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Integral of $\frac{\ln(1+x^2)}{1+x^2}$ Compute $\int\frac{\ln(1+x^2)}{1+x^2}dx$
Attempt:
I tried to do Integration by Parts
$$u=\ln(1+x^2)\Rightarrow du=\frac{2xdx}{1+x^2}\\
dv=\frac{dx}{1+x^2}\Rightarrow v=\arctan x\\
\int\frac{\ln(1+x^2)}{1+x^2}dx=\ln(1+x^2)\arctan x-\int\frac{2x\arctan x dx}{1+x^2}$$
Since it seems that it not done in terms of elementary functions, what about the definite version:
$$\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx$$
|
$$\text{Let }I=\int\dfrac{\ln(1+x^2)}{1+x^2}\mathrm dx=\int\left(\underbrace{\dfrac{i\ln(1+x^2)}{2(x+i)}}_{I_1}-\underbrace{\dfrac{i\ln(1+x^2)}{2(x-i)}}_{I_2}\right)\mathrm dx$$
Solving $2/i\cdot I_1$:
$$\begin{bmatrix}u \\ \mathrm du\end{bmatrix}=\begin{bmatrix}x+i\\ \mathrm dx\end{bmatrix}$$ $$\begin{align}\int\dfrac{\ln(1+x^2)}{x+i}\mathrm dx&=\int\dfrac{\ln((u-i)^2+1)}{u}\mathrm du\\ & =\int\dfrac{\ln(u-2i)}{u}\mathrm du+\dfrac{1}{2}\ln^2(u)\\ & = \int\dfrac{\ln(iu/2+1)}{u}\mathrm du+\ln(-2i)\ln u+\dfrac{1}{2}\ln^2u \\ &= -\mathrm{Li}_2\left(-\dfrac{iu}{2}\right)+\dfrac{1}{2}\ln^2u+\ln(-2i)\ln(u)\\&=-\mathrm{Li}_{2}\left(-\dfrac{i(x+i)}{2}\right)+\dfrac{1}{2} \ln^2(x+i)+\ln(-2i)\ln(x+i)\tag1\end{align}$$
Similarly solve $2/i\cdot I_2$ to get the following result: $$\int\dfrac{\ln(1+x^2)}{x-i}\mathrm dx = -\mathrm{Li}_2\left(-\dfrac{i(x-i)}{2}\right)+\dfrac{1}{2}\ln^2(x-i)+\ln(2i)\ln(x-i)\tag2$$
Computing $i/2\cdot [(1)+(2)]\equiv I_1+I_2$ and simplifying gives you the required antiderivative stated as follows (drum-roll moment): $$I=\dfrac{i}{4}\left[\ln\mid x+i \mid \left(\ln\mid x+i\mid+2\ln(-2i)\right)-\ln\mid x-i\mid\left(\ln\mid x-i\mid +2\ln(2i)\right)\\ +2\mathrm{Li}_2\left(\dfrac{ix+1}{2}\right)-2\mathrm{Li}_2\left(-\dfrac{ix-1}{2}\right)\right]+C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integer solutions to the diophantine equation $a^3+b^3+c^3=6$ I once thought that this equation $x^3+y^3+z^3=6$ has only a few smaller integer solutions. Until somebody told me this $6=192722201207819^3+162765491944499^3+(-225522344776678)^3$ existed. I don't know how it came about.
Question: How to find more integer solutions to the equation $a^3+b^3+c^3={\color{red}{6}}$?
a few smaller integer solutions:
\begin{align*}
(-637)^3+(-205)^3+644^3&=6\\
(-235)^3+(-55)^3+236^3&=6\\
(-58)^3+(-43)^3+65^3&=6\\
(-1)^3+(-1)^3+2^3&=6\\
\end{align*}
Something happened recently
I've also heard that mathematicians have recently solved this problem:
The least integer solutions to the equation $a^3+b^3+ c^3=33$
Triple cubic Diophantine equations
42 is the new 33 - Numberphile
|
Nothing more than brute force computation I'm afraid.
The method used to crack $33$ was to rearrange to $$(a^2-ab+b^2)=\frac{33-c^3}{a+b}$$
The computer can then assign a random $c$ and test all values of $a,b$ such that $(a+b)|(33-c^3)$ which requires much less testing, but still very large amounts.
|
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What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:
*
*$x^2-1=(x+1)(x-1)$
*$\sqrt{4x-4}=2\sqrt{x-1}$
*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{ (x+1)(x-1) }+x-1}{ 2\sqrt{x-1} + (x+1)(x-1) }$
So how do I remove what causes the hole function not to become $\frac{0}{0}$ and solve the limit?
|
I assume:
$$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x\color{red}-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot \sqrt{x+1}+(\sqrt{x-1})^2}{ \sqrt{4x-4}+(\sqrt{x-1})^2(x+1)} =\\
\lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot (\sqrt{x+1}+\sqrt{x-1})}{\sqrt{x-1}\cdot (2+(\sqrt{x-1})(x+1))} =\frac{\sqrt{2}}{2}.$$
|
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Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$ Find values of $a$ such that $x^2+ax+a^2+6a \lt 0$ $\forall$ $x \in (1,2)$
My try: Since $y=x^2+ax+a^2+6a$ is an open upward Parabola, the roots $\alpha,\beta$ should be distinct and satisfy $1 \lt \alpha \lt 2$ and $1 \lt \beta \lt 2$ implies both the roots are distinct and lies between $1$ and $2$.
So we have the following conditions:
$1.$ Since it should cut positive Y axis as both the roots are positive we have
$a^2+6a \gt 0$
$2.$ $D \gt 0$ $\implies$ $a^2+8a \lt 0$
$3.$ $f(1) \gt 0$ $\implies$ $a^2+7a+1 \gt 0$
$4.$ $f(2) \gt 0$ $\implies$ $a^2+8a+4 \gt 0$
How to solve all these four inequalities in a simpler way?
|
Find values of $a$ such that $x^2+ax+a^2+6a<0, ∀x\in (1,2)$.
As commented by Marty Cohen, the roots must be outside the interval:
$$x_1<1 \ \text{and} \ x_2>2 \iff \\
x_1=\frac{-a-\sqrt{-3a^2-24a}}{2}<1 \ \text{and} \ x_2=\frac{-a+\sqrt{-3a^2-24a}}{2}>2 \iff \\
1) \ \sqrt{-3a^2-24a}>-a-2 \ \text{and} \ 2) \ \sqrt{-3a^2-24a}>a+4 \iff \\
1) \begin{cases}-3a^2-24a\ge 0\\ -a-2<0\end{cases} \ \text{or} \begin{cases}-3a^2-24a\ge 0\\ a^2+7a+1<0\end{cases} \Rightarrow a\in \left(\frac{-7-3\sqrt{5}}{2},\frac{-7+3\sqrt{5}}{2}\right) \ \text{and}\\
2) \begin{cases}-3a^2-24a\ge 0\\ a+4<0\end{cases} \ \text{or} \begin{cases}-3a^2-24a\ge 0\\ a^2+8a+4<0\end{cases} \Rightarrow a\in (-4-2\sqrt{3},-4+2\sqrt{3})\\
1) \ \text{and} \ 2) \Rightarrow a\in \left(\frac{-7-3\sqrt{5}}{2},-4+2\sqrt{3}\right).$$
See the Desmos graph.
|
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proving complex Binomial Identity Proving the result $\displaystyle \sum^{\infty}_{n=0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$
what i try
$\displaystyle (1+x)^{n}=\sum^{n}_{r=0}\binom{n}{r}x^r$
$\displaystyle (x+1)^n=\sum^{n}_{r=0}\binom{n}{n-r}x^{n-r}$
Campare coefficient of $x^n$ on left and right
$\displaystyle \binom{2n}{n}= \sum^{n}_{k=0}\binom{n}{k}\binom{n}{n-k}$
How do i solve it Help me please
|
Note that
$$
\begin{align}
\binom{2n}{n}
&=\binom{2n-2}{n-1}\frac{2n(2n-1)}{n^2}\\
&=\binom{2n-2}{n-1}\frac{4n-2}{n}\tag1
\end{align}
$$
Multiply $(1)$ by $4^{-n}$ and set $a_n=4^{-n}\binom{2n}{n}$:
$$
\begin{align}
\overbrace{4^{-n}\binom{2n}{n}}^{\large a_n}
&=4^{-n}\binom{2n-2}{n-1}\frac{4n-2}{n}\\
&=\underbrace{4^{-(n-1)}\binom{2n-2}{n-1}}_{\large a_{n-1}}\frac{n-1/2}{n}\tag2
\end{align}
$$
Therefore, induction and $(2)$ yield
$$
\begin{align}
(-1)^na_n
&=\prod_{k=1}^n\frac{1/2-k}{k}\\
&=\prod_{k=1}^n\frac{-1/2-(k-1)}{k}\\
&=\binom{-1/2}{n}\tag3
\end{align}
$$
Thus,
$$
\binom{2n}{n}=(-4)^n\binom{-1/2}{n}\tag4
$$
Therefore,
$$
\begin{align}
\sum_{n=0}^\infty\binom{2n}{n}x^n
&=\sum_{n=0}^\infty(-4)^n\binom{-1/2}{n}x^n\\[6pt]
&=(1-4x)^{-1/2}\tag5
\end{align}
$$
|
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|
Equality of $2^x + 2^{4-x} \geq 8$
Want to find the value(s) of $x$ for which equality holds in $$2^x + 2^{4-x} \geq 8$$
I've found it by solving $2^x + 2^{4-x} = 8$:
$$2^x + 2^{4-x} = 8 \Rightarrow 2^{2x} - 8 \cdot 2^x + 16 = 0 \Rightarrow (2^x - 4)^2 = 0$$
so clearly $x = 2$.
However, the notes I've been reading goes straight to simply say that equality occurs when $2^x = 2^{4-x}$, i.e. when $x = 4 - x$ and again $x=2$.
I'm probably missing something really simple here, but why does equality hold simply when $$2^x = 2^{4-x}?$$
Edit: to clarify, I want to know why is this is an 'obvious' condition for equality without writing the equals sign and solving the quadratic?
|
You could argue it by the arithmetic-geometric mean inequality:
$$\frac{2^x + 2^{4 - x}}{2} \ge \sqrt{2^x \cdot 2^{4 - x}} = \sqrt{2^4} = 4,$$
with equality if and only if $2^x = 2^{4 - x}$.
|
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|
Prove that $\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$ Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{1}{ab}+\frac{1}{cd} \geq \frac{a^2+b^2+c^2+d^2}{2}$$
I write
$$a^2+b^2+c^2+d^2=16-2\left(ab+cd+\left(a+b\right)\left(c+d\right)\right)$$. Then the inequality is equivalent to
$$\frac{1}{ab}+\frac{1}{cd} +ab+cd+\left(a+b\right)\left(c+d\right) \geq 8$$
But now I don't know how to change $ab,cd$ into the forms of $a+b$ and $c+d$. Moreover, from the last inequality, we have $\frac{1}{ab}+\frac{1}{cd} \geq 4$, whereas $\left(a+b\right)\left(c+d\right) \leq 4$. I cannot handle this. Please help me.
|
Let $f(x)=\frac{1}{x(k-x)}-\frac{x^2+(k-x)^2}{2},$ where $0<x<k$.
Thus, $$f'(x)=-\frac{k-2x}{(kx-x^2)^2}-x-x+k=(2x-k)\left(\frac{1}{(kx-x^2)^2}-1\right)=$$
$$=\frac{(2x-k)(1-kx+x^2)(1+kx-x^2)}{(kx-x^2)^2}.$$
We see that $$1+kx-x^2=1+x(k-x)>0.$$
Consider two cases.
*
*$0<k\leq2.$
Thus, $$1-kx+x^2=\left(x-\frac{k}{2}\right)^2+1-\frac{k^2}{4}\geq0,$$ which says $$x_{min}=\frac{k}{2}$$ and
$$f(x)\geq f\left(\frac{k}{2}\right)=\frac{4}{k^2}-\frac{k^2}{4}.$$
2. $2<k<4.$
In this case we obtain $$\frac{k-\sqrt{k^2-4}}{2}<\frac{k}{2}<\frac{k+\sqrt{k^2-4}}{2},$$
which gives that $f$ gets a minimal value for $kx-x^2=1.$
Id est, $$f(x)\geq\frac{1}{1}-\frac{k^2-2}{2}=2-\frac{k^2}{2}.$$
Now, let $a+b=k\leq2.$
Thus, $c+d=4-k\geq2$ and
$$\frac{1}{ab}+\frac{1}{cd}-\frac{a^2+b^2+c^2+d^2}{2}=\frac{1}{ab}-\frac{a^2+b^2}{2}+\frac{1}{cd}-\frac{c^2+d^2}{2}\geq$$
$$\geq\frac{4}{k^2}-\frac{k^2}{4}+2-\frac{(4-k)^2}{2}=\frac{(2-k)^3(3k+2)}{4k^2}\geq0$$ and we are done!
|
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|
Sum of all real numbers $x$ such that $(\text{A quadratic})^\text{Another quadratic}=1$.
What is the sum of all real numbers $x$ such that
$(x^2-5x+5)^{(x^2-7x+12)}=1$?
So I know that $x^0=1$ and $1^x=1$. So, I can solve for them and find $x$, and add them up.
Solving $x^2-7x+12=0$ for $x^0=1$ gives $x=3, 4$.
Solving $x^2-5x+5=1$ for $1^x=1$ gives $x=1, 4$.
Adding them up gives $1+3+4=8$.
This is wrong. What did I do wrong? Did I miss a case? If so, what case have I missed?
Thanks!
Max0815
|
Three cases:
$$x^0=1$$
$$1^x=1$$
$$(-1)^{\text{even #}}=1$$
Substituting into the original equation gives $$x=3, 4$$ $$x=1, 4$$ $$x=2, 3$$
We don't add in the solutions already accounted for, namely, 3 and 4.
$\sum=3+4+1+2=10$.
|
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|
Expected length of time until the mouse reaches the compost heap. I'm assuming it's a Markov Chain question but I have no idea how to do this. Thanks in advance!
A mouse lives in a mousehole with three exits.
*
*The first exit leads
to a compost heap after 1 minute of scurrying.
*The second exit leads
to a tunnel that returns it to the mousehole after three minutes.
*The third exit leads to a tunnel that returns it to the mousehole after four
minutes.
Assume the mouse has no memory of previous excursions
when it returns to its mousehole, and is always equally likely to choose
any one of the exits, what is the expected length of time until the
mouse reaches the compost heap?
The scan of the original question.
|
More difficult way! Refer to the table, where $X$ is the number of minutes:
$$\begin{array}{c|c|c}
\text{Excursions ($n$)}&X&P(X)&XP(X)\\
\hline
1&1&1/3&1\cdot 1/3\\
\hline
2&3+1&1/3^2&(3+1)\cdot 1/3^2\\
2&4+1&1/3^2&(4+1)\cdot 1/3^2\\
\hline
3&3+3+1&1/3^3&(3+3+1)\cdot 1/3^3\\
3&3+4+1&1/3^3&(3+4+1)\cdot 1/3^3\\
3&4+3+1&1/3^3&(4+3+1)\cdot 1/3^3\\
3&4+4+1&1/3^3&(4+4+1)\cdot 1/3^3\\
\hline
4&3+3+3+1&1/3^4&(3+3+3+1)\cdot 1/3^4
\\
\vdots\end{array}$$
$$\mathbb E(X)=\sum_{n=1}^\infty XP(X)=\sum_{n=1}^\infty ((n-1)\cdot 2^{n-2}\cdot 7+2^{n-1})\cdot \frac1{3^{n}}=8.$$
Wolframalpha answer.
|
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|
A box has $4$ red and $20$ white balls. A person takes $10$ balls. What is the probability that all or none of the red balls were taken? A box has $24$ balls, $4$ red and $20$ white. One person takes $10$ balls and the second the remaining $14$. What is the probability that one of the two people picked up the $4$ red ones?
I don't understand why is this correct.
$$\frac{{20 \choose 6}+{20 \choose 10}}{24 \choose 10} $$
|
You can use the Hypergeometric distribution.
a) The probability that the first person pick up 4 red ones is $\frac{\binom{4}{4}\cdot \binom{20}{6}}{\binom{24}{10}}$
Then the (conditional) probability that the second person does not pick up 4 red ones is just $1.$
b) The probability that the first person does not pick up 4 red ones is $\frac{\binom{4}{0}\cdot \binom{20}{10}}{\binom{24}{10}}$
The (conditional) probability that the second person pick up 4 red one is $\frac{\binom{4}{4}\cdot \binom{10}{10}}{\binom{14}{14}}=1$
So in total we have
$$\frac{\binom{4}{4}\cdot \binom{20}{6}}{\binom{24}{10}}+\frac{\binom{4}{0}\cdot \binom{20}{10}}{\binom{24}{10}}=\frac{ \binom{20}{6}}{\binom{24}{10}}+\frac{ \binom{20}{10}}{\binom{24}{10}}=\frac{\binom{20}{6}+ \binom{20}{10}}{\binom{24}{10}}$$
|
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Convergence of $\sum_{n=1}^\infty\frac{k^2}{k^2-2k+5}$, $\sum_{n=1}^\infty\frac{6\cdot 2^{2n-1}}{3^n}$, and $\sum_{n=1}^\infty\frac{2^n+4^n}{e^n}$ I have three series questions:
1) $$\sum_{n=1}^{\infty} \frac{k^2}{k^2 - 2k + 5}$$
I'm going to use this theorem:
So $$\lim_{x\to \infty} \frac{x^2}{x^2-2x+5}$$
So $$\lim_{x\to \infty} \frac{2x}{2x-2}$$
So $$\lim_{x\to \infty} \frac{2}{2} = 1$$
Is that right?
2) $$\sum_{n=1}^{\infty} \frac{6 \cdot 2^{2n-1}}{3^n}$$
I'm a tad stuck. I can get to here:
$$\sum_{n=1}^{\infty} \frac{2 \cdot 3 \cdot 2^{2n-1}}{3 \cdot 3^{n-1}}$$
$$\sum_{n=1}^{\infty} \frac{2^{2n}}{3^{n-1}}$$
Is this valid:
$$\sum_{n=1}^{+\infty} \frac{2^{n} \cdot 2^{n}}{3^{n-1}}$$
$$\sum_{n=1}^{+\infty} \frac{2^{n} \cdot 2 \cdot 2^{n-1}}{3^{n-1}}$$
$$\sum_{n=1}^{+\infty} 2^{n+1} \frac{2^{n-1}}{3^{n-1}}$$
So using this theorem:
it looks like this could be:
$$a_n = \frac{1}{1-r} = \frac{2^{n+1}}{\frac{1}{3}}$$
which diverges. Is this right? Is there a better way?
3) $$\sum_{n=1}^{\infty} \frac{2^n + 4^n}{e^n}$$
I have no idea where to even start.
|
For your first series $\sum_{k=1}^{\infty}\frac{k^2}{k^2-2k+5}$, apply the Divergence Test as you did: $$\lim_{x\to\infty}\frac{x^2}{x^2-2x+5}=1\implies\sum_{k=1}^{\infty}\frac{k^2}{k^2-2k+5}\space\text{diverges}$$
For your second series, $$\sum_{n=1}^{\infty}\frac{6\cdot2^{2n-1}}{3^n}=6\sum_{n=1}^{\infty}\frac{2^{2n-1}}{3^n}=6\sum_{n=1}^{\infty}\frac{2^{2n}2^{-1}}{3^n}=3\sum_{n=1}^{\infty}\frac{2^{2n}}{3^n}=3\sum_{n=1}^{\infty}\frac{4^n}{3^n}=3\sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^n$$
Thus we have a geometric series with common ratio $|r|=\frac{4}{3}>1$, hence the series diverges.
For the third series, note that $e\approx2.7$ throughout, we have: $$\sum_{n=1}^{\infty}\frac{2^n+4^n}{e^n}=\sum_{n=1}^{\infty}\frac{2^n}{e^n}+\sum_{n=1}^{\infty}\frac{4^n}{e^n}=\sum_{n=1}^{\infty}\left(\frac{2}{e}\right)^n+\sum_{n=1}^{\infty}\left(\frac{4}{e}\right)^n$$
The first series is a convergent geometric series with $|r|=\frac{2}{e}<1$ and the second is a divergent geometric series with $|r|=\frac{4}{e}>1$. Then recall a series that is the sum of a convergent and divergent series is divergent. So the third series diverges.
|
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The limit of functions of two variables The limit of $f(x,y)=\frac{xy^2}{x^2+y^4}$ as $(x,y) \longrightarrow(0,0)$ is doesn't exist, because if we take two paths:
1) Along the path, $x=0$, $y\longrightarrow0$:
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$
2) Along the path, $x=y^2$, $y\longrightarrow0$:
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{y\longrightarrow0}\frac{y^2y^2}{(y^2)^2+y^4}=\frac{1}{2}$$
But if we use polar coordinate methods of evaluating limits of functions of two variable, we get the limit of the above function becomes zero, i.e.,
$x=r\cos(\theta),y=r\sin(\theta)$, we know that $x^2+y^2=r^2$ and this indicates that $r\longrightarrow0$ as $(x,y)\longrightarrow(0,0)$, therefore
$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\longrightarrow0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$
$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$
Please someone help me on this contradictory, why this is so happened?.
|
You need to be careful and consider all paths as well, not only paths with constant $\theta$.
$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$$
$$=\lim_{r\longrightarrow0}\left(r\color{blue}{\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}}\right)$$
Your argument works if the expression in blue stays bounded (near the origin); for all $(r,\theta)$.
Inspired by your own path $x=y^2$, check what happens in polar coordinates ($r \ne 0$):
$$x=y^2 \longrightarrow r\cos\theta=r^2\sin^2\theta \implies r=\frac{\cos\theta}{\sin^2\theta}$$
Along $r=\frac{\cos\theta}{\sin^2\theta}$ you can take $\theta\to\tfrac{\pi}{2}$ to get $r\to 0$, but what happens with your function along that path?
$$\lim_{r\to 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\stackrel{r=\frac{\cos\theta}{\sin^2\theta}}{\longrightarrow}\ldots$$
|
{
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"url": "https://math.stackexchange.com/questions/3170368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$
and
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac{x+y}{2}$ and $B=\frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated
|
The main trick is here:
\begin{align}
\color{red} {x = {x+y\over2} + {x-y\over2}}\\[1em]
\color{blue}{y = {x+y\over2} - {x-y\over2}}
\end{align}
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $\color{red}x$ and $\color{blue}y,\,$ you will obtain
\begin{align}
\sin \color{red} x - \sin \color{blue }y = \sin \left(\color{red}{{x+y\over2} + {x-y\over2} }\right) - \sin \left(\color{blue }{{x+y\over2} - {x-y\over2}} \right) \\[1em]
\end{align}
All the rest is then only a routine calculation:
\begin{align}
\require{enclose}
&= \sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right) +
\sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\\[1em]
&-\left[\sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right) -
\sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\right]\\[3em]
&= \enclose{updiagonalstrike}{\sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right)} +
\sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\\[1em]
&-\enclose{updiagonalstrike}{\sin \left({x+y\over2}\right) \cos\left( {x-y\over2} \right)} +
\sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)
\\[3em]
&=2\sin \left({x-y\over2}\right) \cos\left( {x+y\over2} \right)\\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3171404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Inequality $|\cos(k)| \geq \frac{1}{2^k}$ for $k\geq 0$ My question : Is it true that $|\cos(k)| \geq \frac{1}{2^k}$ for all integers $k\geq 0$ ?
What I tried : I have checked with a computer that the inequality holds for
$0 \leq k \leq 4\times 10^5$. I can also show that the set of $k$ for which the inequality holds is infinite ; in fact, it has density at least $\frac{1}{2}$.
To see that :
Lemma. For $x\in{\mathbb R}$, one has $(*):|\cos(x+1)|+|\cos(x+2)| \geq \frac{3}{4}$.
Corollary of lemma. For $x\in{\mathbb R}$, either $|\cos(x+1)|\geq \frac{|\cos(x)|}{2}$ or $|\cos(x+2)|\geq \frac{|\cos(x)|}{4}$.
Putting $x=k$ in the corollary, we then see that if the inequality holds
for $k$, it will also hold for $k+1$ or $k+2$.
Proof of lemma. Let $f_1(x)=|\cos(x+1)|$ and $f_2(x)=|\cos(x+2)|$. Since
$f_1$ and $f_2$ are $\pi$-periodic, it suffices to show $(*)$ on $[0,\pi]$. Now,
write $[0,\pi]=I_1 \cup I_2 \cup I_3$ where $I_1=[0,\frac{\pi}{2}-1]$, $I_2=[\frac{\pi}{2}-1,\frac{3\pi}{2}-2]$ and $I_3=[\frac{3\pi}{2}-2,\pi]$. Then
$f_1$ and $f_2$ are concave on each of the $I_j$, so it suffices to check (*) on the endpoints, and this is easy.
UPDATE 04/09/2019 : We can use the fact that $\pi$ has irrationality
measure at most $8$ (as explained in the accepted answer to this older question, thanks to i707107 for providing the link). Indeed (in all this section every inequality is meant to hold for all but finitely many $k$), denote by $a_k$ an integer such that $a_k\frac{\pi}{2}$ is closest possible to $k$, so that $$\Big|k-\frac{a_k\pi}{2}\Big| \leq \frac{\pi}{4}\tag{1}$$.
Then $\Big|\pi-\frac{2k}{a_k}\Big| \geq \frac{1}{a_k^8}$. Now $a_k \sim \frac{2k}{\pi}$, so $a_k \gt \frac{k}{\pi}$. It follows that $$\Big|\frac{a_k\pi}{2}-k\Big| \geq \frac{1}{2a_k^7} \geq \frac{\pi^8}{2^7k^7}\tag{2}$$.
