Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What is the area of the triangle here?
We are given that the angle of $BAD$ is $2\alpha$ and the angle of $DAC$ is $\alpha$. $|AC| = 10$, $|BD| = 6$, $|DC| = 5$ units. Find the area of the triangle $ABC$.
The answer would be $33$. We need to show that if we drop an altitude from $A$ to $BC$ at point $E$, $|AE| = 6, |EC| = 8$. Somehow $\triangle AEC$ becomes $6$-$8$-$10$ triangle. Since $|AE| = 6$ and $|BC|=11$, the area becomes $33$.
But how can we prove that?
| Draw $AE$, the angle bisector of $\angle BAD$. Now $\angle A$ is divided into $3$ equal angles. Say, $DE = x$. As $AD$ bisects $\angle CAE$, using angle bisector theorem,
$ \displaystyle AE = 2 DE = 2x$
Using formula for the length of angle bisector,
$AD^2 = AE \cdot AC - DE \cdot CD = 15x$
Now using the fact that $AE$ is angle bisector of $\angle BAD$,
$ \displaystyle \frac{AB}{AD} = \frac{BE}{DE} \implies \frac{AB}{\sqrt{15x}} = \frac{6-x}{x}$
Or, $~AB \cdot AD = 15 (6-x)$
Again using formula for the length of angle bisector, $AE^2 = AB \cdot AD - AD \cdot BE$
Or, $4 x^2 = 90 - 15x - x (6-x)$
Or, $x^2 + 7x - 30 = 0$
Or, $(x-3) (x + 10) = 0$
The only valid solution is $x = 3$. That shows $AE = 2x = 6, CE = 5 + x = 8$ and we conclude $AE \perp BC$
$\therefore [ABC] = 33$
| {
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"timestamp": "2023-03-29T00:00:00",
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The solution to $z^2+(1-i)z-1=0$? I am teaching myself complex numbers, yet I can't solve this exercise:
Find all complex number solutions to $z^2+(1-i)z-1=0,$ and provide them in standard $z=a+bi$ format.
I've tried using the classic formula $\dfrac{-b \pm \sqrt{D} }{2a},$
and have found that $D=4-2i$, which leads to the solution: $$
z=-\frac{1}{2} +\frac{i}{2}\pm\sqrt{1-\frac{i}{2}}.
$$
I am however unable to take the square root of either $1-\dfrac{i}{2}$ or $D=4-2i$, which means I cannot express the solution as $a+bi$.
I have tried inputting this into Wolfram Alpha, Symbolab and other online complex calculators and they all give different answers, none of which are the above answer, nor are they the same as the text gives, which is:
$z=\dfrac{\pm\sqrt{\sqrt{5}+2}-1}{2}+\dfrac{1\pm\sqrt{\sqrt{5}-2}}{2}i.$
I do not see an obvious link between the given equation and the given solution. What am I missing?
| Your answer is correct. Note the principal square root of a complex number is
$$\sqrt{x+iy}=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}+isgn(y)\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} .$$
Applied to your case gives
$$\sqrt{1-i/2}=\frac{1}{2}\sqrt{2+\sqrt{5}}-\frac{i}{2}\sqrt{-2+\sqrt{5}} .$$
So the solution to your quadratic is
$$z=-\frac{1}{2} +\frac{i}{2}\pm\sqrt{1-\frac{i}{2}}\\
=\dfrac{-1\pm\sqrt{2+\sqrt{5}}}{2}+\dfrac{1\mp\sqrt{-2+\sqrt{5}}}{2}i$$
(note the ordering of the plus, minus signs).
| {
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Calculate $\gcd(2a+4b, 2a +8b)$, if $a\equiv b \pmod{\! 5},\ 6a+11b = 5$ Given:
*
*$a$ is even
*$6a+11b=5$
*$a-b=0\pmod 5$
Q: Calculate $\gcd(2a+4b,2a+8b)$
My try:
We know there is some $i$ such that $a=2i$, plus from 3 we know there is some $j$ such that: $a-b=5j$ which means $b=a-5j=2i-5j$. From 2, we get: $34i-55j=5$
So, $$\gcd(2a+4b,2a+8b)=\gcd(12i-20j,20i-40j).$$
I'm stuck here, how to continue?
| This helps with the last step after @Buraian's answer.
Now $\gcd(a,b) \in \{1,5\}$ because there is an integral linear combination of $a$ and $b$ that sums to $5$; in particular, $6a+11b=5$.
Meanwhile, as $a$ is even, it follows that $\frac{a}{2}$ is an integer, and as $\gcd(2,5)=1$, if $5|a$ then $5|\frac{a}{2}$. [In fact, let $k$ be any integer satisfying both $\gcd(k,5)=1$ and $k|a$, such as $k=2$. Then as $\gcd(a,b) \in \{1,5\}$, it follows that $\gcd\left(\frac{a}{k},b\right) = \gcd(a,b)$.] So from this it follows that $\gcd\left(\frac{a}{2},b\right)=\gcd(a,b)$.
So let us now calculate $\gcd(a,b)=\gcd\left(\frac{a}{2},b\right)$. Then from the first answer by @Buraian, $4\times \gcd(a,b) = \gcd(2a+4b,2a+8b)$.
Now, we claim that both $a$ and $b$ divide $5$. [Indeed, let $a \pmod 5 = r$. Then $b \pmod 5 =r$ as well, by the condition that $5|(a-b)$. However, on the one hand, (a) $6a+11b \pmod 5$ is $a \pmod 5 + b\pmod 5$ which is $2r$. As, on the other hand, (b) $6a+11b = 5 \equiv_5 0$ by hypothesis, it follows from putting (a) and (b) together that $2r \equiv_5 0$, so $r$ must be $0$, and thus indeed, both $a$ and $b$ divide $5$.] So $\gcd(a,b)$ must be a multiple of $5$. However, as noted already, $\gcd(a,b)$ is either $1$ or $5$, and so $\gcd(a,b)$ must indeed be exactly $5$.
As noted above, $\gcd(2a+4b,2a+8b)=4\gcd(a,b)$, so $\gcd(2a+4b,2a+8b)$ must be $4 \times 5 = 20$.
| {
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Solving $(x+1)^2-x^2=0$ two ways gives different results In mathematics, there are several method of finding the solution of a particular problem. But in this equation,
$$(x+1)^2-x^2=0$$
Method 1
Using, $(a+b)^2=a^2+2ab+b^2$,
$$x^2+2x+1-x^2=0$$
So, $x=-1/2$ is the solution.
Method 2
$$(x+1)^2-x^2=0$$
$$(x+1)^2=x^2$$
$$x+1=x$$
$$1=0$$
But it is not possible.
So, what am I missing in this question? Please explain to me.
| Another way to see what went wrong is to use the factorization $a^2-b^2=(a+b)(a-b)$ like this:
$$(x+1)^2-x^2=0\\(x+1+x)(x+1-x)=0\\x+1+x=0~~~~\text{ or }~~~~x+1-x=0\\x=-\frac12~~~~\text{ or }~~~~1=0$$
Of the two resulting equations, only the first gives a valid solution for $x$. So in this case there is no $a-b=0$ solution, only an $a+b=0$ one. Using this factorization is of course completely equivalent to going from $a^2=b^2$ to $a=\pm b$ directly.
| {
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How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1$, $$A = \frac{1}{4}$$
However, $$\frac{1}{4(1 + x)} + \frac{1}{2 (1 - x)^2}$$ doesn't seem to equal $$\frac{1}{(1 + x)(1 - x)^2}$$
How do we decompose this into partial fractions?
Thanks
| Hint: $$1=\frac{(1+x)+(1-x)}2.$$ This should make the exercise trivial.
| {
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Integrate $\int_0^{\infty} \frac{\sin^2 x}{\cosh x\>+\>\cos x}\frac{dx}x $ It is known that (see for example)
\begin{align}
&\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4\\
&\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x =\frac\pi8
\end{align}
I am wondering if the similar integral below
$$\int_0^{\infty} \frac{\sin^2 x}{\cosh x+\cos x}\frac{dx}x $$
can be evaluated to a simple close-form as well. I have tried the same approaches for the known integrals above and they do not lead to any results. Alternatively, I have also manipulated the integral and expressed it in the equivalent form
$$\int_0^\infty e^{-x}\cos x\tanh x \>\frac{dx}x
$$
which, though appearing simpler, is not any easier.
| I'm not sure how you showed the two integrals are equivalent, but the following is an evaluation of $$\int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x}.$$
For $\Re(s) >0$, we have
$$ \begin{align} \int_{0}^{\infty} \tanh(t) e^{-st} \, \mathrm dt &= \int_{0}^{\infty} \frac{1-e^{-2t}}{1+e^{-2t}} \, e^{-st} \, \mathrm dt \\ &= \int_{0}^{\infty} (1-e^{-2t})e^{-st} \sum_{n=0}^{\infty} (-1)^{n}e^{-2tn} \, \mathrm dt \\ &= \sum_{n=0}^{\infty} (-1)^{n}\int_{0}^{\infty}\left(e^{-(2n+s)t} -e^{-(2n+s+2)t} \right) \, \mathrm dt \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+s} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2n+s+2} \\ &= \frac{1}{2} \left(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+\frac{s}{2}} - \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+\frac{s}{2}+1} \right) \\ &\overset{(1)}{=} \frac{1}{4} \left(\psi \left(\frac{s}{4} + \frac{1}{2}\right)- \psi \left( \frac{s}{4}\right) - \psi \left(\frac{s}{4}+1 \right) + \psi \left(\frac{s}{4}+\frac{1}{2} \right)\right) \\ &\overset{(2)}{=} \frac{1}{2} \left( \psi \left(\frac{s}{4}+ \frac{1}{2} \right) - \psi \left(\frac{s}{4} \right) - \frac{2}{s} \right). \end{align}$$
Therefore, $ \begin{align} \int_{0}^{\infty} e^{-x} \cos (x) \tanh(x) \, \frac{\mathrm dx}{x} &= \Re\int_{0}^{\infty} e^{-(1+i)x} \frac{\tanh (x)}{x} \, \mathrm dx \\ &= \Re \int_{0}^{\infty}e^{-(1+i)x} \tanh(x) \int_{0}^{\infty} e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \Re \int_{0}^{\infty} \int_{0}^{\infty}\tanh(x) e^{-(1+i+t)x} \, \mathrm dx \, \mathrm dt \\ &= \Re \int_{0}^{\infty} \frac{1}{2}\left(\psi \left(\frac{1+i+t}{4}+ \frac{1}{2} \right) - \psi \left(\frac{1+i+t}{4} \right) - \frac{2}{1+i+t} \right) \, \mathrm dt \\ &= \Re \left(2 \ln\Gamma \left(\frac{3+i+t}{4} \right) -2 \ln \Gamma \left(\frac{1+i+t}{4} \right) -\ln (1+i+t)\right) \Bigg|_{0}^{\infty} \\ &\overset{(3)}= \Re \left(-2 \ln(2) - 2 \ln \Gamma \left(\frac{3+i}{4} \right)+2 \ln \Gamma \left(\frac{1+i}{4} \right) + \ln(1+i)\right)\\ &= - \frac{3 \ln (2)}{2} + 2\Re \left(\ln \Gamma \left(\frac{1+i}{4} \right)- \ln \Gamma \left(\frac{3+i}{4} \right) \right). \end{align}$
$(1)$ For $\Re(z) > 0$, $\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k+z} = \frac{1}{2} \left(\psi \left(\frac{z+1}{2} \right) - \psi \left(\frac{z}{2} \right) \right).$
$(2)$ Recurrence relation of the digamma function
$(3)$ For $a >0$, $\ln \Gamma(z_{1}+ax)- \ln \Gamma(z_{2}+ax) \sim (z_{1}-z_{2}) \ln(ax) + \mathcal{O} \left(\frac{1}{x}\right)$ as $x \to \infty$.
| {
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Limit of some probability of the empirical mean of independent random variables Let $X_{1}, X_{2}, \ldots, X_{n}$ be a sequence of independent, standard Normal, real-valued random variables, and consider the empirical mean $\hat{S}_{n}=\frac{1}{n} \sum_{i=1}^{n} X_{i}$. Since $\hat{S}_{n}$ is again a Normal random variable with zero mean and variance $1 / n$, it follows that for any $\delta>0$,
$$
P\left(\left|\hat{S}_{n}\right| \geq \delta\right) \underset{n \rightarrow \infty}{\longrightarrow} 0,
$$
and for any interval $A$ by CLT:
$$
P\left(\sqrt{n} \hat{S}_{n} \in A\right) \underset{n \rightarrow \infty}{\longrightarrow} \frac{1}{\sqrt{2 \pi}} \int_{A} e^{-x^{2} / 2} d x
$$
Note now that by a change of variable
$$
P\left(\left|\hat{S}_{n}\right| \geq \delta\right)=1-\frac{1}{\sqrt{2 \pi}} \int_{-\delta \sqrt{n}}^{\delta \sqrt{n}} e^{-x^{2} / 2} d x= \frac{1}{\sqrt{2 \pi}} \int_{|x|> \delta \sqrt n} e^{-x^{2} / 2} d x
$$
Now in [Large deviation techniques and applications, Amir Dembo Ofer Zeitouni] it is claimed that
$$
\frac{1}{n} \log P\left(\left|\hat{S}_{n}\right| \geq \delta\right) \underset{n \rightarrow \infty}{\longrightarrow}-\frac{\delta^{2}}{2}
$$
How do you see it?
| Attempting to flesh out Henry's comment:
The left-hand side is $\frac{1}{n} \log [2(1-\Phi(\delta \sqrt{n}))]$, which has the same limit as $\frac{1}{n} \log(1-\Phi(\delta \sqrt{n}))$ (if the limits exist).
From integration by parts, we have the following Mills ratio bounds for $z>0$:
$$\frac{1}{z} - \frac{1}{z^3} < \frac{1-\Phi(z)}{\phi(z)} < \frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5}.$$
[Perhaps this inequality is overkill for your particular question, is there a simpler result?]
This means
$$
-\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) + \frac{1}{n}\log\left(\frac{1}{z} - \frac{1}{z^3}\right)
< \frac{1}{n}\log(1-\Phi(z))
< -\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) + \frac{1}{n} \log\left(\frac{1}{z} - \frac{1}{z^3} + \frac{3}{z^5}\right).$$
That is,
$$
\frac{1}{n} \log(1-z^{-2})
<\frac{1}{n} \log(1-\Phi(z)) - \left(-\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) - \frac{\log z}{n}\right)
< \frac{1}{n} \log(1-z^{-2} + 3 z^{-4})$$
If you plug in $z=\delta\sqrt{n}$ you can show $-\frac{z^2}{2n} + \frac{1}{2n} \log(2\pi) - \frac{\log z}{n} \to -\frac{\delta^2}{2}$ and that the two outer bounds converge to zero.
| {
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Find $m$ such that the inequality has $4$ integer solutions.
Find all values of $m$ such that the inequality: $\log(60x^2+120x+10m-10)>1+3\log(x+1)$ has exactly $4$ integer solutions?
The first thing I did was to gather from the inequality;
$\log(60x^2+120x+10m-10)>1+3\log(x+1)$
$\Leftrightarrow 1+\log(6x^2+12x+m-1)>1+\log(x^3+3x^2+3x+1)$
$\Leftrightarrow 6x^2+12x+m-1>x^3+3x^2+3x+1$
$\Leftrightarrow 0>x^3-3x^2-9x-m+2$
Right here, I have no idea of what to do next, any help would be appreciated. Thank you!
| Consider the inequality
$$m - 2 > x^3 - 3x^2 - 9x. \tag1 $$
As $~x \to -\infty, ~$ the RHS of (1) above goes to $-\infty.$
Therefore, it is tempting, but wrong to assume that regardless of the (fixed) value of $m$ chosen, there will be an infinite number of (negative) integer solutions.
The reason that this is wrong is that the original problem involves logarithms, and you can not take the logarithm of a non-positive number. Therefore, there are two additional (hidden) constraints:
*
*$(x + 1) > 0 \iff x > -1$
*$(60x^2+120x+10m-10) > 0 $
$\iff (6x^2 + 12x + (m-1) > 0$
$\iff (36x^2 + 72x + (6m - 6) > 0 $
$\iff (6x + 6)^2 + (-36 + 6m - 6) > 0 $
$\iff [6(x + 1)]^2 > 42 - 6m \iff $
$\iff 6 \times [(x + 1)^2] > 7 - m$
$\iff m > 7 - \{6 \times [(x + 1)^2]\}$.
Further, given the problem's preCalculus tag, only purely algebraic methods of attack are permitted.
Let $f(x) = x^3 - 3x^2 - 9x.$ Then, the range of $m$ must be identified so that there will be exactly $4$ integer values that satisfy all of the following constraints:
*
*$(m - 2) > f(x).~$ : Constraint-1
*$x > -1.~$ : Constraint-2
*$m > 7 - \{6 \times [(x + 1)^2]\}.~$ : Constraint-3
Then, considering Constraint-2 above,
$f(0) = 0.$
$f(1) = -11$.
$f(2) = -22$.
$f(3) = -27$.
$f(4) = -20$.
$f(5) = 5$.
$f(6) = 54.$
Edit
For $x \geq 6,$ you have that
$f(x+1) - f(x) = (3x^2 + 3x + 1) - 3(2x + 1) - 9$
$= ~3x^2 - 3x - 11 = 3(x^2 - x) - 11 > 0.$
Therefore, for $x \in \Bbb{Z_{\geq 6}}, f(x)$ is strictly increasing.
A natural try is to explore what happens if
$(m - 2) > 0$.
You would also need
$m > 7 - \{6 \times [(x + 1)^2]\}$.
For $x \geq 0$, this will automatically be satisfied if $m > 1.$
Therefore, if $(m - 2) = 0$ then you will have the $4$ solutions of $x \in \{1,2,3,4\}.$
So, the question is, what happens if $m < 2$, or $m > 2 ~$ ? $~~m > 2$ must be excluded, since that would automatically add the solution $x = 0$.
