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We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$.
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .
What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?
| First of all, the equation only makes sense when $a,b,c,d\neq 0$ so you can use this: If
$$\frac{x}{y}=\frac{a}{b}$$
Then
$$\frac{x}{y}=\frac{x+a}{y+b}$$
So your problem solves proving this property, and the only value will be
$$\frac{3(a+b+c+d)}{a+b+c+d}=3$$
But if you get the case $a+b+c+d=0$, then WLOG, $a+b+c=-d$, then you achieve that $a+b+c=-dr$, implying $r=-1$.
Then the only solutions are $r\in \{-1,3\}$, so the solution to the problem would be $3-1=2$.
So to achieve the identity, notice that
$$x(y+b)=y(x+a)$$
$$xy+ab=yx+ya$$
And substracting $xy$ to both sides you get:
$$xb=ya$$
$$\frac{x}{y}=\frac{a}{b}$$
The other side is the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Combinatorial problem - choosing cards Here is the question: Forty cards are placed into a box, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four cards. Four cards are drawn from the box at random and without replacement. Let p be the probability that all four cards bear the same number. Let q be the probability that three of the cards bear a number a and the other bears a number b that is not equal to a. What is the value of q/p?
My solution: For p, we can choose any of the 1~10 cards for all four so we have a probability of $\frac{10}{\binom{40}{4}}.$ For q, we have to choose 2 kinds of cards getting us $\binom{10}{2}$ for the first kind of cards, we choose 3 to get $\binom{4}{3}$ and we choose 1 for the next kind for $\binom{4}{1}.$ Therefore, q = $\frac{45 \cdot 16}{\binom{40}{4}}.$ Thus dividing, we get $\boxed{72}.$
Is my logic correct?
| Your first answer is fine, but your second is not. For the second problem there are $10$ ways to choose the number to be chosen $3$ times, and $\binom43=4$ sets of $3$ cards of that number, so there are $10\cdot4=40$ ways to choose the triplet. For any given triplet there are $9$ possible values of the singleton and $4$ cards of that value, so there are $9\cdot4=36$ ways to choose the singleton. Thus, there are $40\cdot36=1440$ ways to choose a hand of the desired type, and
$$q=\frac{1440}{\binom{40}4}=\frac{1440}{91390}=\frac{144}{9139}\,.$$
| {
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"url": "https://math.stackexchange.com/questions/3784682",
"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2}$ into $\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab}$ I got two algebraic expression, the first should be simplified, with some manipulation I think, to be the second. It's similar to binomial expansion, even though coefficients are a mess.
$$\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2} \tag 1$$
$$\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab} \tag 2$$
It's not exactly a problem related to my studies, but something I'm interested in. I remember something similar, when we add/subtract some terms till we got full expansion or so, but I couldn't do anything. Actually, I'm not sure if this problem is correct, so if there's a proof for this problem as invalid one, it's welcome.
| The key to this problem is in factoring the denominator. We note the following identity:
$$b^4-2a^2b^2+a^4 = (b^2-a^2)^2 = (b-a)^2(b+a)^2 = (b^2+2ab+a^2)(b^2-2ab+a^2)$$
We see that the denominator splits into the goal denominator and another term. We divide the numerator by $b^2-2ab+a^2$ and get the desired numerator. Here is the polynomial division.
$$(b^2-2ab+a^2)(b^3+2ab^2-2a^2b-a^3)\\
=b^5-2ab^4+a^2b^3+2ab^4-4a^2b^3+2a^3b^2-2a^2b^3+4a^3b^2-2a^4b-a^3b^2-2a^4b-a^5\\
=b^5-5b^3a^2+5b^2a^3-a^5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ converges, and its square (formed by Abel's rule) doesn't. Show that the series $\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots$ (say $\sum u_n$) converges, and its square (say $\sum v_n$)(formed by Abel's rule) doesn't.
Abel's rule: given $\sum a_n, \sum b_n$, $\sum_{n=0} ^\infty c_n=\sum_{n=0} ^\infty [\sum_{i=0} ^n a_{n-i}b_i]$ is the infinite series gotten from multiplication of two series.
According to this rule,
$(\frac{1} {\sqrt{1}} -\frac{1} {\sqrt{2}} +\frac{1} {\sqrt{3}} +\dots)^2
=\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}-[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{1}}]+\dots
-[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\dots
+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}
+\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}\dots+ \frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}]
+[\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}+\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}
+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}\dots+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}}]\dots,$
which, if we sum by adding nearby items first, equals
$\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}+\dots
+[(-\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}
-\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
-\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}
-\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}
\dots-\frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}})
+\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}]<\frac{1} {\sqrt{1}}\frac{1} {\sqrt{1}}+\dots
+[
\frac{1} {\sqrt{k+1}} \frac{1} {\sqrt{k+1}}]=\sum_{k=0}^\infty \frac{1}{k+1}$ which diverges.
But in order to prove the series diverges we probably need to prove it's no less than another divergent series.
It's possibly not complicated. I will see how I can modify the proof to make it work. Perhaps I need to use the ratio test $\frac{c_{n+1}}{c_n}=1+\frac{A}{n}+O(\frac{1}{n^2})$, considering the series approximates $\sum_{k=0}^\infty \frac{1}{k+1}$
Context: the significance of the statement is that if it is true, then $\lim_{x\to 1}v_n x^n \neq c_n$, and so (though, for $\sum u_n x^n$ converges absolutely, we have $(\sum u_n x^n)^2=\sum v_n x^n$, letting $x\to 1$,) we don't have $(\sum u_n)^2$ (i.e. the limit of the left side) equals $\sum v_n$.
| This is inspired by an answer.
$|-\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k}}+\frac{1} {\sqrt{1}} \frac{1} {\sqrt{2k+1}}
-\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k-1}}+\frac{1} {\sqrt{2}} \frac{1} {\sqrt{2k}}+\dots
-\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+1}}+\frac{1} {\sqrt{k}}\frac{1} {\sqrt{k+2}}
-\frac{1} {\sqrt{k+1}}\frac{1} {\sqrt{k}}+\frac{1} {\sqrt{k+2}}\frac{1} {\sqrt{k}}
\dots-\frac{1} {\sqrt{2k}}\frac{1} {\sqrt{1}}+ \frac{1} {\sqrt{2k+1}}\frac{1} {\sqrt{1}}|
\\>
2\sum_{m=1}^k \frac{m}{{\sqrt{m(2k+2-m)}}^3}>2\sum_{m=1}^k \frac{m}{(k+1)^3}= \frac{k(k+1)}{(k+1)^3}=\frac{1}{(k+1)}-\frac{1}{(k+1)^2},$
So $\sum_{k=0}^\infty \sum_{m=1}^k (-\frac{1} {\sqrt{m}} \frac{2} {\sqrt{2k+1-m}}+\frac{2} {\sqrt{m}} \frac{1} {\sqrt{2k+2-m}})$ diverges. And so $\sum v_n$ is the sum of two divergent series. Then how to prove that it diverges?
By far we see
$v_n < \frac{1}{k+1}-\frac{1}{(k+1)^2}+\frac{1}{1+k}=-\frac{1}{(k+1)^2}$, and so |$\sum v_n$| is larger than a convergent series, which implies we have not proved the divergence of $\sum v_n$. It seems the problem is more tricky than it looks, were my calculation correct. Perhaps I need to shrink the series $\sum_{k=0}^\infty \sum_{m=1}^k (-\frac{1} {\sqrt{m}} \frac{2} {\sqrt{2k+1-m}}+\frac{2} {\sqrt{m}} \frac{1} {\sqrt{2k+2-m}})$less.
I notice when I use $\frac{1}{\sqrt{ab}}>\frac{2}{a+b}$, the difference between the two sides can be large when a, b is very different, e.g. $\frac{1}{\sqrt{1(2k+1)}}$ differs from $\frac{2}{2k+2}$ as much as $O(\frac{1}{\sqrt{k}})$ differs from $O(\frac{1}{k})$, e.g. k=1/1000,000. But does that matter for knowing the divergence of the series?
Again, at least now I realize it's addition of two divergent series instead of a divergent series and a convergent one, and that put me in the right track.
Addition:
Now I realize it's addition of two divergent series instead of a divergent series and a convergent one. My question is then how can one prove that the series is divergent.
Details of my thoughts about this new question is described in my answer.
The problem is trickier than it seems, and possibly it should be, since we are dealing with multiplication of two convergent (though not absolutely) series, and the most natural result is that we get a convergent series, or if in calculation we separate it into several divergent series, most naturally their sum should be a convergent series like $\sum\frac{1}{n^2}$ in my answer.
Thinking of the context I guess the problem has something to do with the difference, or error $\delta$, between $\sum v_nx^n$ and $\sum v_n$ (possibly, when it is divergent, the former doesn't converge uniformly as $n\to \infty$, and at x=1, it doesn't converges to the latter as $x\to 1$). Also it is possibly caused by the way we do multiplication, where the series tail engages in many more multiplications (with each other of tail terms) than the series head. This could amplify the small 'error' of the tail.
Thus a plausible approach is to calculate $(\sum u_nx^n)^2$ first, and then compare it with $\sum v_n$ and see where the error $\delta$ is mostly ignored (in the proof) (and amplified by multiplication) that leads to incompleteness of the proof. Perhaps this way will be more efficient than trying directly to shrink the series $\sum_{k=0}^\infty \sum_{m=1}^k (-\frac{1} {\sqrt{m}} \frac{2} {\sqrt{2k+1-m}}+\frac{2} {\sqrt{m}} \frac{1} {\sqrt{2k+2-m}})$ less.
Again, it is important to realize what I get above is two divergent series.
Correction:
I see, (https://math.stackexchange.com/a/3787287/577710)
*
*it's $v_n$ doesn't converges to 0 and so $\sum v_n$ diverges.
*And adding adjacent items follows not the rule.
The series possibly diverges between two numbers. It is loosely 'convergent', considering series 'pausing' at n odd and n even, which fits our intuition that convergent series's product is (though not always strictly) 'convergent'.
This example shows not noticing a simplest fact when stuck and a slightly different understanding of the problem from what it is can delay progress.
| {
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"url": "https://math.stackexchange.com/questions/3786298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$ if $a^5+b^5<1$ and $c^5+d^5<1$ then prove that ${a^2}{c^3}+{b^2}{d^3}<1$ given $a,b,c,d$ are non negative real numbers
My try:
It is easy to deduce that $a,b,c,d<1$ ,thus
$$a^2c^3+b^2d^3<a^5c^5+b^5d^5<(a^5+b^5)(c^5+d^5)<1$$
I want to know if there is a more cleaner method to solve this problem and if possible with calculus. I would also want to know if my proof has a mistake anywhere.
| Your proof is wrong in the first step: $$a^2c^3+b^2d^3\geq a^5c^5+b^5d^5.$$
By Holder $$1>(a^5+b^5)^2(c^5+d^5)^3\geq(a^2c^3+b^2d^3)^5.$$
We can prove the last inequality by the following way.
If $abcd=0$, so our inequality is obvious.
Let $abcd\neq0$, $a=xb$ and $c=yd$.
Thus, we need to prove that $$(x^5+1)^2(y^5+1)^3\geq(x^2y^3+1)^5$$ or $f(x)\geq0,$ where $$f(x)=2\ln(x^5+1)+3\ln(y^5+1)-5\ln(x^2y^3+1).$$
But $$f'(x)=\frac{10x^4}{x^5+1}-\frac{10xy^3}{x^2y^3+1}=\frac{10x(x^3-y^3)}{(x^5+1)(x^2y^3+1)},$$
which gives $x_{min}=y$, $$f(x)\geq f(y)=0$$ and we are done!
| {
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"url": "https://math.stackexchange.com/questions/3787529",
"timestamp": "2023-03-29T00:00:00",
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how to solve $\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$
how to solve $$\mathcal{J(a)}=\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$$
i used the differentiation under the integral and got
\begin{align}
\mathcal{J(b)}&=\int _0^1\frac{\ln \left(1+bx\right)}{a^2+x^2}\:\mathrm{d}x
\\[3mm]
\mathcal{J'(b)}&=\int _0^1\frac{x}{\left(a^2+x^2\right)\left(1+bx\right)}\:\mathrm{d}x
\\[3mm]
&=\frac{a^2b}{1+a^2b^2}\int _0^1\frac{1}{a^2+x^2}\:\mathrm{d}x+\frac{1}{1+a^2b^2}\int _0^1\frac{x}{a^2+x^2}\:\mathrm{d}x-\frac{b}{1+a^2b^2}\int _0^1\frac{1}{1+bx}\:\mathrm{d}x
\\[3mm]
&=\frac{ab}{1+a^2b^2}\operatorname{atan} \left(\frac{1}{a}\right)+\frac{1}{2}\frac{\ln \left(1+a^2\right)}{1+a^2b^2}-\frac{\ln \left(a\right)}{1+a^2b^2}-\frac{\ln \left(1+b\right)}{1+a^2b^2}
\end{align}
But we know that $\mathcal{J}(1)=\mathcal{J(a)}$ and $\mathcal{J}(0)=0$
\begin{align}
\int_0^1\mathcal{J'(b)}\:\mathrm{d}b&=a\:\operatorname{atan} \left(\frac{1}{a}\right)\int _0^1\frac{b}{1+a^2b^2}\:\mathrm{d}b+\frac{\ln \left(1+a^2\right)}{2}\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b-\ln \left(a\right)\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b
\\
&-\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b
\\[3mm]
\mathcal{J(a)}&=\frac{1}{2a}\operatorname{atan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{atan} \left(a\right)-\frac{1}{a}\ln \left(a\right)\:\operatorname{atan} \left(a\right)-\underbrace{\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b}_{\mathcal{I}}
\end{align}
but how to calculate ${\mathcal{I}}$, i tried using the same technique but it didnt work
|
Suppose $\left(a,b,c,z\right)\in\mathbb{R}\times\mathbb{R}_{>0}\times\mathbb{R}_{\ge0}\times\mathbb{R}_{>0}$, and set
$$\alpha:=\arctan{\left(\frac{a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$
$$\gamma:=\arctan{\left(\frac{c-a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$$
$$\theta:=\arctan{\left(\frac{z+a}{b}\right)}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$
Then, the following logarithmic integral can be evaluated in terms of Clausen functions using the general result derived here:
$$\begin{align}
\int_{0}^{z}\mathrm{d}x\,\frac{2b\ln{\left(x+c\right)}}{\left(x+a\right)^{2}+b^{2}}
&=\operatorname{Cl}_{2}{\left(2\alpha+2\gamma\right)}-\operatorname{Cl}_{2}{\left(2\theta+2\gamma\right)}+\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-\operatorname{Cl}_{2}{\left(\pi-2\theta\right)}\\
&~~~~~+\left(\theta-\alpha\right)\ln{\left(b^{2}\sec^{2}{\left(\gamma\right)}\right)},\\
\end{align}$$
where the Clausen function (of order 2) is defined for real arguments by the integral representation
$$\operatorname{Cl}_{2}{\left(\vartheta\right)}:=-\int_{0}^{\vartheta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\vartheta\in\mathbb{R}}.$$
Applying the formula to the particular case where $a=0\land c=1\land z=1$, and with a little help from the duplication formula for the Clausen function, we obtain the following delightfully compact result:
$$\begin{align}
\forall b\in\mathbb{R}_{>0}:\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x+1\right)}}{x^{2}+b^{2}}
&=\frac{4\theta\ln{\left(\csc{\left(\theta\right)}\right)}-\operatorname{Cl}_{2}{\left(4\theta\right)}}{4b};~~~\small{\theta:=\arctan{\left(\frac{1}{b}\right)}\in\left(0,\frac{\pi}{2}\right)}.\\
\end{align}$$
It should be clear that Clausen functions are the natural tool to use if you're trying to stick to real methods.
| {
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"url": "https://math.stackexchange.com/questions/3789171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
| Your answer is not correct, please check it.
$x^{2}-8x+17=(x-4)^{2}+1=0$ by completing the square, so we have $(x-4)^{2}=-1$ and thus $x=4+i$ and $x=4-i$ are the required solutions where $i=\sqrt{-1}$.
| {
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"url": "https://math.stackexchange.com/questions/3791855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing $\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$ An advanced sum proposed by Cornel Valean:
$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}$$
$$=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^22\zeta(2)+6\ln2\zeta(2)+\frac16\ln^42-1$$
I managed to find the integral representation of $\ \displaystyle\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}\ $ but not $S$:
Since
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\frac{(2x)^{2n-1}}{n{2n\choose n}}$$
we can write
$$\frac{2\sqrt{x}\arcsin \sqrt{x}}{\sqrt{1-x}}=\sum_{n=1}^\infty\frac{2^{2n}x^{n}}{n{2n\choose n}}$$
now multiply both sides by $-\frac{\ln(1-x)}{x}$ then $\int_0^1$ and use that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we have
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^2{2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\tag1$$
But I could not get the integral representation of $S$. Any idea?
In case you find the integral, I prefer solutions that do not use contour integration or you can leave it to me to give it a try. Thank you.
In case the reader is curious about computing the integral in $(1)$, set $x=\sin^2\theta$ then use the Fourier series of $\ln(\cos \theta)$.
| Following @Felix's idea above:
$$S=\sum_{n=1}^\infty\frac{2^{2n}H_{n+1}}{(n+1)^2{2n\choose n}}=\sum_{n=2}^\infty\frac{2^{2n-2}H_n}{n^2{2n-2\choose n-1}}$$
Note that
$$\frac{{2n+2\choose n+1}}{{2n\choose n}}=\frac{\frac{\Gamma(2n+3)}{\Gamma^2(n+2)}}{\frac{\Gamma(2n+1)}{\Gamma^2(n+1)}}=\frac{\frac{(2n+2)(2n+1)\Gamma(2n+1)}{((n+1)\Gamma(n+1))^2}}{\frac{\Gamma(2n+1)}{\Gamma^2(n+1)}}=\frac{(2n+2)(2n+1)}{(n+1)^2}=\frac{2(2n+1)}{n+1}$$
replace $n$ by $n-1$ we get
$$\frac{1}{{2n-2\choose n-1}}=\frac{2(2n-1)}{n{2n\choose n}}$$
Therefore
$$S=\sum_{n=2}^\infty\frac{2^{2n-1}(2n-1)H_n}{n^3{2n\choose n}}=\sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^2 {2n\choose n}}-\frac12 \sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^3 {2n\choose n}}-1\tag1$$
In the question body we have
$$\sum _{n=1}^{\infty } \frac{2^{2n} H_n}{n^2 {2n\choose n}}=-2\int_0^1 \frac{\arcsin \sqrt{x}\ln(1-x)}{\sqrt{x}\sqrt{1-x}}dx\overset{\sqrt{x}=\sin\theta}{=}-8\int_0^{\pi/2} \theta \ln(\cos\theta)d\theta$$
$$=-8\int_0^{\pi/2}\theta\left(-\ln(2)-\sum_{n=1}^\infty\frac{(-1)^n\cos(2n\theta)}{n}\right)d\theta=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag2$$
and here we already showed
$$\sum_{n=1}^\infty\frac{2^{2n}H_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)\tag3$$
Finally, plug $(2)$ and $(3)$ in $(1)$ we obtain
$$S=4\text{Li}_4\left(\frac12\right)-\frac12\zeta(4)+\frac72\zeta(3)-4\ln^2(2)\zeta(2)+6\ln(2)\zeta(2)+\frac16\ln^4(2)-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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Why does this proof fail? Catalan Constant's Exact Value. A failed attempt at finding an exact value for Catalan's Constant "C"
Definition :
$$ C+ \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{(x)}}{\sqrt{1+\cos{(x)}^2}} = \pi \ln(1+\sqrt{2})
$$
Let
$$I(b) = \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{(bx)}}{\sqrt{1+\cos{(x)}^2}}$$
such that $ I(b=1) $ is our desired integral.
Integrate both sides with respect to b (Ad = antiderivative)
$$ I^{Ad} (b)= -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}} $$
I shall restrict b to be an odd integer greater than $0$
I Examine the values given by the integral :
$$ p = 1 \to \left (\frac{\pi}{4}\right)$$
$$ p = 3 \to \left(2-\frac{3 \pi}{4}\right) $$
$$ p = 5 \to \left(\frac{13 \pi }{4} - 10\right) $$
I divide these values into Pi Numbers ( those with $\pi$) and Rational Numbers (those without it).
The Pi Numbers will follow a known sequence called The Central Delannoy Numbers $D(n)$
$$ D(n) = \sum_{k=0}^{n} \binom{n}{k}^{2} 2^k $$
Putting this into b terms .
$$ \Pi (b) = \frac{\pi}{4} \sum_{k=0}^{\frac{b-1}{2}}\binom{\frac{b-1}{2}}{k}^2\, 2^k $$
For the Rational Numbers $R(b)$ we notice that since b is an integer then the $\cos(bx)$ may be expanded with Chebyschev Polynomials in the form :
$$ \cos(bx) = b \sum_{k=0}^{b} (-2)^{k} \frac{(b+k-1)!}{(b+k)!(2k)!} (1-\cos(x))^{k} $$
Since
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos(x)^{b}}{\sqrt{1+\cos(x)^2}} = \frac{\sqrt{\pi}}{4} \frac{\Gamma(\frac{b}{4}+\frac{1}{4})}{\Gamma(\frac{b}{4}+\frac{3}{4})}$$
for odd integers greater than $0$
We can give $R(b)$ as
$$ \frac{\sqrt{\pi} b}{4} \sum_{k=1}^{\lfloor \frac{b+1}{4} \rfloor} \frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})} \frac{4^{2k-1}}{4k-1}\binom{\frac{b+4k-3}{2}}{{4k-2}} $$
Together :
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}} = \sqrt{2} \cos\left(\frac{\pi}{4}-\frac{\pi b}{2}\right)\left(\frac{\pi}{4} \sum_{k=0}^{\frac{b-1}{4}}\binom{\frac{b-1}{2}}{k}^2\, 2^k - \frac{\sqrt{\pi} b}{4} \sum_{k=1}^{\lfloor \frac{b+1}{4} \rfloor} \frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})} \frac{4^{2k-1}}{4k-1}\binom{\frac{b+4k-3}{2}}{{4k-2}}\right)$$
All that is left is to take the derivative with respect to b and let $b = 1$ ( I am unsure of the constant that is generated when we integrate with respect to b , don't know if it goes to $0$ after we take the derivative again )
I do not know how to take the derivative of b in its current form so i decided to change it into integral form.
The Central Delannoy Numbers can be given with :
$$ D(n) = \frac{1}{\pi} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{1}{\sqrt{-t^2+6t-1}} \frac{1}{t^{n+1}} \,dt $$
The generating function for $R(b) $ is
$$ \frac{\sin^{-1}\left(\frac{4x}{(1-x)^2}\right)}{2\sqrt{x^2-6x+1}} $$
Putting it all together :
$$ -\int_{\frac{\pi}{2}}^{\pi} \frac{\cos{(bx)}}{\sqrt{1+\cos{(x)}^2}}= \cos\left(\frac{\pi}{4}-\frac{\pi b}{2}\right) \left(\frac{1}{2 \sqrt{2}} \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{t^{-b-\frac{1}{2}}}{\sqrt{-t^2+6t-1}} \, dt -\sqrt{2} \,\lim_ {t \to 1} \int \frac{t^{-b-\frac{1}{2}}}{\sqrt{-t^2+6t-1}}\, dt \right) $$
Unfortunately, taking the derivative with respect to b and letting $b=1$ yields
$$ \frac{1}{2} \lim_{t \to 1} \int \frac{\ln{t}}{t} \frac{1}{\sqrt{-t^2+6t-1}} \,dt $$
which cancels the Catalan Constant on the LHS.
Question(s) : Does anybody see another way to proceed which could possibly lead to better results?
Does anyone see any mistakes?
Thank you kindly for your help and time.
| Integrate by parts
\begin{align}
\int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{x}}{\sqrt{1+\cos^2{x}}} dx
=-
\int_{\frac{\pi}{2}}^{\pi} x d(\sinh^{-1}\cos{x})
=\pi \ln(1+\sqrt{2}) - J(1)
\end{align}
where $J(a)= \int^{\frac{\pi}{2}}_{0} \sinh^{-1}(a\cos t) dt$. Evaluate
$$
J’(a)= \int^{\frac{\pi}{2}}_{0} \frac{\cos{t}}{\sqrt{1+a^2\cos^2{t}}} dt
=\frac1a \sin^{-1}\frac a{\sqrt{1+a^2}}= \frac{\tan^{-1}a}a
$$
$$J(1)= \int_0^1 \frac{\tan^{-1}a}ada=-\int_0^1\frac{\ln a}{1+a^2}da=C
$$
Thus
$$ C+ \int_{\frac{\pi}{2}}^{\pi} \frac{x \sin{x}}{\sqrt{1+\cos^2x}} = \pi \ln(1+\sqrt{2})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\text{ }\big(x-\frac{1}x\big)=i\sqrt{2}$. Then compute $\bigg(x^{2187}-\frac{1}{x^{2187}}\bigg)$. Here $i=\sqrt{-1}$
QUESTION: If $\text{ }\big(x-\frac{1}x\big)=i\sqrt{2}$ , $\text{ }$then compute $$\bigg(x^{2187}-\frac{1}{x^{2187}}\bigg)$$ Here $i=\sqrt{-1}$ .