There are then two cases :
If $a_k$ is even then $|\cos(k)|\geq Constant\geq \frac{1}{2^k}$. If, on the other hand, $a_k$ is odd, then then finite increments inequality yields $\Big|\frac{\cos(k)-\cos(\frac{a_k\pi}{2})}{k-\frac{a_k\pi}{2}}\Big| \geq Constant$ so that $|\cos(k)| \geq Constant \times |k-\frac{a_k\pi}{2}|$.
So my inequality holds for all except finitely many $k$.
|
I eventually found a proof, based on i707107's suggestion. By using a computer
program with sufficient precision, one can check that the inequality holds
for $k<316$. So it will suffice to show the inequality for $k\geq 316$.
Denote by $a_k\frac{\pi}{2}$ the integral multiple of $\frac{\pi}{2}$ that's closest to $k$, so that $\varepsilon_k=\big|k-a_k\frac{\pi}{2}\big| \leq \frac{\pi}{4}$. Then $k=a_k\frac{\pi}{2}\pm\varepsilon_k$.
If $a_k$ is even, then $\cos(a_k\frac{\pi}{2})=\pm 1$ and hence $|\cos(k)|=|\cos(\varepsilon_k)| \geq \frac{\sqrt{2}}{2}$ and we are done. So we may assume without loss that $a_k$ is odd.
Then $\cos(a_k\frac{\pi}{2})=0$. Using the finite increments formula, there is
a $\xi \in [a_k\frac{\pi}{2}-\frac{\pi}{4},a_k\frac{\pi}{2}+\frac{\pi}{4}]$ such that $\frac{\cos(k)-\cos(a_k\frac{\pi}{2})}{k-a_k\frac{\pi}{2}}=-\sin(\xi)$. It follows that $\Big|\frac{\cos(k)-\cos(a_k\frac{\pi}{2})}{k-a_k\frac{\pi}{2}}\Big| \geq \frac{\sqrt{2}}{2}$, so that
$$
\begin{array}[lcl]
\big|\cos(k)\big| &\geq& \frac{\sqrt{2}}{2} \Big|k-a_k\frac{\pi}{2}\Big| \\
&=& \frac{\sqrt{2}a_k}{4} \Big|\frac{2k}{a_k}-\pi\Big| \\
&\geq& \frac{\sqrt{2}a_k}{4} \frac{1}{a_k^{42}}=\frac{1}{\sqrt{8}a_k^{41}} \\
&\geq& \frac{1}{\sqrt{8}\big(\frac{2k}{\pi}+\frac{1}{2}\big)^{41}} \\
\end{array}
$$
So, the inequality we want follows from
$$
2^{k} \geq \sqrt{8}\bigg(\frac{2k}{\pi}+\frac{1}{2}\bigg)^{41}
$$
which can be written as $f(k)\geq 0$ where $f(k)=2^{\frac{k-\frac{3}{2}}{41}}-\big(\frac{2k}{\pi}+\frac{1}{2}\big)$. We have
$f'(x)=\frac{\ln(2)}{41}2^{\frac{k-\frac{3}{2}}{41}}-\frac{2}{\pi}$ so that
$f'(x)\geq 0$ for $x\geq 3$. So for $k\geq 316$, we have $f(k)\geq f(316) \gt 0$
which finishes the proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Analytical solution to $\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2} \cdot \sqrt{b^2-x^2}$ Can't find beauty analytical solution to such equation :
$$\sqrt{a^2-x^2} + \sqrt{b^2-x^2} = \sqrt{a^2-x^2}\cdot\sqrt{b^2-x^2}$$
Assuming $a,\ b \in \mathbb{N},\ a \le b,\ x \in \mathbb{R}$
Is it possible to find a solution for general case?
|
Hint:
Let $\sqrt{a^2-x^2}=y$. We have
$$y+\sqrt{b^2-a^2+y^2}=y\sqrt{b^2-a^2+y^2},$$
$$y=(y-1)\sqrt{b^2-a^2+y^2}.$$
Squaring, you will obtain a quartic equation, which doesn't seem to simplify.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3179631",
"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\lim_{x\rightarrow\infty}\frac{x^3}{3^x}$. What is wrong with my calculations? $\frac{x^3}{3^x}=\frac{\exp(3\cdot\ln(x))}{\exp(x\cdot\ln(3))}=\exp(3\ln(x)-x\ln(3))$
$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$
I will use L'Hospital
$\frac{\frac{18(\ln(x))}{x}+2(\ln(3))^2x}{\frac{3}{x}+\ln(3)}$
And this goes to infinity, for $x\rightarrow \infty$
Therefore
$3\ln(x)-x\ln(3)$ goes to $\infty$ and $\exp(3\ln(x)-x\ln(3))$ must also go to infinity
|
This is incorrect:
$$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2+x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$
There should be a minus sign in the numerator:
$$3\ln(x)-x\ln(3)=\frac{9(\ln(x))^2-x^2(\ln(3))^2}{3\ln(x)+x\ln(3)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3181472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-x-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}=-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}$$I know that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) \rightarrow 0$$But I don't know how to show that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) = 0$$ Can you help me?
|
An easy method is to consider the function $f(x)=\arctan{x}+\arccos{\frac{x}{\sqrt{x^2+1}}}$ then show that $f’(x)=0$ and deduce that $f(x)$ is a constant function and therefore $f(x)=f(0)=\pi/2$
|
{
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"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
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|
Is The Series $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$ Divergent This is my solution for $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$. First I let the sequence in the series be labeled $A$, then I constructed a new sequence ($B$) that would have the similar behavior as $A$. This gives be the following:
$$A=\frac{n^4-3n+2}{4n^5+7}$$
$$B=\frac{1}{n}$$
Next I compare the value of the two sequances at $n=1$. This gives me $A=0$ and $B=1$, using this information I can concluded that $B \geq A$. Since I know that the $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent, then so must $\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}$.
Does this solution make sense? I feel like i am missing something
|
For sufficiently large $N_0$, we have $n \ge N_0$, then $$n^4-3n+2\ge \frac12 n^4$$
and $$4n^5+7 \le 5n^5.$$
That is $n \ge N_0$, we have
$$\frac{n^4-3n+2}{4n^5+7}\ge \frac{\frac12n^4}{5n^5}=\frac1{10n}$$
$\sum_{n=1}^{\infty} \frac{n^4-3n+2}{4n^5+7}= \sum_{n=1}^{N_0-1} \frac{n^4-3n+2}{4n^5+7} + \sum_{n=N_0}^{\infty} \frac{n^4-3n+2}{4n^5+7}\ge \sum_{n=1}^{N_0-1} \frac{n^4-3n+2}{4n^5+7} + \frac1{10}\sum_{n=N_0}^{\infty} \frac{1}{n}$
Hence the series diverges.
|
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"url": "https://math.stackexchange.com/questions/3185326",
"timestamp": "2023-03-29T00:00:00",
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|
$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$ Let $x,y,u,v \in \mathbb{R}.$ Prove that
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Proof 1:
$$\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} \geq \sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}$$
Square both side, we have
$$2\sqrt{(x^2+xy+y^2)(u^2+uv+v^2)} \geq 2xu+xv+yu+2yv$$
If $2xu+xv+yu+2yv<0$ then the inequality is true.
If $2xu+xv+yu+2yv \geq 0$ the square both side, we have
$$x^2v^2-2xvyu+y^2u^2 \geq 0$$
$$ \Leftrightarrow (xv-yu)^2 \geq 0$$
which is true.
Thus the inequality is true for all $x,y,u,v \in \mathbb{R}.$
Proof 2 (need help):
Because
$$x^2+xy+y^2 \geq \frac{3}{4}(x+y)^2$$
for all $x,y \in \mathbb{R},$ we have
$$\begin{matrix}
\sqrt{x^2+xy+y^2} + \sqrt{u^2+uv+v^2} & \geq & \frac{\sqrt{3}}{2}(|x+y|+|u+v|)\\
& \geq & \frac{\sqrt{3}}{2}|x+u+y+v|\\
& = & \frac{\sqrt{3}}{2}\sqrt{(x+u+y+v)^2}\\
& = & \frac{\sqrt{3}}{2}\sqrt{(x+u)^2+2(x+u)(y+v)+(y+v)^2}
\end{matrix}$$
I'm stuck here. Do you have any idea? Thank you.
|
By Minkowski (the triangle inequality) we obtain:
$$\sqrt{x^2+xy+y^2}+\sqrt{u^2+uv+v^2}=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2}+\sqrt{\left(u+\frac{v}{2}\right)^2+\frac{3}{4}v^2}\geq$$
$$\geq\sqrt{\left(x+u+\frac{y+v}{2}\right)^2+\frac{3}{4}(y+v)^2}=\sqrt{(x+u)^2+(x+u)(y+v)+(y+v)^2}.$$
|
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|
Prove $ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$
Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$
So I started by combining the two fractions, which gave me:
$$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{2\sin\theta\cos\theta}{1-\cos^2\theta} = \frac{\sin2\theta}{1-\cos^2\theta}$$
I wasn't sure where to go from here considering I'm aiming for $2\cos\theta / \sin\theta$
|
As $\sin^2t=1-\cos^2t=(1-?)(1+?)$
$$\dfrac{\sin t}{1\pm\cos t}=\dfrac{1\mp \cos t}{\sin t}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3187825",
"timestamp": "2023-03-29T00:00:00",
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|
Determine $N$ of series $\sum_{n=1}^{N}\frac{n^n}{(2n+1)!}$ so that it differs from the actual sum by less than $\frac{1}{200}$ I can establish that:
$$\frac{n^n}{(2n+1)!}=\frac{n^n}{(2n)!(2n+1)}\le\frac{n^n}{n!}$$
But
$$\sum_{n=1}^{+\infty}\frac{n^n}{n!}$$
diverges (by the ratio test). And even if it converged I wouldn't have info on the N that I need.
I might say that:
$$\frac{n^n}{(2n+1)!}=\frac{n^n}{(2n)!(2n+1)}\le\frac{n^n}{2^n}=(\frac{n}{2})^n$$
which diverges as well.
I should be able to use the comparison test but I can't see how.
|
Note that\begin{align}\frac{\dfrac{(n+1)^{n+1}}{(2n+3)!}}{\dfrac{n^n}{(2n+1)!}}&=\frac{(n+1)^{n+1}}{n^n}\times\frac{(2n+1)!}{(2n+3)!}\\&=(n+1)\left(1+\frac1n\right)^n\times\frac1{(2n+2)(2n+3)}\\&=\frac12\left(1+\frac1n\right)^n\frac1{2n+3}\\&<\frac1{2n+3}\text{ since $\left(1+\frac1n\right)^n<e<4$}\\&<\frac12.\end{align}So,\begin{align}\left(\sum_{n=1}^\infty\frac{n^n}{(2n+1)!}\right)-\left(\sum_{n=1}^N\frac{n^n}{(2n+1)!}\right)&=\sum_{n=N+1}^\infty\frac{n^n}{(2n+1)!}\\&=\frac{(N+1)^{N+1}}{(2N+3)!}+\frac{(N+2)^{N+2}}{(2N+5)!}+\cdots\\&<\frac{(N+1)^{N+1}}{(2N+3)!}+\frac{(N+1)^{N+1}}{(2N+3)!}\times\frac12+\frac{(N+1)^{N+1}}{(2N+3)!}\times\frac1{2^2}+\cdots\\&=\frac{(N+1)^{N+1}}{(2N+3)!}\left(1+\frac12+\frac1{2^2}+\cdots\right)\\&=2\frac{(N+1)^{N+1}}{(2N+3)!}.\end{align}Since$$2\frac{4^4}{9!}=\frac1{2\,835}<\frac1{200},$$taking $N=3$ will be enough.
|
{
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"url": "https://math.stackexchange.com/questions/3189738",
"timestamp": "2023-03-29T00:00:00",
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|
Conditions for $ax^2+ax+1$ to be a perfect square. I look for the conditions for $ax^2+ax+1$ to be a perfect square, like some sort of recursive algorithm. I have a fixed $a$ and I would like to express $x$ in terms of $a$ or any other form. (My peculiar case is $52x^2+52x+1$). Any help would be appreciated!
|
if $a=4$, $x$ is any integer.
Assume $a \ne 0$,
$ax^2+ax+1 = 1$, solution $x = 0, -1$
$ax^2+ax+1 = 4$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+12}}}-1\right)$
$ax^2+ax+1 = 9$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+32}}}-1\right)$
$ax^2+ax+1 = 16$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+60}}}-1\right)$
$ax^2+ax+1 = 25$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+96}}}-1\right)$
$ax^2+ax+1 = 36$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+140}}}-1\right)$
$ax^2+ax+1 = 49$, solution $x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+192}}}-1\right)$
Any pattern ? 12, 32, 60, 96, 140, 192,...
12 - 0 = 10 + 2
32 - 12 = 20 - 0
60 - 32 = 30 - 2
96 - 60 = 40 - 4
140 - 96 = 50 - 6
192 - 140 = 60 - 8
Seems there is a pattern.
$$x = \frac{1}{2}\left(\frac{1}{\sqrt{\frac{a}{a+\sum_{i=1}^{z}(8i+4)}}}-1\right)$$
where, $z \in \mathbb{N}$
|
{
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"url": "https://math.stackexchange.com/questions/3190477",
"timestamp": "2023-03-29T00:00:00",
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|
Value of $\lim\limits_{n\rightarrow \infty}(a_{1}+a_{2}+\cdots +a_{n})$
If $\displaystyle a_{n}=\bigg(\frac{n!}{1\cdot 3 \cdot 5 \cdot 7\cdot...\cdot (2n+1)}\bigg)^2.$
Then $\displaystyle \lim_{n\rightarrow \infty}\bigg(a_{1}+a_{2}+...+a_{n}\bigg)$ is
Options:
$(a)$ Does not exists
$(b)$ Greater than $\displaystyle \frac{4}{27}$
$(c)$ Less than $\displaystyle \frac{4}{27}$
$(d)$ None of these
My Try: $$a_{n}=\bigg[\frac{n!\cdot 2\cdot 4 \cdot 6 \cdots (2n)}{1\cdot 2 \cdot 3\cdot 4\cdot \cdots (2n)\cdots (2n+1)}\bigg]^2$$
$$a_{n}=\bigg[\frac{n!\cdot 2 \cdot 4 \cdot 6 \cdots (2n)}{(2n+1)!}\bigg]^2$$
Could some help me to solve it , Thanks
|
Using the Wallis formula
$$\dfrac\pi2 = \dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot\dfrac{6\cdot6}{5\cdot7}\dots$$
in the form of
$$a_n=\left(\dfrac{1\cdot2\cdot3\dots n}{3\cdot 5\cdot7\dots(2n+1)}\right)^2 < \frac\pi{(2n+1)2^{2n+1}},$$
one can get
$$a_1+a_2+a_3+a_4+\dots < \frac19+\dfrac\pi{2^5}\left(\dfrac15+\dfrac1{7\cdot4}+\dfrac1{9\cdot4^2}+\dots\right)$$
$$ < \frac19+\dfrac\pi{5\cdot2^5}\cdot\dfrac1{1-\dfrac14}
= \dfrac19+\dfrac\pi{120}\color{brown}{\mathbf{<\dfrac4{27}}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3191145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Checking differentiability of $e^\frac{-1}{x}$
Let $f:\mathbb R\to\mathbb R$ is defined by
$$f(x)=
\begin{cases}
e^\frac{-1}{x}&&x>0\\
0&&x\le0
\end{cases}$$
How do I check differentiability of $f(x)$ at $x=0$?
I have tried to use first principle but cannot proceed.
|
Use L'Hôpital's rule repeatedly as you approach $x = 0$ from the right:
\begin{align*}
\lim\limits_{x \to 0^+}f'(x) &= \lim\limits_{x \to 0^+}\frac{e^{-\frac{1}{x}}}{x^2} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{1}{x^2}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+}\frac{-\frac{2}{x^3}}{e^{\frac{1}{x}} \cdot \big(-\frac{1}{x^2}\big)} \\
&= \lim\limits_{x \to 0^+}\frac{\frac{2}{x}}{e^{\frac{1}{x}}} \quad\text{limit of the form $\frac{\infty}{\infty}$} \\
&= \lim\limits_{x \to 0^+}\frac{-\frac{2}{x^2}}{e^{\frac{1}{x}} \cdot \big(-\frac{1}{x^2}\big)} \\
&= \lim\limits_{x \to 0^+}\frac{2}{e^{\frac{1}{x}}} \\
&= \lim\limits_{x \to 0^+}2e^{-\frac{1}{x}} = 0 \\
\end{align*} Of course, approaching $x = 0$ from the left, $f(x)$ is just the constant function $0$; so $$\lim\limits_{x \to 0^-}f'(x) = \lim\limits_{x \to 0^-} 0 = 0$$ Thus, $f(x)$ is differentiable at $x = 0$ and the value of its derivative there is $0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx$ - correctness of the solution The task is to find the improper integral: $$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx.$$
My solution is (the main part):
$$\int \limits_{-1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx = \int \limits_{-1}^1 \frac{1}{(x-1)^{\frac{2}{3}}} dx + \int \limits_{1}^2 \frac{1}{(x-1)^{\frac{2}{3}}} dx$$
$$
\int \limits_{-1}^1 \frac{1}{(x-1)^{\frac{2}{3}}} dx = \lim \limits_{c \to 1^-} \int_{-1}^c \frac{1}{(x-1)^{\frac{2}{3}}} dx =
\lim \limits_{c \to 1^-} \Big{[} \frac{ (x-1)^{\frac{1}{3}} }{ \frac{1}{3} } \Big{]}_{-1}^c = 3 \lim \limits_{c \to 1^-} \Big{[} (x-1)^{\frac{1}{3}} \Big{]}_{-1}^c =
$$
$$
= 3 \lim \limits_{c \to 1^-} \Big{(} (c-1)^{\frac{1}{3}} - (-2)^{\frac{1}{3}} \Big{)} = - 3 (-2)^{\frac{1}{3}} \approx 1.11012 - 3.27337 \textrm{ i}
$$
Is a complex number a sensible answer here? Shouldn't it be a real number? I am missing something or forgetting about something?
|
Note that your answer $$(-2)^{\frac{1}{3}} =-1.259921...$$ is a real number not a complex one.
In general, one of the cube roots of a real number is always real and in this case we just consider the real answer for the integral.
|
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|
Finding $\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$ $$\lim_{n\to \infty} \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$$
$$\lim_{n\to \infty} \sum_{r=0}^{n-1} \left[\frac{1}{\sqrt{n^2+(r)n}}\right]$$
How can I convert it to Riemann sum because the higher limit is $(n-1)$ not $n$.
|
This is the same as
$$\sum_{r=0}^n\left(\frac1{n\sqrt{1+\frac{r}n}}\right)-\frac1{n\sqrt{1+\frac{r}{n}}}=\frac1n\sum_{r=0}^n\left(\frac1{\sqrt{1+\frac{r}n}}\right)-\frac1{n\sqrt{1+\frac{r}{n}}}$$
Taking the limit gives
$$\lim_{n\to\infty}\left(\frac1n\sum_{r=0}^n\left(\frac1{\sqrt{1+\frac{r}n}}\right)-\overbrace{\frac1{n\sqrt{1+\frac{r}{n}}}}^{0}\right)=\int_0^1\frac{\mathrm{d}x}{\sqrt{1+x}}=[2\sqrt{1+x}]_0^1=\boxed{2(\sqrt{2}-1)}$$
|
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|
Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now, divide both sides by three:
$$x^2 - \frac{4}{3}x = \frac{2}{3}$$
Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:
$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$
Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:
$$
\begin{align}
x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\
\left(x - \frac{2}{3}\right)^2 = \frac{10}{9}
\end{align}
$$
Can anyone please explain how they went from the first step to the second step?
|
Well, $(x-a)^2=(x-a)(x-a)=x^2-2ax+a^2$ with $a=\frac{2}{3}$ gives the last line.
|
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|
Minimum possible values of $\frac{x^2+y^2+z^2+1}{xy+yz+z}$ and $\frac{x^2+y^2+z^2+1}{xy+y+z}$ Let $k$ and $m$ be the minimum possible values of $$\frac{x^2+y^2+z^2+1}{xy+yz+z} \quad \text{and} \quad \frac{x^2+y^2+z^2+1}{xy+y+z}$$ respectively where $x,y,z$ are non-negative real numbers. What is the value of $km+k+m$?
I used the AM-GM inequality to get a minimum value for $x^2+y^2+z^2+1$ but the problem is with getting an upper bound for $xy+yz+z$ and $xy+y+z$.
This problem is from India IMC 2017 team contest.
|
Here is a way to use AM-GM. In the first case, you can find $k$ if you can find suitable $\alpha, \beta$ s.t. the following AM-GMs can achieve equality simultaneously:
$$x^2+\alpha^2 y^2 \geqslant 2\alpha x = kxy\\ (1-\alpha^2)y^2+\beta^2z^2\geqslant 2\sqrt{1-\alpha^2}\beta yz = kyz \\ (1-\beta^2)z^2+1 \geqslant 2\sqrt{1-\beta^2}z=kz$$
as summing above gets $x^2+y^2+z^2+1\geqslant k(xy+yz+z)$. Solving $2\alpha = 2\sqrt{1-\alpha^2}\beta=2\sqrt{1-\beta^2}=k$ to get $k = \sqrt5-1$, with equality when $x=1, y=z=\varphi = \frac12(\sqrt5+1)$, so we have our minimum.
Similar approach in the second case gives $m = \sqrt5-1$ also, but this time with equality when $y=1, x=z=\varphi-1$. Calculate $km+k+m = 4$ to finish.
|
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|
Using a Lyapunov function to classify stability and sketching a phase portrait
Consider the system
$$x' = -x^3-xy^{2k}$$
$$y' = -y^3-x^{2k}y$$
Where $k$ is a given positive integer.
a.) Find and classify according to stability the equilibrium solutions.
$\it{Hint:}$ Let $V(x,y) = x^2 + y^2$
b.) Sketch a phase portrait when $k = 1$
$\it{Hint:}$ What are $x'$ and $y'$ when $y=ax$ for some real number $a$?
a.)
Using $V$, we get $\frac{d}{dt}V=2xx'+2yy'$
Plugging in our system , we get:
$$\frac{d}{dt}V=2x(-x^3-xy^{2k})+2y(-y^3-x^{2k}y)$$
$$=-(x^4+y^4)-x^2y^{2k}-x^{2k}y^2<0$$
I dropped the $2$ since it doesn't matter to determine stability. We see that our own equilibrium is $(0,0)$ since setting $x'=0$ we get
$$y^{2k}=-x^2$$
Which only works for $x=y=0$
Therefore our system is asymptotically stable at the origin.
I am having trouble with b.), mostly because the hint is confusing me.
Let $y=ax$, then our system becomes
$$x'=-x^3-a^2x^3=-x^3(1+a^2)$$
$$y'=-a^3x^3-ax^3=-ax^3(1+a^2)$$
I am not sure what to do with this. Using linearization doesn't work since the Jacobian will be the zero vector at the point of interest. I have never had a problem that asks to draw a phase portrait when linearization doesn't work, so I am hoping someone more clever than me can offer some advice.
|
Phase portraits - a partial offering
Below are phase portraits for $k=1,2,5$. The red lines indicate the null clines where $\dot{y}=0$ and $\dot{y}=0$.
$k = 1$
The linear system is
$$\begin{align}
\begin{split}
\dot{x} &= -x^{3} - xy^{2} = -x \left( x^{2} + y^{2} \right) \\
\dot{y} &= -y^{3} - x^{2}y = -y \left( x^{2} + y^{2} \right)
\end{split}
\end{align}$$
$$ \dot{r} = \frac{x \dot{x} + y \dot{y}}{r} = -r^{3} $$
The lone critical point is the origin.
When $y = a x$, $a\in\mathbb{R}$, we have
$$\begin{align}
\begin{split}
\dot{x} &= -x^{3}\left( 1 + a^{2} \right) \\
\dot{y} &= -a y^{3}\left( 1 + a^{2} \right)
\end{split}
\end{align}$$
$k = 2$
$$\begin{align}
\begin{split}
\dot{x} &= -x^{3} - xy^{4} = -x \left( x^{2} + y^{4} \right) \\
\dot{y} &= -y^{3} - x^{4}y = -y \left( x^{2} + y^{2} \right)
\end{split}
\end{align}$$
$$ \dot{r} = \tfrac{1}{8} r^3 \left(\left(r^2-2\right) \cos (4 \theta )-r^2-6\right) $$
The bounding curves for $\dot{r}$ are when $\cos 4\theta = 1$
$$\dot{r} = -r^{3}$$
and when $\cos 4\theta = -1$
$$\dot{r} = -\tfrac{1}{4} r^3 \left(r^2+2\right)$$
The bounding curves cross at $r=\sqrt{2}$. At no point is $\dot{r}$ ever positive.
|
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|
Show uniform convergence and pointwise convergence for $\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$
Consider the series:
$$\sum_{n=1}^ \infty \frac{z^ {n-1}}{(1-z^n)(1-z^ {n+1})}$$
show this converges to:
(a) $\frac{1}{(1-z)^2}$ for $|z|<1$
(b) $\frac{1}{z(1-z)^2}$ for $|z|>1$
Finally, show that this convergence is uniform for $|z| \leq c <1$ in (a) and for $|z| \geq c >1$ in (b).