Note that $f(1) = -11.$
If $(x = 1)$, then you need that
*
*$(m - 2) > -11$
*$m > 7 - \{6 \times [(2)^2]\} = -17.$
Therefore, the preliminary answer seems to be
$$-9 < m \leq 2 \tag2 $$.
Note that if $m \leq -9$, then $x = 1$ is not a solution, $x = 0$ is not a solution, and $x \geq 5$ is not a solution. Therefore, the range given in (2) above is the final answer.
| {
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Find the power of the matrix. Let $A = \left( {\begin{array}{*{20}{c}}
0&1&1\\
1&0&1\\
1&1&0
\end{array}} \right)$.
I want to find $A^k,$ where $k \in N$. So far I calculated $A^2, A^3, A^4,...$ but I can not see the general formula for $A^k$. Here are $A^2, A^3, A^4, A^5$.
Not sure if this leads to anything but I found the general formula for $B^k$, where $B = \left( {\begin{array}{*{20}{c}}
1&1&1\\
1&1&1\\
1&1&1
\end{array}} \right)$.
${B^k} = \left( {\begin{array}{*{20}{c}}
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}
\end{array}} \right)$
Thanks in advance.
| Hint. From the first few examples you gave, it is easy to conjecture that $A^n = a_nB+(-1)^nI$ for some sequence $(a_n)$ that starts off $1,1,3,5,11,\dots$. If this is the case, then we would have $$a_{n+1}B+(-1)^{n+1}I=A(a_nB+(-1)^nI)=a_nAB+(-1)^nA$$ for all $n$. Rearranging, and substituting the relationship $A+I=B$, this is equivalent to $$(a_{n+1}+a_n+(-1)^{n+1})B=a_nB^2.$$Now it is easy to see that $a_{n+1}+a_n+(-1)^{n+1}=3a_n$, so we have a recurrence for $(a_n)$ which can be solved by standard methods.
| {
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Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.
But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?
As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.
| Since $$\tan\big(\arctan x+\arctan y\big)\equiv\frac{x+y}{1-xy}$$ and $\arctan$ has principal range $\left(-\frac\pi2,\frac\pi2\right),$ thus, for each $(x,y)$ and some $k\in\{-1,0,1\},$ $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)+k\pi.$$
Therefore, \begin{align}&\arctan(x+1) + \arctan(x) + \arctan(x-1) = \arctan(3)\\
\implies&\arctan\left(\frac{x^3-4x}{3x^2-2}\right)=\arctan3+n\pi\quad\text{for some }n\in\{-2,-1,\ldots,2\}\\
\implies&\frac{x^3-4x}{3x^2-2}=3\\
\implies&\cdots\\
\implies&x=-1\:\:\text{or}\:\:5\pm\sqrt{19}\end{align}
$5-\sqrt{19}$ is the only non-extraneous solution. Desmos check.
| {
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Solving integral $\int \frac{\sqrt{x^2 + x}}{x}dx$ (problem 36 in section $6.25$ in Tom Apostol's calculus) Integrals which involve $\sqrt{(cx + d)^2 - a^2}$ could often be simplified if we do a substitution $cx + d = a \sec t$. If we take a concrete example, $\int \frac{\sqrt{x^2 + x}}{x}dx$, then the substitution would be
$$
x + \frac{1}{2} = \frac{1}{2}\sec t \\
dx = \frac{1}{2} \sec t \tan t dt \\
t = arcsec (2x+1)
$$
If I carry on that substitution, I get to
$$
\frac{1}{2} \int \frac{\sqrt{\tan ^ 2t}}{\sec t - 1} \sec t \tan t dt
$$
As far as I understand, that is
$$
\frac{1}{2} \int \frac{|\tan t|}{\sec t - 1} \sec t \tan t dt
$$
Now, there are two cases, $\tan t \ge 0$ and $\tan t < 0$.
$\tan t \ge 0$ case leads me to the solution also written in the book (and here):
$$
\frac{1}{2} \int \frac{1 + \cos t}{\cos ^ 2 t} dt = \\
\frac{1}{2} \tan t + \frac{1}{2} \log{\frac{1 + \tan \frac{t}{2}}{1 - \tan \frac{t}{2}}} + C = \\
\sqrt{x^2 + x} - \frac{1}{2}\log{|2\sqrt{x^2 + x} + 2x + 1|} + C
$$
But if I try the second case (i.e. $\tan t < 0$), I get to the negation of the previous case, so I wonder where did I go wrong? Should I maybe not even consider the negative case?
Thanks!
| You may avoid worrying about square-roots or absolute signs by integrating as follows
\begin{align}
\int \frac{\sqrt{x^2 + x}}{x}dx
=& \int \frac{x+\frac12}{\sqrt{x^2 + x}}\>dx+
\int \frac{\frac12}{\sqrt{x^2 + x}}dx\\
=&\>\sqrt{x^2+x}+\frac12\tanh^{-1}\frac{\sqrt{x^2+x}}{x+\frac12}+C
\end{align}
which is valid for all domain $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given $z^2-1\mid x^2z^2-1$, prove $\frac{x^2z^2-1}{z^2-1}$ is never prime, for $x$, $z$ integers such that $x>z>1$.
Given $z^2-1\mid x^2z^2-1$, prove that $\frac{x^2z^2-1}{z^2-1}$ can never be prime, assuming $x$, $z$ are integers such that $x>z>1$.
So far I have tried taking mod a lot of different numbers, but I cannot find solution. I also tried writing as a quadratic and using quadratic formula, but that doesn't work either. Please help.
| Let $p = \frac{x^2z^2-1}{z^2-1} \implies pz^2 - p = x^2z^2 - 1$
So $z^2(x^2 - p) = 1 - p$.
We know $p > 1$. So this implies $p > x^2$.
Now $p$ divides $xz-1$ or $xz+1$ but not both.
But $p > x^2 > x > z \implies p > xz - 1$ and $p > xz + 1$.
which means $p \nmid xz-1$ and $p \nmid xz+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Let $\left,\left$ be real sequences satisfied as following conditions： $\left<a_i\right>, \left<b_i\right>$ be real sequences with
$$a_1^2+a_2^2+\cdots+a_n^2=1，\\
b_1^2+b_2^2+\cdots+b_n^2=1，\\
a_1b_1+a_2b_2+\cdots+a_nb_n=0.$$
Prove that $(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2\leq n$.
My attempt：
I try to prove it by induction，
As $n=2$，$a_1^2+a_2^2=1，b_1^2+b_2^2=1，a_1b_1+a_2b_2=0$
Because $a_1, a_2, b_1, b_2$ cannot be all $0$，W.L.O.G.，Assume $a_1\neq0$，
$b_1=-\frac{a_2b_2}{a_1}\Rightarrow$ $\frac{a_2^2b_2^2}{a_1^2}+b_2^2=1\Rightarrow b_2^2=a_1^2，b_1^2=a_2^2$
\begin{align}
(a_1+a_2)^2+(b_1+b_2)^2 & = a_1^2+a_2^2+b_1^2+b_2^2+2(a_1a_2+b_1b_2)\\
&=2+2(a_1a_2-\frac{a_2b_2^2}{a_1})\\
&=2+2\cdot\frac{a_2}{a_1}(a_1^2-b_2^2)\\
&=2
\end{align}
And the induction hypothesis is that $n=k$，$a_1^2+a_2^2+\cdots+a_k^2=1，b_1^2+b_2^2+\cdots+b_k^2=1，a_1b_1+a_2b_2+\cdots+a_kb_k=0，$
$(a_1+a_2+\cdots+a_k)^2+(b_1+b_2+\cdots+b_k)^2\leq k$ holds；
As $n=k+1$，$a_1^2+a_2^2+\cdots+a_k^2+a_{k+1}^2=1，b_1^2+b_2^2+\cdots+b_k^2+b_{k+1}^2=1，a_1b_1+a_2b_2+\cdots+a_kb_k+a_{k+1}b_{k+1}=0，$
Assume $a_1\neq 0$，
\begin{align}
& b_1=-\frac{a_2b_2+a_3b_3+\cdots a_{k+1}b_{k+1}}{a_1}\\
\Rightarrow & \frac{(a_2b_2+a_3b_3+\cdots+a_{k+1}b_{k+1})^2}{a_1^2}+b_2^2+\cdots+b_{k+1}^2=1\\
\Rightarrow &
(a_2b_2+a_3b_3+\cdots+a_{k+1}b_{k+1})^2+a_1^2b_2^2+\cdots+a_1^2b_{k+1}^2=a_1^2\\
\Rightarrow & ...
\end{align}
I cannot finish the proof. Please help~
| Set $\vec{e}=(1/\sqrt{n},...,1/\sqrt{n})^t$, $\vec{a}=(a_1,...,a_n)^t$ and $\vec{b}=(b_1,...,b_n)^t$.
WTS: $$(\vec{e}\cdot \vec{a})^2 + (\vec{e}\cdot \vec{b})^2 \leq 1 \, . \tag{1}$$
Proof 1: Setting $\vec{e}\cdot \vec{b}=\cos\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{e}$. Since $\vec{a}$ is orthogonal to $\vec{b}$, it lives in the hyperspace $B_\perp$ with normal vector $\vec{b}$. Since the normal vector of $B_\perp$ is tilted with respect to $\vec{e}$ by angle $\theta$, any projection of a unit-vector in $B_\perp$ onto $\vec{e}$ can be at most $|\sin\theta|$ in magnitude. This is the case when $\vec{a},\vec{b},\vec{e}$ lie in a common plane s.t.
$$\left|\angle(\vec{a},\vec{e}) \pm \angle(\vec{b},\vec{e})\right|=\angle(\vec{a},\vec{b})=90°\\
\text{or}\\
\angle(\vec{a},\vec{e}) + \angle(\vec{b},\vec{e}) = 270° \, .$$
Hence
$$(\vec{e}\cdot \vec{a})^2 + (\vec{e}\cdot \vec{b})^2 \leq \sin^2\theta + \cos^2\theta = 1 \, .$$
Proof 2: More formally, we write $$\vec{e}=R\vec{e}_1$$ for some rotation matrix $R$ and $\vec{e}_1=(1,0,...,0)^t$. Using this in (1), we find
$$\left(\vec{e}_1 \cdot R^t \vec{a}\right)^2 + \left(\vec{e}_1 \cdot R^t \vec{b}\right)^2 = \left(\vec{e}_1 \cdot \vec{a}'\right)^2 + \left(\vec{e}_1 \cdot \vec{b}' \right)^2 \, . \tag{2}$$
Since $R$ is orthogonal, $\vec{a}'$ and $\vec{b}'$ still satisfy all the conditions $$\vec{a}'^2=1 \\
\vec{b}'^2=1\\
\vec{a}'\cdot\vec{b}'=0 \, .$$
Using spherical coordinates, we can write $$\vec{b}'=\begin{pmatrix} \cos\theta \\ \sin\theta \cos\phi_1 \\ \sin\theta \sin\phi_1\cos\phi_2 \\
\vdots \\ \sin\theta \sin\phi_1 \cdots \sin\phi_{n-3} \cos\phi_{n-2} \\
\sin\theta \sin\phi_1 \cdots \sin\phi_{n-3} \sin\phi_{n-2} \end{pmatrix} \, .$$
Since $\vec{a}'$ is orthogonal to $\vec{b}'$, it has the form
$$\vec{a}'=c_\theta \, \partial_\theta \vec{b}' + \sum_{k=1}^{n-2} c_k \, \partial_{\phi_k} \vec{b}' = \begin{pmatrix} -c_\theta\sin\theta \\ c_\theta \cos\theta \cos\phi_1 - c_1 \sin\theta \sin\phi_1 \\ \vdots \end{pmatrix} \, .$$
Furthermore, since $\vec{a}'^2=1$ and the system of basis vectors is orthogonal, we have
$$1=\vec{a}'^2=c_\theta^2 + (\text{stuff } \geq 0) \\
\Rightarrow \quad c_\theta^2 \leq 1$$
Hence it is clear that following (2)
$$a_1'^2 + b_1'^2 = c_\theta^2 \sin^2\theta + \cos^2\theta \leq \sin^2\theta+\cos^2\theta = 1 \, .$$
Proof 3: I found this neat proof, using rotation matrices, that reduces the general case to the case $n=2$. Dropping vectors, $a,b,e$ are as above and assume $n\geq 3$. By rotation with $R$, we could write $e=Re_1$ and we arrived at (2), with $a',b'$ satisfying the conditions as before. We now rotate by $R'$ in the n-1 dimensional subspace s.t. $R'e_1=R'^te_1=e_1$. We define $R'$ s.t.
$$R'a'=R'\begin{pmatrix} a_1' \\ a_2' \\ a_3' \\ \vdots \end{pmatrix}=a''=\begin{pmatrix} a_1' \\ a_2'' \\ 0 \\ \vdots \end{pmatrix}$$
and we can thus write (2) as $$(R'^t e_1\cdot a')^2 + (R'^t e_1 \cdot b')^2 = (e_1\cdot R'a')^2 + (e_1 \cdot R'b')^2 = (e_1\cdot a'')^2 + (e_1 \cdot b'')^2 \, . \tag{3}$$
By construction $$a''^2=a_1'^2 + a_2''^2 = b''^2=b_1'^2 + b_2''^2 + b_3''^2 + ...=1 \quad , \quad a''\cdot b'' = a_1'b_1' + a_2''b_2'' = 0 \, ,$$
from which it is clear that $b_1'^2 + b_2''^2 \leq 1$. Now we define the rotation $R''$ by $$e_{12}=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \\ \vdots\end{pmatrix} = R''e_1$$
i.e. $R''$ is a rotation by $45°$ in the 2-dimensional subspace spanned by $e_1,e_2$.
Thus, (3) then becomes
$$(R''^t e_{12} \cdot a'')^2 + (R''^t e_{12} \cdot b'')^2 = (e_{12} \cdot R''a'')^2 + (e_{12} \cdot R''b'')^2 = (e_{12} \cdot a''')^2 + (e_{12} \cdot b''')^2 \, . \tag{4}$$
Furthermore, we can now define
$$\tilde{a}=\begin{pmatrix} a_1''' \\ a_2''' \end{pmatrix} \quad , \quad \tilde{b}=\frac{1}{\sqrt{b_1'^2 + b_2''^2}}\begin{pmatrix} b_1''' \\ b_2''' \end{pmatrix} \quad , \quad \tilde{e}=\begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} \, ,$$
and because $R''$ is a rotation in the 2-dimensional subspace spanned by $e_1$ and $e_2$, it is plain that
$$\tilde{a}^2=a'''^2 = a''^2 = 1 = \tilde{b}^2 \quad , \quad \tilde{a}\cdot \tilde{b} = a''' \cdot b''' = a'' \cdot b'' = 0$$
and so, continuing with (4), it follows
$$(e\cdot a)^2 + (e\cdot b)^2 \leq (\tilde{e}\cdot\tilde{a})^2 + (\tilde{e} \cdot \tilde{b})^2 = 1 \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Maximum value with inequality This is a problem that my friend and I are working on for olympiad training.
Let $a, b, c$ be real numbers in the interval $[0,1]$ that satisfy $ab+c \leq 1$. What is the maximum value of $a+b+c?$
I'm guessing the maximum is at $a,b=1$ and $c=0$, where we have $a+b+c=2$. As for proving this, I'm not sure how to proceed. Maybe AM-GM? Thanks for the help.
| 1)$ab+c\le 1;$ $a, b, c \in [0,1];$
$d:=a+b+c$
The $2$ expressions are symmetric in $a$ and $b$. It follows that $a=b$ for $d_{max}$.
2)We want to find the maximum of
$d=2a+c,$ with $a^2+c \le 1$.
3)$a^2+d-2a \le 1;$
$(a-1)^2-1 +d \le 1;$
$d \le 2 - (a-1)^2.$
4)Maximal value for $d_{max}=2$ at $a=1;$
Hence $a=b=1$,and $c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k=1$, and for that value of $k$ try to write the integral from $0$ to $2\pi$ as a sum of four integrals from $0$ to $\dfrac{\pi}{2}$ and use the the theorem for first derivative monotony. No luck so far.
Any help would be much appreciated.
Edit 1: I saw the link here about a similarly asked topic. However, this process gets the general case as I perceive it and I am really supposed to use the method described above. I will try this and come back with a definitive answer.
Edit 2: I cannot use the $f''(x) \geq 0$ argument due to the simple fact that I have not been formally taught it as part of the class.
Edit 3: Final proof, with thanks to the contributors below.
Let's start by setting $A = \frac{\pi}{2}$ and $B = \frac{3\pi}{2}$. It follows from basic trigonometry that:
\begin{align}
&\cos x \geq 0, \ x \in [A,B] \ \text{and}\\
&\cos x \leq 0, \ x \in [0, A] \cap [B, 2\pi].
\end{align}
And we also set $L$ to be the line segment such that $L(A) = f(A), \ L(B) = f(B)$. We will prove a basic property of said line in regards to the convex function $f$.
*
*I can take for granted that (we proved this in class)
\begin{align}
L(x) = \dfrac{x-A}{B-A}f(B) + \dfrac{B-x}{B-A}f(A)
\end{align}
so for $x \in [A,B]$ there exists $\lambda \in [0,1]$ such that: $x = \lambda A + (1-\lambda) B$. Taking the aforementioned expression and replacing it on $L(x)$ we get (I omit trivial algebra)
\begin{align}
L(x) = (1-\lambda) f(B) + \lambda f(A).
\end{align}
Since $f$ is convex, we can write
\begin{align}
&f(\lambda A + (1-\lambda) B) \leq \lambda f(A) + (1-\lambda) f(B)\\
\implies &f(x) \leq L(x), \ \forall \ x \in [A,B] \ \text{and} \ \lambda \in [0,1].
\end{align} $\blacksquare$
*We assume that there exists $x \in [0,A]: \ f(x) < L(x)$. Then there exists $A \in [x, B] \ \text{and} \ \lambda \in [0,1]: \ A = \lambda x + (1-\lambda) B$. Then
\begin{align}
&L(A) = \dfrac{A-x}{B-x}f(B) + \dfrac{B-A}{B-x}f(x), \ A \in [x,B]\\
\implies &L(A) = (1-\lambda)L(B)+\lambda L(x)\\
\implies &L(A) = (1-\lambda)f(B) + \lambda L(x).