MY ANSWER: I have done it using the Quadratic formula and De Moivre's Theorem. Let me write down my working before I propose my doubt.. Here's how I did it..
Solving the equation we get $$x^2-(i\sqrt{2})x-1=0$$ $$\implies x=\frac{i\sqrt{2} \pm \sqrt{(i\sqrt{2})^2+4}}{2} $$ $$\implies x=\frac{i\sqrt{2}\pm\sqrt{2}}{2}$$ Take $x=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)=e^{\frac{i\pi}4}$
Now we know that $2187=(273\times8)+3$
$$\therefore x^{2187}=e^{2187\times \frac{i\pi}4}=e^{(273\times 2\pi + \frac{3\pi}4)i}=e^{\frac{{3\pi}}{4}i}=\frac{i-1}{\sqrt{2}}$$
$$\therefore x^{2187}-\frac{1}{x^{2187}}= \frac{i-1}{\sqrt{2}}-\frac{\sqrt{2}}{i-1}$$ $$=\frac{(i-1)^2-2}{(i-1)\sqrt{2}}$$ $$=\frac{2}{\sqrt{2}}\frac{(1+i)}{(1-i)}$$ $$=\frac{\sqrt{2}}{2} (1+i)^2$$ $$=\boxed{\sqrt{2}i}$$
Now my first question is that, the quadratic relation gave us two different values for $x$. One with which I have worked out to reach the answer of $\sqrt {2}i$ and the other, $\big(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\big)$ which I had left behind. Now working with that I find that the angle turns out to be $\frac{\pi}{10}$ and stuff becomes much more complicated after that. The official answer to this one is $\sqrt{2}i$ (which matches with what I have found out).
My doubt is why do we not consider the other value of $x$ ?
And is there any alternative (preferably simpler) method(s) to solve this one?
Thank You so much for your help and support.. :)
| $2187=3^7$. This is a clue. Powers of $3$ are significant. Now
$$\left(x-\frac1x\right)^3=(i\sqrt2)^3=-2i\sqrt2$$
and
$$\left(x-\frac1x\right)^3=x^3-\frac1{x^3}-3\left(x-\frac1x\right)
=x^3-\frac1{x^3}-3i\sqrt2.$$
So
$$x^3-\frac1{x^3}=i\sqrt2.$$
Repeating this,
$$x^9-\frac1{x^9}=i\sqrt2,$$
$$x^{27}-\frac1{x^{27}}=i\sqrt2$$
etc.
Eventually,
$$x^{2187}-\frac1{x^{2187}}=i\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of ways of making a tangential quadrilateral such that its perimeter is $24$ and side lengths are integers from $1$ to $12$ My attempt is as follows:
Label the sides of the quadrilateral as $a$, $b$, $c$ and $d$ in a counterclockwise order. Since the quadrilateral is tangential and its perimeter is $24$, $a+c=b+d=\frac{24}{2}=12$.
The number of noncommutative ways of choosing $a+c=12$ is $11$ ($1+11$, $2+10$, ... $5+7$, $6+6$, $7+5$, ... , $10+2$, $11+1$) and the same applies for $b+d$. The number of choosing two pairs to form a quadrilateral is therefore $11 \cdot 11=121$.
The number $121$ is overcounted since we counted every rotation of the quadrilateral as a new one. For example, $a = 5$,$b = 1$ ,$c = 7$,$d = 11$ describes the same quadrilateral as $a = 11$,$b = 5$,$c = 1$,$d = 7$. There are $4$ rotations, so we counted every quadrilateral $4$ times, exept the case where $a=b=c=d=6$, which we correctly counted only once. The true number is therefore $\frac{121-1}{4}+1=31$
I somehow have a feeling that this number is too low, and I suspect that there is a mistake in my reasoning for the overcounting process. Is there a better way of approaching this problem?
| Indeed you over-divided the numbers. For example $(4,8; 4,8)$ is only counted twice so it should be divided by $2$ while you divided it by $4$.
The systematic way to count this is to let $a$ to be the smallest number.
$a=1: b$ has $11$ choices.
$a=2: b$ has $9$ choices (because $d\geq 2$ as well).
$a=3: b$ has $7$ choices.
And so on until $a=6$ where $b$ has $1$ choices. In this way rotation won't matter and we always get a unique combination.
Therefore totally $11+9+7+...+1=36$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum of $f(x)=\frac{2x\sqrt{(x+1)}}{(9x^2+3)^{\frac{1}{4}}}+\frac{(1-2x)\sqrt{2-2x}}{(9(1-2x)^2+3)^{\frac{1}{4}}}$ on the interval $[0,1/2]$ I would like to find the maxima of the following function in one variable :
$$f(x)=\frac{2x\sqrt{(x+1)}}{(9x^2+3)^{\frac{1}{4}}}+\frac{(1-2x)\sqrt{2-2x}}{(9(1-2x)^2+3)^{\frac{1}{4}}}$$
on the interval $[0,1/2]$. I have already checked that it must occur at $1/3$, but could someone
give me a proper method of proving it (ideally without using computation engines ?). The problem here is obviously that the derivative of this function is not nice-looking at all and i do not wish to "play" with it.
Thanks in advance !
| By C-S $$f(x)=\frac{2x\sqrt{x+1}}{\sqrt[4]{\frac{1}{4}(1+3)(9x^2+3)}}+\frac{(1-2x)\sqrt{2-2x}}{\sqrt[4]{\frac{1}{4}(1+3)(9(1-2x)^2+3)}}\leq$$
$$\geq\frac{2x\sqrt{x+1}}{\sqrt[4]{\frac{1}{4}(3x+3)^2}}+\frac{(1-2x)\sqrt{2-2x}}{\sqrt[4]{\frac{1}{4}(3(1-2x)+3)^2}}=\frac{2\sqrt2x}{\sqrt3}+\frac{\sqrt2(1-2x)}{\sqrt3}=\sqrt{\frac{2}{3}}.$$
The equality occurs for $x=\frac{1}{3}$,which says that we got a maximal value.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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factorization of $a^3 + b^3$ over $\mathbb{C}$ I have been trying to understand how to factorize a basic $a^3 + b^3$ over $\mathbb{C}$. I saw in the book that it is $ \Pi_{i=1}^{3}(a + \omega^{(2i+1)}b)$ where $\omega$ is the cuberoot of $-1$.
I was trying to derive it in the following way and was surprised by something which I cannot explain. $a^3 + b^3 = (a^3 - (\omega b)^3)$, I tried to use the formula for factorization of $a^3 - b^3$. Since we know $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ hence I was trying to factorize the original expression as $a^3 -(\omega b)^3 = (a - \omega b)(a^2 + \omega ab + (\omega b)^2)$.
Then using $ 1 + \omega + \omega^2 = 0$. I substituted $(a -\omega b)(a^2 -(1 + \omega^2) ab + (\omega b)^2) $. This is factored as $(a -\omega b)(a - \omega ^2 b ) (a - b) $. This when expanded is $ (a^2 - \omega^2 ab - \omega ab + \omega^3 b^2) = (a^2 + ab - b^2) (a -b)$. Surely this product does not give back the original expression.
Which assumption am I misunderstanding during substitution which is creating this problem is unclear.
| Not exactly sure what our OP user2714795 is trying to do, but here's my approach:
Does
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)? \tag 1$
we compute
$(a + b)(a^2 - ab + b^2)$
$= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3, \tag 2$
so yes, (1) binds. We also have
$(a + b\omega)(a + b \bar \omega) = a^2 + ab (\omega + \bar \omega) + b^2 \omega \bar \omega \tag 3$
for any
$\omega \in \Bbb C; \tag 4$
if we take
$\omega = -\dfrac{1}{2} + \dfrac{\sqrt 3}{2}i \tag 5$
then (3) becomes
$(a + b\omega)(a + b \bar \omega)$
$= a^2 + ab (\omega + \bar \omega) + b^2 \omega \bar \omega = a^2 - ab + b^2; \tag 6$
combining (1) and (6) yields
$a^3 + b^3 = (a + b)(a + b\omega)(a + b\bar \omega), \tag 7$
and we have factored $a^3 + b^3$ over $\Bbb C$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating to derive relation between binomial co-efficient
If $C_o,C_1...$ are the binomial coefficents in the expansion of $(1+x)^n$ and $$\sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(r+1)^2} = k \sum_{r=0}^{n} \frac{1}{r+1}$$
Find a k such that the above equation is satisfied
My attempt:
$$ (1+x)^n = \sum_{k=0}^{k=n} \binom{n}{k} x^k$$
integrate both sides
$$ \frac{ (1+x)^{n+1} }{n+1} =\sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{k+1} +C$$
$$ x= 0 $$
$$ \implies \frac{-n}{n+1} = C$$
Divide both sides by 'x' and integrate
$$ \int \frac{ (1+x)^{n+1} }{ (x) n+1} dx = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2} - \frac{n}{n+1} \ln(x)+ C'$$
Or,
$$ \int \frac{ (1+x)^{n+1} }{ (x) n+1} dx - \frac{n}{n+1} \ln(x)+ C' = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2}$$
Not so sure what do here, like what to put as bounds. Slightly concerned I may have to evaluate a negative logarithm
| In trying to evaluate
$$S_n = \sum_{r=0}^n (-1)^r {n\choose r} \frac{1}{(r+1)^2}$$
we introduce
$$f(z) = \frac{(-1)^n \times n!}{(z+1)^2}
\prod_{q=0}^n \frac{1}{z-q}$$
which has the property that for $0\le r\le n$
$$\mathrm{Res}_{z=r} f(z) =
\frac{(-1)^n \times n!}{(r+1)^2}
\prod_{q=0}^{r-1} \frac{1}{r-q}
\prod_{q=r+1}^n \frac{1}{r-q}
\\ = \frac{(-1)^n \times n!}{(r+1)^2}
\frac{1}{r!} \frac{(-1)^{n-r}}{(n-r)!}
= (-1)^r {n\choose r} \frac{1}{(r+1)^2}.$$
It follows that
$$S_n = \sum_{r=0}^n \mathrm{Res}_{z=r} f(z).$$
Now residues sum to zero and the residue at infinity of $f(z)$ is zero
by inspection, therefore
$$S_n = - \mathrm{Res}_{z=-1} f(z) =
(-1)^{n+1} \times n! \times
\left. \left( \prod_{q=0}^n \frac{1}{z-q} \right)' \right|_{z=-1}
\\ = (-1)^{n+1} \times n! \times
\left. \prod_{q=0}^n \frac{1}{z-q}
\sum_{q=0}^n \frac{1}{q-z} \right|_{z=-1}
\\ = (-1)^{n+1} \times n! \times
(-1)^{n+1} \prod_{q=0}^n \frac{1}{q+1}
\times \sum_{q=0}^n \frac{1}{q+1}
\\ = n! \times \frac{1}{(n+1)!}
\times \sum_{q=0}^n \frac{1}{q+1}
= \frac{1}{n+1} \sum_{q=0}^n \frac{1}{q+1} = \frac{1}{n+1} H_{n+1}.$$
We see that the desired factor on the sum is $k=\frac{1}{n+1}.$
| {
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"url": "https://math.stackexchange.com/questions/3802161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does the equation with $2 \arctan(x)$ and other Inverse Trigonometric functions have weird conditions? I have these three equations:
$$2\arctan(x) = \arcsin(\frac{2x}{1+x^2}),\left |x\right|\leq1$$
$$2\arctan(x) = \arccos(\frac{1-x^2}{1+x^2}),x\geq0$$
$$2\arctan(x) = \arctan(\frac{2x}{1-x^2}),\left |x\right|<1 $$
I can verify these by simple substitution taking $x = \tan\theta$ for some $\theta$ but I could not understand why these conditions for $x$ are given.
I can see their graphs makes sense but is there an algebraic way of proving/verifying these conditions?
Also the graph is:
| An exercise I found helpful.
Consider the intersection of the line $y = tx$ and the circle $x^2 + y^2 = 2x$
The circle has radius 1 and is centered at $(1,0)$
$x^2 + (tx)^2 = 2x\\
x = \frac {2}{1+t^2}\\
y = \frac {2t}{1+t^2}$
Now lets translate everything one unit to the left. Yes, we could have calculated the intersection of $x^2+ y^2 = 1$ and the line $y = t(x+1)$ but the algebra is simpler this way.
$x = \frac {2}{1+t^2}-1\\
x = \frac {1-t^2}{1+t^2}\\
y = \frac {2t}{1+t^2}$
Do these equations look familiar? What are $x,y,t$ geometrically / trigonometrically?
The line and the circle and the x axis form an angle on the circle, which has half the measure of the central angle to that same arc.
$t = \tan \frac 12 \theta$
And of course,
$x = \cos \theta\\
y = \sin \theta$
Regarding the domain restrictions. If $t> 1$ the line $y = t(x+1)$ will be intersecting the circle in QII.
$\arcsin (\frac {2t}{1+t^2})$ will return a value corresponding to a point in QI
Similarly if $t < 0$
$\arccos (\frac {1-t^2}{1+t^2})$ will return a value for a point in QI or QII while our reference point is actually in QIII or QIV
I found this exercise to more helpful to build intuition than mechanically chug through the trig identities.
Here would be the mechanical approach:
$\sin (\arctan \frac {a}{b}) = \frac {a}{\sqrt {a^2 + b^2}}\\
\cos (\arctan \frac {a}{b}) = \frac {b}{\sqrt {a^2 + b^2}}\\
\theta = 2\arctan t\\
\sin \theta = 2\sin(\arctan t)\cos(\arctan t) = 2\frac {1}{\sqrt {1+t^2}}\frac {t}{\sqrt {1+t^2}} = \frac {2t}{1+t^2}\\
\cos \theta = \cos^2(\arctan t)-\sin^2(\arctan t) = \frac {t^2}{1+t^2} - \frac {1}{1+t^2}= \frac {1-t^2}{1+t^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lower bound for $(1+x)^n$ I am trying to prove the following bound for $n \in N$ and $x \in [-1,0)$, but x closer to 0:
$ e^{nx} - \frac{1}{2}nx^{2}e^{(n-1)x} \leq (1+x)^n $
I have:
$ e^{nx} - \frac{1}{2}nx^{2}e^{(n-1)x} = (1 - \frac{1}{2}nx^{2}e^{-x})e^{nx} \leq 1 - \frac{1}{2}nx^{2}e^{-x} \leq 1 - x(\frac{nx}{2}) $
And looking at some limit representation for $e$ or some $e$ Taylor expansion but I am stuck.
| Fisrt, lets make $x = x_n < 0$ such that $1 - \frac{1}{2}nx^2e^{-x} > 0$ then consider the following taylor expansions:
$(1+x)^n = 1 + xn + \frac{1}{2}nx^2 + (n^3 - 3n^2 -2n)\frac{x^3}{3} + O(x^4)$
$e^{nx} - \frac{1}{2}nx^2e^{(n-1)x} = 1 + xn + \frac{1}{2}nx^2 + (2n^3 - 6n^2 + 6n)\frac{x^3}{3} + O(x^4)$
Substracting them we obtain:
$(1+x)^n - (e^{nx} - \frac{1}{2}nx^2e^{(n-1)x}) = (n^3 - 3n^2 -2n)\frac{x^3}{3} - (2n^3 - 6n^2 + 6n)\frac{x^3}{3} + O(x^4)$
We just have to proove that
$(n^3 - 3n^2 -2n)\frac{x^3}{3} - (2n^3 - 6n^2 + 6n)\frac{x^3}{3} \geq 0$
i.e.
$(n^3 - 3n^2 -2n)\frac{x^3}{3} \geq (2n^3 - 6n^2 + 6n)\frac{x^3}{3}$
i.e.
$(n^3 - 3n^2 -2n) \leq (2n^3 - 6n^2 + 6n) \iff (n^3 - 3n^2 + 8n) \geq 0 $
It should be trivial that for $n \geq 0$ the cubic polynomial images are above $0$, and in $0$ it is exactly $0$.
$\blacksquare$
| {
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Calculate $\int \frac{ 1}{\cos a\>+\> \cos x} \, dx $ where $a$ is a constant How to evaluate
$$\int \frac{ 1}{\cos(a) + \cos(x)} \, dx $$
where $a$ is a constant?
My Attempt:
I substituted in the integral:
$A = \cos(a)$
On solving by usual method, I got:
$$ \frac{2 \arctan(\frac{\sqrt{A-1} . \tan(\frac{x}{2})}{\sqrt{A+1}})}{\sqrt{A^2-1}} + C$$
where $C$ is the constant of integration and $A^2> 1$.
Now that means this solution does not work for $A = \cos(a)$.
How should I proceed with this question in the simplest possible way (beginner's approach) so that it works for $A^2 < 1$?
Thanks in advance!
| Note
\begin{align}
\int \frac{ 1}{\cos a+ \cos x} \, dx
&=\int \frac{ 1}{2\cos \frac{x-a}2 \cos \frac{x+a}2} \, dx\\
&= \frac1{2\sin a}\int\left(\frac{\sin \frac{x+a}2}{ \cos \frac{x+a}2} -\frac{\sin \frac{x-a}2}{ \cos \frac{x-a}2} \right)dx\\
&= \frac1{\sin a} \ln\bigg|\frac{ \cos \frac{x-a}2}{\cos \frac{x+a}2 }\bigg|+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem on Rolle's theorem So there's this question that asks us to prove that, between any two roots of $\tan x=1$ there exists at least one root of $\tan x =-1$. Suppose we assume that $a,b$ are two roots of $\tan x-1=0$, then $f(a)=f(b)=0$, where $f(x)= \tan x-1$. According to the theorem, $f'(c)=0$ where $c \in (a,b)$,i.e., $\sec^2 c =0$....and this is not defined. Either there's something wrong with my understanding or with the problem. Please help.
| Let $f(x)=\tan x -1$, and $g(x)=\tan x +1$.
The roots of $f(x)$ occur in the interval $I=[\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup (\frac{\pi}{2}+\pi k, \frac{5\pi}{4}+\pi k]$ for some $k\in\mathbb{Z}$. The roots of $f(x)$ occur at the endpoints of interval $I$.
The roots of $g(x)$ occur in the interval $J=[-\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup(\frac{\pi}{2}+\pi k, \frac{3\pi}{4}+\pi k]$ for some $k\in\mathbb{Z}$. The roots of $g(x)$ occur at the endpoints of interval $J$.
The interval $(\frac{\pi}{2}+\pi k, \frac{3\pi}{4}+\pi k]$ is contained in $[\frac{\pi}{4}+\pi k, \frac{\pi}{2}+\pi k)\bigcup (\frac{\pi}{2}+\pi k, \frac{5\pi}{4}+\pi k]$ and a root of $g(x)$ occurs at $x=\frac{3\pi}{4}+\pi k$. Roots of $f(x)$ occur at $x=\frac{\pi}{4}+\pi k$ and $x=\frac{5\pi}{4}+\pi k$.
$\frac{\pi}{4}+\pi k<\frac{3\pi}{4}+\pi k<\frac{5\pi}{4}+\pi k $.
So there is at least one root of $g(x)$ between any two roots of $f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify $\frac{4\sin\alpha-5\cos\alpha}{3\cos\alpha-\sin\alpha}$ Simplify $\dfrac{4\sin\alpha-5\cos\alpha}{3\cos\alpha-\sin\alpha}.$
I think we should get $\cot\alpha$ which is $\dfrac{\cos^2\alpha}{\sin^2\alpha}.$ So I tried to write $\sin\alpha$ as $\sqrt{1-\cos^2\alpha}$ but it didn't work.
| Using Weierstrass relations for double angles. Let
$$ \alpha = 2 \beta , t = \tan \beta,\; \sin \alpha= \dfrac{2t}{1+t^2}=S,\; \cos \alpha= \dfrac{1-t^2}{1+t^2}=C;\;$$
When we simplify
$$ \dfrac{4C-5S}{3S-C}$$
we get in terms of tan half angle $t$
$$-\dfrac{5t^2+8t-5}{3t^2+2t-3}$$
but not
$$ \dfrac{1-t^2}{2t}= \cot \alpha $$
as you are expecting.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the maximum value of $\sum_{cyc}\frac{1}{a^2-4a+9}$ s.t. $a+b+c =1$ Problem: Let $a$, $b$, and $c$ be non-negative real numbers such that $a+b+c =1$. Find the maximum value of
\begin{align}
\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}.
\end{align}
By using the Karush–Kuhn–Tucker conditions, I know the maximum occurs at $a=0$, $b=0$, and $c=1$. But I want to know how to solve this problem without using calculus, because it is a math olympiad problem.
| Another way:
By your work we need to prove that:
$$\sum_{cyc}\frac{1}{a^2-4a+9}\leq\frac{7}{18}$$ or
$$\sum_{cyc}\left(\frac{1}{a^2-4a+9}-\frac{1}{5}\right)\leq\frac{7}{18}-\frac{3}{5}$$ or
$$\sum_{cyc}\frac{(2-a)^2}{a^2-4a+9}\geq\frac{19}{18}.$$
Now, by C-S $$\sum_{cyc}\frac{(2-a)^2}{a^2-4a+9}=\sum_{cyc}\frac{(2-a)^2(4-a)^2}{(a^2-4a+9)(4-a)^2}\geq\frac{\left(\sum\limits_{cyc}(2-a)(4-a)\right)^2}{\sum\limits_{cyc}(a^2-4a+9)(4-a)^2}.$$
Id est, it's enough to prove that
$$18\left(\sum\limits_{cyc}(2-a)(4-a)\right)^2\geq19\sum\limits_{cyc}(a^2-4a+9)(4-a)^2,$$
which after homogenization gives:
$$\sum_{sym}(190a^3b+207a^2b^2+813a^2bc)\geq0,$$ which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct? Prove that for all positive real numbers:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$
This is same as this question but a different approach is used there whereas I want to verify my approach to this problem.
My Approach:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\Big(\dfrac{a}{b+c}+1\Big)+\Big(\dfrac{b}{c+a}+1\Big)+\Big(\dfrac{c}{a+b}+1\Big)-3$$
$$=(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3$$
By AM-HM inequality:
$$\dfrac{3}{\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}}\leq\dfrac{2(a+b+c)}{3}\Rightarrow (a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]\geq \dfrac{9}{2}$$
$$(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3\geq \dfrac{3}{2}$$
$\therefore \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\space \forall\ a,b,c\in \mathbb R$ and $a,b,c>0$
Please check this approach and provide suggestions. Also please provide alternative solutions if available.
THANKS
| Another proof$:$
Due to homogeneous, assume $a+b+c=1.$
Let $p=a+b+c=1,q=\dfrac{1-t^2}{3} \quad(\, t\in [\,0,1\,]\,),r=abc.$
Need to prove$:$ $$\frac73\,{t}^{2}+9\,r-\frac13 \geqslant 0$$
Since $$r\geqslant \dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2}$$
We need to prove$:$ $$\dfrac{2}{3} t^2(2-t) \geqslant 0,$$
which is true since $t \in [\,0,\,1\,].$
See also here.
There is also a proof by SS (SOS - Schur) method.
$$\text{LHS}-\text{RHS}={\frac {2\, \left( a-b \right) ^{2} \left( a+b \right) + \left( a
-c \right) \left( b-c \right) \left( a+b+2\,c \right) }{2 \left( b+c
\right) \left( c+a \right) \left( a+b \right) }} \geqslant 0,$$
which is obvious if $c\equiv \min\{a,b,c\}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$ .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a \ge 3\sum a$$
or $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
| By Cauchy-Schwartz ineq.:
$$F=\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab} \ge \frac{(a^{3/2}+b^{3/2}+c^{3/2})^2}{ab+bc+ca}$$
Now, use Mean-Power ineq.: $M_{3/2} \ge M_1:$
$$F \ge \frac{3}{ab+bc+ca} \left(\frac{(a+b+c)^{3/2}}{3} \right)^2=\frac{(a+b+c)^3}{3(+bc+ca)}$$
Finally use $(a+b+c)^2 \ge 3(ab+bc+ca).$
Hence, $$F\ge (a+b+c)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\geqslant \frac{a+b}{b^3+c^3}+\frac{b+c}{c^3+a^3}+\frac{c+a}{a^3+b^3}$ For $a,b,c>0.$ Prove$:$ $$\dfrac{a}{b^3}+\dfrac{b}{c^3}+\dfrac{c}{a^3}\geqslant \dfrac{a+b}{b^3+c^3}+\dfrac{b+c}{c^3+a^3}+\dfrac{c+a}{a^3+b^3}\quad (\text{Tran Quoc Thinh}) $$
It's easy with Buffalo Way and computer so I will not post it.
(Please don't post solution by Buffalo Way, thanks for a real a lot!)
So$,$ we try to find a solution by hand.
