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Proving $\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$ How do I show that:
$$\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24$$
This is actually problem B $4371$ given at this link. Looks like a very interesting problem.
My attempts: Well, I have been thinking about this for the whole day, and I have got some insights. I don't believe my insights will lead me to a $\text{complete}$ solution.
*
*First, I wrote $\sin\frac{5\pi}{14}$ as $\sin\frac{9 \pi}{14}$ so that if I put $A = \frac{\pi}{14}$ so that the given equation becomes, $$\frac{1}{\sin^{2}{A}} + \frac{1}{\sin^{2}{3A}} + \frac{1}{\sin^{2}{9A}} =24$$ Then I tried working with this by taking $\text{lcm}$ and multiplying and doing something, which appeared futile.
*Next, I actually didn't work it out, but I think we have to look for a equation which has roots as $\sin$ and then use $\text{sum of roots}$ formulas to get $24$. I think I haven't explained this clearly.
*
*$\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}$ One post related to this is: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$ I don't know how this will help as I haven't studied anything yet regarding Gauss sums.
| We can prove (below), the roots of $$z^3-z^2-2z+1=0 \ \ \ \ (1)$$
are $2\cos\frac{(2r+1)\pi}7$ where $r=0,1,2$
So,if we set $\displaystyle t=\frac1{\sin^2{\frac{(2r+1)\pi}{14}}}$ where $r=0,1,2$
$\implies 2\cos\frac{(2r+1)\pi}7=2\left(1-2\sin^2{\frac{(2r+1)\pi}{14}}\right)=2\left(1-\frac2t\right)=\frac{2(t-2)}t$ which will satisfy the equation $(1)$
$$\implies \left(\frac{2(t-2)}t\right)^3-\left(\frac{2(t-2)}t\right)^2-2\left(\frac{2(t-2)}t\right)+1=0$$
On simplification we have $$8(t-2)^3-4t(t-2)^2-4t^2(t-2)+t^3=0$$
$$\text{or, }t^3(8-4-4+1)-t^2(8\cdot3\cdot2-4\cdot4-8)+()t+()=0$$
$$\text{or, }t^3-24t^2+()t+()=0$$
Now, use Vieta's Formulas
[
Proof:
Let $7x=\pi$ and $y=\cos x+i\sin x$
Using De Moivre's formula, $y^7=(\cos x+i\sin x)^7=\cos \pi+\sin\pi=-1$
So, the roots of $y^7+1=0\ \ \ \ (1)$
are $\cos \theta+i\sin\theta$ where $\theta=\frac{(2r+1)\pi}7$ where $r=0,1,2,3,4,5,6$
Leaving the factor $y+1$ which corresponds to $r=3,$
we get $y^6-y^5+y^4-y^3+y^2-y+1=0$
Dividing either sides by $y^3,$ $$y^3+\frac1{y^3}-\left(y^2+\frac1{y^2}\right)+y+\frac1y-1=0$$
$$\implies \left(y+\frac1y\right)^3-3\left(y+\frac1y\right)-\{\left(y+\frac1y\right)^2-\}+y+\frac1y-1=0$$
$$\implies z^3-z^2-2z+1=0\ \ \ \ (2)$$ where $z=y+\frac1y=2\cos\theta$
Now, since $\cos(2\pi-A)=\cos A,\cos\left(2\pi-\frac{(2r+1)\pi}7\right)=\cos\left(\frac{(13-2r)\pi}7\right)$ where $r=0,1,2$
So, the roots of equation $(2)$ are $2\cos\frac\pi7=2\cos\frac{13\pi}7, 2\cos\frac{3\pi}7=2\cos\frac{11\pi}7$ and $2\cos\frac{5\pi}7=2\cos\frac{9\pi}7$
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/45144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 5,
"answer_id": 3
} |
Integral with $\sqrt{2x^4 - 2x^2 + 1}$ in the denominator $$\int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}} \: \text{d}x$$
I tried to substitute $x^2=t$ but I am unable to solve it and I also tried to divide numerator and denominator by $x^2$ and do something but could not get anything.
| Well, using mathematica I can see that this function the given answer is the derivative of the integral.
Consider
\begin{align*}
f(x) &= \frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}} \\ &= \frac{\frac{x^{2} \cdot (8x^{3}-4x)}{2 \sqrt{2x^{4}-2x^{2}+1}} - 2x \cdot \sqrt{2x^{4}-2x^{2}+1}}{x^{4}} \quad \ \Bigl[ \text{Note this is} \ f'(x)\Bigr] \\ &= \frac{x^{2} \cdot (4x^{3}-2x) - (2x^{4}-2x^{2}+1) \cdot 2x}{x^{4} \cdot \sqrt{2x^{4}-2x^{2}+1}} \\ &= \frac{4x^{4} -2x^{2} -4x^{4} + 4x^{2}-2}{x^{3} \cdot \sqrt{2x^{4}-2x^{2} +1}} \quad \Bigl[ \text{cancelling out x}\Bigr] \\ &= \frac{2 \cdot (x^{2}-1)}{x^{3} \cdot \sqrt{2x^{4}-2x^{2}+1}}
\end{align*}
So write your integral as $$\int\frac{x^{2}-1}{x^{3}\cdot \sqrt{2x^{4}-2x^{2}+1}} = \frac{1}{2} \int \frac{\text{d}}{\text{dx}}\biggl(\frac{\sqrt{2x^{4}-2x^{2}+1}}{x^{2}}\biggr)\ \text{dx}$$
Using this you can get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/45474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Is $\frac{a_1+\cdots+a_n}{\sqrt{n(b_1+\cdots+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$? For $a_1>0$, $a_2>0,\dots,a_n>0$, and $b_1>0$, $b_2>0,\dots,b_n>0$
I want to prove:
$$\frac{a_1+a_2+\dots+a_n}{\sqrt{n(b_1+b_2+...+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}}
+\cdots+\frac{a_n}{\sqrt{b_n}}\right)$$
| For $n=2$ your inequality reduces to
$$\frac{a_1+a_2}{\sqrt{2(b_1+b_2)}} \leq \frac{1}{2} (\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}})$$
Lets the $a_1=1,a_2=\frac{1}{2},b_1=1,b_2=\frac{11}{16}$ then you have
$$\sqrt{\frac{2}{3}} \leq \frac{1}{2}+\frac{1}{\sqrt{11}}$$
This is false, so the inequality doesn't hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/47616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
| Yet another take. Start with the definition of falling factorial powers:
$\begin{align}
x^{\underline{r}}
= x (x - 1) \dotsm (x - r + 1)
\end{align}$
so that:
$\begin{align}
\Delta n^{\underline{r}}
&= (n + 1)^{\underline{r}} - n^{\underline{r}} \\
&= r n^{\underline{r - 1}} \\
\sum_{0 \le k < n} k^{\underline{r}}
&= \frac{n^{\underline{r + 1}}}{r + 1}
\end{align}$
(the last one is also easy to prove by induction, or otherwise).
Now note that:
$\begin{align}
n^2
= n^{\underline{2}} + n^{\underline{1}}
\end{align}$
so we can write:
$\begin{align}
\sum_{0 \le k \le n} k^2
&= \sum_{0 \le k < n + 1}
\left( k^{\underline{2}} + k^{\underline{1}} \right) \\
&= \frac{(n + 1)^{\underline{3}}}{3}
+ \frac{(n + 1)^{\underline{2}}}{2} \\
&= \frac{(n + 1) n (n - 1)}{3} + \frac{(n + 1) n}{2} \\
&= \frac{(2 n + 1) (n + 1) n}{6}
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/48080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "145",
"answer_count": 32,
"answer_id": 9
} |
Inverse Laplace Transform help Is the information below correct?
Find the inverse Laplace transform of $$ F(s) = \frac{s}{s^2 + 4s + 13}$$
Soln:
a) Complete the squares to simplify our denominator
$$ s^2 + 4s + 13 = (s+2)^2 + 9 = (s+2)^2 + 3^2$$
$$\mathscr{L}^{-1}\left\{F(s)\right\} = \frac{s}{(s+2)^2 + 3^2}. $$ From the table we can deduce that this is
$$\mathscr{L}^{-1}\left\{F(s)\right\} = e^{-2t} \cos(3t).$$
| HINT:
$$
\frac{s}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2 - 2}}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2}}{{(s + 2)^2 + 3^2 }} - \frac{2}{3}\frac{3}{{(s + 2)^2 + 3^2 }}.
$$
Now note that the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2 + \omega ^2 }}$ and $\frac{\omega}{{(s + \alpha)^2 + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/52555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Proving that $ 30 \mid ab(a^2+b^2)(a^2-b^2)$ How can I prove that $30 \mid ab(a^2+b^2)(a^2-b^2)$ without using $a,b$ congruent modulo $5$ and then
$a,b$ congruent modulo $6$ (for example) to show respectively that $5 \mid ab(a^2+b^2)(a^2-b^2)$ and
$6 \mid ab(a^2+b^2)(a^2-b^2)$?
Indeed this method implies studying numerous congruences and is quite long.
| Hint for the divisors 2 and 3.
Note the factorisation $ab(a+b)(a-b)(a^2+b^2)$ which is divisible by $ab(a-b)(a+b)$
Either one of $a,b$ is divisible by 2 (or 3) or not. If not, look at the other factors to complete the proof.
Studying cases does not look too onerous to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
} |
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
| As Willie Wong observed, including an expression of the form $\displaystyle \frac{|\alpha|}{\alpha}$ is a way of ensuring that $\alpha > 0$. (As $\sqrt{|\alpha|/\alpha}$ is $1$ if $\alpha > 0$ and non-real if $\alpha < 0$.)
The ellipse $\displaystyle \left( \frac{x}{7} \right)^{2} + \left( \frac{y}{3} \right)^{2} - 1 = 0$ looks like this:
So the curve $\left( \frac{x}{7} \right)^{2}\sqrt{\frac{\left| \left| x \right|-3 \right|}{\left| x \right|-3}} + \left( \frac{y}{3} \right)^{2}\sqrt{\frac{\left| y+3\frac{\sqrt{33}}{7} \right|}{y+3\frac{\sqrt{33}}{7}}} - 1 = 0$ is the above ellipse, in the region where $|x|>3$ and $y > -3\sqrt{33}/7$:
That's the first factor.
The second factor is quite ingeniously done. The curve $\left| \frac{x}{2} \right|\; -\; \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2}\; -\; 3\; +\; \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}-y=0$ looks like:
This is got by adding $y = \left| \frac{x}{2} \right| - \frac{\left( 3\sqrt{33}-7 \right)}{112}x^{2} - 3$, a parabola on the positive-x side, reflected:
and $y = \sqrt{1-\left( \left| \left| x \right|-2 \right|-1 \right)^{2}}$, the upper halves of the four circles $\left( \left| \left| x \right|-2 \right|-1 \right)^2 + y^2 = 1$:
The third factor $9\sqrt{\frac{\left( \left| \left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right) \right| \right)}{\left( 1-\left| x \right| \right)\left( \left| x \right|-.75 \right)}}\; -\; 8\left| x \right|\; -\; y\; =\; 0$ is just the pair of lines y = 9 - 8|x|:
truncated to the region $0.75 < |x| < 1$.
Similarly, the fourth factor $3\left| x \right|\; +\; .75\sqrt{\left( \frac{\left| \left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right) \right|}{\left( .75-\left| x \right| \right)\left( \left| x \right|-.5 \right)} \right)}\; -\; y\; =\; 0$ is the pair of lines $y = 3|x| + 0.75$:
truncated to the region $0.5 < |x| < 0.75$.
The fifth factor $2.25\sqrt{\frac{\left| \left( .5-x \right)\left( x+.5 \right) \right|}{\left( .5-x \right)\left( x+.5 \right)}}\; -\; y\; =\; 0$ is the line $y = 2.25$ truncated to $-0.5 < x < 0.5$.
Finally, $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like:
so the sixth factor $\frac{6\sqrt{10}}{7}\; +\; \left( 1.5\; -\; .5\left| x \right| \right)\sqrt{\frac{\left| \left| x \right|-1 \right|}{\left| x \right|-1}}\; -\; \frac{\left( 6\sqrt{10} \right)}{14}\sqrt{4-\left( \left| x \right|-1 \right)^{2}}\; -\; y\; =\; 0$ looks like
As a product of factors is $0$ iff any one of them is $0$, multiplying these six factors puts the curves together, giving: (the software, Grapher.app, chokes a bit on the third factor, and entirely on the fourth)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
"answer_count": 10,
"answer_id": 9
} |
Olympiad Inequality Problem Consider three positive reals $x,y,z$ such that $xyz=1$.
How would one go about proving:
$$\frac{x^5y^5}{x^2+y^2}+\frac{y^5z^5}{y^2+z^2}+\frac{x^5z^5}{x^2+z^2}\ge \frac{3}{2}$$
I really dont know even where to begin! It looks a BIT like Nesbitts? Maybe?
| Edit. I posted a new "proof" (now deleted), before I realized I was adddressing the wrong question. I think the original proof is ok, I messed up something in trying to simplify the approach.
Hint.
Since this is an olympiad problem, it is likely that there is a proof without using calculus. I am sketching one here, but I have a feeling it can improved vastly.
Step 1. Make the substitution $x = 1/a$, $y = 1/b$ and $z = 1/c$, so that $abc = 1$. We are now left with the expression
$$
\sum \frac{1}{a^3b^3 (a^2+b^2)} = \sum \frac{c^3}{a^2+b^2} = \frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2}.
$$
Step 2. Assume the order $a \leq b \leq c$ without loss of generality. Then show that
$$
\frac{a^2}{b^2+c^2} \leq \frac{b^2}{c^2+a^2} \leq \frac{c^2}{a^2+b^2}.
$$
Now, by the "Chebyshev sum inequality", we have:
$$
\frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2}
\geq \frac{1}{3} (a+b+c) \cdot \left( \frac{a^2}{b^2+c^2} + \frac{b^2}{c^2+a^2} + \frac{c^2}{a^2+b^2} \right).
$$
Step 3. For any $u,v,w > 0$, prove that
$$
\frac{u}{v+w} + \frac{v}{w+u} + \frac{w}{u+v} \geq \frac{3}{2}.
$$
Step 4. Conclude the inequality by plugging in the third inequality in the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/61289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Limit of difference of two square roots I need to find the limit, not sure what to do.
$\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$
I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
| This answer may be a bit longer, but I've tried to include every step in the process.
\begin{equation} \label{eq1}
\begin{split}
\lim_{x \to \infty} \sqrt{x^2 + ax} - \sqrt{x^2 + bx} & = \lim_{x \to \infty} \left(\sqrt{x^2 + ax} - \sqrt{x^2 + bx}\right)\left(\frac{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}\right) \\
\\
& = \lim_{x \to \infty}\frac{(x^2 + ax) - (x^2 + bx)}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\ \\
& = \lim_{x \to \infty}\frac{ax-bx}{\sqrt{x^2\left(1 + \frac{a}{x}\right)} + \sqrt{x^2\left(1 + \frac{b}{x}\right)}} \\ \\
& = \lim_{x \to \infty}\frac{ax-bx}{|x|\sqrt{1 + \frac{a}{x}} + |x|\sqrt{1 + \frac{b}{x}}} \\ \\
& = \lim_{x \to \infty}\frac{x(a-b)}{|x|\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}\right)} \\ \\
&\text{Recall:} \ \ \ |x| = x \ \ \ \text{as} \ \ \ x \to \infty \\ \\
& = \lim_{x \to \infty}\frac{x(a-b)}{x\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}\right)} \\ \\
& = \lim_{x \to \infty}\frac{a-b}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} \\ \\
& = \frac{a-b}{\sqrt{1 + 0} + \sqrt{1 + 0}} \\ \\
& = \frac{a-b}{2} \\
\end{split}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/62418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can the following be calculated? How can the following series be calculated?
$$S=1+(1+2)+(1+2+3)+(1+2+3+4)+\cdots+(1+2+3+4+\cdots+2011)$$
| Note that $1$ occurs $2011$ times; $2$ occurs $2010$ times; $3$ occurs $2009$ times, and so on, until $2011$ occurs only once. Hence we can rewrite the sum as
$$(2012-1)(1)+(2012-2)(2)+(2012-3)(3)+\cdots+(2012-2011)(2011).$$
Split and regroup terms:
$$2012(1+2+3+\cdots+2011)-(1^2+2^2+3^2+\cdots+2011^2).$$
Now using the two formulas for triangular numbers and square pyramidal numbers, compute
$$2012\frac{2011(2011+1)}{2}-\frac{2011(2011+1)(2\cdot2011+1)}{6}=1357477286.$$
This also evaluates the general sum:
$$1+(1+2)+\cdots+(1+2+\cdots+n)$$
$$=(n+1-1)(1)+(n+1-2)(2)+\cdots+(n+1-n)(n)$$
$$=(n+1)(1+2+\cdots+n)-(1^2+2^2+\cdots+n^2)$$
$$=(n+1)\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=n(n+1)\left[\frac{n+1}{2}-\frac{2n+1}{6}\right]$$
$$=\frac{n(n+1)(n+2)}{6}.$$
One could also use the triangle number formula on each term for a more direct route:
$$\frac{1(1+1)}{2}+\frac{2(2+1)}{2}+\frac{3(3+1)}{2}+\cdots+\frac{n(n+1)}{2}$$
$$=\frac{1}{2}\left[(1+2^2+\cdots+n^2)+(1+2+\cdots+n)\right]$$
$$=\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right]=\frac{n(n+1)}{4}\left[\frac{2n+1}{3}-1\right]$$
$$=\frac{n(n+1)(n+2)}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/65465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Fourier transform integral I'm trying to calculate the 3D fourier transform of this function:
$$\frac{1}{(x^2+y^2+z^2)^{1/2}}$$
Any help would be appreciated, thanks.
| Inserting the Jacobian $r^2\sin\theta$
and $\sqrt{x^2+y^2+z^2}=r$ in polar coordinates gives
\begin{equation}
\int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{i\mathbf{k}\cdot \mathbf{r}}
\end{equation}
\begin{equation}
=
\int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{ikr\cos\theta}
\end{equation}
and with $z=\cos\theta$, $dz=-\sin\theta d\theta$
\begin{equation}
=
2\pi \int_0^\infty r dr \int_0^\pi \sin\theta d\theta e^{ikr\cos\theta}
=
-2\pi \int_0^\infty r dr \int_{1}^{-1} dz e^{ikrz}
=
2\pi \int_0^\infty r dr \int_{-1}^{1} dz e^{ikrz}
\end{equation}
and with $t=ikrz$, $dz=dt/(ikr)$
\begin{equation}
=
2\pi \int_0^\infty r dr \frac{1}{ikr} \int_{-ikr}^{ikr} dt e^t
\end{equation}
\begin{equation}
=
2\pi \int_0^\infty r dr \frac{1}{ikr} [e^{ikr}-e^{-ikr}]
=
4\pi \int_0^\infty r dr \frac{1}{kr} \sin(kr)
=
\frac{4\pi}{k^2} \int_0^\infty kr d(kr) \frac{1}{kr} \sin(kr)
\end{equation}
\begin{equation}
=
\frac{4\pi}{k^2} \int_0^\infty d(kr) \sin(kr)
=
\frac{4\pi}{k^2} \int_0^\infty dz \sin z
\end{equation}
and this exists only in the theory of distributions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/71282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find extremes of function $f(x,y,z) = x^2y + y^2z + x - z$ I am preparing for an exam tuesday morning and I would like to ask you, if someone could please review my solution for the following excercise. I don't have the correct answer so I am unable to check whether it is OK.