\end{align}
Then assuming that $L(x) > f(x)$ we get
\begin{align}
L(A) > (1-\lambda) f(B) + \lambda f(x) \geq f(A), \ \text{assuming convexity}.
\end{align}
Because $L(A) = f(A)$ the above inequality becomes $L(A) > f(A)$ which is a contradiction.
$\blacksquare$
In the same spirit, for $x \in [B, 2\pi]$ doing the exact same replacements and applying the exact same principles we also get
\begin{align}
&f(\lambda B + (1-\lambda) 2 \pi) \leq \lambda f(B) + (1-\lambda) f(2\pi)\\
\implies &f(x) \leq L(x).
\end{align}
$\blacksquare$
We then set $g(x) = f(x) - L(x)$ and from (1) and (2) above it is safe to assume that $\cos x$ and $g(x)$ will have the same sign in the whole domain, that is:
\begin{align}
&g(x) \geq 0, \ \text{where} \ \cos x \geq 0\\
&g(x) \leq 0, \ \text{where} \ \cos x \geq 0.
\end{align}
*We have
\begin{align}
\int_0^{2\pi} g(x) \cos x dx = \int_0^{\frac{\pi}{2}} g(x) \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} g(x) \cos x dx + \int_{\frac{3\pi}{2}}^{2\pi}g(x) \cos x dx \geq 0,
\end{align}
because
\begin{align}
&\text{in} \left[0, \frac{\pi}{2} \right], \ \cos x \geq 0 \implies g(x) \geq 0\\
&\text{in} \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right], \ \cos x \leq 0 \implies g(x) \leq 0\\
&\text{in} \left[\frac{3 \pi}{2}, 2 \pi \right], \ \cos x \geq 0 \implies g(x) \geq 0.\\
\end{align}
*We have that
\begin{align}
&\int_0^{2 \pi}\cos x dx = 0 \ \text{trivial}\\
&\int_0^{2 \pi}x \cos x dx = \int_0^{2 \pi}x (\sin x)' dx = \left[ x \sin x \right]_0^{2\pi} - \int_0^{2 \pi} \sin x dx = 0.
\end{align}
It then follows that
\begin{align}
\int_0^{2 \pi} f(x) \cos x dx = \int_0^{2 \pi} g(x) \cos x dx \geq 0
\end{align}
which was previously proven. For the case $k=1$, the proof is over.
For $k>1$, we have:
\begin{align}
\int_0^{2\pi}f(x) \cos (kx) dx = \sum_{i=0}^{k-1} \int_{\frac{2\pi i}{k}}^{\frac{2\pi (i+1)}{k}} f(x) \cos (kx) dx.
\end{align}
We perform the change of variable
\begin{align}
x = \dfrac{y+2\pi i}{k}\\
\implies \begin{cases}dx = \dfrac{1}{k}dy\\ x = \dfrac{2\pi i}{k} \to y=0\\ x = \dfrac{2\pi (i+1)}{k} \to y = 2\pi \end{cases}.
\end{align}
So the above sum becomes
\begin{align}
\sum_{i=0}^{k-1} \dfrac{1}{k} \int_{0}^{2\pi}f \left(\dfrac{y+2 \pi i}{k} \right) \cos (y+2\pi i)dy.
\end{align}
Because $f$ is convex in $[0, 2\pi]$ there exists $\lambda \in [0,1]$ such that:
\begin{align}
\theta f\left(\frac{y_1 + 2\pi i}{k}\right)
+ (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right)
&\ge f\left(\theta\frac{y_1 + 2\pi i}{k}
+ (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\
&= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right).
\end{align}
Having performed the change of variables:
\begin{align}
x_1 = \dfrac{y_1 + 2\pi i}{k} \ \text{and} \ x_2 = \dfrac{y_2 + 2\pi i}{k}.
\end{align}
So $f \left( \dfrac{y + 2\pi i}{k}\right)$ convex on $[0, 2\pi]$. Using the result from $k=1$ we have
\begin{align}
\int_0^{2\pi} f \left( \dfrac{y + 2\pi i}{k} \right)\cos y dy \geq 0.
\end{align}
$\blacksquare$
| Without loss of generality we may assume that $f$ is twice continuously differentiable on $[0,2\pi].$ Then
$$ \int_0^{2\pi}f(x)\cos(kx)\,dx=\left. f(x)\frac {\sin(kx)} k\right |_0^{2\pi} -\int_0^{2\pi} f'(x)\frac {\sin(kx)} k\,dx= -\int_0^{2\pi} f'(x) \frac {\sin(kx)} k\,dx.$$
Next, again integraiting by parts, this equals
$$\left. f'(x)\frac {\cos(kx)} k\right|_0^{2\pi} - \int_0^{2\pi}f''(x)\cos(kx)\,dx\ge$$ $$f'(2\pi)-f'(1)-\int_0^{2\pi}f''(x)\cdot1\,dx= $$ $$f'(2\pi)-f'(1)-\int_0^{2\pi} f''(x)\,dx=0 $$ since the second derivative of a convex function is nonnegative. The general case is obtained by approximation of $f(x)$ by a smooth function with precision $\epsilon$, making use of convolution, and then by taking the limit as $\epsilon\to 0+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Volume between paraboloid $x^2 +y^2 -4a(z+a)=0$ and sphere $x^2 + y^2 +z^2 =R^2$ I'm trying to obtein the volume via triple integral but think I'm setting the wrong radius. The solid in particular is bounded by the sphere $x^2 + y^2 +z^2 =R^2$ and above the parabolloid $x^2 +y^2 -4a(z+a)=0$ (consedering $R>a>0$). I'm setting cylindrical coordinates and I do eventually get $\theta \in [0,2\pi[$ and $(\rho^2 / 4a) -a\leq z \leq R^2-\rho^2$. I deduce that the radius must be limited in between $0$ and $4Ra-4a^2$ (Intersection is at height $z=R-2a$), so the volume should be:
$$\int_{0} ^{2\pi} \int_0 ^{4Ra-4a^2} \int_{(\rho^2 /4a)-a} ^{R^2 -\rho^2} \mathrm{d}V$$
, but this integral results in a different expression from the original solution. My teacher told us the solution is $2\pi \left ( \frac{a^3}{3} - aR^2 + \frac{2R^3}{3} \right )$.
| Note that it is not advisable to integrate in cylindrical coordinates because the $z$-limit could be either on the sphere or the parapoloid depending on $a$. Instead, integrate in spherical coordinates with the limits $\theta\in (\cos^{-1}\frac{R-2a}R,\pi)$. The volume is then
\begin{align}
V=&\>2\pi\int_{\cos^{-1}\frac{R-2a}R}^\pi
\int_{\frac{2a}{1-cos\theta}}^R \rho^2\sin\theta \>d\rho \>d\theta\\
= &\>\frac{2\pi}3 \int_{\cos^{-1}\frac{R-2a}R}^\pi \left( R^3 - \frac{8a^3}{(1-\cos \theta)^3}\right)\sin\theta \>d\theta\\
= &\>\frac{2\pi}3 \left[ R^3 (1-\cos \theta)+\frac{4a^3}{(1-\cos \theta)^2}\right]_{{\cos \theta \>=\>1-\frac{2a}R}}^{\theta=\pi}\\
=&\> 2\pi \left ( \frac{a^3}{3} - aR^2 + \frac{2R^3}{3} \right )
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $\prod\limits_{k=1}^{n-1}\cot^2(\frac{6k+1}{6n}\pi)=2 - (-1)^n.$
Let $u, v \in \Bbb C \setminus \{0\}$ such that $u^{2n} + v^{2n} − u^n v^n = 0$ where $n \in \Bbb N^+$.
(a) Show that $$u = v \left(\cos\left(\frac{6k+1}{3n}\pi\right)\pm i\sin\left(\frac{6k+1}{3n}\pi\right)\right)$$ for $k=0,1,2,\ldots, n-1.$
(b) i) Using (a), solve the equation $(z+ai)^{2n}+(z-ai)^{2n} - (z^2+a^2)^n=0$ where a is a non-zero real number. ii) Deduce that $$\prod\limits_{k=1}^{n-1}\cot^2\left(\frac{6k+1}{6n}\pi\right) = 2 - (-1)^n$$
For part (a), we can solve the original equation by regarding it as a quadratic equation with the variable u and we have
$$u^n=\frac{v^n±\sqrt{v^{2n}-4v^{2n}}}2=(\frac{1}2-\frac{\sqrt{3}}2i)v^n$$
Hence $u=\sqrt{(\frac{1}2)^2+(\frac{\sqrt{3}}{2})^2} e^{i(\frac{5\pi}{3}+\frac{2k\pi}{n})}v=(\cos(\frac{6k+1}{3n}\pi)±i\sin(\frac{6k+1}{3n}\pi))v$ for $k=0,1,2,\ldots, n-1.$
Therefore
$$u=v(\cos(\frac{6k+1}{3n}\pi)±i\sin(\frac{6k+1}{3n}\pi))$$
for $k=0,1,2,\ldots, n-1$. For part (b) (i), results holds by simply putting $u=z+ai$ and $v=z-ai$
$z=\frac{(\cos(\frac{6k+1}{3n}\pi)-i\sin(\frac{6k+1}{3n}\pi)+1)ai}{\cos(\frac{6k+1}{3n}\pi)-i\sin(\frac{6k+1}{3n}\pi)+1}$
However, I am quite stuck in the last part.
| Here is an answer that I have got. However, it is indeed more complicated than I expected, and it requires a property from one of my previous questions. See $$z^{2n} - 2 a^n z^n \cos (nθ) + a^{2n}= \prod_{k=0}^{n-1}\left[z^2-2az\cos\left(\theta+\frac{2\pi k}{n}\right)+a\right]$$
for any positive integer a.
Now, putting $a=1, \cosθ=1/2$, i.e. $θ=\frac{\pi}3$, we have
$$z^{2n}-z^n+1=\prod\limits_{k=1}^{n-1}[z^2-2z\cos(\frac{\pi}{3}+\frac{2k\pi}{n})+1]=\prod\limits_{k=1}^{n-1}[z^2-2z\cos(\frac{6k+1}{3n}\pi)+1]$$
Now, take $z=1$, we have
$$\prod\limits_{k=1}^{n-1}[2-2\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}[1-\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}2\sin^2(\frac{6k+1}{6n}\pi)$$
$$=1 \ldots\ldots(*)$$
Similarly, take $z=-1$, we have
$$\prod\limits_{k=1}^{n-1}[2+2\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}[1+\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}2\cos^2(\frac{6k+1}{6n}\pi)$$
$$=2-(-1)^n\ldots\ldots(**)$$
Using $\frac{(**)}{(*)}$ we have
$$\frac{2^{n}\prod\limits_{k=1}^{n-1}2\cos^2(\frac{6k+1}{6n}\pi)}{2^{n}\prod\limits_{k=1}^{n-1}2\sin^2(\frac{6k+1}{6n}\pi)}=\prod\limits_{k=1}^{n-1}\frac{\cos^2(\frac{6k+1}{6n}\pi)}{\sin^2(\frac{6k+1}{6n}\pi)}=\prod\limits_{k=1}^{n-1}\cot^2({\frac{6k+1}{6n}\pi})=2-(-1)^n$$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Logic question - just to clear up meaning of 'implies' Okay so;
I have the following statement
$p \lor \neg q \Rightarrow q \lor \neg q$
I simplify the right hand side using the complement law to get
$p \lor \neg q \Rightarrow T $
I think that $\neg q \lor p$ is the same as saying $q \Rightarrow p$ ,
so does q imply p imply true?
Will this statement always be true? I think it does; and my thinking behind it is that since q implies p, then it will either output true or false? And since the other side of the implication is always true then either false of true implies true, correct?
| *
*
so does q imply p imply true?
Note that this phrasing is potentially ambiguous, because the conditional $(\to)$ is not associative, that is, $$(A → B) → C \:\not\equiv\: A → (B → C).$$ \begin{array}{ccc|c@{}c@{}c@{}ccc@{}ccc@{}ccc@{}ccc@{}ccc@{}c@{}c@{}c}
a&b&c&(&(&(&a&\rightarrow&b&)&\rightarrow&c&)&\leftrightarrow&(&a&\rightarrow&(&b&\rightarrow&c&)&)&)\\\hline
1&1&1&&&&1&1&1&&1&1&&\mathbf{1}&&1&1&&1&1&1&&&\\
1&1&0&&&&1&1&1&&0&0&&\mathbf{1}&&1&0&&1&0&0&&&\\
1&0&1&&&&1&0&0&&1&1&&\mathbf{1}&&1&1&&0&1&1&&&\\
1&0&0&&&&1&0&0&&1&0&&\mathbf{1}&&1&1&&0&1&0&&&\\
0&1&1&&&&0&1&1&&1&1&&\mathbf{1}&&0&1&&1&1&1&&&\\
0&1&0&&&&0&1&1&&0&0&&\mathbf{0}&&0&1&&1&0&0&&&\\
0&0&1&&&&0&1&0&&1&1&&\mathbf{1}&&0&1&&0&1&1&&&\\
0&0&0&&&&0&1&0&&0&0&&\mathbf{0}&&0&1&&0&1&0&&&
\end{array}
*Notice that a conditional $X\to Y$ is true whenever its consequent $Y$ is true: \begin{array}{cc|c@{}ccc@{}c}
X&Y&(&X&\rightarrow&Y&)\\\hline
1&1&&1&\mathbf{1}&1&\\
1&0&&1&\mathbf{0}&0&\\
0&1&&0&\mathbf{1}&1&\\
0&0&&0&\mathbf{1}&0&
\end{array} Therefore, since the given proposition $$p \lor \neg q \;\to\; q \lor \neg q$$ has a tautological (always-true) consequent, it is immediately also a tautology, that is, always true. \begin{array}{cc|c@{}c@{}ccc@{}cc@{}c@{}ccc@{}ccc@{}cc@{}c@{}c@{}c}
p&q&(&(&p&\lor&(&\lnot&q&)&)&\rightarrow&(&q&\lor&(&\lnot&q&)&)&)\\\hline
1&1&&&1&1&&0&1&&&\mathbf{1}&&1&1&&0&1&&&\\
1&0&&&1&1&&1&0&&&\mathbf{1}&&0&1&&1&0&&&\\
0&1&&&0&0&&0&1&&&\mathbf{1}&&1&1&&0&1&&&\\
0&0&&&0&1&&1&0&&&\mathbf{1}&&0&1&&1&0&&&
\end{array}
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What's $\Pr(X=0|X=Y)$? Let $X$ and $Y$ be independent random variables that take values in $\{0,1\}$.
My question is: how to solve our $\Pr(X=0 | X=Y)$? We only know that $\Pr(Y=0) = \Pr(Y=1) = 0.5$.
My attempt:
$$\begin{split}
\Pr(X=0|X=Y) &= \frac{
\Pr(X=0)
}{
\begin{split}
&\Pr(X=0,Y=0)\\
+&\Pr(X=1,Y=1)
\end{split}
}\\
&= \frac{
\Pr(X=0)
}{
\begin{split}
&\Pr(X=0)\Pr(Y=0)\\
+&\Pr(X=1)\Pr(Y=1)
\end{split}
}\\
&= \frac{
\Pr(X=0)
}{
\begin{split}
&\Pr(X=0)\Pr(Y=0)\\
+&(1-\Pr(X=0))(1-\Pr(Y=0))
\end{split}
}
\end{split}$$
Then, if I try $\Pr(Y=0)=.5$, I get:
$$\begin{split}
\Pr(X=0|X=Y) &=
\frac{
\Pr(X=0)
}{\Pr(X=0)\times .5 + (1-\Pr(X=0)) \times .5}\\
&=
\frac{
\Pr(X=0)
}{\Pr(X=0)\times .5 + (1 \times .5-\Pr(X=0) \times .5)}\\
&=
\frac{
\Pr(X=0)
}{\Pr(X=0)\times .5 + .5-\Pr(X=0) \times .5}\\
&=
\frac{
\Pr(X=0)
}{.5}\\
&= 2\Pr(X=0)\\
\end{split}$$
Then, if $\Pr(X=0)=.6$, the answer is $2\times.6 = 1.2 > 1$. Madness!
My attempt (NEW):
$$\begin{split}
\Pr(X=0|X=Y)
&= \frac{\Pr(X=0,X=Y)}{\Pr(X=Y)} \\
&= \frac{\Pr(X=0)\Pr(X=Y|X=0)}{\Pr(X=Y)} \\
&= \frac{\Pr(X=0)\Pr(X=Y|X=0)}{\Pr(X=0)\Pr(Y=0)+\Pr(X=1)\Pr(Y=1)} \\
&= \frac{\Pr(X=0)\Pr(Y=0)}{\Pr(X=0)\Pr(Y=0)+\Pr(X=1)\Pr(Y=1)} \\
\end{split}$$
Plugging $\Pr(Y=0) = \Pr(Y=1) = .5$:
$$\begin{split}
\Pr(X=0|X=Y)
&= \frac{.5\Pr(X=0)}{.5\Pr(X=0)+.5\Pr(X=1)} \\
&= \frac{.5\Pr(X=0)}{.5(\Pr(X=0)+\Pr(X=1))} \\
&= \frac{.5\Pr(X=0)}{.5(1)} \\
&= \frac{.5\Pr(X=0)}{.5} \\
&= \Pr(X=0) \\
\end{split}$$
|
Let $X$ and $Y$ be independent random variables that take values in
$\{0,1\}.$
We know that $\Pr(Y=0) = \Pr(Y=1) = 0.5\tag2.$
What is $\Pr(X=0 | X=Y) \;?$
Your new attempt is all good.