I get this SOS$:$
$$\sum {\dfrac { \left( 6\,{a}^{5}+5\,{a}^{4}b+2\,{a}^{4}c+4\,{a}^{3}{b}^{2}+4
\,{a}^{3}bc+8\,{a}^{2}{b}^{3}+6\,{a}^{2}{b}^{2}c+3\,a{b}^{4}+4\,a{b}^{
3}c-2\,{b}^{5}+2\,{b}^{4}c \right) \left( a-b \right) ^{2}}{{a}^{3}{b
}^{3} \left( a+b \right) \left( {a}^{2}-ab+{b}^{2} \right) }} \geqslant 0,$$
By SOS theorem$,$ if $$S_a+S_b+S_c \geqslant 0 ; S_a S_b +S_b S_c +S_cS_a\geqslant 0.$$
Then $$S_a (b-c)^2 +S_b (c-a)^2 +S_c(a-b)^2\geqslant 0.$$
Here$,$ we can prove$:$ $$S_a+S_b+S_c \geqslant 0,$$
but $$S_a S_b +S_b S_c +S_cS_a\geqslant 0$$ is not true!
pqr or $uvw$ technique give a very high degree, I think it is impossble.
| Partial Answer :
I rewrite the inequality like this with $a\geq c \geq b>0$:
$$\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{b^3}+1}\Big)+\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{c^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{a^3}+1}\Big)\geq-\Big(\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{c^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{a^3}+1}\Big)+\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{b^3}+1}\Big)\Big)$$
Or :
$$\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{b^3}+1}\Big)+\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{c^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{a^3}+1}\Big)\geq\Big(\dfrac{b}{c^3}\Big(\frac{1}{2}-\frac{1}{\frac{c^3}{b^3}+1}\Big)+\dfrac{c}{a^3}\Big(\frac{1}{2}-\frac{1}{\frac{a^3}{c^3}+1}\Big)+\dfrac{a}{b^3}\Big(\frac{1}{2}-\frac{1}{\frac{b^3}{a^3}+1}\Big)\Big)$$
Now use rearrangement inequality with the constraint $\frac{c}{b}\geq \frac{a}{c}\geq \frac{b}{a}$ and $\frac{a}{b^3}\geq \frac{b}{c^3}\geq \frac{c}{a^3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Find the maximum value of $2x + 2\sqrt{x(1-x)}$ if $0 \leq x \leq 1.$ Find the maximum value of $2x + 2\sqrt{x(1-x)}$ when $0 \leq x \leq 1.$
I'm trying to find the upper bound of the expression, but I'm having trouble on how to start/how to solve the problem. Can someone help? Thanks in advance!
| Hint : Try to plug $x = cos^{2}(\theta)$ and use some trigonometry to see what can be the maximum value of $sin(A) + cos(A)$.
I hope this helps. If you are stuck, let me know!
Update : Since the OP seems not to follow the hint and work through it, I'll give the complete solution:
Plug in $x = cos^{2}(\theta)$, then we get:
$ 2x + 2\sqrt{x(1-x)} = 2cos^{2}(\theta) + 2 \sqrt{cos^{2}(\theta)(1-cos^{2}(\theta))} = 2cos^{2}(\theta) + 2 \sqrt{cos^{2}(\theta)sin^{2}(\theta)}$ $\mbox{
} $ , using the identity $cos^{2}(\theta) + sin^{2}(\theta) = 1$
$ = 2cos^{2}(\theta) + 2 sin(\theta) cos(\theta) = [1 + cos(2\theta)] + sin(2\theta)$ $\mbox{ }$ (using the identities : $cos(2\theta) = 1 - 2cos^{2}(\theta)$ and $sin( 2\theta) = 2 sin(\theta) cos(\theta)$ )
Now we find maximum of $cos(2\theta) + sin(2\theta)$ :
$cos(2\theta) + sin(2\theta) = \sqrt{2} [1/\sqrt{2} sin(2\theta) + 1/\sqrt{2} cos(2\theta)] = \sqrt{2} [cos(\pi/4) sin(2\theta) + sin(\pi/4) cos(2\theta)] = \sqrt{2}sin( \pi/4 + 2\theta)$
using the identity $sin( A +B) = sin(A)cos(B) + cos(A)sin(B)$
Since maximum value of $sin( \pi/4 + 2\theta) = 1$, $cos(2\theta) + sin(2\theta) \leq \sqrt{2}$.
Thus, $ 2x + 2\sqrt{x(1-x)} \leq 1 + \sqrt{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluating $\int_{-1/2}^{1} \cos^{-1} \frac{1-x^2}{1+x^2} dx$ Let us do the integration by parts $$I=\int_{-1/2}^{1} ~\cos^{-1} \frac{1-x^2}{1+x^2}. 1~ dx =\left.x \cos^{-1} \frac{1-x^2}{1+x^2}\right|_{-1/2}^{1}-\int_{-1/2}^{1} \frac{2x}{1+x^2} dx$$ $$=\frac{\pi}{2}+\frac{1}{2}\cos^{-1} \frac{3}{5}-\left .\ln(1+x^2)\right|_{-1/2}^{1}=\frac{\pi}{2}+\frac{1}{2}\cos^{-1} \frac{3}{5}-\ln\frac{8}{5}.$$
The question is whether something is amiss here and whether the answer is right.
EDit
The correct answer is $$I=\frac{\pi}{2}+\frac{1}{2}\cos^{-1} \frac{3}{5}-\ln\frac{5}{2}.$$
| The error is in the sign that is different for $x\leq0$ and $x\geq0:$
$$
x \frac{d}{dx}\cos^{-1}\frac{1-x^2}{1+x^2}=
\left\{
\begin{alignedat}{2}
-\dfrac{2x}{1+x^2}\quad & -&&1\leq x\leq 0 \\
+\dfrac{2x}{1+x^2}\quad & &&0\leq x\leq 1
\end{alignedat}
\right.
$$
so that
\begin{align}
I
&=\int_{-1/2}^{1}\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx =\\
&=\left.x\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right|_{-1/2}^{1}+\int_{-1/2}^{0} \frac{2x}{1+x^2} dx-\int_{0}^{1} \frac{2x}{1+x^2} dx=\\
&=\frac{\pi}{2}+\frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)-\ln\frac{5}{4}-\ln 2\\
&=\frac{\pi}{2}+\frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)-\ln\frac{5}{2}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating $n$-th power of a matrix I was doing an exercise of a past exam in which one of the things I had to do was calculating the $n$th power of a Jordan matrix
$$J=\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}.$$
I started calculating until the 5th power but I couldn't guess the expression for the $a_{1,2}$, $a_{1,3}$ and $a_{2,3}$ components. When I looked it up in an online calculator it showed that:
$$J^n=\begin{pmatrix}
2^n & \frac{2^n·n}{2} & \frac{2^n·(n^2-n)}{8} \\
0 & 2^n & \frac{2^n·n}{2} \\
0 & 0 & 2^n
\end{pmatrix}.$$
My question is: when the relationships are as difficult as these (especially the $a_{1,3}$ component), are there any tricks for figuring out the$n$th power component? Because these kind of relationships are difficult to think in the middle of an exam.
| $J=2I+A$, where $A=\begin{pmatrix} 0&1&0 \\0&0&1 \\ 0&0&0\end{pmatrix}$. As $A$ and $I$ commute, we can apply the binomial formula.
Observe that $A^2=\begin{pmatrix} 0&0&1 \\0&0&0 \\ 0&0&0\end{pmatrix}$, $A^k=0$ for $k\ge 3$, so the $n$-th power is
$$J^n=2^n I+n2^{n-1}A+\frac{n(n-1) 2^{n-2} }2A^2=2^n I+n2^{n-1}A+n(n-1) 2^{n-3}A^2. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving 2-degree equations in 3 variables. We are given 3 equations:
$x^2+\sqrt3 xy + y^2 = 25$
$y^2 + z^2 = 9$
$x^2 +xz+ z^2 = 16$.
$x,y,z$ are positive real numbers.
Then we have to find value of $xy + 2yz + \sqrt3 xz$.
| Given a triangle with sides $l_1,l_2,l_3$ we have
$$
\cases{
l_1^2=l_2^2+l_3^2-2l_2l_3\cos\theta_1\\
l_2^2=l_1^2+l_3^2-2l_1l_3\cos\theta_2\\
l_2^2=l_1^2+l_2^2-2l_1 l_2\cos\theta_3
}
$$
then making $l_1=x,l_2=y,l_3=z$
$$
\cases{
2\cos\theta_1=0\\
2\cos\theta_2=-1\\
2\cos\theta_3=-\sqrt{3}
}
$$
It is a rectangle. Etc.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\frac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\frac{1}{5}$ For $a,b,c\geqslant 0.$ Prove$:$
$$\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\dfrac{1}{5}$$
I found an AM-GM proof.
Since $$P+\frac{1}{5}\geqslant 0\Leftrightarrow 6\,{a}^{3}+6\,{b}^{3}+8\,{a}^{2}c-2\,a{c}^{2}+8\,{b}^{2}c-2\,b{c}^{2}-19\,abc+3\,{a}^{2}b+3\,a{b}^{2}+{c}^{3} \geqslant 0$$
And by AM-GM$:$
$$2\,a{c}^{2}\leqslant 6{a}^{3}+\frac49{c}^{3},$$
$$2\,b{c}^{2}\leqslant 6{b }^{3}+\frac49{c}^{3},$$
$$19\,abc\leqslant \frac19{c}^{3}+3a{b}^{2}+3{a}^{2}b+8 \,{a}^{2}c+8\,{b}^{2}c.$$
So we are done!
Is there another nice proof$?$ Thanks for a real lot!
| Heres another way. First note:
*
*From the expression, it is enough to consider the case $c \geqslant \max(a, b)$
*From homogeneity, we may set $c=1$. So $a, b \in [0, 1]$.
*Replacing $a, b $ with their arithmetic mean reduces the numerator of LHS, leaving the denominator untouched, as $a^3+b^3, a^2+b^2, -ab$ all become smaller. Hence we it is enough to consider $a=b=t$.
Finally we are left to show for $t \in [0, 1]$:
$$\frac{2t^3+2t(t-1)-5t^2}{(2t+1)^3} \geqslant -\frac15 \iff \frac{(3t-1)^2}{5(2t+1)^2}\geqslant 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with a, b, c about finding minimal and maximal value Find the minimal and maximal value (if they exist) of ${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(a+c)}{a^2+c^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}}$ if are non-negative real numbers, such that at least two of them are positive.
My attempts for the case where all variables are positive:
I tried applying AM-GM on ${\sqrt{a(b+c)}} $ and etc. and then applying $b^2+c^2=>(b+c)^2/2$,but the inequality I received was false. I also tried rewriting $a(b+c)/(b^2+c^2)=(b/a+c/a)/((b/a)^2+(c/a)^2)$ and etc. and then letting $a/b=x$, $b/c=y$ and $c/a=z$ and etc but I was stuck from there. I also tried applying Cauchy Schwarz by squaring the whole expression.
| If $abc = 0,$ example $c = 0$ then
$${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(c+a)}{c^2+a^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}} =\sqrt{\frac ab}+\sqrt{\frac ba} \geqslant 2.$$
Equality occur when $a=b,\,c=0.$
If $abc > 0,$ using the AM-GM inequality, we have
$${\sqrt{\frac{a(b+c)}{b^2+c^2}}} > {\sqrt{\frac{a(b+c)}{(b+c)^2}}} = \sqrt{\frac{a}{b+c}}=\frac{a}{\sqrt{a(b+c)}} > \frac{2a}{a+b+c}.$$
Therefore
$${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(c+a)}{c^2+a^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}} > \frac{2(a+b+c)}{a+b+c}=2.$$
So a minimal value is $2.$
| {
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I am not able to solve the integral $I=\int { \frac { \sqrt { \sqrt { x } +\sqrt { x-1 } } }{ 1+\sqrt { x } } }$ $J=\int { \frac { \sqrt { \sqrt { x } -\sqrt { x-1 } } }{ 1+\sqrt { x } } } $ then I-J
I substituted $x=\sec ^{ 2 }{ \theta } $.
$I-J=\frac { \sqrt { \sec { \theta } +\tan { \theta } } -\sqrt { \sec { \theta } -\tan { \theta } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } \\ I-J=\frac { \sqrt { \sec { \theta } +\tan { \theta } } -\frac { 1 }{ \sqrt { \sec { \theta } +\tan { \theta } } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } \\ I-J=\frac { \frac { \sec { \theta } +\tan { \theta } -1 }{ \sqrt { \sec { \theta } +\tan { \theta } } } }{ 1+\sec { \theta } } \times 2\sec ^{ 2 }{ \theta } \tan { \theta d\theta } $
I am not able to solve further Please help
| There is some simplification available here as
\begin{align}
\left(\sqrt{\sqrt{x}+\sqrt{x-1}}\right)\left(\sqrt{\sqrt{x}-\sqrt{x-1}}\right) &= \sqrt{1}
\end{align}
Since both $I$ and $J$ have these terms in, multiplication of these terms should give you something to work with once you have simplified the integrands using high school methods.
| {
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Finding $\frac{2a^5 - 5a^4 + 2a^3 - 8a^2}{a^2 + 1}$ Suppose $a$ is a root of $x^2 + 3x - 1.$ Find $\frac{2a^5 - 5a^4 + 2a^3 - 8a^2}{a^2 + 1}.$
I was thinking of factoring the fraction a bit first, than letting $a^2 = 1 - 3a.$ However, that leads nowhere.
| A complete solution. Let $a$ be a root of $x^2+3x-1$. Then
$a^2=1-3a$
$a^3=a-3a^2=a-3(1-3a)=10a-3$
$a^4=10a^2-3a=10(1-3a)-3a=10-33a$
$a^5=10a-33a^2=10a-33(1-3a)=-33+109a$
Therefore the numerator:$$2a^5 - 5a^4 + 2a^3 - 8a^2=
-66+218a-5(10-33a)+2(10a-3)-8(1-3a)=-130+427a$$
and the denominator: $a^2+1=2-3a$.
Let $a'$ be another root of $x^2+3x-1=0$. Then $aa'=-1. a+a'=-3$. Multiply the numerator and the denominator by $2-3a'$. The denominator becomes $$(2-3a)(2-3a')=4+9-6(-3)=13-18=-5.$$ The numerator becomes $$(427a-130)(2-3a')=
854a-260+1281+390a'=1021-1170+464a=464a-149.$$
So the fraction is equal to $$\frac{464a-149}{-5}$$ where $$a=\frac{-3\pm \sqrt{13}}{2}.$$
| {
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Taylor series of $\frac{x}{e^x-e^{-x}}$ My textbook says that (when $x$ approaches $0$):
$$\frac{x}{e^x-e^{-x}}=\frac{1}{2}-\frac{x^2}{12}+\cdots$$
It is also said that the result can be deduced by using $e^x=1+x+x^2/2+\cdots$.
But when I sub it in directly:
$$\begin{aligned}\frac{x}{e^x-e^{-x}}&=\frac{x}{(1+x+x^2/2+x^3/6\cdots)-(1-x+x^2/2-x^3/6+\cdots)}\\&=\frac{x}{2x+x^3/3+\cdots}\end{aligned}$$
I cannot get the right expansion. So what is the right approach?
| Another way:
$$\begin{split}\frac{x}{e^x-e^{-x}} &= e^{-x}\frac{x}{1-e^{-2x}} \\ &= e^{-x}\frac{x}{2x-2x^2+4x^3/3+O(x^4)}\\&=e^{-x}\frac{1}{2}\frac{1}{1-(x-2x^2/3+O(x^3))} \\&=\frac12\left(1-x+\frac 12x^2+O(x^3)\right)\left(1+\left(x-\frac23x^2\right)+\left(x-\frac23 x^2\right)^2+O(x^3)\right)\\&=\frac12\left(1-x+\frac 12x^2+O(x^3)\right)\left(1+x+\frac13x^2+O(x^3)\right)\\&=\frac12-\frac1{12}x^2+O(x^3).\end{split}$$ Here, we have used the expansion
$$\frac{1}{1-x}=1+x+x^2+\cdots.$$
| {
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Let $x,y,z$ be positive real numbers with $|x-y|<2, |y-z|<2, |x-z|<2$. Prove that $\sqrt{xy+1}+\sqrt{yz+1}+\sqrt{xz+1}>x+y+z$. Let $x,y,z$ be positive real numbers with $|x-y|<2, |y-z|<2, |x-z|<2$. Prove that $$\sqrt{xy+1}+\sqrt{yz+1}+\sqrt{xz+1}>x+y+z.$$
My approach: I have spent some time and came up with the following idea: Since $|x-y|<2$ then if we multiply by $y>0$ we will get $(y-1)^2<xy+1<(y+1)^2$ then it follows that $\sqrt{xy+1}>|y-1|$. And the same can be done for ech term and we will get that $\sqrt{xy+1}+\sqrt{yz+1}+\sqrt{xz+1}>|x-1|+|y-1|+|z-1|$. But I cannot get the desired inequality.
Can anyone show the solution please?
| $|x-y|<2$ can be squared to give $x^2 -2xy + y^2<4$.
Add $4xy$ which gives $x^2 +2xy + y^2<4 + 4 xy$.
Take the square root: $x+y<2 \sqrt{1 + xy}$.
Adding this up for all 3 inequalities gives the result. $\qquad \Box$
| {
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Compute the arc length of the curve $y = \sqrt{x-x^2}+\sin^{-1}(\sqrt{x})$ Compute the arc length of the curve $y = \sqrt{x-x^2}+\sin^{-1}(\sqrt{x})$ from for $0 \leq x \leq 1$
This problem is pretty brutal! I'd appreciate if somebody could hold my hand through this integral and really lay out the details for me... I've been struggling with it for awhile now and can't get it down!!
Basically we know that: $$L = \int_0^1 \sqrt{1+(\frac{dy}{dx})^2}$$
Where $$\frac{dy}{dx}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}$$
If somebody could help me simplify and integrate this that would be great... Thank so you much!!
| You're practically there, just simplify where you left off:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-2x}{2\sqrt{x-x^2}}+\frac{1}{2\sqrt{1-x}\sqrt{x}}=\frac{2-2x}{2\sqrt{x-x^2}}=\frac{1-x}{\sqrt{x}\sqrt{1-x}}=\sqrt{\frac{1-x}{x}}$$
Therefore,
$$L = \int_0^1 \sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \; \mathrm{d}x= \int_0^1 \sqrt{\frac{1}{x}} \; \mathrm{d}x=2\sqrt{x} \bigg \rvert_0^1 = \boxed{2}$$
| {
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Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $
But can anyone attack this problem in a different & more standard way?
| We have:
\begin{align}
n\left (\frac{1}{n} - \frac{1}{n+1} + \cdots \right ) &= n \left ( \frac{1}{n(n+1)} + \frac{1}{(n+2)(n+3)} + \cdots \right ) \\
&\le n \left ( \frac{1}{n^2} + \frac{1}{(n+2)^2} + \cdots \right ) \\
&\le n \int_{n-2}^{\infty} \frac{1}{2x^2}dx = \frac{n}{2(n-2)}
\end{align}
Similarly,
\begin{align}
n\left (\frac{1}{n} - \frac{1}{n+1} + \cdots \right ) \ge \frac{n}{2(n+1)}
\end{align}
Thus, by letting $n$ tends to infinity, we obtain
\begin{align}
\lim_{n\to \infty} {n \left ( \frac{1}{n} - \frac{1}{n+1} + \cdots \right )} = \frac{1}{2}
\end{align}
| {
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Induction to prove that $\log_2 2^k \leq 2^{k/2}$ Prove by induction that for every $k\geq4,$ $\log_2(2^k)\leq 2^{k/2}$.
Then prove that $\log_2n\leq\sqrt{2n}$ for all n>=1 by using $\log_2 n$ is at most $\log_2(2^k)$ for the smallest integer k such that $ n<=2^k$.
I tried the first part by induction: $\log_2(2^k)\leq 2^{k/2}$.
Base case: k=4, $\log_2(2^4)\leq2^{4/2}$ gives $4\leq4,$ which is true.
Induction Hypothesis: Assume true for (k-1): $\log_2(2^{(k-1)})\leq 2^{(k-1)/2}$
Inductions Step: Show that it is true for k: $\log_2(2^k)\leq 2^{k/2}$
At this step, I tried to make the inequality look like the one in the inductive hypothesis: $\log_2(2^{(k-1)}2^1)\leq 2^{k/2}$, which reduces to $\log_2(2^{k-1})+1\leq2^{k/2}$. Then we can write it like this: $\log_2(2^{k-1})\leq2^{k/2}-1$. I am stuck here.
| $\log_2 2^k = \log_2 2\cdot 2^{k-1} = \log_2 2^{k-1} + \log_2 2=\log_2 2^{k-1} + 1\le 2^{\frac{k-1}2} +1$
Now $(2^{\frac{k-1}2} +1)^2 = 2^{k-1} + 2\cdot 2^{\frac{k-1}2} + 1=$
$2^{k-1} + 2^{\frac {k-1}2+1} + 1=2^{k-1}(1 + 2^{-\frac {k-1}2 + 1} + \frac 1{2^{k-1}})$
$k-1 \ge 4$ so $\frac 1{2^{k-1}} < \frac 14$. And $-\frac {k-1}2 + 1\le -1$ so $2^{-\frac {k-1}2 + 1}\le \frac 12$. so
$1 + 2^{-\frac {k-1}2 + 1} + \frac 1{2^{k-1}} < 1 + \frac 12 + \frac 14 < 2$.
So $(2^{\frac{k-1}2} +1)^2 < 2^{k-1}\cdot 2 = 2^k$.
So $2^{\frac{k-1}2} +1 < (2^k)^{\frac 12} = 2^{\frac k2}$ an
$\log_2 2^k < 2^{\frac k2}$.
.....
Which seems like a lot of work just to avoid pointing out $\log_2 2^k = k$ and this whole thing is equivalent to proving $k \le 2^{\frac k2}$ for $k\ge 4$. Which is probably easier.
If $k-1 \le 2^{\frac {k-1}2}$ then
$(k-1)^2 \le 2^{k-1}$ and as $2k-1 > 0$ then
$k^2 < k^2 - 2k+1 =(k-1)^2 \le 2^{k-1} < 2^k$ so $k < 2^{\frac k2}$.
| {
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How to find the cartesian equation of a plane given the vector equation? The vector equation is given by $r=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$
To find the cartesian equation, I have to consider a point on the plane:
$$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$$
which gives:
$$\frac{x+3}{\:6}=\:\frac{y+5}{6}=\frac{z+1}{-3}$$
Now this is the part where I don't understand. Up till now my working follows with the answers provided by my book, but then the book did something that puzzle me:
$$\frac{x+3}{6}+\frac{y+5}{6}=2(\frac{z+1}{-3})$$
How did they reached here?
| Maybe you did not report the whole question.
This solution gives the equation of some plane that contains the given line. Indeed,
$$a=b=c\implies a+b=2c,$$ but there are infinitely many other combinations, like
$$a=b=c\implies 7a-6b=c$$ or $$a=b=c\implies a=b,$$
all giving planes.
| {
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question from South Korean selection exam 1998, about proving that an inequality holds true if $a+b+c=abc$ I was just doing the following question:
If $a,b,c>0$ such that $a+b+c=abc$, prove that:
$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le \frac{3}{2}$
I think that this question can be solved through the use of homogenization, something which I attempted to do in the following way:
We have that $\frac{a+b+c}{abc}=1$ and hence also $\sqrt{\frac{a+b+c}{abc}}=1$. So $\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}=\frac{3}{2}$.
So all we have to prove now is $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}$ which are homogenized and hence we do not need the original equality any more. This is where I couldn't continue from, and got stuck. Could you please explain to me how I could finish it off like this, or tell me why it can't and how it can be done using homogenization?
| Multiply the first term top & bottom by $\sqrt{bc}$ & use the constraint
\begin{eqnarray*}
\frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}}.
\end{eqnarray*}
Now use AM-GM
\begin{eqnarray*}
\frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}}<\frac{1}{2}\left( \frac{b}{a+b} +\frac{c}{a+c} \right).
\end{eqnarray*}
| {
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Arc length of $\alpha(t)=(t^6,t^9)$ I have to evaluate the arc length of the curve
$$\alpha(t)=(t^6,t^9)$$
for $t\in[-\sqrt[3]{2},\sqrt[3]{2}]$.
But when I do it, the final integral is
$$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt$$
which value is zero. What should I do?
| Remember that $\sqrt{x^2}=|x|$.
So as mentioned by Arctic Char it should be $|t|^5$. Then you have $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3|t|^{5}\sqrt{4+9t^{6}}dt=$$
$$=\int_{0}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt-\int_{-\sqrt[3]{2}}^{0}3t^{5}\sqrt{4+9t^{6}}dt=$$
$$=\big[\frac{1}{27}(9t^6+4)^{\frac{3}{2}}\big]_{t=0}^{t=\sqrt[3]{2}}-\big[\frac{1}{27}(9t^6+4)^{\frac{3}{2}}\big]_{t=-\sqrt[3]{2}}^{t=0}$$
$$=\frac{1}{27}[40^{\frac{3}{2}}-8]-\frac{1}{27}[8-40^{\frac{3}{2}}]$$
$$=\frac{1}{27}[160\sqrt{10}-16]=\frac{16}{27}[10\sqrt{10}-1] \space(\approx18.14683)$$
(where $40^{\frac{3}{2}}=40\sqrt{40}=80\sqrt{10}$).
