}$ are both positive integers and we can set $\sqrt{a} = c$ for some arbitrary integer $c$ and $\sqrt{b} = d$ for some arbitrary integer $d$. Therefore, $a = c^2$ and $b = d^2$.
| Let $a = \Pi_{i=1}^{r} p_i^{d_i}$ and $b=\Pi_{j=r+1}^{r+s} p_j^{d_j}$ be the unique prime factorizations of $a$ and $b$, respectively. Since $gcd(a,b)=1$, none of the $r$ primes which are divisors of $a$ is equal to any of the $s$ primes which are divisors of $b$. Hence, the product $ab$ can be written as $ab=\Pi_{i=1}^{r+s} p_i^{d_i}$, where the $p_i$'s are distinct.
Since $ab$ is a perfect square, each $d_i$ is even. Suppose $d_i = 2e_i$ for $i=1,\ldots, r+s$. Then, $a=x^2$ and $b=y^2$, where $x = \Pi_{i=1}^r p_i^{e_i}$ and $y=\Pi_{j=r+1}^{r+s} p_j^{e_j}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix.
| Note that solving $XA=0$ is equivalent to solving $A^\top Y=0$. Since
$$
\DeclareMathOperator{rref}{rref}\rref A^\top =
\left[\begin{array}{rrrr}
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
it follows that all solutions to $A^\top \vec y=\vec 0$ are of the form
$$
\vec y=\lambda
\left[\begin{array}{r}
1 \\
1 \\
-1 \\
0
\end{array}\right]
$$
Now, let $\lambda_1,\dotsc,\lambda_4$ be your four favorite numbers. Then
$$
Y=
\begin{bmatrix}
\lambda_1\vec b & \lambda_2\vec b & \lambda_3\vec b & \lambda_4\vec b
\end{bmatrix}
$$
is a $4\times 4$ matrix that solves $A^\top Y=0$ where
$$
\vec b=
\left[\begin{array}{r}
1 \\
1 \\
-1 \\
0
\end{array}\right]
$$
Taking $X=Y^\top$ then solves your original problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite series of $\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$ Find Value of $$S=\lim_{n \to \infty}\sum_{r=n^2}^{(n+2)^2} \frac{1}{\sqrt{r}}$$
$$S=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+4n+4}}$$
But here i cannot express above sum in form of $$\lim_{n \to \infty} \frac{1}{n} \times \sum_{r=1}^{n}f(\frac{r}{n})$$
| On the interval $r\in[n^2,(n+2)^2]$ the terms $\frac{1}{\sqrt{r}}$ have a quite small variation: they are approximately all equal to $\frac{1}{n}$; since there are $4n+5$ terms, the limit is expected to be $\color{red}{4}$.
In a more rigorous way,
$$ \frac{4n+5}{n+2}\leq \sum_{r=n^2}^{(n+2)^2}\frac{1}{\sqrt{r}} \leq \frac{4n+5}{n} $$
and the previous claim follows by squeezing.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to get the inverse of this function, when we have qudratic? $$f(x)=-x^2+6x-5$$
How do I find the inverse? I tried by making it
$$y=-x^2+6x-5$$
Then swapping $y$ with $x$, and then solve it for y, but I got $y^2$.
The domain is $x$ greater or equal to $m$, and in this case $m=5$.
After that we need to find the domain of the inverse.
| Note that
$$
\begin{align}
y
&=-x^2+6x-5\\
&=-(x^2-6x+9-9)-5\\
&=-(x^2-6x+9-9)-5\\
&=-((x-3)^2-9)-5\\
&=-(x-3)^2+4
\end{align}
$$
by completing the square so that
$$
4-y=(x-3)^2\iff \sqrt{4-y}=\lvert x-3\rvert
$$
But $x\geq 5$ so what can you conclude? Solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Three quadratic equations have positive roots The three quadratic equations $ax^2-2bx+c=0$, $bx^2-2cx+a=0$ and $cx^2-2ax+b=0$ has both roots positive. Then which of the following is/are true
A) $a^2=bc$
B) $b^2=ac$
C) $c^2=ab$
D) $a=b=c$
I assumed $a>0$ so parabola is open upwards and hence it should definitely cut positive $Y $axis as it has both roots positive and hence $c \gt 0$.
Since $c \gt 0$ the third parabola is open upwards and so $b \gt 0$ and finally with same reasoning $a \gt 0$. Now each Discriminant is Non negative so
$$b^2 \ge ac$$
$$c^2 \ge ab$$
$$a^2 \ge bc$$
adding all
$$a^2+b^2+c^2-ab-bc-ca \ge 0$$ $\implies$
$$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0$$
if we take equality above i will get $a=b=c$ so fourth option is valid. But how to check whether other options a is/are true
| I would like to warn you that the condition you've got, i.e.
$$
(a-b)^2+(b-c)^2+(c-a)^2 \ge 0
$$
is a necessary condition and not a sufficient one. It means that the testing of D against it just ensures that $a,b,c$ may satisfy D, but does not ensure that they do. Moreover, it is pretty useless condition as a sum of squares is always non-negative for real numbers, so it has, in fact, no information about $a,b,c$ at all.
We can start by saying that for the three equations to have two root each we need to have $a,b,c$ nonzero. Then we continue by writing down all three solutions
\begin{align}
&\frac{b}{a}\pm\frac{\sqrt{b^2-ac}}{a}\ge 0,\\
&\frac{c}{b}\pm\frac{\sqrt{c^2-ab}}{b}\ge 0,\\
&\frac{a}{c}\pm\frac{\sqrt{a^2-bc}}{c}\ge 0.
\end{align}
Clearly $\frac{b}{a}>0$, $\frac{c}{b}>0$ and $\frac{a}{c}>0$, that is $a,b,c$ have the same sign. Since the roots does not change after multiplying the equations by an arbitrary number, we can assume WLOG that $a,b,c>0$. Hence, we know that
\begin{align}
b^2-ac\ge 0\quad\Leftrightarrow\quad b^3\ge abc,\\
c^2-ab\ge 0\quad\Leftrightarrow\quad c^3\ge abc,\\
a^2-bc\ge 0\quad\Leftrightarrow\quad a^3\ge abc.\\
\end{align}
Adding together and dividing by $3$ we get
$$
\frac{a^3+b^3+c^3}{3}\ge abc=\sqrt[3]{a^3b^3c^3}.
$$
The AM-GM inequality gives then that it is, in fact, equality, therefore $a^3=b^3=c^3$, which implies $a=b=c$, so D is true.
It makes A,B,C to be trivially true as well.
| {
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Parametrization of two surfaces $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ and $\frac{x^2}{p}+\frac{y^2}{q}=2z$. Can someone please help me to parametrize the following surfaces in terms of hyperbolic(for second it might not be possible but i need some more convenient set of parametric equation than mine ) and trigonometric functions
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 $$ and $$\frac{x^2}{p}+\frac{y^2}{q}=2z$$
I have tried to do but the set of parametric equations I got were too complicated as I have to use those in some further calculation which makes the result very ugly.