Find the extremes of polynomial function $f(x,y,z) = x^2y + y^2z + x - z$
So here is how i solved it:
$$
\begin{align*}
f'_x &= 2xy + 1 \hspace{10mm} f'_y = x^2 + 2yz \hspace{10mm}f'_z = y^2 - 1\\
\end{align*}
$$
$$
\begin{align*}
f''_{xx} = 2y \hspace{10mm} f''_{xy} = 2x \hspace{10mm} f''_{xz} = 2y\\
f''_yy = 2z \hspace{10mm} f''_{yz} = 2y \hspace{10mm} f''_{zz} = 0\\
\end{align*}
$$
So now I need to solve this system:
$$
\begin{align*}
2xy + 1 &= 0\\
x^2 + 2yz &= 0\\
y^2 -1 &= 0
\end{align*}
$$
So I get the points:
$A = [-\frac{1}{2}, 1, -\frac{1}{8}]$
$B = [\frac{1}{2}, -1, \frac{1}{8}]$
For point A:
$$
\begin{align*}
f''{xx}(-\frac{1}{2}, 1, -\frac{1}{8}) = 2 \hspace{10mm}
f''{xy}(-\frac{1}{2}, 1, -\frac{1}{8}) = -1 \hspace{10mm}
f''{xz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\
f''{yy}(-\frac{1}{2}, 1, -\frac{1}{8}) = -\frac{1}{4} \hspace{10mm}
f''{yz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 2 \hspace{10mm}
f''{zz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\
\end{align*}
$$
$$
H =
\begin{pmatrix}
2 & -1 & 0 \\
-1 & -\frac{1}{4} & 2 \\
0 & 2 & 0
\end{pmatrix}
$$
Subdeterminants are: $2, -\frac{3}{2}, -8$
Therefore we don't know whether point A is maxima or minima.
For point B:
$$
\begin{align*}
f''{xx}(-\frac{1}{2}, 1, -\frac{1}{8}) = -2 \hspace{10mm}
f''{xy}(-\frac{1}{2}, 1, -\frac{1}{8}) = 1 \hspace{10mm}
f''{xz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\
f''{yy}(-\frac{1}{2}, 1, -\frac{1}{8}) = \frac{1}{4} \hspace{10mm}
f''{yz}(-\frac{1}{2}, 1, -\frac{1}{8}) = -2 \hspace{10mm}
f''{zz}(-\frac{1}{2}, 1, -\frac{1}{8}) = 0\\
\end{align*}
$$
$$
H =
\begin{pmatrix}
-2 & 1 & 0 \\
1 & \frac{1}{4} & -2 \\
0 & -2 & 0
\end{pmatrix}
$$
Subdeterminants are: $2, -\frac{3}{2}, 8$
Therefore point B is a local maximum.
So please - is it correct? Thanks in advance.
| You wrote $f''_{xz} = 2y$, which is incorrect; it should be 0. However in your later calculations you correctly treat the term as zero.
For your second Hessian, the first subdeterminant is -2, not 2. Even if it were 2, H would not be negative-definite: Sylvester's Criterion states that a matrix is positive-definite (and the critical point a local minimum) if all of the subdeterminants are positive, and negative-definite (and the critical point a local maximum) if the subdeterminants alternate sign, starting with negative. Your second H satisfies neither criterion, and isn't singular (since the determinant is non-zero) so the critical point is neither a minimum nor a maximum.
| {
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"question_score": "5",
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Find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$ I want to find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$.
My thoughts so far: I want to find $p$ such that $ \left( \frac{15}{p} \right) = 1$. By multiplicativity of the Legendre symbol, this is equivalent to $ \left( \frac{5}{p} \right) \left( \frac{3}{p} \right) = 1 $. Using the Law of Quadratic Reciprocity, this is equivalent to finding $p$ such that $ - \left( \frac{p}{5} \right) \left( \frac{p}{3} \right) = 1$.
So there are two cases:
*
*$ \left( \frac{p}{5} \right) = -1, \left( \frac{p}{3} \right) = 1$.
*$ \left( \frac{p}{5} \right) = 1, \left( \frac{p}{3} \right) = -1 $.
For case (1.), the quadratic residues modulo $5$ are $1$ and $4$, so for $ \left( \frac{p}{5} \right ) = -1$, we must have that $p$ is $2$ or $3$ modulo $5$. We must also have that $p$ is $1$ modulo $3$ from the other condition. One of these pairs is incompatible, and we can solve to give $p$ is $13$ modulo $15$.
Similarly for case (2.)
Is this the correct approach? I'm unsure if each step in my working is an "if and only if". If $p$ is $13$ modulo $15$, is $15$ necessarily a quadratic residue modulo $p$?
Thanks!
| Note that there is no such thing as conditions mod 3 and mod 5 being incompatible; see The Chinese Remainder Theorem. In particular, 7 is 2 mod 5 and 1 mod 3.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Relations between coefficient and exponent of Proth prime form $k\cdot 2^n+1$? Definition: Proth number is a number of the form :
$$k\cdot 2^n+1$$
where $k$ is an odd positive integer and $n$ is a positive integer such that : $2^n>k$
My question : If Proth number is prime number are there some other known relations in addition to $2^n>k$ , between exponent $n$ and coefficient $k$ ?
| There are many simple relationships that involve congruences. They have a flavour much like results you have mentioned in earlier posts.
For example, if $n>1$ is odd, then $k$ must be of the form $6a-1$ or $6a+3$. If $n$ is even, then $k$ must be of the form $6a+1$ or $6a+3$. The arguments are the familiar ones. For example, if $n$ is odd, then since $2\equiv -1\pmod 3$, it follows that $2^n\equiv (-1)^n =-1\pmod{3}$. But if $k$ is of the form $6a+1$, it follows that $k2^n \equiv -1 \pmod{3}$, and therefore $k2^n \equiv 0\pmod{3}$. If $n>1$, this means that $k2^n+1$ cannot be prime, since it is greater than $3$ and divisible by $3$.
We can obtain similar restrictions by working modulo primes greater than $3$. For example, suppose that $n \equiv 2\pmod {4}$, that is, $n$ is of the form $4a+2$. Then $2^n \equiv 4 \pmod {5}$, and therefore if $k \equiv 1\pmod {5}$, we have $k2^n +1\equiv 4+1=5 \pmod{5}$. This is impossible unless $n=2$ (and therefore $k=1$). So we conclude that if $n$ is of the form $4a+2$, and $k2^n+1$ is prime, then $k$ cannot be of the form $10b+1$ except in the case $n=2$, $k=1$. We can obtain similar restrictions on $k$ if $n$ is of the form $4a$, also $4a+1$, also $4a+3$.
Added: In a comment, the OP asks for a proof that if $k\equiv 1\pmod 3$ and $k\equiv 1\pmod{10}$ (and $k2^n+1$ is prime), then $n\equiv 0\pmod 4$.
This is not absolutely true, take $n=2$, $k=1$, but let's not worry about isolated exceptions at the beginning. We need to rule out the other possibilities for $n$.
By the contents of the answer above, $n\equiv 2\pmod{4}$ is ruled out apart from a single exception. Now we need to rule out $n \equiv 1$ and $n\equiv 3$ (modulo $4$). That would make $n$ odd. In that case, again by the answer above, $2^n \equiv -1\pmod{3}$. So if $k \equiv 1\pmod{3}$, we find that $k2^n+1\equiv 0\pmod{3}$. With the single exception of $k=1$, $n=1$, this means that $k2^n+1$ cannot be prime.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Help with convergence in distribution $Y$ is a random variable with $$M(t) = \frac{1}{(2-\exp(t))^s}.$$
Does $$\frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$$ converge in distribution as $s$ tends to infinity?
I let $Z = \frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$.
Differentiating the MGF of $Y$ and let $t = 0$ we have $$E(Y) = s,$$$$E(Y^2) = s^2 +2s,$$$$\operatorname{Var}(Y) = 2s.$$
Thus Mgf of Z:
\begin{align*}E(\exp(tz)) &= E\left(\exp\left(\frac{t(y-s)}{\sqrt{2s}}\right)\right)\\
&=E\left(\exp\left(\frac{ty}{\sqrt{2s}}\right)\exp\left(\frac{-ts}{\sqrt{2s}}\right)\right)\\
&=E\left(\exp\left(\frac{ty}{\sqrt{2s}}\right)\exp\left(\frac{-ts}{\sqrt{2s}}\right)\right)\\
&=\exp\left(\frac{-ts}{\sqrt{2s}}\right)\frac{1}{\left(2-\exp\left(\frac{t}{\sqrt{2s}}\right)\right)^s}.
\end{align*}
But I'm not sure how to proceed - seems to me it tends to $0$?
| As you have determined:
$$
M_Z(t) = \mathbb{E}(\exp(t Z ) ) = \frac{\exp\left(-t \frac{s}{\sqrt{2s}}\right)}{\left( 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) \right)^s} = \left( \frac{\exp\left(-t \frac{1}{\sqrt{2s}}\right)}{ 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) } \right)^s
$$
In order to determine large $s$ behavior, use Taylor series:
$$
\frac{\exp\left(-t \frac{1}{\sqrt{2s}}\right)}{ 2 - \exp\left(\frac{t}{\sqrt{2s}} \right) } = \frac{1- \frac{t}{\sqrt{2s}} + \frac{t^2}{4 s} + o\left(\frac{t^2}{s}\right)}{ 1- \frac{t}{\sqrt{2s}} - \frac{t^2}{4 s} + o\left(\frac{t^2}{s}\right) } = 1 + \frac{t^2}{2s} + o\left(\frac{t^2}{s}\right)
$$
Thus
$$
M_Z(t) = \left( 1 + \frac{t^2}{2s} + o\left(\frac{t^2}{s}\right) \right)^s
$$
The large $s$ limit is now easy to find:
$$
\lim_{s\to \infty} M_Z(t) = \exp\left( \frac{t^2}{2} \right)
$$
This is the mgf of the standard normal variable, as expected by CLT.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find the area of the region, bounded by various curves?
Find the area of the region bounded by the curves $y=x^2$ and $y=x$.
Find the area of the region bounded by the curves $y=x^2+1$ and $y=2$
I have a ton of questions like this and I have been graphing them and then splitting them into intervals and adding them up but this is giving me an answer thats a little off and its taking forever....is there a faster way? Also I am stuck on $y=x^2+1$ and $y=2$
because I dont know what region they want..I see $y=2$ as a line intersecting $x^2+1$, when I graph it.
| First we will find the area of the region bounded by the curves:
$y = x^2$ ... (i)
and $y = x $ ... (ii)
To determine the shaded area between these two curves, we need to sketch these curves on a graph.
Now, we will find the area of the shaded region from O to A.
Area of Shaded Region Between Two Curves :
$A = \displaystyle \int _a^b [f(x)-g(x)] \;dx$
Where, $f(x)$ is the top curve
$g(x)$ is the bottom curve
$a$ (Lower limit) = $x$ coordinate of extreme left intersection point of the area to be found.
$b$ (Upper limit) = $x$ coordinate of extreme right intersection point of the area to be found.
So, $f(x) = y = x$
$g (x) = y = x^2$
We need to find the limits, $a$ and $b$.
How to find the limits ?
Since limits, $a$ and $b$, are the $x$ coordinates of the intersection points, So, we will find the intersection points of the given curves.
Put the value of $y$ from equation (ii) into equation (i)
$x=x^2$
$x^2-x=0$
$x(x-1)=0$
$ x=0, x = 1 $
Put these values in equation (ii)
$y = 0, \; y = 1$
Thus, the points of intersection are $O(0,0)$ and $A (1,1)$
$\therefore \;a=0, \;b=1$
Area between Curves :
The are will be, $A = \displaystyle\int _a^b [f(x)-g(x)] dx$
$A=\displaystyle \int_0^1\; (x-x^2)\;dx$
$A=\displaystyle \int_0^1\; x\;dx-\displaystyle \int_0^1\; x^2\;dx$
$
=\left ( \dfrac {x^2}{2}\right)_0^1-\left ( \dfrac {x^3}{3}\right)_0^1
$
On putting limits,
$
=\left ( \dfrac {1}{2}-0\right)-\left ( \dfrac {1}{3}-0\right)
$
$=\dfrac {1}{2}-\dfrac {1}{3}$
$A=\dfrac {1}{6}$
(II) Now, we will find the shaded area bounded by the curves:
$y = x ^2 + 1$ ... (iii)
$y = 2$ ... (iv)
Curves on Graph :
We will find the area of the shaded region from $A$ to $B$
Here,
$ f(x) = y = 2$
$g(x) = y = x^2+1 $
Finding the limits using intersection points :
Put the value of $y$ in equation (iii)
$2 = x^2+1 $
$ x^2 = 1 $
$ x = -1, x = 1$
Put these values in equation (iii)
$y = 2, y = 2 $
Thus, the points of intersection are $A (-1, 2)$ and $B (1, 2)$
$ \therefore \;a=-1, \;b=1$
Area between Curves :
The area will be,
$A = \displaystyle\int _a^b [f(x)-g(x)]dx $
$A=\displaystyle \int_{-1}^{1}\; [2-(x^2+1)]\;dx$
$A=\displaystyle \int_{-1}^{1}\; 1\;dx-\displaystyle \int_{-1}^{1}\; x^2\;dx$
$=\left ( x \right)_{-1}^{1}-\left ( \dfrac {x^3}{3}\right)_{-1}^{1}$
On putting limits, we get,
$= (1+1)- \left( \dfrac {1}{3}+\dfrac {1}{3}\right)$
$=2-\dfrac {2}{3}$
$A=\dfrac {4}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Help with integrating $\int \frac{t^3}{1+t^2} ~dt$ What am I doing wrong on this integration problem?
$$
\begin{align*}
\int\frac{t^3}{1+t^2} &=
\frac14 t^4 (\ln(1+t^2) (t+\frac13 t^3))
\\ &= \frac14 t^4(t \ln(1+t^2)+\frac13 t^3 \ln(1+t^2)
\\ &= \frac14 t^5 \ln(1+t^2)+\frac{1}{3}t^7 \ln(1+t^2)
\end{align*}$$
Answer should be $\frac{1}{2}(t^2-\ln(t^2+1))$. I'm way off
Any help appreciated.
Thanks!
| Your integral is:
$$\int \frac{t^3}{1+t^2} dt$$
Substitute: $x = 1+t^2$ and thus $dx = 2t dt$. Then the above transforms to:
$$\int \frac{t^3}{1+t^2} dt = \frac{1}{2} \int \frac{t^2 \ 2t dt}{(1+t^2)}$$
Using the transformation suggested earlier, we can re-write the right hand side as:
$$\frac{1}{2} \int \frac{(x-1) \ dx}{x}$$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed.
Can anyone help me solve it?
| Let $\arctan{\sqrt{x}} = \theta$. Then we have $x = \tan^2 (\theta)$. Hence, $$\frac{x-1}{x+1} = \frac{\tan^2 (\theta)-1}{\tan^2 (\theta)+1} = \sin^2(\theta) - \cos^2(\theta) = - \cos(2 \theta)= \sin \left(2 \theta - \frac{\pi}{2} \right)$$
Hence, $$2 \arctan{\sqrt{x}} = 2 \theta = \arcsin \left(\frac{x-1}{x+1} \right) + \frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/89759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to find the equation of the tangent line to $y=x^2+2x-4$ at $x=2$?
I'm given a curve $$y=x^2+2x-4$$
How do I find the tangent line to this curve at $x = 2$?
| General formulas
$y=f(x)=x^2+2x-4$
$m=f'(x)=2x+2$ Slope equation
Point-Slope equation of a line
$(y-y_0)=m_0\cdot(x-x_0)$
Compute actual values
$x_0 = 2$
$y_0=f(x_0)=f(2)=(2)^2+2(2)-4=4+4-4=4$
$m_0=f'(x_0)=f'(2)=2(2)+2=4+2=6$
Substitute values into the line equation
$y-4=6\cdot(x-2)$
$y=6\cdot(x-2)+4$
$y=6x-8$
General Formula for a tangent line to a Parabola @ point $x_0$ on the Parabola
$y=f'(x_0)\cdot(x-x_0)+f(x_0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/92165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
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Proving that $ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+...+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1°)}$ I would like to show that the following trigonometric sum
$$ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+\cdots+\frac{1}{\sin(133°)\sin(134°)}$$
telescopes to $$\frac{1}{\sin(1°)}$$
We have:
$$\begin{align}
\sin(45°)\sin(46°)&=\frac{1}{2}(\cos(1°)+\sin(1°))\\
\sin(47°)\sin(48°)&=\frac{1}{2}(\cos(1°)+\sin(5°))\\
\sin(49°)\sin(50°)&=\frac{1}{2}(\cos(1°)+\sin(9°))\\
&\ \vdots\\
\sin(133°)\sin(134°)&=\frac{1}{2}(\cos(1°)+\sin(177°))
\end{align}$$
So the sum is:
$$\begin{align}
\sum_{k=0}^{44} &\frac{2}{\cos(1°)+\sin(1+4k)}
=\frac{2}{\cos(1°)+\sin(1°)}+\frac{2}{\cos(1°)+\sin(5°)}+\\
&\kern2.5in +\frac{2}{\cos(1°)+\sin(9°)}+\cdots+\frac{2}{\cos(1°)+\sin(177°)}.
\end{align}$$
Although I don't think this new expression simplifies the problem.
| $$\frac{\sin(1^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=\frac{\sin((x+1)^\circ-x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}=$$
$$\frac{\sin((x+1)^\circ) \cos (x^\circ)}{\sin(x^\circ) \sin(x+1)^\circ}-\frac{\sin(x^\circ) \cos(x+1)^\circ}{\sin(x^\circ) \sin(x+1)^\circ}= \cot(x^\circ)-\cot(x+1)^\circ$$
Add them and you get your telescopic sum ;)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Show $|x|<|x^2+b|$ for real $x$ and $b>\frac{1}{4}$ How do I show $|x|<|x^2+b|$ for real $x$ and $b>\frac{1}{4}$ ?
I mean it's clear for $x=0$ and $|x|>1$ but what is if $0<|x|<1$?
Please help!
| You only have to prove it for $x\geq 0$, since if true for $x$ it is true for $-x$.
If $x\geq 0$, then $|x|=x$ and $|x^2+b|=x^2+b$ and you only need to prove that $x^2+b>x$.
But $$x^2-x+b = x^2 - x + \frac{1}{4} + (b-\frac{1}{4}) = (x-\frac{1}{2})^2 + (b-\frac{1}{4})$$
Since $b>\frac{1}{4}$, this means that $x^2-x+b > 0$, so $x^2+b > x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit $\lim_{n\to+\infty} (1-\frac1{2^2})(1-\frac1{3^2})\cdot \cdots \cdot(1-\frac{1}{n^2})$ and series $\sum_{n=2}^{\infty} \ln(1-\frac1{n^2})$
Possible Duplicate:
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$
Compute:
\begin{align*}
\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})
\end{align*}
Well, I do so:
$$\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\lim_{n\to+\infty}\prod_{j=2}^n (1-\frac{1}{j^2})$$
let $$a_n = \prod_{j=2}^n (1-\frac{1}{j^2})\quad\Rightarrow\quad \ln a_n = \ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)$$
so:
$$\ln\left(\prod_{j=2}^n (1-\frac{1}{j^2})\right)=\sum_{j=2}^{\infty} \ln (1-\frac{1}{j^2})$$
consider
$$\sum_{n=2}^{\infty} \ln(1-\frac{1}{n^2})$$
but as you study this series??
| If you write it like
$$
\lim_{n\to \infty}\prod_{j=2}^n \frac{(j+1)(j-1)}{j^2}
$$
you get
$$
\lim_{n\to \infty}\frac{1}{2}\frac{(n+1)!(n-1)!}{(n!)^2},
$$
where factorials cancel in the limit and give $\frac{1}{2}$, because a $2$ is missing in the numerator to get $(n+1)!$.
| {
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"question_score": "1",
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Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
| Note that
$$I=\int_{0}^{1}\frac{x^4+1}{x^6+1}\,dx\quad\stackrel{\large x\,\mapsto\,\frac{1}{x}}\Longrightarrow\quad I=\int_{1}^{\infty}\frac{x^4+1}{x^6+1}\,dx\quad\Longrightarrow\quad I=\frac{1}{2}\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,dx$$
Using (proof can be seen here)
$$\int_0^\infty\dfrac{x^{\large a-1}}{1+x^b}\ dx=\frac{\pi}{b}\csc\left(\frac{a\pi}{b}\right)$$
then
$$I=\frac{1}{2}\int_{0}^{\infty}\frac{x^4+1}{x^6+1}\,dx=\frac{1}{2}\left[\frac{\pi}{6}\csc\left(\frac{5\pi}{6}\right)+\frac{\pi}{6}\csc\left(\frac{\pi}{6}\right)\right]=\frac{\pi}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$.