Or since the integrand is an even function you have $$\int_{-\sqrt[3]{2}}^{\sqrt[3]{2}}3|t|^{5}\sqrt{4+9t^{6}}dt$$
$$=2\int_{0}^{\sqrt[3]{2}}3t^{5}\sqrt{4+9t^{6}}dt$$
$$=2\cdot\frac{1}{27}[80\sqrt{10}-8]=\frac{16}{27}[10\sqrt{10}-1].$$
| {
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Prove that $\frac{AB}{AE} + \frac{AD}{AG} = \frac{AC}{AF}$ in parallelogram $ABCD$, where $E$, $F$, $G$ are points on a line intersecting the sides
Let $ABCD$ be a parallelogram. A line meets segments $AB$, $AC$, $AD$ at points $E$, $F$, $G$, respectively. Prove that $\frac{AB}{AE} + \frac{AD}{AG} = \frac{AC}{AF}$.
So recently I've been assigned a few problems, and this is one of them. So far, I've thought of extending line $EG$ and diagonal $BD$ to meet at a point, which we can call point $X$ and making point $O$ as the intersection of the diagonals in the paralellogram. And from here, I thought that maybe using Menelaus's theorem on triangle $BEX$ with line $AC$, which gets us $\frac{BA}{AE} \cdot \frac{EF}{FX} \cdot \frac{XO}{OB} = 1$. And similarly on triangle $DGX$ with line $AC$, we would get $\frac{DA}{AG} \cdot \frac{GF}{FX} \cdot \frac{XD}{OD} = 1$. But I'm not sure how to proceed from here and to relate these back to the original problems. Does anyone have any ideas on how I could do so?
| Let $K\in DO$ and $M\in BO$ such that $GK||AO$ and $EM||AO.$
Also, let $AO=a$, $GK=x$, $FO=y$ and $EM=z$.
Let $x>z$ and $L\in GK$ such that $LKME$ be parallelogram, $LE\cap FO=\{N\}.$
Thus, $$\frac{x-y}{y-z}=\frac{x-z}{y-z}-1=\frac{EL}{EN}-1=\frac{KM}{OM}-1=\frac{KO+OM}{OM}-1=\frac{KO}{MO}=$$
$$=\frac{\frac{KO}{DO}}{\frac{MO}{DO}}=\frac{\frac{DO-DK}{DO}}{\frac{BO-BM}{BO}}=\frac{1-\frac{x}{a}}{1-\frac{z}{a}}=\frac{a-x}{a-z},$$
which gives $$(x+z-2y)a+xy+yz-2xz=0.$$
Now, we need to prove that: $$\frac{AB}{AE}+\frac{AD}{AG}=\frac{2AO}{AF}$$ or
$$\frac{AB}{AB-BE}+\frac{AD}{AD-DG}=\frac{2a}{a-y}$$ or
$$\frac{1}{1-\frac{BE}{AB}}+\frac{1}{1-\frac{DG}{AD}}=\frac{2a}{a-y}$$ or
$$\frac{1}{1-\frac{z}{a}}+\frac{1}{1-\frac{x}{a}}=\frac{2a}{a-y}$$ or
$$\frac{1}{a-z}+\frac{1}{a-x}=\frac{2}{a-y}$$ or
$$(x+z-2y)a+xy+yz-2xz=0$$ and we are done!
| {
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For $x>1$, evaluate $\lim_{n\to \infty} \prod _{k=0} ^n \left(1+\frac{2}{x^{2^k} + x^{-2^k}}\right)$ Simplifying the part after the product function
For $k=0$
$$\frac{x+x^{-1} +2}{x+x^{-1}}$$
$$=\frac{(x^{\frac 12}+x^{-\frac 12})^2}{x+x^{-1}}$$
Performing a similar simplification for all values of $k$, then end result is
$$\frac{1}{x^{2^k}+x^{-2^k}} (x^{\frac 12}+x^{-\frac 12})^2 (x+x^{-1})(x^2+x^{-2})...(x^{2^{k-1}}+x^{-2^{k-1}})$$
I am not able to simplify it further. How should I proceed?
| Hint :
Multiply and divide by $(x^{\frac{1}{2}}-x^{−\frac{1}{2}})$
and use $(a+b)(a-b)=a^2-b^2$.
| {
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How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016) I was just doing the following question:
Find a prime number which divides the number $A=14^7+14^2+1$.
I solved it by finding the result which is $A=105413504+196+1=105413701$ and then trying out all prime numbers till I found that 211 divides it. However, obviously this is extremely tedious. I hence looked at the solution which says that $x^7+x^2+1=(x^2+x+1)(x(x-1)(x^3+1)+1)$ and from here by saying that $x=14$ we get the solution. However I can't seem to think of how to intuitively turn $x^7+x^2+1$ into $(x^2+x+1)(x(x-1)(x^3+1)+1)$. I realize that from the question it is obvious to go looking for factors of A and hence trying to factorize $14^7+14^2+1$, but I can't work out how to go about factorizing it, what are the steps which you need to take in order to factorize a given polynomial. Could you please explain to me how to go about factorizing such an expression and how to intuitively think of each step?
| By the factor theorem, $x^7+x^2+1$ has no rational roots and no linear factors over $\Bbb Z$, so we look for quadratic factors. To avoid non-integer coefficients, the only such factors are of the form $x^2+ax+1$. The case $a=1$ looks especially promising because, modulo $x^2+x+1$,$$x^3+1\implies x^7=x\implies x^7+x^2+1=0.$$This is essentially the reasoning in @TheSilverDoe's answer, but saves us having to work with explicit roots. As for finding the quotient, note$$\begin{align}x^7+x^2+1&=x(x^6-1)+x^2+x+1\\&=x(x^3-1)(x^3+1)+x^2+x+1\\&=(x^2+x+1)(x(x^3-1)(x+1)+1).\end{align}$$
| {
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How to know which values of $x$ can be used to find the unknowns $A$, $B$ and $C$? I'm having trouble understanding what my maths textbook says, and I've not been able to find help. I would appreciate an explanation in simple (high-school) terms.
Worked example
Question
Given that $5x^2 - 7x + 3 = A(x - 1)(x - 2) + B(x - 1) + C$ for all values of $x$, find the values of $A$, $B$ and $C$.
Answer
Since $5x^2 - 7x + 3 = A(x - 1)(x - 2) + B(x - 1) + C$ for all values of $x$, we can select any value of $x$ to find the unknowns $A$, $B$ and $C$.
Let $x$ = $1$, then $5(1)^2 - 7(1) + 3 = A(1 - 1)(1 - 2) + B(1 - 1) + C$,
i.e. $C = 1$.
Let $x$ = $2$, then $5(2)^2 - 7(2) + 3 = A(2 - 1)(2 - 2) + B(2 - 1) + C$,
i.e. $B = 8$.
Let $x$ = $0$, then $5(0)^2 - 7(0) + 3 = A(0 - 1)(0 - 2) + B(0 - 1) + C$,
i.e. $A = 5$.
$\therefore$ The values of $A$, $B$ and $C$ are $5$, $8$ and $1$ respectively.
Question
*
*Why where the values $x = 0, 1, 2$ chosen?
*Can we use other numbers such as $x = -3, 5$ and $7$ to find the unknowns?
I'm not sure about the answers to the two questions above.
For 1, I think the answer is that since any value of $x$ could be chosen, these were 3 random values.
I think that the answer to question 2 is also yes, but I was unable to find the values of $A$, $B$ and $C$ when $x = -3, 5$ and $7$.
Please could someone tell me if I'm right, or if not, where my reasoning is wrong?
| For 1, the values $1$ and $2$ are chosen such that $(x-1)=0$ and $(x-2)=0$ respectively. This simplifies the calculations. $x=0$ simplifies the polynomial on the left hand side.
For 2, you can with $-3,5, 7$, but you need to solve the system:
$$\begin{cases}
69=20A-4B+C\\
93=12A+4B+C\\
199=30A+6B+C
\end{cases}$$
which is harder to solve than
$$\begin{cases}
3=2A-B+C\\
1=C\\
9=B+C
\end{cases}$$
obtained by using the textbook's choice $0,1,2$.
EDIT: To solve the first system, we subtract the first and second equations from the third, obtaining the equations:
$$\begin{cases}
130=10A+10B\\
106=18A+2B
\end{cases}$$
Now with only two unknowns, it is easy to show that $A=5$, $B=8$.
Finally substitute the values back to any equation and we will have $C=1$, as desired.
| {
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"url": "https://math.stackexchange.com/questions/3842516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exploitation of Equality in Inequality. I want to ask:
How do I exploit equalities in the solution of an inequality when equality holds for variables equal to each other?
One can take this example:
For positive real numbers that sum to $3$, prove that$$\sqrt{a + \sqrt{b^2 + c^2}} +
\sqrt{b + \sqrt{c^2 + a^2}} +
\sqrt{c + \sqrt{a^2 + b^2}} \ge
3\sqrt{\sqrt{2} + 1}$$
As this one which is a little tough for me, one can almost easily say that (though I am not so sure)
Equality holds for $a=b=c=1$.
But how do I exploit this in the solution? So far, I have squared both sides, but ended up only with a giant square root term, and still I can not see what can I do with this equality case.
This is not only for this one, but every time I solve an inequality and in the solution see that the equality holds for variables equal to each other, I can't still observe what happened in my solution from this equality.
Any explanation will be thankfully welcome!
| The first step does not change something:
$$\sum_{cyc}\left(a+\sqrt{b^2+c^2}+2\sqrt{(a+\sqrt{b^2+c^2})(b+\sqrt{a^2+c^2})}\right)\geq9\sqrt2+9$$ or
$$\sum_{cyc}\left(\sqrt{b^2+c^2}+2\sqrt{(a+\sqrt{b^2+c^2})(b+\sqrt{a^2+c^2})}\right)\geq9\sqrt2+6.$$
The second step:
By C-S $$\sqrt{b^2+c^2}=\frac{1}{\sqrt2}\sqrt{(1^2+1^2)(b^2+c^2)}\geq\frac{b+c}{\sqrt2}$$ saves the case of the equality occurring: $a=b=c=1$.
It's enough to prove that:
$$\sum_{cyc}\left(\sqrt{a^2+b^2}+2\sqrt{\left(a+\frac{b+c}{\sqrt2}\right)\left(b+\frac{a+c}{\sqrt2}\right)}\right)\geq9\sqrt2+6.$$
The rest is the same: any step saves the case of the equality occurring, which gives the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Approximating $\int_0^1 x\sin xdx$ Check whether $$\int_0^1 x\sin xdx>0.3$$
The anti-derivative of $x\sin x$ is $\sin x-x\cos x$
So the integral is $$I\approx\sin(57^\circ)-\cos(57^\circ)$$
Since $\sin x $ is increasing in the first quadrant and $\cos x$ is decreasing,
$$I<\sin(60^\circ)-\cos(60^\circ)$$
$$I<\frac{\sqrt3-1}{2}$$
$$I<0.37$$
Which also means $$I>0.3$$
Is there any other, preferably graphical, method?
| $I < 0.37$ does not necesarrily imply that $I > 0.3$. By Taylor series again, you would have:
$$I = \sin 1 - \cos 1 \approx 1 - \frac{1^3}{3!} + \frac{1^5}{5!} - \frac{1^7}{7!} - \left(1 - \frac{1^2}{2!} + \frac{1^4}{4!} - \frac{1^6}{6!}\right) = \frac{253}{840} \approx 0.3012.$$
and the maximum error of a Taylor series is given by the next term:
$$\text{Maximum error} = \left|\frac{1}{9!}\right| + \left|\frac{1}{8!}\right| = \frac{1}{36288}$$
but $\frac{253}{840} - \frac{1}{36288} > 0.3$. Hence $\sin 1 - \cos 1 = \int_0^1 x \sin x \ dx > 0.3$.
The series approximation gives $0.3011629 \cdots$, where the actual value of $\sin 1 - \cos 1$ is $0.3011687 \cdots$, off by only $6 \cdot 10^{-6}$.
| {
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Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$
*
*Using AM-GM Inequality, we have:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$
*
*Therefore, $\min P = -8$
Is this solution correct? If not, then why?
| Let $x = a - b, y = b-c, z = c-a$. Then we have $x^2 + y^2 + z^2 = 12$ and $x + y + z = 0$.
Also, we have $xy + yz + zx = \frac{(x+y+z)^2 - (x^2+y^2+z^2)}{2} = -6$.
We need to find the minimum of $xyz$.
Let $p = x + y + z = 0, q = xy + yz + zx = -6$ and $r = xyz$.
It is easy to prove that
$$-4p^3r+p^2q^2+18pqr-4q^3-27r^2 = (x-y)^2(y-z)^2(z-x)^2 = 27(32-r^2)\ge 0$$
which results in $-4\sqrt{2} \le r \le 4\sqrt{2}$. Also, when $x = \sqrt{2}, y = -2\sqrt{2}, z= \sqrt{2}$ (e.g. $a = -\sqrt{2}, b= -2\sqrt{2}, c = 0$),
we have $xyz = -4\sqrt{2}$. Thus, the minimum of $xyz$ is $-4\sqrt{2}$.
Remark: Actually, $-4p^3r+p^2q^2+18pqr-4q^3-27r^2$ is the discriminant of $u^3 - pu^2 + qu - r = 0$.
| {
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Integral $\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt$ I am preparing for a master's degree entrance exam. One of the questions from the past exam asks to show the following formula. Could someone provide some hint?
$$\int_0^\infty \frac{t\sin{tx}}{(1+t^2)(4+t^2)}dt = \frac{\pi}{6} (e^{-x} - e^{-2x})$$ for $x \geq 0.$
Thank you!
| 1st Solution. Following @Claude Leibovici's suggestion, let us utilize complex-analytic technique. First, symmetrize the integral to write
$$ I
= \frac{1}{2}\int_{-\infty}^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}\,\mathrm{d}t
= \frac{1}{2}\operatorname{Im}\biggl(\int_{-\infty}^{\infty}\frac{te^{itx}}{(1+t^2)(4+t^2)}\,\mathrm{d}t\biggr). $$
Now let $R > 2$, and let $\mathcal{C}_R = L_R \cup \Gamma_R $ be the contour where
*
*$L_R = [-R, R]$ is the line segment traced from left to right, and
*$\Gamma_R = \{ Re^{i\theta} : 0 \leq \theta \leq \pi \}$ is the upper semicircular arc traversed counter-clockwise.
Then by the residue theorem, for $x \geq 0$,
\begin{align*}
\int_{\mathcal{C}_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z
&= 2\pi i \left( \mathop{\underset{z=i}{\mathrm{Res}}} \frac{ze^{izx}}{(1+z^2)(4+z^2)} + \mathop{\underset{z=2i}{\mathrm{Res}}} \frac{ze^{izx}}{(1+z^2)(4+z^2)}\right) \\
&= 2\pi i \left( \left. \frac{ze^{izx}}{(2z)(4+z^2)} \right|_{z=i} + \left. \frac{ze^{izx}}{(1+z^2)(2z)} \right|_{z=2i} \right) \\
&= \frac{\pi i}{3} \left( e^{-x} - e^{-2x} \right).
\end{align*}
On the other hand,
$$ \left| \int_{\Gamma_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z \right|
\leq \int_{\Gamma_R} \frac{R}{(R^2 - 1)(R^2 - 4)} \, \left|\mathrm{d}z\right|
= \frac{\pi R^2}{(R^2 - 1)(R^2 - 4)} $$
and so, the contour integral along $\Gamma_R$ vanishes as $R\to\infty$. Therefore
\begin{align*}
I
&= \frac{1}{2}\operatorname{Im}\biggl(\lim_{R\to\infty} \int_{L_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z \biggr) \\
&= \frac{1}{2}\operatorname{Im}\biggl(\lim_{R\to\infty} \int_{\mathcal{C}_R} \frac{ze^{izx}}{(1+z^2)(4+z^2)} \, \mathrm{d}z \biggr) \\
&= \frac{\pi}{6} \left( e^{-x} - e^{-2x} \right).
\end{align*}
2nd Solution. Using the decomposition $ \frac{t}{(1+t^2)(4+t^2)} = \frac{4}{3t(4+t^2)} - \frac{1}{3t(1+t^2)} $, we may write
$$ I = \frac{4}{3} \int_{0}^{\infty} \frac{\sin(xt)}{t(4+t^2)} \, \mathrm{d}t - \frac{1}{3} \int_{0}^{\infty} \frac{\sin(xt)}{t(1+t^2)} \, \mathrm{d}t. $$
In order to compute this, let $a>0$ and consider
$$ J(x) := \int_{0}^{\infty} \frac{\sin(xt)}{t(a^2+t^2)} \, \mathrm{d}t. $$
Then by direct computation, we find that $J$ solves
$$ J''(x) - a^2 J(x) = - \int_{0}^{\infty} \frac{\sin(xt)}{t} \, \mathrm{d}t = -\frac{\pi}{2}. $$
Among the general solution $ \frac{\pi}{2a^2} + c_1 e^{ax} + c_2 e^{-ax} $ of this ODE, the unique solution that satisfies $J(0) = 0$ and remains bounded as $x\to\infty$ is
$$ J(x) = \frac{\pi}{2a^2}(1 - e^{-ax}). $$
Therefore
$$ I = \frac{\pi}{6}(1 - e^{-2x}) - \frac{\pi}{6}(1 - e^{-x}) = \frac{\pi}{6}(e^{-x} - e^{-2x}). $$
| {
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Show that $-1^2+3^2-5^2\mp ...+(2^n-1)^2=2^{2n-1}$ Playing with numbers, I construct following expression.
Can it be shown that
$$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$
attempt
We can construct following, using finite calculus as
$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$
$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$
Let $4k-1=2^n-1$ so we can write above claim as
$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$
Which is equivalent to show
$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$
Here I'm stuck. Thanks
| Hint: What is $\sum_{k=1}^N k^2 i^k$, where $i^2=-1$? Then take its imaginary part.
Solution:
The sum in question is $-\sum_{k=1}^N k^2 i^k=\frac12 ((i-1)i^N N^2 -2i^N-i^{N+1}+i)$.
When $N$ is a multiple of $4$ we have $i^4=1$ and so the imaginary part is $\frac12 N^2$.
When $N=2^{n}$ (so that the last term is $2^n-1$), the sum is thus $\frac12 2^{2n} = 2^{2n-1}$.
| {
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Geometry problem, finding missing angles One of my students showed me a problem that she says is similar to what they would do in high school in her home country (which i am attaching here.
. The goal of the problem is to find the measure of $\angle DEC$ using the given angle measures provided. I've tried working on this to see what other angle measures I could deduce, and I'm including that here This is where I'm stuck. I've tried:
*
*Labeled one unknown angle as x and determined all other unknown angles in terms of $x$, but it's fully consistent and nothing seems to simplify to indicate what $x$ is.
*Drawn in lines parallel to the sides through various points, and use what I know about parallel lines cut by transversals, but it doesn't seem to get me any closer to the target.
I suspect I might need to draw in some additional line or extend the diagram in some way but I can't figure out what. Any help would be greatly appreciated. Thanks!
| Here's the proof with sine rule. This will be the shortest solution, but it is inelegant, as is the case with all "World's Hardest Easy Geometry Problem" type problems. And as with all those questions, there will be a nice solution, (hopefully.) Consider the "simplified" diagram below.
I messed up the naming so we are looking for $\theta = \angle EDC$.
In $\triangle CDE: \dfrac {EC}{\sin \theta} = \dfrac {DC}{\sin (180^\circ - 80^\circ - \theta)} = \dfrac {DC}{\sin (80^\circ + \theta)}$
In $\triangle ACE: \dfrac {EC}{\sin 40^\circ} = \dfrac {AC}{\sin 80^\circ}$
In $\triangle ACD: \dfrac {AC}{\sin 150^\circ} = \dfrac {DC}{\sin 10^\circ}$
Hence:
$$\frac {\sin (80^\circ + \theta)}{\sin \theta} = \frac {DC}{EC} = \frac {AC \sin 10^\circ}{\sin 150^\circ} \cdot \frac {\sin 80^\circ}{AC \sin 40^\circ} = \frac {\sin 10^\circ \sin 80^\circ}{\sin 150^\circ \sin 40^\circ}$$
$$\frac{\sin (80^\circ + \theta)}{\sin \theta} = \frac {\sin 80^\circ \cos \theta + \cos 80^\circ \sin \theta}{\sin \theta} = \sin 80^\circ \cot \theta + \cos 80^\circ$$
Thus we have:
\begin{align}\theta &= \cot^{-1} \left(\frac {\sin 10^\circ}{\sin 150^\circ \sin 40^\circ}-\cot 80^\circ\right)\\
& = \cot^{-1} \left(\frac {\sin 10^\circ}{\sin 30^\circ \sin 40^\circ}-\tan 10^\circ\right)\\
& = \cot^{-1} \left(\frac {2\sin 10^\circ}{4\sin10^\circ \cos 10^\circ \cos 20^\circ}-\frac{\sin 10^\circ}{\cos 10^\circ}\right)\\
& = \cot^{-1} \left(\frac {1}{2\cos 10^\circ \cos 20^\circ}-\frac{\sin 10^\circ \cos 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\
& = \cot^{-1} \left(\frac {\sin 30^\circ - \sin 10^\circ \cos 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\
& = \cot^{-1} \left(\frac {\cos 10^\circ \sin 20^\circ}{\cos 10^\circ \cos 20^\circ}\right)\\
& = \cot^{-1} \tan 20^\circ\\
& = 70^\circ
\end{align}
The last steps are only possible if we know in advance the solution is nice.
| {
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How can we prove that there are no other integers with $\phi(n)=2$ besides 3,4,6? Where $\phi(n)$ is Euler's phi function that counts the number of relatively prime integers less than or equal to n.
I've been able to compute that 3,4,6 all have only 2 relatively prime integers less than or equal to them, however I'm unsure how to prove that there are indeed no others. While I am certain that this is the case, how can this be proved rigorously?
| First prove that if $n = \prod p_i^{k_i}$ is the unique prime factorization of $n$ then $\phi(n) = \prod p_i^{k_i - 1} \prod (p_i - 1)$. You should have already proven this. The statement is totally equivalent to: 1) if $\gcd(a,b)= 1$ then $\phi(ab) = \phi(a)\phi (b)$ and $\phi(p^k)=p^{k-1}(p-1)$ if $p$ is prime and 2) $\phi( n )= n\prod_{p|n}(1-\frac 1p)$.
Then we have $2 = \prod p_i^{k_i -1} \prod (p_i-1)$. The only way that can happen is if either i) $\prod p^{k_i-1} = 2$ and $\prod (p_i-1) = 1$ or if ii) $\prod p_i^{k_i-1} = 1$ and $\prod (p_i-1) = 2$.
If i) then $\prod (p_i-1) = 1$ imples $\{p_i\} = \{2\}$ and $\prod p^{k_i-1} = 2^1$ implies $\{p_i\} = 2$ and $k_i = 2$ so $n = 2^2 = 4$.
if ii) then $\prod p_i^{k_i-1} = 1$ implies $k_i =1$ of all the $p_i$ and $n = \prod p_i$, a square-free number. Then $\prod (p_i - 1) =2$ implies that one of the $p_i-1 = 2$ and all the other $p_j$ (if any) are $p_j -1 = 1$. One of the prime factors is $3$ and if there is any other prime factor it can only be $2$. but there need not be any other prime factor. So we could have $n=3$ or $n=2\cdot 3 = 6$.
....
If that was too hand-wavy, here is a coffin with a few dozen nails:
Suppose $p> 3$ is a prime divisor of $n=\prod p_i^{k_i}$. Then $\phi(n) = \prod p_i^{k_i - 1} \prod (p_i - 1)$ so $p-1|\phi(n)$. But $p-1 > 2$ so $\phi(n) > 2$. So if $\phi(n) = 2$ then $n$ has no prime divisors greater than $3$.
So $n = 2^a$ or $n = 3^b$ or $n=2^a3^b$ or $n = 1$.
If $n = 2^a$ then $\phi(n) = 2^{a-1}(2-1) = 2^{a-1}=2$. So $a-1 =1$ and $a=2$ and $n = 2^2 = 4$.
If $n=3^b$ then $\phi(n) = 3^{b-1}(3-1) = 2\cdot 3^{b-1} =2$. So $b-1 =0$ and $b=1$ and $n = 3^1=3$.
If $n = 2^a3^b$ then $\phi(n) = 2^{a-1}3^{b-1}(2-1)(3-1) = 2\cdot 2^{a-1}3^{b-1} = 2^a3^{b-1} =2$. So $a =1$ and $b-1 =0$ and $b=1$ and $n = 2^1\cdot 3^1 = 6$.
And of course if $n= 1$ then $\phi(n)=\phi(1) =1\ne 2$
So $4,3,6$ are the only three options for $\phi(n) =2$.
| {
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Prove that $\sum\limits_{j=0}^k\,j\,\binom{n}{j}\,\binom{n-j}{2k-2j}\,2^{2k-2j}=n\binom{2n-2}{2k-2}$ I encountered this in my homework. I derived two ways to solve the problem and the answer which I have tested using programming, seem to be the same, but I am not sure how to prove this equation.