This is my first course in complex analysis and I'm struggling a bit with seeing why this is true. Question (a) suggests recognising the product of two geometric series $$\frac{1}{1-z} \cdot \frac{1}{1-z}= \sum z^ k \cdot \sum z^ k $$
But I do not see how to rewrite the expression. I have no idea how one would derive the expression in question $(b)$ any hints on that would be greatly appreciated. Once I know how to derive these expressions I will maybe also know a way to apply the Weierstrass M-test for uniform convergence. Since the uniform limit must equal the pointwise limit all we need to show that the series converges uniformly on some domain of definition. Can someone provide some guidance on how to tackle this question, drop some small hints so I can maybe proceed?
|
Use partial fraction decomposition and exploit a telescoping sum to obtain the pointwise convergence results in (a) and (b). For example, we have
$$\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} = \frac{z^{n-1}}{z^n - z^{n+1}}\left(\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}} \right) \\ = \frac{1}{z(1-z)}\left(\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}} \right)$$
I'll leave the remainder of the details to you. The technique is illustrated in more detail below in the argument for uniform convergence.
Regarding uniform convergence, note that
$$\begin{align}\left|\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} \right| = \frac{|z|^{n-1}}{|1-z^n||1-z^{n+1}|} \leqslant \frac{|z|^{n-1}}{(1-|z|^n)(1-|z|^{n+1})} \end{align}$$
where the last inequality follows from application of the the reverse triangle inequality, $|1 - z^m| \geqslant 1 - |z|^m$, to the expressions in the denominator.
If $|z| \leqslant c < 1$, then we have
$$\tag{*}\left|\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} \right| \leqslant \frac{c^{n-1}}{(1- c^n)(1- c^{n+1})} = \frac{c^{n-1}}{c^n -c^{n+1}} \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right) \\ = \frac{1}{c(1-c)} \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right)$$
The series with terms appearing on the RHS of (*) is telescoping and convergent
$$\sum_{n=1}^\infty \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right)= \lim_{m \to \infty}\sum_{n=1}^m \left(\frac{1}{1-c^n}-\frac{1}{1 - c^{n+1}}\right) = \frac{1}{1-c} - \lim_{m\to \infty}\frac{1}{1 - c^{m+1}} \\ = \frac{1}{1-c}-1$$
Therefore, we have uniform convergence for $|z| \leqslant c < 1$, by the Weierstrass test.
For the case where $|z| \geqslant c > 1$, rearrange as
$$\left|\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} \right| = \frac{|z|^{-(n+2)}}{|1 - z ^{-n}||1- z^{-(n+1)}|} \leqslant \frac{|z|^{-(n+2)}}{(1 - |z |^{-n})(1- |z|^{-(n+1)})}, $$
and proceed in a similar way as before to apply the Weierstrass test.
|
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|
How prove with using mathematical induction $\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$?
Prove the identity
\begin{align}
\prod_{i=0}^n \left(1+q^{2^i}\right) = \frac{1-q^{2^{n+1}}}{1-q}
\end{align}
for each nonnegative integer $n$.
To begin with, I cannot verify the equality itself.
$\prod_{i=0}^n 1+q^{2^i} = \frac{1-q^{2^{n+1}}}{1-q}$
If $n=1$ then I have next expressions
$(1+q)(1+q^2) = \frac{1-q^{2^{1+1}}}{1-q} = $
$ = (1+q)(1+q^2) = \frac{1-q^{4}}{1-q}$
Well played and calculated =(.Speak me where i s mistaked.
|
Let the induction statement be such that
$$P(n):\prod_{i=0}^{n} 1 + q^{2^i}=\dfrac{1-q^{2^{n+1}}}{1-q}$$
We show the base $(n=0)$ case, which states
$$P(0):\prod_{i=0}^{0} 1 + q^{2^0}=\dfrac{1-q^{2^{0+1}}}{1-q}$$
Working from the right hand side, we have
$$\dfrac{1-q^{2^{0+1}}}{1-q}=\dfrac{1-q^{2}}{1-q}=\dfrac{(1-q)(1+q)}{1-q}=1+q=\prod_{i=0}^{0} 1 + q^{2^0}$$
With the base case verified, assume $P(n)$ is true and we must show $P(n+1)$ which states
$$P(n+1):\prod_{i=0}^{n+1} 1 + q^{2^i}=\dfrac{1-q^{2^{(n+1)+1}}}{1-q}$$
This should be straight forward starting from the left hand side.
$$\begin{align}
\prod_{i=0}^{n+1} 1 + q^{2^i}&=\Bigg(\prod_{i=0}^{n} 1 + q^{2^i}\Bigg)\cdot(1+q^{2^{n+1}})
\\
&=\Bigg(\dfrac{1-q^{2^{n+1}}}{1-q}\Bigg)\cdot(1+q^{2^{n+1}})
\\
&=\dfrac{1-q^{2^{n+1}}q^{2^{n+1}}}{1-q}
\\
&=\dfrac{1-q^{2^{n+1}+2^{n+1}}}{1-q}
\\
&=\dfrac{1-q^{2\cdot2^{n+1}}}{1-q}
\\
&=\dfrac{1-q^{2^{n+2}}}{1-q}
\\
&=\dfrac{1-q^{2^{(n+1)+1}}}{1-q}
\end{align}$$
Thus,
$$\prod_{i=0}^{n+1} 1 + q^{2^i}=\dfrac{1-q^{2^{(n+1)+1}}}{1-q}$$
And $P(n+1)$ is true for all $n \in \mathbb{N} \cup \{0\}$.
|
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|
$\sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ Show that $\alpha = \sqrt[3]{3} + \sqrt[3]{9}$ is algebraic over $\mathbb{Q}$ by presenting a polynomial $p$ from $\mathbb{Q}[X]$ with $p(\alpha)$.
It seems like $$\mathbb{Q}(\sqrt[3]{3} + \sqrt[3]{9}) = \{a + b\sqrt[3]{3} + c\sqrt[3]{9},~a,b,c\in\mathbb{Q} \},$$ but I'm stuck proving that it's closed under inverse. If that is true, then the minimal polynomial should be of 3rd degree. True or not, I can't find any such polynomial.
Solution: As @José Carlos Santos mentioned below:
$$\alpha^3=12+9\sqrt[3]3+9\sqrt[3]9=12+9\alpha,$$
from which it follows
$$(x - \sqrt[3]3-\sqrt[3]9)(x^2 + x(\sqrt[3]3+\sqrt[3]9) + (\sqrt[3]3+\sqrt[3]9)^2) - 9(x - \sqrt[3]3 - \sqrt[3]9)= x^3 - 12 - 9(\sqrt[3]3+\sqrt[3]9) -9x + 9(\sqrt[3]3+\sqrt[3]9) = x^3 - 9x - 12 = p(x);~~ p(\alpha) = 0.$$
|
If $\root 3 \of 3$ and $\root 3 \of 9$ are algebraic, then so is $\root 3 \of 3 + \root 3 \of 9$. The polynomial for $\root 3 \of 3$ is obviously $x^3 - 3$, and $x^3 - 9$ for $\root 3 \of 9$.
I wonder if maybe we can just add up the polynomials? We thus get $2x^3 - 12$. By the fundamental theorem of algebra, this has three solutions.
Those solutions are $\root 3 \of 6$, $\omega \root 3 \of 6$ and $\omega^2 \root 3 \of 6$, where $\omega$ is the complex cubic root of 1 with positive imaginary part. None of those numbers are equal to $\root 3 \of 3 + \root 3 \of 9$, so this seems to be a dead end.
However, it does suggest the idea of cubing $\root 3 \of 3 + \root 3 \of 9$ and seeing if that suggests a polynomial, which was probably the thought process behind Jose's answer.
|
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|
If $z=e^{\frac{2\pi i}m}$, then $\sum_{k=1}^mz^k=0$ ($m\neq1$) Suppose $z=e^{\frac{2\pi i}m}$ for $m\in\mathbb N$ and $m\neq1$.
Is the following equality hold?
$$\sum_{k=1}^mz^k=0\tag{1}$$
$(1)$ seems trivial geometrically ;
it says that the sum of all vectors with equal magnitudes and uniform angle differences should be zero.
If $m$ is even ($m=2l$), then
$$z^l=e^{i\pi}=-1$$
and
$$z^k+z^{k+l}=z^k(1+z^l)=0$$
for every $k=1,2,\cdots,l$.
So $(1)$ holds.
If $m$ is odd, then $(1)$ seems nontrivial.
For $m=3$, the question becomes straightforward. If $m=5$, it is quite sophisticated but solvable by using elementary mathematics ; Let $\theta=\frac25\pi$.
Then $(1)$ is equivalent to $(2)$ and $(3)$ where
$$\sum_{k=1}^5\cos k\theta=0\tag{2}$$
and
$$\sum_{k=1}^5\sin k\theta=0\tag{3}$$
$(3)$ is trivial from $\sin(2\pi-\phi)=-\sin\phi$.
For $(2)$, we can use $\cos(2\pi-\phi)=\cos\phi$ and compute as follows
\begin{align*}
\sum_{k=1}^5\cos k\theta
&=\cos\theta+(2\cos^2\theta-1)+(2\cos^2\theta-1)+\cos\theta+1\\
&=4\cos^2\theta+2\cos\theta-1
\tag{4}
\end{align*}
On the other hand, (denote $s=\sin\frac\pi{10}$ and $c=\cos\frac\pi{10}$)
$$2sc=\sin\frac\pi5=\cos\left(\frac\pi2-\frac\pi5\right)=\cos\frac{3\pi}{10}=4c^3-3c$$
Dividing both sides by $c(\neq0)$ yields
\begin{gather*}
2s=4c^2-3\\
4s^2+2s-1=0\tag{5}
\end{gather*}
Since $s=\cos\theta$, the last expression of $(4)$ becomes zero, and this proves $(2)$.
So $(1)$ holds for every even $m$ and $m=3,5$.
But what if $m$ is an odd number grater than $5$?.
Is there a general way to explain $(1)$?
And if it is the case, can I conclude $(2)$ and $(3)$, $m$ in place of $5$?
i.e. do
$$\sum_{k=1}^m\cos\frac{2\pi k}m=0\tag{2*}$$
and
$$\sum_{k=1}^m\sin\frac{2\pi k}m=0\tag{3*}$$
hold for every $m\in\mathbb N\setminus\{1\}$?
|
$\sum\limits_{k=0}^{m-1}z^{k}=\frac {z^{m}-1}{z-1}=0$ because $z^{m}=1$. Now just multiply both sides by $z$.
|
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|
Given that $d$ divides $n$, what is $d + \frac{n}{d} \pmod 4$ As the question titles states, how can I efficiently find whether $d + \frac{n}{d} \equiv 0 \pmod 4$, given that $d$ divides $n$? An example would be with $n = 35$ and $d = 5$.
$5 + \frac{35}{5} = 5+7 = 12 \equiv 0 \pmod 4$, so it "passes". I tried $d + \frac{n}{d} \equiv 0 \pmod 4$, so $\frac{n}{d} \equiv -d \pmod 4$. This means that $n$ must equal $-d^{2}$ (mod 4). Can someone let me know if this is right? If not, what is a good way/are some preconditions of $d$ and $n$ so that $d + \frac{n}{d} \equiv 0 \pmod 4$.
|
If $n$ is odd, then $d$ is odd. In that case $d+(n/d)\equiv 0$ iff $x^2\equiv -n$ because $d$ is a unit $\bmod 4$. But for odd $d$ we must have $d^2\equiv 1$, so the ordered pair $(n,d)$ passes for odd $n$ iff $n\equiv 3\bmod 4$.
If $n$ is even, then $d$ and $n/d$ must both be even (meaning $n$ is a multiple of $4$). Assuming this parity requirement is met, then either both $d$ and $n/d$ must be twice an odd number so $d+(n/d)\equiv 2+2$, like $n=12, d=2$, or else both $d$ and $n/d$ must be multiples of $4$. Among even values of $n$, all multiples of $16$ will have at least one $d$ meeting this criterion, and all remaining multiples of $4$ that are not multiples of $8$ will also have at least one such $d$.
All told, $7/16$ of all $n$ values ($n\in \{0,3,4,7,11,12,15\}\bmod 16$) will have at least one passing value of $d$.
|
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|
Trigonometry substitution issue with sign When solving an integral such as $\displaystyle\int\frac{dx}{\sqrt{x^2+4}}$, you eventually end up with
$$ \ln\lvert\sec\theta+\tan\theta\rvert+C.$$
The next step is to rewrite this in terms of $x$. My book does the following: $x=2\tan\theta$, so by drawing a triangle, it can be seen that $x$ is the opposite side, $2$ the adjacent one, and finally, $\sqrt{2^2+x^2}$ the hypotenuse. Therefore, $$\sec\theta=\frac{\sqrt{2^2+x^2}}2.$$ The problem that I see however is that $\sqrt{2^2+x^2}$ is the magnitude of the hypotenuse, so this has lost a possible negative sign since it is not necessarily true that
$$ \sec\theta = \frac{|r|}{2}.$$
How can this be omitted?
|
When you make the substitution $x=2\tan \theta$, you have to be careful to specify the domain of $\theta$: the substitution is only valid if $\theta$ has a small enough domain for $\tan \theta$ to be continuous. The simplest possible choice of domain is probably $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. Note that the range of $2\tan \theta$ on this domain is the entire real line, so taking $\theta$ in this domain doesn't lose any generality.
But when $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we always have $\sec \theta > 0$. So in fact, if you make this choice of domain, it is always true that $2 \sec \theta=\sqrt{x^2+4}$, without any sign issues.
It's instructive to think about what happens if you choose a different domain for $\theta$. If $\sec \theta > 0$ on that domain, nothing will change. If $\sec \theta < 0$ on that domain, then
$$\int \frac{dx}{\sqrt{x^2+4}}=\int \frac{\sec^2 \theta \,d\theta}{-\sec \theta}=-\int \sec \theta \,d\theta
$$
because $\sqrt{x^2+4}$ is still positive. So the integral in terms of $\theta$ evaluates to $-\ln|\sec \theta +\tan \theta|+C$. Then, when we rewrite in terms of $x$, we again have $\sec \theta=-\sqrt{x^2+4}$, so the integral in terms of $x$ is
$$-\ln\left|-\sqrt{x^2+4}+x\right|+C=-\ln\left(\sqrt{x^2+4}-x\right) +C\, ,$$
because $\sqrt{x^2+4} > x$ for all $x$.
But then
\begin{align*}
-\ln\left(\sqrt{x^2+4}-x\right)&=\ln\left(\frac{1}{\sqrt{x^2+4}-x}\right)\\
&=\ln\left(\frac{\sqrt{x^2+4}+x}{(\sqrt{x^2+4})^2-x^2}\right)&\text{(multiplying by the conjugate)}\\
&=\ln\frac{\sqrt{x^2+4}+x}{4}\\
&=\ln(\sqrt{x^2+4}+x)-\ln 4 \, .
\end{align*}
So we get the nearly same result whatever domain we choose, but the constant term may be different.
|
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|
Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$ How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$
Rationalise the denominator
$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$
This is still not simplify.
|
Hint. Note that
$$(2\pm\sqrt{2})=\sqrt{(2\pm\sqrt{2})^2}=\sqrt{6\pm 4\sqrt{2}}.$$
|
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|
Prove $(c^2-1)/4 \pm 1$ is not divisible by $2, 3, $ or $5$ if $c>3$ is an odd prime In this question asked the other day, RTn conjectured that, if $c>3$ is an odd prime,
then $k=\dfrac{c^2\pm4-1}4=\dfrac{c^2-1}4\pm1$ is prime. As the answers and comments there show,
this conjecture is not true; some numbers of that form are multiples of $7$, $11$, $13$, etc.
I suspect, however, that RTn was misled because $k$ cannot be divisible by $2$, $3$, or $5$.
This question is how to prove this.
I'm taking the option of answering my own question, but I'd appreciate feedback or other answers.
|
If $c>3$ is an odd prime, then $c$ is not divisible by $2, 3, $ or $5,$
unless $c=5, $ in which case the claim can be easily verified.
If $c$ is not divisible by $2,$ then $c^2-1=(c+1)(c-1)$ is the product of two consecutive
even numbers so divisible by $8,$ so $\dfrac {c^2-1}4$ is even, so $\dfrac{c^2-1}4\pm1$ is not divisible by $2$.
If $c$ is not divisible by $3$, then $c+1$ or $c-1$ is, so $c^2-1$ is, so $\dfrac{c^2-1}4$ is (since $4\equiv1\mod3),$
so $\dfrac{c^2-1}4\pm1$ is not.
Finally, if $c$ is not divisible by $5,$
then $c^2\equiv 1$ or $4\mod5$, so $c^2-1\equiv 0$ or $3\mod 5$;
since $4\equiv-1\mod 5,$ that means $\dfrac{c^2-1 }4\equiv0 $ or $2\mod 5$, so $\dfrac{c^2-1}4\pm1\not\equiv0\mod5.$
|
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|
Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to
$$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^3-1)}{n(n-2)!}\right)$$
But this didn't give me any clues to solve this.
I just guess telescoping is the way to solve because this is finite-sum, but I don't have any idea.
|
We have
$$100! \times (1 + \sum_{n=1}^{100} \frac{(-1)^n(n^2 + n + 1)}{n!})$$
$$= 100! \times (1 + \sum_{n=1}^{100} (-1)^n (\frac{n}{(n-1)!} + \frac{n+1}{n!}))$$
because $\frac{n^2 + n + 1}{n!} = \frac{n}{(n-1)!} + \frac{n+1}{n!}$. Expanding, we have
$$= 100! \times (1 + [-\frac{1}{(1-1)!} - \frac{1+1}{1!} + \frac{2}{(2-1)!} + \frac{2+1}{2!} - \dots - \frac{99}{(99-1)!} - \frac{99+1}{99!} + \frac{100}{(100-1)!} + \frac{100 + 1}{100!}])$$
You should be able to see that all terms are cancelled out apart from the first and the last (telescoping, as you suspected).
$$= 100! \times (1 + [-\frac{1}{0!} + \frac{101}{100!}])$$
$$= 100! \times (1 - 1 + \frac{101}{100!})$$
$$= 100! \times \frac{101}{100!}$$
$$= 101$$
|
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|
Calculating $\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$ Calculate $$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
Here is my attempt:
$$\lim_{n\to\infty}\left( \frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}= \left(\frac{4\infty^2+5\infty-6}{4\infty^2+3\infty-10}\right)^{3-4\infty}$$
$$=\left(\frac{\infty(4\infty+5)}{\infty(4\infty+3)}\right)^{-4\infty}=\left(\frac{4\infty}{4\infty}\right)^{-4\infty} = 1^{-4\infty} = \boxed{1}$$
However, when I try to graph the function, I can't reliably get my answer due to precision limitations, and I feel that this method of calculating limits is less than ideal. How can I confirm that this is indeed the limit?
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$$\lim_{n\to\infty}\left(\frac{4n^2+5n-6}{4n^2+3n-10}\right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(\frac{4n^2+3n-10+(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
From $\lim_{n\to\infty} \frac{2n+4}{4n^2+3n-10}=0$,
$$\Rightarrow \lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{3-4n}$$
$$=\lim_{n\to\infty}\left(1+\frac{(2n+4)}{4n^2+3n-10} \right)^{\left(\frac{4n^2+3n-10}{2n+4}\right)\times \left(\frac{2n+4}{4n^2+3n-10}\right) \times 3-4n}$$
$$=\lim_{n\to\infty} e^{\frac{(2n+4)(3-4n)}{4n^2+3n-10}}$$
$$=e^{-2}$$
|
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|
Find the basis of GF(2) I do past papers and I stumbled upon this question in one of the papers. I know what is GF(2), but I have no idea how to find the basis from the given data.
\begin{bmatrix}1\\1\\0\\0\end{bmatrix}
\begin{bmatrix}1\\0\\1\\0\end{bmatrix}
\begin{bmatrix}1\\0\\0\\1\end{bmatrix}
The question asks me to find the basis of GF(2)^4 with the given 3 elements above. I tried to find information online, but could not find any examples upon GF(2). Basically, I need to find one more element, but I have no idea how to do it.
|
Unsophisticated method:
There are $16$ elements of GF$(2)^4:$
$\begin{bmatrix}0\\0\\0\\0\end{bmatrix}
\begin{bmatrix}0\\0\\0\\1\end{bmatrix}
\begin{bmatrix}0\\0\\1\\0\end{bmatrix}
\begin{bmatrix}0\\0\\1\\1\end{bmatrix}
\begin{bmatrix}0\\1\\0\\0\end{bmatrix}
\begin{bmatrix}0\\1\\0\\1\end{bmatrix}
\begin{bmatrix}0\\1\\1\\0\end{bmatrix}
\begin{bmatrix}0\\1\\1\\1\end{bmatrix}
\begin{bmatrix}1\\0\\0\\0\end{bmatrix}
\begin{bmatrix}1\\0\\0\\1\end{bmatrix}
\begin{bmatrix}1\\0\\1\\0\end{bmatrix}
\begin{bmatrix}1\\0\\1\\1\end{bmatrix}
\begin{bmatrix}1\\1\\0\\0\end{bmatrix}
\begin{bmatrix}1\\1\\0\\1\end{bmatrix}
\begin{bmatrix}1\\1\\1\\0\end{bmatrix}
\begin{bmatrix}1\\1\\1\\1\end{bmatrix}
$
Linear combinations of your $3$ elements yield the following $8$:
$\begin{bmatrix}0\\0\\0\\0\end{bmatrix}
\begin{bmatrix}1\\1\\0\\0\end{bmatrix}
\begin{bmatrix}1\\0\\1\\0\end{bmatrix}
\begin{bmatrix}1\\0\\0\\1\end{bmatrix}
\begin{bmatrix}0\\1\\1\\0\end{bmatrix}
\begin{bmatrix}0\\1\\0\\1\end{bmatrix}
\begin{bmatrix}0\\0\\1\\1\end{bmatrix}
\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$
Take an element on the first list that's not on the second list, and you're done.
|
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|
Finding center and rotation angle of ellipse that contains three points Given three points $p_1, p_2, p_3 \in \mathbb{R}^2$, and an ellipse with shape parameters $(a,b)$ (the semi-major and semi-minor), is it possible to determine, if they exist, a center $c \in \mathbb{R}^2$ and a rotation angle $\theta \in [0, \pi]$, such that the ellipse centered at $c$ rotated by $\theta$ contains $p_1, p_2, p_3$?
In other words, let
$$E(p, \theta)=\dfrac{(p_x\cos{\theta} + p_{y}\sin{\theta})^2}{a^2} + \dfrac{(p_x\sin{\theta} -p_y\cos{\theta})^2}{b^2}$$
I want to determine $c\in \mathbb{R}^2$ and $\theta \in [0, \pi]$, such that:
\begin{equation}
E(p_1-c, \theta) = 1\\
E(p_2-c, \theta) = 1\\
E(p_3-c, \theta) = 1
\end{equation}
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Let the three points be $A,B,C$. The idea of this solution is to transform $\triangle ABC$ into an equilateral triangle with vertices $(0,0), (1,0), (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) $
There are three possible arrangements for the vertices in terms of the pre-images of the equilateral triangle vertices:
$A , B , C$
$A , C , B$
$B , C , A$
The affine transformation is
$f(P) = T (p - V_1)$
where $V_1$ is the first vertex.
Take for example the first arrangement, then
$ T \begin{pmatrix} B- A && C - A \end{pmatrix} = \begin{pmatrix} 1 && \dfrac{1}{2} \\ 0 && \dfrac{\sqrt{3}}{2} \end{pmatrix} $
From which, we can calculate the transformation matrix $T$.
Now, we'll pass an ellipse in standard orientation through the three vertices of the equilateral triangle, let the center of the ellipse be $ C = (1/2 , c )$, then the equation of ellipse
$ \dfrac{(x - (1/2))^2}{ a^2 } + \dfrac{(y - c)^2} {b^2} = 1 $
The points through which this ellipse passes are $(0,0), (1,0), (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} ) $
Substituting the first is the same as substituting the second, they both give us
$ \dfrac{1}{4 a^2} + \dfrac{c^2}{ b^2} = 1 $
Substituting the third point,
$ \dfrac{ (\sqrt(3)/2 - c ) ^2}{ b^2 }= 1 $
So that $ b = sqrt(3)/2 - c $
Now define the matrix
$Q = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix}$
The pre-transformation Q is obtained as follows:
The position vectors are related by $r = T (r' - A)$ where $r'$ is the position vector of the original points, and $r$ is the position vector after applying the transformation.