For first equation the set of parametric equations is: $$x=a\sqrt{1+\frac{u^2}{c^2}}\cos v, \ \ y=b\sqrt{1+\frac{u^2}{c^2}}\sin v \ \ z=u$$
and for second: $$x=\sqrt{2pu} \cos v ,\ \ y=\sqrt{2qu} \sin v, \ \ z=u $$
|
$$
\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1
$$
Try
\begin{align}
x &=& a\cosh u \\
y &=& b\sinh u \cos v \\
z &=& c\sinh u \sin v
\end{align}
Then
\begin{eqnarray}
\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} &=& \cosh^2 u -(\sinh^2u\cos^2v + \sinh^2u\sin^2v) \\
&=& \cosh^2 u - \sinh^2 u \\
&=& 1
\end{eqnarray}
$$
\frac{x^2}{p} + \frac{\color{red}{y}^2}{q} = 2z
$$
Try
\begin{align}
x &=& \sqrt{2up}\cos v \\
y &=& \sqrt{2uq}\sin v \\
z &=& u
\end{align}
In this case
\begin{eqnarray}
\frac{x^2}{p} + \frac{y^2}{q} &=& \frac{2 u p}{p}\cos^2v + \frac{2 u q}{q}\sin^2 u = 2u \\
&=& 2z
\end{eqnarray}
| {
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"question_score": "2",
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The square of any odd number is $1$ more than a multiple of $8$ I'm taking a single variable calculus course and asked following :
Say wether the following is a valid proof or not :
the square of any odd number is 1 more than a multiple of $8$.
Proof : By the division theorem any number can be expressed in one of the forms $4q , 4q+1 , 4q+2 , 4q+3$
Squaring each of these gives :
$$(4q+1)^2=16q^2+8q+1=8(2q^2+q)+1$$
$$(4q+3)^2=16q^2+24q+9=8(2q^2+3q+1)+1$$
My answer :
I think this proof is invalid as it does not prove ' the square of any odd number is $1$ more than a multiple of $8$. ' is true for any odd number as it does not prove for the odd number $(2n+1)$ or $(2n-1)$. Is my assertion correct ?
| The proof is valid and your comment isn't: the proof correctly states that an odd number can be represented as either $4q+1$ or $4q+3$ and argues, in both cases, that its square minus $1$ is divisible by $8$.
A different approach might be with congruences: an odd number is congruent modulo $8$ to $1$, $3$, $5$ or $7$; now
\begin{align}
1^2=1&\equiv 1\pmod{8}\\
3^2=9&\equiv 1\pmod{8}\\
5^2=25&\equiv 1\pmod{8}\\
7^2=49&\equiv 1\pmod{8}
\end{align}
Another approach. Let $n=2q+1$ be odd; then
$$
n^2-1=(n-1)(n+1)=2q(2q+2)=4q(q+1)
$$
and $q(q+1)$ is even, so $4q(q+1)$ is divisible by $8$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here).
So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations.
Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated
| Hint : Use :
$$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$$
| {
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"source": "stackexchange",
"question_score": "2",
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If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?
| $$
\cos^8 x + 2\cos^6 x + \cos^4 x = (\cos^2 x(1+\cos^2 x))^2
=(\sin x(1+\sin x))^2 = 1
$$
| {
"language": "en",
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Show that $a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4$, Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4.$ Show that $$a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4.$$
| Let $f(x,y,z) = x+y+z + \sqrt{\dfrac{x^{2} + y^{2}+z^{2} }{3}}$ and $g(x,y,z) = x+y+z + xyz$. By Lagrange's Method of Undetermined Multipliers, we will maximise $f$ subject to the constraint that $g=4$. (The set $D=\{(x,y,z): g(x,y,z)=4\}$ is closed and bounded and since $g$ is continuous, by the Extreme Value Theorem it attains its bounds on $D$.)
By Lagrange's Method,
$\nabla f= \lambda \nabla g$
$\Rightarrow f_{x} = \lambda g_{x}$
$\Rightarrow f_{y} = \lambda g_{y}$
$\Rightarrow f_{z} = \lambda g_{z}$
$\Rightarrow 1 + \dfrac{2x}{\sqrt{3(x^{2} + y^{2}+z^{2})}} = \lambda(2x + yz)$ $\cdots (1)$
$\Rightarrow 1 + \dfrac{2y}{\sqrt{3(x^{2} + y^{2}+z^{2})}} = \lambda(2y + xz)$
$\Rightarrow 1 + \dfrac{2z}{\sqrt{3(x^{2} + y^{2}+z^{2})}} = \lambda(2z + yx)$
From the above $3$ equations, subtracting one the other yields the system below:
$(x-y)\left[\dfrac{2}{\sqrt{3(x^{2} + y^{2}+z^{2})}}-\lambda(2-z)\right] =0$
$(y-z)\left[\dfrac{2}{\sqrt{3(x^{2} + y^{2}+z^{2})}}-\lambda(2-x)\right] =0$
$(z-x)\left[\dfrac{2}{\sqrt{3(x^{2} + y^{2}+z^{2})}}-\lambda(2-y)\right] =0$
$\Rightarrow x=y=z$ or
${\lambda}(z-2)={\lambda}(x-2)=\lambda(y-2)$
Clearly $\lambda \neq 0$ since if that were true, plugging back in $(1)$ would result in
$\Rightarrow 1 + \dfrac{2x}{\sqrt{3(x^{2} + y^{2}+z^{2})}} =0$, which has no solutions when $x,y,z \in \mathbb{R}^{+}$.
Thus in both cases we see that $x=y=z=a$, which after plugging into $g(x,y,z)=4$ gives:
$a^{3}+3a^{2}-4 =0$
$\Rightarrow (a-1)(a+2)^{2}=0$
$\Rightarrow a=1$ since $a \in \mathbb{R}^{+}$
$\Rightarrow f$ has an extremum at $x=y=z=1$ at which $f(x,y,z)=4$. Clearly this is a point of maximisation since if it were a point of minimisation it would be contradicted by the fact that $f\left(\dfrac{1}{2}, \dfrac{1}{2},\dfrac{1}{2}\right) = \dfrac{7}{8} <4$. This proves the desired inequality.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2271863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Number of real solutions $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $ The number of real solutions of $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $ is
(1) 0
(2) 1
(3) 2
(4) infinitely many .
My work :
$$
1\geq \cos\left(\frac{x^2+x}{2}\right)=\frac{2^x+2^{-x} }{2}\geq 1
\qquad \text{by (AM-GM).}
$$
So $\frac{x^2+x}{2}=2n\pi$ for all $n\in \mathbb{Z}$ . Now discriminant $=1+2n\pi$ is always positive for $n\geq0$ . But the equation is a quadratic so it has only two solution . Hence the answer must be 2 .
PS: I'm aware that the this problem is already on the site but i posted as there were no complete solution .
| Answer is 1. You wrote that it should be definitely $2^x + 2^{-x} = 2$. Let $y = 2^x$, so we are intereseted in $y^2 - 2y + 1 = (y - 1)^2 = 0$. The only solution comes when $y = 1$ or $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
\frac{x^2-y^2}{2} = 0 \\
xy = 1\\ \end{cases} \\
=\begin{cases}
x^2-y^2 = 0 \\
x = \frac{1}{y}\\ \end{cases} \\
=\begin{cases}
\frac{1}{y}-y^2 = 0 \\
x = \frac{1}{y}\\ \end{cases} \\=
\begin{cases}
\frac{1-y^3}{y} = 0 \\
-\\ \end{cases}
\\=
\begin{cases}
y^3 = 1 \\
-\\ \end{cases}
\\=
\begin{cases}
y = 1 \\
x =1\\ \end{cases} $$
And so $\sqrt{2i} = 1+i$, but my book states the solution is $\sqrt{2i} = 1+i$ and $\sqrt{2i} = -1-i$.