Let $n$ and $k$ be nonnegative integers with $k\leq n$. Prove that $$\sum\limits_{j=0}^k\,j\,\binom{n}{j}\,\binom{n-j}{2k-2j}\,2^{2k-2j}=n\binom{2n-2}{2k-2}\,.$$
The original problem is the following:
A shoe rack has n pairs of shoes. Of those, 2k individual shoes are chosen at random, k ≤ n. Calculate the expected number of matching shoes among 2k chosen shoes.
The left hand side is from directly calculating expectation, while the right hand side is using sum of indicator variables of each pair being chosen. The expectation is just the equation divided by $\binom{2n}{2k}$.
| Using coefficient extractors we present a minor variation and seek to
prove
$$\sum_{j=1}^k {n-1\choose j-1} {n-j\choose 2k-2j} 2^{2k-2j}
= {2n-2\choose 2k-2}$$
or alternatively
$$\sum_{j=0}^{k-1} {n-1\choose j} {n-j-1\choose 2k-2j-2} 2^{2k-2j-2}
= {2n-2\choose 2k-2}.$$
The LHS is
$$\sum_{j=0}^{k-1} {n-1\choose j} 2^{2k-2j-2}
[z^{2k-2j-2}] (1+z)^{n-j-1}
\\ = 2^{2k-2} [z^{2k-2}] (1+z)^{n-1}
\sum_{j=0}^{k-1} {n-1\choose j} (1+z)^{-j} z^{2j} 2^{-2j}.$$
The coefficient extractor enforces the upper limit of the sum:
$$ 2^{2k-2} [z^{2k-2}] (1+z)^{n-1}
\sum_{j\ge 0} {n-1\choose j} (1+z)^{-j} z^{2j} 2^{-2j}
\\ = 2^{2k-2} [z^{2k-2}] (1+z)^{n-1}
\left(1+\frac{z^2}{4(1+z)}\right)^{n-1}
\\ = 2^{2k-2} [z^{2k-2}]
\left(1+z+\frac{z^2}{4}\right)^{n-1}
= [z^{2k-2}]
\left(1+2z+z^2\right)^{n-1}
\\ = [z^{2k-2}] (1+z)^{2n-2} = {2n-2\choose 2k-2}.$$
This is the claim.
| {
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Proving $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$ using definition I got a question regarding my answer of proving limit using epsilon-delta, here's the question
Prove $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$
Here's the answer I've come up so far
let $f(x) = \frac{x+1}{x-2} + x$
by algebra manipulation, we get
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1| $ $=|\frac{x^2 - 1}{x-2}|$ $=|\frac{(x-1)(x+1)}{x-2}|$
let $|x-1| < 1$, by triangle inequality we get $|x| < 2$, then
$|x + 1| < 3$ and $|x - 2| < 1$
now, using the definiton of limit,
for every $\epsilon > 0$, there exist $\delta = min\{1, \frac{\epsilon}{3}\}$ such that
if $0 < |x - 1| < \delta$ then,
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1|$ $=|\frac{(x-1)(x+1)}{x-2}|$ $=|\frac{1 \cdot 3}{1}|$ $< \epsilon$
Is this correct? honestly I'm not sure on getting the upper bound of $|x-2|$, so I used the assumption of $|x-1| < 1$
Any tips would help, thanks beforehand.
| Let $\epsilon > 0$. Then choose $\delta < \text{min}\{1,\epsilon/2\}$. If $|x-1|<\delta$, then we also have that $|x-2|> 1$ and $|x+1|<2$. Then
$$
|f(x)-1| = \frac{|x-1|\cdot|x+1|}{|x-2|}< \frac{\delta|x+1|}{|x-2|}< \frac{2\delta}{|x-2|}<\frac{2\delta}{1} < \epsilon
$$
| {
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Find the Condition Number for $w\left(a+b\right)=a^2+b^2$ This is what I got:
$$w\left(a+b\right)=a^2+b^2$$
$$ \alpha =a\left(1+E_1\right),\:\beta \:=b\left(1+E_2\right)$$
$$ \frac{\left|w\left(\alpha ,\beta \right)-w\left(a,b\right)\right|}{\left|w\left(a,b\right)\right|}=\frac{\left|a^2\left(1+E_1\right)^2+b^2\left(1+E_2\right)^2-\left(a^2+b^2\right)\right|}{\left|a^2+b^2\right|}=\frac{\left|a^2\left(2E_1+E_1^2\right)+b^2\left(2E_2+E_2^2\right)\right|}{\left|a^2+b^2\right|} $$
Let $E\::=max\left(E_1,\:E_2\right)$, then:
$$\le \frac{\left|a^2\left(2E+E^2\right)+b^2\left(2E+E^2\right)\right|}{\left|a^2+b^2\right|}=2E+E^2$$
So we got:
$$ \frac{\left|w\left(\alpha ,\beta \right)-w\left(a,b\right)\right|}{\left|w\left(a,b\right)\right|} \le 2E+E^2 $$
For $E\::=max\left(E_1,\:E_2\right)$
But I don't know what to do further on, how to actually find $cond\left(w,\:a,\:b\right)$.
I know that the basic definition is:
$$\frac{\left|\phi \left(d+\Delta d\right)-\phi \left(d\right)\right|}{\left|\phi \left(d\right)\right|}\le cond\left(\phi ,\:d\right)\:\frac{\left|\Delta d\right|}{\left|d\right|}$$
But what I got is completely independent of our "d" here (a and b). And also, in this example, how would $\frac{\left|\Delta d\right|}{\left|d\right|}$ look like?
| As $w$ is bivariate the inequality becomes $$\frac{|w(a+\epsilon_a,\beta+\epsilon_b)-w(a,b)|}{|w(a,b)|}\le\kappa(a,b)\frac{\|(\epsilon_a,\epsilon_b)\|}{\|(a,b)\|}$$ where $\epsilon_a,\epsilon_b>0$ and $\|\cdot\|$ is a norm on the codomain of $w$ (which is $\Bbb R_{\ge0}$).
As $\epsilon_a,\epsilon_b\to0^+$, taking the Euclidean norm yields \begin{align}\kappa(a,b)&\ge\frac{|\epsilon_a^2+\epsilon_b^2+2(a\epsilon_a+b\epsilon_b)|}{|a^2+b^2|}\sqrt{\frac{a^2+b^2}{\epsilon_a^2+\epsilon_b^2}}\end{align} as a necessary condition. Note that \begin{align}\frac{|\epsilon_a^2+\epsilon_b^2+2(a\epsilon_a+b\epsilon_b)|}{|a^2+b^2|}\sqrt{\frac{a^2+b^2}{\epsilon_a^2+\epsilon_b^2}}&=\frac1{\sqrt{a^2+b^2}}\left|\frac{\epsilon_a^2+\epsilon_b^2+2(a\epsilon_a+b\epsilon_b)}{\sqrt{\epsilon_a^2+\epsilon_b^2}}\right|\\&\le\frac1{\sqrt{a^2+b^2}}\left(\sqrt{\epsilon_a^2+\epsilon_b^2}+2\left|\frac{a\epsilon_a+b\epsilon_b}{\sqrt{\epsilon_a^2+\epsilon_b^2}}\right|\right)\\&\le2\cdot\frac{\max\{|a|,|b|\}}{\sqrt{a^2+b^2}}+E\end{align} for a small $E>0$. Therefore, choosing $$\kappa(a,b)=2\cdot\frac{\max\{|a|,|b|\}}{\sqrt{a^2+b^2}}+1$$ suffices.
It does not matter whether we use $\epsilon_a$ or $E_a:=a\epsilon_a$ as both tend to zero as $\epsilon_a\to0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need to proof by induction (calculus1 n and n+1) what needs to be proven
I need to prove that for every natural n that happens and the prove needs to be by induction,
I wont prove for n=1 because its obvious,so lets move to n and n+1
$$ \frac{(2n!)}{(n!)^2} ≥ \frac{(4^n)}{(2n+1)} $$
the assumption for n+1
$$ \frac{(2n+2)!}{(n+1)!^2} ≥ \frac{(4^{n+1})}{2n+3} $$
now I dont now how to progress,I tried this for n+1
$$ \frac{(2n+2)(2n+1)(2n!)}{(n+1)!^2} ≥ \frac{(4^{n}4^1)}{(2n+1)+2)} $$
and had a suggestion to use
$ 2n! = (n!) \prod _{n=n}^{2n}\:n\cdot )! $
but cant see how it helps,ty.
| $$\frac{(2n+2)!}{((n+1)!)^2}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+1)(n!)^2}=\frac{2(2n+1)(2n)!}{(n+1)(n!)^2}\geq\frac{4^{n+1}}{2n+3}$$
Assuming that $\dfrac{(2n)!}{(n!)^2}\geq\dfrac{4^{n}}{2n+1}$
Let's show
$$\frac{2(2n+1)}{n+1}\frac{4^n}{2n+1}>\frac{4^{n+1}}{2n+3}$$
$$\frac{2}{n+1}>\frac{4}{2n+3}$$
$$4n+6>4n+4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$ The question is this:
If $a\ge b\ge c\ge 0$ and $a^2+b^2+c^2=3$, then prove that
$$abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$$
For my work on this inequality, I have proved already under constraints that it is true.
Proof for: $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0.$
$$
\sqrt{3}abc +
\sqrt{2}a -
\sqrt{3} -
\sqrt{2}c
\geqslant 0
$$
$$
a\left(
\sqrt{3}bc +
\sqrt{2}
\right)
+ (-1)\left(
\sqrt{3} +
\sqrt{2}c
\right) \geqslant 0
$$
$$
(1 + 1)(a\left(
\sqrt{3}bc +
\sqrt{2}
\right)
+ (-1)\left(
\sqrt{3} +
\sqrt{2}c
\right)) \geqslant 0
$$
By Chebyshev,
$$
(a - 1)
(\sqrt{3}bc +
\sqrt{2} +
\sqrt{3} +
\sqrt{2}c
)\geqslant0
$$
$$
a \geqslant 1
$$
Chebyshev Inequality requires the sequences to be monotonous. As $a+1>0$, we need to have the other sequence also in the same order, hence the condition: $\sqrt{3}bc + \sqrt{2} \geqslant\sqrt{3} + \sqrt{2}c$. The sequences are $(a,-1)$ and $(\sqrt{3}bc + \sqrt{2} ,\sqrt{3} + \sqrt{2}c)$.
I have tried another way but that was untrue. I have reached this far. The constraint $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0$ isn't true always. Try $(a,b,c) = (\sqrt{3},0,0)$.
Thanks for extensions or other solutions too are welcome!
| Let $f(a,b)=abc+(a-c)k$ where $c^2=3-a^2-b^2$ and $k=\sqrt{2/3}$. Assuming $c\ne0$, $$f_a=bc+(1-c_a)k=0$$ for critical points. Now $c_a=-a/c$ so $bc^2+(c+a)k=0$, contradicting $c\ne0$. This means that either $c=0$, or any other solutions must lie on the boundaries of the constraints, which are:
*
*$a=b$ which yields $f(a)=(a^2-k)\sqrt{3-2a^2}+ak$;
*$b=c$ which yields $f(a)=a(3-a^2)/2+(a-\sqrt{(3-a^2)/2})k$.
When $c=0$ we have $a^2+b^2=3$ such that $a\ge b$ so $a\ge\sqrt{3/2}$ and $f(a,b)=0+ak\ge1$.
For the first case we have $a\ge c\implies a\ge1$ so the domain of $f$ is $[1,\sqrt{3/2}]$. Notice that $f(1)=f(\sqrt{3/2})=1$ and $f(a)-1$ is positive. Similarly, for the second case we also have $a\ge1$. Note that $f(1)=1$ and calculus yields $f(a)\ge1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Use the division algorithm to establish that, The cube of any integer is either of the form $9k ,9k + 1, 9k + 8$.
Use the division algorithm to establish that, The cube of any integer is either of the form $9k ,9k + 1, 9k + 8$.
Let $a$ is an integer, write $a = 9k + r, 0 \le r < 9$ hence $r = \{0,1,2,3,4,5,6,7,8\}$ then $$a^3 =(9k + r)^3 = 9(9k^3 + 3kr(9k +r )) + r^3, 0 \le r³ < 9$$
when $r=0 \to r^3 = 0$
when $r=1 \to r^3 = 1$
when $r=2 \to r^3 = 8$
when $r=3 \to r^3 = 27$
Since $0\le r^3 < 9$ above $8$ values cannot be accepted. Hence $r^3\in{0,1,8}$. Hence $a^3$ can express in $a^3 = 9k, a^3 = 9k + 1, a^3 = 9k + 8$ forms. Therefore cube of any integer is form $9k, 9k + 1,$ or $9k + 8$.
Is this correct? Are there other solutions?
| *
*$a=3n\Longrightarrow a^3=(3n)^3=9\cdot(3n^3)\equiv0\pmod 9$
*$a=3n+1\Longrightarrow a^3=(3n)^3+3(3n)^2+3(3n)+1=9\cdot(3n^3+3n^2+n)+1\equiv1\pmod9$
*$a=3n+2\Longrightarrow a^3=(3n)^3+6(3n)^2+12(3n)+8=9\cdot(3n^3+6n^2+4n)+8\equiv8\pmod9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for every natural number $n$ there exists some power of 2 whose final $n$ digits are all ones and twos. Here's the problem :
Prove that for every natural number $n$ there exists some power of $2$ whose final $n$ digits are all ones and twos.
My attempt :
Since the final digit of a power of $2$ can not be $1$ , it has to be $2$ , which happens for numbers of the form $2^{4k+1}$.
For the second last digit , it has to be $1$ , as the number would be divisible by $4$ (for $n\ge 2$). But I couldn't observe any fixed pattern for that .
I am not sure whether this approach is taking me anywhere towards to solution at all .
Could someone please help me solve this problem ?
Thanks!
| Lemma : For any positive integer $x$ with $n$ digits (leading zeroes allowed), $x$ is the last $n$ digits of infinitely many powers of $2$ if and only if $2^n \mid x$ and $5 \nmid x$.
Proof of Lemma : The only if condition is trivial. For arbitrarily large powers of $2$, we must have $2^n$ as a factor, and thus we need $2^n \mid x$. Moreover, no power of $2$ is divisible by $5$, and hence $5 \nmid x$. Next, we count the number of $x$ that are the last $n$ digits of infinitely many powers of $2$. We can see that starting from $2^n$, all powers of $2$ have last $n$ digits divisible by $2^n$. By the pigeonhole principle, the last $n$ digits of powers of $2$ starting from $2^n$ must be a periodic sequence. Thus, the period must be $k-n$, where $k$ is the smallest positive integer $>n$ such that $2^k \equiv 2^n \pmod{10^n}$. This is the same as $2^{k-n} \equiv 1 \pmod{5^n}$. By Lifting the Exponent Lemma, the smallest such $k-n$ is:
$$k-n=4 \cdot 5^{n-1}$$
and thus, this is the period. Thus, there are $4 \cdot 5^{n-1}$ strings of last $n$ digits that occur infinitely often as last $n$ digits of powers of $2$.
To prove the if condition, it suffices to show that the number of $x$ such that $2^n \mid x$ and $5 \nmid x$ is also $4 \cdot 5^{n-1}$. Since $2^n \mid x$, we must have $x=2^nq$ for $q <5^n$. Since $q$ is any non-negative integer coprime to $5$, we have $4 \cdot 5^{n-1}$ choices, as required.
Now, it suffices to show that we can use $1$s and $2$s as the last $n$ digits to form a number divisible by $2^n$ but not by $5$. The last part is obvious since the last digit is only $1$ or $2$. For the first part, we use induction. The base case is trivial. Now, if you can fill last $n$ digits to be divisible by $2^{n}$, let us say the digits are $x$, we can either have $10^n+x$ or $2 \cdot 10^n + x$ as the last $n+1$ digits. We can see that both these numbers are incongruent modulo $2^{n+1}$ but are divisible modulo $2^n$. Hence, one of them must be divisible by $2^{n+1}$, as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sum _ { n = 1 } ^ { \infty } n \binom {2 n} {n}( - z )^n$ According to wolframalpha,
$$\sum _ { n = 1 } ^ { \infty } n \binom {2 n} {n} ( - z )^n = - \frac { 2 z } { ( 4 z + 1 ) ^ { 3 / 2 } } \text { when } 4 | z | < 1$$ However I have been unable to find how to get this result.
| $$n\binom{2n}n=n\cdot\dfrac{(2n!)}{n! n!}=2\binom{2n-1}{n-1}$$
Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $.
comparing the expansion of $(1+x)^m$ with $$2\sum_{n=1}^\infty\binom{2n-1}{n-1} (-z)^{n-1}$$
$$mx=2\binom31(-z)^{2-1}\text{ and }\dfrac{m(m-1)}2x^2=2\binom52(-z)^{3-1}$$
to find $m=-\dfrac32, x=4z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using an auxiliary triangle to find the range of a trig function such as $f(x)=a\sin x\pm b\cos x$ I read online that when finding range of a trigonometric function such as
$$f(x) = a\sin x \pm b\cos x$$
as shown in the image, we make a triangle and assign those values of constant in it.
*
*I did not understand that why is $a$ the base and $b$ perpendicular, and why not the opposite way.
*How did we predict by looking at that equation #3 and say the range?
| You can also do the other way as you stated.
Now,
$\cos\theta = \frac{b}{\sqrt{a^2+b^2}}$ and $\sin\theta = \frac{a}{\sqrt{a^2+b^2}}$
$\Rightarrow a= \sqrt{a^2+b^2}\sin\theta$ and $b= \sqrt{a^2+b^2}\cos\theta$
$\Rightarrow f(x) = \sqrt{a^2+b^2}\left[\sin\theta\sin x \pm \cos\theta\cos x\right] = \sqrt{a^2+b^2}\left[\mp\cos(x\pm\theta)\right]$
The range of $\cos\alpha $ is $[-1,1]$
So, range of $f(x)$ is $[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}] $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying expression involving inverse trigonometric functions ${\left( {\frac{1}{{{y^2}}}{{\left( {\frac{{\cos \left( {{{\tan }^{ - 1}}y} \right) + y\sin \left( {{{\tan }^{ - 1}}y} \right)}}{{\cot \left( {{{\sin }^{ - 1}}y} \right) + \tan \left( {{{\sin }^{ - 1}}y} \right)}}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} = \_\_\_\_\_$.
The official answer is $1$
My approach is as follow
${\tan ^{ - 1}}y = \gamma ;{\sin ^{ - 1}}y = \lambda \Rightarrow \tan \gamma = y\& \sin \lambda = y$
$ \Rightarrow {\left( {\frac{1}{{{y^2}}}{{\left( {\frac{{\cos \gamma + y\sin \gamma }}{{\cot \lambda + \tan \lambda }}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {\frac{1}{{{y^2}}}{{\left( {\frac{{\cos \gamma + \tan \gamma \sin \gamma }}{{\cot \lambda + \tan \lambda }}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} \Rightarrow {\left( {\frac{1}{{{{\tan }^2}\lambda }}{{\left( {\frac{1}{{\cos \gamma \left( {\cot \lambda + \tan \lambda } \right)}}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {{{\left( {\frac{1}{{\cos \gamma \left( {1 + {{\tan }^2}\lambda } \right)}}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} \Rightarrow {\left( {{{\left( {\frac{{{{\cos }^2}\lambda }}{{\cos \gamma }}} \right)}^2} + {y^4}} \right)^{\frac{1}{2}}} \Rightarrow {\left( {\frac{{{{\cos }^4}\lambda }}{{{{\cos }^2}\gamma }} + {y^4}} \right)^{\frac{1}{2}}}$
$ \Rightarrow \tan \gamma = y\& \sin \lambda = y$
$ \Rightarrow {\left( {\frac{{{{\cos }^4}\lambda }}{{{{\cos }^2}\gamma }} + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}}$
$ \Rightarrow \tan \gamma = \sin \lambda \Rightarrow {\tan ^2}\gamma = {\sin ^2}\lambda \Rightarrow {\sec ^2}\gamma - 1 = {\sin ^2}\lambda \Rightarrow \frac{1}{{{{\cos }^2}\gamma }} = 1 + {\sin ^2}\lambda $
$ \Rightarrow {\left( {\left( {1 + {{\sin }^2}\lambda } \right){{\cos }^4}\lambda + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}} \Rightarrow {\left( {{{\sin }^2}\lambda {{\cos }^4}\lambda + {{\cos }^4}\lambda + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {{{\sin }^2}\lambda {{\cos }^4}\lambda - 2{{\sin }^2}\lambda {{\cos }^2}\lambda + 2{{\sin }^2}\lambda {{\cos }^2}\lambda + {{\cos }^4}\lambda + {{\sin }^4}\lambda } \right)^{\frac{1}{2}}}$
$ \Rightarrow {\left( {{{\sin }^2}\lambda {{\cos }^4}\lambda - 2{{\sin }^2}\lambda {{\cos }^2}\lambda + 1} \right)^{\frac{1}{2}}}$
How do I proceed from here.
| We can have the following reasoning as well:
If $\tan^{-1}y=A, -\dfrac\pi2<A<\dfrac\pi2$
$\tan(\tan^{-1}y)=\tan A=y$
$\cos(\tan^{-1}y)=+\dfrac1{\sqrt{\tan^2A+1}}=?$
$\sin(\tan^{-1}y)=?$
Similarly choose, $\sin^{-1}y=B, -\dfrac\pi2\le B\le\dfrac\pi2$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$ Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$
I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:
$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$
I'm not sure how to carry on. I factored out the $\sin x$, but ended up with
$$\frac{1+\sin x}{1-\sin x}$$
I haven't been taught that as equal to $tan x$.
| \begin{align}
\frac{1-\cos 2x + \sin 2x}{1+\cos 2x + \sin 2x} &=\frac{1-(1-2\sin^2 x) + 2\sin x \cos x}{1+(2\cos^2 x -1) + 2\sin x \cos x} \\
&=\frac{2\sin^2x +2\sin x \cos x}{2\cos^2x + 2\sin x \cos x}
\end{align}
Can you simplify from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$ Given $$C=1+r\cos{\theta}\ +\frac{r^2\cos{2\theta}}{2!}\ + \frac{r^3\cos{3\theta}}{3!}\ + \dotsb$$ and $$S = r\sin{\theta}\ + \frac{r^2\sin{2\theta}}{2!}\ + \frac{r^3\sin{3\theta}}{3!}\ + \dotsb$$
Show the following$$C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$$
I am currently solving the problems given in Differential Calculus for Beginners by Joseph Edwards. As a beginner I am completely clueless about the approach to the above question. I tried to find the answer keys to this book, but sadly non exist on the internet.
| We have
$$\begin{align}
C+iS&=1+r(\cos\theta+i\sin\theta)+\frac{r^2}{2!}(\cos2\theta+i\sin2\theta)+\frac{r^3}{3!}(\cos3\theta+i\sin3\theta)+\cdots\\
&=1+re^{i\theta}+\frac{1}{2!}(re^{i\theta})^2+\frac{1}{3!}(re^{i\theta})^3+\cdots\\
&=e^{re^{i\theta}}
\end{align}$$
using De Moivre's Theorem and Euler's relation in the second line and the Maclaurin series expansion for $e^x$ in the final line. Using $C+iS$ is a common approach when finding closed form formulae for trigonometric summations (where $C$ is the series for $\cos$ and $S$ is a similar series for $\sin$), whether they be infinite or finite. I can make the working more complete if you like.
ADDED SECTION
This is also equivalent to
$$\begin{align}
(e^{e^{i\theta}})^r &=(e^{\cos\theta+i\sin\theta})^r\\
&=(e^{\cos\theta}\cdot e^{i\sin\theta})^r\\
&=e^{r\cos\theta}(\cos(\sin\theta)+i\sin(\sin\theta))^r\\
&=e^{r\cos\theta}(\cos(r\sin\theta)+i\sin(r\sin\theta))\\
&=e^{r\cos\theta}\cos(r\sin\theta)+ie^{r\cos\theta}\sin(r\sin\theta)
\end{align}$$
So we have
$$C=e^{r\cos\theta}\cos(r\sin\theta)$$
$$S=e^{r\cos\theta}\sin(r\sin\theta)$$
FINAL SECTION
So, if
$$C=e^{r\cos\theta}\cos(r\sin\theta)$$
$$S=e^{r\cos\theta}\sin(r\sin\theta)$$
then $$\frac{dC}{dr}=e^{r\cos\theta}\cos\theta\cos(r\sin\theta)-e^{r\cos\theta}\sin\theta\sin(r\sin\theta)=e^{r\cos\theta}\cos(\theta+r\sin\theta)$$
$$\frac{dS}{dr}=e^{r\cos\theta}\cos\theta\sin(r\sin\theta)+e^{r\cos\theta}\sin\theta\cos(r\sin\theta)=e^{r\cos\theta}\sin(\theta+r\sin\theta)$$
So
$$\begin{align}
C\frac{dC}{dr}+S\frac{dS}{dr}&=e^{2r\cos\theta}\cos(r\sin\theta)\cos(\theta+r\sin\theta)+e^{2r\cos\theta}\sin(r\sin\theta)\sin(\theta+r\sin\theta)\\
&=e^{2r\cos\theta}\cos\theta.