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
| We have
$$
\begin{align}
n^3+6n^2+11n+6
&=6\binom{n}{3}+18\binom{n}{2}+18\binom{n}{1}+6\binom{n}{0}\\
&=6\left(\binom{n}{3}+3\binom{n}{2}+3\binom{n}{1}+\binom{n}{0}\right)
\end{align}
$$
so $6\mid(n^3+6n^2+11n+6)$ for all $n\in\mathbb{Z}$.
Of course, since $3\mid 6$, we have $3\mid(n^3+6n^2+11n+6)$, as requested.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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about Pythagorean quadruples Respected Mathematicians,
I would like to prepare a function, which will generate Pythagorean quadruples (a, b, c, d) = $d^2$ = $a^2$ + $b^2$ + $c^2$...-> (1). How far I am correct I don't know. For some set of integers a and b, consider $a^2$ + $b^2$ = m and d = c + ε. Now by (1), c = (m - $ε^2$)/2ε)....>(2) Now I will discuss the generation of such quadruples.
For a is even and b is odd or b is even and a is odd;
(1) Let us take a = 12 and b = 15 then m = 369 = $3^2$ X 41
then ε = 1, $3^2$ but ε = 41 is not possible as $ε^2$ > m.
Now for c = 184 and d = 185 for ε = 1; for c = 16 and d = 25 when $3^2$ = ε;
So the primitive quadruples for a = 12 and b = 15 are (12, 15, 184, 185) and also (12, 15, 16, 25).
(2) For a = 2 X 3 X 5 X 7 = 210 and b = $3^3$ X 5 = 135 then m = $3^2$ X $5^2$ X 277 = 62325 then ε = 1, $3^2$, $5^2$, $3^2$ X $5^2$. Other combination of ε are not possible due to $ε^2$ > m.
If I am correct with the cited examples, let me know the generalization of the Pythagorean quadruples.
Note: I considered a and b have common factors $p_1$, ...$p_n$.
Thanking you.
BABA
| Let us only deal with primitive triples where the forms are
$A=2n+1\quad B=4n\quad CC=4n=1$. We begin with Euclid's formula shown here as $ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2.\quad$
Now we choose any odd number greater that one $(15?)$ for the $A$-value and solve the $A$-function for $k$, testing a defined range of $m$-values to see which ones yield integers.
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\qquad\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=15\implies \lfloor\sqrt{15+1}\rfloor=4\le m \le \frac{15+1}{2} =8\\
\land\quad m\in\{4,8\}\implies k \in\{1,7\} $$
$$F(4,1)=(15,8,17)\qquad \qquad F(8,7)=(15,112,113) $$
Now we can use the same procedure to find new triples where $A=17$ or $A=113.\quad$ Doing so we find $F(9,8)=(17,144,145)\implies 15^2+8^2+144^2=145^2\quad $ or
$\qquad \quad F(57,56)=(113,6384,6385)\implies 15^2+112^2+6384^2=6385^2.$
The logic is that side $A$ of the new triple is replaced by $\sqrt{A^2+B^2}$ of the original triple, changing the new triple into a quadruple.
| {
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Factorizing polynomial $x^5+x+1$ I'm given a problem to factorize
$$ P(x)=x^5+x+1 $$
I've done the following:
$$ P(x)=(x^5+x^4+x^3)-(x^4+x^3+x^2)+(x^2+x+1)= (x^2+x+1)(x^3-x^2+1)$$
Is it possible to prove that this cannot be factorized any further?
| It depends what you are factoring over. Notice that if we use the quadratic formula, $x^2 + x + 1$ has roots $-\frac{1}{2} \pm \frac{i\sqrt{3}}{2}$, so it can be factorized as
$$(x^2 + x + 1) = (x - (-\frac{1}{2} + \frac{i\sqrt{3}}{2}))(x - (-\frac{1}{2} - \frac{i\sqrt{3}}{2}))$$
If you are factoring over just the real numbers, finding the roots shows that this polynomial (at least the $x^2+x+1$ portion) cannot be reduced further.
The $x^3-x^2+1$ portion can be factored further over the reals because it has one real root, but it is not further reducible over just the rational numbers.
So the point at which a polynomial has no further factors depends on the set over which you are factoring.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere.
Can anybody provide a hint as to how I should be looking at this problem?
| The extended Euclidean algorithm gives
$$
9 = (2 n + 1)(n^3+1)+(-2 n^2 - n + 4)(n^2+2)
$$
and so $\gcd(n^3 + 1, n^2 + 2)$ divides $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Square on the curve A square is drawn on the curve $y = x^3 + 27\cdot x^2 + 8\cdot x + 91$. What would be the area of the square. All the points of the square should lie on the curve.
All four points lie on the curve (not the sides)
| Time for some reverse engineering. Suppose we start with a square; translate it so that it is centred at the origin. Then for some $s$ and $\theta$, the vertices are $(s \cos(\theta), s \sin(\theta))$, $(s \sin(\theta), -s \cos(\theta))$, $(-s \cos(\theta), -s \sin(\theta))$ and $(-s \sin(\theta), s \cos(\theta))$, and the area is $2 s^2$. The cubic that interpolates these points is $$f(x) = \frac{4}{s^2 \sin(4 \theta)} x^3 - \frac{3 + \cos(4 \theta)}{\sin(4 \theta)} x$$
Note that for $b \ge \sqrt{8}$, $b = (3 + \cos(4 \theta))/\sin(4 \theta)$ has real solutions, namely $$\sin(4 \theta) = \frac{3b \pm \sqrt{b^2-8}}{b^2 - 1}$$ Note also that these solutions have $\sin(4 \theta) > 0$. We can then take $s=2/\sqrt{\sin(4\theta)}$ to make the cubic $f(x) = x^3 - b x$, and the area is
$$2s^2 = \frac{8}{\sin(4\theta)} = 3 b \mp \sqrt{b^2 - 8}$$ The given problem has $b = 235$ and area $705 \mp \sqrt{55217}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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solving some complex equations I want to solve the following three equations for $x,y,z$:
$$\begin{eqnarray}k_1&=&\frac{x+y+z}{yz}\\
k_2&=&\frac{x^2+y^2+z^2}{y^2z^2}\\
k_3&=&\frac{x^3+y^3+z^3}{y^3z^3}\end{eqnarray}$$
where $k_1,k_2,k_3$ are constants.
Is there any kind of standard methods?
| If you set $\frac{x}{yz} = p$, $\frac{1}{y} = q$, $\frac{1}{z} = r$, then your equations are
$p + q + r = k_1$
$p^2 + q^2 + r^2 = k_2$
$p^3 + q^3 + r^3 = k_3$
Using $(p+q+r)^2 - (p^2 + q^2 + r^2) = 2(pq + qr + rp)$ we get
$pq + qr + rp = (k_1^2 - k_2)/2$
Similarly we can find the value of $pqr$.
Thus we can find the polynomial $(x-p)(x-q)(x-r)$, by finding its coefficients. Once we have the polynomial, since it is a cubic, it is solvable by standard methods.
Once we find $p,q,r$, finding $x,y,z$ should not be difficult (keep in mind the permutations possible).
See: Newton's identities for a way to find the coefficients.
See: Roots of cubic for a formula for the roots of cubic.
| {
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Using double angle formulas in integration, trouble following an example. I have just started looking at integration and I am having trouble understanding what has been done in one of the examples in the book I am working through.
It involves using the double angle formula for $\sin(2\theta)$ to provide a rearrangement for which an indefinite integral can then be found.
The double angle formula provided is $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ and the example is as follows:
$$\int\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)dx=\int\frac{1}{2}\sin\left(x\right)dx$$
$$=-\frac{1}{2}\cos\left(x\right)+c$$
The part of this example I am specifically stuck with is the first line where $\cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)$ is rewritten as $\frac{1}{2}\sin\left(x\right)$ using the previously stated double angle formula.
| Consider $\theta = \frac{1}{2}x$. Then
$$ \cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)
= \cos \theta \sin \theta$$
Notice that the double angle formula could be written:
$$ \sin \theta \cos \theta = \frac{1}{2}\sin(2\theta)$$ So the integrand is now:
$$ \cos\left(\frac{1}{2}x\right)\sin\left(\frac{1}{2}x\right)
= \cos \theta \sin \theta = \frac{1}{2} \sin(2\theta) = \frac{1}{2}\sin\left(2\cdot \frac{1}{2}x\right) = \frac{1}{2}\sin x$$
Hope this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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How can one prove that $\sum_{k=1}^{\infty}\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}=\frac{\pi}{4}-1$? A question from some Russian book, about different summations an integrals, by Prudnikhov, Brichkhov and Marichev.
Page 746, 20, they write:
$$ \sum_{k=1}^{\infty}\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}=\frac{\pi}{4}-1 $$
Without proof, actually the whole book is with no single proof.
How could I start proving that, what is product of $\prod_{m=1}^{2k}\text{ctg}\frac{m\pi}{2k+1}$?
Thanks!
| I believe we can show that
$$\prod_{m=1}^{2k} \cot \frac{m\pi}{2k+1} = (-1)^k \frac{1}{2k+1}$$
Using Chebyshev Polynomials $\displaystyle T_n(x)$ (of degree $\displaystyle n$, and leading coefficient $\displaystyle 2^{n-1}$), which satisfy
$$T_{n}(\cos x) = \cos (n x)$$
and
$$T_{2n+1}(\sin x) = (-1)^k\sin ((2n+1) x)$$
We get the product of roots of $\displaystyle T_{2k+1}(x) = -1$, to find $\displaystyle \prod_{m=1}^{2k+1} (-1)^k\cos \frac{m\pi}{2k+1}$
(Note that its roots are $\displaystyle \cos (\frac{(2r+1)\pi}{2k+1})$ and $\displaystyle -\cos (\frac{2r\pi}{2k+1})$)
which gives us $$\prod_{m=1}^{2k} \cos \frac{m\pi}{2k+1} = \frac{(-1)^k}{2^{2k}}$$
To find $\displaystyle \prod_{m=1}^{2k} \sin \frac{m\pi}{2k+1}$, we need to find the product of roots of $\displaystyle \frac{T_{2k+1}(x)}{x} = 0$.
We can prove that the coefficient of $\displaystyle x$ in $\displaystyle T_{2k+1}(x)$ is $\displaystyle 2k+1$ and that would give us
$$\prod_{m=1}^{2k} \sin\frac{m\pi}{2k+1} = \frac{2k+1}{2^{2k}}$$
Thus
$$\prod_{m=1}^{2k} \cot\frac{m\pi}{2k+1} = \frac{(-1)^k}{2k+1}$$
Since $$1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} = \int_{0}^{1} \frac{1}{1+x^2} \text{ dx} = \frac{\pi}{4}$$
we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate using Partial Fraction decomposition, completing the square The given problem is $\int{x\over x^3-1}dx$.
I know this equals
$${1\over3}\int {1\over x-1}-{x-1\over x^2+x+1}dx,$$
which can be separated into
$${1\over3}\int {1\over x-1}dx - {1\over3}\int{x+(1/2)-(3/2)\over x^2+x+1}dx.$$ This can further be separated into $${1\over3}\int{1\over x-1}dx - {1\over3}\int{x+(1/2)\over x^2+x+1}dx + {1\over2}\int{1\over(x+(1/2))^2+(3/4)}dx.$$
I know the integral of ${1\over3}\int {1\over x-1}dx$ is $(1/3)\ln(x+1)+C$ where $C$ is an arbitrary constant. Using u-subsitution, where $u=x^2+x+1$ and $du=(2x+1)dx$ and $(1/2)du=(x+(1/2))dx$, I know the integral of $-{1\over3}\int{ x+(1/2)\over x^2+x+1 }dx$ is $(1/6)\ln(x^2+x+1)+C$. I need to get the last part, $${1\over2}\int{1\over(x+(1/2))^2+(3/4)}dx$$ to be some form of arctan. I can use u-substitution where $u=x+(1/2)$ and $du=dx$, but I don't know where to go from there.
| The purpose is to get an integral of the form $\displaystyle\int \frac{1}{v^{2}+1}\,dv=\arctan v$. Using the substitution indicated by you $u=x+1/2$, $du=dx$,
we obtain
$$\begin{eqnarray*}
\frac{1}{2}\int \frac{1}{\left( x+1/2\right) ^{2}+3/4}dx &=&\frac{1}{2}\int
\frac{1}{u^{2}+3/4}\,du \\
&=&\frac{1}{2\cdot 3/4}\int \frac{1}{\frac{u^{2}}{3/4}+1}\,du \\
&=&\frac{2}{3}\int \frac{1}{\left( \frac{u}{\sqrt{3}/2}\right) ^{2}+1}\,du,
\end{eqnarray*}$$
where we have manipulated the integrand so that it is of the form $
1/(v^{2}+1)$, with $v=\dfrac{u}{\sqrt{3}/2}$. Using this new substitution,
since $dv=\frac{du}{\sqrt{3}/2}$, the integral becomes
$$\begin{eqnarray*}
\frac{2}{3}\int \frac{1}{\left( \frac{u}{\sqrt{3}/2}\right) ^{2}+1}\,du &=&%
\frac{2}{3}\int \frac{1}{v^{2}+1}\cdot \frac{\sqrt{3}}{2}\,dv \\
&=&\frac{\sqrt{3}}{3}\int \frac{1}{v^{2}+1}\,dv \\
&=&\frac{\sqrt{3}}{3}\arctan v \\
&=&\frac{\sqrt{3}}{3}\arctan \left( \frac{u}{\sqrt{3}/2}\right) \\
&=&\frac{\sqrt{3}}{3}\arctan \left( \frac{x+1/2}{\sqrt{3}/2}\right) \\
&=&\frac{\sqrt{3}}{3}\arctan \left( \frac{\sqrt{3}}{3}\left( 2x+1\right)
\right) +\text{Const.}
\end{eqnarray*}$$
Added. We could have used the single substitution $v=\dfrac{x+1/2}{\sqrt{3}/2}$,
resulting from both substitutions $v$ and $u$.
$$\begin{eqnarray*}
\frac{1}{2}\int \frac{1}{\left( x+1/2\right) ^{2}+3/4}dx &=&\frac{1}{2}\int
\frac{2}{3}\frac{\sqrt{3}}{v^{2}+1}\,dv \\
&=&\frac{1}{3}\sqrt{3}\arctan v=\frac{1}{3}\sqrt{3}\arctan \left( \frac{x+1/2
}{\sqrt{3}/2}\right) +\text{ Const.}
\end{eqnarray*}$$
| {
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Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
| Note: Using Newton's identities, we can calculate the below expressions more easily, following the easy recursive definition.
But, your idea of writing as roots of third degree polynomial works I believe, but requires some work and we show that here:
Let $\displaystyle x,y,z$ be roots of $\displaystyle t^3 + at - b = 0$. We have that $\displaystyle a = xy+yz+zx$ and $\displaystyle b = xyz$.
Since $\displaystyle t^3 = b - at$, multiply by $\displaystyle t^4$ we get $\displaystyle t^7= bt^4 - a t^5$.
Setting $\displaystyle t=x,y,z$ in turn and adding gives us
$\displaystyle x^7 + y^7 + z^7 = b(x^4 + y^4 + z^4) - a(x^5 + y^5 + z^5)$
Similar to above, we get $\displaystyle t^5 = bt^2 - a t^3$, setting $\displaystyle t=x,y,z$ and adding gives us
$\displaystyle x^5 + y^5 + z^5 = b(x^2 + y^2 + z^2) - a(x^2 + y^3 + z^3)$.
Similarly we get
$\displaystyle x^3 + y^3 + z^3 = 3b$
We also have $\displaystyle (x+y+z)^2 = 0$, giving us
$\displaystyle x^2 + y^2 + z^2 = -2a$.
Thus
$\displaystyle x^5 + y^5 + z^5 = -2ab - 3ab = -5ab$.
Now $\displaystyle t^4 = bt - at^2$ and in a similar fashion we get
$\displaystyle x^4 + y^4 + z^4 = -a(x^2+y^2 +z^2) = 2a^2$
Thus $\displaystyle xyz(x^4 + y^4 + z^4) = 2a^2 b$
and $\displaystyle x^7 + y^7 + z^7 = 2a^2b - (-5a^2b) = 7a^2 b$.
Thus the given expression is $\displaystyle \frac{7}{2}$.
This approach can be used to generate identities.
For instance, show that
$\displaystyle 10(x^7 + y^7 + z^7) = 7(x^2 + y^2 + z^2)(x^5 + y^5 + z^5)$
| {
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Zeroes of a Particular Function I'm looking for the zeroes of $f(k) = e^{\sqrt{k}}[\frac{s}{k} - \frac{d}{\sqrt{k}}] - 1$ on the set $k > 0$. Is there a nice way to describe the set of solutions for given $s$ and $d$? Thanks!
| When the ratio $s/d$ is small, we can use the Lagrange inversion formula (see, e.g., my answer here) to solve $e^x(s-dx) = x^2$ (as per Arturo's comment) for $x$ as a power series in either $s$ or $d$:
$$\begin{align}x &= \frac{s}{d}-\frac{s^2}{d^3}+ \left(\frac{1}{d^4}+\frac{2}{d^5}\right) s^3-\left(\frac{1}{2 d^5}+\frac{5}{d^6}+\frac{5}{d^7}\right) s^4 + \cdots \\ &= \frac{s}{d}-\frac{s^2}{d^3}+\frac{s^3}{d^4}+\frac{4 s^3-s^4}{2 d^5} -\frac{30 s^4-s^5}{6 d^6} + \cdots\end{align}$$
Replacing $x$ by $\sqrt{k}$ will then yield solutions to the original problem.
If $s$ and $d$ are related in some way then perhaps more appropriate estimates could be obtained.
| {
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How to simplify this square $(3 \times 4 + 2)^2$? If I have this $(3 \times 4 + 2)^2$,
How can I simplify it with out the final result.
Do I distribute the $^2$ over each number like this:
$(3^2 \times 4^2 + 2^2)$?
What is the rule?
| $(3 \times 4 + 2)^2 = (12 + 2)^2 = 14^2 =196$ while $(3^2 \times 4^2 + 2^2) = 9 \times 16 +4 = 144+4 = 148$, so that does not work.
If you want a rule for squares of sums, try: $$(x+y)^2 = x^2 + 2 x y +y^2.$$
| {
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For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$
Find the roots of:
$z^2 - 3z + (3 - i) = 0$
$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$
$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$
So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I proceed?
| $$\begin{cases}
x^2-y^2-3x+3=0 \\
2xy-3y-1=0
\end{cases}$$
$$x = \frac {3y+1}{2y} \Rightarrow \left(\frac {3y+1}{2y}\right)^2-y^2-3 \cdot \frac {3y+1}{2y}+3=0 \Rightarrow$$
$$\Rightarrow (3y+1)^2-4y^4-3(3y+1)\cdot 2y+12y^2=0 \Rightarrow$$
$$\Rightarrow 4y^4-3y^2-1=0$$
Substitute $~y^2=t~$ and solve for $t$ over Reals .
| {
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"url": "https://math.stackexchange.com/questions/119626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do we get the result of the summation $\sum\limits_{k=1}^n k \cdot 2^k$?