What did I forget?
| In complex numbers, square roots (and cube roots, and fourth roots, etc) are no longer uniquely defined; a square root of a complex number is any value that, when squared, yields the number.
So if one value of $\sqrt{z}$ is $w$, then another value is $-w$, since $w^2 = (-w)^2 = z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Completing the square of $x^2 - mx = 1$ is not giving me the right answer. This is my attempt
$$
\begin{align}
x^2 - mx &= 1 \\
x^2 - mx - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4} - \frac{m^2}{4}\right) - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - \frac{4}{4} &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2 - 4}{4} &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) &= \frac{m^2 - 4}{4} \\
\left(x - \frac{m}{2}\right)^2 &= \frac{m^2 - 4}{4} \\
\sqrt{\left(x - \frac{m}{2}\right)^2} &= \sqrt{\frac{m^2 - 4}{4}} \\
x - \frac{m}{2} &= \pm \frac{\sqrt{m^2 - 4}}{\sqrt{4}} \\
x &= \frac{m}{2} \pm \frac{\sqrt{m^2 - 4}}{2} \\[20pt]
x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 - 4}}{2} \\
x_1 &= \frac{m - \sqrt{m^2 - 4}}{2} \\[16pt]
x_2 &= \frac{m}{2} + \frac{\sqrt{m^2 - 4}}{2} \\
x_2 &= \frac{m + \sqrt{m^2 - 4}}{2} \\
\end{align}
$$
However, the correct answer according to the text is:
$$
\begin{align}
x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 + 4}}{2} \\
x_2 &= \frac{m + \sqrt{m^2 + 4}}{2} \\
\end{align}
$$
Why $\sqrt{m^2 + 4}$ instead of $\sqrt{m^2 - 4}$ ???
| it is $$x^2-2\frac{m}{2}x+\frac{m^2}{4}-\frac{m^2}{4}-1=0$$ and from here we get
$$\left(x-\frac{m}{2}\right)^2=\frac{m^2}{4}+1$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof involving altitude of a right triangle and circle Let $AD$ be the altitude corresponding to the hypotenuse $BC$ of the right triangle $ABC$. The circle of diameter $AD$ intersects $AB$ and $M$ and $AC$ at $N$. Prove that $\displaystyle\frac{BM}{CN} = \left(\frac{AB}{AC}\right)^3$.
This is what I have so far:
Power of a point(B) = $BM = \displaystyle\frac{BD^2}{AB}$ and Power of a point(C) = $CN = \displaystyle\frac{CD^2}{AC}$.
Using the Altitude Theorem we in $\triangle ABC$ with altitude $AD$ we obtain: $AD^2 = BD \cdot CD$ and therefore $BD^2 = \displaystyle\frac{AD^4}{CD^2}$ and $CD^2 = \displaystyle\frac{AD^4}{BD^2}$.
Plugging this into the equations for $BM$ and $CN$ we get:
$BM = \displaystyle\frac{AD^4}{CD^2} \cdot \frac{1}{AB}$ and $CN = \displaystyle\frac{AD^4}{BD^2} \cdot \frac{1}{AC}$.
I am not sure how to obtain $\displaystyle\frac{AB^3}{AC^3}$. If someone could provide me with a hint as to where to go from here, or if what I have done so far is not the right way to approach the proof please guide me in the right direction.
| $$BM=\frac{AD^4}{CD^2}\cdot\frac1{AB}\quad CN=\frac{AD^4}{BD^2}\cdot\frac1{AC}$$
$$\frac{BM}{CN}=\frac{BD^2}{CD^2}\cdot\frac{AC}{AB}$$
Now since
$$\frac{AB}{AC}=\frac{BD}{AD}=\frac{AD}{CD}$$
we have
$$\frac{AB^2}{AC^2}=\frac{BD}{CD}$$
$$\frac{AB^4}{AC^4}=\frac{BD^2}{CD^2}$$
$$\frac{BM}{CN}=\frac{AB^4}{AC^4}\cdot\frac{AC}{AB}=\left(\frac{AB}{AC}\right)^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Modulus problem (Complex Number) If complex number $z(z\neq2)$ satisfies the equation : $z^2=4z+\lvert z\rvert^2+\frac{16}{\lvert z\rvert^3}$, then what is the value of $\lvert z\rvert^4$?
My try: I tried to take $z=x+iy$ and solved for the value of $\lvert z\rvert^4$ but everytime ended up getting a value greater than $9$.
Hint- The answer lies between $0$ and $9$ , both included.
| This question CAN be solved by assuming $z=x+iy$, where $x$ and $y$ are reals.
The given equation can be written as $$z^2-4z= |z|^2+ \frac{16}{|z|^3}$$
Plugging in $z=x+iy$ and
$$x^2-y^2 +2ixy -4x-4iy=x^2+y^2 + \frac{16}{(\sqrt{x^2+y^2})^3}$$
taking the complex terms to one side, we obtain$$2iy(x-2)=2y^2+4x+\frac{16}{(\sqrt{x^2+y^2})^3}$$
But if $x$ and $y$ are real numbers, then the LHS cannot be equal to the RHS, as one is a real number and the other is a complex number . This means the LHS must be zero.
$\implies y=0$ or $x=2$.
First lets check for $y=0$. The equation reduces to
$$4x+\frac{16}{|x|^3}=0 \implies x=-\sqrt2$$
Hence
$z=-\sqrt2$
$|z|^4 = 4$
NOTE: $x$ cannot be equal to $2$, because then, the expression can have no real value for $y$.(you can check this yourself)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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If $ x^4+x^3+x^2+x+1=0$ then what is the value of $x^5$ If $$x^4+x^3+x^2+x+1=0$$ then what's the value of $x^5$ ??
I thought it would be $-1$ but it does not satisfy the equation
| $x^4+x^3+x^2+x+1 = 0$
Multiply everything by $x$
$x^5+x^4+x^3+x^2+x = 0$
Add $1$ to both sides:
$x^5+x^4+x^3+x^2+x+1 = 1$
$x^5+(x^4+x^3+x^2+x+1) = 1$
$x^5+(0) = 1$
$x^5=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove and how to find this value?(without derivative)
idea 1
let $\tan{\dfrac{x}{2}}=t$,then we have
$$f=\dfrac{(t+1)^2}{\sqrt{t^4+2t^3+6t^2+2t+5}}$$
Therefore,it suffices to prove that
$$\dfrac{(t+1)^4}{t^4+2t^3+6t^2+2t+5}\le\dfrac{32}{25}$$
or$$32(t^4+2t^3+6t^2+2t+5)-25(t+1)^4\ge 0$$
or
$$ (t-3)^2(7t^2+6t+15)\ge 0$$
But this method if we without derivative,we don't known the maximum is $\dfrac{4\sqrt{2}}{5}$.
idea 2
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{(\sin{x}+1)+2(\cos{x}+1)}}$$
Let $u=\sin{x}+1,v=\cos{x}+1$,then $(u-1)^2+(v-1)^2=1$,find the maximum of the
$$\dfrac{u}{\sqrt{u+2v}}$$
| Change variables so that $\sin(x)$ is $x$ and $\cos(x)$ is $\sqrt{1-x^2}$. The range of $x$ is $[0,1]$, of course.