Note that armed only with the information in the first line but without statement $(2),$ $X$ and $Y$ attaining the same value, i.e., $\{X=Y\},$ and $\{X=0\}$ are not necessarily independent events. (After all, the probability that $Y$ equals $x$ generally does depend on whether $x$ equals $0.)$ Verify this by setting $P(X=0)P(X=Y)=P(X=0=Y).$
These two events are independent precisely when $\{X=0\}$ is impossible or certain or $Y$ is uniformly distributed. Here, statement $(2)$ informs us of the latter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429398",
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"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
why is $ 2 = \frac{5}{1+\frac{8}{4+\frac{11}{7 + \frac{14}{10 + \dots}}} } $ Why is $ 2 = \cfrac{5}{1+\cfrac{8}{4+\cfrac{11}{7 + \cfrac{14}{10 + \ddots}}} } $
where the sequences $5,8,11,14,\dots$ and $1,4,7,10,\dots$ are of the form $5 + 3 n$ and $1 + 3n$.
(This converges on both even and uneven iterates)
I was surprised this is an integer.
Maybe it would help to rewrite this generalized continued fraction into a "normal" simple continued fraction.
But I believe that would give us coefficients that generalized the double factorial to a sort of " triple factorial " meaning $ f(n) = n \cdot f(n-3) \cdot f(n-6) \cdot f(n-9) \cdots $ and I have almost no skills or understanding of those.
Maybe some transformation formula's make this easy, but Im not seeing it.
| Assume the RHS of the equation as $A_n$ so we have the following recurrence relation:
$$
A_n = \frac{3n + 2}{3n - 2 + A_{n + 1}}.
$$
Claim:
$$A_n = \frac{n + 1}{n}.$$
Proof:
Step of the induction:
$$A_{n + 1} = \frac{n + 2}{n + 1} \Rightarrow A_n = \frac{3n + 2}{3n - 2 + \frac{n + 2}{n + 1}} = \frac{n + 1}{n}.$$
Base of the induction:
Limit of the $A_n$ in infinity according to our closed form solution:
$$A_{\infty} = \lim_{n \to \infty}{\frac{n + 1}{n}} = 1.$$
Limit of the $A_n$ in infinity according to the given recurrence relation:
$$ x = \lim_{n \to \infty}{A_n} = \lim_{n \to \infty}{\frac{3n + 2}{3n - 2 + A_{n + 1}}} = \lim_{n \to \infty}{\frac{3n + 2}{3n - 2 + x}} = 1.$$
So the base of the induction is correct as well.
So $A_n = \frac{n + 1}{n}$.
Your question asks about the value of $A_{n = 1}$ and we have $ A_{1} = \frac{n + 1}{n} = \frac{1 + 1}{1} = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435750",
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"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Explanation for a solution: Howard Anton, Elementary Linear Algebra I am currently working with the 1st edition of Howard Anton's "Elementary Linear Algebra". I tried the following problem:
Excercise Set 1.2 (p. 17), Problem 12:
For which values of $a$ will the following system have no solutions? Exactly one solution? Infinitely many solutions?
$$\begin{array}{rccccl}
x &+& 2y &-& 3z &=& 4 \\
3x &-& y &+& 5z &=& 2 \\
4x &+& y &+& (a^2 - 14)z &=& a + 2\end{array}$$
By using Gauss-Jordan-Elimination (and Gaussian for a double check), I found the following solution set:
$$\begin{align*} x &= \frac{8}{7} + \frac{-a+4}{a^2-16} \\
y &= \frac{10}{7} + \frac{2a-8}{a^2-16} \\
z &= \frac{a-4}{a^2-16}\end{align*}$$
The solution in the textbook says, that the system has no solution if $a=-4$ an one solution if $a\neq\pm4$. This part I understand, since $z=\frac{a-4}{a^2-16}$ as well as others terms in the formulas for $y$ and $z$ are not defined for $a=-4$ but the formulas for $x,y,z$ will yield unambiguous values for $a\neq\pm4$. However, the solution also says, that the system has infinitely many solutions for $a=4$ and this is the point, which I don't understand. Isn't for instance $z=\frac{a-4}{a^2-16}$ still undefined for $a=-4$ or can I simply put in $z = \frac{0}{0} = 0$. And if I can, isn't $z=0$ still a unique value. I can find no room for different values of $x,y,z$ to satisfy the system of equations.
Can somebody explain this to me? Thanks in advance!
| You cannot just look at the final product if you did not carefully note steps in which you were assuming facts about the value of $a$. So let us take a careful look at the Gaussian elimination process.
Starting from
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
3 & -1 & 5 & 2\\
4 & 1 & a^2-14 & a+2
\end{array}\right)$$
we first subtract three times the first row from the second, and four times the first row from the third row. We get:
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
0 & -7 & 14 & -10\\
0 & -7 & a^2-2 & a-14
\end{array}\right)$$
Then subtracting the second row from the third row, we obtain:
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
0 & -7 & 26 & -10\\
0 & 0 & a^2-16 & a-4
\end{array}\right).$$
At this point: if $a=-4$, then the last row becomes
$$\left(\begin{array}{rrr|r}
0 & 0 & 0 & -8
\end{array}\right).$$
So the system has no solutions.
If $a=4$, on the other hand, the last row is
$$\left(\begin{array}{ccc|c}
0 & 0 & 0 & 0
\end{array}\right)$$
and your matrix has rank $2$, giving you infinitely many solutions.
And if $a\neq 4$ and $a\neq -4$, then you have a matrix of rank $3$, so you will get exactly one solution. This matches what the solutions say.
I suspect what happened is that you proceeded to divide the second row by $-7$ (no problem there):
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
0 & 1 & -\frac{26}{7} & \frac{10}{7}\\
0 & 0 & a^2-16 & a-4
\end{array}\right)$$
and then divided the last row by $a^2-16$. But this last step requires the assumption that $a^2-16\neq 0$. Thus, you are implicitly saying "and by the way, $a\neq 4$ and also $a\neq -4$." Nothing you get after that can be used in the case where $a=4$ or where $a=-4$. You need to consider those cases separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Consider $E = \mathbb{Q}(\sqrt{2}, \sqrt{7} )$. Here are the following questions and my respective answers follow below. I hope you have suggestions or if there are any mistakes I hope you help me fix them.
*
*Find a basis for $E$ over $\mathbb{Q}(\sqrt{2}).$
*Find a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}.$
*Write the general form of elements of $E.$
*Determine whether the mapping $\phi: E \rightarrow E$ such that $\phi(\sqrt{2}) = \sqrt{7}, \phi(\sqrt{7}) = \sqrt{2},$ and $\phi(q) = q, \forall q \in \mathbb{Q},$ is an automorphism of $E$ or not.
For #1: Let $\alpha = \sqrt{7}.$ Then $\alpha^2 = 7 \Rightarrow \alpha^2 - 7 = 0.$ Thus, the degree of $\alpha$ over $\sqrt{7}$ is 2. Therefore, a basis for $E$ over $\mathbb{Q}(\sqrt{2})$ is $\{1, \sqrt{7} \}.$
For #2: $x^2 - 7$ is the irreducible polynomial of $\sqrt{2}$ in $\mathbb{Q}[x].$ Thus, a basis for $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is $\{ 1, \sqrt{2} \}.$
For #3: The general form of the elements of $E$ is $a+b\sqrt{2} + c\sqrt{7} + d\sqrt{14},$ where $a,b,c,d \in \mathbb{Q}.$
For #4:
Take the elements $1 - \sqrt{2}$ and $\sqrt{2} - \sqrt{7}$ of $\mathbb{Q}(\sqrt{2},\sqrt{7}).$ By the definition of $\phi$, $\phi(1 - \sqrt{2}) = 1 - \sqrt{7},$ and $\phi(\sqrt{2} - \sqrt{7}) = \sqrt{7} - \sqrt{2}.$ So, $\phi(1 - \sqrt{2}) \phi(\sqrt{2} - \sqrt{7}) = \sqrt{7} - \sqrt{2} - 7 + \sqrt{2}\sqrt{7} $.
But $(1-\sqrt{2})(\sqrt{2} - \sqrt{7}) = \sqrt{2} - 2 - \sqrt{7} + \sqrt{2}\sqrt{7} \Rightarrow \phi(\sqrt{2} - 2 - \sqrt{7} + \sqrt{2}\sqrt{7} ) = \sqrt{7} - 2 - \sqrt{2} + \sqrt{7}\sqrt{2}.$
Thus, $\phi((1 - \sqrt{2})(\sqrt{2} - \sqrt{7})) \neq \phi(1 - \sqrt{2}) \phi(\sqrt{2} - \sqrt{7}).$ Therefore, $\phi$ is not an automorphism of $E = \mathbb{Q}(\sqrt{2}, \sqrt{7} )$.
Are there any comments? Are there any mistakes or can there be any improvements in my solutions?
Thank you!
| 1, 2, 3 look good.
Regarding 4, you can simplify by noticing that $p(x)=x^2-2 \in \mathbb Q[x]$ is a fixed polynomial under $\phi$ having $\sqrt 2$ as a root, while $\sqrt 7$ is not a root.
Or if you want to avoid using polynomials...
You have $\left(\sqrt 2\right)^2 - 2 = 0$. So applying $\phi$ you should have
$$0= \phi(0)=\phi(\left(\sqrt 2\right)^2 - 2) = \left(\phi(\sqrt 2)\right)^2 - 2 = \left(\sqrt 7\right)^2 - 2 =5$$
A contradiction implying that such automorphism can't exist.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding pattern in matrix inverse I want to see if any pattern that could be formulated exist across the rows (or equivalently , columns) of the matrix inverse (in the middle , let's name it $A^{-1}$) so that an analytical formula that relates $o_n$ and $b_n$ (perhaps along with other b's) exists (for any positive integer n )
$$
\begin{pmatrix}
o_1 \\
o_2 \\
: \\
: \\
: \\
o_n
\end{pmatrix} =
\begin{pmatrix}
1 & 0 & 0 & 0 & ... & 0 \\
w & 1 & 0 & 0 & \ddots & 0 \\
i & w & 1 & 0 & \ddots & 0 \\
i & i & w & 1 & \ddots & 0 \\
: & \ddots & \ddots & \ddots & \ddots & \ddots \\
i & i & i & ... & w & 1
\end{pmatrix}^{-1}
\begin{pmatrix}
b_1 \\
b_2 \\
: \\
: \\
: \\
b_n
\end{pmatrix}
$$
The matrix inverse $A^{-1}$ is a lower triangular matrix with diagonal full of 1 , subdiagonal full of w and others are i .
when $n = 6$ , the matrix inverse becomes
$$ \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ -w & 1 & 0 & 0 & 0 & 0 \\ w^2-i & -w & 1 & 0 & 0 & 0 \\ -w^3+2 i w-i & w^2-i & -w & 1 & 0 & 0 \\ w^4-3 i w^2+2 i w+i^2-i & -w^3+2 i w-i & w^2-i & -w & 1 & 0 \\ -w^5+4 i w^3-3 i w^2-3 i^2 w+2 i w+2 i^2-i & w^4-3 i w^2+2 i w+i^2-i & -w^3+2 i w-i & w^2-i & -w & 1 \end{array} \right) $$
I recommend matrix calculator https://www.dcode.fr/matrix-inverse to quickly find matrix inverse symbolically .
Since in this case it doesn't matter whether we look at rows or columns , by looking at the first column , I could only write down
$$ \left\{ \begin{array}{ccc}
a_1 & = & 1 \\
a_2 & = & a_1 (-w) \\
a_3 & = & a_2 (-w) - i \\
a_4 & = & a_3 (-w) + iw - i\\
a_5 & = & a_4 (-w) + iw - iw^2 - i + i^2\\
a_6 & = & a_5 (-w) + iw - iw^2 +iw^3 - 2i^2w - i + 2i^2\\
\end{array} \right.
$$
Edit : After reading @metamorphy 's answer
Below is my note ,
since the "1"-diagonal of $S^k$ will lie on the $k^{th}$ subdiagonal , we may visualize $A^{-1}$ and write down
$$
o_n = ( \alpha_+\lambda_{+}^0 + \alpha_- \lambda_{-}^{0} ) b_n +
( \alpha_+\lambda_{+}^{1} + \alpha_- \lambda_{-}^{1} ) b_{n-1} + ... +
( \alpha_+\lambda_{+}^{n-1} + \alpha_- \lambda_{-}^{n-1} ) b_{1}
$$
$$
= \sum_{k=0}^{n-1} ( \alpha_+\lambda_{+}^{k} + \alpha_- \lambda_{-}^{k} ) b_{n-k}
$$
with
$$ \alpha_+ = \frac{\lambda_{+} - 1 }{\lambda_{+} - \lambda_{-}} ,\qquad \alpha_- = \frac{\lambda_{-} - 1 }{\lambda_{-} - \lambda_{+}}
$$
| Let $S$ be the $n\times n$ matrix with $S_{ij}=\begin{cases}1,&i=j+1\\0,&\text{otherwise}\end{cases}$.
The matrix being inverted is $A=I+wS+iS^2(I-S)^{-1}$, hence $A^{-1}=f(S)$ with $$f(z)=\frac{1-z}{1-(1-w)z-(w-i)z^2}.$$
Next, we find the partial fraction decomposition of $f(z)$, say $$f(z)=\frac{\alpha_+}{1-\lambda_+ z}+\frac{\alpha_-}{1-\lambda_- z},\qquad\lambda_\pm=\frac12\left(1-w\pm\sqrt{(1+w)^2-4i}\right)$$ (this is possible if $(1+w)^2\neq4i$; use continuity or symbolic argument otherwise). Then $$A^{-1}=\sum_{k=0}^{n-1}(\alpha_+\lambda_+^k+\alpha_-\lambda_-^k)S^k$$ because $S^k=0$ for $k\geqslant n$; this gives a closed form of the elements on $k$-th diagonals of $A^{-1}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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If $\cos\frac \pi{n+1}$ is a root of the equation $8x^3+4x^2-4x-1=0$, then find n $(n\in\mathbb N)$
If $\cos\frac \pi{n+1}$ is a root of the equation $8x^3+4x^2-4x-1=0$, then find n $(n\in\mathbb N)$
My Attempt:
Let $\theta=\frac\pi{n+1}$, therefore,
$$8\cos^3\theta+4\cos^2\theta-4\cos\theta-1=0$$
Also, $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\cos2\theta=2\cos^2\theta-1$, so,
$$2(\cos3\theta+3\cos\theta)+2(1+\cos2\theta)-4\cos\theta-1=0\\\implies\cos3\theta+\cos2\theta+\cos\theta=-\frac12\\\implies\frac{\sin\frac{3\theta}2}{\sin\frac\theta2}\cos2\theta=-\frac12\\\implies\frac{\sin\frac{3\pi}{2(n+1)}}{\sin\frac\pi{2(n+1)}}\cos\frac{2\pi}{n+1}=-\frac12$$
How to proceed from here?
| If you know that a polynomial has a root of the form $\cos\left(\frac{\pi}{n+1}\right)$ then you have to think in roots of unity and cyclotomic polynomials. The idea is to find the right change of variable that transforms your polynomial into a cyclotomic polynomial. Since $8x^3+4x^2-4x-1=(2x)^3+(2x)^2-2(2x)-1$ it seems resonable to look for a change of variable of the form $2x=\dots$ and some standard change can be $2x=y+\frac{1}{y}$. Now, try to make this change of variable and see what you get, you should be able to finish the exercise from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{2(1+4a^2)}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$ for $0\leq a<\frac13$ and $x>\frac{(1+a)^2}8$
How to prove the inequality below？
$$\frac{2(1+4a^2)}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$$ holds for all $0\leqslant a<\frac{1}{3}$ and $x>\frac{(1+a)^2}{8}$.
This question has been bothering me for a while. Using numerical experiment, we found that $\frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}-\frac{2(1+4a^2)}{(12x-1)^3}$ is strictly increasing in $x\in (\frac{(1+a)^2}{8},+\infty)$, but it seems not easy to prove it.
Thanks to everyone!
| Note that
$$(1 - a)^4 + (1 + a)^4 - 2(1 + 4a^2) = 4a^2 + 2a^4 \ge 0.$$
It suffices to prove that
$$\frac{(1 - a)^4 + (1 + a)^4}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$$
or
\begin{align*}
&\frac{1}{(12x - 1)^3}\\
\le\,& \frac{(1 - a)^4}{(1 - a)^4 + (1 + a)^4}\cdot \frac{1}{[12x - (1 - a)^2]^3}
+ \frac{(1 + a)^4}{(1 - a)^4 + (1 + a)^4}\cdot \frac{1}{[12x - (1 + a)^2]^3}.
\end{align*}
Note that $u\mapsto \frac{1}{(12x - u)^3}$ is convex on $0 \le u \le 8x$.
By Jensen's inequality,
we have
$$\mathrm{RHS} \ge \frac{1}{(12x - A)^3} \ge \frac{1}{(12x - 1)^3}$$
where
$$A = \frac{(1 - a)^4}{(1 - a)^4 + (1 + a)^4}(1 - a)^2
+ \frac{(1 + a)^4}{(1 - a)^4 + (1 + a)^4}(1 + a)^2,$$
and we have used
$$A - 1 = \frac{(1 - a)^6 + (1 + a)^6 - (1 - a)^4 - (1 + a)^4}{(1 - a)^4 + (1 + a)^4} \ge 0.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Laurent series $\ \ \ \ \ $ Trying to compute first few terms of the Laurent series for:
$$ \frac{e^z}{z^2(z^2+1)} =\sum_{n=-2}^\infty c_n z^n = \quad ?$$
I know the expansion of $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$ and I could have $\frac{e^z}{z^2}$ expanded. But how do I fit all together and be able to see when it converges.