Possible Duplicate:
Formula for calculating $\sum_{n=0}^{m}nr^n$
Can someone explain step by step how to derive the following identity? $$\sum_{k=1}^{n} k \cdot 2^k = 2(n \cdot 2^n - 2^n + 1) $$
| Although you got great answers, I will add another way to look at the expression you are looking for
Let us denote
$$ S = \sum_{k=1}^{n} k \cdot 2^k $$
This can be expanded as
$$
\begin{align*}
S &= 2 + 2 . 2^2 + 3 . 2^3 + \cdots n 2^n\\
2S &= \hspace{8pt}+1.2^2+2.2^3 + \cdots (n-1) . 2^n+n . 2^{n+1}
\end{align*}
$$
If you notice, by multiplying by $2$ and writing under similar terms and then subtracting, we get
$$
\begin{align*}
S = n . 2^{n+1} - \left(2+2^2+2^3+\cdots+2^n \right) &= n .2^{n+1}-2^{n+1}+2\\
&= 2(n . 2^n-2^n+1)
\end{align*}
$$
| {
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Using antiderivative to calculate complex integral $$\int_{1}^{3}(z-2)^3 dz $$
I get the following -
$$\frac{1}{4}[(3-2)^3 - (1-2)^3] = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
However the answer sheet I have just show it reduced to $\frac{1}{4} - \frac{1}{4} = 0$
Cant see how they are getting a - instead of a +...what am I missing?
| The integral of $(z-2)^3$ is $\frac14 (z-2)^4+c$ not $\frac14 (z-2)^3+c$, so you should get $$\tfrac{1}{4}(3-2)^4 - \tfrac{1}{4}(1-2)^4 =0.$$
| {
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Prove for any positive real numbers $a,b,c$ $\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$ Since the problem sheets says I should use Cauchy-Schwarz inequality, I used
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3}$
$\geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$
I first multiplied each term by $a,b,c$ to get a perfect square on top like
$\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$
$=\frac{a^4}{a(a^2+ab+b^2)}+\frac{b^4}{b(b^2+bc+c^2)}+\frac{c^4}{c(c^2+ca+a^2)}$
$\hspace{120pt}\geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} $
But I am still stuck for few hours. This is not a homework, but is a set of problems to prepare for AMC/USAMO.
(Note: I started off with a general question and got closed as off topic. I have another genuine Math Question now, and I am just trying to see if math.se is going to be useful)
| Let $S$ be the sum on the left hand side and $T$ be the sum:
$\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}$
Then $S-T = (a-b)+(b-c)+(c-a)=0$
So it suffices to show $S+T \geq \frac{2(a+b+c)}{3}$. We can achieve this by showing $\frac{a^3+b^3}{a^2+ab+b^2} \geq \frac{a+b}{3}$
The inequality is equivalent to $a^2+b^2\geq 2ab$.
| {
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The net signed area between $t=0, y=0, t=x$, and $y=f(t)$ f(t) is continuous function.So
I know that $\int _0^x {f(t) dt}=$ "The net signed area between $t=0, y=0, t=x$, and $y=f(t)$"
And I can find the same result with endless small rectangulars areas method.
"The net signed area between $t=0, y=0, t=x$, and $y=f(t)$"=$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})$
Therefore,
$$\int _0^x {f(t) dt}=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})$$
Is there any other analytical method (such as transform methods or any other one) to prove that the equation is correct?
EDIT:During my attempts to proof the equality, I have noticed that the equality can be proved via power series expression of $f(x)$. I did not think that way before asking the question. I would like to share it with you. It is also welcome your comments about my approach and waiting for your different methods to show a proof via another method as well.
$ f(x) =f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^2}{2!}+.....=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $
$$\int _0^x {f(t) dt}=\int _0^x(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n)dt=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\int _0^x t^n dt)=\sum_{n=0}^{\infty} (\frac{f^{(n)}(0)}{n!}\frac{x^{n+1}}{n+1})$$
$$\int _0^x {f(t) dt}=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^{n+1}}{(n+1)!}$$ $$(1)$$
$$f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m$$
$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.
More information about summation http://en.wikipedia.org/wiki/Summation
$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n \sum_{m=0}^{\infty} \frac{f^{(m)}(0)}{m!} (\frac{kx}{n})^m=\lim_{n\to\infty} \frac{x}{n}\sum_{m=0}^{\infty} \frac{x^m}{n^m} \frac{f^{(m)}(0)}{m!} \sum \limits_{k=1}^n k^m=\lim_{n\to\infty} \frac{x}{n}[f(0)n+\frac{f'(0)x}{n 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^2}{n^2 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^3}{n^3 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^4}{n^4 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]=
\lim_{n\to\infty} [f(0)x+\frac{f'(0)x^2}{n^2 1!}(\frac{n^2}{2}+\frac{n}{2})+ \frac{f''(0)x^3}{n^3 2!}(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})+\frac{f'''(0)x^4}{n^4 3!}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+\frac{f^{(4)}(0)x^5}{n^5 4!}(\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30})+...... ]= [f(0)x+\frac{f'(0)x^2}{ 2!}+ \frac{f''(0)x^3}{ 3!}+\frac{f'''(0)x^4}{ 4!}+\frac{f^{(4)}(0)x^5}{ 5!}+...... ]$$
$$\lim_{n\to\infty} \frac{x}{n}\sum \limits_{k=1}^n f(\frac{kx}{n})=\sum_{m=0}^{\infty} \frac{f^{(m)}(0)x^{m+1}}{(m+1)!}$$
$$(2)$$
Equation $(1)$ and equation $(2)$ are equal to each other. The proof is completed.
| I suppose you're assuming $f$ is continuous. The statement is easily seen to be true for $f(t)=t$, and by subtracting a multiple of this we can assume $f(0) = f(x)$. By the Stone-Weierstrass theorem, trigonometric polynomials with period $x$ are uniformly dense in the continuous functions with period $x$, and so it suffices to prove the result for $f(t) = \cos(2 \pi j t/x)$ and $\sin(2 \pi j t/x)$ for each integer $j$. But for those it can be done explicitly: $\sum_{k=1}^n \sin(2 \pi j k/n) = 0$ for all integers $j$, $\sum_{k=1}^n \cos(2 \pi j k/n) = n$ if $j$ is a multiple of $n$, $0$ otherwise...
| {
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Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
How do you find the sum of these and prove it by induction? Can someone help me get through this?
| The answer from lhf has already dealt with the second question posed here. Here is a Wikipedia article about it.
Here's a probabilistic approach to the first question. When you toss a coin $n$ times, the probability that a "head" appears exactly $k$ times is $\dbinom n k (1/2)^n$. The average number of times a "head" appears is therefore
$$
\left( \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n \right) \left(\frac 1 2 \right)^n.
$$
But the average number of times a "head" appears is obviously $n/2$. Therefore
$$
\left( \binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n \right) \left(\frac 1 2 \right)^n = \frac n 2.
$$
Consequently
$$
\binom n 0 \cdot 0 + \binom n 1 \cdot 1 + \binom n 2 \cdot 2 + \cdots + \binom n k \cdot k + \cdots + \binom n n \cdot n = n 2^{n-1} .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show $ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$ Show $$ I = \int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = \frac{\pi}{\sqrt 2}$$
| Based on KV Raman,$\displaystyle{\frac{1}{1+\cos^2 x}}$ is even function.
$$\int_0^{\pi} \frac{\mathrm{d}x}{1+\cos^2 x} = 2 \int_0^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+\cos^2 x}$$
$$
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{1}{1+\cos^2 x} \,\mathrm{d}x &=\int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{\sin^2 x+2\cos^2 x}\\
&= \int_0^{\frac{\pi}{2}}\frac{\sec^2 x\mathrm{d}x}{\tan ^2 x+2}\\
&=\int_0^{\frac{\pi}{2}}\frac{\mathrm{d}\tan x}{\tan ^2 x +2}\\
\end{align*}
$$
below is the same
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| You are almost done: If $n \equiv 0, \pm 1 \pmod 5$ we are done (as we have $5 \mid n$, $5 \mid n-1$ or $5 \mid n+1$ then). So suppose $n \equiv \pm 2 \pmod 5$, but then $n^2 + 1 \equiv (\pm 2)^2 + 1 \equiv 0 \pmod 5$, hence $5 \mid n^2 + 1$, and so $5 \mid n^5 -n$ also in this case.
| {
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How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
| Look at $(x-1)(x-4)$ and $(x-2)(x-3)$ they multiply as $(x^{2}-5x+4)$ and $x^{2}-5x+6$. Now put $t= x^{2}-5x$ and reduce it to a quadratic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
| If $f(x) = \frac{x}{2} + \frac{1}{4}\sin(2x)$, then $f'(x) =$ $\frac{1}{2} + \frac{cos(2x)}{2}$. If you want $\int{f(x)}$, then we have $\frac{x^2}{4} - \frac{cos(2x)}{8}+C$. Here, we differentiate/integrate $\frac{x}{2}$ and $\frac{1}{4}\sin(2x)$ separately. Do you have an initial condition?
| {
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Tiling pythagorean triples with minimal polyominoes Given a Pythagorean triple $(a,b,c)$ satisfying $a^2+b^2=c^2$, how to calculate the least number of polyominoes of total squares $c^2$, needed, such that both the square $c^2$ can be build by piecing them together, as well as the two separate squares of side length $a$ and $b$?
| WLOG let $(a, b, c)=1$. Then there is an upper bound of $2+a+b-c$. This bound is sharp for the pair $(3, 4, 5)$, and all the other pairs I've tested. It is attainable as follows:
Let one piece be a $a\times a$ square, and another be a $b\times b$ square with a $(a+b-c)\times (a+b-c)$ square removed from a corner. Now, note that in the $c\times c$ square, there are $2$ blocks remaining, each $(c-a)\times (c-b)$. In the pair of smaller squares, there is a $(a+b-c)\times (a+b-c)$ square remaining.
Now let $(c-a, c-b)=d$. Thus $d^2|2(c-a)(c-b)=(a+b-c)^2\implies d|a+b-c$. So $d|a, d|b\implies d=1$. This, together with $2(c-a)(c-b)=(a+b-c)^2$, means that $c-a, c-b$ are, in some order, $2p^2$ and $q^2$ for $(p, q)=1$ and $2pq=a+b-c$. Now each of the $(c-a)\times (c-b)$ blocks can be dissected into $pq$ equally sized blocks, each $2p\times q$ in dimension. These can be reassembled into a $2pq\times 2pq$ block, as desired.
This gives a total of $2+2pq=2+a+b-c$ blocks. In the example of $a=8, b=15, c=17$, this method produces the following set:
-1 $8\times 8$ block
-1 $15\times 15$ block with an upper corner of $6 \times 6$ missing
-6 $2\times 3$ blocks
For a total of $2+(8+15-17)=8$ blocks.
Note: If $(a, b, c)=d>1$, then this upper bound is just $2+\frac{a+b-c}{d}$.
Note: $(a_1, a_2, ...)$ denotes the $\gcd$ of $a_1, a_2, ...$.
| {
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Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that $a_n$ converges to $c$. Let $a_{n}$ be a sequence such that $(a_{n})^{2}=ca_{n-1}$ where ($c>0,a_{1}>0$).Prove that the sequence converges to $c$.
My first problem was to find some terms of the sequence to verify that point and show that converges and converges to that point using some convergence criterion.
Could someone help me through this problem?
| Note that calculating some values gives:
$$\Large \eqalign{
& {a_1} = a \cr
& {a_2} = \sqrt c \sqrt a \cr
& {a_3} = \sqrt c \root 4 \of c \root 4 \of a \cr
& {a_4} = \sqrt c \root 4 \of c \root 8 \of c \root 8 \of a \cr} $$
In general you can prove that
$$\Large {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots+ \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^{n-1}}}}}}$$
Since on has
$$\Large\mathop {\lim }\limits_{n \to \infty } \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \frac{1}{{{2^{n - 1}}}} = 1$$
and
$$\Large\mathop {\lim }\limits_{n \to \infty } \root n \of a = 1$$
for any real $a>0$, it is immeadiate that
$$\Large\lim {a_n} = {c^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \frac{1}{{{2^{n - 1}}}}}}{a^{\frac{1}{{{2^n}}}}} = {c^1} = c$$
| {
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Curve In a Closed Interval with an Infinite Length How can I show that the length of the curve
$$f(x)=\begin{cases}
\sqrt{x}\cos(\pi/x) & \text{ if } x\neq0\\
0 & \text{ if } x=0
\end{cases}$$
is infinite on $[0,1]$?
I tried using the arc length formula, but ended up with the very nasty integral:
$$L=\int_0^1\sqrt{1+\left(\frac{x\cos(\pi/x)+2\pi\sin(\pi/x)}{2x^{3/2}}\right)^2}dx.$$
Perhaps the way to go here is to use the definition of arc length I defined in this thread, but I am not certain.
Do you guys have any ideas?
| Choosing $x$ so that
$$\pi/x \;\; = \;\; \frac{3 \pi}{2}, \; \frac{4 \pi}{2},\; \frac{5 \pi}{2},\; \frac{6 \pi}{2},\; \frac{7 \pi}{2},\; \frac{8 \pi}{2},\; \frac{9 \pi}{2},\; \frac{10 \pi}{2}, \; \frac{11 \pi}{2},\; \frac{12 \pi}{2},\; \frac{13 \pi}{2},\; ...$$
so that
$$x \;\; =\;\; \frac{2}{3}, \; \frac{2}{4}, \; \frac{2}{5}, \; \frac{2}{6}, \; \frac{2}{7}, \; \frac{2}{8}, \; \frac{2}{9}, \; \frac{2}{10}, \; \frac{2}{11}, \; \frac{2}{12}, \; \frac{2}{13}, \; ... $$
we get
$$\sqrt{x} \cos\left(\pi/x\right) \; \; = \;\; 0, \;\; \sqrt{\frac{1}{2}}, \;\; 0, \;\; -\sqrt{\frac{1}{3}}, \;\; 0, \;\; \sqrt{\frac{1}{4}}, \;\; 0, \;\; -\sqrt{\frac{1}{5}}, \;\; 0, \;\; \sqrt{\frac{1}{6}}, \;\; 0, \; ...$$
Now, if you plot these points (not necessarily to scale, just make a rough hand sketch), it will be clear that the sum of the lengths of the inscribed segments corresponding to the partition
$$\cal{P}_{n} \;\; =\;\; \left\{1, \;\frac{2}{3}, \; \frac{2}{4}, \; \frac{2}{5}, \; ..., \; \frac{2}{2n}, \; 0\right\}$$
is greater than
$$\sqrt{\frac{1}{2}} \;\; + \;\; \sqrt{\frac{1}{3}} \;\; + \;\; ... \;\; + \;\; \sqrt{\frac{1}{n}}$$
Note that this last sum diverges to $+\infty$ as $n \rightarrow \infty$, as it corresponds to the $p$-series from elementary calculus for $p = \frac{1}{2}.$
| {
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The roots of $t^5+1$ Just a quick question, how do we go about finding the roots of $t^5 +1$?
I can see that since $t^5=-1$ that an obvious root is $\sqrt[5]{-1}$.
I am assuming that since there is a $-1$ involved, some of the factors will be complex?
Any help would be welcome.
| Since no one has yet pointed this out, it may be of interest to show how J.M.'s comment leads to a purely algebraic solution (i.e. no use made of trigonometric or exponential function ideas).
Using
$$t^5 + 1 \;\; = \;\; \left(t+1\right)\left(t^4 - t^3 + t^2 - t + 1\right),$$
it follows that the solutions to $t^5 + 1 = 0$ are $t=-1$ along with the solutions to
$$t^4 - t^3 + t^2 - t + 1 \;\; = \;\; 0$$
Dividing both sides of this last equation by $t^2$ gives
$$t^2 \; - \; t \; + \; 1 \; - \; \frac{1}{t} \; + \; \frac{1}{t^2} \;\; = \;\; 0$$
Rearranging terms gives
$$\left(t^2 \; + \; \frac{1}{t^2} \right) \; - \; \left(t \; + \; \frac{1}{t} \right) \; + 1 \; = \; 0$$
This is a reciprocal equation. (An equation with the property that if $t=r$ is a solution, then $t = \frac{1}{r}$ is also a solution. The phrase recurring equation was often used in the early and mid 1800s English literature.) Therefore, we make the substitution $u = t + \frac{1}{t}$ (note that $u^2 = t^2 + 2 + \frac{1}{t^2}$), which leads to
$$(u^2 - 2) \; - \; u \; + \; 1 \; = \; 0$$
$$u^2 - u - 1 \; = \; 0$$
Now use the quadratic formula to solve for $u$:
$$ u \;\; = \;\; \frac{-(-1) \; \pm \; \sqrt{(-1)^2 \; - \; 4(1)(-1)}}{2(1)} \;\; = \;\; \frac{1 \; \pm \; \sqrt{5}}{2}$$
This leads to the following two equations:
$$t \; + \; \frac{1}{t} \;\; = \;\; \frac{1 \; + \; \sqrt{5}}{2}$$
and
$$t \; + \; \frac{1}{t} \;\; = \;\; \frac{1 \; - \; \sqrt{5}}{2}$$
These two equations can be rewritten as:
$$2t^2 \; - \; \left(1 + \sqrt{5}\right)t \; + \; 2 \;\; = \;\; 0$$
and
$$2t^2 \; - \; \left(1 - \sqrt{5}\right)t \; + \; 2 \;\; = \;\; 0$$
Using the quadratic formula, the first of these equations has the following solutions:
$$ t \;\; = \;\; \frac{(1 + \sqrt{5}) \; \pm \; \sqrt{(6 + 2\sqrt{5}) \; - \; 16}}{4}$$
$$ t \;\; = \;\; \frac{(1 + \sqrt{5}) \; \pm \; i\sqrt{(10 - 2\sqrt{5}}}{4}$$
$$ t \;\; = \;\; \frac{1}{4}\left( 1 + \sqrt{5}\right) \; \pm \; \left(\frac{1}{4} \sqrt{10 - 2\sqrt{5}}\right)i$$
In the same way, the second of the these equations has the following solutions:
$$ t \;\; = \;\; \frac{1}{4}\left( 1 - \sqrt{5}\right) \; \pm \; \left(\frac{1}{4} \sqrt{10 + 2\sqrt{5}}\right)i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$ So the question I am trying to work through is:
Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$
for convergence.
The solution (using D'Alembert's ratio test) is:
$$u_n=\frac{n^3}{3^n}\;,$$
so
$$\begin{align*}
\frac{|u_{n+1}|}{|u_n|} &=\frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3}\;.
\end{align*}$$
How do we get from there to...
$$=\frac{n^3+3n^2+3n+1}{3n^3}$$
What happens with $3^n$ in the numerator and power of $n+1$ in the denominator? How do they cancel out?
Also, in the very next step that all goes to being equal to
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=\frac{1}{3}<1\;,$$
which means the series is convergent.
But how do we get to $\dfrac{1}{3}$?
| *
*You have a factor of $3^n$ in the numerator, and a factor of $3^{n+1}$ in the denominator. So
$$\frac{3^n(\text{stuff})}{3^{n+1}(\text{other stuff})} = \frac{3^n(\text{stuff})}{3\times 3^{n}\text{(other stuff)}} = \frac{\text{stuff}}{3(\text{other stuff})}.$$
Since $3^{n+1}=3\times 3^n$.