Now, we have the ellipse that passes through the three vertices of the equilateral triangle is given by
$ (r - C)^T Q (r - C) = 1 $
where $C = (1/2, c )$
Substitute $r$
$(T (r' - A) - C )^T Q (T (r' - A) - C) = 1 $
And this simplifies to
$ (r' - A - T^{-1} C ) ^T T^T Q T ( r' - A - T^{-1} C ) = 1 $
so our $Q' = T^T Q T$
Expanding,
$Q' = \begin{bmatrix} (1/a^2) t_1^2 + (1/b^2) t_3^2 && (1/a^2) t_1 t_2 +(1/b^2) t_3 t_4 \\(1/a^2) t_1 t_2 +(1/b^2) t_3 t_4 && (1/a^2) t_2^2 + (1/b^2) t_4^2 \end{bmatrix} $
where $T = \begin{bmatrix} t_1 && t_2 \\ t_3 && t_4 \end{bmatrix} $
we want the eigenvalues of $Q'$ to be $ 1/a_e^2 $ and $ 1/b_e^2 $ where $a_e, b_e$ are the given semi-axes lengths of the ellipse.
so using the trace and the determinant, we obtain
$ \dfrac{1}{a^2} (t_1^2 + t_2^2) + \dfrac{1}{b^2} (t_3^2 + t_4^2 ) = \dfrac{1}{a_e^2} + \dfrac{1}{b_e^2} \hspace{30pt}(1) $
and for the determinant expression, it simplifies to
$ \dfrac{1}{a b} | t_1 t_4 - t_2 t_3 | = \dfrac{1}{a_e b_e} \hspace{30pt} (2) $
Now recall that
$ \dfrac{1}{4 a^2} + \dfrac{c^2}{ b^2} = 1 $
and that
$ c = \dfrac{\sqrt{3}}{2} - b $
substituting this last equation in the previous one, gives
$ \dfrac{1}{4 a^2} + \dfrac{3}{4 b^2} - \dfrac{\sqrt{3}}{b} = 0 \hspace{30pt}(3)$
This system of equations $(1), (2), (3)$ in the two unknowns $a , b$ is overdetermined. So the strategy to solve it is to solve $(1), (2)$ and then check that $(3)$ is satisfied. Once all three equations are satisified then we have our ellipse, and we can compute the center and the $Q'$ matrix from which by diagonalizing we can compute $\theta $.
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|
Calculate a limit: $\lim_{t \to 0+} {1\over 2}\cdot({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}) $ Please calculate the limit $\lim\limits_{t \to 0+} {1\over 2}\cdot \left({\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}} - {1\over t^2}\right)$ and provide the corresponding procedure.
The answer is $\pi^2 \over 6$.
I tried L'Hospital's Rule but I failed.
The following is not necessary to read:
The background of this problem:
this problem arose from a problem related to the Fourier Transform. I tried to use Poisson's summation formula to get $\sum\limits_{n = -\infty}^\infty \frac{1}{n^2 + t^2} = {\pi \over t}\cdot\frac{1+e^{-2\pi t}}{1 - e^{-2\pi t}}$ . (This can be proved by let $f(x) = \frac{1}{x^2+t^2}$, where $t>0$ is a parameter. Then $\hat{f}(n) = {\pi \over t}\cdot e^{-2\pi |n| t}$.) Then $\sum\limits_{n = 1}^\infty {1\over n^2} = \frac{\pi^2}{6} $ should be a corollary.
|
Use Taylor series of an exponential function so that a direct computation shows
\begin{align*}&\frac{1}{2} \bigg(\frac{\pi}{t}\frac{1+e^{-2\pi t} }{1-e^{-2pi t} } -
\frac{1}{t^2} \bigg)
\\&= \frac{1}{2t} \bigg\{ \pi \frac{2+(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }{
(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }(-1) - \frac{1}{t} \bigg\}
\\&= \frac{1}{4 t^2 } \bigg\{ \frac{
2+(-2\pi t) + \frac{(-2\pi t)^2}{2}+\cdots }{ 1 + \frac{(-2\pi
t)^1}{2} + \frac{(-2\pi t)^2}{3!}+\cdots }-2 \bigg\}
\\&= \frac{1}{4t^2}
\frac{ \frac{(-2\pi t)^2}{6}+\cdots }{ 1+ \frac{(-2\pi t)}{2}+\cdots }\end{align*}
|
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|
Name for this method of factoring quadratic and are there any textbooks that describe it? I remember learning this method of factoring quadratics in middle school or high school, but looking for a name or more information on it leads me to dead ends.
Given:
$ax^2+bx+c=0$
$d*e=a*c$
$d+e=b$
Then the factorization of the quadratic is:
$(x+\frac{e}{a})*(x+\frac{d}{a})$
Proof:
$(x+\frac{e}{a})*(x+\frac{d}{a})=0$
$x^2+\frac{ex+dx}{a}+\frac{ed}{a^2}=0$
$x^2+\frac{x(e+d)}{a}+\frac{ed}{a^2}=0$
Via substitution of the given above:
$x^2+\frac{bx}{a}+\frac{ac}{a^2}=0$
$x^2+\frac{bx}{a}+\frac{c}{a}=0$
$a*(x^2+\frac{bx}{a}+\frac{c}{a})=a*(0)$
$ax^2+bx+c=0$
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This is method is known to me as middle term factor.
Lets take an example $f(x)=x^2+6x+8$. We have to find two numbers such that their sum is their product is $8$ and the sum is $6$. So, factors are $(x+4)(x+2)$.
In general, $ax^2+bx-c$ here constant term $ac$ is negative so we have to find two numbers such that their difference is $b$ and the product is $ac$.
Sometimes finding what to add or subtract might be difficult in that case we can use quadratic formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$. You will get two solutions from here $x=\alpha,\beta$. Hence your required factors will be $(x-\alpha)(x-\beta)$.
|
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|
Solve intersection of the straight line with equation $y=x-a$ and the circle equation $x^2 + y^2 = 4$
Find the coordinates of the points of intersection of the straight line with equation $y=x-a$ and the circle with equation $x^2 + y^2 = 4 $. Give the conditions on a for:
I did see the answer from the practice paper: the answer is *see the below
$$\left(\frac{a + \sqrt{8-a^2}}{2}, \frac{-a + \sqrt{8-a^2}}{2} \right); \left(\frac{a - \sqrt{8-a^2}}{2}, \frac{-a - \sqrt{8-a^2}}{2} \right) $$
but I am struggling on how to solve it. can you please explain to me how to solve it. thanks :)
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Substituting $y = x - a$ into $x^2 + y^2 = 4$ gives $x^2 + (x - a)^2 = 4$, which expands to $2x^2 - 2ax + a^2 - 4 = 0$. This is a quadratic equation in $x$ that you can then use the quadratic formula to solve to get $x = \frac{2a \pm \sqrt{4a^2 - 4(2)(a^2 - 4)}}{4} = \frac{2a \pm \sqrt{-4a^2 + 32}}{4} = \frac{a \pm \sqrt{8 - a^2}}{2}$, and then get $y$ from $x - a$, giving those $2$ answers you mention.
As for the conditions on $a$, note that you need $8 - a^2 \ge 0$ for $\sqrt{8 - a^2}$ to be a real value, so $a^2 \le 8$ giving $-2\sqrt{2} \le a \le 2\sqrt{2}$ as the range of valid values of $a$.
|
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Proof-Verification:$\int x[3+\ln(1+x^2)]\arctan x{\rm d}x$. $$\begin{aligned}
&\int x[3+\ln(1+x^2)]\arctan x{\rm d}x\\
=&\int 3x\arctan x{\rm d}x+\int x\ln(1+x^2)\arctan x{\rm d}x\\
=&\int \arctan x{\rm d}\left(\frac{3x^2}{2}\right)+\int \ln(1+x^2)\arctan x{\rm d}\left(\frac{x^2}{2}\right)\\
=&\int \arctan x{\rm d}\left(\frac{3x^2}{2}\right)+\int \arctan x{\rm d}\left[\frac{1}{2}(1+x^2)\ln(1+x^2)-\frac{1}{2}x^2\right]\\
=&\int \arctan x {\rm d} \left[ \frac{3x^2}{2}+\frac{1}{2}(1+x^2)\ln(1+x^2)-\frac{1}{2}x^2 \right]\\
=&\int \arctan x {\rm d} \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]
\end{aligned}$$
Integrating by parts, we obtain
$$\begin{aligned}
&\int \arctan x {\rm d} \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\int \frac{1}{1+x^2}\cdot\left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]{\rm d}x\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\int \frac{x^2}{1+x^2}{\rm d}x-\frac{1}{2}\int \ln(1+x^2){\rm d}x\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\int \frac{x^2}{1+x^2}{\rm d}x-\frac{1}{2}x\ln(1+x^2)+\int \frac{x^2}{1+x^2}{\rm d}x\\
=&\arctan x \left[x^2+\frac{1}{2}(1+x^2)\ln(1+x^2)\right]-\frac{1}{2}x\ln(1+x^2)+C.\\
\end{aligned}$$
Please correct me if I'm wrong.
|
We can verify your solution by simply finding its derivative. If $F(x)$ is the antiderivative of $f(x)$, then we have $F'(x)=f(x)$.
$$F'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan(x)\left(x^2+\frac{1}{2}\left(x^2+1\right)\ln\left(x^2+1\right)\right)-\frac{1}{2}x\ln\left(x^2+1\right)+C\right]$$
Let us find the derivative of the last two terms first. The derivative of $C$, the constant of integration, is obviously zero.
$$\frac{\mathrm{d}}{\mathrm{d}x}\left[-\frac{1}{2}x\ln\left(x^2+1\right)+C\right]=-\frac{1}{2}\left(\frac{2x^2}{x^2+1}+\ln\left(x^2+1\right)\right)=-\frac{x^2}{x^2+1}-\frac{\ln\left(x^2+1\right)}{2}$$
Next, let us find the derivative of the remaining portion of $F(x)$. For simplicity and ease of calculation, allow for the functions $g(x)=\arctan(x)$ and $h(x)=x^2+\frac{1}{2}\left(x^2+1\right)\ln\left(x^2+1\right)$. Apply the product rule.
$$F'(x)=g(x)h'(x)+h(x)g'(x)-\frac{x^2}{x^2+1}-\frac{\ln\left(x^2+1\right)}{2}$$
First, let us find $h'(x)$. The derivative of $g(x)$ should come to mind quickly.
$$h'(x)=2x+\frac{1}{2}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left[x^2\ln\left(x^2+1\right)+\ln\left(x^2+1\right)\right]=2x+\frac{x^3+x}{x^2+1}+x\ln\left(x^2+1\right)$$
Factor $x$ from each term of $h'(x)$ above and simplify. Multiply this result with the function $g(x)$ and we find that $g(x)h'(x)=x\left(3+\ln\left(x^2+1\right)\right)\arctan(x)$. This is clearly the original integrand $f(x)$, but we are not quite finished yet. Let us find $h(x)g'(x)$ first. We are almost finished verifying your solution.
$$h(x)g'(x)=\frac{1}{x^2+1}\left(x^2+\frac{\left(x^2+1\right)\ln\left(x^2+1\right)}{2}\right)=\frac{x^2}{x^2+1}+\frac{\ln\left(x^2+1\right)}{2}$$
Return to $F'(x)$. At this moment, only insert $h(x)g'(x)$. This will simplify very nicely for us.
$$F'(x)=g(x)h'(x)+\frac{x^2}{x^2+1}+\frac{\ln\left(x^2+1\right)}{2}-\frac{x^2}{x^2+1}-\frac{\ln\left(x^2+1\right)}{2}=g(x)h'(x)$$
Finally, insert $g(x)h'(x)$. You have solved the problem correctly.
$$F'(x)=f(x)=x\left(3+\ln\left(x^2+1\right)\right)\arctan(x)$$
|
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|
If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x = a + \sqrt{b}, z = c$. We can find $y$ as
$$ y = \sqrt{c^{2}- (a + \sqrt{b})^{2}} $$
so we have
$$ \cos(18) = \frac{y}{z} = \frac{\sqrt{c^{2}- (a + \sqrt{b})^{2}}}{c}$$
I also found out that
$$b = (c \sin(18) - a)^{2} = c^{2} \sin^{2}(18) - 2ac \sin(18) + a^{2}$$
I got no clue after this.
The solution says that $$ \sin(18) = \frac{-1 + \sqrt{5}}{4} $$
I gotta intuition that we must find $A,B,C$ such that
$$ A \sin(18)^{2} + B \sin(18) + C = 0 $$
then $\sin(18)$ is a root iof $Ax^{2} + Bx + C$, and $a = -B, b = B^{2} - 4AC, c = 2A$.
Totally different. This question is not asking to prove that $sin(18)=(-1+\sqrt{5})/4$, that is just part of the solution.
|
Let $A = 18°$
$$5A = 90°$$
$$⇒ 2A + 3A = 90˚$$
Taking sine on both sides, we get
$$\sin 2A = \sin (90˚ - 3A) = \cos 3A $$
$$⇒ 2 \sin A \cos A = 4 \cos^3 A - 3 \cos A$$
$$⇒ 2 \sin A \cos A - 4 \cos^3A + 3 \cos A = 0 $$
$$⇒ \cos A (2 \sin A - 4 \cos^2 A + 3) = 0 $$
Dividing both sides by cos A
$$⇒ 2 \sin A - 4 (1 - sin^2 A) + 3 = 0$$
$$⇒ 4 \sin^2 A + 2 \sin A - 1 = 0$$
After solving this quadratic
$$ \sin18°=\frac{-1+√5}{4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in... Let $\{a_n\}$ be a sequence of positive real numbers such that
$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.
Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
(A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.
Solution attempt:
Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n>0$, the other root is rejected).
We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$, then $\lim a_n \to \infty$.
Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$. (Again, the other root is rejected due to the same reason).
Hence, $(a_n)$ increases monotonically.
Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$
Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}< 1.3$).
So, option $(A)$ is the correct answer.
Is the procedure correct?
I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.
Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?
|
$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$ $a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$ $1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$ $a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$ $\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$ $\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$ $\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$
$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$
|
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|
A problem on number of solutions of a functional equation: $ f ( x ) = 8 f ( 2 x + 1 ) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that $ f ( 0 ) = 1 $ and for all $ x \ne - 1 $,
$$ f ( x ) = 8 f ( 2 x + 1 ) \text . $$
I have found only one solution: $ \frac 1 { ( x + 1 ) ^ 3 } $. The method was by iterated substitution of $ 2 x + 1 $ for $ x $ and finding a pattern in the resulted sequence.
|
Let $g(x) = f(x-1)$. Then we obtain $g(x) = f(x-1) = 8f(2x-1) = 8g(2x)$, and $g(-1)=1$. Note that values of $g$ for positive $x$ cannot be related to negative $x$. Clearly, $g(0)$ is arbitrary.
To find solutions for positive $x$, we let $h(x) = g(2^x)$, giving $h(x) = 8h(x+1)$. This has solutions $8^{-x} h_1(x)$ for any function $h_1$ with period $1$. Substituting back into $g$ we obtain $g(x) = 8^{-\log_2(x)} h_1(\log_2(x)) = \frac{1}{x^3}h_1(\log_2(x))$, for positive $x$.
The solutions to negative $x$ is handled similarly, now letting $h(x) = g(-2^x)$, and obtaining solutions $h(x) = 8^{-\log_2(-x)} h_2(x)$, for any function $h_2$ with period 1. So we obtain $g(x) = \frac{-1}{x^3}(\log_2(-x))$. We can remove the negative sign by replacing $h_2$ with $-h_2$, since it is an arbitrary function.
So we use $f(x) = g(x+1)$ to find all solutions:
$$f(x) =
\begin{cases}
\frac{1}{(x+1)^3}h_1(\log_2(x+1)) & \text{if $x>-1$}\\
r & \text{if $x=-1$}\\
\frac{1}{(x+1)^3}h_2(\log_2(-x-1)) & \text{if $x<-1$}\\
\end{cases}$$
for all reals $r$, and all functions $h_1, h_2$ with period $1$.
|
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|
complete set of values of $a$ having modulus and linear terms
If $(9-x^2)>|x-a|$ has at least one negative real solution for $a\in\mathbb{R}.$ Then complete set of values of $a$ is
Plan
If $x>a$ Then $9-x^2>x-a\Rightarrow x^2+x-(a+9)<0$
If $x\leq a$ Then $9-x^2>a-x\Rightarrow x^2-x+a-9<0$
How do i solve these inequalityHelp me please
|
For the case $x>a$:
$$x^{2} + x - (a+9) < 0 $$
The roots are:
$$\frac{-1 \pm \sqrt{4a+37}}{2} $$
Notice $a \ge -37/4$. Clearly the solution is:
$$\frac{-1 - \sqrt{4a+37}}{2} < x < \frac{-1 + \sqrt{4a+37}}{2} $$
The least possible value for $x$ is therefore $-1/2$, and clearly $-1/2 > -37/4$, so $a \ge -37/4$ does not violates $x>a$. So, this solution has at least one negative value, $x=-1/2$, that is when $a=-37/4$. Notice that for this case it is always $\frac{-1 - \sqrt{4a+37}}{2} < 0$. So all feasible values of $a$ is $a \ge \frac{-37}{4}$.
For the case $x \le a$, the method is similar.
|
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|
positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
|
Let $a=\sqrt{3}$.
\begin{align*}
\text{Then}\;\;&x^2+x-3-\sqrt{3}=0\\[4pt]
\iff\;&x^2+x=3+\sqrt{3}\\[4pt]
\iff\;&x^2+x=a^2+a\\[4pt]
\end{align*}
So by inspection, we get the solution $x=a$.
By Vieta's formula, the sum of the roots is $-1$, so the other root is $-1-a$, which is negative.
Thus, $x=\sqrt{3}$ is the only positive root.
Alternatively,
\begin{align*}
&x^2+x=a^2+a\\[4pt]
\iff\;&(x^2-a^2)+(x-a)=0\\[4pt]
\iff\;&(x-a)(x+a+1)=0\\[4pt]
\iff\;&x=a\;\,\text{or}\;\,x=-1-a\\[4pt]
\iff\;&x=\sqrt{3}\;\,\text{or}\;\,x=-1-\sqrt{3}\\[4pt]
\end{align*}
so as before, $x=\sqrt{3}$ is the only positive root.
|
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|
Number of onto functions from $Y$ to $X$ (JEE Advanced 2018)
Let $X$ be a set with $5$ elements and $Y$ be a set with $7$ elements. If $\beta$ is the number of onto functions from $Y$ to $X$ then the value of $\dfrac{\beta}{5!}$ is?
My approach is:
First I give each element of $Y$ one element of $X$ which then leaves me with two possibilities for the remaining two elements of $Y$.
1) Both can take same value
2) Both take different value from X.
So number of onto functions = $\beta = {^7}C_5\times 5! \times(5+ 5\times4)$ but this is wrong.
What's my mistake?
Also, I know how to solve the problem using principle of inclusion exclusion and that gives the right answer but we are not given calculators in exam and the calculation involved there is very lengthy.
|
Your method counts each surjective function that maps three elements of $Y$ to one element of $X$ three times, once for each way you could designate one of those three elements as the element of $Y$ that maps to that element of $X$.
For example, consider the surjective function $f: Y \to X$ defined by $f(1) = 1$, $f(2) = 2$, $f(3) = 3$, $f(4) = 4$, $f(5) = f(6) = f(7) = 5$. You count this function three times:
\begin{array}{c | c}
\text{designated ordered pairs} & \text{additional ordered pairs}\\ \hline
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) & (6, 5), (7, 5)\\
(1, 1), (2, 2), (3, 3), (4, 4), (6, 5) & (5, 5), (7, 5)\\
(1, 1), (2, 2), (3, 3), (4, 4), (7, 5) & (5, 5), (6, 5)\\
\end{array}
where we have written $(y, x)$ if $f(y) = x$.
You count each surjective function that maps two elements of $Y$ to one element of $X$ and two other elements of $Y$ to a different element of $X$ four times, twice for each pair of elements of $Y$ that map to a single element of $X$.
For example, consider the surjective function $f: Y \to X$ defined by $f(1) = 1$, $f(2) = 2$, $f(3) = 3$, $f(4) = f(6) = 4$, $f(5) = f(7) = 7$. You count this function four times:
\begin{array}{c | c}
\text{designated ordered pairs} & \text{additional ordered pairs}\\ \hline
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5) & (6, 4), (7, 5)\\
(1, 1), (2, 2), (3, 3), (6, 4), (5, 5) & (4, 4), (7, 5)\\
(1, 1), (2, 2), (3, 3), (4, 4), (7, 5) & (6, 4), (5, 5)\\
(1, 1), (2, 2), (3, 3), (6, 4), (7, 5) & (4, 4), (5, 5)\\
\end{array}
Let's correct your count.
Surjective functions in which exactly three elements of $Y$ map to a single element of $X$: Choose which three of the seven elements of $Y$ map to a single element of $X$. Choose that element of $X$. Since the function is surjective, the remaining four elements must map to distinct elements from the remaining four elements of $X$. The number of such functions is
$$\binom{7}{3}\binom{5}{1}4!$$
Surjective functions in which exactly two elements of $Y$ map to a single element of $X$ and exactly two other elements of $Y$ map to a different element of $X$: Choose which two of the five elements of $X$ are each the images of two elements of $Y$. Choose which two of the seven elements of $Y$ map to the smaller of these elements of $X$. Choose which two of the remaining five elements of $Y$ map to the larger of these elements of $X$. Since the function is surjective, the remaining three elements of $Y$ must map to distinct elements from the remaining three elements of $X$. The number of such functions is
$$\binom{5}{2}\binom{7}{2}\binom{5}{2}3!$$
Total: Since the two cases above are mutually exclusive and exhaustive, the number of surjective functions from $Y$ to $X$ is
$$\binom{7}{3}\binom{5}{1}4! + \binom{5}{2}\binom{7}{2}\binom{5}{2}3!$$
|
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|
prove $\sum_{n=1}^\infty \frac{H_n^2}{n^4}=\frac{97}{24}\zeta(6)-2\zeta^2(3)$ this series was evaluated by Cornel Valean here using series manipulation.
I took a different path as follows:
using the identity:$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
multiply both sides by $\ln^3x/x$ then integrate
$$-6\sum_{n=1}^\infty \frac{H_n^2-H_n^{(2)}}{n^4}=\int_0^1\frac{\ln^2(1-x)\ln^3x}{x(1-x)}\ dx$$
I was able here to find
\begin{align}
\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}&=\frac43\zeta^2(3)-\frac23\sum_{k=1}^\infty\frac{H_k^{(3)}}{k^3}\\
&=\zeta^2(3)-\frac13\zeta(6)
\end{align}
as for the integral, it seems very tedious to calculate it using the derivative of beta function.
can we find it with or without using beta function?
|
This solution is by Cornel Valean.
Using the follwing identity: ( see Lemma $2(b)$ in this paper)
$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$
and since $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k}\int_0^1x^{n+k-1}\ dx=2\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}$$
Then $$\sum_{k=1}^\infty\frac{H_{k-1}}{k(n+k)}=\frac{H_n^2+H_n^{(2)}}{2n}\tag{1}$$
Divide both sides by $n^3$ then sum both sides from $n=1$ to $\infty$, we get
\begin{align}
S&=\color{blue}{\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^4}+\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}}=\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{n=1}^\infty\frac{1}{n^3(n+k)}\right)\\
&=\sum_{k=1}^\infty\frac{H_{k-1}}{k}\left(\sum_{n=1}^\infty\left[\frac{1}{k^3}\left(\frac{1}{n}-\frac{1}{n+k}\right)-\frac1{n^2k^2}+\frac1{n^3k}\right]\right)\\
&=\sum_{k=1}^\infty\left(\frac{H_k}{k}-\frac{1}{k^2}\right)\left(\frac{H_k}{k^3}-\frac{\zeta(2)}{k^2}+\frac{\zeta(3)}{k}\right)\\
&=\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\sum_{k=1}^\infty\frac{H_k}{k^5}-\zeta(2)\sum_{k=1}^\infty\left(\frac{H_k}{k^3}-\frac1{k^4}\right)+\zeta(3)\sum_{k=1}^\infty\left(\frac{H_k}{k^2}-\frac1{k^3}\right)\\
&=\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\left(\frac74\zeta(6)-\frac12\zeta^2(3)\right)-\zeta(2)\left(\frac14\zeta(4)\right)+\zeta(3)\left(\zeta(3)\right)\\
&=\color{blue}{\sum_{k=1}^\infty\frac{H_k^2}{k^4}-\frac{35}{16}\zeta(6)+\frac32\zeta^2(3)}
\end{align}
Rearranging the blue sides, we get
$$\sum_{k=1}^\infty\frac{H_k^2}{k^4}=\frac{35}{8}\zeta(6)-3\zeta^2(3)+\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}\\
=\frac{97}{24}\zeta(6)-2\zeta^2(3)$$
where we used $\ \displaystyle\sum_{k=1}^\infty\frac{H_k^{(2)}}{k^4}=\zeta^2(3)-\frac13\zeta(6)\ $ (can be found in the same paper I linked or here)
|
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Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$
Prove that you can't write $\sqrt[3]{4}$ like $a+b\sqrt[3]{2}$, with $a,b \in \Bbb Q$
How I can prove it? I tried elevating to cube but then?
$4 = a^3+ 3a^2b \sqrt[3]{2} + 3ab^2\sqrt[3]{4}+ 2b^3$
|
Assume for contradiction that $\sqrt[3] 4 = a + b \sqrt[3] 2$. We then cube both sides, to obtain $4 = a^3 + 3a^2b \sqrt[3] 2 + 3ab^2 \sqrt[3] 4 + 2b^3$. We have that $4, a^3, 2b^3$ are all rational, so we also have $3a^2b \sqrt[3] 2 + 3ab^2 \sqrt[3] 4 = q$ for some rational $q$. But $\sqrt[3] 4 = (\sqrt[3] 2)^2$, so we obtain a quadratic in $\sqrt[3] 2$ with rational coefficients, implying that $\sqrt[3] 2$ can be written as some $c \pm \sqrt d$. Clearly $d$ is nonzero since $\sqrt[3] 2$ is irrational, so cubing both sides again we obtain $c^3 \pm c^2 \sqrt d + cd \pm d\sqrt d = 2$, implying that $(c^2+d) \sqrt 2$ is rational. But $c^2+d$ is rational, and $\sqrt 2$ is not, yielding a contradiction.
|
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|
Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$ Evaluate $\int_0^{\infty} x^2 e^{-x^2}dx$
The original problem is :
$$\text{Evaluate} \iint _R ye^{-x^2-y^2}dxdy$$
Where $R=\left\{ (x,y) \vert x\geq0,y\geq0\right\}$
I sub-ed $x=r\cos\theta, y=r\sin\theta$ and this changed to
$\left(\int_0^{\frac{\pi}{2}}\sin\theta d\theta \right)\left(\int_0^{\infty} r^2 e^{-r^2}dx\right)$
So I must calculate $\int_0^{\infty} r^2 e^{-r^2}dr$, is there an easy way?
|
Let $z=\sqrt{2}r$ so that $dz=\sqrt{2}dr$ and $r^2=\frac{1}{2}z^2$. Then:
$$
\int_0^\infty r^2e^{-r^2}dr=\int_0^\infty\frac{1}{2}z^2e^{-\frac{1}{2}z^2}\frac{dz}{\sqrt{2}}=\frac{1}{2\sqrt{2}}\int_0^{\infty}z^2e^{-z^2/2}dz=\frac{1}{4\sqrt{2}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz.
$$
The last equality above uses symmetry. Next, we note:
$$
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty z^2e^{-z^2/2}dz=1
$$
because it is the second-moment of a standard normal distribution. And so:
$$
\int_0^\infty r^2e^{-r^2}dr=\frac{1}{4\sqrt{2}}\sqrt{2\pi}=\frac{\sqrt{\pi}}{4}.
$$
|
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|
Convergence of moving point inside of unit disk. Suppose Set D :
$$D=\left\{(x, y)\vert x^2 + y^2 \leq 1 \right\}$$
And Point $P(0, 0)$ on coordinate plane.