From the comment I have:
$$ \frac{e^z}{1+z^2} = \sum_{n=0}^\infty \frac{z^n}{n!}
\sum_{n=0}^\infty (-1)^n z^{2n}
$$
And using Caucy product: $$ c_n =\sum_{k=0}^n \frac{z^k}{k!} (-1)^{n-k} z^{2(n-k)} $$
It follows that:
$$ \frac{e^z}{z^2(z^2+1)} = \frac{1}{z^2}\sum_{n=0}^\infty \sum_{k=0}^n \frac{z^k}{k!} (-1)^{n-k} z^{2(n-k)} $$
| Via the Cauchy product,
\begin{align}
\frac{e^z}{z^2+1}
&= e^z\cdot\frac{1}{1-(-z^2)} \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (-z^2)^n\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty (iz)^{2n}\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{1+(-1)^n}{2}(iz)^n\right) \\
&= \left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{i^n+(-i)^n}{2}z^n\right) \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{1}{k!} \cdot\frac{i^{n-k}+(-i)^{n-k}}{2}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{i^{n-k}+(-i)^{n-k}}{2(k!)}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{i^k+(-i)^k}{2((n-k)!)}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{1+(-1)^k}{2}\cdot\frac{i^k}{(n-k)!}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{i^{2k}}{(n-2k)!}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^n. \tag1
\end{align}
So the desired Laurent series is
\begin{align}
\frac{e^z}{z^2(z^2+1)}
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^{n-2} \\
&= \sum_{n=-2}^\infty \left(\sum_{k=0}^{(n+2)/2} \frac{(-1)^k}{(n+2-2k)!}\right) z^n.
\end{align}
In hindsight, I took the long way to obtain $(1)$. Here's a shorter approach:
\begin{align}
\frac{e^z}{z^2+1}
&= \frac{1}{1-(-z^2)}\cdot e^z \\
&=
\left(\sum_{n=0}^\infty (-z^2)^n\right)
\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)
\\
&=
\left(\sum_{n=0}^\infty (iz)^{2n}\right)
\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)
\\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} i^{2k}\cdot \frac{1}{(n-2k)!}\right) z^n \\
&= \sum_{n=0}^\infty \left(\sum_{k=0}^{n/2} \frac{(-1)^k}{(n-2k)!}\right) z^n.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Analytic solution inverse kinematics - different solutions with different calculation steps I have the following problem:
$$5\cos \theta_1+3\sqrt{3}\sin \theta_1=4\qquad\qquad\textbf{(I)}$$
$$5\sin \theta_1-3\sqrt{3}\cos \theta_1=6\qquad\qquad\textbf{(II)}$$
I have two seemingly correct paths to a solution, however only one actually computes the correct one.
Path 1, yielding the correct solution (I$\cdot$5, II $\cdot-3\sqrt{3}$):
$$25 \cos\theta_1 + 15 \sqrt{3} \sin\theta_1
= 20$$
$$-15 \sqrt{3} \sin\theta_1 + 27 \cos\theta_1
= -18\sqrt{3}$$
Add both:
$$52 \cos\theta_1 = 20 - 18\sqrt{3}$$
$$\theta_1=\cos^{-1}\left(\frac{20-18\sqrt{3}}{52}\right)\approx1.7874$$
Path 2, resulting in a different solution, that doesn't even satisfy the system of equations(I $\cdot3\sqrt{3}$, II $\cdot5$):
$$15 \sqrt{3} \cos\theta_1 + 27 \sin\theta_1
= 12 \sqrt{3}$$
$$25 \sin\theta_1 - 15 \sqrt{3} \cos\theta_1
= 30$$
Add both:
$$52 \sin\theta_1 = 30 + 12\sqrt{3}$$
$$\theta_1 = \sin^{-1} \left( \frac{15 + 6\sqrt{3}}{26} \right)
\approx 1.3542$$
$$5\cos(1.3542)+3\sqrt{3}\sin(1.3542)=6.1493\neq4$$
I don't see any errors in both solutions, however only the first path calculates the correct value. How can this be? How is this generalizable (or not) to the solutions of systems of equations with sin/cos in them?
| For this system, I would set $\sigma=\sin \theta_1,\chi=\cos\theta_1$ and work those out rather than throwing yourself into inverse trig functions of irrationals so early.
$$3\sqrt 3 \sigma +5\chi =4 \tag I$$$$5\sigma-3\sqrt 3 \chi=6\tag {II}$$
Applying $(I)* 3\sqrt 3, (II)*5$ as in your second choice, we get: $$27\sigma+15\sqrt 3\chi=12\sqrt3 \tag{III}$$$$25\sigma-15\sqrt3 \chi =30 \tag{IV}$$
yielding $\sigma =\frac{6\sqrt 3+15}{26}$ and $\chi=\frac{10-9\sqrt 3}{26}$
You can calculate that $\sigma^2+\chi^2=1$ indeed.
You can then solve and see that picking the second solution of $\sin \theta= \frac{6\sqrt3+15}{26}$, where $\theta \approx1.787$ gives the correct answer here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $c$ is a root of the equation $6x^3-6sx^2+3(s^2-t)x+3st-s^3-2u=0$
(a) Let $p,q$ and $r$ be real numbers. Given that there are numbers $a$ and $b$ such that $$a+b=p, ~~~ a^2+b^2=q, ~~~ a^3+b^3=r~~~~~~~~~~~~~~~(*)$$
Show that $3pq - p^3=2r$
Workings: $3pq-p^3 = 3(a+b)(a^2+b^2)-(a+b)^3 = 2a^3+2b^3=2r$
(b) Conversely, you are given that the real numbers $p,q$ and $r$ satisfy $3pq-p^3=2r$. By considering the equation $2x^2-2px+(p^2-q)=0$, show that there exist numbers $a$ and $b$ such that the three equations $(*)$ hold
Workings: Suppose that the roots of the quadratic are $$2x^2-2px+(p^2-q)=0 \iff (x-a)(x-b)=0\\ \therefore a+b=p, ~~~ ab=\frac{p^2-q}{2} \\ a^2+b^2=(a+b)^2-2ab \implies a^2+b^2=q \\ a^3+b^3 = (a+b)^3-3ab(a+b) \implies a^3+b^3= p^3-3(\frac{p^2-q}{2})(p)=\frac{1}{2}(3pq-p^3)=r$$
Hence, if $a,b$ satisfy the quadratic equation, then we have shown that it also satisfies the three equations in $(*)$
(c) Let $s,t,u$ and $v$ be real numbers. Given that there distinct numbers such that $a,b$ and $c$ such that $$a+b+c=s,~~~ a^2+b^2+c^2=t,~~~a^3+b^3+c^3=u,~~~ abc=v$$
Show, using part (a), that $c$ is a root of the equation $$6x^3-6sx^2+3(s^2-t)x+3st-s^3-2u=0$$
I am stuck on this part of the problem. I used the same approach as (a) to solve this but scored 0 marks. As I didn't use part (a). Could I have an explanation on how to show this using part (a)?
| In order to make use of (a), note that
$$\begin{aligned}
a+b&=s-c=:p\\
a^2+b^2&=t-c^2=:q\\
a^3+b^3&=u-c^3=:r
\end{aligned}$$
so that
$$ 3pq - p^3=2r$$
translates to
$$3(s-c)(t-c^2)-(s-c)^3=2(u-c^3). $$
Expand and rearrange this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Method checking to get the range of the three variables under two constraints Suppose a,b,c are real numbers and $a+b+c= 6$ , and $ab+bc+ca = 9$ , also $a<b<c$ find range of a , b, c .
My method was on eliminating c we get $a^2 + b^2 -6a - 6b +ab + 9$ = 0 so making a quadratic in a in terms of b we get a = $\frac {6-b \pm √[(b-6)^2 - 4(b-3)^2 ]}{2}$ , substituing this into the inequality of $a<b$ , and solving for range of $b$ we get it to be $(3,4)$ , but answer is its range is $(1,3)$ , and $c$ is $(3,4)$ , $a$ is $(0,1)$ respectively , may anyone tells whats going wrong here ?in my method and how to get thr ranges of $a$ and $c$ respectively from my method only .
Note : i once tried with interchanging and making quadratic in $b, c$ i got $b=f(c)$ then solve for $f(c)<c$ it lead me value of $c$ to be in $(3,4)$ , now solving for range of $b$ would be just checking the values $f(c)$ gets in the domain of $(3,4)$ but the problem is $f(c)$ is not actually a function because it has two values for same c so how will we solve it? Although putting the domain give the range of b to be $(1,3)$ and then $0<6-b-c =a<1$ ,but how to check for b values as f is not a function , also whats wrong with the $a =f(b)$ part which was giving wrong range for b ?
| We have a plane $ a + b + c = 6 $ and the hyperboloid of two sheets $ab + ac + bc = 9 $
The intersection of the two is a circle of center $(2,2,2)$ and radius $\sqrt{6}$
Hence, the vector $v = (a,b,c) $ can be described parametrically as
$ v = (2,2,2) + \sqrt{6} \left( \cos(t) \dfrac{(1, -1, 0)}{\sqrt{2}} + \sin(t) \dfrac{(1, 1, -2)}{\sqrt{6}} \right) \\= (2 + \sqrt{3} \cos(t) + \sin(t), 2 - \sqrt{3} \cos(t) + \sin(t) , 2 - 2 \sin(t) )$
Since we want $ a \lt b \lt c $, then we want
$ \sqrt{3} \cos(t) \lt - \sqrt{3} \cos(t) $
and
$ - \sqrt{3} \cos(t) + \sin(t) \lt - 2 \sin(t) $
The first equation implies that $ t \in (\dfrac{\pi}{2}, \dfrac{3 \pi}{2}) $
And the second eqauation implies
$ 3 sin(t) \lt \sqrt{3} \cos(t) $
Since, from the first inequality, $\cos(t) \lt 0 $, then
$ \tan(t) \gt \dfrac{1}{\sqrt{3}} $
whose solution is $ t \in [ \dfrac{7\pi}{6} , \dfrac{3\pi}{2} ] $
and this is the overall range. What remains is to find the range of the functions $a,b,c$ over this interval.
The easiest is $c = 2 - 2 \sin(t) $ , so its range is $[ 3 , 4 ]$
Then we have $ a = 2 + \sqrt{3} \cos(t) + \sin(t) = 2 + 2 \cos(t - \dfrac{\pi}{6} )$, therefore, its range is $[0, 1] $
And finally, $ b = 2 - \sqrt{3} \cos(t) + \sin(t) = 2 - 2 \cos( t + \dfrac{\pi}{6} )$, therefore, its range is $[1, 3] $
| {
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"url": "https://math.stackexchange.com/questions/4445237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the limit of the cubic formula approach the quadratic one as the cubic coefficient goes to $0$? The formula for solving a cubic equation of the form $ax^3+bx^2+cx+d=0$ does not seem to yield the quadratic formula for the limit $\lim _{a \rightarrow 0} \text{(cubic formula)}$.
But, if one tries the same thing with the quadratic formula the limit exists for the right choice of the square root sign.
My question is, is there a way to take the limit $\lim _{a \rightarrow 0} \text{(cubic formula)}$ and produce the quadratic formula? or does the limit simply not exist? Finally, if the limit does not exist, is there a technical reason for that?
Any input is very much appreciated. Thanks
Edit:
I was trying to tackle the simple case of $b=0$. Doing so, the cubic formula reduces to
$$\left(-\frac{d}{2a} - \left(\frac{4c^3 + 27ad^2}{108a^3}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}} + \left(-\frac{d}{2a} + \left(\frac{4c^3 + 27ad^2}{108a^3}\right)^{\frac{1}{2}}\right)^{\frac{1}{3}}$$
If one only considers the first term cubed, then it can be written as
$$\frac{1}{a}\left(-\frac{d}{2} - \left(4c^3+27ad^2\right)^{\frac{1}{2}}\right)$$
and I don't see how is it possible to find a finite limit as $a\rightarrow 0$. Am I missing something trivial?
| When $a \to 0$ one of the roots goes to infinity, which complicates the algebraic manipulations. Instead, it is easier to show that when $d=0$ the non-zero roots reduce to the quadratic formula.
Assume WLOG $\,a=1\,$, then with the wikipedia notations for the cubic formula in the case $d=0\,$:
$$
\begin{align}
\Delta_{0} = b^{2}-3ac &= b^2 - 3c
\\ \Delta_{1} = 2b^{3}-9abc+27a^{2}d &= 2b^3 - 9bc
\\ C = {\sqrt[{3}]{\frac{1}{2}\left(\Delta _{1}\pm {\sqrt {\Delta _{1}^{2}-4\Delta _{0}^{3}}}\right)}} &= \sqrt[3]{\frac{1}{2}\left(2b^3 - 9bc \pm 3c\sqrt{-3(b^2 - 4c)}\right)} \tag{1}
\end{align}
$$
The radicals in $(1)$ can be denested, since it is straightforward to verify that:
$$
\left(-b \pm \sqrt{-3(b^2-4c)}\right)^3 = 4\left(2b^3 - 9bc \pm 3c\sqrt{-3(b^2 - 4c)}\right) \tag{†}
$$
It follows that:
$$
\require{cancel}
\begin{align}
C &= \frac{1}{2}\left(-b \pm \sqrt{-3(b^2-4c)}\right) \tag{2}
\\ \frac{\Delta_0}{C} = \frac{2(b^2-3c)}{-b \pm \sqrt{-3(b^2-4c)}} &= \frac{1}{2}\left(-b \mp \sqrt{-3(b^2-4c)}\right) \tag{3}
\end{align}
$$
Finally, the roots are $\displaystyle\, x = -{\frac {1}{3}}\left(b + \omega C+{\frac {\Delta_0}{\omega C}}\right)\,$ where $\,\omega^3 = 1\,$, and $\,\omega = 1\,$ gives the root $\,x=0\,$. Otherwise, with $\,\omega = \dfrac{-1 \pm i\sqrt{3}}{2}\,$ a complex cube root of unity, using $(2)$ and $(3)\,$:
$$
\begin{align}
x &= -\frac{1}{3}\left(b + \omega\,\frac{-b \pm \sqrt{-3(b^2-4c)}}{2} + \overline\omega\,\frac{-b \mp \sqrt{-3(b^2-4c)}}{2}\right)
\\ &= -\frac{1}{3}\left(b \cdot \left(1-\frac{\omega + \overline\omega}{2}\right) \pm \sqrt{-3(b^2-4c)} \cdot \frac{\omega - \overline\omega}{2} \right)
\\ &= -\frac{1}{3}\left(\frac{3}{2}\, b \pm \frac{i \sqrt{3}}{2} \sqrt{-3(b^2-4c)} \right)
\\ &= \frac{-b \pm \sqrt{b^2-4c}}{2}
\end{align}
$$
The latter matches the quadratic formula, as expected.
[ EDIT ] $\;$ The denesting of $\,C = {\sqrt[{3}]{\frac{\Delta _{1}}{2}\pm {\sqrt {\frac{\Delta _{1}^{2}}{4} - \Delta _{0}^{3}}}}}\,$ in $\,(\dagger)\,$ follows from my answer here.
a sufficient condition for $\,\sqrt[3]{m \sqrt{p} \pm n\sqrt{q}}\,$ to denest is for $\,m^2 \cdot p - n^2 \cdot q\,$ to be the cube of a rational $\,r\,$, and for the cubic $\,p\, t'^{\,3} - 3r\, t' - 2m\,$ to have an appropriate rational root
Replacing "rational" with "rational expressions in the coefficients", the sufficient conditions are satisfied in this case, since:
*
*$\,m^2 \cdot p - n^2 \cdot q = \Delta_0^3\,$ is a perfect cube;
*$\,t'^{\,3} - 3 \Delta_0 t' - \Delta_1 = t'^{\,3} - 3(b^2-3c) t' - b(2b^2-9c)\,$ has the root $\,t'=-b\,$.
| {
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"source": "stackexchange",
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Find the values of a and b from arithmetic and geometric series The $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of an arithmetic series are $a, b, a^2$, where $a$ is a negative number. The $1^{st}$, $2^{nd}$ and $3^{rd}$ terms of a geometric series are $a, a^2,b$.
Find the values of $a$ and $b$ and find the sum to infinity of the geometric series.
What I have tried:
By equating the geometric series to the following:
$a^2=ar, b=ar^2 \implies r=a$
Then equating for the arithmetic series.
$a^3-a = a^2-a^3 \implies 2a^3-a^2-a=0 \implies a(a-1)(2a+1) \\ a=0, a=1, a=-\frac{1}{2} \\
\implies b=0, b=1, b=-\frac{1}{8}$
After plugging in the values of $a$ for $b=ar^2$ when $r=a$.
This produces the following series:
$$(0,0) = 0,0,0 ... \\ (1,1) = 1, 1, 1, ... \\ (-\frac{1}{2},-\frac{1}{8})=-\frac{1}{2},\frac{1}{4},-\frac{1}{8}...$$
The first series is equal to zero, the second is equal to zero, and the third has the following geometric series:
$$\sum_{n=0}^{\infty}(-1)^n \left(\frac{1}{2}\right)^n =\frac{2}{3}$$
Are my calculations correct for this?
| Looks mostly OK – note that $a$ is negative, so $a=-\frac 1 2$, and $b=-\frac 1 8$ (you don't need the other two cases).
The sum is $\frac{-\frac 1 2}{1 + \frac 1 2} = -\frac 1 3$, though. The formula for a geometric series is $\frac{a}{1 - r}$, and here, $r=a$. Your sum should be from $n=1$; as it stands, it includes a term of $1$ at the beginning, which is why your answer is too large by $1$.
One other thing – the sum of the series $1,1,1,\dots$ is not $0$ – it is undefined.
| {
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Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have $$a^2+b^2+c^2-ab-bc-ca=3\implies (a-b)^2+(b-c)^2+(c-a)^2=6$$
So, we get $$a^3-c^3=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7$$ $$\implies a^3-c^3+2c^3-2abc-11=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc-11.$$
And $$3 a^3 - 2 a^2 b - 4 a^2 c + 2 a b^2 + 4 a c^2 - 2 b^2 c + 2 b c^2 - 3 c^3 - 7 = 0 $$
So, it's enough to show that $$[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc=6b $$
or show that $$2 a^3 - 4 a^2 c + 4 a c^2 + 2 b^3 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
or show that $$2 (a^3+b^3) - 4 a^2 c + 4 a c^2 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
I got something ahead, but failed. Any solutions?
| Hint: When things look cyclic, first thing you do is add them. Try adding the three cyclic-looking equations, and recall the relation
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). $$
| {
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Which integrating technique should I use? Just some context: In the mathematical course, I have undertaken this year, I've just learnt how to integrate using partial fractions, substitution(not trig though, just a variable) and integrating derivatives of inverse trig functions(arcsin, arccos, arctan)
Here's is where I am confused:
When it comes to integrating fractions; particularly those with quadratics in their denominators, how do I know which option to pick correctly? With exams coming up, I do not have the time to play around with these to find the correct method. So is there a strategy that you may have?