Square the function to get rid of the radical, so that now your function is:
$(1+x)^2/(3+x+2\sqrt{1-x^2})$.
Try to prove by direct manipulation that $(1+x)^2/(3+x+2\sqrt{1-x^2})]\leq1$ in $[0,1]$ , i.e. get all but the radical on one side, then get rid of the radical and arrive at a polynomial inequality. You will be able to factor it and show that it is strictly negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sin x = \sin 2x$ The equation is: $\sin x = \sin 2x$
I recognize that $\sin 2x$ is a double angle and use the double angle identity, so it becomes:
$$\begin{array}{rrcl}
&\sin x &=& 2 \sin x \cos x \\
\implies& \sin x - 2 \sin x \cos x &=& 0
\end{array}$$
Then I am stuck... Not sure how to proceed.
| $$\sin x = \sin 2x \implies \sin x = 2\sin x \cos x \implies \sin x - 2\sin x \cos x = 0 \implies \sin x (1-2\cos x) = 0$$
That means $\sin x = 0$ or $\cos x = \frac{1}{2}$
$\sin x = 0$ is true when $x = k\pi$
$\cos x = \frac{1}{2}$ is true when $x = \frac{\pm (1+ 2k\pi)}{3}$ or $x = \frac{\pm (5+ 2k\pi)}{3}$
So this statement holds when $x = k\pi$ or $x = \frac{\pm (1+ 2k\pi)}{3}$ or $x = \frac{\pm (5+ 2k\pi)}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
express $a$ in terms of $b$ and $c$
Given that $$c=\frac{\sqrt{a+3b}}{a-3b}$$ express $a$ in terms of $b$ and $c$
My attempt,
\begin{align}c^2(a^2-6ab+9b^2)&=a+3b\\
c^2a^2+(-6bc^2-1)a+9b^2c^2-3b&=0\\
a&=\frac{-(-6bc^2-1)\pm \sqrt{(-6bc^2-1)^2-4c^2(9b^2c^2-3b)}}{2c^2}\\
a&=\frac{6bc^2+1\pm \sqrt{24bc^2+1}}{2c^2}\end{align}
My question: Is my answer correct?
| For convenience, let $d:=a-3b$, and
$$c=\frac{\sqrt{d+6b}}d,$$ giving
$$c^2d^2-d-6b=0,$$
$$d=\frac{1\pm\sqrt{1+24bc^2}}{2c^2}.$$
But to ensure $cd\ge0$, the sign of the numerator must be that of $c$ and
$$d=\begin{cases}c>0\to\begin{cases}-1\le24bc^2<0\to\dfrac{1\pm\sqrt{1+24bc^2}}{2c^2}\\24bc^2>0\to\dfrac{1+\sqrt{1+24bc^2}}{2c^2}\end{cases}\\c<0\to\begin{cases}-1\le24bc^2<0\to\text{no solution}\\24bc^2>0\to\dfrac{1-\sqrt{1+24bc^2}}{2c^2}\end{cases}\end{cases}.$$
When $c=0$, $d=-6b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Darboux Integral Partition Question I looking at a sample problem to use the Darboux Integral to find the integral of $f(x)=x^2$.
Textbook Solution:
For a partition P $={0=t_0<t_1<t_2<...<t_n=b}$ we have,
$U(f,P) = \sum^n_{k=1}sup(x^2:x \in [t_{k-1},t_k])*(t_k-t_{k-1})$
(I get lost here) If we choose $t_k=\frac{kb}{n}$ then we have:
$U(f,P) = \sum^n_{k=1} \frac{k^2b^2}{n^2}(\frac{b}{n}) $...
My question is how was $t_k=\frac{kb}{n}$ "picked". There is no explanation for it in the book.
| What is being done is we are partitioning the interval $[0,b]$ into $n$ subintervals of equal length. So each subinterval is $1/n$ the size of the full interval, so each length is $b/n$. So the interval endpoints are
$$ 0 < \frac{b}{n} < \frac{2b}{n} < \frac{3b}{n} < \cdots< \frac{(n-1)b}{n} < b $$
This is what is meant by the "picked" values for the partition subinterval endpoints. Rather than choosing an arbitrary partition, it is just the standard $n$-partition $\mathcal{P}^{(n)}$, the partition that splits the interval into $n$ subintervals of equal length.
Since $f(x) = x^2$ is a continuous function, the supremum on a subinterval is a maximum on the subinterval. Since $f(x) = x^2$ is increasing on $[0,b]$, the maximum value occurs at the right endpoint of the subinterval. So, denoting the partition endpoints as $x_k$ and the $k$-th subinterval as $[\Delta x_k] = [x_{k-1},x_k]$, each with length $\Delta x = b/n$ we see:
\begin{align*}
U(\mathcal{P}^{(n)}) &= \sum_{k=1}^n \left( \sup_{x \in [\Delta x_k]} f(x)\right) \Delta x \\
&= \sum_{k=1}^n \left( \max_{x \in [\Delta x_k]} x^2\right) \frac{b}{n} \\
&= \frac{b}{n} \sum_{k=1}^n \left( k \cdot \frac{b}{n} \right)^2 \\
&= \frac{b^3}{n^3} \sum_{k=1}^n k^2 \\
&= \frac{b^3}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \\
&= b^3 \cdot \frac{(n+1)(2n+1)}{6n^2} \\
\lim_{n \to \infty} U(\mathcal{P}^{(n)}) &= \lim_{n \to \infty} \left[b^3 \cdot \frac{(n+1)(2n+1)}{6n^2} \right] \\
\int_0^b x^2 \, dx &= b^3 \cdot \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} \\
&= b^3 \cdot \frac{2}{6} \\
&= \frac{b^3}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A series for $\frac{355}{113}-\pi$
Series for sums $\pi+\dfrac{p_n}{q_n}$
Lehmer's interesting series relating $\pi$ to its early convergents $3$, $\dfrac{22}{7}$ and $\dfrac{355}{113}$ may be written as follows.
$$\begin{align}
\sum_{n=1}^\infty \dfrac{n2^n}{\dbinom{2 n}{n}}&=\pi+3\tag{A.1}\\
\dfrac{2}{7}\sum_{n=1}^\infty \dfrac{n^22^n}{\dbinom{2 n}{n}}&=\pi+\dfrac{22}{7}\tag{A.2}\\
\dfrac{2}{35}\sum_{n=1}^\infty \dfrac{n^32^n}{\dbinom{2 n}{n}}&=\pi+\dfrac{22}{7}\tag{A.3}\\
\dfrac{1}{113}\sum_{n=1}^\infty \dfrac{n^42^n}{\dbinom{2 n}{n}}&=\pi+\frac{355}{113}\tag{A.4}\\
\end{align}$$
Series for differences $\pi-\dfrac{p_n}{q_n}$ or $\dfrac{p_n}{q_n}-\pi$
Series that prove the sign of the error when approximating $\pi$ by its convergents include:
$$\begin{align}
\pi-3&=4·24\sum_{k=0}^\infty \frac{(4k+1)!}{(4k+6)!}(k+1)\tag{B.1}\\
\frac{22}{7}-\pi&=4^2·240\sum_{k=0}^\infty \frac{(4k+3)!}{(4k+11)!}(k+1)(k+2)\tag{B.2}\\
\frac{22}{7}-\pi&=4^3·285120\sum_{k=0}^\infty \frac{(4k+1)!}{(4k+14)!}(k+1)(k+2)(k+3)\tag{B.3}\\
\end{align}$$
A series to prove $\frac{22}{7}-\pi>0$
Series and integrals for inequalities and approximations to $\pi$
Changing the initial value for summations leads to results as in A.1, A.2 and A.3.