Define 'Movement' :
For a point P, select any direction and move $\frac{1}{2^n}$ to straight (Sorry, my english is poor.) When 'Movement' Execute $n$-th times
For example :
$1$st time, Suppose I selected positive-x-axis direction
Then, Point $P(0,0)$ moves to $P\left(\frac{1}{2},0\right)$
$2$nd time, Suppose I selected Positive-y-axis direction
Then, Point $P\left(\frac{1}{2}, 0\right)$ moves to $P\left(\frac{1}{2},\frac{1}{4} \right)$
If I take $n\longrightarrow\infty$, Arbitrary point $A(x, y)\in D$ can be expressed by 'Movement'?
I Know that perimeter of $x^2+y^2=1$ can be expressed by 'Movement' because $$\sum_{n=1}^\infty \frac{1}{2^n}=1$$
But how about inside of $x^2 + y^2 =1$?
Example
The point with $1, 2, 3, 4, \cdots$ is a point which moved by $n$-th 'Movement'
The angle(direction) is free.
the problem is : If $n \longrightarrow\infty$, arbitrary points in $D$ can be expressed by 'Movement'?
|
Short answer: Not only you can do this, it is even possible to move on one line connecting the origin and your point of interest. Maybe not surprisingly, this is related to the binary representation of a number.
Consider a point $p=re^{i\theta}=x+iy$, and let us limit ourselves to movement at angle $\theta$ only (i.e. on the line connecting the origin and $p$). Thus, this problem essentially boils down to the question if every number $-1\leq z\leq 1$ can be written as
$$z=\sum_{n=1}^{\infty}\dfrac{a_{n}}{2^{n}}=\sum_{n=1}^{\infty}\dfrac{\frac{1+a_{n}}{2}}{2^{n-1}}-\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}=\sum_{n=1}^{\infty}\dfrac{b_{n}}{2^{n-1}}-1=\sum_{n=0}^{\infty}\dfrac{c_{n}}{2^{n}}-1$$
with $a_{n}\in\left\{-1,1\right\}$, $b_{n}=\frac{1+a_{n}}{2}\in\left\{0,1\right\}$ and $c_{n}=b_{n+1}\in\left\{0,1\right\}$. Let $q=z+1$, so $0\leq q\leq 2$ and the above condition is
$$q=\sum_{n=0}^{\infty}\dfrac{c_{n}}{2^{n}}$$
with $c_{n}\in\left\{0,1\right\}$. But this is just binary representation.
Example 1: Want to get back to the origin? No problem. Take $z=0$, thus $q=1$ and $c_{n}=\left(1,0,0,\dots\right)\Longrightarrow a_{n}=\left(1,-1,-1,\dots\right)$. In this case $\theta$ is arbitrary so let's walk on the $x$ axis. This means you should go
$$\left(0,0\right)\rightarrow\left(\frac{1}{2},0\right)\rightarrow\left(\frac{1}{2}-\frac{1}{4},0\right)\rightarrow\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8},0\right)\rightarrow\dots\rightarrow\left(\frac{1}{2}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2^{n}},0\right)=\left(0,0\right)$$
Example 2: Want to go to $\left(\frac{1}{2},0\right)$? Take $\theta=0$ and $z=\frac{1}{2}$, thus $q=\frac{3}{2}$ and $c_{n}=\left(1,1,0,0,\dots\right)\Longrightarrow a_{n}=\left(1,1,-1,-1,\dots\right)$. This time your path is
$$\left(0,0\right)\rightarrow\left(\frac{1}{2},0\right)\rightarrow\left(\frac{1}{2}+\frac{1}{4},0\right)\rightarrow\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8},0\right)\rightarrow\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{16},0\right)\rightarrow\dots\rightarrow\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{2^{n}},0\right)=\left(\frac{1}{2},0\right)$$
|
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|
Show integral is 2 Pi Consider $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{0}^{\infty}\biggr(\frac{1}{\sqrt{(1-x)^2+y^2+z^2}}-\frac{1}{\sqrt{x^2+y^2+z^2}}\biggr)^2dxdydz$$
A numerical study suggest the integral converges to $2\pi$. See: Link to numerical studyHowever I am unable to show this. When y,z are zero, we have a singular point at x=1. There the integrand behaves like $$\biggr(\frac{1}{(1-x)}-1\biggr)^2$$ right? And even in the Principal sense, the integral of this doesn't converge.
I was wondering if someone could help me show the value of the integral analytically.
Update: I just noticed that we have two singular points: Have another at x=0 when y,z=0 so maybe these two are canceling in the principal sense leading to a finite value for the integral.
Update 2: Made a mistake on integral boundaries: x goes from 0 to infinity and have made the corrections above as per comment below.
Thanks.
|
It's not a problem that the integral does not converge on one line $y=z=0$, because a contribution from a single line is negligible for an integral over a volume.
Let us make a change of coordinates:
$$ x=r\cos\theta, \qquad y = r\sin\theta\cos\phi, \qquad z=r\sin\theta\sin\phi$$
$$ 0\le r\le\infty, \qquad 0\le\theta\le\pi, \qquad 0\le\phi\le \pi$$
we have $$ x^2+y^2+z^2 = r^2 $$
$$ (1-x)^2+y^2+z^2 = 1 - 2r\cos\theta + r^2$$
$$ dxdy dz = r^2\sin\theta\, dr d\theta d\phi$$
so
\begin{align} I &= \int_0^\infty dr \int_0^{\pi}d\theta \int_0^{\pi}d\phi\, r^2\sin\theta\left(\frac{1}{\sqrt{1 - 2r\cos\theta + r^2}} - \frac{1}{r}\right)^2 =^{\eta=\cos\theta} \\
&= \pi \int_0^\infty dr \int_{-1}^{1}d\eta \, \left(\frac{r^2}{1 - 2r\eta + r^2} - \frac{2r}{\sqrt{1 - 2r\eta + r^2}} +1\right) = \\
&= \pi \int_0^\infty dr \left.\left(-\frac{r}{2} \ln(1-2r\eta + r^2) +2\sqrt{1 - 2r\eta + r^2} +\eta\right)\right|_{\eta=-1}^{\eta=1} = \\
&= \pi \int_0^\infty dr \left(-r \ln |1-r| + r \ln(1 + r) +2|1-r| - 2(1 + r)+12\right) = \\
&= \pi \int_0^1 dr \left(-r \ln (1-r) + r\ln(1 + r) - 4r + 2\right) + \\
&\quad + \pi \int_1^\infty dr \left(-r \ln (r-1) + r\ln(1 + r) - 2\right) = \\
&= \pi \Big(3r-2r^2 - \frac12(1-r^2)\ln\frac{1+r}{1-r}\Big)\Big|_{r=0}^{r=1} + \pi \Big(-2r - \frac12(1-r^2)\ln\frac{r+1}{r-1}\Big)\Big|_{r=1}^{r=\infty} = \\
&=\pi + \pi = 2\pi \end{align}
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|
How to expand $\sum_{k=0}^\infty \frac{k^k}{k!}e^{-k\lambda}$ I know that $\sum_{k=0}^\infty \dfrac{k^k}{k!}\dfrac{e^{-k-k\tfrac{s}{n}}}{\sqrt{n}}$ can be expanded in powers of $\frac{1}{\sqrt{n}}$ to yield:
$\frac1{\sqrt{2s}}+\frac{1}{3\sqrt{n}}+\frac{\sqrt{2s}}{12n}+\dotso$
My question is twofold: (i) is there an actual proof of this (beyond numerical),
(ii) how does this expansion continue (general formula, or
at least the next few terms)?
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We will assume $s > 0$ or the series diverges.
Recall for $|z| \le \frac1e$, the $0$-branch of the Lambert W function has following expansion:
$$W_0(z) = \sum_{k=1}^\infty \frac{(-k)^{k-1}}{k!} z^k$$
Pick any $\lambda_0 > 1$ and substitute $z$ by $-e^{-\lambda}$ for $\lambda > \lambda_0$. Above expansion converges uniformly for such $\lambda$. So does the series of its derivatives for each individual term. In this situation, it is legal to exchange the order of summation and taking derivatives. This implies
$$f(\lambda) \stackrel{def}{=}\sum_{k=0}^\infty \frac{k^k}{k!} e^{-\lambda k}
= 1 + \sum_{k=1}^\infty \frac{k^k}{k!} e^{-\lambda k}
= 1 + \frac{d}{d\lambda} W_0(-e^{-\lambda})
$$
Let $u = W_0(-e^{-\lambda})$, we have $u e^u = -e^{-\lambda}$. Differentiate against $\lambda$, we get
$$(u+1)e^u \frac{du}{d\lambda} = e^{-\lambda} = -u e^u
\implies \frac{du}{d\lambda} = - \frac{u}{1+u}$$
This leads to
$$f(\lambda) = 1 + \frac{du}{d\lambda} = \frac{1}{1+u} = \frac{1}{1 + W_0(-e^{-\lambda})}$$
It is easy to see the series at hand evaluates to $\displaystyle\;\frac{1}{\sqrt{n}}f\left(1 + \frac{s}{n}\right)$
For simplicity, let's assume $n = 1$. Let $\lambda = 1 + s$ and $w = 1 + u$. We have
$$f(1 + s) = \frac{1}{w}\quad\text{ and }\quad ue^u = -e^{-\lambda}
\iff (1-w) e^w = e^{-s}$$ This leads to
$$s = -(\log(1-w) + w) = \frac{w^2}{2} + \frac{w^3}{3} + \frac{w^4}{4} + \cdots
$$
For small $s$, the leading approximation of $w$ is clearly $\sqrt{2s}$ and hence
$f(1+s) \sim \frac{1}{\sqrt{2s}}$.
To process further, write $s$ as $\frac{t^2}{2}$ and expanse $w$ as a power series in $t = \sqrt{2s}$. By matching coefficients of different powers of $t$ using a CAS, we get
$$\begin{align}w &=
t-\frac{t^2}{3}+\frac{t^3}{36}+\frac{t^4}{270}+\frac{t^5}{4320}-\frac{t^6}{17010} + \cdots\\
\implies
\frac1w &=
\frac{1}{t}+\frac{1}{3}+\frac{t}{12}+\frac{2t^2}{135}+\frac{t^3}{864}-\frac{t^4}{2835}-\frac{139 t^5}{777600} + \cdots
\end{align}
$$
Putting $n$ back, we get
$$\frac{1}{\sqrt{n}}f\left(1 + \frac{s}{n}\right)
= \frac{1}{\sqrt{2s}} + \frac{1}{3n^{\frac12}}
+ \frac{\sqrt{2s}}{12n}
+\frac{4\,s}{135\,{n}^{\frac{3}{2}}}
+\frac{\sqrt{2s}^3}{864 n^2}
-\frac{4\,{s}^{2}}{2835\,{n}^{\frac{5}{2}}} + \cdots
$$
It looks like there is a typo in the third term of the expansion you get.
|
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|
Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?
Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing around, I also found a solution at $(3,2,2,2)$. Then I checked $a=5,b=4,c=4$ and found that it also worked.
It seems that there is a family of solutions at $(2n-1,2n-2,2n-2,n)$. I can prove this easily:
\begin{align}
LHS&=2^{2n-1}+2^{2n-2}+2^{2n-2}\\
&=2^{2n-1}+2^{2n-1}\\
&=2^{2n}\\
&=4^n\\
&=RHS
\end{align}
Is this the only solution? If it is, how do I go about proving it?
|
We have
$$
\eqalign{
& \left( {a,b,c,d} \right)\quad \Rightarrow \quad 2^{\,a} + 2^{\,b} + 2^{\,c} = 4^{\,d} \quad \Rightarrow \cr
& \Rightarrow \quad 2^{\,2n} \left( {2^{\,a} + 2^{\,b} + 2^{\,c} } \right) = 2^{\,2n} 4^{\,d} \quad \Rightarrow \cr
& \Rightarrow \quad 2^{\,a + 2n} + 2^{\,b + 2n} + 2^{\,c + 2n} = 4^{\,d + n} \quad \Rightarrow \cr
& \Rightarrow \quad \left( {a + 2n,b + 2n,c + 2n,d + n} \right) \cr}
$$
So the problem shifts to find the minimal quadruples.
In binary notation it is clear how it shall be
$$
\eqalign{
& 2^{\,a} \quad \quad \quad \quad = 0,0, \cdots ,1,0,0 \cr
& 2^{\,b} \quad \quad \quad \quad = 0,0, \cdots ,1,0,0 \cr
& 2^{\,a} + 2^{\,b} \quad \quad = 0,0, \cdots ,0,1,0 \cr
& 2^{\,c} \quad \quad \quad \quad = 0,0, \cdots ,0,1,0 \cr
& 2^{\,a} + 2^{\,b} + 2^{\,c} = 0,0, \cdots ,0,0,1 \cr
& 4^{\,d} = 2^{\,2d} \quad \;\; = \underbrace {0,0, \cdots ,0,0,1}_{0,1, \cdots ,\;2d - 2,\;2d - 1,\;2d} \cr}
$$
So
- $2^k$ has only one $1$, and same has $4^d$;
- in summing the $2$'s, to reach to have only one $1$, we cannot leave any $1$ behind;
- so the (only) one of $2^b$ must be just below that of $2^a$
- ... and we must end on bit $2d$.
The conclusion is clear.
|
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|
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expression can be reduced as sum of sine functions as:
$$(a-c) \sin^2 \theta + \dfrac b2 \sin 2 \theta + c \tag{*} $$
Now, there is one way to take everything as a function of $\theta$ and get the expression in the form of $ a \sin \theta + b \cos \theta = c$ and dividing it by $ \sqrt{ a^2 + c^2} $ both sides, but square in sine function is a big problem, also both have different arguments.
Other way, I can think of is taking $ \tan \dfrac \theta 2 = t$ and getting sine and cosine function as $ \sin \theta = \dfrac{ 2t}{1+t^2} $ while cosine function as $ \dfrac{ 1-t^2} {1+t^2}$ solving. So getting $(*)$ as a function of $t$, and simplifying we get,
$$ f(t) = \dfrac{2 Rt + 2 R t^3 + R_0 t - R_0 t^3}{1+t^4 + 2t^2} + c\tag{1}$$
For $R_0 = 2b, R = (a-c)$ , but this is where the problem kicks in!, The Range of given fraction seems $ (-\infty,+ \infty)$ and is not bounded!
So what's the problem here? Can it be solved?
Thanks :)
Edit : I'd like to thank @kaviramamurthy for pointing out that as $t \rightarrow \pm \infty, f(t) \rightarrow c$. That's a mistake here.
|
This problem is equivalent to
$$
\min(\max) a x^2+b x y + c y^2 \ \ \mbox{s. t.}\ \ x^2+y^2=1
$$
this is an homogeneous problem so calling $y = \lambda x$ and substituting we have equivalently
$$
\min(\max) f(\lambda) = \frac{a+\lambda b+\lambda^2c}{1+\lambda^2}
$$
and the extremals condition is
$$
f'(\lambda) = 0\Rightarrow 2 \lambda (c-a)-b \lambda ^2+b = 0
$$
giving
$$
\lambda = \frac{c-a\pm\sqrt{(a-c)^2+b^2}}{b}
$$
now substituting into $f(\lambda)$ we have
$$
\frac{1}{2} \left(-\sqrt{(a-c)^2+b^2}+a+c\right)\le f(\lambda)\le \frac{1}{2}
\left(\sqrt{(a-c)^2+b^2}+a+c\right)
$$
|
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|
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1}$$
upon expanding them we get:
$$f_1(x)=1 + x + x^2$$
$$f_2(x)=1 + 2 x + 3 x^2 + 4 x^3 + 3 x^4 + 2 x^5 + x^6$$
$$f_3(x)=1 + 3 x + 6 x^2 + 10 x^3 + 15 x^4 + 18 x^5 + 19 x^6 + 18 x^7 +
15 x^8 + 10 x^9 + 6 x^{10} + 3 x^{11} + x^{12}$$
$$f_4(x)=1 + 4 x + 10 x^2 + 20 x^3 + 35 x^4 + 56 x^5 + 80 x^6 + 104 x^7 +
125 x^8 + 140 x^9 + 146 x^{10} + 140 x^{11} + 125 x^{12} + 104 x^{13} +
80 x^{14} + 56 x^{15} + 35 x^{16} + 20 x^{17} + 10 x^{18} + 4 x^{19} + x^{20}$$
$$\vdots$$
$$f_{n-1}(x)=1 + ?x + ?x^2 + ?x^3 +?x^4+ ?x^5+\cdots+?x^{n(n-1)}$$
I'm wondering how to determine the coefficients for the n-th order? I can observe that the coefficients are symmetric.
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We can think at the problem as a stars and bar problem, that uses the inclusion-exclusion principle, as in the answer cited by IV_ and this one.
We interprete
\begin{equation}(1+x+x^2 + \cdots + x^{n})^{n-1} = \underbrace{(1+x+x^2 + \cdots + x^{n})(1+x+x^2 + \cdots + x^{n})\cdots (1+x+x^2 + \cdots + x^{n})}_{n-1\text{ times }}\end{equation}
as follows. There are $n-1$ boxes, separated by $n-2$ bars. Since we are interested in the coefficient of $x^k$, we want to put $k$ indistiguishable balls in those $n-1$ boxes. You are allowed to take $n$ balls from each term of the product above, but no more.
Suppose you want to evaluate the coefficient of $x^{15}$ in
$$(1+x+x^2+x^3+x^4+x^5+x^6)^5,$$
without expanding the whole product. There are $5$ boxes separated by $5-1=4$ bars, in which we have to put $k=15$ balls, with the restriction that we may pick from one of the five terms in the product at most $6$ balls (the highest term is $x^6$). One example of that is
$$******\vert****\vert**\vert*\vert**,$$
which would be the contribution of $x^6$, $x^4$, $x^2$, $x$ and $x^2$ from the first, second, third, fourth and fifth term of the product. The number of ways to put $15$ indistinguishable balls in the $5$ boxes is
$${15 + 5-1\choose 5-1} = 3876.$$
Now you have to subtract the number of ways to put $15$ indistinguishable balls in $5$ boxes, so that in at least one box more than $6+1=7$ balls appear. You have ${5\choose 1}$ ways to choose the box in which to put the $7$ balls. The other $15-7 = 8$ balls can be placed in ${8 + 5-1\choose 5-1}$ ways. In total you have
$${5\choose 1}{8 + 5-1\choose 5-1} = 2475 $$
ways to put $15$ indistinguishable balls into $5$ boxes, so that in at least one box more than $7$ balls appear.
Now, here comes the inclusion exclusion principle in play, you also did subtract the cases, where in more than $1$ container there were at least $7$ balls. So you add the number of ways to put $15$ indistinguishable balls into $5$ boxes, where in more than $2$ container there were at least $7$ balls. You have ${5\choose 2}$ ways to choose the containers where to put the $7$ balls and ${1 + 5 -1 \choose 5-1}$ ways to put the remaining ball in the other containers. So in total there are
$${5 \choose 2}{1+5-1\choose 5-1} = 50$$
ways to put the $15$ indistinguishable balls into $5$ boxes, where in more than $2$ container there were at least $7$ balls.
The coefficient of $x^{15}$ in $(1+x+x^2+x^3+x^4+x^5+x^6)^5$ is therefore
$${15 + 5-1\choose 5-1} - {5\choose 1}{8 + 5 -1 \choose 5-1} + {5\choose 2}{1+5-1\choose 5-1} = 3876-2475+50 = 1451.$$
|
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|
$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$.
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$
Before you complain, this problem is adapted from a recent competition. I have put my solution down below, there might be more practical and correct answers. In that case, please post them.
|
We have that $$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)}$$
$$ = \left[\frac{xy}{z^2(x + y)} + \frac{1}{z}\right] + \left[\frac{yz}{x^2(y + z)} + \frac{1}{x}\right] + \left[\frac{zx}{y^2(z + x)} + \frac{1}{y}\right] - \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)$$
$$ = \frac{xy + yz + zx}{z^2(x + y)} + \frac{xy + yz + zx}{x^2(y + z)} + \frac{xy + yz + zx}{y^2(z + x)} - \frac{xy + yz + zx}{xyz}$$
$$ = (xy + yz + zx)\left[\frac{1}{z^2(x + y)} + \frac{1}{x^2(y + z)} + \frac{1}{y^2(z + x)} - 2\right]$$
Now we have to prove that $$\frac{1}{z^2(x + y)} + \frac{1}{x^2(y + z)} + \frac{1}{y^2(z + x)} \ge 3$$
It is evident that
$$\frac{1}{z^2(x + y)} + \frac{1}{x^2(y + z)} + \frac{1}{y^2(z + x)} = \frac{2xyz}{z^2(x + y)} + \frac{2xyz}{x^2(y + z)} + \frac{2xyz}{y^2(z + x)}$$
$$2\left(\frac{xy}{zx + yz} + \frac{yz}{xy + zx} + \frac{zx}{yz + xy}\right)$$
$$ = 2\left[(xy + yz + zx)\left(\frac{1}{xy + yz} + \frac{1}{yz + zx} + \frac{1}{zx + xy}\right) - 3\right]$$
$$ \ge 2\left[(xy + yz + zx)\frac{9}{2(xy + yz + zx)} - 3\right] = 3$$
The equality sign occurs when $x = y = z = \sqrt[3]{\dfrac{1}{2}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.
|
Since by PM $$\left(\frac{x^n+1}{2}\right)^k\geq\left(\frac{x^k+1}{2}\right)^n$$ is true for all $x\geq0$ and $n\geq k>0,$ it's enough to prove that
$$\frac{x^{m+1}+1}{x^m+1}\geq\sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$ or $f(x)\geq0,$ where
$$f(x)=\ln\left(x^{m+1}+1\right)-\ln\left(x^m+1\right)-\frac{1}{2m+1}\ln\left(x^{2m+1}+1\right)+\frac{\ln2}{2m+1}.$$
Indeed, $$f'(x)=\frac{(m+1)x^m}{x^{m+1}+1}-\frac{mx^{m-1}}{x^m+1}-\frac{x^{2m}}{x^{2m+1}+1}=$$
$$=\frac{x^{m-1}\left(mx^{2m+2}-(m+1)x^{2m+1}+(m+1)x-m\right)}{\left(x^m+1\right)\left(x^{m+1}+1\right)\left(x^{2m+1}+1\right)}.$$
Let $g(x)=mx^{2m+2}-(m+1)x^{2m+1}+(m+1)x-m.$
Thus, $$g'(x)=m(2m+2)x^{2m+1}-(m+1)(2m+1)x^{2m}+m+1;$$
$$g''(x)=2m(m+1)(2m+1)x^{2m}-2m(m+1)(2m+1)x^{2m-1}=$$
$$=2m(m+1)(2m+1)x^{2m-1}(x-1)\geq0,$$
which says $$g'(x)\geq g'(1)=0,$$ which says
$$g(x)\geq g(1)=0,$$ which gives
$$f(x)\geq f(1)=0$$ and we are done!