*Dividing numerator and denominator by $n^3$, we have
$$\begin{align*}
\lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{3n^3} &= \lim_{n\to\infty}\frac{\frac{1}{n^3}(n^3+3n^2+3n+1)}{\frac{1}{n^3}(3n^3)}\\
&= \lim_{n\to\infty}\frac{1 + \frac{3}{n} + \frac{3}{n^2}+\frac{1}{n^3}}{3}\\
&= \frac{\lim\limits_{n\to\infty}(1 + \frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3})}{\lim\limits_{n\to\infty}3}\\
&= \frac{1 + 0 + 0 + 0}{3} = \frac{1}{3}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Equation with trigonometry To solve $\sin^3x+\cos^3x=1$
So I just thought of a solution like:
Let $\sin x=t$. Then we have:
$$t^3+(1-t^2) \sqrt{1-t^2} =1 \\ (1-t^2) \sqrt{1-t^2} =1-t^3 \\ (1-t^2)^3=(1-t^3)^2 \\ (1-t)^3(1+t)^3=(1-t)^2(1+t+t^2)^2 \\ (1-t)^2\left[ (1-t)(1+t)^3-(1+t+t^2)^2 \right]=0$$
Which is followed by $\sin x=0$ or $\sin x=1$. However, this solution seems very easy to make a silly mistake in (with all the squares and cubes of differences and sums). Is there any easier solution to this?
| Note that $\sin x$ and $\cos x$ are always at most $1$, so $\sin^3 x+\cos^3 x\leq \sin^2 x+\cos^2 x=1$ with equality only $\sin x,\cos x$ are $0$ or $1$.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determining the Smith Normal Form
Consider the integral matrix
$$R = \left(\begin{matrix} 2 & 4 & 6 & -8 \\ 1 & 3 & 2 & -1 \\ 1 & 1 & 4 & -1 \\ 1 & 1 & 2 & 5 \end{matrix}\right).$$
Determine the structure of the abelian group given by generators and relations.
$$A_r = \{a_1, a_2, a_3, a_4 | R \circ \vec{a} = 0\}$$
I know you have to row/column reduce the matrix however am unsure what to do next.
| We do want to use a kind of Gaussian elimination, but you have to be careful since you should not multiply a row by anything other than $1$ and $-1$, and you should not add non-integer multiples of one row to another row. So we can get started simply enough:
$$\begin{align*}
\left(\begin{array}{rrrr}
2 & 4 & 6 & -8 \\
1 & 3 & 2 & -1 \\
1 & 1 & 4 & -1 \\
1 & 1 & 2 & 5
\end{array}\right)
&\to
\left(\begin{array}{rrrr}
1 & 1& 2 & 5\\
1 & 3 & 2 & -1\\
1 & 1 & 4 & -1\\
2 & 4 & 6 & -8
\end{array}\right)
&&\to \left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 2 & 2 & -18
\end{array}\right)\\
&\to \left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 2 & -12
\end{array}\right)
&&\to\left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 0 & -6
\end{array}\right)\\
&\to\left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 0 & 6
\end{array}\right).
\end{align*}$$
This uses only elementary row operations.
From this we see that the relations on your group are equivalent to:
$$\begin{array}{rcccccccl}
r_1&+&r_2&+&2r_3&+&5r_4 &= & 0\\
& &2r_2 & & & - & 6r_4 &=& 0\\
& & & & 2r_3 & - &6r_4 & = & 0\\
& & & & & &6r_4 & = & 0
\end{array}$$
These elementary row operations replace the relations on our original set of generators with a new set of relations which are equivalent to the original, in the sense that if the generators satisfy these relations, then they satisfy the original relations and vice-versa.
We can now use elementary column operations, which also correspond to performing certain base changes. For example, subtracting five times the first column from the fourth column is equivalent to replacing $r_1$ with $r_1-5r_4$, which does not change the subgroup generated by $r_1,r_2,r_3,r_4$. Etc. Performing those elementary column operations, since the $(1,1)$ entry is the gcd of the entries on the first row (and similarly for the rest of the rows), we can eliminate all nondiagonal entries and end up with the diagonal matrix
$$\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 6
\end{array}\right),$$
from which you can read off the structure of the group in question.
The column operations corresponds to replacing our set of generators with a new set that generates the same group. For example, the operations we performed with the first column, subtracting the first column from the second, twice the first column from the third, and five times the first column from the fourth, correspond to replacing the original generator $r_1$ with the generator $r_1-r_2-r_3-5r_4$. This does not change the subgroup, because $$\langle r_1,r_2,r_3,r_4\rangle = \langle r_1-r_2-r_3-5r_4,r_2,r_3,r_4\rangle.$$
Similarly with the other operations. In the end we will have an abelian group generated by elements $a,b,c,d$, where, in terms of the original generators, we have
$$\begin{align*}
a & = r_1-r_2-r_3-5r_4\\
b &= r_2-3r_4\\
c &= r_3-3r_4\\
d &= r_4,
\end{align*}$$
which yields the relations
$$\begin{align*}
a&=0\\
2b&=0\\
2c&=0\\
6d&=0
\end{align*}$$
from which we can just read off the abelian group structure as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/144589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Interesting Determinant Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1<x_2<\cdots<x_n$.
Define \begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & \cdots & x_{n-1}-x_{1} & x_{n}-x_{1} \\
x_{2}-x_{1} & 0 & \cdots & x_{n-1}-x_{2} & x_{n}-x_{2} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
x_{n-1}-x_{1} & x_{n-1}-x_{2} & \cdots & 0 & x_{n}-x_{n-1} \\
x_{n}-x_{1} & x_{n}-x_{2} & \cdots & x_{n}-x_{n-1} & 0%
\end{bmatrix}%
\end{equation*}
Could you determine the determinant of $A$ in term of $x_1,x_2,\ldots,x_n$?
I make a several Calculation:
For $n=2$, we get
\begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} \\
x_{2}-x_{1} & 0%
\end{bmatrix}%
\text{ and}\det (A)=-\left( x_{2}-x_{1}\right) ^{2}
\end{equation*}
For $n=3$, we get
\begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & x_{3}-x_{1} \\
x_{2}-x_{1} & 0 & x_{3}-x_{2} \\
x_{3}-x_{1} & x_{3}-x_{2} & 0%
\end{bmatrix}%
\text{ and}\det (A)=2\left( x_{2}-x_{1}\right) \left( x_{3}-x_{2}\right)
\left( x_{3}-x_{1}\right)
\end{equation*}
For $n=4,$ we get
\begin{equation*}
A=%
\begin{bmatrix}
0 & x_{2}-x_{1} & x_{3}-x_{1} & x_{4}-x_{1} \\
x_{2}-x_{1} & 0 & x_{3}-x_{2} & x_{4}-x_{2} \\
x_{3}-x_{1} & x_{3}-x_{2} & 0 & x_{4}-x_{3} \\
x_{4}-x_{1} & x_{4}-x_{2} & x_{4}-x_{3} & 0%
\end{bmatrix} \\%
\text{ and} \\ \det (A)=-4\left( x_{4}-x_{1}\right) \left( x_{2}-x_{1}\right)
\left( x_{3}-x_{2}\right) \left( x_{4}-x_{3}\right)
\end{equation*}
Finally, I guess that the answer is $\det(A)=2^{n-2}\cdot (x_n-x_1)\cdot (x_2-x_1)\cdots (x_n-x_{n-1})$. But I don't know how to prove it.
| Expanding Robert solution.
Let $det(A) = P(x)$. Let the polynomial on the right is a multi-variable polynomial $P(x)$.
If $x_1 = x_2$ then $det(A) = 0$ i.e $P(x) = 0$ i.e. $(x_1 - x_2)$ is a factor of $P(x)$.
If $x_2 = x_3$ then $det(A) = 0$ i.e $P(x) = 0$ i.e. $(x_2 - x_3)$ is a factor of $P(x)$.
etc. We calculate possible factors of $P(x)$. Have we calculated all possible factors of $P(x)$?
Let $Q(x) = (x_1 - x_2) (x_2 - x_3) \ldots (x_{n} - x_{1}) $
What we know about the degree of $P(x)$? It is $n$, equal to that of $Q(x)$. Thus $Q(x)$ multiplied by some constant factor should give us $P(x)$ i.e. we already have all possible factors of $P(x)$.
A Robert has already mentioned, we should calculate this constant factor.
It follows that if for any $i$, $x_i = x_{i+1}$, then $P(x) =0$ i.e. $det(A) = 0$. Since you alretady have constraints such as $x_1 >x_2 \ldots x_n$, $det(A) \ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/144818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\cos(x)$ and $\arccos(x)$ couple limit
Find the value of the following limit:
$$\lim_{n\to\infty} \frac {\cos 1 \cdot \arccos \frac{1}{n}+\cos\frac
{1}{2} \cdot \arccos \frac{1}{(n-1)}+ \cdots +\cos \frac{1}{n} \cdot
\arccos{1}}{n}$$
| What follows is a little hand-wavy, and I wish I had more rigorous demonstration, but the post is too big for a comment.
$$ \begin{eqnarray}
\frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) &=& \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right)
\end{eqnarray}
$$
We can now split the sum in two parts, $1 \leqslant k \leqslant \lfloor\frac{n}{2}\rfloor$ and $\lfloor\frac{n}{2}\rfloor < k \leqslant n$. In each of these parts, either $\sin$, or $\arcsin$ will be small, and in the limiting value will be $\frac{\pi}{2}$:
In[28]:= Table[
N[1/n Sum[Cos[1/k] ArcCos[1/(n - k + 1)], {k, 1, n}],
50], {n, {5000, 10000, 100000}}] // N
Out[28]= {1.56861, 1.56963, 1.57066}
More rigorously:
$$
\frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) = \frac{\pi}{2} - \frac{\pi}{n} \sum_{k=1}^n \sin^2 \frac{1}{k} - \frac{1}{n} \sum_{k=1}^n \arcsin\frac{1}{k} + \frac{2}{n} \sum_{k=1}^n \sin^2\left( \frac{1}{k}\right) \arcsin\left(\frac{1}{n+1-k}\right)
$$
Now:
$$
0 \leqslant \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sin^2\left(\frac{1}{k}\right) \leqslant
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k^2} = 0
$$
$$
0 \leqslant \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \arcsin\left(\frac{1}{k}\right) \leqslant \lim_{n \to \infty} \frac{\pi}{2n} \sum_{k=1}^n \frac{1}{k} = \lim_{n \to \infty} \frac{\pi}{2 n} \ln(n) = 0
$$
$$
0 \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \sin^2 \left( \frac{1}{k} \right) \arcsin\left(\frac{1}{n+1-k} \right) \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} \frac{1}{n+1-k} \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$ I can not get the correct answer.
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\int \sec \theta$$
$$\ln| \sec \theta + \tan \theta|$$
Then I solve for $\theta$:
$x = 4 \tan \theta$
$x/4 = \tan \theta$
$\arctan (\frac{x}{4}) = \theta$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$
This is wrong and I do not know why.
| Cosmetically nicer taking $$ x = 4 \sinh t \; , $$
so that $$ dx = 4 \cosh t \; dt $$
The quadratic formula is enought ot give us
$$ t = \log \left( \frac{x + \sqrt{x^2 + 16}}{4} \right) = \log \left( x + \sqrt{x^2 + 16} \right) - \log 4 $$
Then we get
$$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \int 1 dt = t + C = \log \left( x + \sqrt{x^2 + 16} \right) - \log 4 + C \; , $$ or
$$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \log \left( x + \sqrt{x^2 + 16} \right) + C_2 \; . $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$
Now I am not sure what to do.
| Suppose $\frac{5x+1}{(2x+1)(x-1)} = \frac{A}{2x+1} + \frac{B}{x-1}$. Then $5x+1 = A(x-1)+B(2x+1) = (A+2B)x + (B-A)$. So by comparing coefficients, we get $A+2B = 5$ and $B-A = 1$. Solving this gives $A = 1$ and $B = 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
how many 5-digit numbers satisfy the following conditions How many five-digit numbers divisible by 11 have the sum of their digits equal to 30?
I am able to get the 5-digit numbers divisible by 11
and
I am also able to get the five-digit numbers whose sum of their digits equal to 30.
But i am not able to get how i can get the count of 5 digit numbers satisying both the condition.
Thanks in advance.
Thanks in advance.
combinatorics permutations
| An integer written in ordinary base ten notation is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. That is, if the digits of a five-digit number are $d_1,d_2,d_3,d_4,d_5$ from left to right, the number is a multiple of $11$ if and only if $d_1-d_2+d_3-d_4+d_5$ is a multiple of $11$. Let $x=d_1+d_3+d_5$ and $y=d_2+d_4$; you need to choose the digits so that $x+y=30$ and $x-y$ is a multiple of $11$. Since digits must be between $0$ and $9$ inclusive, the smallest possible value of $x-y$ is $-18$ (or $-17$ if you don’t allow leading zeroes), and the largest possible value is $27$. Thus, the only possible multiples of $11$ are $-11,0,11$, and $22$. We’ll consider each of them separately.
If $x-y=-11$, we must solve the system $$\begin{cases}x+y=30\\x-y=-11\;;\end{cases}$$ adding the equations yields $2x=19$, which is clearly impossible, since $x$ and $y$ must be integers. A similar problem arises if $x-y=11$, so we abandon that possibility as well.
If $x-y=22$, adding the equations yields $2x=52$, so $x=26$, and $y=4$. The only way to have three digits that sum to $26$ is to have two $9$’s and an $8$; two digits that sum to $4$ must be a $0$ and $4$, $1$ and $3$, or a pair of $2$’s. There are $3$ ways to arrange the first, third and fifth digits: $8\_9\_9$, $9\_8\_9$, and $9\_9\_8$. The other two digits can be chosen and arranged in $5$ ways: $\_0\_4\_$, $\_4\_0\_$, $\_1\_3\_$, $\_3\_1\_$, and $\_2\_2\_$. Thus, there are $3\cdot5=15$ numbers of the desired type in this case.
If $x-y=0$, we have $x=y=15$. To finish the problem, you need only count the number of ways to pick three digits adding up to $15$ for the first, third, and fifth slots, and two more adding up to $15$ for the remaining two slots. Clearly there are just $4$ ways to do the latter: $\_6\_9\_$, $\_9\_6\_$, $\_7\_8\_$, and $\_8\_7\_$. Can you finish it off from here?
| {
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Equivalent of $ u_{n}=\sum_{k=1}^n (-1)^k\sqrt{k}$ I'm trying to show that $$ u_{n}=\sum_{k=1}^n (-1)^k\sqrt{k}\sim_{n\rightarrow \infty} (-1)^n\frac{\sqrt{n}}{2}$$ when $n\rightarrow\infty$
How can I first show that $$u_{2n}\sim_{n\rightarrow \infty} \frac{\sqrt{2n}}{2}$$ and then deduce the equivalent of $u_{n}$?
| Pair consecutive terms together:
$$u_{2n} = \sum_{k=1}^{2n} (-1)^k \sqrt{k} = \sum_{k=1}^n \left( \sqrt{2k}-\sqrt{2k-1}\right) .$$
Since $\displaystyle \sqrt{1 - \frac{1}{2k}} = 1 - \frac{1}{4k} + \mathcal{O}(k^{-2})$ and $\displaystyle \sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + \mathcal{O}(n^{p-1})$ we get
$$u_{2n}= \sum_{k=1}^n \left(\frac{\sqrt{2}}{4\sqrt{k}} + \mathcal{O}(k^{-3/2})\right)= \frac{\sqrt{2n}}{2} + \mathcal{O}(1).$$
Now $u_{2n+1} = u_{2n} - \sqrt{2n+1}$ so the result that actually holds is $\displaystyle u_n \sim (-1)^n \frac{\sqrt{n}}{2}.$
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
| Your integral is OK.
$${\ln |\sec x| + \frac{{\cos 2x}}{4} + C}$$
$${\ln \left|\frac 1 {\cos x}\right| + \frac{{1+\cos 2x}}{4} + C}-\frac 1 4$$
$${-\ln \left| {\cos x}\right| + \frac 1 2\frac{{1+\cos 2x}}{2} + K}$$
$${-\ln \left| {\cos x}\right| + \frac 1 2 \cos ^2 x + K}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Formula to estimate sum to nearly correct : $\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$ Estimate the sum correct to three decimal places :
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$
This problem is in my homework. I find that n = 22 when use Maple to solve this. (with some programming) But, in my homework, teacher said find the formula for this problem.
Thanks :)
| With a little bit of approximation, we can achieve the desired $5\times 10^{-4}$ accuracy with fewer terms.
Notice
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^3} = -\left[\sum_{n=1}^{\infty}\frac{1}{n^3}- \sum_{n=1}^\infty\frac{2}{(2n)^3}\right] = -\frac34\sum_{n=1}^\infty \frac{1}{n^3} = -\frac34\zeta(3)$$
If we want to estimate the sum at the left accurate up to $5\times 10^{-4}$,
it just suffices to estimate $\zeta(3)$ accurate up to $6.67\times 10^{-4}$.
In the sum of $\zeta(3)$, if we pick a $m$ and replace $\displaystyle\;\frac{1}{n^3}$ by $\displaystyle\;\frac{1}{n^3-n}$ for $n \ge m$, the error introduced $\mathcal{E}_m$ is
$$\mathcal{E}_m =_{def} \sum_{n=m}^\infty\left(\frac{1}{n^3-n} - \frac{1}{n^3}\right)
= \sum_{n=m}^\infty \frac{1}{(n^2-1)n^3} \le \frac{m^2}{m^2-1}\sum_{n=m}^\infty \frac{1}{n^5}
$$
Since $\displaystyle\;\frac{1}{x^5}$ is a convex function, we have
$$\frac{1}{n^5} \le \int_{n-1/2}^{n+1/2} \frac{dx}{x^5}
\quad\implies\quad \mathcal{E}_m \le \frac{m^2}{m^2-1}\int_{m-1/2}^\infty \frac{dx}{x^5}
= \frac{m^2}{4(m^2-1)(m-1/2)^4}
$$
For $m = 5$, we have $\displaystyle\;\mathcal{E}_5 = \frac{25}{39366} \approx 6.35\times 10^{-4}\;$. This is already good enough for our purposes. Notice
$$\sum_{n=m}^{\infty}\frac{1}{n^3-n}
= \sum_{n=m}^{\infty}\frac{1}{(n-1)n(n+1)}
= \sum_{n=m}^{\infty}\left(\frac{1}{2(n-1)n}-\frac{1}{2n(n+1)}\right)
= \frac{1}{2(m-1)m}
$$
We find the original sum is approximately equal to
$$-\frac34 \left( 1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{2\cdot 4\cdot5} \right) = -\frac{10391}{11520} \approx -0.90199653$$
with an error smaller than $5\times 10^{-4}$.
Compare this with the exact value of the sum $\approx -0.90154268$, the difference
$\approx -4.5385 \times 10^{-4}$ is indeed smaller than $5 \times 10^{-4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is wrong.
I must be setting up the $A, B, C$ thing wrong but I do not know why.
| I suppose that you could have also used trigonometric substitutions. Completing the square as you have done, we get that
$$\displaystyle \int \frac{dx}{x^2-2x} = \int \frac{dx}{(x-1)^2 - 1}.$$
We now let $x-1 = \sec \theta.$ Notice then that $dx = \tan \theta \sec \theta \ d\theta.$ We now have
$$ \int \frac{dx}{(x-1)^2 - 1} = \int \frac{\tan \theta \sec \theta \ d\theta}{\sec^2\theta-1} = \int \frac{\tan \theta \sec \theta \ d\theta}{\tan^2 \theta} = \int \frac{\sec \theta \ d\theta}{\tan \theta} = \int \csc \theta \ d\theta. $$
And,
$$ \int \csc \theta \ d\theta = \ln (\csc \theta - \cot \theta) + C = \ln \left(\frac{\sqrt{x-2}}{\sqrt{x}} \right) + C,$$ as desired.
(This is because from $x-1 = \sec \theta,$ we get $1/(x-1) = \cos \theta$ and thus $\sin \theta = (\sqrt{x^2-2x})/(x-1)$, and we just make the necessary substitutions in $\csc \theta - \cot \theta$).
| {
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Another Congruence Proof I've been asked to attempt a proof of the following congruence. It is found in a section of my textbook with Wilson's theorem and Fermat's Little theorem. I've pondered the problem for a while and nothing interesting has occurred to me.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2} (\text{mod} p)$
| Here $p$ must be an odd prime.
There are two cases to consider, $p\equiv 1 \pmod 4$ and $p\equiv 3 \pmod{4}$. We deal with the first case.