\end{align}$$
on using the compound angle formulae for Cosine.
And finally,
$$(C^2+S^2)\cos\theta=(e^{2r\cos\theta}\cos^2(r\sin\theta)+e^{2r\cos\theta}\sin^2(r\sin\theta))(\cos\theta)=e^{2r\cos\theta}\cos\theta$$
as required!!
Thanks so much for giving me the opportunity to answer this question, I have hugely enjoyed it!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$ Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.
I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.
Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.
This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.
| We can use also your idea, but in another writing:
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a(a+b)}}{a+b}>\sum_{cyc}\frac{\sqrt{a\cdot a}}{a+b}>\sum_{cyc}\frac{a}{a+b+c}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Rate if change of surface area when volume of cylinder is increasing. I get the same answer as the text book for part a) of the following question but cannot agree with it for part b).
The radius of the base of a right circular cylinder is r cm and its height is 2r cm. Find a) the rate at which its volume is increasing, when the radius is 2 cm and is increasing at 0.25 cm/s. b) the rate at which the total surface area is increasing when the radius is 5 cm and the volume is increasing at $5\pi$ cubic cm per second.
I have said let V = volume and A = area.
a)
$V = 2 \pi r^3$ and $A = 4 \pi r^2$
$dV/dt = \frac{dV}{dr}.\frac{dr}{dt}$
$= \frac{6 \pi r^2}{4} = 6 \pi$ when r = 2.
b)
$dV/dt = \frac{dV}{dr}.\frac{dr}{dt}$
$5 \pi = 6 \pi r^2.\frac{dr}{dt}$
$\frac{dr}{dt} = \frac{5 \pi}{6 \pi r^2} = \frac{5}{6r^2}$
$dA/dt = \frac{dA}{dr}.\frac{dr}{dt}$
$dA/dt = 8 \pi r.\frac{5 \pi}{6 \pi r^2} = \frac{4 \pi}{3}$
But the book says $2 \pi$
Is the book wrong?
| $A=6 \pi r^2$. Four of them come from the cylindrical surface, which is $2 \times 2\pi$ and two from the ends, which are $\pi r^2$ each.
| {
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If $a$ is sufficiently large as compared with $b,$ and $\sqrt \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the value of $k?$
If $a$ is sufficiently large as compared with $b,$ and $\sqrt
\frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the
value of $k?$
Query
Can we find it using particular value?,like putting $a=4,b=2,$ then we get
$k=4\sqrt2+4\sqrt \frac{2}{3}-8$
This problem is given in Multiple Choice Question exercise,the possible answers given are
A.$\frac{2}{3}$
B.$\frac{3}{4}$
C.$\frac{4}{5}$
D.$\frac{5}{6}$
None of the answer mathches with mine....
Please give some hint to determine value of $k?$
| Use the Negative Binomial Theorem.
$(1-a/b)^{-1/2}+(1+a/b)^{-1/2}=2+2\left(\frac{a}b\right)^2\left[\binom{-1/2}2+\binom{-1/2}4\left(\frac{a}b\right)^2+\binom{-1/2}6\left(\frac{a}b\right)^4+...\right]$
Since we want $k$ (constant) to be independent of $a/b$, we reject the higher order terms in the bracket on the ground that $a>>b$, giving $k=2\binom{-1/2}2=(-1/2)\times(-1/2-1)=3/4$.
Note that with the obtained $k$, $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}\approx2+k(a/b)^2$ i.e. $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ is not actually equal to $2+k(a/b)^2$ in general. As you have seen, for equality $k$ will have to be taken as a function of $a,b$ and not constant. Thus plugging in specific values of $a,b$ to find $k$ (which gives the exact value of $k$) will not yield the desired answer.
If you wanted to take a risk, you could check which option for $k$ yields the closest answer to the value of $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ for the selected $a,b$. But remember that you still might not get $k=3/4$ since it is not necessary that $3/4$ gives the closest approximation of $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ for all $a,b$.
| {
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Arranging word FACETIOUS but vowels must be in order The word FACETIOUS is the most common word in the English language which has each of the five vowels in order from left to right. How many ways can the letters in the word FACETIOUS be arranged such that all five vowels are in order from left to right?
I am wondering if there is an easier way than bashing and counting all the possiblities.
| There are nine spaces, $5$ vowels that must be in order, and $4$ are consonants. You can pick any letter first and choose where to place it so lets choose to place the $4$ consonants first.
We can put the $F$ in any of the $9$ places. We can put the $C$ in any of the remaining $8$. We can place the $T$ and the $S$ in any of the remaining $7$ and $6$ places. That's $9\times 8\times 7 \times 6$ ways to place the consonants.
The remaining $5$ spots must be filled with the vowels in order. There is only one way left to do that.
So there are $9\times 8\times 7 \times 6$ ways.
....
Alternatively. There are ${9\choose 4}$ ways to pick spots for the consonants, or ${9\choose 5}$ ways to pick spots of the vowels. For any one of the vowel/consonants placements we can arrange-- in the four spots set for the consonants-- the consonants in $4!$ ways. So there are ${9\choose 4}\times 4!$ or ${9\choose 5} \times 4!$ ways.
......
Or there are $9!$ ways total to do arrange the nine letters. However the vowels can be ordered in $5!$ ways and we can only accept $1$ of the the $5!$ orders. So of all the $9!$ ways to arrange the nine letters we can only accept $1$ out of $5!$ of them. SO there are $\frac{9!}{5!}$ ways.
===
Method 1: gives us $9\cdot 8 \cdot 7\cdot 6=3024$ ways.
Method 2a: gives us ${9\choose 4}\cdot 4! = \frac {9!}{5!4!}4!= \frac {9!}{5!} = \frac {1\cdot 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4\cdot5}=6\cdot7\cdot8\cdot9 =3024$ ways. Method 2b: and ${9\choose 5} = \frac {9!}{4!5!} = {9\choose 4}$ gives us the same answer.
Method 3: gives us $\frac {9!}{5!} = \frac {1\cdot 2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4\cdot5}=6\cdot7\cdot8\cdot9 =3024$ ways.
| {
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Prove that the recursively defined sequence is Cauchy. Define a sequence recursively by $x_1=1$, $x_{n+1}=\frac{5+5x_n}{5+x_n}$. Prove that this sequence is Cauchy. Then find the limit.
We call a sequence $\{a_n\}$ Cauchy, if for any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m\geq M$, $|a_n-a_m|<\epsilon$.
| $$
x_{n+1} - \sqrt 5 = \frac{5+5x_n}{5+x_n} - \sqrt{5} = \frac{(5-\sqrt{5})(x_n-\sqrt 5)}{5+x_n}
$$
$$
x_{n+1} + \sqrt 5 = \frac{5+5x_n}{5+x_n} + \sqrt{5} = \frac{(5+\sqrt{5})(x_n+\sqrt 5)}{5+x_n}
$$
Therefore
$$
\frac{x_{n+1}-\sqrt 5}{x_{n+1}+\sqrt 5}=\frac{5-\sqrt 5}{5+\sqrt 5} \cdot \frac{x_n-\sqrt 5}{x_n+\sqrt 5}\\
\Rightarrow \frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{5-\sqrt 5}{5+\sqrt 5}\right)^{n-1} \cdot \frac{1-\sqrt{5}}{1+\sqrt{5}}
$$
then you can get a closed form solution for $x_n$ and the rest should be straightforward.
Please check this post: Recursive sequence depending on the parameter
I just learned about Möbius transformation the other day.
| {
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Find the limit of the function without L'Hôpital's rule I have a problem, such as:
$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$
How do I solve this problem without using L'Hôpital's rule or small-o?
Thanks!
| Composing Taylor series
$$y= \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}\implies\log(y)=\frac{1}{x^2}\big[\log(\cos(ax))-\log(\cos(bx))\big]$$
Using
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$
$$\cos(cx)=1-\frac{c^2 x^2}{2}+\frac{c^4 x^4}{24}+O\left(x^6\right)$$
$$\log(\cos(ax))=-\frac{c^2 x^2}{2}-\frac{c^4 x^4}{12}+O\left(x^6\right)$$
Use it twice (for $c=a$ and $c=b$ and subtract to get
$$\log(\cos(ax))-\log(\cos(bx))=\frac{1}{2} x^2 \left(b^2-a^2\right)+\frac{1}{12} x^4
\left(b^4-a^4\right)+O\left(x^6\right)$$
$$\log(y)=\frac{1}{2} \left(b^2-a^2\right)+\frac{1}{12} x^2
\left(b^4-a^4\right)+O\left(x^4\right)$$
$$y=e^{\log(y)}=e^{\frac {b^2-a^2}2}\left(1+\frac{1}{12} x^2 \left(b^4-a^4\right)+O\left(x^4\right)\right)$$ which shows the limit and also how it is approached.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$ Here is my attempt:
Let $(x, y)$ be any such point.
Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.
Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$
Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.
Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$
Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?
| $$\dfrac{x-4}{\cos t}=\dfrac{y-2}{\sin t}=r\ge0$$
So, the point $P(4+r\cos t,2+r\sin t)$$
We need $$(4+r\cos t)^2=(2+r\sin t)^2=r^2$$
$$r(1+\cos t)=4$$ as $r(1-\cos t)\ge0$
Similarly, $r(\sin t+1)=2$
Using https://en.m.wikipedia.org/wiki/Weierstrass_substitution#The_substitution, $$\dfrac42=\dfrac2{(1+p)^2}$$
where $p=\tan\dfrac t2$
$\implies(1+p)^2=1\implies p=0,-2$
If $p=0,r=2$
If $p=-2,r=?$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to transform quadratic terms to something like $u^2+v^2$? I am told that by the substitution of $x = (a\cos \theta)u - (b\sin \theta) v$ and $y = (a\sin \theta)u + (b\cos \theta)v$, we can reduce $3x^2 + 2xy + 3y^2$ to $u^2 + v^2$.
I tried to plug in the substitutions, but I did not see how the quadratic term is reduce to $u^2 + v^2$. What I got was something complicated containing $a$, $b$, $\cos \theta$ and $\sin \theta$.
I am asked to reduce $3x^2 + 2xy + 3y^2 - x - 2y$ to $u^2 + v^2$ by similar substitutions. I do not know how to construct this substitution. Could you explain how do find this kind of substitutions in general?
| The answer is $x=\frac{-1}{2}$, $b=-\frac{1}{\sqrt{2}}$, $\theta=\frac{\pi}{4}$ and
$$
x=(a\cos\theta)u-(b\sin\theta)v\textrm{, }y=(a\sin\theta)u+(b\cos\theta)v.\tag 1
$$
Setting this into $3x^2+2xy+3y^2$ you get $u^2+v^2$.
By demanding $kx^2+lxy+my^2$ to be $u^2+v^2$, you get after some calculations
$$
a=\frac{\pm1}{\sqrt{k\cos^2 \theta+l\cos \theta \sin \theta+m \cos^2\theta}}
$$
$$
b=\frac{\pm1}{\sqrt{m\cos^2 \theta-l\cos \theta \sin \theta+k\sin^2 \theta}}
$$
$$
l\cos 2\theta+(m-l)\sin 2\theta=0
$$
| {
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Why this recursion is the same as OEIS sequence A005252 I have the recursion:
$$\left\{ \begin{array}{l}
f(0) = f(1) = f(2) = f(3) = 1\\
f(n) = f(n - 1) + 1 + \sum_{i = 1}^{n - 4}f(i) \text{ if $n \geqslant 4$}
\end{array} \right.$$
So we have e.g.
$$\begin{array}{lll}
f(4) & = & f(3) + 1 = 2\\
f(5) & = & f(4) + 1 + f(1) = 2 + 1 + 1 = 4\\
f(6) & = & f(5) + 1 + f(2) + f(1) = 4 + 1 + 1 + 1 = 7\\
\ldots & &
\end{array}$$
I noticed this sequence is the same as A005252 - OEIS, which is defined as:
$$a(n) = \sum_{k=0..\lfloor n/4 \rfloor} \binom{n-2k}{2k}$$
I'm not sure how to get there? I tried to use a generating function, but not sure how to manipulate it.
| Let $F(z)=\sum_{n \ge 0} f_n z^n$ be the generating function. Then the recurrence relation implies that
\begin{align}
F(z)
&= \sum_{n=0}^3 1 z^n + \sum_{n \ge 4} \left(f_{n - 1} + 1 + \sum_{i = 1}^{n - 4}f_i\right)z^n \\
&= \sum_{n\ge 0} z^n + z \sum_{n \ge 4} f_{n - 1}z^{n-1} + \sum_{i \ge 1} f_i \sum_{n\ge i+4} z^n \\
&= \frac{1}{1-z} + z \left(F(z)-\sum_{n=0}^2 f_n z^n\right) + \sum_{i \ge 1} f_i \frac{z^{i+4}}{1-z} \\
&= \frac{1}{1-z} + z F(z)-z\sum_{n=0}^2 z^n + \frac{z^4}{1-z}\left(F(z)-1\right), \\
\end{align}
so
$$F(z)=\frac{\frac{1}{1-z} -z(1+z+z^2) - \frac{z^4}{1-z}}{1-z- \frac{z^4}{1-z}}
=\frac{1-z}{(1-z+z^2)(1-z-z^2)},$$
which matches the generating function shown in OEIS.
| {
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Convergence of the improper integral $\int_0^2\frac x{(16-x^4)^{1/3}}\,\mathrm dx$ Need your help in the checking convergence of this improper integral :$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}$$I tried my luck and somehow was able to establish a relationship between $(1-t^4)$ and $t^4$ i.e $$(1-t^4)\geq t^4$$
$$\frac{1}{1-t^4} \leq \frac{1}{t^4}$$
$$\Rightarrow \frac{1}{(1-t^4)\frac{1}{3}} \leq \frac{1}{t^\frac{4}{3}}$$ $$\Rightarrow \frac{t}{(1-t^4)\frac{1}{3}} \leq \frac{t}{t^\frac{4}{3}}$$
for some $$t\in\left[0,\frac{1}{2^\frac{1}{4}}\right]$$ It can be shown that$$\int_0^2\frac{xdx}{(16-x^4)^\frac{1}{3}}=2^\frac{2}{3}\int_0^1\frac{tdt}{(1-t^4)^\frac{1}3} $$
Therefore ,
$$F(x) =\int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{(1-t^4)^\frac{1}3} \leq \int_0^\frac{1}{2^\frac{1}{4}}\frac{tdt}{t^\frac{4}{3}} = g(x)$$ For this particular interval I was able to prove its convergence as $g(x)$ converges but I am unable to comment on the remaining interval. Please help and also provide some insight which might help me in future. Thanks in advance...
| Since$$16-x^4=(2-x)(8+4x+2x^2+x^3),$$you have$$\frac x{(16-x^4)^{1/3}}=\frac x{\sqrt[3]{2-x}\sqrt[3]{8+4x+2x^2+x^3}}$$and therefore$$\lim_{x\to2^-}\frac{\frac x{(16-x^4)^{1/3}}}{\frac1{\sqrt[3]{2-x}}}=\frac1{\sqrt[3]4}\ne0.$$So, since the integral$$\int_0^2\frac1{\sqrt[3]{2-x}}\,\mathrm dx$$converges, then so does your integral.
| {
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Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational? $$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$
Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.
I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
Sorry if this is a stupid question.
| Writing $c = \sqrt[3]{2-\sqrt{5}}, d = \sqrt[3]{2+\sqrt{5}}, a=c+d$, cubing the expression gives:
\begin{align}(c+d)^3 &= c^3+3c^2d+3cd^2+d^3\\
&=(2-\sqrt 5) + 3cd(c+d) + (2+ \sqrt 5)\\
&=4+3a\sqrt[3]{(2-\sqrt5)(2+\sqrt 5)}\\
&=4 -3a=a^3
\end{align}
Hence $a^3+3a-4 = 0 = (a-1)(a^2+a+4)$.
The quadratic equation $a^2+a+4=0$ has no solutions in the reals. Hence $a=1$.
Hopefully this answer can give some insight on how you can approach the general problem for $\sqrt[3]{x\pm\sqrt y}$.
Update: Since the other answer has proceeded to partially solve your second question, I may as well prove the sufficient condition.
Writing $a = \sqrt[3]{x-\sqrt y}, b =\sqrt[3]{x+\sqrt y}, c=a+b$:
\begin{align}(a+b)^3 &= a^3+3a^2b+3ab^2+b^3\\
&=(x-\sqrt y) + 3ab(a+b) + (x+ \sqrt y)\\
&=2x+3c\sqrt[3]{(x-\sqrt y)(x+\sqrt y)}\\
&=4 - 3c\sqrt[3]{x^2-y}=c^3
\end{align}
From $c^3 + 3c\sqrt[3]{x^2-y} - 2x = 0$ and $c=1$ being a solution we must have $$y= x^2 + \left(\dfrac {2x-1}3\right)^3$$ where $x \equiv 2 \pmod 3$ for $y$ to be an integer.
For positive integer values of $y$ we have $3\sqrt[3]{x^2-y} = 1-2x$. Hence the cubic equation reduces to
$$c^3+(1-2x)c-2x= 0=(c-1)(c^2+c+2x)$$
For $x > \dfrac18$, $c^2+c+2x$ has no solution in the reals. Hence $c=1$, showing that this condition is indeed necessary and sufficient.
One could, of course, obtain the explicit form of $\sqrt[3]{x\pm\sqrt y}$ with our condition, by observing:
$$8 (x \pm \sqrt y) = 1^3 \pm 3\sqrt {\frac {8x-1}3} +3 \sqrt {\frac {8x-1}3}^2 \pm \sqrt {\frac {8x-1}3}^3 = \left(1 \pm \sqrt{\frac {8x-1}3}\right)^3$$
Which gives:
$$\sqrt[3]{x\pm\sqrt y} = \frac{1 \pm \sqrt{\frac {8x-1}3}}2$$
and indeed sums up to $1$, as desired.
| {
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Solving $a(a + 1) + b(b + 1) = 12, (a+ 1)(b+1) = 4$, is there any trick? What are all (possibly complex) solutions to the following two equations?
$a(a + 1) + b(b + 1) = 12$
$(a+ 1)(b+1) = 4$
Clearly one can substitute the second equation into the first then solve the quartic. But I am wondering if there is a shorter/sneakier way to solve this.
| How sneaky would you like to be? If we interpret these expressions as the curve equations $ \ x(x + 1) + y(y + 1) \ = \ 12 \ $ and $ \ ( x + 1)(y+1) \ = \ 4 \ \ , $ we can write the first as the circle equation $ \ \left(x + \frac12 \right)^2 + \left(y + \frac12 \right)^2 \ = \ \frac{25}{2} \ \ $ , with center at $ \ \left(-\frac12 \ , \ -\frac12 \right) \ $ and the second as a "translated" rectangular hyperbola with its center at $ \ (-1 \ , \ -1) \ \ . $ This may not seem helpful initially, but the location of the centers suggests that there is a "diagonal symmetry" for the solutions.
Two of these can be "picked off" by inspection from looking at the intercepts of the curves. Setting $ \ x \ = \ 0 \ $ gives us $ \ 1·(y + 1) \ = \ 4 \ \Rightarrow \ y \ = \ 3 \ $ for the hyperbola, and we see that this also applies to the circle: $ \ 0·(0 + 1) + 3·(3 + 1) \ = \ 12 \ \ . $ The symmetry indicates that there is also a solution $ \ x \ = \ 3 \ , \ y \ = \ 0 \ \ . $ So the two curves intersect one another on the coordinate axes.
Are there any more solutions? This symmetry about the line $ \ y = x \ $ suggests that we might look at $ \ (x + 1)^2 \ = 4 \ $ and $ \ 2·x(x + 1) \ = \ 12 \ \ . $ The first of these produces $ \ x \ = \ -1 \ \pm \ 2 \ = \ 1 \ , \ -3 \ \ ; \ $ the point $ \ (1 \ , \ 1) \ $ is where the diagonal line meets the hyperbola, but it is certainly not on the circle, while the other intersection is $ \ (-3 \ , \ -3) \ \ . $ The intersections of the diagonal line and circle are found from $$ \ x·(x + 1) \ = \ 6 \ \Rightarrow \ x^2 + x - 6 \ = \ (x + 3)·(x - 2) \ = \ 0 \ \ . $$ The point $ \ (2 \ , \ 2) \ $ is not on the hyperbola, but $ \ (-3 \ , \ -3) \ $ definitely is. So the circle, hyperbola, and diagonal line all meet at $ \ (-3 \ , \ -3) \ \ . $
The circle and hyperbola in fact just intersect tangentially there, which tells us that this solution has an even multiplicity. In regard to the quartic equation you mention and which player3236 develops (a biquadratic), we can conclude that we have found all four of its solutions: $ \ (0 \ , \ 3) \ \ , \ \ (3 \ , \ 0) \ \ , $ and $ \ (-3 \ , \ -3) \ $ with multiplicity $ \ 2 \ \ . $
We can check up on this by using the "rotated" coordinates $ \ u \ = \ x + y \ , \ v \ = \ x - y \ \ , $ which transform the curve equations into $ \ u^2 + 4u - v^2 \ = \ 12 \ , \ u^2 + 2u + v^2 \ = \ 24 \ \ . $ Adding these together yields
$$ 2u^2 \ + \ 6u \ - 36 \ \ = \ \ 2·(u + 6)·(u - 3) \ \ = \ \ 0 \ \ . $$
Solving each of the transformed curve equations for $ \ v^2 \ $ gives us
$$ v^2 \ \ = \ \ u^2 + 4u - 12 \ \ = \ \ (u + 6)·(u - 2) \ \ \Rightarrow \ \ u \ = \ -6 \ \rightarrow \ v^2 \ = \ 0 \ \ , \ \ u \ = \ 3 \ \rightarrow \ v^2 \ = \ 9 $$
and
$$ v^2 \ \ = \ \ 24 - u^2 - 2u \ \ = \ \ -(u + 6)·(u - 4) \ \ \Rightarrow \ \ u \ = \ -6 \ \rightarrow \ v^2 \ = \ 0 \ \ , $$ $$ u \ = \ 3 \ \rightarrow \ v^2 \ = \ 9 \ \ . $$
So we find three solution points:
$$ (u \ , \ v) \ \rightarrow \ \left(x = \frac{u + v}{2} \ , \ y = \frac{u - v}{2} \right) \ : \ \ \ (3 \ , \ 3 ) \ \rightarrow \ (3 \ , \ 0 ) \ \ , \ \ (3 \ , \ -3 ) \ \rightarrow \ (0 \ , \ 3 ) \ \ , $$
$$ (-6 \ , \ 0 ) \ \rightarrow \ (-3 \ , \ -3 ) \ \ . $$
| {
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"url": "https://math.stackexchange.com/questions/3902088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Drawing numbers without replacement So if I were to draw 11 unique numbers from 0-99, (0 included), what is the probability that the last digit of all numbers drawn contain 0-9. (i.e a good outcome would be 10,11,12,13,14,15,16,17,18,19,20 as the last digits of 0-9 are selected).
Would I need to use the inclusion exclusion principle?
Is the answer to this question $1- \dfrac{\binom {10}{1} \binom {90}{11}- \binom {10}{2} \binom {80}{11}+ \binom {10}{3} \binom {70}{11}- \binom {10}{4} \binom {60}{11}+ \binom {10}{5} \binom {50}{11} - \binom {10}{6} \binom {40}{11} + \binom {10}{7} \binom {30}{11} - \binom {10}{8} \binom {20}{11} + \binom{10}{9} \binom{10}{11}}{\binom{100}{11}}$?
| Your answer is correct, but there is really no need to use the devious route of inclusion-exclusion.
There are $10$ ways to choose the digit that will occur twice,
and the rest is just application of the "normal" hypergeometric formula, thus
$10\times\left[\frac{\dbinom{10}2\dbinom{10}{1}^9}{\dbinom{100}{11}}\right]$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?
| This is just a hint, which may be useful, not a full answer.
$$(2^k-1)(2^k+1)=(2^k-x)(2^k+x)t$$
From this we see that $x$ should be odd. And easy to prove that the $$(2^k-x,2^k+x)=1 $$
As the $2^k+x > 2^k-1$ and $2^k+x > 2^k+1$ for the $x>1$ then $$(2^k+x,2^k-1)=a > 1$$ and $$(2^k+x,2^k+1)=b > 1$$ and $(a,b)=1$ and $ab=2^k+x$.
Let assume that $$(2^k-1,2^k-x)=c$$ and $$(2^k+1,2^k-x)=d$$ then obviously $(a,c)=1$, $(a,d)=1$, $(b,c)=1$, $(b,d)=1$, $(d,c)=1$ and $cd=2^k-x$.
And $a,b,c,d$ are odd.
$$ac | 2^k-1$$
$$bd | 2^k+1$$
$$ab+cd=2^{k+1}$$
$$ab-cd=2x$$
From this it seems that there should be some solution, but probably for the $k$-s for which the $2^k-1$ and $2^k+1$ has enough divisors.