$$\begin{align}
\pi+3&=4·24\sum_{k=-1}^\infty \frac{(4k+1)!}{(4k+6)!}(k+1)\tag{C.1}\\
\frac{22}{7}+\pi&=-4^2·240\sum_{k=-3}^\infty \frac{(4k+3)!}{(4k+11)!}(k+1)(k+2)\tag{C.2}\\
\frac{22}{7}+\pi&=-4^3·285120\sum_{k=-3}^\infty \frac{(4k+1)!}{(4k+14)!}(k+1)(k+2)(k+3)\tag{C.3}\\
\end{align}$$
Let us assume there is a series for the fourth convergent
$$\sum_{k=0}^\infty \frac{q}{\Pi_{i}(4k+d_i)}=\frac{355}{113}-\pi\tag{B.4}$$
for some positive rational $q$ and positive integers $d_i$.
Question:
Is there a technique to obtain the difference series (B.1, B.2, B.3) from their sum counterparts (A.1, A.2, A.3) that allows to determine B.4 from A.4?
| Since the second series for $\frac{22}{7}-\pi$ comes from
$$\frac{1}{28}\int_0^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2} dx = \frac{22}{7}-\pi$$
the following integral could be a starting point.
$$\frac{1}{99440} \int_0^1 \frac{x^{11}(1-x)^{12}( 858416(1+x^2)+6215x)}{1+x^2}dx = \frac{355}{113}-\pi$$
Maybe smaller coefficients would help such as in
$$\frac{1}{3164}\int_0^1 \frac{x^9(1-x)^8(791x+50(1+x^2))}{1+x^2} dx = \frac{355}{113} -\pi,$$
but the series obtained applying the technique in https://math.stackexchange.com/a/1657416/134791 is
$$\frac{4561920}{113} \sum_{k=0}^\infty \frac{ (81 k^2 + 648 k + 1145) (4 k + 16)}{ (4 k + 10)_{13} } = \frac{355}{113} - \pi$$
not as simple as the ones in the question.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Prove that $T(n) = T(n/3) + T2n/3) + 5n$ is $O(n log n)$ I'm doing some research about time complexity of algorithms and stumbled upon the following problem that I'm not able to solve:
Let $T(n) = T(n/3) + T(2n/3) + 5n$. prove that $T(n) = O(n log n)$
First, I made a recursion tree, which is the same as the one in the question: Recursion tree T(n) = T(n/3) + T(2n/3) + cn
I found out that each level costs $5n$ and that the leaves have different depths (The left path is the shortest with height $log_3 n$ and the one on the right is the longest with height $log_{\frac{3}{2}}n$).
Now, since we need an upper bound, I took the longest path of height $log_{\frac{3}{2}}n$. This gives total costs $5n \cdot log_{\frac{3}{2}}n$.
Now I want to prove with induction on $n$ that this is true. I proceeded in the following way:
Assume as induction hypothesis that $T(k) \leqslant 5k \cdot log_{\frac{3}{2}}k$ for $k < n$. Then:
$T(n) \leqslant T(n/3) + T(2n/3) + 5n$
$=^{IH} \frac{5}{3}n log_{\frac{3}{2}}(\frac{n}{3}) + \frac{10}{3}n log_{\frac{3}{2}}(\frac{2n}{3}) + 5n$
$= \frac{25}{3}n log_{\frac{3}{2}} n - \frac{5}{3}n log_{\frac{3}{2}} 3 - \frac{10}{3}n log_{\frac{3}{2}} 3 + 5n$
$= \frac{25}{3}n log_{\frac{3}{2}} - 5n log_{\frac{3}{2}} 3 + 5n$
And this is certainly not smaller then or equal to $5n \cdot log_{\frac{3}{2}}n$. I've spend an entire day now on solving this recurrence relation, but nothing solved it so far. Could you help me solving this problem?
|
You seem way too attached to the constant 5 and the base $3/2$. You don't need this exact constant to prove what you want, so why make your life miserable by clinging to it instead of picking "anything that works"?
Short advice: do not commit in advance to anything that makes your life harder if you can avoid it.
As an example, in your case let your induction hypothesis be
$$
T(k) \leq C k \ln k, \qquad \forall k < n \tag{$\dagger$}
$$
($\ln$ is the natural logarithm) for some absolute constant $C>0$ that we will pick momentarily.
Then you get
$$\begin{align}
T(n) &= T\left(\frac{n}{3}\right)+T\left(\frac{2n}{3}\right)+5n \\
&\leq_{(\dagger)} C \frac{n}{3}\ln \frac{n}{3} + C\frac{2n}{3}\ln \frac{2n}{3} + 5n\\
&\leq Cn \ln n- C\frac{n}{3}\ln 3 - C\frac{2n}{3}\ln \frac{3}{2} + 5n \\
&\leq Cn \ln n
\end{align}$$
where the last inequality holds as long as we can choose $C$ such that
$$
C\frac{n}{3}\ln 3 + C\frac{2n}{3}\ln \frac{3}{2} \geq 5n
$$
for all $n\geq 1$. This is satisfied for any $C$ such that
$$
C \geq \frac{5}{\frac{1}{3}\ln 3+\frac{2}{3}\ln \frac{3}{2}} \simeq 7.8
$$
so we can "retroactively" choose, say, $C\stackrel{\rm def}{=} 8$: the proof now goes through.
| {
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"timestamp": "2023-03-29T00:00:00",
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From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$
Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{3^3} { 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 }$$
$$a_n = \frac{3^{n-1}}{ \frac {(2n+1)!}{3!} }$$
$$a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}$$
Proof by induction:
$$a_1 = \frac {3! \cdot 3^{0}} {3!} =1$$
$$a_{n+1} = \frac {3! \cdot 3^{n}} {(2n+3)!}$$
Very grateful for the feedback given before. I am new to this. I am a bit stock at the end, what is the most efficient way to reach back to $a_n$?
| Plug it back to formula $P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$. By induction hypothesis, $a_n = \frac {3! \cdot 3^{n-1}} {(2n+1)!}$, so
$$a_{n+1}= \frac{3 \cdot 3! \cdot 3^{n-1}}{(2n+2)(2n+3)\cdot (2n+1)!} = \frac{3! \cdot 3^{n}}{(2n+3)!}$$
which is what you need to complete the induction.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given two real numbers $a$ and $b$ such that $aLet $a$ and $b$ be two given real numbers such that $a < b$, and let $\left\{x_n\right\}$ and $\left\{ y_n \right\}$ be the sequences defined as follows:
Let us choose $x_1$ and $y_1$ such that $$a < x_1 < b, \qquad a < y_1 < b$$ arbitrarily, and then let
$$x_2 = \frac{a+x_1}{2}, \qquad y_2 = \frac{y_1 + b}{2},$$
$$x_3 = \frac{a + x_1 + x_2 }{3}, \qquad y_3 = \frac{ y_1 + y_2 + b}{3},$$
and so on
$$ x_n = \frac{a+ x_1 + \cdots + x_{n-1} }{n}, \qquad y_n = \frac{ y_1 + \cdots + y_{n-1} + b}{n} $$
for $n= 3, 4, 5, \ldots$. Then what can we say about the convergence of these sequences?