Here is an example that may help clarify:
I want to integrate the following: $\int{\frac{1}{x^2+8x+4}} dx$
*
*How would one start, by always trying to factor out(might be a long process though if it does not work out)?
*This integral could be an arctan function but it could be a natural logarithm as well
Then what if there is a linear function in the denominator? Would substitution always work?
I have so many questions, and I need answers(or rather strategies).
|
How would one start, by always trying to factor out (might be a long process though if it does not work out)?
I would not recommend it precisely because of the reasons you mention, it doesn't always work out, and even if it does, you then have to do partial fractions which can be a very long and tedious process. The most direct and (in my opinion) easiest way to solve these integrals where you have a quadratic in the denominator is:
*
*If you have a linear factor in the numerator, split the integrals so that you end up in a situation where you just have constants in the numerator.
*Once you have an integral involving only a quadratic in the denominator, complete the square and make the necessary substitutions to get something of the form $\int\frac{\mathrm{d}s}{s^2 \pm 1}$ or $\int\frac{\mathrm{d}s}{s^2}$, and then solve these as an arctan, logarithm or $u$-substitution in their corresponding case.
Yeah, but how do I do that exactly?
Any integral of the form $\int \frac{\mathrm{d}x}{ax+b}$, i.e. with a linear function in the denominator, can always be solved like this
$$
\int \frac{\mathrm{d}x}{ax+b} \overset{\color{blue}{u = ax+b}}{=} \frac{1}{a} \int\frac{\mathrm{d}u}{u} = \frac{1}{a} \ln|u|+C = \frac{1}{a} \ln|ax+b|+C
$$
Any integral of the form $\int \frac{Ax+B}{ax^2+bx +c}\mathrm{d}x$, i.e. with a linear function in the numerator and a quadratic in the denominator, can also always be solved using substitution:
\begin{align}
\int \frac{Ax+B}{ax^2+bx +c}\mathrm{d}x&= \int \frac{\frac{A}{2a}(2ax+b)+B - \frac{Ab}{2a}}{ax^2+bx +c}\mathrm{d}x\\
& = \frac{A}{2a} \int\frac{2ax +b}{ax^2+bx +c}\mathrm{d}x + \left( B - \frac{Ab}{2a}\right) \int\frac{1}{ax^2 +bx +c}\, \mathrm{d}x\\
& \overset{\color{blue}{u = ax^2 +bx+c}}{=}\frac{A}{2a} \int\frac{1}{u}\mathrm{d}u + \left( B - \frac{Ab}{2a}\right) \int\frac{1}{ax^2 +bx +c}\, \mathrm{d}x
\end{align}
The first integral falls into the linear function in the denominator case we just covered, so that part is solved. For the second integral we do the following:
Integrals of the form $\int\frac{1}{ax^2 +bx +c}\, \mathrm{d}x$ where you don't have a linear term in the numerator, you should always start by completing the square. i.e., rewriting the integrand as follows:
\begin{align}
\int\frac{\mathrm{d}x }{ax^2 +bx +c} &= \frac{1}{a} \int\frac{\mathrm{d}x }{\underbrace{x^2 + \color{purple}{2}\frac{b}{\color{purple}{2}a}x +\color{green}{\left( \frac{b}{2a}\right)^2}}_{\left(x + \frac{b}{2a}\right)^2}+ \frac{c}{a}-\color{green}{\left( \frac{b}{2a}\right)^2}} \\
& \overset{\color{blue}{u =x+ \frac{b}{2a}}}{=} \frac{1 }{a}\int \frac{\mathrm{d}u}{u^2 +\left(\frac{c}{a}-\color{green}{ \frac{b^2}{4a^2}}\right)}
\end{align}
And here you have $3$ cases:
*
*If $\frac{c}{a}- \frac{b^2}{4a^2}>0$ then we can take it's square root as follows:
\begin{align}
& =\frac{1 }{a}\int \frac{\mathrm{d}u}{u^2 + \left(\sqrt{\frac{c}{a}- \frac{b^2}{4a^2}}\right)^2}\\
& \overset{\color{blue}{s \sqrt{\frac{c}{a} - \frac{b^2}{4a^2}}= u}}{=}\frac{1}{a\sqrt{\frac{c}{a} - \frac{b^2}{4a^2}}} \int \frac{\mathrm{d}s}{s^2 +1}
\end{align}
and since you mention you know derivatives of inverse functions, you should immediately recognize that $\int \frac{\mathrm{d}s}{s^2 +1} = \arctan(s)$, so after reversing the previous substitutions you're done.
*If $\frac{c}{a}- \frac{b^2}{4a^2}<0$ then we can't take its square root, but we can take the square root of $\frac{b^2}{4a^2} - \frac{c}{a}>0$. So this is what we do:
\begin{align}
& =\frac{1 }{a}\int \frac{\mathrm{d}u}{u^2 \mathbin{\color{red}{-}} \left(\sqrt{ \frac{b^2}{4a^2}- \frac{c}{a}}\right)^2}\\
& \overset{\color{blue}{s \sqrt{ \frac{b^2}{4a^2}-\frac{c}{a}} = u}}{=}\frac{1}{a\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}} \int \frac{\mathrm{d}s}{s^2 -1}
\end{align}
And once you get to this point, you should remember that there's a very easy factorization of $\frac{1}{s^2 -1} = \frac{1}{(s+1)(s-1)}$, so you can apply partial fractions. This leads you to $\frac{1}{s^2 -1} = \frac{1}{2}\frac{1}{s-1} - \frac{1}{2}\frac{1}{s+1}$ and thus
\begin{align}
& =\frac{1}{a\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}} \int \frac{\mathrm{d}s}{s^2 -1}\\
& =\frac{1}{a\sqrt{\frac{b^2}{4a^2}-\frac{c}{a}}} \left[\frac{1}{2} \int \frac{\mathrm{d}s}{s -1} - \frac{1}{2} \int \frac{\mathrm{d}s}{s +1}\right]
\end{align}
and each resulting integral is just a linear factor in the denominator, which we already know how to solve from the beginning.
*If $\frac{c}{a}- \frac{b^2}{4a^2}=0$ then
\begin{align}
& =\frac{1 }{a}\int \frac{\mathrm{d}u}{u^2 +0}\\
& =\frac{1 }{a}\int u^{-2}\, \mathrm{d}u\\
& =-\frac{1 }{a u} + C
\end{align}
Hope this helps!
| {
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"url": "https://math.stackexchange.com/questions/4450994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding minimum value of $x^2+y^2+xy+x-4y+9$
What is the minimum value of $f(x,y)=x^2+y^2+xy+x-4y+9$ ?
I tried completing squares,
$$x^2+y^2+xy+x-4y+9=\frac12(x^2+2xy+y^2+x^2+2x+1+y^2-8y+16+1)=\frac12[(x+y)^2+(x+1)^2+(y-4)^2+1]$$But not sure how to continue.
| Your work is already very useful. Since the function is continuous and bounded from below (your calculations show that the expression is greater or equal than $\frac 12$), it has a global minimum. Since this is a differentiable function in an open set, this global minimum must be a stationary point, i.e.
$$
2x +y +1=0, \quad 2y + x-4 = 0.
$$
So, the global minimum is attained at $(-2,3)$. The minimum is 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What did I do wrong in solving $\int\sec^{-1} x\,{dx}$? I used integration by parts:
let u=$\sec^{-1}\,x$, dv=dx,
then du=$\frac{1}{|x|\sqrt{x^2-1}}$, v=x.
I = $x\sec^{-1}\,x\;-\;\int\frac{x}{|x|\sqrt{x^2-1}}dx\\$
Integration of $\int\frac{x}{|x|\sqrt{x^2-1}}dx$:
Let x=$\sec\theta$, then dx = $\sec\theta\tan\theta\,d\theta$. $\theta\in(0, \frac{\pi}2)\cup(\frac{\pi}2, \pi)\\$.
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{|\sec\theta|\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta\,d\theta\\\qquad\qquad\qquad=\,\int\frac{\sec\theta}{|\sec\theta|}\frac{\tan\theta}{|\tan\theta|}\sec\theta\,d\theta$
When $\theta\in(0, \frac{\pi}2),\;\sec\theta=x\text{, which is}\gt0$. $\tan\theta\gt0\text{, and }\tan\theta=\sqrt{x^2-1}$,
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{\sec\theta}\frac{\tan\theta}{\tan\theta}\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\int\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\ln|\sec\theta+\tan\theta|+c\\\qquad\qquad\qquad=\,\ln|x+\sqrt{x^2-1}|+c$.
Then, I = $x\sec^{-1}\,x\;-\;\ln|x+\sqrt{x^2-1}|+c$.
I know that this is the right answer, but as I continue, I get a different answer for $\theta\in(\frac{\pi}2, \pi)$.
When $\theta\in(\frac{\pi}2, \pi)$, $\sec\theta=x\text{, which is }\lt0$, $\tan\theta\lt0\text{, and }\tan\theta=-\sqrt{x^2-1}$,
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{-\sec\theta}\frac{\tan\theta}{-\tan\theta}\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\int\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\ln|\sec\theta+\tan\theta|+c\\\qquad\qquad\qquad=\,\ln|x-\sqrt{x^2-1}|+c\\\qquad\qquad\qquad=\,-\ln|x+\sqrt{x^2-1}|+c$
Then, I = $x\sec^{-1}\,x\;\boldsymbol{+}\;\ln|x+\sqrt{x^2-1}|+c$
The answer is $x\sec^{-1}\,x\;\boldsymbol{-}\;\ln|x+\sqrt{x^2-1}|+c\\$, what did I do wrong in the last part?
| The anti-derivative valid over all domain $|x|\ge 1$ is obtained as follows
\begin{align}
\int \sec^{-1}x \ dx
=&\ x\sec^{-1}x - \int \frac x{|x|\sqrt{x^2-1}}dx\\
=&\ x\sec^{-1}x - \int \frac 1{\sqrt{|x|^2-1}}d(|x|)\\
=&\ x\sec^{-1}x - \cosh^{-1}(|x|)\
=x\sec^{-1}x - \ln\left(|x|+\sqrt{x^2-1}\right)
\end{align}
Your or the cited answer is only valid for a sub-domain.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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There are $3$ urns $A,B$ and $C$. Urn $A$ contains $4$ red balls and $3$ black balls. Urn $B$ contains $5$ red balls and $4$ black balls. Urn $C$ There are $3$ urns $A,B$ and $C$. Urn $A$ contains $4$ red balls and $3$ black balls. Urn $B$ contains $5$ red balls and $4$ black balls. Urn $C$ contains $4$ red balls and $4$ black balls. One ball is drawn from each of these urns . What is the probability that $3$ balls drawn consists of $2$ red balls and a black ball?
My solution goes like this:
Considering events $A,B$ and $C$ as picking up a red ball from urn $A$, picking up a red ball from urn $B$,picking up a red ball from urn $C$. The probability of picking up two balls of red color from urn $A$ and $B$ and a black ball from urn $C$ is $P(A)P(B)P(\overline{C})=\frac{4.5.4}{7.9.8}$ . Now, this selection can be made in three different ways so the total probabiblity in this case is $P(A)P(B)P(\overline{C})=\frac{4.5.4.3}{7.9.8}$. Now, the probability of picking up two balls of red color from urn $A$ and $C$ and a black ball from urn $B$ is $P(A)P(C)P(\overline{B})=\frac{4.4.4}{7.8.9}$. This selection can be made in three different ways so the total probabiblity in this case is $P(A)P(C)P(\overline{B})=\frac{4.4.4.3}{7.8.9}$. The probability of picking up two balls of red color from urn $B$ and $C$ and a black ball from urn $A$ is $P(B)P(C)P(\overline{A})=\frac{5.4.3}{9.8.7}$.This selection can be made in three different ways so the total probabiblity in this case is $P(B)P(C)P(\overline{A})=\frac{5.4.3.3}{9.8.7}$. So, the total probability is $\frac{4.5.4.3}{7.9.8}+\frac{4.4.4.3}{7.8.9}+\frac{5.4.3.3}{9.8.7}$.
However, this is not a valid probability as you can see as the probability is greater than $1$. Where the problem is occuring ? Why this method is not valid? I am not getting it? Is it because the question does not support taking all arrangements in consideration ? Does the question is made for only the specific order i.e first drawing two red balls and then drawing a black ball? Does the question only stand valid for the previous case mentioned?Should the problem specify this that the order in which the balls of a particular color apre chosen follows a soecific order i.e say we must choose first from urn $A$ then urn $B$ and then urn $C$?
| "Now, this selection can be made in three different ways"... No. It cannot. You can select red from A, red from B, and black from C in one way (each with a probability of success). So $$\mathsf P(A, B, \overline C)=\dfrac{4\cdot 5\cdot 4}{7\cdot 9\cdot 8}$$
The "three ways" a black ball may be drawn from an urn is accounted for in your summation.$${\mathsf P(A,B,\overline C)+\mathsf P(A,\overline B,C)+\mathsf P(\overline A, B, C)\\=\dfrac{4\cdot 5\cdot 4}{7\cdot 9\cdot 8}+\dfrac{4\cdot 4\cdot 4}{7\cdot 9\cdot 8}+\dfrac{3\cdot 5\cdot 4}{7\cdot 9\cdot 8}\\=\dfrac {17}{42}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of this weird function $ \int \frac{1}{(x^2+x+1)^2}dx $ I put this equation into Symbolab and it produced me a very complex result
Basically this is the result but I believe there is a simpler way to solve this question
$$
\frac{2}{3\sqrt{3}}\left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)+\sin \left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)\right)\right)+C
$$
This was my question
$$
\int \frac{1}{\left(x^2+x+1\right)^2}dx
$$
| Substitute $y=x+\frac12$\begin{align}
&\int \frac{1}{(x^2+x+1)^2}dx\\
=& \int \frac{1}{(y^2+\frac34)^2}dy
= \int \frac{2}{3y}\ d\bigg( \frac{y^2}{y^2 +\frac34}\bigg)
\overset{ibp}=\frac{2y}{3(y^2+\frac34)}+\frac23\int \frac1{y^2+\frac34}dy\\
=&\ \frac{8y}{3(4y^2+3)}+ \frac4{3\sqrt3}\tan^{-1}\frac{2y}{\sqrt3}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\frac{3\pi}{2}-\alpha\right)$ Prove that $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\dfrac{3\pi}{2}-\alpha\right)$$
The RHS is equal to $\tan\alpha,$ so we are to show $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\tan\alpha$$
My try for simplifying the LHS: $$\dfrac{2\sin2\alpha\cos2\alpha}{1+\cos^22\alpha-\sin^22\alpha}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{1+\cos^2\alpha-\sin^2\alpha}=\\=\dfrac{4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)}{1+(\cos^2\alpha-\sin^2\alpha)^2-(2\sin\alpha\cos\alpha)^2}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{1+\cos^2\alpha-\sin^2\alpha}\\=\dfrac{4\sin\alpha\cos\alpha(\cos\alpha-\sin\alpha)(\cos\alpha+\sin\alpha)}{1+\sin^4\alpha-2\cos^2\alpha\sin^2\alpha+\cos^4\alpha-4\sin^2\alpha\cos^2\alpha}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{2\cos^2\alpha}$$
| Recall that $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$, and $\cos(x) = \frac{e^{ix}+e^{-ix}}{2},$ where $i = \sqrt{-1}$. Then:
$$X = \frac{\sin(4\alpha)}{1+\cos(4\alpha)}\cdot\frac{\cos(2\alpha)}{1+\cos(2\alpha)}=
\frac{1}{i}\cdot\frac{(e^{i4\alpha}-e^{-i4\alpha})(e^{i2\alpha}+e^{-i2\alpha})}{(2 + e^{i4\alpha}+e^{-i4\alpha})(2+e^{i2\alpha}+e^{-i2\alpha})}.$$
Let $s = e^{i2\alpha}$. Then:
$$X=\frac{1}{i}\cdot\frac{(s^2-s^{-2})(s + s^{-1})}{(2 + s^2+s^{-2})(2+s+s^{-1})} = \ldots \text{some boring algebra}\ldots = \\=\frac{1}{i}\frac{s-1}{s+1} = \frac{1}{i}\frac{e^{i2\alpha}-1}{e^{i2\alpha} + 1} = \\
\frac{1}{i}\frac{e^{i\alpha}(e^{i\alpha}-e^{-i\alpha})}{e^{i\alpha}(e^{i\alpha} + e^{-i\alpha})} = \frac{1}{i}\frac{e^{i\alpha}-e^{-i\alpha}}{e^{i\alpha} + e^{-i\alpha}}\\
= \displaystyle\frac{\frac{e^{i\alpha}-e^{-i\alpha}}{2i}}{\frac{e^{i\alpha} + e^{-i\alpha}}{2}} = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha).\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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The integral $\int_{0}^{1} \frac{ \log (1-x)}{1+x^2}dx$ Recently a very interesting result
$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
has been proved in a more than elegant way. See Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
Here we find the value of the integral
$I=\int_{0}^1 \frac{\log(1-x)}{1+x^2} dx$
Let $x=\tan t$, then
$I=\int_{0}^{\pi/4} \log(1-\tan t) dt.$
By IVth property, we get
$I=\int_0^{\pi/4} \log (1-\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1-\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log (2 \tan x)~dx-\int_{0}^{\pi/4}\log(1+\tan x)dx=\frac{\pi}{4} \log 2+J-K.$
$J=\int_{0}^{\pi/4} \log \tan x ~dx=-C,$ see Definite integral $\int_0^{\pi/4}\log\left(\tan{x}\right)\ dx$
Let us work for $K=\int_{0}^{\pi/4} \log(1+\tan x) dx$, by IV property, again we get
$K=\int_0^{\pi/4} \log (1+\tan(\pi/4-t)) dt=\int_{0}^{\pi/4} \log \left( 1+\frac{1-\tan x}{1+\tan x}\right) dt=\int_{0}^{\pi/4} \log 2~ dx-K \implies K=\frac{\pi}{8} \log 2.$
Finally, we get $I=\frac{\pi}{8}\log 2-C,$ where $C$ is the Catalan constant.