In this scope it is also interesting to consider the Bang's theorem
| {
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Limit of $(1-\frac{1}{2^n})^n$. What is the limit of the sequence $(1-\frac{1}{2^n})^n$?
Let $a_n:=(1-\frac{1}{2^n})^n$ then $\ \ln \ a_n=n\ \ln(1-\frac{1}{2^n})$
$\implies \lim \limits_{n\rightarrow \infty} \ln\ a_n=\lim\limits_{n \rightarrow \infty} n\ \ln(1-\frac{1}{2^n})$
$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\ln(1-\frac{1}{2^n})}{\frac{1}{n}}$
$\implies \ln \lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} \frac{\frac{ln2}{2^n(1-2^n)}}{\frac{-1}{n^2}} [\frac{0}{0} \ \ \text{form}]$
I am stuck here, what should I do after this?
| Continuing where you left:
$$ \frac{ \frac{\ln2}{2^n (1 -2^n)} }{\frac{-1}{n^2}} = - \frac{n^2 \ln2}{2^n(1-2^n)} = \frac{n^2 \ln2}{2^n(2^n - 1)}$$
It's pretty obvious / not hard to prove that this tends to $0$ as $n$ tends to infinity ("exponent beats polynomial") and so your limit should go to $e^0 = 1$
Another approach is to "fix" the exponent so it would suit the limit of $e$:
$$ \lim_{n \rightarrow \infty}( 1 - \frac{1}{2^n})^n = \lim_{n \rightarrow \infty}( \overbrace{(1 + \frac{-1}{2^n})^{2^n}}^{\text{1/e}})^{ \overbrace{\frac{n}{2^n}}^{\text{Tends to 0}}}$$
| {
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Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$
I've calculated several values of this function, and it seems to have a limit of $1$.
| Hint: Using the fact that $\frac {\sin x} x \to 1$ as $x \to 0$ verify that $ x\ln (\frac 1 2 \sqrt x) <x\ln (2\sin\sqrt x+\sqrt x \sin (\frac 1 x)) <x \ln (3\sqrt x)$. Conclude that the limit is $e^{0}=1$.
[$x \ln x \to 0$ as $x \to 0+$]
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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find the value $AB$, for which $\log_{A} B =\log_{B} A$ Suppose $A$ and $B$ are positive real numbers for which $\log_{A} B = \log_{B} A$ . If neither $A$ nor $B$ is $1$ and $A$ $\not\equiv B$, find the value of $AB$.
We can rewrite as
$A^{\log_{A} B} = B$.
$B^{\log_{B} A} = A$.
$(B^{\log_{B} A})^{A^{\log_{A} B}} =B$.
$B^{2 (\log_{B} A)} =B$.
$B^{2 (\frac{1}{2})} =B$.
I'm unable to deduce a correct solution with $\frac{1}{2}$.
I wanted to try a different approach by converting the powers to multiplication but can't figure out the base. i.e.
${\log_{A} B (\log_{?}A)} $.
| To change from the base $b$ to the base $c$ in the logarithm below
$$\log_{b}a$$
we can use the following formula
$$\log_{b}a=\frac{\log_{c}a}{\log_{c}b}$$
So, to change from the base $B$ to base $A$:
$$\log_{B}A=\frac{\log_{A}A}{\log_{B}A}$$
and, since $\log_{A}A=1$, we get
$$\log_{B}A=\frac{1}{\log_{A}B}$$
and
$$\log_{B}A=\log_{A}B$$
then
$$(\log_{A}B)^{2}=1\Rightarrow\log_{A}B=\pm1$$
Therefore,
$$A=B$$
or
$$A=B^{-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a hard limit by approximating the sine function I think I'm supposed to solve this limit using approximations for the behavior when $x$ is close to $0$, but I don't know how to apply the logic for the aproximation.
$$\lim_{x \rightarrow 0} (\frac{1}{sin^2(x)} - \frac{1}{x^2})$$
In the context of this exercise the textbook mentions the following method for approximating:
$$f(x) = \frac{1}{1+x+x^2}$$
Since $\lim_{x \rightarrow 0} f(x) = 1$, then $lim_{x \rightarrow 0} [f(x) - 1] =0$; but $f(x)-1 = -x\frac{x+1}{1+x+x^2}$ so, as $x \rightarrow 0$, $f(x) \approx 1-x$. Also for $x$ near $0$ $f(x) - 1 + x = x^3\frac{1}{1+x+x^2}$ so $f(x) \approx 1-x + x^2$.
I think I grasp the logic for it, but I can't apply this to solving limits like the one with $\frac{1}{sin^2(x)}$.
Note: at this point we haven't seen Taylor Polinomials yet, just this method for approximation near $0$.
| $$y= \frac{1}{\sin^2(x)} - \frac{1}{x^2}=\frac{x^2-\sin^2(x)}{x^2\,\sin^2(x)}$$ Now, using Taylor series
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$
$$y=\frac{-\frac{x^4}{3}+O\left(x^6\right) } {x^2\left(x^2-\frac{x^4}{3}+O\left(x^6\right) \right) }=\frac{1}{3}+O\left(x^2\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the formula of combination number $$
\sum_{k=0}^n{\left( \begin{array}{c}
n\\
k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{x+k}=\frac{n!}{x\left( x+1 \right) \cdots \left( x+n \right)}}
\\
$$
I recently read an article and found this formula above. I tried to prove it, but only proved the following result. I don't know how to prove this more general combinatorial equation.
$$
\sum_{k=0}^n{\left( \begin{array}{c}
n\\
k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{m+k+1}=\int_0^1{\sum_{k=0}^n{\left( \begin{array}{c}
n\\
k\\
\end{array} \right) \left( -1 \right) ^kx^{m+k}\text{d}x}=\int_0^1{x^m\left( 1-x \right) ^n}\text{d}x}=\frac{m!n!}{\left( m+n+1 \right) !}}
$$
| By partial fractions, we know
$$\frac{n!}{x(x+1) \cdots (x+n)} = \frac{a_0}{x} + \frac{a_1}{x+1} + \dots + \frac{a_n}{x+n}$$
for some set of numbers $a_0, a_1, \dots , a_n$. To find $a_k$ we multiply both sides of the equation by $x+k$ and then set $x=-k$ in the resulting equation. The result is
$$a_k = (-1)^k \binom{n}{k}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving complex numbers identity using, $\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$ Show that $\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$ can be expressed in the form $\alpha$ or $\alpha^2$, Hence prove $$(a+b\alpha+c\alpha^2)(a+b\alpha^2+c\alpha)=\frac12[(a-b)^2+(b-c)^2+(c-a)^2]$$
My try:
I wrote the expression as,$$y=\sqrt{-1-\sqrt{-1-{\sqrt{-1...}}}}$$
$$y^2+y+1=0$$
$$y=\frac{-1+\sqrt{3}i}{2}\text{ or } y=\frac{-1-\sqrt{3}i}{2}$$
This is in the form of $\alpha$ and $\alpha^2.$I have trouble solving the second part. Can someone please explain it to me? Thank you in advance.
| The LHS is $a^2+b^2 \alpha^3+c^2 \alpha^3+ab(\alpha^2+\alpha)+ac(\alpha^2+\alpha)+bc(\alpha^4+\alpha)$.
Use the fact that $\alpha^4=\alpha$,$\alpha^2+\alpha=-1$,$\alpha^3=1$.
So The LHS becomes $$a^2+b^2+c^2-ab-ac-bc\\=\frac 1 2\left[a^2+b^2-2ab+b^2+c^2-2bc+a^2+c^2-2ac\right]\\=\frac 1 2\left[(a-b)^2+(b-c)^2+(a-c)^2\right]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the given system of equations
$$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$
I got $x^3 + y^3 = 6xy = (x+y)^2$ which I simplified to $x^2 - xy + y^2 = x + y$ and I'm not sure what to do now. Please help.
| We are given that $$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$
Consider the expression $(x+y)\left(\frac{x}{y}+\frac{y}{x}\right)$:
$$(x+y)\left(\frac{x}{y}+\frac{y}{x}\right)=x+y+\frac{x^2}{y}+\frac{y^2}{x}$$
so
$$(x+y)\left(\frac{x}{y}+\frac{y}{x}-1\right)=\frac{x^2}{y}+\frac{y^2}{x}$$
But using the initial conditions we see that
$$3(x+y)=6\implies x+y=2\implies x=y-2$$
which can be subsituted into your equations to provide solutions for $x$ and $y$.
This is a bit more of an experimental solution than the other answers; the expression considered intially is not necessarily obvious. Often though some fiddling around with expressions like that is useful.
I hope that helps :)
| {
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How to prove that $a^3 + b^3 \geq a^2b + ab^2$? Um I am solving problems in Arthur Engels book "Problem Solving Strategies". I was doing a problem from inequalities chapter, and I stumbled across a problem which I managed to condense and simplify into this:--
For $a + b> 0$, $$a^3 + b^3 \geq a^2b + ab^2$$
I have no idea how to begin but this is what I did.
By A.M-G.M,
$$\frac{a^3 + a^3 + b^3}{3} \geq \sqrt[3]{a^3a^3b^3}$$
$$\implies a^3 + a^3 + b^3 \geq 3a^2b$$
...(i)
$$$$
$$\frac{b^3 + b^3 + a^3}{3} \geq \sqrt[3]{b^3b^3a^3}$$
$$\implies b^3 + b^3 + a^3 \geq 3ab^2$$
...(ii)
Now adding (i) and (ii) we have,
$$3a^3 + 3b^3 \geq 3a^2b + 3ab^2$$
Dividing everything by 3 we have,
$$a^3 + b^3 \geq a^2b + ab^2$$
Which is exactly what I wanted, but I have no idea whether this is correct. Please check it for me and please also tell if there are other methods to prove this.
(Also could you please invite I am very new to stackexchange and would like increase my reputation. Please.)
| Hint:
$$a^3+b^3-(a^2b+ab^2)=(a+b)(a^2-ab+b^2)-ab(a+b)=(a+b)(a-b)^2\ge0$$ if $a+b\ge0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x) d \theta}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} } x \in \Bbb{R}$ Compute
$$ f(x) = \int_0^{\pi} \frac{ \sin \theta ( \cos \theta - x)}{ [(\cos \theta - x)^2 + \sin^2 \theta]^{1.5} }
\,\,d \theta\,,\quad x \in \Bbb{R}$$
The denominator makes it a bit annoying. We can evaluate it as
$ [(\cos \theta - x)^2 + \sin^2 \theta = (1 - 2 \cos \theta x + x^2)^{1.5}$ but it doesn't make it any easier. We can also expand $(1 + x^2 - 2 \cos \theta x )^{1.5}$ using $(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} x^n$ but it's still difficult.
The denominator also has some geometric meaning (cubed distance from $(x,0)$ to $(\cos \theta, \sin \theta)$ ) but does it help?
| Write $r = r(\theta) = \sqrt{1 - 2x\cos\theta + x^2}$. Then we easily check that
$$ \frac{\mathrm{d} r^{\alpha}}{\mathrm{d}\theta} = \alpha r^{\alpha-2} x\sin\theta, \qquad\text{and so},\qquad \int r^{\beta}\sin\theta \, \mathrm{d}\theta = \frac{r^{\beta+2}}{(\beta+2)x}. $$
Now denoting the integral by $f(x)$ and assuming that $x \neq \pm 1$, we have
\begin{align*}
f(x)
&= \int_{0}^{\pi} r^{-3}\sin\theta(\cos\theta - x) \, \mathrm{d}\theta \\
&= \left[ - \frac{r^{-1}(\cos\theta - x)}{x} \right]_{0}^{\pi} - \frac{1}{x} \int_{0}^{\pi} r^{-1}\sin\theta \, \mathrm{d}\theta \\
&= \frac{1}{x}\left( \frac{1-x}{r(0)} + \frac{1+x}{r(\pi)} \right) - \left[ \frac{r}{x^2} \right]_{0}^{\pi}.
\end{align*}
By using $r(\pi) = \left| 1 + x \right|$ and $r(0) = \left| 1 - x \right|$, we get
\begin{align*}
f(x) &= \frac{\operatorname{sgn}(1+x) + \operatorname{sgn}(1-x)}{x} - \frac{\left| 1 + x\right| - \left|1 - x \right|}{x^2} \\
&= \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{\left| 1 + x\right| - \left|1 - x \right|}{x} \right).
\end{align*}
Noting that
$$ \frac{\left| 1 + x\right| - \left|1 - x \right|}{x} = \begin{cases}
2/x, & x > 1, \\
2, & -1 < x < 1, \\
-2/x, & x < -1,
\end{cases} $$
it follows that
$$ f(x) = \begin{cases}
-2/x^2, & x > 1, \\
0, & -1 < x < 1, \\
2/x^2, & x < -1.
\end{cases} $$
Given this form, I suspect that $f$ comes from an electromagnetism problem of computing the electric field arising from a charged metal sphere or something like that.
Remark. It can be checked separately that
$$ f(1)
= -\frac{1}{2\sqrt{2}} \int_{0}^{\pi} \frac{\sin\theta}{\sqrt{1-\cos\theta}} \, \mathrm{d}\theta
= -\frac{1}{\sqrt{2}} \left[ \sqrt{1-\cos\theta} \right]_{0}^{\pi}
= -1 $$
and
$$ f(-1)
= \frac{1}{2\sqrt{2}} \int_{0}^{\pi} \frac{\sin\theta}{\sqrt{1+\cos\theta}} \, \mathrm{d}\theta
= -\frac{1}{\sqrt{2}} \left[ \sqrt{1+\cos\theta} \right]_{0}^{\pi}
= 1. $$
Therefore the complete answer is as follows:
$$ f(x) = \begin{cases}
-\frac{2}{x^2}, & x \in (1, \infty), \\
-1, & x = 1, \\
0, & x \in (-1, 1), \\
1, & x = -1, \\
\frac{2}{x^2}, & x \in (-\infty, -1).
\end{cases} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$ I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:
$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$
It's rather easy to show
$$
\frac{4xy}{x^2 + y^2} \le 2
$$
and obviously
$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$
but combining these two does not give me a sufficient low bound.
| How about this method?
$$\frac{2(x + y)²}{2x² + y²}-3=\frac{2x^2+4xy+2y^2-(6x^2+3y^2)}{2x^2+y^2}=-\frac{4x^2+4xy+y^2}{2x^2+y^2}=-\frac{(2x-y)^2}{2x^2+y^2}\le0..$$
| {
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How to find the SVD of a matrix with one zero eigenvalue? I am trying to find the SVD of the matrix \begin{pmatrix}1&2\\3&6\end{pmatrix}
$A^TA=\begin{pmatrix}1&3\\2&6\end{pmatrix}\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}10&20\\20&40\end{pmatrix}$
$\lambda_1=50$
$\lambda_2=0$
$\lambda_1>\lambda_2>=0$
$\Sigma=\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}$
When $\lambda_1=50$:
$v_1=\begin{pmatrix}1\\2\end{pmatrix}$
Normalize:
$u_1=\begin{pmatrix}\frac{1}{\sqrt{5}}\\\frac{2}{\sqrt{5}}\end{pmatrix}$
When $\lambda_1=0$:
$v_2=\begin{pmatrix}-2\\1\end{pmatrix}$
Normalize:
$u_2=\begin{pmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{pmatrix}$
$V=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{-2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$V^T=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$A=U\Sigma V^T$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}u_1&u_2\\u_3&u_4\end{pmatrix}\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}5\sqrt{2}u_1&0\\5\sqrt{2}u_3&0\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}\sqrt{10}u_1&2\sqrt{10}u_1\\\sqrt{10}u_3&2\sqrt{10}u_3\end{pmatrix}$
$u_1=\frac{1}{\sqrt{10}}$
$u_3=\frac{3}{\sqrt{10}}$
How can I find $u_2$ and $u_4$?
| Note that the given matrix is rank-$1$ and that in $\Bbb R^2$ it is easy to find orthogonal vectors. Hence,
$$\begin{aligned} {\rm A} := \begin{bmatrix} 1 & 2 \\ 3 & 6\end{bmatrix} = \begin{bmatrix} 1 \\ 3\end{bmatrix} \begin{bmatrix} 1 \\ 2\end{bmatrix}^\top &= \begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} 1 & -2\\ 2 & 1\end{bmatrix}^\top \\ &= \left( \frac{1}{\sqrt{10}} \begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix} \right) \begin{bmatrix} \sqrt{50} & 0\\ 0 & 0\end{bmatrix} \left( \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -2\\ 2 & 1\end{bmatrix} \right)^\top = {\rm U} \Sigma {\rm V}^\top\end{aligned}$$
Lastly, do note that the nonzero singular value is $\sqrt{\mbox{tr} \left( {\rm A}^\top {\rm A} \right) } = \| {\rm A} \|_{\text{F}} = \sqrt{50} = \| {\rm A} \|_2$.
matrices svd singular-values
| {
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} |
Prove the determinant is $0$ This is Problem 16.17 from the book Exercises in Algebra by A. I. Kostrikin.
Prove that
$$
\left|\begin{array}{ccccc}
\dfrac{1}{2 !} & \dfrac{1}{3 !} & \dfrac{1}{4 !} & \cdots & \dfrac{1}{(2 k+2) !} \\
1 & \dfrac{1}{2 !} & \dfrac{1}{3 !} & \cdots & \dfrac{1}{(2 k+1) !} \\
0 & 1 & \dfrac{1}{2 !} & \cdots & \dfrac{1}{(2 k) !} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \dfrac{1}{2 !}
\end{array}\right|=0, \quad k \in \mathbb{N}
$$
My Attempt: I tried to expand it by the first column, but it seemed to be more complicated when I did that. I also tried to add edges to the determinant(in the hope that it will be easier to calculate), but I still failed to work it out.
So, My Question is, how to calculate this determinant?
| Finally I found a direct way to work this out, that is, to use Generating Functions.
From this article we get the following proposition:
Proposition. Consider the following infinite matrix with $1$s in the super diagonal.
$$
D=\left[\begin{array}{ccccc}
b_{0} & 1 & 0 & 0 & \cdot \\
b_{1} & c_{1} & 1 & 0 & . \\
b_{2} & c_{2} & c_{1} & 1 & . \\
b_{3} & c_{3} & c_{2} & c_{1} & . \\
b_{4} & c_{4} & c_{3} & c_{2} & . \\
\cdot & . & . & . & .
\end{array}\right]
$$
Let $B(x)=\sum_{n=0}^{\infty} b_{n} x^{n},$ and $C(x)=\sum_{n=1}^{\infty} c_{n} x^{n}$ be the generating functions for the sequences $b_{0}, b_{1}, b_{2}, \ldots$ and $c_{1}, c_{2}, c_{3}, \ldots,$ respectively. If
$$
A(x)=\frac{B(x)}{1+C(x)}=\sum_{n=0}^{\infty} a_{n+1} x^{n}
$$
then $a_{n}=(-1)^{n-1} D_{n}$ and $1+x A(-x)$ is the generating function of $D_{n}$.
For the original problem, it is equivalent to calculate the determinant:
$$
D=\left| \begin{matrix}
\dfrac{1}{2!}& 1& 0& \cdots& 0\\
\dfrac{1}{3!}& \dfrac{1}{2!}& 1& \cdots& 0\\
\dfrac{1}{4!}& \dfrac{1}{3!}& \dfrac{1}{2!}& \cdots& 0\\
\vdots& \vdots& \vdots& \ddots& \vdots\\
\dfrac{1}{\left( 2k+2 \right) !}& \dfrac{1}{\left( 2k+1 \right) !}& \dfrac{1}{\left( 2k \right) !}& \cdots& \dfrac{1}{2!}\\
\end{matrix} \right|
$$
We can see that $\displaystyle B(x)=\sum _{i=0}^{\infty } \frac{x^{i}}{(i+2)!}= \frac{-x+e^x-1}{x^2}$, $\displaystyle C(x)=\sum _{i=1}^{\infty } \frac{x^i}{(i+1)!} = \frac{-x+e^x-1}{x}$, so $A(x)=\dfrac{1}{x}+\dfrac{1}{1-e^x}$, and finally the generating function for $D_n$ is $D(x)=\dfrac{x}{e^x-1}+x$.
Notice the order $N$ of original matrix $D$ is always odd (since $N=(2k+2)-2+1=2k+3$), so the original claim is equivalent to when $n>4$ and $n$ is odd, $[x^n]D(x)=0$. It's easy to see that $D(x) -1 - x/2$ is an even function, which implies its series only contains terms of the form $x^{2k}$, thus we finished our proof.
I think this method really requires a good understanding of linear equations and Cramer's Law, along with the knowledge of generating functions.
| {
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"answer_count": 2,
"answer_id": 1
} |
An Interesting Number Theory Problem: Solve $2520=(x+y+xy)^2+2xy+2y-3x, x, y \in \mathbb Z$ Solve:
$2520=(x+y+xy)^2+2xy+2y-3x$
Please help, I tried Solving SFFT and Quadratic, but couldn't end up with a result.
Edit: I solved this much.I factorized it
$((x+1)(y+1)-1)^2+2((x+1)(y+1)-1)-5x$
Then substituted $t=((x+1)(y+1)-1)$
so $t^2+2t-5x=2520$
and after this i am stuck
Edit 2: Solve for Integers
| Denote $z=x+1, w=y+1$ then your result $$t^2+2t-5x=2520 \implies (t+1)^2-5(z-1)=2519 \\\implies w^2z^2-5z=2516=z(w^2 z - 5) \implies z|2516=2^2\cdot 17\cdot 37$$
If $z$ is even then $w^2z-5$ is odd, so $4|z, z=4u, 4u^2w^2=5u+629 \implies u\equiv -1 \pmod 4$, and $u|17\cdot 37$, therefore $u=-1, -17, -37, -629$, none of which yields a perfect square for $5u+629$.
If $z$ is odd then $wz$ is odd, $(wz)^2 \equiv 1 \pmod 4$, so $z\equiv 1 \pmod 4, z=1, 17, 37, 629$. We have the following four cases:
$z=1, w^2-5=2516$, no solution;
$z=17, 17w^2-5=148 \implies w=\pm 3 \implies x=16, y=2 \text{ or } -4;$
$z=37, 37w^2-5=68$, no solution;
$z=629, 629w^2-5=4$, no solution.
So the only solutions are $(16, 2)$ and $(16, -4)$.
| {
"language": "en",
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} |
Slightly confused about the commutator subgroup Let's look at the group of all permutations of a set $\left\{ 1,2,3 \right\}$. There are exactly 6 elements in this group: $\left( \begin{matrix}
1 & 2 & 3 \\
1 & 2 & 3 \\
\end{matrix} \right)$, $\left( \begin{matrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{matrix} \right)$, $\left( \begin{matrix}
1 & 2 & 3 \\
2 & 1 & 3 \\
\end{matrix} \right)$, $\left( \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{matrix} \right)$, $\left( \begin{matrix}
1 & 2 & 3 \\
3 & 1 & 2 \\
\end{matrix} \right)$, $\left( \begin{matrix}
1 & 2 & 3 \\
3 & 2 & 1 \\
\end{matrix} \right)$. It is usually called the symmetric group $S_{3}$. Just out of curiosity I tried to find its commutator subgroup $\left[ {{S}_{3}},{{S}_{3}} \right]$. As a home assignment, I've already proven that the commutator subgroup is a normal subgroup, hence $\forall x\in {{S}_{3}}:x\left[ {{S}_{3}},{{S}_{3}} \right]=\left[ {{S}_{3}},{{S}_{3}} \right]x$. Except for the neutral element $\left( \begin{matrix}
1 & 2 & 3 \\
1 & 2 & 3 \\
\end{matrix} \right)$, I couldn't find a single permutation $p$ with the property: $\forall x\in {{S}_{3}}:xp=px$. For instance, $\left( \begin{matrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{matrix} \right)\left( \begin{matrix}
1 & 2 & 3 \\
2 & 1 & 3 \\
\end{matrix} \right)=\left( \begin{matrix}
1 & 2 & 3 \\
2 & 3 & 1 \\
\end{matrix} \right)\ne \left( \begin{matrix}
1 & 2 & 3 \\
3 & 1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
1 & 2 & 3 \\
2 & 1 & 3 \\
\end{matrix} \right)\left( \begin{matrix}
1 & 2 & 3 \\
1 & 3 & 2 \\
\end{matrix} \right)$.
Am I missing something?
| HINT:
$$\forall x\in {{S}_{3}}:x\left[ {{S}_{3}},{{S}_{3}} \right]=\left[ {{S}_{3}},{{S}_{3}} \right]x$$
does not mean that $xp = px$ for $p \in \left[ {{S}_{3}},{{S}_{3}} \right]$. It means, for $p \in \left[ {{S}_{3}},{{S}_{3}} \right]$, we have $xp = qx$ for some $q \in \left[ {{S}_{3}},{{S}_{3}} \right]$, not necessarily $q = p$.
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
differential of homogeneous function I am trying to calculate the dy of:
$y+xy=x$
I know the formula $\frac{dy}{dx}=f(x)'$
My logic says to do so: $y(1+x)=x$ ->$y=\frac{x}{\left(1+x\right)}$
$dy=(\frac{x}{\left(1+x\right)})'dx$ -> $dy=\frac{1}{\left(1+x\right)^2}dx$
But my textbook has the answer:
$dy=\frac{1-y}{1+x}dx$
Where am I wrong?