We know that $(p-1)!\equiv -1\pmod{p}$. Rearrange the numbers from $1$ to $p-1$, so that we get the odd numbers from $1$ on going up, interleaved with the even numbers from $p-1$ going down. For example, if $p=13$, we arrange the numbers from $1$ to $12$ in the order $1$, $12$, $3$, $10$, $5$, $8$, $7$, $6$, $9$, $4$, $11$, $2$.
In general the listing is $1$, $p-2$, $3$, $p-3$, $5$, $p-5$, and so on until at the end we get to $p-2$, followed by $p-(p-2)$. Now take the product, in that order, noting that $p-k\equiv -k \pmod{p}$.
We get that
$$(p-1)!\equiv (1)(-1)(3)(-3)(5)(-5)\cdots (p-2)(-(p-2))\equiv -1\pmod{p}.\tag{$1$}$$
The number of even numbered entries in the product is $(p-1)/2$. These have minus signs in front of them. Gather the minus signs together. We get
$$(-1)^{(p-1)/2} 1^23^25^2\cdots (p-2)^2 \equiv -1\pmod{p}.$$
Note that $(p-1)/2$ is even. So we get that
$$1^23^25^2\cdots (p-2)^2 \equiv -1\pmod{p}.$$
Now we are finished, since $-1=(-)^{(p+1)/2}$.
The argument for $p\equiv 3\pmod{4}$ is essentially the same. In fact the two arguments could be gathered into one. The main difference is that in $(1)$, the number of minus signs, which is $(p-1)/2$, now turns out to be odd. So we get
$$-1^23^25^2\cdots (p-2)^2 \equiv -1\pmod{p}.$$
or equivalently
$$1^23^25^2\cdots (p-2)^2 \equiv 1\pmod{p}.$$
Since in this case we have $(-1){(p+1)/2}=1$, again the result follows.
Remark: The above shows the useful result that if $p\equiv 1\pmod{4}$, then there is an $x$ such that $x^2\equiv -1\pmod{p}$. Indeed it gives an explicit expression for such an $x$. Regrettably, the expression is not computationally useful if $p$ is large.
| {
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Given $3$ dice, which number is the most likely to appear Given $3$ dice, what is the value of the sum of the number of the $3$ dice most likely to appear? I know that by symmetry, there would always be $2$ different values with the same probability to appear, however I don't know which pairs appear with the highest chance.
| Note that there are a total of $6 \times 6 \times 6 = 216$ options. The possible sums are from $3$ to $18$.
Also note that the distribution has to be symmetric since if a roll gives $x,y,z$ adding to $n$, then $7-x,7-y,7-z$ adds to $21-n$.
Hence, the number of ways of getting $n$ is same as the number of ways of getting $21-n$. We would hence expect the maximum to occur for $10$ and $11$.
Number of ways to get $3$ is $1$ i.e. $\dbinom{2}{2}$, which is the same as the number of ways to get $21-3=18$.
Number of ways to get $4$ is $3$ i.e. $\dbinom{3}{2}$, which is the same as the number of ways to get $21-4=17$.
Number of ways to get $5$ is $6$ i.e. $\dbinom{4}{2}$, which is the same as the number of ways to get $21-5=16$.
Number of ways to get $6$ is $10$ i.e. $\dbinom{5}{2}$, which is the same as the number of ways to get $21-6=15$.
Number of ways to get $7$ is $15$ i.e. $\dbinom{6}{2}$, which is the same as the number of ways to get $21-7=14$.
Number of ways to get $8$ is $21$ i.e. $\dbinom{7}{2}$, which is the same as the number of ways to get $21-8=13$.
Number of ways to get $9$ is $25$ i.e. $\dbinom{8}{2}-3$, which is the same as the number of ways to get $21-9=12$.
Number of ways to get $10$ is $27$ i.e. $\dbinom{9}{2} - 3 - 6$, which is the same as the number of ways to get $21-10=11$.
Now by symmetry you can get the number of ways to get the sum between $11$ and $18$.
Hence, the distribution peaks at $10$ and $11$ as expected.
As a sanity check, we have $2 \left( 1 + 3 + 6 +10 +15 + 21 + 25 + 27\right) = 216$.
Below is the distribution of the number of times a number occurs as a sum of three dices, where the $X$-axis the sum of the three dice and $Y$-axis is the number of times the sum occurs.
| {
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A linear differential equation Find the general solution to:
$$x'' + 2c \; x' + \left( \frac{2}{\cosh^2 t} - 1\right)x =0$$
where $c$ is a constant.
| When $c=0$ we can guess that $x_1 = \frac{1}{\cosh t}$ is a solution. Then by inserting $x = x_1 \cdot u$ into equation it simplifies to:
$$u'' - 2 \tanh t \;u' = 0$$
From which we conclude that: $x(t) = \frac{C_1}{\cosh t} + C_2 \left(\sinh t + \frac{t}{\cosh t} \right)$.
For more general case when $c \neq 0$ we use substitution $x = w\,e^{-ct}$. The equation transforms to:
$$w'' = \left(1 + c^2 - \frac{2}{\cosh^2 t} \right)w \quad \quad (1)$$
the trick is to use Darboux theorem. Let's say we have two functions $y$ and $z$ that are solutions to:
$$y'' = p(t)y, \quad z'' = \left(p(t) - \lambda \right)z$$
then the function:
$$\overline{y} = y' - \frac{yz'}{z}$$
satisfies:
$$\overline{y}'' = \left( p - 2 \frac{d}{dt} \left(\frac{z'}{z} \right) \right)\overline{y} \quad\quad(2)$$
Taking $z=\cosh t$ and $p=1+c^2$ equation $(2)$ turns out to be $(1)$.
We solve:
$$y'' = (1+c^2) y \Rightarrow y = C_1 e^{\sqrt{1+c^2} t} + C_2 e^{-\sqrt{1+c^2} t}$$
then we plug in for $\overline{y} = w$ and $x$ which yields the final result:
$$x = C_1 \left(\sqrt{1+c^2} - \tanh t \right)e^{\sqrt{1+c^2} t - ct} + C_2 \left(\sqrt{1+c^2} + \tanh t \right)e^{-\sqrt{1+c^2} t - ct}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the limit at x=0 Find the limit of $f(x)=\frac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x}-\sqrt{a+x}}$ (at x=0) so that $f(x)$ becomes continuous for all $x$. My answer is $2\sqrt{a}$. Am I right?
Sorry to state the question incorrectly. We have to define $f(x)$ at $x=0$ such that $f(x)$ is continuous for all $x$. In that case my answer is $2\sqrt{a}$.
Let f(x)=$[x]$+$[-x]$ be a function where [.] stands for the greatest integer not greater than x. For any integer $m$, what can we say about $lim_{x \to m}$. Is $f(x)$ contiuous at $x=m$. Sorry for asking such vague question but I am forgetting the exact wording and the options given. This was a question asked in a class test.
| The answer should be $\sqrt{a}$ assuming $a \geq 0$. Compute the limit of $f(x)$ as $x \to 0$. So for the function to be continuous at $x=0$, the functional value must be set equal to $\lim_{x \to 0} f(x)$.
Move your mouse over the gray area for the complete answer.
Note that $f(x)$ is not defined at $x=0$. Assuming $x \neq 0$. $$\begin{align} f(x) & = \dfrac{\sqrt{a^2-ax+x^2}-\sqrt{a^2+ax+x^2}}{\sqrt{a-x} - \sqrt{a+x}}\\ & = \dfrac{(a^2-ax+x^2)-(a^2+ax+x^2)}{(a-x) - (a+x)} \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\\ & = \dfrac{-2ax}{-2x} \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}}\\ & = a \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} \end{align} $$ We will also assume that $a \geq 0$. Hence, $$\lim_{x \to 0} f(x) = \lim_{x \to 0} a \times \dfrac{\sqrt{a-x}+\sqrt{a+x}}{\sqrt{a^2-ax+x^2}+\sqrt{a^2+ax+x^2}} = a \times \dfrac{2 \sqrt{a}}{2a} = \sqrt{a}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?
| $5^3 = 125 \equiv 3 \mod 61$, so $5^5 \equiv 25 \times 3 = 75 \equiv 14 \mod 61$,
$5^{10} \equiv 14^2 = 196 \equiv 13 \mod 61$, $5^{20} \equiv 13^2 = 169 \equiv 47 \mod 61$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Some doubt on Linear Diophantine equation We know $ax+by=c$ is solvable iff $(a,b)|c$ where $a,b,c,x,y$ are integers.
If $x=2$, $a=\dfrac{k(k+5)}{2}$, $y=k$, $b=k+3$ and $c=2k$, where $k$ is any integer, then
$$2 \frac{k(k+5)}{2} - k(k+3) = 2k.$$
So, $\left(\dfrac{k(k+5)}{2}, (k+3)\right) | 2$ for all integral value of $k$.
But $k+3$ cannot be even unless $k$ is odd.
Where I am going wrong?
| $2k(k+5)/2 - k(k+3)=2k$ with $a=k(k+5)/2$, $x=k$, $y=k$, would mean $b=k+3$, not $k(k+3)$.
The conclusion is that $(k(k+5)/2,k+3) | 2k$, not $2$. For example, with $k=3$, $(24,6)=6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Let $a,b,c>0$ how to prove that :
$$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$$
I find that
$$\ \frac{ab}{a^{2}+3b^{2}}=\frac{1}{\frac{a^{2}+3b^{2}}{ab}}=\frac{1}{\frac{a}{b}+\frac{3b}{a}} $$
By AM-GM
$$\ \frac{ab}{a^{2}+3b^{2}} \leq \frac{1}{2 \sqrt{3}}=\frac{\sqrt{3}}{6} $$
$$\ \sum_{cyc} \frac{ab}{a^{2}+3b^{2}} \leq \frac{\sqrt{3}}{2} $$
But this is obviously is not working .
| From the question it is obvious that equality is achieved when $a=b=c$, this hints at the use of AM-GM. Firstly, use the fact that $ab \le \frac{1}{2} (a^2+b^2)$, we just need to show
$$ \frac{a^2+b^2}{a^2+3b^2} + \frac{b^2+c^2}{b^2+3c^2} + \frac{c^2+a^2}{c^2+3a^2} \le \frac{3}{2},$$ which is equivalent to (after removing a constant 1 from each term in LHS)
$$ \frac{b^2}{a^2+3b^2} + \frac{c^2}{b^2+3c^2} + \frac{a^2}{c^2+3a^2} \ge \frac{3}{4},$$
by AM-GM, $a^2 + 3b^2 \ge 4 \sqrt[4]{a^2b^6} = 4\sqrt{ab^3}$, hence we just need
$$ \frac{\sqrt{b}}{\sqrt{a}} + \frac{\sqrt{c}}{\sqrt{b}} + \frac{\sqrt{a}}{\sqrt{c}} \ge 3, $$ but this is obviously true by one more application of AM-GM.
| {
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How many natural numbers $x, y$ are possible if $(x - y)^2 = \frac{4xy}{(x + y - 1)}$. How many natural numbers $x$, $y$ are possible if $(x - y)^2 = \frac{4xy}{x + y - 1}$.
Does this system has infinite solutions which can be generalized for some integer $k \geq 2?$
$(x - y)^2(x + y) = (x + y)^2 ;$
since $(x + y)$ can not be $0$ ;
$(x - y)^2 = (x + y);$
$x^2 - x(2y + 1) + y^2 - y = 0$;
For $x$ to be integer, discriminant($D$) should be perfect square;
$D = 8y + 1;$
$y = k(k + 1)/2$;
$(x, y) = (\frac{k(k + 1)}{2},\frac{k(k - 1)}{2})$ or vice versa;
infinite possibilities
| It is easy to verify that $(x, y) = (\frac{k(k + 1)}{2},\frac{k(k - 1)}{2})$ is solution of $(x - y)^2 = \frac{4xy}{x + y - 1}$.
we substitute $x$ and $y$ in $(x - y)^2 = \frac{4xy}{x + y - 1}$.
$$\left(\frac{k(k+1)}{2}-\frac{k(k-1)}{2}\right)^2 = \frac{4.[k(k-1)/2][k(k+1)/2]}{[k(k-1)/2]+[k(k+1)/2]-1}$$
after simplification we get $$k^2=k^2$$
$$LHS=RHS$$
therefore it has infinitely many solution for each values of $k$.
| {
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How to prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$ Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
| $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
Since
$(a-bc)^2\geq 0$,
$(b-ac)^2\geq 0$,
$(c-ab)^2\geq 0$,
then
$a^2+b^2c^2\geq 2abc$,
$b^2+a^2c^2\geq 2abc$,
$c^2+a^2b^2\geq 2abc$.
By of collectted through for through three inequalities last will be obtained
$a^2+b^2c^2+b^2+a^2c^2+c^2+a^2b^2\geq6abc$,
or
$a^2+a^2b^2+b^2+b^2c^2+c^2+a^2c^2\geq6abc$.
| {
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A question about arithmetic progressions, a.k.a. arithmetic sequences What is the relation between the general formula of the sum of $n$ terms of an arithmetic progression, $a n^2+b n+c$, and the first term $a+b+c$ and the common difference?
| Suppose that the terms we are adding up are $a_1,a_2,a_3$, and so on. Let $s_1=a_1$, $s_2=a_1+a_2$, $s_3=a_1+a_2+a_3$, and so on.
If $a_1,a_2,a_3,\dots$ is an arithmetic sequence, let $d$ be the common difference. So $a_1=a_1$, $a_2=a_1+d$, $a_3=a_1+2d$, and so on. We therefore have $a_k=a_1+(k-1)d$. It follows that
$$s_n=a_1+(a_1+d)+ (a_1+2d)+\cdots +(a_1+(n-1)d).$$
The above sum is equal to
$$na_1+d(1+2+\cdots +(n-1)).$$
By a standard result, $1+2+\cdots+(n-1)=\frac{1}{2}n(n-1)$. Thus
$$s_n=na_1+\frac{1}{2}(n)(n-1)d=\frac{d}{2}n^2 +\left(a_1-\frac{d}{2}\right)n.$$
Comparing with the proposed formula $s_n=an^2+bn+c$, we get that $c=0$ and $d=2a$.
Conversely, suppose that $c=0$, and the sum of the first $n$ terms is given by the formula $s_n=an^2+bn$. The $k$-th term $a_k$ is then $(ak^2+bk)-(a(k-1)^2+b(k-1))$. This simplifies to $a_k=2ak +b$. Write $d$ instead of $2a$. Then
$a_k=kd+b=b-d+(k-1)d$, exactly the expression for the $k$-th term of an arithmetic sequence with first term $b-d$ and common difference $d$.
Another way: We are given that $s_n=an^2+bn+c$. We will show that the sequence $a_1,a_2,a_3,\dots$ is an arithmetic sequence iff $c=0$, and will determine the common difference.
The first term (where we take first to mean $n=1$) is $a+b+c$.
So $a_1=a+b+c$.
The sum $s_2$ of the first $2$ terms is $4a+2b+c$. So $a_2$ must be $(4a+2b+c)-(a+b+c)$, that is, $3a+b$. Thus the difference $a_2-a_1$ between the second term and the first must be $(3a+b)-(a+b+c)$, which is $2a-c$. This is the common difference, if the sequence really is an arithmetic sequence. Let's check whether it really is.
Let $k\ge 2$. We have $s_k=ak^2+bk+c$, and $s_{k-1}=a(k-1)^2+b(k-1)+c$. Thus
$$a_k=s_k-s_{k-1}=(ak^2+bk+c)-(a(k-1)^2+b(k-1)+c)=2ak-a+b.$$
It follows that $a_{k+1}=2a(k+1)-a+b$. The difference between $a_{k+1}-a_k$ between the $(k+1)$-th term and the $k$-th term is therefore $2a$ when $k \ge 2$.
We saw that the difference $a_2-a_1$ is $2a-c$, and that all further differences are $2a$. We therefore have an arithmetic progression iff $c=0$. Then the common difference is $2a$.
Remark: If we are willing to take it on faith that the expression $an^2+bn+c$ does give the sum of an arithmetic sequence, say with first term $w$ and common difference $d$, then we can operate more concretely. Perhaps the following is intended.
We have $a+b+c=w$, $4a+2b+c=w+(w+d)=2w+d$, and $9a+3b+c=w+(w+d)+(w+2d)=3w+3d$.
Use the first two equations to eliminate $c$, also the last two equations. We get $3a+b=w+d$ and $5a+b=w+2d$. Subtract to eliminate $b$. We get $2a=d$. From $3a+b=w+d$ we then get $w=3a-2a+b=a+b$. Then from $a+b+c=w$ we get $c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating :$\int \frac{1}{x^{10} + x}dx$ $$\int \frac{1}{x^{10} + x}dx$$
My solution :
$$\begin{align*}
\int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\
&=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\
&=\ln|x|-\frac{1}{9}\ln|x^9+1|+C
\end{align*}$$
Is there completely different way to solve it ?
| Not really different, but even simpler:
$$\begin{align}
\int\frac{1}{x^{10}+x} dx=&\int\frac{x^{-10}}{1+x^{-9}} dx
=-\frac 1 9 \log |1+x^{-9}| + C
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Proving $\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$ for $a,b>1$ I'm trying to prove rigorously the following:
$$\left\lfloor \frac{x}{ab} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{x}{a} \right\rfloor}{b} \right\rfloor$$ for integers $a,b \geq 1$ and real $x$.
So far I haven't gotten far. It's enough to prove this instead:
$$\left\lfloor \frac{z}{c} \right\rfloor = \left\lfloor \frac{\lfloor z \rfloor}{c} \right\rfloor$$ for integers $c \geq 1$ and real $z$
since we can just put $z=\lfloor x/a \rfloor$ and $c=b$.
| Let $\lfloor x/a \rfloor$ = c(say)=>x=ca+d 0≤d
Let $\lfloor c/b \rfloor$=e(say)=>c=be+f 0≤f
=>x=ca+d=(be+f)a+d=abe+af+d.
$\frac{x}{a}$=be+f+ $\frac{d}{a}$
$\frac{\frac{x}{a}}{b}$=e+$\frac{f}{b}$+$\frac{d}{ab}$
=>$\lfloor x/a/b \rfloor$ = e = $\lfloor c/b \rfloor$ = $\lfloor \lfloor x/a \rfloor /b \rfloor$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
What can be the possible value of $a+b+c$ in the following case? What can be the possible value of $a+b+c$ in the following case?
$$a^{2}-bc=3$$
$$b^{2}-ca=4$$
$$c^{2}-ab=5$$
$0, 1, -1$ or $1/2$?
After doing $II-I$, $III-I$ and $III-II$, I got,
$$(a+b+c)(b-a)=1$$
$$(a+b+c)(c-a)=2$$
$$(a+b+c)(c-b)=1$$
I'm unable to solve further, please help.
| You have
$$\left\{
\begin{align*}
&(a+b+c)(b-a)=1\\
&(a+b+c)(c-a)=2\\
&(a+b+c)(c-b)=1\;.
\end{align*}\right.\tag{1}$$
These clearly imply that $a+b+c\ne 0$, so the first and third of these imply that $b-a=c-b$. In other words, $\langle a,b,c\rangle$ is an arithmetic progression. (The second equation of $(1)$ confirms this.) Set $d=b-a$; then $b=a+d$ and $c=a+2d$, so $a+b+c=3(a+d)$, and each of the equation in $(1)$ reduces to $3d(a+d)=1$.