To generalize this problem a little further, let $\left\{ r_n \right\}$ be a given sequence of positive real numbers, and let us now define
$$x_2 = \frac{r_1 x_1 + a r_2}{r_1 + r_2}, \qquad y_2 = \frac{r_1 y_1 + r_2 b}{r_1 + r_2},$$
$$x_3 = \frac{r_1 x_1 + r_2 x_2 + r_3 a }{r_1 + r_2 + r_3}, \qquad y_3 = \frac{ r_1 y_1 + r_2 y_2 + r_3 b}{r_1 + r_2 + r_3 },$$
and so on
$$ x_n = \frac{r_1 x_1 + \cdots + r_{n-1} x_{n-1} + r_n a }{r_1 + \cdots + r_n }, \qquad y_n = \frac{ r_1 y_1 + \cdots + y_{n-1} + r_n b}{r_1 + \cdots + r_n} $$
for $n= 3, 4, 5, \ldots$. Then what can we say about the convergence of these sequences?
What if we proceed as follows?
Let $r_0 > 0$ be given, and let
$$x_2 = \frac{r_0a+ r_1 x_1}{r_0 + r_1 }, \qquad y_2 = \frac{ r_1 y_1 + r_0 b}{r_1 + r_0},$$
$$x_3 = \frac{r_0 a + r_1 x_1 + r_2 x_2 }{r_0 + r_1 + r_2}, \qquad y_3 = \frac{ r_2 y_2 + r_1 y_1 + r_0 b}{r_2 + r_1 + r_0},$$
and so on
$$ x_n = \frac{r_0 a + r_1 x_1 + \cdots + r_{n-1} x_{n-1} }{r_0 + \cdots + r_{n-1} }, \qquad y_n = \frac{r_{n-1} y_{n-1} + \cdots + r_1 y_1 + r_0 b}{r_{n-1} + \cdots + r_0} $$
for $n= 3, 4, 5, \ldots$. What can we say about the convergence of these sequences now?
I can handle the situation only if we have only unit weights and only average of two terms is involved at a time, but I simply have no idea of what happens in this case!!
So, I would be really grateful for a detailed answer!
| This problem is in fact much easier than it seems at first. For $n \geq 2$, we have $$x_n = \dfrac{r_0a+r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1}}{r_0+r_1+\cdots+r_{n-1}} \\\Rightarrow r_0a+r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1} = (r_0+r_1+\cdots+r_{n-1})x_n.$$
Thus $$x_{n+1} = \dfrac{r_0a+r_1x_1+r_2x_2+\cdots+r_{n}x_{n}}{r_0+r_1+\cdots+r_{n}} = \dfrac{(r_0+r_1+\cdots+r_{n-1})x_n+r_nx_n}{r_0+r_1+\cdots+r_n} = x_n.$$
Thus $$\frac{r_0a+r_1x_1}{r_0+r_1} = x_2=x_3=\cdots.$$ An analogous result holds for $\{y_i\}$.
If instead you use the formula $$x_n = \dfrac{r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1}+r_na}{r_1+r_2+\cdots+r_n},$$ the solution needs to be modified a bit. We have $$r_1x_1+r_2x_2+\cdots+r_{n-1}x_{n-1} = (r_1+r_2+\cdots+r_n)x_n-r_na.$$
Then $$x_{n+1}=\dfrac{r_1x_1+r_2x_2+\cdots+r_{n}x_{n}+r_{n+1}a}{r_1+r_2+\cdots+r_{n+1}}=\dfrac{(r_1+r_2+\cdots+r_n)x_n-r_na+r_nx_n+r_{n+1}a}{r_1+r_2+\cdots+r_{n+1}}\\=x_n-\dfrac{r_{n+1}x_n+r_na-r_nx_n-r_{n+1}a}{r_1+r_2+\cdots+r_{n+1}}=x_n-\dfrac{(r_{n+1}-r_n)(x_n-a)}{r_1+r_2+\cdots+r_{n+1}}.$$
I'm not completely sure where to proceed from here but if I solve it I will come back and edit my solution in.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the Jordan form for a $4\times4$ matrix $$A:= \begin{bmatrix}4 & -4 & -11 & 11\\3 & -12 & -42 & 42\\ -2 & 12 & 37 & -34 \\ -1 & 7 & 20 & -17 \end{bmatrix}$$
I'm struggling with this matrix: it has $p_A(x) = (x-3)^4 $, yet $\ker(A - 3I)^3 $ is already the whole space $ \mathbb C^4 $. I read on another post that (for the $ \mathbb C^3 $ case of a similar matrix) I should take this to be the size of the biggest Jordan block, thus leaving me with $1$ standard eigenvector and a Jordan chain (image of the image of $\ldots$ ) of a vector $v$ not in $\ker(A - 3I)^2$. But this doesn't work for any $v$.
What steps should I follow to find the Jordan form of $A$?
| It maybe of interest to have the explicit decomposition
$A=PJP^{-1}$ with
$$J=\left(\begin{array}{rrr|r}3 & 1 & 0 & 0\\
0 & 3 & 1 & 0\\
0 & 0 & 3 & 0\\ \hline
0 & 0 & 0 & 3\end{array}\right) \ \ \text{and} \ \ P=\begin{pmatrix}
\ \ 3 & 3 & 14 & 0\\
\ \ 9 & 0 & 0 & 0\\
-6 & 6 & 1 & 1\\
-3 & 6 & 0 & 1\end{pmatrix}.$$
Let $B=A-3I_4$. We have $\ker(B) \subsetneqq \ker(B^2) \subsetneqq
\ker(B^3)$ with:
$\ker(B)=\langle \begin{pmatrix}
0\\0\\1\\1\end{pmatrix},\begin{pmatrix}
2\\6\\-1\\1\end{pmatrix}\rangle, \ \ \ker(B^2)=\langle \begin{pmatrix}
0\\0\\1\\1\end{pmatrix},\begin{pmatrix}
2\\6\\-1\\1\end{pmatrix},\begin{pmatrix}
1\\0\\0\\0\end{pmatrix}\rangle, \ \ \ker(B^3)=\mathbb{R}^4.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the simplified form of $\frac {1}{\cos x + \sin x}$ Find the simplified form of $\dfrac {1}{\cos x + \sin x}$.
a). $\dfrac {\sin (\dfrac {\pi}{4} +x)}{\sqrt {2}}$
b). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{\sqrt {2}}$
c). $\dfrac {\sin (\dfrac {\pi}{4} + x)}{2}$
d). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{2}$
My Attempt:
$$=\dfrac {1}{\cos x + \sin x}$$
$$=\dfrac {1}{\cos x + \sin x} \times \dfrac {\cos x - \sin x}{\cos x - \sin x}$$
$$=\dfrac {\cos x - \sin x}{\cos^2 x - \sin^2 x}$$
$$=\dfrac {\cos x - \sin x}{\cos 2x}$$
| $$\frac{1}{\cos(x) + \sin(x)} = \frac{\sqrt2/2}{\sqrt2/2*\cos(x) + \sqrt2/2*\sin(x)} = \frac{\sqrt2/2}{\sin(\pi/4+x)} = \frac{\csc(\pi/4 + x)}{\sqrt2}$$
| {
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Show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$. Let $x_n \ge 0$ for all $n \in \mathbf{N}$ and $x>0$, show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$.