Thank you
| Same as your answer...
$\frac{1}{(1+x)^2}=\frac{1}{1+x} \cdot \frac{1}{1+x}$
and in the equation $y+xy=x$ we have:
Subtract $xy$ on both sides: $y=x-xy$
Factor the right hand side: $y=x(1-y)$
Divide both sides by $1-y$: $\frac{y}{1-y}=x$
So we can write your answer as:
$\frac{1}{(1+x)^2}=\frac{1}{1+x} \cdot \frac{1}{1+x}$
$\frac{1}{(1+x)^2}=\frac{1}{1+\frac{y}{1-y}} \cdot \frac{1}{1+x}$
$\frac{1}{(1+x)^2}=\frac{1-y}{1-y+y} \cdot \frac{1}{1+x}$ (I multiplied first fraction in product by $1-y$)
$\frac{1}{(1+x)^2}=\frac{1-y}{1} \cdot \frac{1}{1+x}$
$\frac{1}{(1+x)^2}=\frac{1-y}{1+x}$
--------------Also if you wanted the book's answer------------
Just differentiate as is instead of putting it in explicit form and then differentiating.
$y+xy=x$
$y'+xy'+1y=1$ I did the $xy$ term using product rule...
Now let's solve for $y'$
$y'(1+x)+y=1$
$y'(1+x)=1-y$
$y'=\frac{1-y}{1+x}$
| {
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Proving $\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+1}{n+1})}k>\sum_{k=1}^n\frac{\sin(k\pi\frac{2m+3}{n+1})}k$ A few days ago I asked a question about an interesting property of the partial sums of the series $\sum\sin(nx)/n$.
Here's the link:
Bound the absolute value of the partial sums of $\sum \frac{\sin(nx)}{n}$
The proof given there left some details for me, in particular I have to prove that
$$
\sum_{k = 1}^{n}{1 \over k}
\sin\left(k\pi\frac{2m+1}{n+1}\right) >
\sum_{k=1}^{n}{1 \over k}\sin\left(k\pi\frac{2m+3}{n+1}\right)
$$
for every natural $n$ and $m$ such that $m = 0, 1, \ldots, \left\lfloor\left(n-1\right)/2\right\rfloor$. I don't think this is relatively simple; anyway I tried a few expansions with sum-to-product and viceversa formulas, but everything I tried leads to $$\sum_{k=1}^{n}\frac{\cos(2(m+1)x_k)\sin(x_k)}{k} < 0$$
where $x_k = k\pi/(n+1)$. From here I don't see any further semplification. Also I noted that $\sin(x_k)$ is always positive since $k$ is at most $n$, but we can't say much about $\cos$.
Then I tried to pair terms since for $k' = n+1-k$ we have $\sin(x_k) = \sin(x_k')$, which gives $$\sum_{k=1}^{n/2}\left[\frac{\cos(2(m+1)x_k)}{k}+\frac{\cos(2(m+1)x_{k'})}{n+1-k}\right]\sin(x_k) < 0$$ for $n$ even. Then I hoped that the term between square brackets were always negative, but it's not always the case... How can I do?
| You have already determined the local maxima of a partial sum $S_n(x)$ in $[0,\pi]$ are at
$x_{n,m} = \frac{2m+1}{n+1} \pi$ for $m = 0,1, \ldots ,\lfloor\frac{n-1}{2} \rfloor$.
We first show that the vertical distance between successive peaks is diminishing in that
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right) > \ldots $$
Note that $\frac{2m+3}{n+1}\pi = \frac{2m+1}{n+1}\pi + \frac{2\pi}{n+1}$ and
$$S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x+ \frac{n\pi}{n+1} \right) \cos \left(\frac{n+1}{2}x+\pi\right)}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ =\frac{\sin \frac{n}{2}x \cos \frac{n+1}{2}x}{\sin \frac{x}{2}} - \frac{\sin \left(\frac{n}{2}x- \frac{\pi}{n+1} \right) \cos \frac{n+1}{2}x}{\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)} \\ = \frac{\cos \frac{n+1}{2}x}{\sin \frac{x}{2}\sin \left(\frac{x}{2}+ \frac{\pi}{n+1}\right)}\left[\sin \frac{x}{2}\sin \left(\frac{n}{2}x+ \frac{\pi}{n+1}\right) -\sin \frac{x}{2}\sin \left(\frac{n}{2}x- \frac{\pi}{n+1}\right) \right]$$
Applying angle addition identities and simplifying we eventually get
$$\tag{1}S_n'\left(x\right) - S_n'\left(x+ \frac{2\pi}{n+1} \right) = \frac{\sin \frac{\pi}{n+1}\sin (n+1)x}{\cos \frac{\pi}{n+1} - \cos \left(\frac{\pi}{n+1} +x \right)}$$
Now consider the points
$$\frac{2m+1}{n+1}\pi \leqslant \frac{2m+2}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+2}{n+1} \leqslant \frac{2m+2}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+3}{n+1}\pi, \\ \frac{2m+3}{n+1}\pi \leqslant \frac{2m+4}{n+1}\pi-\frac{y}{n+1} \leqslant \frac{2m+4}{n+1} \leqslant \frac{2m+4}{n+1}\pi+\frac{y}{n+1} \leqslant \frac{2m+5}{n+1}\pi$$
Using (1) with $x_- = \frac{2m+2}{n+1}\pi-\frac{y}{n+1}$, $x'_-= x_- + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi-\frac{y}{n+1}$, $x_+ = \frac{2m+2}{n+1}\pi+\frac{y}{n+1}$, and$x'_+=x_+ + \frac{2\pi}{n+1} = \frac{2m+4}{n+1}\pi+\frac{y}{n+1}$we get
$$G'(y) = \frac{d}{dy} \left[S_n \left(x_-) - S_n \left(x'_-\right) \right] - S_n(x_+) - S_n(x'_+)\right] \\ = \frac{\sin \frac{\pi}{n+1} \sin y}{n+1}\left[\frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi-y}{n+1}\right)} - \frac{1}{\cos \frac{\pi}{n+1}- \cos \left(\frac{(2m+3)\pi+y}{n+1}\right)} \right] $$
Notice that $G(0) = 0$ and $G'(y) > 0$ for $0 < y < \pi$, so $G(\pi) > 0$ which implies
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right)> S_n\left(\frac{2m+3}{n+1}\pi \right) - S_n\left(\frac{2m+5}{n+1}\pi \right)$$
When $m = \lfloor \frac{n-1}{2}\rfloor -1 $ we have $\frac{2m+3}{n+1} < \pi < \frac{2m+5}{n+1} = \pi + \delta$. Hence,
$$S_n\left(\frac{2m+5}{n+1}\pi \right) = S_n(\pi + \delta) = S_n(\pi + \delta - 2\pi) = S_n(\delta - \pi)= - S_n(\pi - \delta) < 0,$$
Since $\frac{2m+3}{n+1}$ is the last relative extremum point before $\pi$, $S_n(\pi) = 0$, and $\frac{2m+3}{n+1} < \pi - \delta < \pi$, it follows that
$$S_n\left(\frac{2m+3}{n+1}\pi \right) > 0 > S_n\left(\frac{2m+5}{n+1}\pi \right)$$
Thus, for all $m = 0,1,\ldots, \lfloor\frac{n-1}{2}\rfloor -1$ we have
$$S_n\left(\frac{2m+1}{n+1}\pi \right) - S_n\left(\frac{2m+3}{n+1}\pi \right) > 0$$
| {
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Is the expected value of determinant times random matrix proportional to expected value of random matrix. Let $\mathbf{X}$ be a random matrix. Is it true that
$$
E[\det(\mathbf{X})^h \mathbf{X} ] \propto E[\mathbf{X}],
$$
where $h$ is a scalar and $\det(.)$ denotes the determinant?
| This is unfortunately not true. Consider the following counterexample for $h=1$:
\begin{align*}
X(\omega_{1}) &= \begin{pmatrix} 2 & \ 1 \\ 0 & \ 2 \end{pmatrix} \\
X(\omega_{2}) &= \begin{pmatrix} 1 & -1 \\ 0 & \ 1\end{pmatrix}
\end{align*}
with $P(\omega_{1}) = P(\omega_{2}) = \frac{1}{2}$.
Then we have
$$ \mathbb{E}[X] = \begin{pmatrix} \frac{3}{2} & 0 \\ 0 & \frac{3}{2} \end{pmatrix}.$$
On the other hand, you can see that $\det(X)(\omega_{1}) = 4$ and $\det(X)(\omega_{2}) = 1$, so
\begin{align*}
[\det(X) \cdot X](\omega_{1}) &= \begin{pmatrix} 8 & \ 4 \\ 0 & \ 8 \end{pmatrix} \\
[\det(X) \cdot X](\omega_{2}) &= \begin{pmatrix} 1 & -1 \\ 0 & \ 1\end{pmatrix}
\end{align*}
As such, we have
$$ \mathbb{E}[\det(X) \cdot X] = \begin{pmatrix} \frac{9}{2} & \frac{3}{2} \\ 0 & \frac{9}{2}\end{pmatrix}. $$
| {
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Find limit of formula with binomial coefficcients The Problem
Given:
$$f(n) = \frac{\sum_{i=1}^{n/2} \left[ \binom{n/2}{i} \sum_{j=0}^{i} \binom{n/2}{j} \right]}{2^n}$$
I want to prove that $\lim_{n \rightarrow \infty}f(n) = \frac{1}{2}$.
This relates to some research work I am doing and I am pretty sure it converges to $1/2$ from 'above' (or at least the data suggests that) but I would really like to have some solid proof about this limit.
What I've tried
I've tried to sandwich the limit by substituting the term $\sum_{j=0}^{i} \binom{n/2}{j}$ with a minimum value, i.e. $\binom{n/2}{0}+\binom{n/2}{1}=1+\frac{n}{2}$ (common to all subterms), which leads to $\lim_{n \rightarrow \infty}f(n) = 0$ and with a maximum value, i.e. $\sum_{j=0}^{n/2} \binom{n/2}{j}=2^{n/2}$ (adding extra subterms), which leads to $\lim_{n \rightarrow \infty}f(n) = 1$, but well, I need something better :) I've tried using the some other related formulas but couldn't find something better than that. Maybe it can be solved more straightforward though?
| Let $m=n/2$ for ease of notation and let a symbol $\mathbf x$ in bold denote $\binom mx$. Let's write out the numerator for $m=4$:
$$\begin{align}&\mathbf{1(0+1)}+\\
&\mathbf{2(0+1+2)}+\\
&\mathbf{3(0+1+2+3)}+\\
&\mathbf{4(0+1+2+3+4)}
\end{align}$$
By the symmetry of binomial coefficients, the above expression is equal to
$$\begin{align}\mathbf{(0+1+2+3+4)0}&\\
+\mathbf{(1+2+3+4)1}&\\
+\mathbf{(2+3+4)2}&\\
+\mathbf{(3+4)3}&
\end{align}$$
If we add the two together we get
$$\begin{align}
&\mathbf{0(0+1+2+3+4)}+\\
&\mathbf{1(0+1+2+3+4+1)}+\\
&\mathbf{2(0+1+2+3+4+2)}+\\
&\mathbf{3(0+1+2+3+4+3)}+\\
&\mathbf{4(0+1+2+3+4)}
\end{align}$$
$$=(\mathbf{0+1+2+3+4})^2+\mathbf 1^2+\mathbf 2^2+\mathbf 3^2=2^{2×4}+\mathbf 1^2+\mathbf 2^2+\mathbf 3^2$$
It is easy enough to derive the generalisation:
$$f(m)=\frac{(1/2)(2^{2m}+\sum_{k=1}^{m-1}\mathbf k^2)}{2^{2m}}$$
As $n\to\infty$ (and hence $m\to\infty$ too), $\sum_{k=1}^{m-1}\mathbf k^2$ does not grow as fast as $2^{2m}$ – it is smaller by a factor of $\sqrt{\pi m}$ – so it becomes negligible:
$$\lim_{m\to\infty}f(m)=\lim_{m\to\infty}\frac{(1/2)2^{2m}}{2^{2m}}=\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$? Question
Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.
My attempt
Let $f(x) = 3x^3 + 7x^2 - 48x + 49$
$\text{Using Remainder Theorem,}$
\begin{align}
f(-k) &= f(k) \\
3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}
$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$
\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}
$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $
My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$
| By the factor theorem, $3x^3+7x^2-48x+49 = (ax+b)(x^2 - k^2) + c$. In other words, dividing by a quadratic term leaves a linear quotient and a constant remainder, for the powers on both sides to match.
Observe that there is only one way to make the terms of $x^3, x^2, x$. Thus $ax^3 = 3x^3 \implies x= 3$ and hence $-ak^2 = -48 \implies k^2 = 16, k = ±4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\prod_{n=1}^{6} (g(a_n))$ Question:
Let $f(x)=x^6-2x^3-8$ and $g(x)=x^2+2x+4$. Let $a_1$ through $a_6$ be its roots. Find the value of $\displaystyle \prod_{n=1}^{6} (g(a_n))$
My process:
I first thought of setting $z=a_i^2+2a_i+4$ and then putting the value back into $f(z)$ and then from there finding the product using Vieta's. But the process turned incredibly long and tedious and I found it to be impossible to do by hand.
So I just went to to the brute force method, expanding the whole expression out. Writing all the terms would be really tedious so I would like to abbreviate it. Let $P= a_1 \cdot a_2 \cdots \cdot a_6$ and let $e_n$ denote the sum of roots taken $n$ at a time.
Here is the expression I got:
$$\prod_{i=1}^{6} (a_i^2+2i+4)=P^2+2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P+2^5*4*e_5+2^4*4^2*e^4+2^3*4^3*e_3+2^2*4^4*e_2+2*4^5*e_1+4_6+4(\sum_{cyc}(a_1a_2a_3a_4a_5)^2))^+4^2(\sum_{cyc}(a_1a_2a_3a_4)^2)+4^3\sum_{cyc}(a_1a_2a_3)^2+4^4\sum_{cyc}(a_1a_2)^2+4^5\sum_{cyc}(a_1)^2$$
Which is a monster in its own right. However, I was extremely interested in the patterns that kept popping up. Like the $2e_5P+2^2e_4P+2^3e_3P+2^4e_2P+2^5e_1P+2^6P$ where the power of 2 and the index of $e$ keeps increasing and decreasing with each other. Therefore I wondered if there could be any general formula for any $f(x)$ and $g(x)$.
And that is my question, is there any way to find the answer efficiently without all this unnecessary calculation; and also can this process be generalised to any polynomials $f(x)$ and $g(x)$. And if it cannot be generalised, then is there any sort of algorithm or methodical way that one can take while solving this type of problem?
| Note that $(x^2+2x+4)(x-2)=x^3-8$. So the product of the $g(a_i)$ is $P/Q$, where
$$P=\prod(a_i^3-8)$$
$$Q=\prod(a_i-2)$$
As @mathcounterexamples.net notes, the roots of $f$ are the cube roots of $4$ and $-2$; so $a_i^3-8$ takes the values $-4,-4,-4,-10,-10,-10$. Hence $P=64000$.
And the product of $a_i-2$ taken over all roots of $f(x)$ is the product of $a_i$ taken over all roots of
$$f(x+2)=(x+2)^6-2(x+2)^3-8$$
which is simply the constant term $2^6-2.2^3-8=40$.
So we get $P/Q=1600$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Complex coefficients giving real roots? Edit: I figured it out, I just miscalculated. No, there are no complex solutions to k.
For what values of $k$ does the equation $kx^2+3kx+k=-9-kx$ have real roots? Just to simplify it, I'll write the equation as $kx^2+4kx+k+9=0$. Now the obvious solution would probably be to plug it into the quadratic formula and try to solve such that the discriminant is greater than zero. In the quadratic formula, $a = k$, $b = 4k$, and $c = k + 9$. Plugging these values in, I get
$$\begin{align}
&\frac{-4k±\sqrt{(4k)^2-4(k)(k+9)}}{2k}\\
&=\frac{-4k±\sqrt{16k^2-4k^2-36k}}{2k}\\
&=\frac{-4k±\sqrt{12k^2-36k}}{2k}\\
&=-2±\frac{\sqrt{12k^2-36k}}{2k}\\
&=-2±\frac{\sqrt{3k^2-9k}}{k}
\end{align}$$
Since the only thing that matters is finding real values, $-2$ can just be ignored as well as the $\pm$ sign, since nothing really changes anyways. Now the logical thing to do here would be to take the discriminant, set it to be greater than or equal to zero, and get the inequality for all the real values of $k$ (in case you're wondering, $k ≤ 0$ and $k ≥ 3$). But the more interesting part of the question would be finding the complex solutions to $k$. While attempting this, I use assumptions:
First, the square root of a complex number with an imaginary part is always another complex number with an imaginary part. I don't see how this isn't true and I haven't found anything that says otherwise yet, so it seems good to me.
Second, if an equation with real and imaginary parts sums to $0$, then the real and imaginary parts alone will also sum to zero. I don't see how this is false either, since imaginary parts can't contribute to real parts, so both must just equal zero.
So in order to solve this problem, I set $k$ to equal $a+bi$ such that $a$ and $b$ are real numbers. Therefore,
$$\begin{align}
&-2±\frac{\sqrt{3k^2-9k}}{k}\\
&=-2±\frac{\sqrt{3(a+bi)^2-9(a+bi)}}{a+bi}\\
&=-2±\frac{\sqrt{3(a^2+2abi-b^2)-9a-9bi)}}{a+bi}\\
&=-2±\frac{\sqrt{3a^2+6abi-3b^2-9a-9bi}}{a+bi}\\
\end{align}$$
Now if the numerator is equal to the denominator, then the fraction would equal 1 so I set it to that.
$$\begin{align}
\sqrt{3a^2+6abi-3b^2-9a-9bi} &= a+bi\\
3a^2+6abi-3b^2-9a-9bi &= (a+bi)^2\\
3a^2+6abi-3b^2-9a-9bi &= a^2+2abi-b^2\\
2a^2+4abi-2b^2-9a-9bi &= 0
\end{align}$$
Taking imaginary parts only:
$$\begin{align}
4abi-9bi &= 0\\
4abi &= 9bi\\
4ab &= 9b
\end{align}$$
Assuming that $b ≠ 0$:
$$\begin{align}
4a &= 9\\
a &= \frac{9}{4}
\end{align}$$
Now taking real parts only:
$$\begin{align}2a^2-2b^2-9a &= 0\\
2\left(\frac{9}{4}\right)^2-2b^2-9\left(\frac{9}{4}\right) &= 0\\
2\left(\frac{9}{4}\right)^2-9\left(\frac{9}{4}\right) &= 2b^2\\
\frac{486}{16} &= 2b^2\\
\frac{243}{16} &= b^2\\
±\frac{9\sqrt{3}}{4} &= b
\end{align}$$
Therefore $k = \frac{9}{4} + i\frac{9\sqrt{3}}{4}$ should be a solution. Whether or not cases like $k = \frac{9}{2} + i\frac{9\sqrt{3}}{2}$ that are multiples of the original answer would work don't really matter, but this seems like it should work. Another case that should also work would be setting the numerator of the original fraction to equal to zero. This also produces a complex solution for $k$.
The question is, can complex coefficients create real roots for quadratic equations, or did I calculate/assume/do the question wrong? If it is possible, did I miss any solutions? Any help would be appreciated.
| You mention that we should be able to set the numerator of $\frac{\sqrt{3k^2 - 9k}}{k}$ to $0$. This would give $3k^2 = 9k$ and so $k = 3$, (discarding $k = 0$). This isn't complex so it isn't providing a solution to your question.
Other than that, your proposed solutions for $k$ do not seem to provide real solutions to the quadratic given.
Note that you can rewrite your quadratic as $k ( x^2 + 4x +1) = -9$. So we have
$$k = \frac{-9}{ x^2 + 4x +1}$$
So if we plug in real numbers for $x$ (which would be the proposed solutions to your quadratic), then $k$ will also be real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Weird geometry question ($AG^3=CE^2\times AB$) $ABC$ is a triangle with $AB = 2AC$ and $E$ is the midpoint of $AB$. The point $F$ lies on the line $EC$ and the point $G$ lies on the line $BC$ such that $A, F, G$ are collinear and $FG = AC$. Show that $AG^3=CE^2\times AB$
| $AC=\frac{1}{2}BC=FG$ therefore $G$ coincides with $B$ and $F$ with $E$ (see the picture below).
From cosine law in triangle $AEC$ we have $CE^2=x^2+x^2-2x^2\cos 2t$
Applying the given relationship $AG^3=CE^2\cdot AB$ we get
$$(2 x)^3= \left(-2 x^2 \cos (2 t)+x^2+x^2\right)(2 x)$$
$$8x^3=8 x^3 \sin ^2 t\to \sin t=1\to t=90°\to 2t=180°$$
which is clearly impossible.
$$
...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$
I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$
From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$
Hence $1+9abc\ge 4(ab+bc+ac)$.
This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?
| WLOG, assume $c = \min(a, b, c)$.
Let $P(x) = a^3 + b^3 + c^3 + abc x$ and $Q(x) = 2(\frac{a+b}{2})^3 + c^3 + (\frac{a+b}{2})^2 c x$ where $x \ge 0$.
We have
\begin{align}
P(x) - Q(x) &= a^3 + b^3 - 2(\tfrac{a+b}{2})^3
- ((\tfrac{a+b}{2})^2 - ab)c x \\
&= \frac{3}{4}(a+b)(a-b)^2 - \frac{1}{4}(a-b)^2 cx\\
&= \frac{1}{4}(a-b)^2(3a + 3b - cx)\\
&\ge \frac{1}{4}(a-b)^2(3a + 3b - \tfrac{a+b}{2} x)\\
&= \frac{1}{8}(a-b)^2(a+b)(6 - x). \tag{1}
\end{align}
Also, since $a + b + c = 1$, we have $Q(x) = \frac{1}{4}(x+3)c^3 + \frac{1}{4}(-2x+3)c^2 + \frac{1}{4}(x-3)c + \frac{1}{4}$.
Now, we split into two cases:
If $d > \frac{15}{4}$, by (1), we have
\begin{align}
a^3 + b^3 + c^3 + abcd &\ge a^3 + b^3 + c^3 + \frac{15}{4} abc\\
&= P(\tfrac{15}{4}) \\
&\ge Q(\tfrac{15}{4}) \\
&= \frac{1}{4}(\tfrac{15}{4} +3)c^3 + \frac{1}{4}(-2\cdot \tfrac{15}{4} + 3)c^2 + \frac{1}{4}(\tfrac{15}{4}-3)c + \frac{1}{4}\\
&= \frac{3}{16}c(3c-1)^2 + \frac{1}{4}\\
&\ge \frac{1}{4}\\
&= \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).
\end{align}
If $d\le \frac{15}{4}$, by (1), we have
\begin{align}
a^3 + b^3 + c^3 + abcd &= P(d) \\
&\ge Q(d)\\
&= \frac{1}{4}(d+3)c^3 + \frac{1}{4}(-2d+3)c^2 + \frac{1}{4}(d-3)c + \frac{1}{4}\\
&= \frac{1}{9} + \frac{1}{27}d + \frac{1}{108}(3c-1)^2(3cd + 9c - 4d + 15)\\
&\ge \frac{1}{9} + \frac{1}{27}d\\
&= \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).
\end{align}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$ To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?
| $$\int\frac{dx}{2+x+\sqrt{1-x^2}}$$
Substitute $x= \sin 2u;\;dx=2\cos 2u$
$$\int \frac{2\cos (2 u)}{2+\sin (2 u)+\cos (2 u)} \, du$$
$$\int \frac{2\cos ^2 u-2\sin ^2 u}{2-\sin ^2 u+\cos ^2 u+2 \sin u \cos u}\,du$$
$$\int \frac{2\cos ^2 u-2\sin ^2 u}{\sin ^2 u+3 \cos ^2 u+2 \sin u \cos u}\,du$$
divide numerator and denominator by $\cos^2 u$
$$\int \frac{2-2\tan ^2 u}{\tan ^2 u +2 \tan u +3}\,du$$
substitute $\tan u = t\to dt=\frac{du}{1+u^2}$
$$\int \frac{2-2t^2}{\left(t^2+1\right) \left(t^2+2 t+3\right)}\,dt$$
using partial fraction
$$\int \left(\frac{1-t}{t^2+1}+\frac{t-1}{t^2+2 t+3}\right)\,dt$$
$$\frac{1}{2} \left(-\log \left(t^2+1\right)+\log \left(t^2+2 t+3\right)+2 \arctan t-2 \sqrt{2} \arctan\left(\frac{t+1}{\sqrt{2}}\right)\right)+C$$
$t=\tan u$ and $x=\sin 2u$ we have $t=\frac{1-\sqrt{1-x^2}}{x}$
$$\frac{1}{2} \log \left(\sqrt{1-x^2}+x+2\right)+\arctan\left(\frac{1-\sqrt{1-x^2}}{x}\right)-\sqrt{2} \arctan\left(\frac{-\sqrt{1-x^2}+x+1}{\sqrt{2} x}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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