By the way, we see that your inequality is true for all reals $x\geq0$ and $m>0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve First Order Linear PDE with Method of Characteristic Given a problem
$x^2\,u_x+xy\,u_y=y\,u$
Then, the characteristic equation is
$\dfrac{dx}{x^2}=\dfrac{dy}{xy}=\dfrac{du}{yu}$
From the first two of those equation, i got
$C_1=\dfrac{y}{x}$
How to obtain the other $C_2$? So that i have implicit solution $F(C_1)=C_2$?
|
$$\dfrac{dx}{x^2}=\dfrac{dy}{xy}=\dfrac{du}{yu}$$
From the first two ratio
\begin{align}
\dfrac{dx}{x^2} &= \dfrac{dy}{xy} \\
\implies \dfrac{dx}{x} &= \dfrac{dy}{y} \\
\implies \log x - \log y &= \log c_1 \\
\implies\frac{x}{y} &= c_1
\end{align}
From the last two
\begin{align}
\dfrac{dy}{xy} &= \dfrac{du}{yu} \\
\implies \dfrac{dy}{x} &= \dfrac{du}{u} \\
\implies \dfrac{dy}{c_1y} &= \dfrac{du}{u} \\
\implies \log y &= c_1(\log u - \log c_2) \\
\implies \log y &= \frac{x}{y}\log \left(\frac{u}{c_2} \right) \\
\implies \frac{u}{c_2} &= y^{\frac{y}{x}} \\
\implies c_2 &= uy^{-\frac{y}{x}}
\end{align}
So the implicit solution is $$F \left(\frac{x}{y} \right)=uy^{-\frac{y}{x}}$$
|
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|
What's the mistake in $\sin(\frac{a}{2})=\pm\sqrt{\frac{1\pm \cos(a)}{2}}$? When I was encountered with the formula of $\sin(\frac{a}{2})$, I tried to derive it. First, I tried from the formula of $\cos(2a)$ and I successfully did that but after a while I was curious about if that can be derived from sine's double angle formula and for which I came up with the equation above in the title, totally wrong, as that is kind of combination of sine's and cosine's half angle formula. Here is the derivation:-
$$\sin(2a)=2\sin(a)\cos(a)$$
or, $$\sin(a)=2\sin(\frac{a}{2})\cos(\frac{a}{2}) $$
or, $$\sin^2(a)=4\sin^2(\frac{a}{2})(1-\sin^2(\frac{a}{2})) $$
or, $$\sin^4(\frac{a}{2})-\sin^2(\frac{a}{2})=-\frac{\sin^2(a)}{4}$$
which by completing the square or by applying quadratic formula I got:-
$$\sin^2(\frac{a}{2})= \frac{1\pm \sqrt{1- \sin^2(a)}}{2}$$
or, $$\sin^2(\frac{a}{2})=\frac{1\pm \cos(a)}{2}$$
or, $$\sin(\frac{a}{2})=\pm \sqrt{\frac{1\pm \cos(a)}{2}}$$
Even though I haven’t broke the algebraic rules, I know there is some mistakes and that's what I'm trying to trace. Where is the problem actually? How's the 'minus' thing became 'plus-minus' thing?? Help me please.
Thanks in advance--
|
I think, it's better
$$\left|\sin\frac{\alpha}{2}\right|=\sqrt{\frac{1-\cos\alpha}{2}}$$ and
$$\left|\cos\frac{\alpha}{2}\right|=\sqrt{\frac{1+\cos\alpha}{2}}$$
Your mistake is that $\sqrt{x^2}=\pm x$ is wrong.
The right identity it's:
$$\sqrt{x^2}=|x|.$$
For example, after
$$\sin^4\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\frac{\sin^2\alpha}{4}=0$$ we obtain:
$$\sin^2\frac{\alpha}{2}=\frac{1+\sqrt{1-\sin^2\alpha}}{2}$$ or
$$\sin^2\frac{\alpha}{2}=\frac{1-\sqrt{1-\sin^2\alpha}}{2},$$ which is
$$\sin^2\frac{\alpha}{2}=\frac{1+|\cos\alpha|}{2}$$ or
$$\sin^2\frac{\alpha}{2}=\frac{1-|\cos\alpha|}{2},$$ which is
$$\sin^2\frac{\alpha}{2}=\frac{1+\cos\alpha}{2}$$ or
$$\sin^2\frac{\alpha}{2}=\frac{1-\cos\alpha}{2},$$ which
does not give a mistake, but also, does not give any sense.
Also, squaring of the both sides in $\sin\alpha=2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$ gave a wrong case.
Because $a=b$ is not equivalent to $a^2=b^2.$
|
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|
What property was used in this sine transformation? I have this expression:
$$
ψ(χ) = A\sin^3(\frac{πχ}{α})
$$
And somehow the book i read equalizes the previous equation to this one:
$$
ψ(χ) = \frac{A}{4}[3\sin(\frac{πχ}{α}) - \sin(\frac{3πχ}{α})]
$$
What trigonometric identity was used to make this possible?
|
The identity is
$$\sin^3x=\frac{3\sin x-\sin(3x)}{4}$$
Proof:
\begin{align*}
4\sin^3x-3\sin x &= \sin x(4\sin^2x-3)
\\
&=\sin x(1-4\cos^2x) \quad \text{Pythagorean identity}
\\
&=\sin x(2\cos^2x-\cos(2x)-4\cos^2x) \quad \text{Double angle cosine}
\\
&=-\sin x(2\cos^2x+\cos(2x))
\\
&=-[(2\sin x\cos x)\cos x+\sin x\cos(2x)]
\\
&=-[\sin (2x)\cos x+\sin x\cos(2x)]\quad \text{Double angle sine}
\\
&=-\sin (2x+x)\quad \text{identity sum sine}
\\
&=-\sin(3x)
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate the maximum value of $\sum_{cyc}\frac{1}{\sqrt{a^2 + b^2}}$ where $a, b, c > 0$ and $abc = a + b + c + 2$.
$a$, $b$ and $c$ are positives such that $abc = a + b + c + 2$. Caculate the maximum value of $$\large \frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}}$$
This problem is adapted from a recent competition. Here's what I got, and I am pretty proud of it. (Actually not.)
$$abc = a + b + c + 2$$
$$\iff abc + (ab + bc + ca) + (a + b + c) + 1 = ab + bc + ca + 2(a + b + c) + 3$$
$$\iff (a + 1)(b + 1)(c + 1) = (a + 1)(b + 1) + (b + 1)(c + 1) + (c + 1)(a + 1)$$
$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 1$$
And I'm done.
|
Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x+y}{z}$ and by AM-GM twice we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}=\sum_{cyc}\frac{xy}{\sqrt{y^2(y+z)^2+x^2(x+z)^2}}\leq$$
$$\leq\sum_{cyc}\frac{xy}{\sqrt{2xy(x+z)(y+z)}}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq$$
$$\leq\frac{1}{2\sqrt2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2\sqrt2}.$$
The equality occurs for $x=y=z=1,$ which says that we got a maximal value.
|
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|
Matrix complicated equation Let $$A = \begin{bmatrix}
1 & 3 & 4\\
3 & 6 & 9\\
1 & 6 & 4
\end{bmatrix},$$
$B$ be a $3\times 3$ matrix and $$A \cdot A^{T} \cdot A +3B^{-1} =0$$
What would be the value of
$ \det( \operatorname{adj} (A^{-1}(B^{-1}){2B^{T}}))$ ?
|
$B$ is a red herring here and might be replaced by any invertible $3\times3$-matrix.
Since $\operatorname{adj}(C)=\det(C)C^{-1}$ for invertible $C$ we have
$$\det(\operatorname{adj}(C)=\det(C^{-1}\cdot\det(C))=(\det(C))^3\det(C^{-1})=(\det(C))^2$$
if $C$ is of type $3\times3$.
Now happily compute
\begin{align}\det( \operatorname{adj} (A^{-1}B^{-1}{2B^{T}})
&=\bigl(\det(A^{-1}B^{-1}2B^{T})\bigr)^2\\
&=\bigl(2^3\cdot\det(A^{-1})
\underbrace{\det(B^{-1})\det(B^{T})}_{=1}\bigr)^2\\
&=\left(2^3\cdot\frac{1}{9}\right)^2
=\frac{64}{81}
\end{align}
as $\det(A)=9$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimal Rook Difference Grids In the below grid all 18 orthogonal differences are distinct, with a difference of 18 missing.
Could the highest number be 18? The resulting graph would have valence 4, making it an Eulerian Graceful graph with edges(mod 4)=2. Rosa (1967) proved Eulerian Graceful graphs must have edges(mod 4)=0 or 3, so 18 is impossible.
Thus the minimal $3\times3$ rook difference grid has $rdg(3,3)=19$.
For $rdg(1,n)$ see Golomb Ruler.
$rdg(2,3)=9$ and $rdg(2,4)=16$, as shown below.
What are values for larger grids?
|
I used a depth first search written in C to find the following:
$rdg(3,4)=30$, so the $3\times4$ rook graph is graceful.
\begin{array}{|c|c|c|c|}
\hline
0 & 1 & 9 & 30 \\
\hline
16 & 29 & 2 & 19 \\
\hline
22 & 3 & 27 & 7 \\
\hline
\end{array}
$rdg(4,4)=48$, so the $4\times4$ rook graph is also graceful.
\begin{array}{|c|c|c|c|}
\hline
0 & 1 & 23 & 47 \\
\hline
19 & 44 & 9 & 2 \\
\hline
37 & 42 & 3 & 11 \\
\hline
48 & 4 & 36 & 32 \\
\hline
\end{array}
$rdg(3,5)=46$. This is not graceful. Like the $3\times3$, Rosa (1967) shows this is the minimum possible.
\begin{array}{|c|c|c|c|c|}
\hline
0 & 1 & 10 & 26 & 46 \\
\hline
23 & 45 & 37 & 5 & 8 \\
\hline
42 & 14 & 44 & 38 & 3\\
\hline
\end{array}
Misha Lavrov's answer, gives a graceful labeling of the $2\times5$ rook graph, but larger $2\times n$ rook graphs cease being graceful:
$rdg(2,6)=38$. The $2\times6$ rook graph has $36$ edges.
\begin{array}{|c|c|c|c|c|c|}
\hline
0 & 1 & 10 & 16 & 34 & 38 \\
\hline
35 & 32 & 24 & 37 & 5 & 12\\
\hline
\end{array}
$rdg(2,7)=53$. The $2\times7$ rook graph has $49$ edges.
\begin{array}{|c|c|c|c|c|c|c|}
\hline
0 & 6 & 16 & 24 & 38 & 41 & 53 \\
\hline
31 & 52 & 3 & 51 & 12 & 8 & 1 \\
\hline
\end{array}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This sum was proposed by Cornel and I solved it using integration but can we solve it using series manipulation?
The integral representation of the sum is $\ \displaystyle\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx$ in case it is needed.
|
$$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\
\\
let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\
\\
\therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\
\\
let\ \ G=\int_{0}^{\infty }\frac{ln^2(x))}{1+a^2x^2}dx\ \ \ \ ,\ \ but \ we \ know\ \\
\\
G(a)=\int_{0}^{\infty }\frac{x^p}{(1+x^2)a^{p+1}}dx=\frac{\pi }{2}\frac{sec\frac{\pi p}{2}}{a^{p+1}}\\
\\
\therefore \frac{\partial^2 G(a)}{\partial^2 p}=\frac{\pi }{2}[tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})+ln(a)sec(\frac{\pi p}{2})ln(a)a^{-p-1}]+\frac{1}{a^{p+1}}[\frac{\pi ^{2}}{4}tan^2(\frac{\pi p}{2})sec(\frac{\pi p}{2})+sec^3(\frac{\pi p}{2})-\frac{\pi }{2}tan(\frac{\pi p}{2})sec(\frac{\pi p}{2})ln(a)]\\$$
$$now\ \ take\ \ p=0\ \\
\\
\therefore \frac{\partial ^2 G(a)}{\partial p^2}_{p=0}=\frac{\pi }{2}[\frac{ln^2(a)}{a}+\frac{\pi ^{2}}{4a}]=\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}dx\ ,\ \ \ \ take\ a=1\\
\\
\therefore \int_{0}^{1 }\frac{ln^2(x)dx}{1+x^2}=\frac{\pi ^{3}}{8}, \ \ \ \ \ now\ going\ to \ I\\
\\
\therefore I(a)=\int_{0}^{1}(\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2(x)}{1+x^2})dx=\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a}\\
\\
\\
\therefore I'(a)=\frac{2a}{1-a^2}(\frac{\pi ^{3}}{8}(\frac{1-a}{a})+\frac{\pi ln^2(a)}{2a})$$
$$\therefore I(1)=\frac{\pi ^{3}}{4}\int_{0}^{1}\frac{dx}{1+x}+\pi \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx\\
\\
\therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1-x}dx+\frac{1}{2}\int_{0}^{1}\frac{ln^2(x)}{1+x}dx\\
\\
\therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{1}{2}[-Ln^2(x)ln(1-x)\tfrac{1}{0}+ln^2(x)ln(1+x)\tfrac{1}{0}+2\int_{0}^{1}\frac{ln(1-x)lnx}{x}dx-2\int_{0}^{1}\frac{ln(1+x)lnx}{x}dx]\\
\\
\\
\therefore \int_{0}^{1}\frac{ln^2(x)}{1-x^2}dx=\frac{7}{4}\zeta (3)\\
\\
\therefore I=\int_{0}^{\infty }\frac{ln^2(x)ln(1+x^2)}{1+x^2}dx=\frac{\pi ^{3}}{4}ln(2)+\frac{7\pi }{4}\zeta (3)$$
|
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|
How to get the value of the root? I have this statement:
If $\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3}$, Which of
the following values are the closest to $\sqrt{21}$ ?
A) 68/15 B) 14/3 C) 19/4 D) 55/12 E) 9/2
My development was:
$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{35}-\sqrt{21}}{2} \approx \frac{2}{3}$
My idea was to treat the sign $\approx$ as a sign $=$ and thus eliminate roots and clear $\sqrt21$, with which I have obtained $55/12$ but I do not know if this is correct.
$\frac{\sqrt{35}-\sqrt{21}}{2} = \frac{2}{3}$
$35 = (\frac{4}{3} +\sqrt{21})^2$
$\sqrt{21} = \frac{110}{9} * \frac{3}{8} = 55/12$
So my doubt is: Can I treat a $\approx$ sign as a $=$ sign to work like an normal equation?
|
From
$$\left(\frac23\right)^2\approx\left(\frac{\sqrt7}{\sqrt5+\sqrt3}\right)^2$$
we get
$$\frac49\approx\frac{7}{5+3+2\sqrt{15}},$$
hence $\sqrt{15}\approx31/8$.
Now we have
$$\frac{2\sqrt{3}}3\approx
\frac{\sqrt7\sqrt3}{\sqrt5+\sqrt3}=\frac{\sqrt{21}}{\sqrt5+\sqrt3}.$$
Using the approximate value for $\sqrt{15}$ we arrive in
$$\sqrt{21}\approx\frac{55}{12}.$$
|
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|
If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
I have provided a (dumbfounding) solution down below if you want to check out. There should be simpler solutions, I believe so.
|
1) When $S=x^2-2xy-y$ and $T = xy-2y^2-x$, then $$ 5|S-T =
(x-y)(x-2y+1) \ (a)$$
and $$ 5| S+T = (x-2y-1)(x+y)\ (b)
$$
Hence I know that $(x-2y+1)-(x-2y-1)=2$, by mathlove's comment.
Hence at least one of $(x-2y+1),\ (x-2y-1)$ can not be divided by $5$. So by $(a),\ (b)$ we have $5|x^2-y^2$.
2) When $U=2x^2+y^2+2x+y$, then $$ U+S+2T= 3(x^2-y^2) $$
So we complete the proof.
|
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|
How to prove this inequality for $a,b,c>0$? How to prove the inequality for $a,b,c>0$ :
$$\frac{2a-b-c}{2(b+c)^2}+\frac{2b-a-c}{2(a+c)^2}+\frac{2c-b-a}{2(b+a)^2}\geq 0$$
?
|
The triples $(2a-b-c,2b-a-c,2c-a-b)$ and $\left(\frac{1}{(b+c)^2},\frac{1}{(a+c)^2},\frac{1}{(a+b)^2}\right)$ are the same ordered.
Thus, by Chebyshov we obtain:
$$\sum_{cyc}\frac{2a-b-c}{(b+c)^2}\geq\frac{1}{3}\sum_{cyc}(2a-b-c)\sum_{cyc}\frac{1}{(b+c)^2}=0.$$
|
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|
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\
&=\frac{n+2}{(n+1)^2} \\
&=\frac{n+2}{n^2 + 2n + 1} \\
&<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\
&=\frac{n+2}{n(n+2)}\\
&=\frac{1}{n} \end{align} $$
I am just wondering if there is a simpler way of doing this.
|
Just for the fun of it, you can re-write your inequality as:
$$\frac{1}{n}-\frac{1}{n+1}>\frac{1}{(n+1)^2}.$$
This can be re-writen as:
$$\int_{n}^{n+1}\frac{1}{x^2}dx>\frac{1}{(n+1)^2},$$
or, if $f(x)=\dfrac{1}{x^2}$:
$$\int_{n}^{n+1}f(x)dx>f(n+1),$$
which is obvious if you consider that $f$ is strictly decreasing and positive on $(0,+\infty)$ and that the right-hand side of the inequality is the area of a rectangle with height $f(n+1)$ and a basis of $1$ - see the figure below.
|
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|
sum of series $\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $
The sum of series
$$\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $$
My attempt $$\displaystyle \sum^{n}_{k=1}\frac{1}{(3k-2)(2k+1)}=\frac{1}{7}\sum^{n}_{k=1}\bigg[\frac{3}{3k-2}-\frac{2}{2k+1}\bigg]$$
$$=\frac{1}{7}\sum^{n}_{k=1}\int^{1}_{0}\bigg(3x^{3k-3}-2x^{2k}\bigg)dx$$
$$=\frac{1}{7}\int^{1}_{0}\bigg(\sum^{n}_{k=1}3x^{3k-3}-2x^{2k}\bigg)dx$$
$$=\frac{1}{7}\int^{1}_{0}\bigg[\frac{3(1-x^{3n})}{1-x^3}-\frac{2(1-x^{2n+2})}{1-x^2}\bigg]dx$$
How do i solve it Help me please
|
The approximate value of sum of terms in $HP$ is $S_n\approx \frac{1}{d}\ln(\frac{2a+2(n-1)d}{2a-d})$ here $a$=reciprocal of first term $(a_1=4,a_2=3)$ respectively. $d$=difference between the reciprocal of two terms $(d_1=3,d_2=2)$ respectively
Note that the two HPs are $3+3(\frac{1}{4}+\frac{1}{7}+....)-2(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)$
|
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|
Modular arithmetic with Legendre symbol Let $n\in\mathbb{Z}_{>0}$ and let $p\neq3$ be a prime divisor of $n^2+n+1$. Show that $p\equiv1\mod3$.
I thought of trying to prove that $\left(\frac{p}{3}\right)=1$, since 1 is the only element of $\mathbb{F}_3$ that is a square modulo 3. I am supposed to use quadratic reciprocity, which leads to $\left(\frac{p}{3}\right)=\left(\frac{3}{p}\right)(-1)^{\frac{p-1}{2}}$. However, I don't know how to proceed from here.
|
Since $4(n^2+n+1)=(2n+1)^2+3$ is divisible by $p$, we have $-3$ is a quadratic residue mod $p$. So
\begin{align*}
1=\left(\frac{-3}{p}\right)&=\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)\\
&=\left(\frac{3}{p}\right)(-1)^{(p-1)/2}\\
&=\left(\frac{p}{3}\vphantom{\frac3p}\right)(-1)^{\frac{p-1}{2}\cdot\frac{3-1}{2}}(-1)^{\frac{p-1}{2}}&&\text{Quadratic Reciprocity}\\
&=\left(\frac{p}{3}\vphantom{\frac3p}\right)
\end{align*}
So $p\equiv 1\pmod{3}$.
|
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|
mod cancellation: compute $\, n/2\bmod 6\, $ from $\,n \bmod m\,$ for even $n$ I am using the C++ language.
I want to calculate these 2 expressions:-
In our case, $x= 100000000000000000$
Expression(1)
$$((3^x-1)/2)\mod7$$
The numerator $3^x-1$ is always divisible by $2$(basic number theory)
I calculated the above expression using Extended-Euclid-Gcd algorithm.
The above algorithm only works when gcd(denominator,mod-value)=1...in our case $\gcd(2,7)=1$ . So we were able to calculate it using the above algorithm.
Expression(2)
$$((3^x-1)/2)\mod6$$
The numerator $3^x-1$ is always divisible by $2$(basic number theory-again)
Now, how do I calculate the above expression as $\gcd(2,6)=2$ which is not equal to $1$ ?
|
Dividing by $\boldsymbol{q}$ when the modulus is divisible by $\boldsymbol{q}$
Note that
$$
aq\equiv bq\pmod{pq}
$$
precisely when
$$
a\equiv b\pmod{p}
$$
Thus, for $n\gt0$,
$$
3^n-1\equiv\left\{\begin{array}{}2&\text{if $n$ is odd}\\8&\text{if $n$ is even}\end{array}\right.\pmod{12}
$$
implies
$$
\frac{3^n-1}2\equiv\left\{\begin{array}{}1&\text{if $n$ is odd}\\4&\text{if $n$ is even}\end{array}\right.\pmod{6}
$$
A More Detailed Answer
This can be solved mod $6$ by solving mod $3$ and mod $2$, then applying the Chinese Remainder Theorem.
For $n\gt0$, $3^n\equiv0\pmod3$; therefore,
$$
\frac{3^n-1}2\equiv1\pmod3
$$
To compute
$$
\frac{3^n-1}2\pmod2
$$
we can to look at
$$
3^n-1\equiv(-1)^n-1\pmod4
$$
to get
$$
\frac{3^n-1}2\equiv\frac{(-1)^n-1}2\equiv\left\{\begin{array}{}0&\text{if $n$ is even}\\1&\text{if $n$ is odd}\end{array}\right.\pmod2
$$
Then the Chinese Remainder Theorem says we can combine the results mod $3$ and mod $2$ to get
$$
\frac{3^n-1}2\equiv\left\{\begin{array}{}4&\text{if $n$ is even}\\1&\text{if $n$ is odd}\end{array}\right.\pmod6
$$
|
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|
Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ . Problem. Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove:
$$a^{3}b+ b^{3}c+ c^{3}a\leqq 8$$
My solution in M&Y : (and I'm looking forward to seeing a nicer one(s), thanks for your interests !)
Because of the invariant of this circular permutation in $a, b, c$, suppose that $a\equiv \max\{\!a,\,b,\,c\!\}> 1$ :
$$\therefore\,ab+ bc+ ca= \frac{(a+ b+ c)^{2}- (a^{2}+ b^{2}+ c^{2})}{2}= \frac{3^{2}- 5}{2}= 2$$
Again, we have $5- a^{2}= b^{2}+ c^{2}\leqq (b+ c)^{2}= (3- a)^{2}$, therefore $a\geqq 2$ or $a\leqq 1\,(\!impossible\,\,!\!)$ .
$$\begin{align}
\therefore\,b,\,c\leqq 1\,\therefore\,a^{3}b+ b^{3}c+ c^{3}a & \leqq a^{3}b+ bc+ ca= ab(a^{2}- 1)+ 2\\
& \leqq \frac{1}{4}(a^{2}+ 4\,b^{2})(a^{2}- 1)+ 2\\
& \leqq \frac{1}{4}(5+ 3\,b^{2})(4- b^{2})+ 2\\
& = 8- \frac{1}{4}(1- b^{2})(4- 3\,b^{2})\leqq 8
\end{align}$$
The equality condition $\{\!a= 2,\,b= 1,\,c= 0\!\}\bigcup\{\!a= 1,\,b= 0,\,c= 2\!\}\bigcup\{\!a= 0,\,b= 2,\,c= 1\!\}$/q.e.d
|
There is another solution by Rearrangement and AM-GM.
Indeed, let again $\{a,b,c\}=\{x,y,z\},$ where $x\geq y\geq z$.
Thus, $x+y+z=3,$ $xy+xz+yz=2$ and $$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x^2+2xz+z^2)=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+2\cdot\frac{x+z}{2}}{3}\right)^3=4,$$
which gives $$a^2b+b^2c+c^2a\leq4-abc.$$
Id est,
$$\sum_{cyc}a^3b=(a+b+c)(a^2b+b^2c+c^2a)-\sum_{cyc}(a^2b^2+a^2bc)=$$
$$=3(a^2b+b^2c+c^2a)-\sum_{cyc}(a^2b^2+2a^2bc)+\sum_{cyc}a^2bc=$$
$$=3(a^2b+b^2c+c^2a)-4+3abc\leq3(4-abc)-4+3abc=8.$$
|
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|
Show that $ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$ Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$
LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\frac{n}{\sqrt {n+1}}+\frac{n}{\sqrt {n+2}}.......\frac{n}{\sqrt {2n}}] = \infty$
Let $a_n=\frac{n}{\sqrt{2n}}$ then $a_n = \frac{1}{\sqrt 2}\sqrt{n}$
hence $\lim_{n\to\infty}a_n = \infty$
I continue the problem therefore using Cauchy's first theorem on limits.
But the way I took $a_n$ is it correct?
Does it not make $a_1 = \frac{1}{\sqrt{2}}$
If my way of taking $a_n$ is wrong, please suggest me the right way. Thank you
|
You could have quite good approximations using generalized harmonic numbers
$$S_n=\sum_{i=0}^n \frac 1 {\sqrt{n+i}}=H_{2 n}^{\left(\frac{1}{2}\right)}-H_{n-1}^{\left(\frac{1}{2}\right)}$$
Using the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{p}\right)^{1/2}-\frac{1}{24}
\left(\frac{1}{p}\right)^{3/2}+O\left(\frac{1}{p^{7/2}}\right)$$ Using it twice and continuing with Taylor series
$$S_n=2 \left(\sqrt{2}-1\right) \sqrt{n}+\frac{\left(2+\sqrt{2}\right)}{4 \sqrt n}
+O\left(\frac{1}{n^{3/2}}\right)$$ Just trying for $n=100$, the value should be $\approx 8.369654$ while the above truncated series gives $\frac{3}{40} \left(267 \sqrt{2}-266\right) \approx 8.369627$
|
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|
Prove the following identity of Gamma : $\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$
Prove
$$\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$$
We know that :
$\Gamma(x+\frac{1}{2})=\frac{(2x)!\sqrt{π}}{4^{x}x!}$
So : $\Gamma(\frac{7}{8})=\Gamma(\frac{1}{2}+\frac{3}{8})$ !!
|
By the reflection formula, if $z\notin\mathbb{Z}$,
$$\Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$$
so you can evaluate $\Gamma(\frac{1}{8})\Gamma(\frac{7}{8})$ and $\Gamma(\frac{3}{8}) \Gamma(\frac{5}{8})$ and verify that
$$\frac{\Gamma(\frac{1}{8})\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})\Gamma(\frac{5}{8})}=1+\sqrt{2}.$$
P.S. See also Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$ or the Silver ratio (due to user 493905).
|
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|
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . .
If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that
$$a^2 + b^2 + c^2=d$$
then $a+b+c+d$ equals?
$$(A)\ 0 \quad (B)\ 1 \quad (C)\ 2 \quad (D)\ 4$$
Here is what I tried: assume the common difference is $x$.
Thus,
$$a^2 + a^2 + 2ax + x^2 +a^2 +4ax +4x^2=a + 3x$$
That is,
$$3a^2 + 6ax + 2x^2=a + 3x$$
First I assumed it as a quadratic in $a$ and solved for it.
$$a=\frac{(1-6x) \pm \sqrt{12x^2-48x +1}}{6}$$ but to no avail.
Then I assumed it as a quadratic in $x$ and got:
$$x=\frac{(3-6a) \pm \sqrt{12a^2 -36a +81}}{6}$$
And I noticed that I'm stuck. Please help. Thanks in advance!
Edit: I made a calculation error as pointed out by multiple answers, and so the equation should have been $$3a^2 + 6ax + 5x^2=a + 3x$$ which leads to different solutions.
|
Let $x$ be the common difference.
From $a^2+b^2+c^2=d$, we get $d > 0$,
Since $a,b,c,d$ are distinct, we get $x\ne 0$.
If $x < 0$, then $c > d$, hence
$$c^2 > d^2\ge d=a^2+b^2+c^2 > c^2$$
contradiction.
Hence $x > 0$.
\begin{align*}
\text{Then}\;\;&a^2+b^2+c^2=d\\[4pt]
\implies\;&(b-x)^2+b^2+(b+x)^2=b+2x\\[4pt]
\implies\;&3b^2+2x^2=b+2x\\[4pt]
\implies\;&b(1-3b)=2x(x-1)\\[4pt]
\implies\;&b(1-3b)\ge 0\\[4pt]
\implies\;&b=0\;\;\\[4pt]
\implies\;&2x(x-1)=0\\[4pt]
\implies\;&x=1\\[4pt]
\implies\;&a,b,c,d=-1,0,1,2\\[4pt]
\implies\;&a+b+c+d=2\\[4pt]
\end{align*}
|
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|
$\lim_{n\to \infty} \ \frac{1}{n} \Bigl[(a+\frac{1}{n})^2+(a+\frac{2}{n})^2+\cdots+(a+\frac{n-1}{n})^2\Bigr]$ without L'Hopital Find limit:
$$a_n = \lim_{n\to \infty} \ \frac{1}{n} \Bigl[\Bigl(a+\frac{1}{n}\Bigr)^2+\Bigl(a+\frac{2}{n}\Bigr)^2+\cdots+\Bigl(a+\frac{n-1}{n}\Bigr)^2\Bigr]$$
I tried limiting it with $$n\cdot\frac{1}{n}\Bigl(a+\frac{1}{n}\Bigr)^2\leqslant a_n \leqslant n \cdot \frac{1}{n}\Bigl(a+\frac{n-1}{n}\Bigr)^2$$
but that got me to the answer that left side limits to $a^2$ and right side to $(a+1)^2$
and so I didn't squeeze it the right way. Help, and keep it simple.
|
If you expand the squares, then you get
$$\begin{split} \sum_{k=1}^{n-1} \Big(a+\frac{k}{n}\Big)^2&=\sum_{k=1}^{n-1} \Big(a^2+2a\frac{k}{n}+\frac{k^2}{n^2}\Big)
\\
&=(n-1)a^2+2a\cdot \frac{1}{n}\cdot\frac{n(n-1)}{2}+\frac{1}{n^2}\cdot\frac{n(n-1)(2n-1)}{6}.
\\
&=(n-1)a^2+a(n-1)+\frac{(n-1)(2n-1)}{6n},
\end{split}$$
and thus
$$\begin{split}
\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} \Big(a+\frac{k}{n}\Big)^2 &=\lim_{n\to\infty} \Big((1-1/n)a^2+a(1-1/n)+\frac{(n-1)(2n-1)}{6n^2}\Big)
\\
&=a^2+a+\frac{1}{3}.
\end{split}$$
|
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|
Closed form for $f(x)=\ _3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$
I am seeking a closed form for the function $$f(x)=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$$
I expect there to be one, because of this post and Wolfram. The Wolfram link produces closed forms involving $\mathrm{Li}_2$ for any value of $x$ that I've tried so far, so I can only assume that a general closed form exists.
I've started my attempts by noticing that
$$f(x)=\frac12\int_0^1 \frac{_2F_1(\tfrac12,\tfrac12;\tfrac32;xt)}{\sqrt{t}}dt,$$
because
$$\frac12\int_0^1 \frac{(xt)^n}{\sqrt{t}}dt=\frac{x^n}{2n+1}$$
which would introduce another factor of $$\frac{n+1/2}{n+3/2}$$
when computing the ratio of the terms. Similarly,
$$_2F_1\left(\tfrac12,\tfrac12;\tfrac32;x\right)=\frac12\int_0^1 \frac{_1F_0(\tfrac12;;xt)}{\sqrt{t}}dt.$$
The last hypergeometric I was able to recognize as $$_1F_0\left(\tfrac12;;xt\right)=\frac1{\sqrt{1-xt}}.$$
So, all in all,
$$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu,$$
which looks like the Beta function's evil cousin.
I do not know how to turn this integral into something containing $\mathrm{Li}_2$ and I need some help. Thanks!
|
For $x>0$ we have:$$f(x)=\frac14\int_0^1\int_0^1 \frac{1}{\sqrt{vu}\sqrt{1-xvu}}dvdu\overset{vu=t}=\frac14\int_0^1\frac{1}{u}\int_0^u \frac{1}{\sqrt{t}\sqrt{1-xt}}dtdu$$
$$=\frac12\int_0^1 \frac{1}{u}\frac{\arcsin \sqrt{xt}}{\sqrt{x}}\bigg|_0^udu=\frac1{2\sqrt x} \int_0^1 \frac{\arcsin\sqrt{xu}}{u}du\overset{xu=t}=\frac{1}{2\sqrt x}\int_0^x \frac{\arcsin \sqrt t}{t}dt$$
$$\overset{t=y^2}=\frac{1}{\sqrt x}\int_0^\sqrt x \frac{\arcsin y}{y}dy\overset{IBP}=\frac{1}{\sqrt x} \ln \sqrt x \arcsin\sqrt x-\frac{1}{\sqrt x}\int_0^\sqrt x \frac{\ln y}{\sqrt{1-y^2}}dy$$
$$\overset{y=\sin z}=\frac{1}{\sqrt x} \ln \sqrt x \arcsin\sqrt x-\frac{1}{\sqrt x}\int_0^{\arcsin \sqrt x} \ln(\sin z)dz$$
$$=\frac{1}{\sqrt x} \ln \sqrt x \arcsin\sqrt x+\frac{\arcsin \sqrt x}{\sqrt x}\ln 2+\frac{1}{2\sqrt x}\operatorname{Cl}_2(2\arcsin \sqrt x)$$
$$=\boxed{\frac{1}{\sqrt x} \left(\arcsin \sqrt x \ln(2\sqrt x) +\frac12\operatorname{Cl}_2(2\arcsin \sqrt x) \right)}$$
In terms of Clausen function, which of course is the imaginary part of some dilogarithms.
For $x<0$ we can work with $x=-y$, $y>0$ and using:
$$\frac{\arcsin \sqrt{-z}}{\sqrt{-z}}=\frac{\operatorname{arcsinh} \sqrt{z}}{\sqrt{z}}$$
We will arrive at:
$$f(-y)=\frac{1}{2\sqrt y} \int_0^1 \frac{\operatorname{arcsinh}\sqrt{yu}}{u}du\overset{yu=t}=\frac{1}{2\sqrt y}\int_0^y \frac{\operatorname{arcsinh}\sqrt{t}}{t}dt$$
$$\overset{t=v^2}=\frac{1}{\sqrt y} \int_0^\sqrt y \frac{\operatorname{arcsinh}v}{v}dv\overset{IBP}=\frac{1}{\sqrt y} \ln \sqrt y \operatorname{arcsinh}\sqrt y-\frac{1}{\sqrt y} \underbrace{\int_0^\sqrt y\frac{\ln v}{\sqrt{1+v^2}}dv}_{=J}$$
For $J$ put $v=\frac{1-t^2}{2t}$ then:
$$J=\int_1^{\sqrt{1+y}-\sqrt y}\ln\left(\frac{2t}{1-t^2}\right)\frac{dt}{t}=\left(\frac12\ln^2(2t) +\frac12 \operatorname{Li}_2(t^2)\right)\bigg|_1^{\sqrt{1+y}-\sqrt y}$$
$$=\frac12\left( \ln^2[2\sqrt{1+y}-2\sqrt y]-\ln^2 2 +\operatorname{Li}_2[(\sqrt{1+y}-\sqrt y)^2]-\frac{\pi^2}{6}\right)$$
$$\small \Rightarrow \boxed{f(-y)=\frac{1}{\sqrt y} \ln \sqrt y \operatorname{arcsinh}\sqrt y+\frac{1}{2\sqrt y}\left(\ln^2 2 -\ln^2[2\sqrt{1+y}-2\sqrt y] - \operatorname{Li}_2[(\sqrt{1+y}-\sqrt y)^2]+\frac{\pi^2}{6}\right)}$$
|
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|
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{7^{k+1}}+5^{7^{k+1}}\equiv r \pmod{7^{k+2}}$$
so $3^{7^{k+1}}+5^{7^{k+1}}=7^{k+2}*q_2+r$
Now if we subtract them we get that:
$$3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}=7^{k+1}(7q_2-q1)+r-1$$
From Euler's theorem we know that $a^{\phi(7^{k+1})}\equiv 1 \pmod{7^{k+1}}$
and $\phi(7^{k+1}) = 6\cdot7^{k}$ so $3^{6\cdot7^{k}}\equiv 1 \pmod{7^{k+1}}$ and $5^{6 \cdot 7^{k}}\equiv 1 \pmod{7^{k+1}}$.
So $3^{7^{k+1}}+5^{7^{k+1}}-3^{7^k}-5^{7^k}\equiv3^{7^k}\cdot3^{6\cdot{7^{k}}}+5^{7^k}\cdot5^{6\cdot{7^{k}}}-3^{7^k}-5^{7^k}\equiv 0 \pmod{7^{k+1}}$
Finally I get that $r\equiv1\pmod{7^{k+1}}$.
Here I got stuck. I don't know how to show that $r=1$.
|
In this answer I will try to consider generalizations, and organize things better. My first answer was a proof discovered by ad-hoc methods.
Lemma
If $x \equiv y \pmod p$, then $x^{p^n} \equiv y^{p^n} \pmod {p^{n+1}}$ for all $n \geq 0$.
Proof
The case $n = 0$ is trivial. We now prove that if $x \equiv y \pmod {p^k}$ for some $k \geq 1$, then $x^p \equiv y^p \pmod {p^{k+1}}$. This follows from writing $x = m p^k + y$ and using the binomial theorem: $$x^p \equiv (m p^k + y)^p \equiv \sum\limits_{i = 0}^p {p \choose i} p^{ik} m^i y^{p - i} \equiv y^p \pmod {p^{k+1}}$$ because all terms with $i \geq 1$ are divisible by $p^{k+1}$.
I used the fact that ${p \choose i}$ is divisible by $p$ for $1 \leq i \leq p - 1$, and the fact that $kp \geq k + 1$ when $k \geq 1$ for the case $i = p$.
The result follows by induction on $n$.
Proposition
Suppose that $p \equiv 1 \pmod 6$ is a prime. Suppose $a$ and $b$ are the two distinct elements of order $6$ modulo $p$. Then for all $n \geq 0$, we have $(a + b)^{p^n} \equiv a^{p^n} + b^{p^n} \equiv 1 \pmod {p^{n+1}}$.
Note
The elements of order $6$ modulo $p$ must be roots of the $6$-th cyclotomic polynomial $X^2 - X + 1$, which has degree $\varphi(6) = 2$, so there are at most $2$ of them. If $p$ does not divide $6$, it has no multiple root, so the roots are distinct. The roots exist if and only if $p$ is odd and $-3$ is a square $\pmod p$, which means $p \equiv 1 \pmod 6$ by quadratic reciprocity.
The original question is a special case where $p = 7$. You can check that $3$ and $5$ are in fact the elements of order $6$ modulo $7$. One simple way to see this is to note that $3+5 \equiv 8 \equiv 1 \pmod 7$ and $3\cdot 5 \equiv 15 \equiv 1 \pmod 7$, so Vieta's formulas imply that they are roots of $X^2 - X + 1$.
Proof
We know that $a^6 \equiv b^6 \equiv 1 \pmod p$. Using the lemma, we see that $\left(a^{p^n}\right)^6 \equiv \left(b^{p^n}\right)^6 \equiv 1 \pmod {p^{n+1}}$.
It follows that $a^{p^n}$ and $b^{p^n}$ are roots of the equation $X^6 - 1 \equiv (X^3 - 1)(X+1)(X^2-X+1) \equiv 0 \pmod {p^{n+1}}$.
We know that $a^{p^n}$ and $b^{p^n}$ cannot be roots of the first two factors modulo $p$, because that would imply that the order of $a$ and $b$ modulo $p$ is not $6$ (here we use the fact that $p$ does not divide $6$, so the order of $a^{p^n}$ modulo $p$ is the same as the order of $a$ modulo $p$).
Since none of the other factors are divisible by $p$, we must have that $a^{p^n}$ and $b^{p^n}$ are roots of $X^2 - X + 1 \equiv 0 \pmod {p^{n+1}}$.
Note that both $a$ and $a^{-1}$ have order $6$ modulo $p$ (and are distinct), so in fact $b \equiv a^{-1} \pmod p$. That means $ab \equiv 1 \pmod p$, and the lemma implies that $a^{p^n}b^{p^n} \equiv (ab)^{p^n} \equiv 1 \pmod {p^{n+1}}$.
This means $a^{p^n}$ and $b^{p^n}$ are inverses modulo $p^{n+1}$.
If $a^{p^n} + b^{p^n} \equiv s \pmod {p^{n+1}}$, then $a^{p^n}$ and $b^{p^n}$ are roots of $\left(X - a^{p^n}\right)\left(X - b^{p^n}\right) = X^2 - sX + 1$. Since they are also roots of $X^2 - X + 1$, they are roots of the difference, $(s - 1)X = 0$. So $(s - 1)a^{p^n} \equiv 0 \pmod {p^{n+1}}$. But $a^{p^n}$ is invertible modulo $p^{n+1}$, so we conclude $s \equiv 1 \pmod {p^{n+1}}$.
Finally we see that $a^{p^n} + b^{p^n} \equiv 1 \pmod {p^{n+1}}$.
Another generalization
Let $p$ be a prime and $r$ a positive integer not divisible by $p$. Suppose there are $m = \varphi(r)$ distinct elements $a_1, \ldots, a_m$ of order $r$ modulo $p$. Then for all $n \geq 0$, we have $a_1^{p^n} + \cdots + a_m^{p^n} \equiv c \pmod {p^{n+1}}$, where $c$ is the negative of the coefficient of $X^{m-1}$ in the $r$-th cyclotomic polynomial $\Phi_r(X)$.
Note
Since $a_1, \ldots, a_m$ are roots of $\Phi_r(X)$ modulo $p$, their sum is $a_1 + \cdots + a_m \equiv c \pmod p$. The lemma then implies that $(a_1 + \cdots + a_m)^{p^n} \equiv c^{p^n} \pmod {p^{n+1}}$. So the theorem shows that whenever $c^{p^n} \equiv c \pmod {p^{n+1}}$, we have $(a_1 + \cdots + a_m)^{p^n} \equiv a_1^{p^n} + \cdots + a_m^{p^n} \equiv c \pmod {p^{n+1}}$.
Proof
The lemma essentially allows lifting the roots of the $r$-th cyclotomic polynomial modulo $p$ to roots of the $r$-th cyclotomic polynomial modulo $p^{n+1}$. This comes from the fact that all roots of $\Phi_r(X) \pmod p$ have order $r$ modulo $p$, and raising to the $p^n$-th power preserves the order modulo $p$. So all $a_i^{p^n}$ have order $r$ modulo $p$, and are also roots of $X^r - 1$ modulo $p^{n+1}$ (by the lemma!). If they were not roots of $\Phi_r(X)$ modulo $p^{n+1}$, then they would be roots (modulo $p$) of a cyclotomic polynomial of smaller order dividing $r$, and so their order modulo $p$ would not be $r$.
So the $a_i^{p^n}$ are common roots modulo $p^{n+1}$ of the polynomials $\left(X - a_1^{p^n}\right)\cdots\left(X - a_m^{p^n}\right)$ and $\Phi_r(X)$ which both have degree $m$. They are also distinct, because if $a_i^{p^n} \equiv a_j^{p^n} \pmod {p^{n+1}}$ then this is also true modulo $p$, and then Fermat's little theorem implies $a_i \equiv a_j \pmod p$ so $i = j$.
We can use Euclidean division of $\Phi_r(X)$ by each $X - a_i^{p^n}$ (we work over $\mathbb{Z}/p^{n+1}\mathbb{Z}$, which is not a field, but since each factor is monic, the procedure still works). At each step, the degree of $\Phi_r(X)$ is reduced by $1$, and each remainder term is a constant which must be zero since the $a_i^{p^n}$ are roots. Since we repeat the step as many times as $m$, at the end there is nothing left. Therefore $\left(X - a_1^{p^n}\right)\cdots\left(X - a_m^{p^n}\right) \equiv \Phi_r(X) \pmod {p^{n+1}}$.
If follows from Vietas formulas that the sum $a_1^{p^n} + \cdots + a_m^{p^n}$ is the negative of the coefficient of $X^{m-1}$ in $\Phi_r(X)$.
|
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|
Other methods for a limit I know that I can compute the limit
$$
\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2}=\frac{N(N+1)}{2}
$$
using L'Hospital's rule (not one but two times) but I am looking for other ways. Are there any of them?
p.s.: the limit follows from a shortcut used in order to find the value of $$
\sum_{k=1}^Nkx^k\Big|_{x=1}.
$$
|
You can simplify your fraction as
\begin{align}
\frac{Nx^{N+1}-Nx^N+1-x^N}{(1-x)^2} &=\frac{-Nx^N(1-x)+(1-x)(1+x+\dots+x^{N-1})}{(1-x)^2} \\
& = \frac{-Nx^N+1+x+\dots+x^{N-1}}{1-x} \\
& = \frac{1+x+\dots+x^{N-1}+x^N-(N+1)x^N}{1-x}
\end{align}
For $x \to 1$ the numerator tends to $0$, so this is still a form $0/0$ and you can apply de l'Hôpital:
\begin{align}
\lim_{x \to 1} -(1+2x+3x^2+\dots+Nx^{N-1}-N(N+1)x^{N-1})& =-\sum_{k=1}^Nk+N(N+1) \\
& = N(N+1)/2.
\end{align}
|
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|
$a_n=\min\limits_{x+y=1}(x^n+y^n)=\frac{1}{2^{n-1}}$ We must prove :$\min\limits_{x+y=1}(x^n+y^n)=\frac{1}{2^{n-1}} $ for all $n \in \mathbb{N}_{>0}$
And to prove this we can use the inequality: $\frac{x+y}{2}\leq (\frac{x^n+y^n}{2})^{\frac{1}{n}}$
where equality is satisfied if $x=y=1/2$
My question is how we can prove inquality: $\frac{x+y}{2}\leq (\frac{x^n+y^n}{2})^{\frac{1}{n}}$ ?
|
This is just convexity of the function $f(x)=x^{n}$. The inequality simply says $f(\frac {x+y} 2) \leq \frac {f(x)+f(y)} 2$.
|
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|
Prove $\sum_\text{cyc}\frac{1}{3-ab}\le\frac{3}{2}$
Let $a,b,c>0$ such that $a^2+b^2+c^2=1$. Prove that $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le\frac{3}{2}$$
By Am/Gm we have: $$ {1\over 3-ab}\leq {2\over 6-a^2-b^2} = {2\over 5+c^2}<{2
\over 5}$$
so
$$\sum_{cyc} {2\over 5+a^2} < {6\over 5} <{3\over 2}$$
So I get much better bound. Where did I go wrong?
Edit: If nothing is wrong then what is a smallest constant $m$ such $$\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\le m$$ For $a=b=c=\sqrt{3}/3$ we get $m\geq {9\over 8}$.
|
Your calculation is correct: $m = \frac 65$ is an upper bound. It is not the best upper bound though, because the estimates $\frac{2}{5+c^2} < \frac 25$ are not sharp for all three variables simultaneously.
The best upper bound is $\color{red}{m=\frac 98}$. Proof: The function $f(t) = \frac{1}{3-\sqrt t}$ is increasing and concave on $[0, 1]$:
$$
f'(t) = \frac{1}{2 \sqrt t (3-\sqrt t)^2} \ge 0 \, ,\\
f''(t) = -\frac{3(1-\sqrt t)}{4 t^{3/2}(3-\sqrt t)^3} \le 0 \, .
$$
Therefore Jensen's inequality gives
$$
\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} = f(a^2b^2)+f(b^2c^2)+ f(c^2a^2) \\
\le 3 f\left( \frac{a^2b^2+b^2c^2+c^2a^2}{3}\right)
\underset{(*)}{\le} 3 f\left( \bigl( \frac{a^2+b^2+c^2}{3} \bigr)^2\right)
= 3 f(\frac 19) = \frac 98 \, .
$$
At $(*)$ we have used that
$$
\frac{a^2b^2+b^2c^2+c^2a^2}{3} \le \left( \frac{a^2+b^2+c^2}{3} \right)^2 \, ,
$$
which is MacLaurin's inequality applied to
$$
(x_1, x_2,x_3) = (a^2, b^2, c^2) \, .
$$
So $m= \frac98$ is an upper bound, and it is best possible since equality holds for $a=b=c=1/\sqrt 3$.
|
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|
Showing $f(x) = \frac{x^2}{\sin(x)}$ is analytic near $0$ Problem
Show the function
$$
f(x) = \frac{x^2}{\sin(x)}
$$
is an analytic about $x=0$.
Try
We have
$$
f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }
$$
Letting $f(x) = \sum_{n=0}^\infty a_n x^n $, we have
$$
a_0 = 0, a_1 = 1, a_2 = 0, a_3 = 1/6, a_4 = 0, a_5 = 7/360, \cdots
$$
thus $f(x) = x - x^3/6 + \cdots$.
Now we have to decide if $\sum_{n=0}^\infty a_n x^n$ is convergent near zero.
But I cannot see the rules of $a_0, a_1, a_2, \cdots$, so how should I proceed?
It is known that
Assume $f: = \sum_{n=0}^\infty b_n x^n$ and $g:= \sum_{n=0}^\infty c_n x^n$ be convergent near zero with radius of convergence $\rho>0$.
If $g(0) \neq 0$, we have
$$
\frac{f}{g} = \sum_{n=0}^\infty d_n
$$
for some radius of convergence $\le \rho$. So relating this fact to my problem,
$$
f(x) = \frac{x}{1 - x^2/3! + x^4/5! - \cdots }
$$
may be convergent with radius of convergence $\le \infty$, which is the radius of $x^2$(poylnomial) and $\sin(x)$
|
We have $f(x) = \frac{x^2}{x - x^3/3! + x^5/5! - \cdots }=\frac{x}{1 - x^2/3! + x^4/5! - \cdots }.$
The nominator and the denominator are power series with radius of convergenc $= \infty$. The power series in the denominator is $ \ne 0$ at $x=0.$ Hence there is $ r>0$ such that
$f(x)= \sum_{n=0}^\infty d_nx^n$ for $|x|<r.$
We have $r= \pi$. Why ?
|
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|
How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-\log_{3}(3)} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-1} + 8\\
\log_{3}(x) & = -\log_{3}(x) + 8\\
2\log_{3}(x) & = 8\\
\log_{3}(x) & = 4\\
x & = 4
\end{align*}
What am I doing wrong?
|
$\log_3 x=4\implies x=3^4=81,$ not $x=4$
|
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|
An equality from the line through the centroid of a triangle $G$ is the centroid of $\triangle\mathit{ABC}$. $X$, $Y$, and $Z$ are points on the sides (or lines through the sides) $\overline{\mathit{BC}}$, $\overline{\mathit{AC}}$, and $\overline{\mathit{AB}}$, respectively, such that $G$, $X$, $Y$, and $Z$ are collinear. If $Y$ is between $G$ and $Z$,
\begin{equation*}
\frac{1}{\mathit{GX}} = \frac{1}{\mathit{GY}} + \frac{1}{\mathit{GZ}} .
\end{equation*}
How can I show this without using vectors (directed line segments) and Menelaus' Theorem?
A solution that I have seen assumes that X is a point on $\overline{\mathit{BC}}$ and constructs points U and V that trisect the side; U is half as far from B as it is from C. (We assume that X is distinct from both U or V so that the line through G and X intersects the other two sides.) Since G is half as far from the endpoint of the median on $\overline{\mathit{BC}}$ as it is from A, and since U is half as far from B as is $\overline{\mathit{GU}}$ is parallel to side $\overline{\mathit{AB}}$. Likewise, $\overline{\mathit{GV}}$ is parallel to side $\overline{\mathit{AC}}$.
\begin{equation*}
\frac{GX}{GZ} = \frac{UX}{BU}
\qquad \text{and} \qquad
\frac{GX}{GY} = \frac{VX}{CV} .