Going back to the original equations, we see that
$$\begin{align*}
3&=a^2-bc=a^2-(a+d)(a+2d)\\
&=-3ad-2d^2=-3d(a+d)+d^2\\
&=d^2-1\;,
\end{align*}$$
or $d^2=4$. Thus, $d=\pm 2$, $1=3d(a+d)=\pm6(a+d)$, and $a+b+c=3(a+d)=\pm\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept
A circle is tangent to the $y$-axis at $y=3$ and has one $x$-intercept at $x=1$. Find the other $x$-intercept
Like previously mentioned, I'm not all too familiar with circles. So, I plotted the two points and I do not know the next step. I would guess to find the slope which is $-\dfrac{3}{1}$. But that doesn't seem right. I'm just taking a shot in the dark. If someone can tell me what to do next, or at least how to find the center that would be helpful. Please do not give me the answer.
| If (a,b) be the centre of the circle and radius=r and
clearly the circle passes through (1,0)
then $r^2=(1-a)^2+b^2$
The equation of the circle $(x-a)^2+(y-b)^2=(1-a)^2+b^2$
The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{(a-x)}{(y-b)}$
The gradient of the circle at (0,3) =$ \frac{(a-0)}{(3-b)}$
As y-axis is tangent to the circle at (0,3) and its gradient is ∞, so b=3.
(i)The equation of the circle becomes $(x-a)^2+(y-3)^2=(1-a)^2+3^2$.
As the circle passes through (0,3), $(0-a)^2+(3-3)^2=(1-a)^2+3^2$ =>a=5.
Or (ii) As the circle passes through (0,3),
$r^2=(0-a)^2+(3-b)^2$, but $r^2=(1-a)^2+b^2$ =>a=3b-4 =>a=5
As any intersection of x-axis & the circle, (x,0)
=>$(x-a)^2+(0-b)^2=(1-a)^2+b^2$
=>x=1,2a-1.
So, the other x-intercept is 2(5)-1=9
Alternatively,
the circle passes through (0,3), (1,0).
Let the equation of the circle : $x^2+y^2+2gx+2fy+c=0$
9+6f+c=0 and 1+2g+c=0.
Let the expected x-intercept be t, so the third point on the circle (t,0).
So, $t^2+2gt+c=0$
So, t,1 are the roots of $s^2+2gs+c=0$
=>t+1=-2g and t.1=c
=>2g=-t-1 and c=t
As 9+6f+c=0, 2f=-$\frac{9+c}{3}$
So, the equation of the circle becomes $x^2+y^2-(t+1)x-\frac{9+t}{3}y+t=0$
The gradient of the circle at (x,y) = $\frac{dy}{dx} = \frac{3(2x-t-1)}{(9+t-6y)}$
The gradient of the circle at (0,3) =-$\frac{t+1}{t-9}$
As y-axis is tangent to the circle at (0,3) and its gradient is ∞, so t=9.
So, the other x-intercept is 9
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Compute integral $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ I want to solve $\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x$ but I get the wrong results:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x =
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
$$
$$
= \left[ \frac{(4e^{4x} + 8e^{2x} + 4x)2}{e^{2x}} \right]_{-6}^6 =
\left[ \frac{8e^{4x} + 16e^{2x} + 8x}{e^{2x}} \right]_{-6}^6
$$
$$
= (\frac{8e^{24} + 16e^{12} + 48}{e^{12}}) - (\frac{8e^{-24} + 16e^{-12} - 48}{e^{-12}})
$$
$$
= e^{-12}(8e^{24} + 16e^{12} + 48) - e^{12}(8e^{-24} + 16e^{-12} - 48)
$$
$$
= 8e^{12} + 16 + 48e^{-12} - (8e^{-12} + 16 - 48e^{12})
$$
$$
= 8e^{12} + 16 + 48e^{-12} - 8e^{-12} - 16 + 48e^{12})
$$
$$
= 56e^{12} + 56e^{-12}
$$
Where am I going wrong?
| You had these steps ok:
$$
\int_{-6}^6 \! \frac{(4e^{2x} + 2)^2}{e^{2x}} \, \mathrm{d} x =
\int_{-6}^6 \! \frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} \, \mathrm{d} x
$$
After that, there are a number of choices. It looks like you forgot to integrate the solution.
You could do this:
$$\int_{-6}^6 {\frac{16e^{4x} + 16e^{2x} + 4}{e^{2x}} dx}$$
$$= \int_{-6}^6 { \left( 16e^{2x} + 16 + 4e^{-2x} \right) dx}$$
$$= \left[ { 8e^{2x} + 16x - 2e^{-2x} } \right]_{-6}^6$$
The integration is directly above. Plugging in the values then gives:
$$ \left( 8e^{2(6)} + 16(6) - 2e^{-2(6)} \right) - \left( 8e^{2(-6)} + 16(-6) - 2e^{-2(-6)} \right) $$
$$= \left( 8e^{12} + 96 - 2e^{-12} \right) - \left( 8e^{-12} -96 - 2e^{12} \right) $$
$$= 10e^{12} + 192 - 10e^{-12} $$
$$\approx 1.62774*10^6$$
To get the hyperbolic sine ($\sinh$), note that
$$ \sinh(x) = \frac{ e^{x} - e^{-2x} } {2}$$
$$ \sinh(12) = \frac{ e^{12} - e^{-12} } {2}$$
$$20\sinh(12) = 10 \left( e^{12} - e^{-12} \right)$$
So we have
$$ 10e^{12} - 10e^{-12} + 192 $$
$$= 20\sinh(12) + 192 $$
$$= 4 \left( 5 \sinh(12) + 48 \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Evaluating complicated sum Evaluate for a fixed $m\neq 1$ ( $m\in \mathbb{N}$ )
$$\sum _{k=1}^{n}\left[\left( \sum _{i=1}^{k}i^{2}\right) \left(\sum _{k_{1}+k_{2}+...+k_{m}=k}\dfrac {\left( k_{1}+k_{2}+\ldots +k_{m}\right) !} {k_{1}!k_{2}!...k_{m}!}\right)\right]$$
| As Shaktal showed earlier it is remains to find closed form for
$$
\sum\limits_{k=1}^n\frac{m^k k^3}{3}
$$
Consider the following equality
$$
\sum\limits_{k=1}^n x^k=\frac{x(x^n-1)}{x-1}
$$
After triple differentiation we get
$$
\sum\limits_{k=1}^n kx^{k-1}=\frac{d}{dx}\frac{x(x^n-1)}{x-1}\\
\sum\limits_{k=1}^n k(k-1)x^{k-2}=\frac{d^2}{dx^2}\frac{x(x^n-1)}{x-1}\\
\sum\limits_{k=1}^n k(k-1)(k-2)x^{k-3}=\frac{d^3}{dx^3}\frac{x(x^n-1)}{x-1}
$$
Now let's make a small trick
$$
\sum\limits_{k=1}^n\frac{k^3 x^k}{3}=
\frac{1}{3}\sum\limits_{k=1}^n (k (k-1)(k-2)+3k(k-1)+k)x^k=
\frac{1}{3}\left(x^3\sum\limits_{k=1}^n k (k-1)(k-2)x^{k-3} +3x^2\sum\limits_{k=1}^n k(k-1)x^{k-2} +x\sum\limits_{k=1}^n k x^{k-1} \right)=
\frac{1}{3}\left(x^3\frac{d^3}{dx^3}\frac{x(x^n-1)}{x-1} +3x^2\frac{d^2}{dx^2}\frac{x(x^n-1)}{x-1} +x\frac{d}{dx}\frac{x(x^n-1)}{x-1} \right)
$$
It is remains to simplify this expression (i.e. differentiante) and then substitute $x=m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$ This is the sequel of my previous question
$$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.
Any hint?
| Thanks for the nice question.
The answer is
$$
I(a) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}}
$$
The sketch of the proof: expand $\arctan$ in series, and integrate term-wise (can do this for small enough $a$, since the sine is bounded):
$$
\arctan\left(a \sin^2(x)\right) = \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{2n+1} \sin^{4n+2}(x)
$$
This gives
$$
\int_0^\infty \frac{\sin^{4n+2}(x)}{x^2} \mathrm{d} x = \frac{1}{\binom{2n}{\tfrac{1}{2}}} = \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma(2n+\frac{1}{2})}{(2n)!}
$$
The summation is easy, since the summand is a hypergeometric term:
$$
I(a) = \frac{\sqrt{\pi}}{2} \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{2n+1} \frac{\Gamma(2n+\frac{1}{2})}{(2n)!} = \frac{\pi a}{2} \cdot {}_2F_1\left(\frac{1}{4}, \frac{3}{4}; \frac{3}{2}; -a^2\right) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}}
$$
Added: The hard part is to prove that $S_n = \int_0^\infty \frac{\sin^{4n+2}(x)}{x^2} \mathrm{d} x$ is a hypergeometric term as claimed above. This can be done using:
$$\begin{eqnarray}
\sin^{4n+2}(x) &=& \left(\frac{\mathrm{e}^{ix} - \mathrm{e}^{-i x}}{2i}\right)^{4n+2} = -\frac{1}{4} \cdot \frac{1}{16^n} \sum_{m=0}^{4n+2} \binom{4n+2}{m} (-1)^m \mathrm{e}^{i (4n+2-2m)x} \\
&\stackrel{\text{symmetry}}{=}& -\frac{1}{4} \cdot \frac{1}{16^n} \sum_{m=0}^{4n+2} \binom{4n+2}{m} (-1)^m \underbrace{\cos((4n+2-2m)x)}_{1-2 \sin^2((2n+1-m)x)} \\
&=& \frac{1}{2} \cdot \frac{1}{16^n} \sum_{m=0}^{4n+2} \binom{4n+2}{m} (-1)^m \sin^2((2n+1-m)x) \\
&\stackrel{\text{symmetry}}{=}& \frac{1}{16^n} \sum_{m=0}^{2n} \binom{4n+2}{m} (-1)^m \sin^2((2n+1-m)x)
\end{eqnarray}
$$
Now:
$$\begin{eqnarray}
S_n &=& \frac{1}{16^n} \sum_{m=0}^{2n} \binom{4n+2}{m} (-1)^m \int_0^\infty \frac{\sin^2\left((2n+1-m) x\right)}{x^2} \mathrm{d} x \\
&=& \frac{1}{16^n} \sum_{m=0}^{2n} \binom{4n+2}{m} (-1)^m \frac{\pi}{2} \left(2n+1-m\right) \\ & \stackrel{m \to 2n-m}{=}& \frac{1}{16^n} \frac{\pi}{2} \sum_{m=0}^{2n} \binom{4n+2}{2n+2+m} (-1)^m \left(m+1\right)
\end{eqnarray}
$$
The latter sum readily yields to telescoping method, establishing the claim:
$$
S_n = \frac{\pi}{2} \cdot \frac{n+1 }{4 n+1} \cdot \frac{1}{16^n} \binom{4 n+2}{2 n+2} =
\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(2n+\frac{1}{2}\right)}{(2n)!}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
What values can $n$ assume, when $z^2+n= x^2+y^2$? $x^2+y^2 =n +z^2$ where $x,y,z$ are different natural numbers. What values can $n$ assume?
What if x,y,z>0 considering the confusion of 0 as natural number?
| Any odd number : $0^2+(n+1)^2=(2n+1)+n^2$
Any positive even number : $1^2+(n+1)^2=(2n+2)+n^2$
Any negative odd number : $0^2+n^2=-(2n+1)+(n+1)^2$
Any negative even number : $1^2+n^2=-2n+(n+1)^2$
If you really need to have different numbers x y z just remark that
$5^2+11^2=2+12^2$
and $3^2+5^2= -2 + 6^2$
If you need x, y, z to be non zero for odd numbers :
$2^2+(n+1)^2=(2n-3)+n^2$ for example...
$2^2+n^2=-(2n-3)+(n+1)^2$
and for the case 1 (all different) $4^2+7^2=1+8^2$ and $4^2+8^2=-1+9^2$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine the sum $T=a_0+a_1+a_2+...+a_{2012}$ Let ${a_n}$, $n \ge 0$ be a sequence of positive real numbers, given by
$a_0=1$ and
$a_m<a_n$ for all $m,n \in \mathbb{N}, m<n$ with
$a_n=\sqrt{a_{n+1}a_{n-1}}+1$ and $4\sqrt{a_n}=a_{n+1}-a_{n-1}$ for all $n \in \mathbb{N}, n\neq 0$.
Help me, determining the sum $T=a_0+a_1+a_2+...+a_{2012}$.
| $(a_{n+1}+a_{n-1})^2=(a_{n+1}-a_{n-1})^2+4a_{n+1}.a_{n-1}$
$=16a_n+4(a_n-1)^2=4(a_n+1)^2$
=>$a_{n+1}+a_{n-1}=2(a_n+1)$ as $a_n$ is increases with n and $a_0=1$
Or, $a_{n+1}-2 a_n +a_{n-1} -2 =0$
Putting n=m+1 and m,
$a_{m+2}-2 a_{m+1} +a_{m} -2 =0$ and
$a_{m+1}-2 a_m +a_{m-1} -2 =0$
Subtracting $a_{m+2}-3a_{m+1}+3a_m-a_{m-1}=0$
If $a_m=b^m$, then $(b-1)^3=0$
So, $a_m=(Am^2+Bm+C)1^m=(Am^2+Bm+C)$
1=$a_0=C$
Following avatar, 4 = $a_1=A+B+C$ and 9 = $a_2=4A+2B+C =>(A,B,C)=(1,2,1)$
=>$a_m=(m^2+2m+1)=(m+1)^2$
Now, it should not be too tough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do you solve this radical equation $\sqrt{2x+5} + 2\sqrt{x+6} = 5$? I have a radical equation
$$
\sqrt{2x+5} + 2\sqrt{x+6} = 5
$$
and I am having trouble calculating an answer. I keep on getting weird numbers that are not correct as my answer. How do you solve this? A step-by-step procedure would be highly appreciated.
| First we require $x \geq -\frac{5}{2}$ (for both radicals to be well defined). Now we have that $\sqrt{2x+5}=5-2\sqrt{x+6}$.
If $5-2\sqrt{x+6}\ge 0\Rightarrow -\frac{5}{2}\le x\le \frac{19}{4}$ then
\begin{equation}\sqrt{2x+5}^2=(5-2\sqrt{x+6})^2\Leftrightarrow 2x+5=25+4(x+6)-20\sqrt{x+6}\end{equation}
Can you continue from here? Do you understand why we make the hypothesis?
Here is some of the rest:
\begin{equation} 2x+5=25+4(x+6)-20\sqrt{x+6}\Leftrightarrow 20\sqrt{x+6}=44+2x\Leftrightarrow 10\sqrt{x+6}=x+22\end{equation}
Since $x+22\ge 0$ (remember that $-\frac{5}{2}\le x\le \frac{19}{4}$) we have that
\begin{equation} (10\sqrt{x+6})^2=(x+22)^2\Leftrightarrow 100(x+6)=x^2+44x+484\Leftrightarrow x^2-56x-116=0\end{equation}
Solve the last equation and choose only the $x$ that lie in $\left[-\frac{5}{2}, \frac{19}{4}\right]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt{2}}{4} + \frac{1}{2} }$
*$\sin(72^\circ) = \sqrt{ \frac{\sqrt{5}}{8} + \frac{5}{8} }$
*$\sin(75^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
*?
Is there is a list of known exact values of $\boldsymbol \sin$ somewhere?
Found a related post here.
| $\sin 3^\circ=\frac{(\sqrt{3}+1) (\sqrt{5}-1)}{8 \sqrt{2}}-\frac{(\sqrt{3}-1) \sqrt{5+\sqrt{5}}}{8}$.
Solving a cubic equation you can get a huge expression for $\sin 1^\circ$ in radicals, and therefore, for any $\sin n^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/176889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
recurrence solution to gambler's ruin From DeGroot 2.4.2, let $a_i$ be the conditional probability that the gambler wins all $k$ given gambler is at $i$.
$a_i = pa_{i+1} + (1 - p)a_{i-1} $
It's not clear from the text what steps are taken to solve for the general form $a_i$ (maybe Gaussian elimination). How do you solve the recurrence equation for $a_i$? If this can be found through the characteristic equation of a matrix $A$, then how do you construct $A$? As the problem is described, the first row has $a_0 = 0$, and last with $a_k = 1$.
| This is best solved using generating functions. Let $f(x) = \sum_{k=0}^\infty a_k x^k$. Now multiply the recurrence equation with $x^k$ and perform the summation:
$$
f(x) - a_0 = \sum_{k=1}^\infty a_k x^k = \sum_{k=1}^\infty (p a_{k+1} + (1-p) a_{k-1}) x^k =
\frac{p}{x} \sum_{k=1}^\infty a_{k+1} x^{k+1} + (1-p) x \sum_{k=1}^\infty a_{k-1}) x^{k-1} = \frac{p}{x} \left( f(x) - a_0 - a_1 x\right) + (1-p) x f(x)
$$
Now solve for $f(x)$:
$$
f(x) = a_0 \frac{p-x}{(1-x)(1-(1-p)x)} + a_1 \frac{ p x}{(1-x)(1-(1-p)x)} \\ = \left(\frac{\left(a_1-a_0\right) p}{1-\frac{1-p}{p} x}+\frac{a_0 (1-p)-a_1 p}{
1-x}\right) \frac{1}{1-2p} \\ =
\frac{1}{1-2p} \sum_{n=0}^\infty \left( \left(a_1-a_0\right) \frac{(1-p)^n}{p^{n-1}} + \left(a_0 (1-p)-a_1 p\right) \right)x^n \\
= \sum_{n=0}^\infty \left( a_0 \frac{p^{n-1}(1-p) - (1-p)^{n}}{(1-2p) p^{n-1}} + a_1 \frac{(1-p)^n - p^n}{(1-2p)p^{n-1}} \right) x^n
$$
where geometric series has been used. The case of $p =\frac{1}{2}$ requires taking a limit. Using L'Hospital's rule:
$$
a_k\left(\frac{1}{2}\right) = a_1 n + a_0 (1-n) = a_0 + (a_1-a_0) n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$.
Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$ where
a) $f(x)=2x+3,$
b) $f(x)=\frac{1}{x+1},$
c) $f(x)=x^2.$
I believe that if anyone can help me out with the first one, the other two might come clearer to me. But I started out this problem by plugging in $f(x)$ and got:
$$\dfrac{(2x+3)+f(h)-(2x+3)}{h}$$
I have no idea what to do after this. Because the $2x+3$'s can cancel out and leave me with just $\frac{f(h)}{h}$ but that doesn't make sense to me. Please help.
EDIT: The first one is solved and now for the second one, this is what I got:
$$\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}$$ Now, to subtract fractions the denominator has to be the same and they are the exact same except for the first fraction has an $h$ while the other one does not. How would I go about this?
For the third problem, this is all of my work:
$$\begin{align*}\dfrac{(x+h)^2-x^2}{h}&=
\dfrac{x^2+2hx+h^2-x^2}{h}\\
&= \dfrac{2hx+h^2}{h}\\
&= \dfrac{h(2x+h)}{h}\\
&= 2x+h\end{align*}$$
Is this all correct?
| $f(x+h)$ means replace $x$ with $x+h$ in your function definition. If $f(x) = 2x+3$, then $f(x+h) = 2(x+h)+3$, not $2x+3+f(h)$.
Therefore, $$\frac{f(x+h)-f(x)}{h} = \frac{2(x+h)+3-\left(2x+3\right)}{h} = \frac{2x+2h+3-2x-3}{h} = \cdots$$
Edit: To answer your second question, how do you handle just $\frac{1}{x+h+1}-\frac{1}{x+1}$? One simple way: multiply the first term above and below by the second term's denominator, and vice-versa: $$\frac{1}{x+h+1}\frac{x+1}{x+1}-\frac{1}{x+1}\frac{x+h+1}{x+h+1} = \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$? How to construct a $2\times 2$ real matrix $A$ not equal to Identity such
that $A^3=I$?
There is a correspondence between the ring of complex numbers and the
ring of $2\times2$ matrices (0 matrix
is included!) i.e.,$$a+ib\leftrightarrow\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$
Can I apply this result and construct such matrix?
| Find the three cube roots of $1$. Let $a+bi$ be one of those. They are solutions of $x^3=1$. Hence $x^3-1=0$.