What could be other interesting ways of finding $I$?
| Perhaps a cleaner way is to consider the integral
$$
\int_1^\infty \frac{\ln(u-1)}{1+u^2}du.
$$
First, perform the substitution $u\rightarrow (u+1)/(u-1) = 1 + 2/(u-1)$ to get
$$
\int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_1^\infty \frac{\ln\left(\frac{2}{u-1}\right)}{1+u^2}du = \frac{\pi}{4}\ln2 -\int_1^\infty \frac{\ln(u-1)}{1+u^2}\Longrightarrow \int_0^1\frac{\ln(u-1)}{1+u^2} = \frac{\pi}{8}\ln 2.
$$
Now use the substitution $u\rightarrow x^{-1}$ to get
$$
\int_1^\infty \frac{\ln(u-1)}{1+u^2}du = \int_0^1\frac{\ln(x^{-1}-1)}{1+x^2}dx = \int_0^1\frac{\ln(1-x)}{1+x^2}dx -\int_0^1\frac{\ln x}{1+x^2}dx = \frac{\pi}{8}\ln 2.
$$
Rearranging gives us an expression for the original integral:
$$
\int_0^1\frac{\ln(1-x)}{1+x^2}dx = \frac{\pi}{8}\ln 2 + \int_0^1\frac{\ln x}{1+x^2}dx = \frac{\pi}{8}\ln 2 - C,
$$
where $\int_0^1\ln(x)/(1+x^2)dx = -C$ is apparently a well-known integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given a triangle ABC inscribed in the unit circle , the 3 vertices could be described via 3 complex number, namely, $a$, $b$, and $c$.
Now $AD$ is an altitude, $D$ is the foot of $AD$ on $BC$.
Prove: $$D = \frac{a+b+c}2 - \frac{bc}{2a}$$
So far my progress is --
*
*the circumcentre of triangle $ABC$ is just $O = 0$.
*its centroid $G = \frac{a+b+c}3$.
*by Euler line, the orthocentre $H = a+b+c$.
*also, we can see that the centre of the nine point circle $N =
\frac{a+b+c}2$.
But then I'm a bit stuck. $D$ is on the line $AH$ and $BC$, but I couldn't reach the conclusion to be proved.
| Instead of treating the points as arbitrary $\mathbb{R}^2$ vectors, let's make use of them being complex numbers. If $z_1, z_2 \in \mathbb{C}$ are interpreted as vectors, then:
*
*They are perpendicular if the quotient between them is a pure imaginary number.
*They are parallel (or overlapping) if the quotient between them is a real number.
So,
$$AD \perp BC \implies \Re(\frac{d - a}{c - b}) = 0$$
$$BD \parallel BC \implies \Im(\frac{d - b}{c - b}) = 0$$
Also recall that $\Re(z) = \frac{z + \overline{z}}{2}$, $\Im(z) = \frac{z - \overline{z}}{2i}$, and conjugation is distributive over all of the four basic operations. Therefore, from the first equation:
$$\Re(\frac{d - a}{c - b}) = 0$$
$$\frac{d - a}{c - b} + \frac{\overline{d} - \overline{a}}{\overline{c} - \overline{b}} = 0$$
$$(d - a)(\overline{c} - \overline{b}) + (\overline{d} - \overline{a})(c - b) = 0$$
$$\overline{c}d - \overline{b}d - a\overline{c} + a\overline{b} + c\overline{d} - b\overline{d} - \overline{a}c + \overline{a}b = 0$$
$$(\overline{c} - \overline{b})d + (c - b)\overline{d} = a\overline{c} + \overline{a}c - a\overline{b} - \overline{a}b$$
And from the other equation:
$$\Im(\frac{d - b}{c - b}) = 0$$
$$\frac{d - b}{c - b} - \frac{\overline{d} - \overline{b}}{\overline{c} - \overline{b}} = 0$$
$$(d - b)(\overline{c} - \overline{b}) - (\overline{d} - \overline{b})(c - b) = 0$$
$$\overline{c}d - \overline{b}d - b\overline{c} + b\overline{b} - c\overline{d} + b\overline{d} + \overline{b}c - b\overline{b} = 0$$
$$(\overline{c} - \overline{b})d + (b - c)\overline{d} = b\overline{c} - \overline{b}{c}$$
Adding these two equations together gives us:
$$2(\overline{c} - \overline{b})d = a\overline{c} + \overline{a}c - a\overline{b} - \overline{a}b + b\overline{c} - \overline{b}{c}$$
Now, to get rid of the conjugation operators. Maybe there's something clever we can multiply everything by. Hmm...
Ooh, I've got an idea! If $z$ is on the unit circle, then $z\overline{z} = |z|^2 = 1$. So let's just multiply everything by $abc$, and all of the conjugates will have something to cancel with.
$$2a(b - c)d = a^2b + bc^2 - a^2c - b^2c + ab^2 - ac^2$$
$$d = \frac{a^2b + bc^2 - a^2c - b^2c + ab^2 - ac^2}{2a(b - c)}$$
$$d = \frac{a^2(b - c) - bc(b - c) + a(b^2 - c^2)}{2a(b - c)}$$
$$d = \frac{a^2 - bc + a(b + c)}{2a}$$
$$d = \frac{a^2 + ab + ac}{2a} - \frac{bc}{2a}$$
$$d = \frac{a + b + c}{2} - \frac{bc}{2a}$$
Q.E.D.
| {
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"url": "https://math.stackexchange.com/questions/4464012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Why Lagrange Multiplier Doesn't Work? Question: Find maximum value $f(x,y,z) = xy + zy + xz - 4xyz$ subject to constraint $x + y + z = 1$ and $x,y,z \geq 0$.
$$
g(x,y,z) = x + y + z - 1
$$
and
$$
\nabla g(x,y,z) \neq 0,
\qquad
\nabla g(x,y,z) = \langle 1,1,1 \rangle.
$$
When we apply Lagrange Multiplier Method, we find
$f \bigl( \frac{1}{2}, \frac{1}{4}, \frac{1}{4} \bigr) = \frac{3}{16}$, the maximum value, but the answer is
$f \bigl( 0, \frac{1}{2}, \frac{1}{2} \bigr) = \frac{1}{4}$.
Why does that happen? Lagrange Multiplier Method has a only $1$ rule: $\nabla g(x,y,z) \neq 0$, there is no rule break.
EDİT: I find f(0,1/2,1/2) with another method and f(1/2,1/4,1/4) is not both a global maxima and a global minima point but Lagrange Multiplier gives that result, and f(1/2,1/4,1/4) is not a local minimum f(0,0,1) is lower than that.
| You may use KKT conditions instead.
Or you may apply Lagrange Multiplier for the following equivalent problem:
Find the maximum of $F(a, b, c) = a^2b^2 + b^2c^2 + c^2a^2 -4a^2b^2c^2$ subject to $a^2+b^2+c^2 = 1$.
(Note: We have only one constraint. )
Let
$$L(a, b, c) := a^2b^2 + b^2c^2 + c^2a^2 -4a^2b^2c^2 + \lambda (a^2 + b^2 + c^2 - 1).$$
We have
\begin{align*}
\frac{\partial L}{\partial a}
&= -2a(4b^2c^2 - b^2 - c^2 - \lambda) = 0,\\
\frac{\partial L}{\partial b}
&= -2b(4c^2a^2 - a^2 - c^2 - \lambda ) = 0, \\
\frac{\partial L}{\partial c}
&= -2c(4a^2b^2 - a^2 - b^2 - \lambda) = 0, \\
\frac{\partial L}{\partial \lambda} &= a^2 + b^2 + c^2 - 1 = 0.
\end{align*}
Let us solve this system of equations.
If $abc \ne 0$, we have
\begin{align*}
4b^2c^2 - b^2 - c^2 - \lambda &= 0, \tag{1}\\
4c^2a^2 - a^2 - c^2 - \lambda & = 0, \tag{2}\\
4a^2b^2 - a^2 - b^2 - \lambda &= 0, \tag{3}\\
a^2 + b^2 + c^2 - 1 &= 0. \tag{4}
\end{align*}
We can easily get all four solutions of $(a^2, b^2, c^2, \lambda)$:
$$(1/4, 1/4, 1/2, -1/4),
(1/2, 1/4, 1/4, -1/4),
(1/4, 1/2, 1/4, -1/4),
(1/3,1/3, 1/3, -2/9).$$
(Note: We actually need $a^2, b^2, c^2$. We don't need $a, b, c$.)
If $a=b=0$, we have $c^2 = 1, \lambda = 0$.
If $a=c=0$, we have $b^2 = 1, \lambda = 0$.
If $b=c=0$, we have $a^2 = 1, \lambda = 0$.
If $a = 0, bc \ne 0$, we have
$b^2 = c^2 = 1/2$ and $\lambda = -1/2$.
If $b = 0, ac \ne 0$, we have
$a^2 = c^2 = 1/2$ and $\lambda = -1/2$.
If $c = 0, ab \ne 0$, we have
$a^2 = b^2 = 1/2$ and $\lambda = -1/2$.
It is easy to get the maximum $1/4$ when $a = 0, b^2 = c^2 = 1/2$ or $b = 0, c^2 = a^2 = 1/2$ or $c = 0, a^2 = b^2 = 1/2$.
Come back to the original problem. The maximum is $1/4$ when $x = 0, y = z = 1/2$ or $y = 0, x = z = 1/2$ or $z = 0, x = y = 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Where is my mistake in evaluating $\lim_{x \to -\infty} \frac{\sqrt{x^2+4}}{x}$? Given is $\lim_{x \to -\infty} \dfrac{\sqrt{x^2+4}}{x}$
I divide numerator and denominator by x to the largest degree in the denominator and I get
$$\lim_{x \to -\infty} \frac{\sqrt{1+\frac{4}{x^2}}}{1}=\frac{\sqrt{1}}{1}=1.$$
But the answer should be -1. Where did I make a mistake?
| The mistake is when you are shortening out $x$. $x$ is negative, but the square root is positive. Thus:
$$\begin{align}
\lim_{x \to-\infty} \frac{\sqrt{x^2+4}}{x}
&= \lim_{x \to-\infty} \frac{|x|\sqrt{1+4/x^2}}{-|x|} \\
&= \lim_{x \to-\infty} \frac{\sqrt{1+4/x^2}}{-1} \\
&\to -1
\end{align}$$
where $x=-|x|$ in the denominator because $x < 0$.
Alternative way to see the limit is substituting $x= -y$:
$$\begin{align}
\lim_{x \to-\infty} \frac{\sqrt{x^2+4}}{x}
&= \lim_{y \to\infty} \frac{\sqrt{(-y)^2+4}}{-y} \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can we show that $\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\mathrm{d}t=\frac{\pi}{2a^2}\cdot\ln^2(1+a)$ without complex analysis? $\newcommand{\d}{\,\mathrm{d}}\newcommand{\Res}{\operatorname{Res}}\newcommand{\Arg}{\operatorname{Arg}}$Consider the function: $$\begin{align}I:(0,\infty)&\to(0,\infty)\\a&\mapsto\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t\end{align}$$
Using complex analysis we can evaluate this integral:
Let $R\gt a$ be any real number. Note that $\int_0^R\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t=\frac{1}{2}\int_{-R}^R\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t=:\frac{1}{2}J_R(a)$ by evenness.
Define the complex contour $\gamma_R$ to be the union of the line segment $[-R,R]$ and the arc $[-\pi,0]\ni t\mapsto Re^{it}\in\Bbb C$, $C_R$. We may rewrite (with the principal branch of the logarithm, modified to $\Arg(z)\in[-\pi,\pi)$): $$\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}=\Im\left[\frac{\ln^2(1+it)}{t(t^2+a^2)}\right]=:\Im(f(t))$$Then: $$\begin{align}\left|\int_{C_R}f(z)\d z\right|&\le\int_0^\pi\frac{\ln^2(|1+iRe^{it}|)+\Arg^2\left(\frac{R\cos(t)}{1-R\sin(t)}\right)}{|(Re^{it})^2+a^2|}\d t\\&\le\frac{1}{4}\int_0^\pi\frac{\ln^2(1+R^2-2R\sin(t))+4\pi^2}{R^2-a^2}\d t\\&\le\frac{\pi}{4}\cdot\frac{\ln^2(1+R^2)+4\pi^2}{R^2-a^2}=:\varepsilon_R\end{align}$$Due to the clockwise orientation, and the fact that there is only one enclosed pole at $z=-ia$, we find: $$\begin{align}\int_{\gamma_R}f(z)\d z&=-2\pi i\cdot\Res_{z=-ia}f(z)\\&=-2\pi i\cdot\frac{\ln^2(1+a)}{-2a^2}\\&=i\cdot\frac{\pi}{a^2}\cdot\ln^2(1+a)\end{align}$$Then, taking imaginary parts: $$\begin{align}0\le\frac{1}{2}\lim_{R\to\infty}\left|J_R(a)-\frac{\pi}{a^2}\cdot\ln^2(1+a)\right|&\le\frac{1}{2}\lim_{R\to\infty}\varepsilon_R=0\\\left|I(a)-\frac{\pi}{2a^2}\cdot\ln^2(1+a)\right|&=0\end{align}$$And we can conclude: $$I(a)=\int_0^\infty\frac{\arctan(t)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t=\frac{\pi}{2a^2}\cdot\ln^2(1+a),\,a\gt0$$
I am interested in ways to evaluate this integral using real-analytic techniques. I had two ideas (well, tips from someone else...) involving differentiating under the integral sign with $(1+t^2)^s$, to produce $\ln(1+t^2)$, and letting also $s\to-1$ to obtain the $\arctan$ term via integration by parts. That failed quite miserably... A different attempt, also using differentiating under the integral sign, was a bit more successful:
Put $I(a,b)=\int_0^\infty\frac{\arctan(bt)\cdot\ln(1+t^2)}{t(t^2+a^2)}\d t$. Then: $$\begin{align}\frac{\partial}{\partial b}I(a,b)&=\int_0^\infty\frac{\ln(1+t^2)}{(1+b^2t^2)(t^2+a^2)}\d t\\&=\frac{1}{a^2-\frac{1}{b}}\left[\int_0^\infty\frac{\ln(1+t^2)}{1+b^2t^2}\d t-\frac{1}{b^2}\int_0^\infty\frac{\ln(1+t^2)}{t^2+a^2}\d t\right]\\&=\frac{1}{a^2-\frac{1}{b}}\left[\frac{1}{b}\int_0^\infty\frac{\ln(b^2+t^2)-2\ln(b)}{1+t^2}\d t-\frac{\pi}{ab^2}\cdot\ln(1+a)\right]\\&=\frac{\pi}{a^2b-1}\cdot\ln(1+1/b)-\frac{\pi}{a^3b^2-ab}\cdot\ln(1+a)\end{align}$$And clearly $I(0)=0$, so we obtain (abusing notation slightly): $$I(a,1)=\pi\int_0^1\frac{\ln(1+1/x)}{a^2x-1}\d x-\frac{\pi}{a}\cdot\ln(1+a)\int_0^1\frac{1}{a^2x^2-x}\d x$$
And all numerical experiments suggests that the above equation is wrong.
I don't know what went wrong with my real analytic approach, hopefully someone can point it out! I used this result.
Question: Can we evaluate $I$ in any (hopefully clean...) way without using complex analysis?
Motivation: I need to learn to get better at "normal" integration :)
Pointing out what went wrong with my approach is just a P.S., it is not my main interest.
| Utilize the integral
\begin{align}
&\int_0^\infty\frac{\ln x}{t^2+(x+1)^2}\overset{x\to\frac{1+t^2}x} {dx}\\
=&\int_0^\infty \frac{\ln(1+t^2)-\ln x}{t^2+(x+1)^2}dx
=\frac12 \int_0^\infty\frac{\ln (1+t^2)}{t^2+(x+1)^2}dx
=\frac1{2t}\tan^{-1}t\ln(1+t^2)
\end{align}
to integrate
\begin{align}
&\int_0^\infty\frac{\tan^{-1}t\ln(1+t^2)}{t(t^2+a^2)}dt\\
=& \ 2\int_0^\infty \int_0^\infty \frac{\ln x}{(t^2+a^2)(t^2+(x+1)^2)}dxdt
=\frac{\pi}a \int_0^\infty \frac{\ln x}{a(x+1)+(x+1)^2}
\overset{x\to \frac{1+a}x}{dx}\\
=& \ \frac{\pi}a \int_0^\infty \frac{\ln(1+a)-\ln x}{a(x+1)+(x+1)^2}dx
= \frac{\pi \ln(1+a)}{2a} \int_0^\infty \frac{1}{a(x+1)+(x+1)^2}dx\\
=& \ \frac{\pi \ln(1+a)}{2a} \cdot \frac{\ln(1+a)}{a}
= \frac{\pi}{2a^2}\ln^2(1+a)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Given a permutation $\sigma = (13)(254)$, state $\sigma^2$. Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.
$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425)
$
Or, in two row format, get:
$$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$
Then, $$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
$$
Due to disjoint cycles, they can commute as no affect on map produced.