My textbook does the following proof:
Let $\epsilon >0$, we must find an $N$ such that $n \ge N$ implies $|\sqrt{x_n} - \sqrt{x}|< \epsilon$ for all $n \ge N$.
\begin{align}
|\sqrt{x_n} - \sqrt{x}| &= |\sqrt{x_n} - \sqrt{x}|\left(\frac{\sqrt{x_n} + \sqrt{x}}{\sqrt{x_n} + \sqrt{x}}\right) \\
& = \frac{|x_n - x|}{\sqrt{x_n} + \sqrt{x}} \\
& \le \frac{|x_n - x|}{\sqrt{x}} \ \ \ \ \ \ \cdots \ \ \ \ \ \ (1)
\end{align}
Since $(x_n) \rightarrow x$ and $x>0$, we can choose $N$ such that $|x_n - x| < \epsilon\sqrt{x}$ whenever $n \ge N$ and so for all $n \ge N$, $|\sqrt{x_n} - \sqrt{x}| < \frac{\epsilon \sqrt{x}}{\sqrt{x}} = \epsilon$.
What I'm wondering is, in Eqn.$(1)$, why is $\sqrt{x}$ kept in the denominator? Couldn't one just have simply $\le |x_n - x|$ and choose $N$ such that $|x_n - x| < \epsilon$ for $n \ge N$ and the rest follows?
| They keep the denominator for the case that $x < 1$. If you assume $x\ge 1$ you can use your assumption.
| {
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Simplifying the sum $\sum\limits_{t=1}^{n} t(t+1)v^t$ For the sum:
$$\sum_{t=1}^{n} t(t+1)v^t,$$
would there be a "simple" formula for this? Such as:
$$S=\sum_{t=1}^{n} tv^t = \frac{\frac{1-v^n}{1-v} - n v^n}{\frac{1}{v} - 1}$$?
(The second sum is the present value of an annuity increasing by \$1 every period, for anyone curious).
I understand that it'll boil down to simplifying the sum to
$$\sum_{t=1}^{n}t^2v^t + S $$
but I'm not sure if there's a nice formula for the first part.
| Let $S = \sum_{t = 1}^{n} t^2 v^t$
$S = v + 4v^2 + 9v^3 + ... + n^2 v^n$
$Sv = \ \ \ \ \ v^2 + 4v^3 + ... + n^2 v^{n+1}$
$S(1-v) = v + 3v^2 + 5v^3 + ... (n^2 - (n-1)^2)v^n - n^2 v^{n+1}$
$S(1-v)v = \ \ \ \ \ v^2 + 3v^3 + ... ((n-1)^2 - (n-2)^2)v^n + (n^2 - (n-1)^2)v^{n+1} - n^2 v^{n+2}$
$S(1-v)^2 = v + 2v^2 + 2v^3 + ... 2v^n - (n^2 + 2n - 1)v^{n+1} + n^2 v^{n+2}$
$S(1-v)^2 = v + 2v^2(1 + v + v^2 + ... v^{n-2}) - (n^2 + 2n - 1)v^{n+1} + n^2 v^{n+2}$
$S = \frac{v}{(1-v)^2} - \frac{2v^2 (v^{n-1} - 1)}{(1-v)^3} - \frac{(n^2 + 2n -1 )v^{n+1}}{(1-v)^2} + \frac{n^2v^{n+2}}{(1-v)^2}$
EDIT: This can be further simplified to:
$S = -\frac{n^2v^{n+1}}{(1-v)} - \frac{2nv^{n+1}}{(1-v)^2} + \frac{v(1-v^n)(1+v)}{(1-v)^3}$
Verify: $n = 0$, gives $S = 0$
$n = 1$; $S = -\frac{v^2}{(1-v)} - \frac{2v^2}{(1-v)^2} + \frac{v(1-v)(1+v)}{(1-v)^3} = v$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus
Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus.
I tried to express $$f(x)=3x^4-16x^3+18x^2+5=A(ax^2+bx+c)^2+B(ax^2+bx+c)+C $$ which is a quadratic in $ax^2+bx+c$ which itself is quadratic in $x$. Comparing coefficients, we get
$$Aa^2=3 \tag{1}$$
$$2abA=-16$$
$$A(b^2+2ac)+aB=18$$
$$2bcA+bB=0$$
$$Ac^2+Bc+C=5$$
But I felt its very lengthy to solve these equations. Any hints?
| Let us consider the minimum of $f(x)$ as $k$. Then $f(x)-k$ has at least one repeated root. That is
$$f(x)-k=3x^4-16x^3+18x^2+5- k=3(x-a)^2(x^2+bx+c)$$
Expand and equate coefficients, you get $k=5-3a^2 c$, and
$$3(b-2a)=-16 \\ 3(c+a^2-2ab)=18 \\ a(ab-2c)=0$$
From third equation if $a=0$, then $k=5$. For $a \neq 0$ combinig three equations gives $a^2=4a-3$. Thus $a=1, 3$.
$a=1$ leads to $c=-\frac{5}{3}$, and $k=5-3(-\frac{5}{3})=10$.
$a=3$ gives $c=1$, and $k=5-27=-22$.
Therefore minimum is $k=-22$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorize :$f(x)=x^n+x+1 \ \ \ \ \ \ : n=3k+2 ,k\in \mathbb{N}$ How to factorize :
$$f(x)=x^n+x+1 \ \ \ \ \ \ : n=3k+2 ,k\in \mathbb{N}$$
And :
$$g(x)=x^n+x-1 \ \ \ \ \ \ : n=3k+2 ,k\in2m-1 \ \ \ , \ \ m\in\mathbb{N}$$
My try :
$$f(x)=x^n+x+1=x^{3k+2}+x+1$$
$$=(x^{3k+2}+x^{3k+1}+x^{3k})-(x^{3k+2}+x^{3k+1}+x^{3k})+(x+1)$$
Now what ?
| $$x^{3k+2}+x+1=x^{3k+2}-x^2+x^2+x+1=x^2((x^3)^k-1)+x^2+x+1$$
and use $x^3-1=(x-1)(x^2+x+1)$.
For $n=6m-1$ we have
$$x^n+x-1=x^n+x^2-x^2+x-1=x^2(x^{n-2}+1)-(x^2-x+1)$$
and use $x^3+1=(x+1)(x^2-x+1).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Largest prime number $p$ that cannot be written as the sum of three odd composite numbers? I consider the sequence of composite odd integers: 9, 15, 21, 25, 27, 33, 35, 41, ...
I observe that there are certain large gaps between the composite odd integers and this may contribute towards the solution.