Since $1$ is one of the solutions, $x-1$ must be one of the factors, thus:
$$
x^3-1 = (x-1)(\cdots\cdots\cdots).
$$
Fill in the blanks by doing long division. You should get
$$
(x-1)(x^2+x+1).
$$
So the equation is
$$
(x-1)(x^2+x+1) = 0.
$$
That implies
$$
x-1=0\quad\text{or}\quad x^2+x+1 = 0.
$$
Solve the quadratic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Hypergeometric functions & integral I'm having difficulty re-deriving a result a calculation from a paper. The integral is
$$\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12d\theta d\phi,$$
where $\eta$ and $c$ are parameters such that $\sinh^2\eta = 2$ and $c\sinh^2\eta < 1.$
From what I've read about hypergeometric functions, one could potentially evaluate this using elliptic integrals of the first and second kind,
\begin{align*}
K(z) &= \int_0^{\pi/2} (1-z^2\sin^2 t)^{-1/2} dt \\
E(z) &= \int_0^{\pi/2} (1-z^2\sin^2 t)^{1/2} dt.
\end{align*}
These can be written in hypergeometric form as
\begin{align*}
K(z) &= \frac{\pi}{2}F(\frac{1}{2},\frac{1}{2};1;z^2)\\
E(z) &= \frac{\pi}{2}F(-\frac{1}{2},\frac{1}{2};1;z^2).
\end{align*}
The expected result of the integral is
$$S = 8\pi^2 \frac{b^2}{a} G_1(c/a).$$
where $a = \sqrt{c^2+b^2}$ and
$$G_1(x) = F(3/2,1/2,1;x^2)+x^2/2F(3/2,3/2,2;x^2).$$
I've tried doing this, but I am inexperienced in these calculations and it's going to take me a while. I'll keep working at it, but I was wondering if anybody could take a look and tell me whether I am even going about this in the right way.
| Going off Peter's comment, note that your integral is separable, and can thus be factored into a product of two one-dimensional integrals:
$$\begin{split}&\int_0^{2\pi} \int_0^{2\pi} \frac{\sinh\eta}{(\cosh\eta-\cos\theta)^2}\left(1-c\sinh^2\eta\sin\phi\right)^\frac12\mathrm d\theta\mathrm d\phi=\\&\quad\left(\color{green}{\sinh\eta\int_0^{2\pi} \frac{\mathrm d\theta}{(\cosh\eta-\cos\theta)^2}}\right)\left(\color{blue}{\int_0^{2\pi} \sqrt{1-c\sinh^2\eta\sin\phi} \mathrm d\phi}\right)\end{split}$$
The first integral is elementary:
$$\color{green}{\sinh\eta\int_0^{2\pi} \frac{\mathrm d\theta}{(\cosh\eta-\cos\theta)^2}}=\frac{2\pi\cosh\,\eta}{\sinh^2\eta}$$
while the second requires the services of the incomplete elliptic integral of the second kind $E(\phi\mid m)$ (see Byrd and Friedman, formula 288.01 for the required identity):
$\displaystyle\begin{split}&\color{blue}{\int_0^{2\pi} \sqrt{1-c\sinh^2\eta\sin\phi} \mathrm d\phi}=\\&\Tiny 2\left(\sqrt{\frac{1+c\sinh^2\eta}{1-c\sinh^2\eta}}\;E\left(\arcsin\left(\sqrt{\frac{1+c\sinh^2\eta}{2}}\right)\mid\frac{2c\sinh^2\eta}{1+c\sinh^2\eta}\right)-\frac{c\sinh^2\eta}{\sqrt{1-c\sinh^2\eta}}+2\sqrt{1+c\sinh^2\eta}\;E\left(\frac{\pi}{4}\mid\frac{2c\sinh^2\eta}{1+c\sinh^2\eta}\right)\right)\end{split}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why is the last digit of $n^5$ equal to the last digit of $n$? I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.
| $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=n(n^2-1)(n^2-4)+5n(n^2-1)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\cdot \underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$
Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer
So, $(n-2)(n-1)n(n+1)(n+2)$ is divisible by $5!=120$ and $(n-1)n(n+1)$ is divisible by $3!=6$
$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 7,
"answer_id": 4
} |
Evaluating a simple definite integral I'm currently teaching myself calculus and am onto the Mean Value Theorem for Integration.
I am finding the value of $f(c)$ on the function $f(x)=x^3-4x^2+3x+4$ on the interval $[1,4]$.
So, with the equation $(b-a)\cdot f(c)=\int_1^4f(x)dx $, you get
$(4-1)\cdot f(c)=\int_1^4(x^3-4x^2+3x+4)dx$
Now my book says that this equals $3f(c)=\frac{57}{4}$.
I've been racking my brain and can't figure out how $\int_1^4(x^3-4x^2+3x+4)dx=\frac{57}{4}$
So how did the author evaluate that integral to get the answer?
| We have the following string of equalities:
$$\begin{align*}\int_1^4(x^3-4x^2+3x+4)dx &= \left.\frac{x^4}{4}-\frac{4x^3}{3}+\frac{3x^2}{2}+4x\right|_{x=1}^4\\
&= \left(\frac{4^4}{4}-\frac{4\cdot4^3}{3}+\frac{3\cdot4^2}{2}+4\cdot4\right)-\left(\frac{1^4}{4}-\frac{4\cdot1^3}{3}+\frac{3\cdot1^2}{2}+4\cdot1\right)\\
&= \left(64-\frac{256}{3}+24+16\right)-\left(\frac{1}{4}-\frac{4}{3}+\frac{3}{2}+4\right)\\
&= \left(\frac{56}{3}\right)-\left(\frac{53}{12}\right)\\
&=\frac{57}{4}.\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$, find the closest pair of integers, a and b, such that, $a \lt t \lt b$. $t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$
find the closest pair of integers, a and b, such that, $a \lt t \lt b$.
$30=1+29$
$(1+29)^{65}=(1+29)(1+29)^{64}$
| Let $A,B,a,b,n$ be positive numbers.
$\frac{A^{n+1}-B^{n+1}}{A^n-B^n}-A=\frac{B^n(A-B)}{A^n-B^n}$ will be $>0$ if $A>B$ and $n ≥ 1$
Putting $A=a+1,B=a$, $\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}>(a+1)$ for $n ≥ 1$.
Alternatively,
$\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}-(a+1)$
$=\frac{(a+1)^{n+1}-a^{n+1}-(a+1)^{n+1}+(a+1)a^n}{(a+1)^n-a^n}$
$=\frac{a^n}{(a+1)^n-a^n}>0$ for $n ≥ 1$
$=>\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}>(a+1)$ for $n ≥ 1$
Now, $\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}-(a+1+1)$
$=\frac{a^n}{(a+1)^n-a^n}-1 $
$=\frac{2a^n-(a+1)^n}{(a+1)^n-a^n}$
$=\frac{2-(1+\frac{1}{a})^n}{(1+\frac{1}{a})^n-1}$
But $(1+\frac{1}{a})^n=1+n\cdot\frac{1}{a}+\frac{n(n-1)}{2}\frac{1}{a^2}+...$ will be $>2$ if $n>a$
So, $2-(1+\frac{1}{a})^n<0$ if $n>a$
$\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}<(a+1+1)$ if $n>a$
So, $(a+1)<\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}<(a+1+1)$ if $n>a$
Here in this problem, $n=64,a=29$
| {
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$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
| Using Hölder's inequality we have:
$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2/3} (a(a+b)+b(b+c)+c(c+a))^{1/3}\geq a+b+c.$$
i.e.
$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2} \geq \frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca}.$$
We have to prove that:
$$\frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{9}{2}.$$
i.e.
$$2(a+b+c)^3\geq9\left(a^2+b^2+c^2+ab+bc+ca\right). \tag{1}$$
Let $p=a+b+c$ and $q=ab+bc+ca$ and using that $abc=1$ and $AM-GM$ we obtain that $q \geq 3$.
Inequality $(1)$ is equivalent to:
$$2p^3 \geq 9\left(p^2-2q+q\right) \Leftrightarrow 2p^3+9q \geq 9p^2.$$
Applying $AM-GM$ we obtain
$$2p^3+9q \geq 2p^3+27=p^3+p^3+27 \geq 3\cdot \sqrt[3]{27p^6}=9p^2,$$ as required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?
I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.
Are they just a coincidence or is there a theory behind?
Is it about modular arithmetic?
| Let $d=a-b$. By the binomial theorem, $a^n = (d+b)^n = dt+b^n$. So $a^n-b^n=dt=(a-b)t$.
| {
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Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$ For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
| This is problem J57 of mathematical reflections, vol.2007, issue 4;
$a^2+b^2+c^2 \ge ab+bc+ca =1$
$2\left(\frac ab+\frac bc+\frac ca\right)\ge6$ (as BenjaLim does)
It remains
$\frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} \ge 9$ which follows by (use AGM) $\frac 3{(abc)^{2/3}}\ge9$ or $(abc)^{2/3}\le \frac13.$ This is a consequence of $ab+bc+ca=1$ since always by AGM $1= ab+bc+ca\ge 3(abc)^{2/3}$
| {
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Inequality. $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq 3+n$ I can't find any solution for this inequality which can be found here
Exercise 1.3.5 Let $a,b,c$ be positive real numbers such that $a+b+c=ab+bc+ca$ and $n \leq 3$. Prove that
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{3n}{a^2+b^2+c^2} \geq 3+n$$
Thanks :)
| Making use of given hint,
$$ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} = {(a + b +c)(a^2 + b^2 + c^2)\over (a+b+c)} = a^2 + b^2 +c^2$$
The minimum value of $a^2+b^2+c^2 \geq a+b+c \geq 3 $ for given criteria.
Let ${a^2 + b^2 + c^2 = z}$ then $f(z, n) = z + {3n \over z}$ for $n \leq 3$ and $z \geq 3$
$$f(z,n) = z + {3n \over z} \geq 6 \text{ for n=3}$$
For $n < 3$ the minimum value occurs at $f'(z) = 0 \implies z = \sqrt{3n } < 3 \leq z$ so further more $f'(z \geq 3) > 0$ suggest that $f(z)$ is increasing. So, the min-value must occur at $z=3$ which gives the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
solving for a coefficent term of factored polynomial.
Given: the coefficent of $x^2$ in the expansion of $(1+2x+ax^2)(2-x)^6$ is $48,$ find the value of the constant $a.$
I expanded it and got
$64-64\,x-144\,{x}^{2}+320\,{x}^{3}-260\,{x}^{4}+108\,{x}^{5}-23\,{x}^{
6}+2\,{x}^{7}+64\,a{x}^{2}-192\,a{x}^{3}+240\,a{x}^{4}-160\,a{x}^{5}+
60\,a{x}^{6}-12\,a{x}^{7}+a{x}^{8}
$
because of the given info $48x^2=64x^2-144x^2$ solve for $a,$ $a=3$.
Correct?
P.S. is there an easier method other than expanding the terms?
I have tried using the bionomal expansion; however, one needs still to multiply the terms. Expand $(2-x)^6$ which is not very fast.
| You need not expand completely as only low powers play a role:
$(2-x)^6 = 2^6-6\cdot 2^5\cdot x + \frac{6\cdot5}2\cdot 2^4\cdot x^2+\ldots = 64-192x+240x^2+\ldots$
where the dots represent anything involving $x^3$ or even higher powers.
After this
$(1+2x+ax^2)(2-x)^6 = (64-192x+240x^2+\ldots) + (128x-384x^2+\ldots)+ (64 a x^2+\ldots) = 64 -64 x + (-144+64a)x^2+\ldots$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can I get better approximation of $\sum_{k=1}^{n} k^k$ Is it possible to get approximation$f(n)$ of $\sum_{k=1}^{n} k^k$ with
\begin{align}
\lim_{n\to +\infty }\left(f(n)-\sum_{k=1}^{n} k^k\right)=0
\end{align}
Thanks for your attention!
| I have no idea how $f(n)$ should look like, and actually I suspect that there is no such simple formula for $f(n)$. But at least we can improve Sasha's asymptotics. Indeed, a moment of thought gives that for any fixed $m$, we have
\begin{align*}
\frac{1}{n^n} \sum_{k=1}^{n} k^k
&= \sum_{k=0}^{n-1} \left(1 - \frac{k}{n}\right)^{n-k}\frac{1}{n^k} \\
&= \sum_{k=0}^{n-1} \frac{1}{(en)^k}\exp\bigg\{k \sum_{j=1}^{\infty} \frac{(k/n)^j}{j(j+1)} \bigg\} \\
&= \sum_{k=0}^{m} \frac{1}{(en)^k}\exp\bigg\{k \sum_{j=1}^{m-k} \frac{(k/n)^j}{j(j+1)} \bigg\} + \mathcal{O}(n^{-(m+1)})
\end{align*}
Plugging $m = 1$, we have
$$ \frac{1}{n^n} \sum_{k=1}^{n} k^k = 1 + \frac{1}{en} + \mathcal{O}(n^{-2}) $$
With aid of Mathematica, for $m = 4$ we have
\begin{align*}
\frac{1}{n^n} \sum_{k=1}^{n} k^k
&= 1 + \frac{1}{en} + \left( \frac{1}{e^2}+\frac{1}{2e}\right) \frac{1}{n^2} + \left(\frac{1}{e^3}+\frac{2}{e^2}+\frac{7}{24 e}\right)\frac{1}{n^3} \\
&\qquad + \left(\frac{1}{e^4}+\frac{9}{2e^3}+\frac{10}{3 e^2}+\frac{3}{16 e}\right)\frac{1}{n^4} + \mathcal{O}(n^{-5})
\end{align*}
| {
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An intriguing definite integral: $\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$ I need some hints, suggestions for the following integral
$$\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$
Since it's a high school problem, I thought of some variable change, integration by parts, but I can't see yet how to make them work. I don't know where I should start from. Thanks!
| Let $u = 2\tan^{-1}\left(\frac{x}{4}\right) - x$. Then
$$ du = -\frac{x^2 + 8}{x^2 + 16}\,dx. $$
Now since
$$ \begin{align*}
\frac{x^2 + 8}{(x^2 - 16)\sin x + 8x \cos x}
&= \frac{\frac{x^2 + 8}{x^2 + 16}}{\frac{8x}{16+x^2} \cos x - \frac{16-x^2}{16+x^2}\sin x} \\
&= \frac{\frac{x^2 + 8}{x^2 + 16}}{\sin\left(2\tan^{-1}\left(\frac{x}{4}\right)\right) \cos x - \cos\left(2\tan^{-1}\left(\frac{x}{4}\right)\right)\sin x} \\
&= \frac{\frac{x^2 + 8}{x^2 + 16}}{\sin\left(2\tan^{-1}\left(\frac{x}{4}\right) - x\right)} \\
&= -\frac{1}{\sin u}\frac{du}{dx},
\end{align*}$$
it remains to take integration by substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| The real solution:
$$x^3+1=0<=>$$
$$x^3=-1<=>$$
Take cube roots of both sides:
$$x=-1$$
Complex solution:
$$x^3+1=0<=>$$
$$x^3=-1<=>$$
$$x^3=|-1|e^{arg(-1)i}<=>$$
$$x^3=e^{(\pi +2\pi k)i}<=>$$
(with k is the element of Z)
$$x=\left(e^{(\pi +2\pi k)i}\right)^{\frac{1}{3}}<=>$$
$$x=e^{\left(\frac{1}{3}\pi +\frac{2}{3}\pi k\right)i}$$
(with k goes from 0 to 3 -> k=0-2)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 8,
"answer_id": 5
} |
More and more limits for sequences So here goes a bit of homework:
$$\lim_{n\to\infty}{\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}}$$
Well, this would trivially lead to:
$$\lim_{n\to\infty}{\left(\frac{3+\frac{2}{n}+\frac{1}{n^2}}{3+\frac{5}{n^2}}\right)^{\frac{n\left(1+\frac{2}{n^2}\right)}{2+\frac{1}{n}}}}$$
Which is clearly an indetermination of type "$1^\infty$". Now, I can't really get through this step... Any hints?
| $$\frac{3n^2+2n+1}{3n^2-5}=1+\frac{2n+6}{3n^2-5}=1+\frac{2}{\frac{3n^2-5}{n+3}}\Longrightarrow$$
$$\Longrightarrow\left(\frac{3n^2+2n+1}{3n^2-5}\right)^{\frac{n^2+2}{2n+1}}=\left[\left(1+\frac{2}{\frac{3n^2-5}{n+3}}\right)^{\frac{3n^2-5}{n+3}}\right]^\frac{n^3+3n^2+2n+6}{6n^3+3n^2-10n-5}\Longrightarrow$$
Well, now the inner limit must be well-known, and for the exterior one just check the exponent behaves as $\,1/6\,$ for large values of $\,n\,$ . The limit indeed is $\,e^{1/3}\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/194468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
| If $10x=\pi$
$\sin 2x=\cos 3x$ as $2x+3x=5x=\frac{\pi}{2}$
$\implies2\sin x \cos x=4\cos^3x-3\cos x$
$\implies 2\sin x=4\cos^2x-3$ as $\cos x≠0$
If $\sin x=t, 2t=4(1-t^2)-3\implies 4t^2+2t-1=0$
$$\implies t=\frac{-1±\sqrt{5}}{4}$$, but $\sin x>0$ as $0<x<\pi$
$$\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$$
(1)So, $$\cos \frac{\pi}{10}=\sqrt{1-(\sin \frac{\pi}{10})^2}=\frac{\sqrt{10+2\sqrt5}}{4}$$
So, $$\tan \frac{\pi}{10}=\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt5}}{4}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt5}}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt5}}$$
$$=\sqrt{\frac{(\sqrt 5 -1)^2}{10+2\sqrt5}}=\sqrt{\frac{3-\sqrt 5}{\sqrt 5(\sqrt 5+1)}}=\sqrt{\frac{(3-\sqrt 5)(\sqrt 5 -1)}{\sqrt 5(\sqrt 5+1)(\sqrt 5 -1)}}=\sqrt{\frac{\sqrt 5-2}{\sqrt 5}}$$
Or(2) $$\cos \frac{\pi}{5}=1-2(\frac{\sqrt{5}-1}{4})^2=\frac{\sqrt 5 + 1}{4}$$
We know $$\cos2y=\frac{1-\tan^2y}{1+\tan^2y}\implies \tan^2y=\frac{1-\cos2y}{1+\cos2y}$$
So, $$\tan^2 \frac{\pi}{10}= \frac{1-\frac{\sqrt 5 + 1}{4}}{1+\frac{\sqrt 5 + 1}{4}}=\frac{3-\sqrt 5}{\sqrt 5(\sqrt 5+1)}$$ which we have already encountered in (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/196067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
$\int\frac{x^3}{\sqrt{4+x^2}}$ I was trying to calculate
$$\int\frac{x^3}{\sqrt{4+x^2}}$$
Doing $x = 2\tan(\theta)$, $dx = 2\sec^2(\theta)~d\theta$, $-\pi/2 < 0 < \pi/2$ I have:
$$\int\frac{\left(2\tan(\theta)\right)^3\cdot2\cdot\sec^2(\theta)~d\theta}{2\sec(\theta)}$$
which is
$$8\int\tan(\theta)\cdot\tan^2(\theta)\cdot\sec(\theta)~d\theta$$
now I got stuck ... any clues what's the next substitution to do?
I'm sorry for the formatting. Could someone please help me with the formatting?
| $$
\begin{aligned}
\int \frac{x^{3}}{\sqrt{4+x^{2}}} d x &=\int x^{2} d \sqrt{4+x^{2}} \\
&\stackrel{IBP}{=} x^{2} \sqrt{4+x^{2}}-\int \sqrt{4+x^{2}} d\left(x^{2}\right) \\
&=x^{2} \sqrt{4+x^{2}}-\frac{2}{3}\left(4+x^{2}\right)^{\frac{3}{2}}+C \\
&=\frac{\sqrt{4+x^{2}}}{3}\left[3 x^{2}-2\left(4+x^{2}\right)\right]+C \\
&=\frac{\sqrt{4+x^{2}}}{3}\left(x^{2}-8\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/197744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
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