\end{equation*}
That is fine.
Now the solution says that $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ so that
\begin{equation*}
\frac{\mathit{GX}}{\mathit{GZ}}
= \frac{\mathit{UX}}{\mathit{BU}}
= \frac{\mathit{UZ} - \mathit{XZ}}{\mathit{BU}}
\end{equation*}
and
\begin{equation*}
\frac{\mathit{UZ}}{\mathit{BU}}
= \frac{\mathit{GX}}{\mathit{GZ}} + \frac{\mathit{XZ}}{\mathit{BU}} .
\end{equation*}
Likewise, since $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$,
\begin{equation*}
\frac{\mathit{VZ}}{\mathit{CV}}
= \frac{\mathit{GX}}{\mathit{GY}} + \frac{\mathit{XZ}}{\mathit{CV}} .
\end{equation*}
I would appreciate an elementary explanation of these last two equalities - involving only similar triangles and proportions - without mentioning $\mathit{UX} = \mathit{UZ} - \mathit{XZ}$ and $\mathit{VX} = \mathit{VZ} - \mathit{XZ}$.
|
Let $Z$ be placed on the line $AB$ such that $A$ is placed between $Z$ and $B$, $X$ be placed on the side $BC$,
$Y$ be placed on the side $AC$, $BD$ be a median of $\Delta ABC$, $GX=x$, $GY=y$ and $GZ=z$.
Thus, $$[Z,Y,G,X]=[A,Y,D,C],$$ which says
$$\frac{ZG}{ZX}:\frac{YG}{YX}=\frac{AD}{AC}:\frac{YD}{YC}$$ or
$$\frac{z}{x+z}\cdot\frac{x+y}{y}=\frac{\frac{AC}{2}}{AC}\cdot\frac{YC}{YC-\frac{1}{2}AC}$$ or
$$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{AC}{YC}}.$$
Now, let $GE||BC$ and $E\in DC$.
Thus, $$\frac{EC}{CY}=\frac{GX}{XY}$$ or
$$\frac{\frac{1}{3}AC}{CY}=\frac{x}{x+y},$$ which says $$\frac{AC}{CY}=\frac{3x}{x+y}.$$
Id est,
$$\frac{(x+y)z}{(x+z)y}=\frac{1}{2-\frac{3x}{x+y}}$$ or
$$\frac{z}{(x+z)y}=\frac{1}{2y-x}$$ or
$$2yz-xz=xy+yz$$ or
$$\frac{1}{x}=\frac{1}{y}+\frac{1}{z}$$ and we are done!
Actually. By definition, for four points $A$, $B$, $C$ and $D$ which they are placed on the same line we have:
$$[A,B,C,D]=\frac{AC}{AD}:\frac{BC}{BD}.$$
You can read about this here: https://en.wikipedia.org/wiki/Cross-ratio
|
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|
Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum where x = 1.8 and y = 1.6 and a distance of $\frac{4\sqrt{5}}{5}$.
For P = (2,0), the local minimum was larger than the distance between P and the ellipse at y = 0., which was equal to 1. So there is some threshold for $a$ for which the local minimum is not truly the smallest distance.
Basic strategy:
1) Solve the ellipse formula for y and put that into the distance formula for the distance between the ellipse and the point (Pythagorean theorem).
2) Take the derivative.
3) Set the derivative to 0 and solve for x.
4) Use the ellipse formula to solve for y.
5) Put in x and y in the distance formula and get the distance.
However, at some threshold the distance between the point and the ellipse where y = 0 is even shorter. So the shortest distance will have one of two cases, one between 0 and just below the threshold and one for the threshold value and above.
I figured that I could apply the same method I did for the previous cases of P = (1,0) and P = (2,0) and see if I can find some way to find this threshold.
Use the ellipse formula and solve for y:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1 \Rightarrow y = \sqrt{4\Big(1-\frac{x^2}{9}\Big)}$$
Write down distance formula
$$D = \sqrt{(x-a)^2 + y^2}$$
Put in y from the ellipse formula
$$D = \sqrt{(x-a)^2 + \Big(\sqrt{4\Big(1-\frac{x^2}{9}\Big)}\Big)^2}$$
Simplify:
$$D = \sqrt{\frac{5}{9} \cdot x^2 - 2ax +4}$$
Take the derivative:
$$D' = \frac{\frac{10}{9}x-2a}{2\sqrt{\frac{5}{9}x^2 - 2ax +4}}$$
Set the derivative to zero and solve for x:
$$\frac{10}{18}x -a = 0 \Rightarrow x = 1.8a$$
Use ellipse function and solve for y:
$$y = \sqrt{4\Big(1-\frac{(1.8a)^2}{9}\Big)} = a\sqrt{0.76}$$
Putting in x and y to the distance formula
$$D = \sqrt{(1.8a-a)^2 + (a\sqrt{0.76})^2} = \sqrt{0.64a^2 + 0.76a^2} = \sqrt{1.4a^2} = a\sqrt{1.4}$$
...but I feel like I am not getting anywhere. This is not the correct distance and the threshold is nowhere to be seen.
Expected answer is:
$$\sqrt{4-\frac{4}{5}a^2}$$
for $0 < a < \frac{5}{3}$ and
$$|a-3|$$
if $a \geq \frac{5}{3}$
|
Hint: Write $$d=\sqrt{(x-a)^2+y^2}$$ where $$y^2=4-\frac{4}{9}x^2$$
|
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|
Why do the digits of a number squared follow a similar quotient? I realized that, any number $k$ with $n$ digits, and when $k$ is squared, i.e $k^2$ will have $2n-1$ or $2n$ digits.
Per example, let $k = 583$, thus $n = 3$ and the digits of $583^2$ are $2n$.
But I started thinking, how can I narrow the result to be more precise?
But, I asked myself. How can I determine when I will have $2n-1$ or $2n$ digits?
What I did was the following:
$1)$ The last number of $1$ digit whose square has $2n-1$ digits, is $3$ since $3^2 = 9$, and the next number $4^2 = 16$ have $2n$ digits. And the quotient between the max numbers of $1$ digit($9$) and the last number of $1$ digit to the pow of $2$ with $2n-1$ digits(3) is: $9/3 = 3$
$2)$ The last number of $2$ digits whose square has $2n-1$ digits, is $31$, since $31^2 = 961$. The quotient here is $3.19354839$
$3)$ The last number of $3$ digits whose square has $2n-1$ digits, is $316$ since $316^2 = 99856$. The quotient here is $3.16139241$
$4)$ The last number of $8$ digits whose square has $2n-1$ digits, is $31622776$, since $31622776^2 = 9.9999996\cdot10^{14}$. The quotient here is $3.16227768871$
The quotient is each time smaller and closer to $3.16$
$i)$ Why does the quotient between the largest number with $n$ digits and the last number with $n$ digits squared that have $2n-1$ follow this "pattern" closer and closer to $3.16$?
$i)$ With this I can assure for all numbers that: If i have a number, per example $k = 7558$ and $k$ have $4$ digits and the quotient between $9999/7558 < 3.2$, then $k$ have $2n = 8$ digits?
That more generally, I can assure you this?:
If i have a number $k$ with $n$ digits, this number have
$2n$ digits if $\frac{10^{n}-1}{k} \leq 3.2$ otherwise it will have $2n-1$
digits
|
A (natural) number $k$ has $m$ digits if an only if $10^{m-1} \le k < 10^m$
So $10^{2m-2} \le k^2 < 10^{2m}$.
If $10^{2m-2} \le k^2 < 10^{2m-1}$ then $k^2$ will have $2m-1$ digits.
Taking the square roots we see This happens when $10^{m-1} \le k < \sqrt{10^{2m-1}}$
$\sqrt{10^{2m-1}} = \sqrt{10*10^{2m-2}} = 10^{m-1}*\sqrt {10}$ which is never an integer.
Now... thing for a moment. If $\sqrt{10} \approx 3.1622776601683793319988935444327....$, then $10^{m-1}*\sqrt{10}$ will be $3.1622776601683793319988935444327...$ "shifted over $m$ decimal places".
So the largest possible natural number less than $10^{m-1}*\sqrt {10}$ will be the first $m$ digits of $\sqrt{10}$.
So a $m$ digit number when square will have $2m-1$ digits if $k < 10^{m-1}*\sqrt 10$ and will have $2m$ digits if $k > 10^{m-1}*\sqrt 10$. And the way can tell if $k <$ or $k > 10^{m-1}*\sqrt 10$ is by checking if the digits match the digits of $31622776601683793319988935444327...$ up to $m$ plaes.
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Find all pairs $(k, n)$ of positive integers such that $k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$ Find all pairs $(k, n)$ of positive integers such that
$$k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$$
I tried to solve this problem but only found one solution $(1,1)$. Please help me to solve this problem.
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Via Legendre's formula, we have for any prime number $p$
$$ \nu_p(k!)=\sum_{i=1}^{\infty}\left \lfloor \frac{k}{p^i} \right \rfloor
\le \sum_{i=1}^{\infty} \frac{k}{p^i} =\frac{k}{p-1} $$
Note that $\nu_2(2^n-2^i)=i$, and hence
$$\nu_2(k!)=\nu_2\left(\prod_{i=0}^{n-1}({2^n-2^i})\right)=\sum_{i=0}^{n-1}i=\frac{n(n-1)}{2}$$
Thus
$$\frac{n(n-1)}{2}\le k \tag{1}$$
Note that $\nu_3(2^n-2^i)=\nu_3(2^{n-i}-1)$ and
$3\nmid 2^m-1$ if $m$ is odd. In addition $$\nu_3(2^{2m}-1)=\nu_3(m)+1$$
which is a corollary of lifting the exponent lemma.
Thus
$$\nu_3(k!)=\nu_3\left(\prod_{i=0}^{n-1}({2^n-2^i})\right)
= \left [ \frac{n}{2} \right] + \sum_{i=1}^{\left [ \frac{n}{2} \right]} \nu_3(i)
=\nu_3 (\left[ \frac{n}{2} \right]!) + \left [ \frac{n}{2} \right]
\le \frac{\frac{n}{2}}{2} +\frac{n}{2} =\frac{3n}{4}$$
Note that $\nu_3(k!)\ge \left[ \frac{k}{3} \right] >\frac{k}{3}-1$. Thus $$\frac{k}{3}-1<\frac{3n}{4} \tag{2}$$
From $(1)$ and $(2)$ we know $n \in \{1,2,3,4,5,6\}$. Via verification we get the solutions are $(k,n)=(1,1)$ or $(3,2)$
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|
Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$
Prove that
$$\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx =\frac{\pi}{4} \ln2$$
I tried to use King's rule and to scale by $2$ and then to add the integrals, to get product of terms and use the result $$\int_{0}^{\frac{\pi}2} \ln(\sin{x})dx=\int_{0}^{\frac{\pi}2} \ln(\cos{x})dx=-\frac{\pi}2\ln2$$
but it didnt work. Any help?
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Here is a different solution using complex method. We can notice that
\begin{align*}
I = &\int_0^{\frac \pi 4} \ln\left(1+\tan x + \sqrt{2\tan x} \right) dx + \int_0^{\frac\pi 4} \ln(\cos x) dx \\
= &I_1 + I_2.
\end{align*}
Evaluation of $I_1$ : Let $\displaystyle\omega = e^{\frac {\pi i} 4} =\frac{1+i}{\sqrt 2}$ and make substitution $\tan x= y^2 $ to find that
\begin{align*}
I_1 =& \int_0^1 \log\left((1+\omega y)(1+\bar \omega y)\right)\frac{2y }{1+y^4}dy \\
=& 4 \Re \int_0^1 \frac{y \log\left(\frac{1+\omega y}2 \right)}{1+y^4} dy +\frac {\pi\ln 2}2.
\end{align*} Let $\omega y = z$. Then using $\omega^2 = i, \omega^4 = -1$,
\begin{align*}
\Re \int_0^1 \frac{y \log\left(\frac{1+\omega y}2 \right)}{1+y^4} dy =& \Re \int_0^\omega \frac{ \frac{z}{\omega} \log\left(\frac{1+z}{2}\right)}{1+ (\frac z \omega)^4} \frac{dz}\omega \\
=& \Im \int_0^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz \\
=& \frac 1 2 \Im \int_{-\omega}^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz.
\end{align*} Note that $\displaystyle f(z) = \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} $ is analytic in $|z|<1$, and continuous in $|z|\le 1$ except at points $\displaystyle -1, \pm i$. So by choosing a contour $z = e^{i\theta}, -\frac \pi 4\le \theta \le \frac \pi 4$ and using $\log(1+e^{i\theta}) = \log\left(2\cos\left(\frac \theta 2\right)\right) + \frac {i\theta}{2}$, etc, we get
\begin{align*}
\Im \int_{-\omega}^\omega \frac{z \log\left(\frac{1+z}{2}\right)}{1-z^4} dz =&\Im \int_{-\frac \pi 4}^{\frac \pi 4} \frac{ie^{i2\theta}\left(\log(\cos (\theta/2)) + i\theta/2\right)}{1-e^{i4\theta}} d\theta\\
=&-\Im \int_{-\frac \pi 4}^{\frac \pi 4} \frac{\log(\cos (\theta/2)) + i\theta/2}{2\sin(2\theta)} d\theta\\
=& -\frac 1 4\int_{-\frac \pi 4}^{\frac \pi 4} \frac{\theta}{\sin(2\theta)} d\theta\\
=& -\frac 1 4 \int_{0}^{\frac \pi 4} \frac{\theta}{\sin\theta \cos \theta} d\theta\\
=:& -\frac 1 4 J
\end{align*} (In fact $J = G$, the Catalan's number.) This gives
$$
I_1 = -\frac 1 2 J +\frac {\pi \ln 2}2.
$$
Evaluation of $I_2$ : By integrating by parts, we have
\begin{align*}
I_2 =& x\ln(\cos x)|^{\frac \pi 4}_0 +\int_0^{\frac \pi 4} \frac {x \sin x}{\cos x} dx \\
=&-\frac{\pi \ln 2}{8} +K.
\end{align*} We observe that
\begin{align*}
K-J = & -\int_0^{\frac \pi 4} \frac{x\cos x}{\sin x} dx\\
=& -x\log(\sin x)|^{\frac \pi 4}_0 + \int_0^{\frac \pi 4} \log(\sin x) dx\\
=& \frac{\pi \ln 2}{8} + \int_{\frac \pi 4}^{\frac\pi 2} \log(\cos x) dx \tag{$\frac \pi 2 - x\mapsto x$}\\
=&-\frac{3\pi \ln 2}{8} -I_2
\end{align*} thus $\displaystyle K = -\frac{3\pi \ln 2}{8} -I_2 +J$ and
$$
I_2 = -\frac{\pi \ln 2}{2} -I_2 +J \Longrightarrow I_2 = \frac 1 2 J -\frac{\pi \ln 2}{4}.
$$ Therefore, we have
$$
I = I_1 + I_2 = \left(-\frac 1 2 J +\frac{\pi\ln 2}2\right)+\left(\frac 1 2 J -\frac{\pi \ln 2}{4}\right) = \frac {\pi \ln 2}4.
$$
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|
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
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No need to find the intersection, that is no need for quadratics. The normal to $y=-\frac34x+3$ through the origin is given by the equation $y=\frac43x$; determine the intersection point of both to find that the distance from the line to the origin is $\frac{3}{\sqrt{1+\left(-\frac34\right)^2}}=\frac{12}{5}$. Hence the distance from the circle to the line is $\frac{12}{5}-2$.
NB: In general the distance from $y=mx+b$ to the origin is $\frac{|b|}{\sqrt{1+m^2}}$
|
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|
A lucky proof for the Basel problem. I'll modify this part since I want the proof to be here.
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac43\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=-\frac43\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac43\int_0^1 \frac{\ln x}{x^2-1}dx$$
$$\int_0^1 \frac{\ln x}{x^2-1}dx\overset{x\rightarrow \frac{1}{x}}=\int_1^\infty \frac{\ln x}{x^2-1}dx\Rightarrow \sum_{n=1}^\infty \frac{1}{n^2}=\frac23 \int_0^\infty \frac{\ln x}{(x+1)(x-1)}dx$$
$$=\frac23 I(1,-1)=\frac23 \frac{\ln^2 (1)-\ln^2(-1)}{2(1-(-1))}=\frac23 \frac{\pi^2}{4}=\frac{\pi^2}{6}$$
Where we considered the following integral:
$$I(a,b)=\int_0^\infty \frac{\ln x}{(x+a)(x+b)}dx\overset{x\rightarrow \frac{ab}{x}}=\int_0^\infty \frac{\ln\left(\frac{ab}{x}\right)}{(x+a)(x+b)}dx$$
Summing up the two integrals from above gives:
$$2I(a,b)=\ln(ab)\int_0^\infty \frac{1}{(x+a)(x+b)}dx=\frac{\ln(ab)}{a-b}\ln\left(\frac{x+b}{x+a}\right)\bigg|_0^\infty $$
$$\Rightarrow I(a,b)=\frac{\ln(ab)}{2}\frac{\ln\left(\frac{a}{b}\right)}{a-b}=\frac{\ln^2 a-\ln^2 b}{2(a-b)}$$
From here we know that:
$$\int_0^\infty \frac{\ln x}{(x+a)(x-1)}dx=\frac{\ln^2 a+\pi^2}{2(a+1)} $$
Also by plugging $b=-1$ in $I(a,b)$ we get:
$$I(a,-1)=\int_0^\infty \frac{\ln x}{(x+a)(x-1)}dx=\frac{\ln^2a -\ln^2 (-1)}{2(a-(-1))}=\frac{\ln^2 a+\pi^2 }{2(a+1)}$$
We already know that this is true from the linked post, but let's ignore it, since the linked post uses the Basel problem to prove the result.
Can someone prove rigorously that we are allowed to plug in $b=-1$ in order to get the correct result?
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Fixing $a >0$, we see that the integral
\begin{align*}
f(z) := \int_0^\infty \frac{\log x}{(x+a)(x+z)} dx.
\end{align*} converges absolutely for $z$ away from $(-\infty,0]$. So $f$ defines an analytic function on $\mathbb C \setminus (-\infty,0]$ (which can be checked by Morera's theorem and Fubini's theorem, for example.) This implies
$$
f(z) = \frac{\log^2 a - \log^2 z}{2(a-z)}
$$ holds not only on $(0,\infty)$ but also $\mathbb C \setminus (-\infty,0]$ by analytic continuation.
Now we need to check the continuity of $f$ at $z=-1$. Let
$$
F(x) = \int_1^x \frac{ \log t}{(t+a)(t-1)} dt,\quad x\ge 0
$$ so that $F(\infty) - F(0) $ is the desired integral $\displaystyle I(a) = \int_0^\infty \frac{\log x}{(x+a)(x-1)} dx$. By integration by parts,
\begin{align*}
f(z) =& \left[F(x)\frac{ x-1}{x+z}\right]^\infty _0 - (1+z)\int_0^\infty \frac{F(x)}{(x+z)^2} dx\\
=& \left(F(\infty) +\frac{F(0)}{z} \right)- (1+z)\int_0^\infty \frac{F(x)}{(x+z)^2} dx,\quad z\neq 0
\end{align*} The first term tends to $I(a)$ as $z\to -1$. For $z=-1+it, t>0$, the second term can be estimated as
\begin{align*}
|1+z|\left|\int_0^\infty \frac{F(x)}{(x+z)^2} dx\right| \le& t \int_0^\infty \frac{|F(x)|}{|x-1+it|^2} dx \\
\le& \int_0^\infty \frac{t}{(x-1)^2 + t^2} |F(x)| dx\\
&\xrightarrow{t\to 0} \pi |F(1)| = 0.
\end{align*} Thus
$$
I(a)= \lim_{t\to 0^+} f(-1+it) =\frac{\log^2 a +\pi^2}{2(1+a)} .
$$
Note that this argument only works for $z=-1$ since $\displaystyle F(x) =\int_c^x \frac{\log t}{(t+a)(t-c)} dt$ is not well-defined (due to the singularity at $t=c$) for $c>0, c\ne 1$.
|
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|
Is the result for $3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}$ known in the literature? I was able to get the following result
$$3\sum\limits_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^6}+\sum\limits_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^5}=11\zeta(3)\zeta(6)+\frac52\zeta(4)\zeta(5)-\frac{13}{6}\zeta^3(3)-2\zeta(2)\zeta(7)-5\zeta(9)$$
where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.
based on a nice identity and some manageable Euler sums. Is this result known in the literature? Can we evaluate the terms separately?
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In answer to your question, can the sums be evaluated separately? Yes they can. The results for each of these two Euler sums can be found in the 2016 paper Euler sums and integrals of polylogarithm functions by Ce Xu et al.
The results are:
$$\sum_{n = 1}^\infty \frac{H_n H^{(2)}_n}{n^6} = \frac{17}{6} \zeta (3) \zeta (6) + \frac{173}{72} \zeta (9) + \frac{1}{4} \zeta (4) \zeta (5) - 3 \zeta (2) \zeta (7) - \frac{2}{3} \zeta^3 (3) \quad \text{(See Eq. 3.18)}$$
and
$$\sum_{n = 1}^\infty \frac{H_n H^{(3)}_n}{n^5} = \frac{679}{24} \zeta (9) - 11 \zeta (2) \zeta (7) - \frac{1}{2} \zeta (3) \zeta (6) - \frac{29}{4} \zeta (4) \zeta (5) - \frac{1}{6} \zeta^3 (3).$$
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|
Why $xy = 100$ does not represent a direct variation? $xy = 100 $
$y = 100/x$
$x$ is not equal to $0$ and can be represented as $1\cdot x$
However I still do not understand on why $100/x$ does not represent a direct variation. Is it because of the use of division within the right side of the equation is not valid?
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The relation is between the two variables $x$ and $y$. A direct variation means that when $x$ increases $y$ will increase too by an amount specified by the equation, for example:
$y=x$. If $x = 1$ then $y = 1$ and if $x$ increases to be $x=2$ then $y=2$ this is a direct variation and it may be represented like the $f(x) = x$.
Let's change the formula quite a bit $f(x) = x + 1$ this is the same as $y = x + 1$. If $x = 1$ then $y = 1 + 1 = 2$ and if $x = 2$ then $y = 2 + 1 = 3$. This is a direct variation.
Let's change the formula again $f(x) = x - 1$. If $x = 1$ then $y = 1 - 1 = 0$ and if $x = 2$ then $y = 2 - 1 = 1$. This is also a direct variation.
Again $f(x) = x \times 2$. If $x = 1$ then $y = 1 \times 2 = 2$ and if $x = 2$ then $y = 2 \times 2 = 4$. This is also a direct variation.
But what about $f(x) = \frac{1}{x} : x \ne 0$. If $x = 1$ then $y = \frac{1}{1} = 1$ and if $x = 2$ then $y = \frac{1}{2} = 0.5$. This is an inverse variation, as $x$ increases $y$ decreases or their product is a constant.
In the end a direct variation means that as $x$ increases $y$ increases, and an inverse variation means that as $x$ increases $y$ decreases.
Seeing graphs helps in understand so this is the graph of your equation $f(x) = \frac{100}{x}$:
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Show that $2x^6+12x^5+30x^4+60x^3+80x^2+30x+45=0$ has no real roots I tried solving the above question but was unable to prove it.
I used Descartes rule of sign, factorisation techniques, and many other things but could not figure out the solution.
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Factor $x^4$ and separate a non-negative part of the expression covering completely the terms $x^6$ and $x^5,$ then factor $x^2,$ ...
$$\begin{aligned}P(x)=&2x^6+12x^5+30x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+12x^4+60x^3+80x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+30x^2+30x+45\\=&2x^4(x^2+6x+9)+3x^2(4x^2+20x+25)+15(2x^2+2x+3)\end{aligned}$$ which is strictly positive for any real $x.$
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If $\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$, compute: $\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$ Be $f(x)$ a polynomial in $\Bbb R$ such that:
If$$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3$$
Compute: $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}$$
I noticed that, if the first limit exists, then $f(x)+2 = (x-1)P(x)$, where $P(x)$ is another polynomial.
Then $$\lim \limits_{x \to 1} P(x) = 3$$
I tried to use that in the second limit, but i can't proceed further.
Any hints?
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Given that $$\lim \limits_{x \to 1} \frac{f(x)+2}{x-1} = 3\tag1$$
Take $~f(x)=3x-5~$, then the above limit is satisfied.
Now $$\lim \limits_{x \to -1} \frac{(f(-x))^2-4}{x^2-1}=\lim \limits_{x \to -1} \frac{(-3x-5)^2-4}{x^2-1}$$
$$=\lim \limits_{x \to -1} \frac{(3x+3)(3x+7)}{x^2-1}$$
$$=\lim \limits_{x \to -1} \frac{3(3x+7)}{x-1}$$
$$=\frac{3(-3 +7)}{-1-1}=-6$$
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Tips on solving $a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$ $$a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
This is true with $a>b>0$, according to Wolfram Alpha, but I am not able to prove this.
I to simplify using the fact that $a^2(a+2b)=a^3+2ba^2>3b^3$, and then proving a stronger result:
$$3b^3(a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
$$\iff 3(a + b^2 + b)>(a^2 + a + b)$$
$$\iff 3a + 3b^2 + 3b>a^2 + a + b$$
But this is not true, e.g. $a=100, b=1$.
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Try moving all the terms raised to the power of 3/2 onto one side of the equation only, and move all other terms to the other side of the equation. Then square both sides so that the power of 3/2 becomes 3, and use algebra from there.
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.