So I start by considering some sums first:
9 + 9 + 9 = 27, 9 + 9 + 15 = 33. So this means that 31 is potentially such a prime number.
Then I consider other sums and manage to obtain 39, 43 and 45. So now 41 becomes the potential contender.
But this method is clearly just trial and error. Is there a more elegant method?
| For every prime $p\gt3$, either $p-25$ or $p-35$ is divisible by $6$. (Both numbers are even, and $p-25\equiv p-1$ mod $3$ while $p-35\equiv p-2$ mod $3$.). If $p\ge53$, the difference ($p-25$ or $p-35$) is at least $18$, hence can be written in the form $6(a+b+1)=3(2a+1)+3(2b+1)$, with $a,b\ge1$. Thus every prime $p\gt47$ can be written as the sum of two odd multiples of $3$ and either $25$ or $35$.
Finally, $p=47$ cannot be written as a sum of three composite odd numbers: If $47=x+y+z$ with $x\le y\le z$ from the list of odd composites, then, since $47/3\lt16$, $x$ must be either $9$ or $15$. If $x=9$, then $y+z=38$, implying $y$ is also either $9$ or $15$ (since $38/2=19$), neither of which works. If $x=15$, then $y+z=32$, again implying $y$ is either $9$ or $15$, neither of which works. (Alternatively, since $25$ and $35$ are the only odd composites less than $47$ that are not multiples of $3$, and since $25+25\gt47$, we would have to write $47=x+y+35$. But $x+y=12$ has no solutions in odd composites.)
| {
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"timestamp": "2023-03-29T00:00:00",
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What does the group of the geometric images of z, such that $z+\overline{z}+|z^2|=0$ define?
What does the group of the geometric images of z, such that
$z+\overline{z}+|z^2|=0$ define?
A) A circumference of center (0,-1) and radius 1.
B) Two lines, of equations $y=x+2$ and $y=x$.
C)A circumference of center (1,0) and radius 1.
D)A circumference of center (-1,0) and radius 1.
I tried to simplify the expression:
$$z+\overline{z}+|z^2|=0 \Leftrightarrow \\
x+yi+x-yi+|(x+yi)^2|=0\Leftrightarrow \\
2x+\sqrt{(x^2+y^2)^2+(2xy)^2}=0 \Leftrightarrow \\
\sqrt{(x^2+y^2)^2+(2xy)^2}=-2x \Leftrightarrow \\
(x^2+y^2)^2+(2xy)^2=4x^2 \Leftrightarrow \\
x^4 - 2 x^2 y^2 + y^4+4x^2y^2 = 4x^2 \Leftrightarrow \\
x^4+y^4+2x^2y^2 = 4x^2 \Leftrightarrow \\
???$$
How do I continue from here?
I have also been thinking that if the sum of those three numbers is zero then they could be the vertices of a triangle. I rewrote the expression:
$$\rho \cdot cis(\theta)+\rho cis(-\theta)+|\rho^2 \cdot cis(2\theta)|=0 \Leftrightarrow \\
\rho \cdot cis(\theta)+\rho cis(-\theta)+\rho^2 =0 \Leftrightarrow \\
\rho(cis(\theta)+cis(-\theta)+\rho) = 0 \Leftrightarrow \\
\rho = 0 \lor cis(\theta)+cis(-\theta)+\rho = 0 \Leftrightarrow \\
cis(\theta)+cis(-\theta) = -\rho \Leftrightarrow \\
\cos(\theta)+\sin(\theta)i+\cos(-\theta)+\sin(-\theta)i = -\rho \Leftrightarrow \\
\cos(\theta)+\cos(\theta) = -\rho \Leftrightarrow \\
2\cos(\theta) = -\rho \Leftrightarrow \\
\rho = -2\cos(\theta)$$
This means that $\rho$ will be between 0 and 2. If $|z^2|=\rho^2 = \rho^2 cis(0)$, then one of the vertices is $4$.
But what do I do next? How do I solve this?
| That solution is very laboured: $|z^2|=x^2+y^2$ so the curve's equation is
$$x^2+y^2+2x=0$$
or
$$(x+1)^2+y^2=1.$$
I'm sure you can identify the curve now.
| {
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Does $\sum_{k=1}^\infty \frac{k^2}{3^k}$ converge or diverge? I am looking at the series $$\sum_{k=1}^\infty a_k $$
where $$a_k = \frac{k^2}{3^k}$$
The terms denominator grow much faster (Exponential) than the numerators (Quadratic) so I am guessing it will converge, but I also want to find the value of it since I am almost sure it does converge
| Your intuition is correct - the sum converges. I will show this by showing that the sum, taken from the fifth term onward, converges. This is sufficient because the first four terms are finite.
Notice that for all $k>4$, $$2^k > k^2 \qquad \implies \qquad \frac{2^k}{3^k} > \frac{k^2}{3^k}$$
This implies $$\sum_{k=5}^\infty\, \left( \frac{2}{3}\right)^k \,>\, \sum_{k=5}^\infty \,\frac{k^2}{3^k}$$
But the left hand side is a geometric series with common ratio $2/3$, which we know converges. Therefore the right hand side (your sum) must converge.
Now that we know that it converges, suppose you'd like to find the exact value of the sum.
Let $S$ be the value to which your sum converges.
\begin{align}
S&=\sum_{k=1}^\infty \,\frac{k^2}{3^k}\\\\
S&=0 + \frac{1}{3} +\frac{4}{9} + \frac{9}{27}+\frac{16}{81}+\frac{25}{243}+ \cdots\\\\
3S&= 1 + \frac{4}{3} +\frac{9}{9} + \frac{16}{27}+\frac{25}{81}+\frac{36}{243}+ \cdots
\end{align}
Now subtract the second-to-last equation above from the last equation. Let $R = 2S$ be this new series.
\begin{align}
3S - S &= 1 + \frac{3}{3} + \frac{5}{9} + \frac{7}{27} + \frac{9}{81} + \frac{11}{243} + \cdots \qquad \\\\
R &= 1 + \frac{3}{3} + \frac{5}{9} + \frac{7}{27} + \frac{9}{81} + \frac{11}{243} + \cdots\\\\
3R &= 6 + \frac{5}{3} + \frac{7}{9} + \frac{9}{27} + \frac{11}{81} + \frac{13}{243} + \cdots
\end{align}
Now subtract the second-to-last equation above from the last equation. Let $T = 2R = 4S$ be this new series.
\begin{align}
3R - R &= 5 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} + \frac{2}{243} + \cdots \qquad \\\\
T &= 3 + 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + \frac{2}{81} + \frac{2}{243} + \cdots\\\\
T &= 3 + 2\left(\sum_{k=0}^\infty \frac{1}{3^k}\right)
\end{align}
We've reached a geometric series with first term $1$ and common ratio $1/3$. This converges to $3/2$, so we are able to go back up the rabbit hole to find the value of $S$, at long last:
\begin{align}
4S &= 2R = T = 3 + 2(3/2) = 6 \qquad\qquad \\\\
S&=\boxed{\frac{3}{2}}
\end{align}
It's worth noting that series like the intermediate $R$ encountered here are called arithmetico-geometric series because the terms' numerators are in arithmetic progression, while the denominators are in geometric progression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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