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
$$
My answer is:
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 1 & 3 \end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
= e.\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
= \begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
$$
But, the answer is different: $(245)$?
| We have
$$\begin{align}
\sigma^2&=(13)(254)(13)(254)\\
&=(13)(13)(254)(254)\\
&=(254)^2\\
&=(245)
\end{align}$$
because
$$\begin{align}
1&\xrightarrow{(254)}1\xrightarrow{(254)}1,\\
2&\xrightarrow{(254)}5\xrightarrow{(254)}4,\\
3&\xrightarrow{(254)}3\xrightarrow{(254)}3,\\
4&\xrightarrow{(254)}2\xrightarrow{(254)}5,\\
5&\xrightarrow{(254)}4\xrightarrow{(254)}2.\\
\end{align}$$
In row format,
$$\begin{align}
\sigma^2&=\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 5 & 1 & 2 & 4
\end{pmatrix}^2\\
&=\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 5 & 1 & 2 & 4
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 5 & 1 & 2 & 4
\end{pmatrix}\\
&=\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
1 & 4 & 3 & 5 & 2
\end{pmatrix},
\end{align}$$
which gives $(245)$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x)^2}=\sum_{n=1}^\infty (-1)^{n+1}nx^{n-1}=\sum_{n=0}^\infty(-1)^n(n+1)x^n$$
From these,
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{4}+\sum_{n=0}^\infty \frac{(-1)^nx^n}{4}+\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)x^n}{2} \\ =\sum_{n=0}^\infty\frac{x^n+(-1)^nx^n+(-1)^{n+1}(n+1)x^n}{4} \\ =\sum_{n=0}^\infty\frac{(1+(-1)^n+(-1)^{n+1}(n+1))x^n}{4}$$
$f^{(2020)}(0)=(2020)!\times a_{2020}$ where $f(x)=\sum a_nx^n$
In this case, $a_n = 1+(-1)^n+(-1)^{n+1}(n+1), a_{2020}=-2019$ but there is no same answer.
Where did I make a mistake?
| We have
$$f(x) = \frac14 \cdot \frac{1}{1-x}
+ \frac14 \cdot \frac{1}{1 + x}
+ \frac12 \cdot \left(\frac{1}{1 + x}\right)'.$$
By observing the first several derivatives to see the pattern:
$$\left(\frac{1}{1-x}\right)' = \frac{1}{(1-x)^2},
\left(\frac{1}{1-x}\right)'' = \frac{2}{(1-x)^3},
\left(\frac{1}{1-x}\right)''' = \frac{3}{(1-x)^4}, \cdots$$
$$\left(\frac{1}{1+x}\right)' = \frac{-1}{(1+x)^2},
\left(\frac{1}{1+x}\right)'' = \frac{2}{(1+x)^3},
\left(\frac{1}{1+x}\right)''' = \frac{-3}{(1+x)^4}, \cdots$$
we have
$$\left(\frac{1}{1-x}\right)^{(n)}
= \frac{n!}{(1-x)^{n+1}}$$
and
$$\left(\frac{1}{1+x}\right)^{(n)}
= \frac{(-1)^n n!}{(1+x)^{n+1}}.$$
Thus, we have
$$f^{(n)}(x) = \frac14 \cdot \frac{n!}{(1-x)^{n+1}}
+ \frac14 \cdot \frac{(-1)^n n!}{(1+x)^{n+1}} + \frac12 \cdot \frac{(-1)^{n+1} (n+1)!}{(1+x)^{n+2}} $$
and
$f^{(2020)}(0) = -1010 \cdot 2020!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| Alternatively, let $x_n = a^n + b^n$. Then $x_{n+2}=7x_{n+1}-2x_n$ and so we get
$x_0=2$
$x_1=7$
$x_2=7x_1-2x_0=45$
$x_3=7x_2-2x_1=301$
$x_4=7x_3-2x_2=2017$
$x_5=7x_4-2x_3=13517$
$x_6=7x_5-2x_4=90585$
This is easy to do by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 5
} |
Showing that switching the order of variables gives same sum so as to simplify the required expression for exact sum If $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$ , then if we want to show that $S(x,y) = S(y,x)$ so as to get the simplification by adding both to get the exact sum . I tried to evaluate S(x,y)-S(y,x) , which gives $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)(5^y - 5^x)}{(5^{x+y}(5^x +5^y))}$ now how to do we show its value is zero ?
| $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$
By interchanging $x$ and $y$ we get
$S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^y(5^x +5^y))}$
$\implies 2S = \sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x +5^y)}\left(\frac{1}{5^x}+\frac{1}{5^y}\right)=\sum_{y=0}^{\infty}\sum_{x=0}^{\infty}\frac{(x+y +xy)}{(5^x 5^y)}$
$\implies 2S=\sum_{y=0}^{\infty}\sum_{x=0}^{\infty}\frac{(x+y +xy)}{(5^x 5^y)}$
Note that $\sum_{k=0}^{\infty} z^k=\frac{1}{1-z}$ and $\sum_{k=0}^{\infty}kz^k=\frac{z}{(1-z)^2},$ if $|z|<1$, then
$2S=2 \sum_{k=0}^{\infty} 5^{-k} \sum_{k=0}^{\infty}k 5^{-k}+\left(\sum_{k=0}^{\infty} k 5^{-k}\right)^2=\frac{25}{32}+\frac{25}{256}\implies S=\frac{223}{512}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is this relationship between trigonometric and hyperbolic function? In the following, I don't understand how they put $\tan{\phi} = \sinh{\frac{\psi}{\sqrt{2}}}$.
Is there a relationship?
| This isn't asserting a relationship. It's simply setting up a substitution that trades circular functions for hyperbolics in order to escape the square root. It's really no different than having $\sqrt{4-x^2}$ and setting $x=2\sin\theta$.
In the exercise, first note that the expression (ignoring $d\theta$) can be rewritten as
$$a^2\left((1+\tan^2\phi)+\tan^2\phi\right)^{1/2} \sec^2\phi\, d\phi = a^2\left(1+2\tan^2\phi\right)^{1/2} \sec^2\phi\,d\phi$$
The author is suggesting that the form $1+2\tan^2\phi$ calls for the substitution
$$\tan\phi = \frac{\sinh\psi}{\sqrt{2}} \qquad\qquad \sec^2\phi\,d\phi=\frac{\cosh\psi}{\sqrt{2}}\,d\psi$$
(note that the entire $\sinh\psi$ is divided by $\sqrt{2}$, not just $\psi$) so that we have
$$\begin{align}
a^2\left(1+2\cdot\frac{\sinh^2\psi}{2}\right)^{1/2} \frac{\cosh\psi\,d\psi}{\sqrt{2}} &= \frac1{\sqrt{2}} a^2\left(1+\sinh^2\psi\right)^{1/2} \cosh\psi\,d\psi \\[4pt]
&= \frac1{\sqrt{2}}a^2\left(\cosh^2\psi\right)^{1/2} \cosh\psi\,d\psi \\[4pt]
&= \frac1{\sqrt{2}}a^2\cosh^2\psi\,d\psi \tag1
\end{align}$$
Alternatively, one could make a standard circular substitution thusly:
$$\tan\phi = \frac{\tan\omega}{\sqrt{2}} \qquad\qquad
\sec^2\phi\,d\phi = \frac{\sec^2\omega}{\sqrt{2}}d\omega$$
to give
$$\begin{align}
a^2\left(1+2\cdot\frac{\tan^2\omega}{2}\right)^{1/2} \frac{\sec^2\omega d\omega}{\sqrt{2}} &= \frac1{\sqrt{2}}a^2\left(1+\tan^2\omega\right)^{1/2} \sec^2\omega\,d\omega \\[4pt]
&= \frac1{\sqrt{2}}a^2\left(\sec^2\omega\right)^{1/2} \sec^2\omega\,d\omega \\[4pt]
&= \frac1{\sqrt{2}}a^2\sec^3\omega\,d\omega \tag2
\end{align}$$
Expression $(1)$ has fairly straightforward integral, which in turn has a fairly straightforward back-substitution into $\phi$-form. Expression $(2)$ is perhaps a little messier on both counts. (The author may have anticipated this, or the author may have had some other reason to favor the hyperbolic substitution.) The end result in both cases is the same.
By the way, a "third" way to proceed is with this substitution:
$$\tan\phi = \frac{u}{\sqrt{2}} \qquad\qquad \sec^2\phi\,d\phi=\frac{du}{\sqrt{2}}$$
This gives
$$\begin{align}
a^2\left(1+2\cdot\frac{u^2}{2}\right)^{1/2} \frac{du}{\sqrt{2}} &= \frac1{\sqrt{2}}a^2\left(1+u^2\right)^{1/2}du \tag3
\end{align}$$
From here, substitution $u=\sinh\psi$ reverts to $(1)$, and $u=\tan\omega$ reverts to $(2)$, so this isn't necessarily a different approach. On the other hand, integration by parts and back-substituting are pretty straightforward from here, too.
| {
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Primes representable by either $x^2+36y^2$ or $4x^2+9y^2$ Is there a simple criterion for primes that are representable by either $x^2 + 36 y^2$ or $4x^2 + 9y^2$?
This is not my area of expertise, so any pointers appreciated.
I had a look in the Cox book "Primes of the form $x^2 + n y^2$", but if those specific forms are there, I missed it.
| Sure, for $p \equiv 1 \pmod 4,$ there is an expression $u^2 + 36 v^2 = p$ precisely when $x^4 + 3$ factors into four linear factors $\pmod p,$ that is four distinct roots. Let me check whether it is enough to have one root. This is from Kenneth S. Williams and D. Liu, Tamkang Journal of Mathematics, volume 25, number 4, Winter 1994; pages 321-334.
Once $p \equiv 1 \pmod {12},$ it is enough to have one root of $x^4 + 3 \pmod p$
Meanwhile, this can probably be done with biquadratic reciprocity, as in Ireland and Rosen.
More description: we need only primes $p \equiv 1 \pmod {12}$ Now, we can always find a number $a$ such that $$ (x^2 + a) ( x^2 - a) \equiv x^4 + 3 \pmod p$$
Indeed, we demanded $(-3|p) = 1,$ we are taking $a^2 \equiv -3 \pmod p$
If $a$ is not a square $\pmod p$ then $p = 4 u^2 + 9 v^2 $ In this case the two quadratics displayed are irreducible $\pmod p.$
If we can take $a = b^2, $ so that
$$ (x^2 + b^2) ( x^2 - b^2) \equiv x^4 + 3 \pmod p$$
then $ p = u^2 + 36 v^2 .$ Note how we can see the condition $b^4 \equiv -3 \pmod p$
Let's see, $( -1 | p) + 1,$ there is an $i$ with $i^2 \equiv -1 \pmod p,$ we may define $c = bi$ and write
$$ (x^2 -c^2) ( x^2 - b^2) \equiv x^4 + 3 \pmod p$$ to display the four roots
| {
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Proofs for couple of details when proving that $n^7 + 7$ is not a perfect square for any integer $n$.
Prove that $n^7 + 7$ is not a perfect square for any integer $n$.
The proof given for the problem is the following
Assume $n > 0$ since $n = 0$ and $n = -1$ are easy and for $n \le -2$ the expression is negative. Suppose $n^7 + 7 = a^2$. Then $$n^7+2^7 = a^2+11^2.$$ Taking modulo $4$ gives $n ≡ 1 \pmod 4$, but $n + 2 \mid a^2 + 11^2$, and $n + 2 ≡ 3 \pmod4$. Note that $a^2 + 11^2$ has no $3$ mod $4$ prime factors except possibly an $11^2$, by Fermat’s Christmas theorem. Since $n + 2 ≡ 3 \pmod 4$ we would need to have $\nu_{11}(n + 2) = 1$ as a result, since $\nu_{11}(n + 2)$ should be odd and at most $2$. However, we then get $$\nu_{11}(a^2+11^2)=\nu_{11}(n^7+2^7)=\nu_{11}(n+2)+\nu_{11}(7)=\nu_{11}(n+2)=1$$
by the exponent lifting lemma, which is impossible.
There are few parts I don't understand. The first one is the line
Note that $a^2 + 11^2$ has no $3$ mod $4$ prime factors except possibly an $11^2$, by Fermat’s Christmas theorem.
I know that we have since $n+2 \mid a^2+11^2$ and $n+2 \equiv 3 \pmod{4}$ the term $n+2$ must have a prime divisor $p$ such that $p \equiv 3 \pmod 4$ and now by Fermat’s Christmas theorem $p \mid n+2 \implies p \mid a^2 +11^2 \implies p \mid a$ and $p \mid 11$ implying that $p = 11$. But why is this stating that it could have $11^2$ as a factor and no other factors which are $3$ mod $4$?
The second fact I don't understand is that
Since $n + 2 ≡ 3 \pmod 4$ we would need to have $\nu_{11}(n + 2) = 1$ as a result, since $\nu_{11}(n + 2)$ should be odd and at most $2$.
Why should $\nu_{11}(n+2)=1$ and what makes $\nu_{11}(n+2)$ odd and at most $2$?
| We first show that $\nu_{11}(a^2+11^2)$, and thus, $\nu_{11}(n+2)$, is at most $2$. Now, on the one hand, for $\nu_{11}(a^2+11^2)$ to be positive at least $2$ in the first place, $a^2$ and thus $a$ itself must also be a multiple of $11$, say $a=11c$ for some integer $c$. So $a^2+11^2$ can be written $a^2+11^2 = (11c)^2+11^2 = 11^2(c^2+1)$. However, on the other hand, for $11^3$ to divide $11^2(c^2+1)$, it follows that the integer $c$ must must be such $11$ that divides $c^2+1$, which gives $c^2 \equiv_{11} -1$, or equivalently, $-1$ a square in $\mathbb{Z}/11\mathbb{Z}$, which is impossible, as $11 \pmod 4 = 3$. So $11$ cannot divide $c^2+1$ after all, and thus $11^3$ cannot divide $11^2(c^2+1)=a^2+11^2$ after all, and thus indeed, $\nu_{11}(a^2+11^2)$ is at most $2$.
Finally, $\nu_{11}(n+2)$ can be at most $2$ as well, as $\nu_{11}(a^2+11^2)$ is at most $2$ and $n+2$ divides $a^2+11$.
To see that $11$ is the only prime $p$ satisfying $p \pmod 4 =3$ dividing $a^2+11^2$, note that $p|(a^2+11^2)$ gives $a^2 \equiv_{p} -(11^2),$ which implies either that
*
*Either $-1$ is a square in $\mathbb{Z}/p\mathbb{Z}$ [impossible for all primes $p$ such that $p \pmod 4 = 3$];
*Or $11^2 \equiv_p 0$ which gives $11=p$.
We now show that $\nu_{11}(n+2)$ must be odd. Now, as shown above, $11$ is the only prime $p$ satisfying $p \pmod 4 =3$ that may divide $n+2$. As $n+2$ is odd, it follows that every other prime $q$ dividing $n+2$ satisfies $q \pmod 4 = 1$, and thus, $$n+2 \pmod 4 = 11^{\nu_{11}(n+2)} \pmod 4.$$
As $n+2 \pmod 4$ is $3$ and $11^e \pmod 4$ is $1$ for all even nonnegative integers $e$, it follows that $\nu_{11}(n+2)$ must indeed be odd.
| {
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Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The problem is that, when graphed, these 2 graphs do not intersect at all, so there should be no real solutions. As for the given complex solutions, neither were considered by wolfram alpha.
Knowing that there are no real solutions, I'm confused as to how I might get complex solutions through means not already attempted and explained above.
| I would recommend you to notice that
\begin{align*}
x^{2} + 5x + 8 = \left(x^{2} + 5x + \frac{25}{4}\right) + \frac{7}{4} = \left(x + \frac{5}{2}\right)^{2} + \frac{7}{4} \geq \frac{7}{4}
\end{align*}
Hence we conclude that $-x^{2} - 5x - 8 < 0$ for every possible value of $x$.
Based on such considerations, we conclude the proposed equation has no solutions in $\mathbb{R}$.
Hopefully this helps!
| {
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Find another method of solving the shaded area. The diagram shows a square of side length $10\;\rm cm$. A quarter circle, of radius $10\;\rm cm$, is drawn from each vertex of the square. Find the exact area of the shaded region.
And This is my answer.
The answer is right, but I am searching for other ways.
Thanks.
|
Area of closed figure contains square $EFGH$ and 4 segments congruent to $EF$
Area of segment can be found as difference of sector and triangle:
$$S_{seg}=\frac12 AF^2 \frac{\pi}{6}-\frac12 AF^2 \sin \frac{\pi}{6}=\frac{\pi-3}{12}AF^2$$
Square side can be found in many ways, for example with cosine rule
$$AB^2=AG^2+BG^2-2AG\cdot BG\cos \frac{5\pi}{6}=(2+\sqrt{3})AG^2$$
$$FG^2=\frac{1}{2+\sqrt{3}}AB^2=(2-\sqrt{3})AF^2$$
$$S=4S_{seg}+FG^2=\frac{\pi-3}{3}AF^2+(2-\sqrt{3})AF^2=\left(1-\sqrt{3}+\frac{\pi}{3}\right)AF^2$$
| {
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Find the measure of the relationship $\frac{1}{r_1} - \frac{1}{r} $in the figure below In In a right triangle $ABC$ ($A=90°$) with inradio $r$, cevian $AD$ is drawn in such a way that the inradium of $ABD$ and $ADC$ are equal to $r1$.If $AD=2$, calculate $\frac{1}{r1}-\frac{1}{r}$ (Answer:0,5).
My progress:
$\triangle CED \sim \triangle CAB \\
\frac{CE}{AC}=\frac{DE}{AB}=\frac{CD}{BC}\\
\triangle BDL \sim \triangle BCA\\
\frac{DL}{AC}=\frac{BD}{BC}=\frac{LB}{AB}\\
CE = CI\\
BK = BL$
but I still haven't found the necessary relationship to finalize
| Here's what seems to be an unnecessarily-complicated solution.
Define $b:=|AC|$, $c:=|AB|$, $d:=|AD|$, $p:=|BD|$, $q:=|CD|$. Let $r$ be the inradius of $\triangle ABC$, and let $s$ be the common inradius of $\triangle ABD$ and $\triangle ACD$.
We know $$\text{inradius}\cdot \text{perimeter} = 2\,\text{area}$$ so we can write
$$\begin{align}
s(c+d+p) &= 2|\triangle ABD|=\frac{p}{p+q}\cdot 2|\triangle ABC| = \frac{p}{p+q}\, r (b+c+p+q) \tag1\\[8pt]
s(b+d+q) &= 2|\triangle ACD|=\frac{q}{p+q}\cdot 2|\triangle ABC| = \frac{q}{p+q}\, r (b+c+p+q) \tag2
\end{align}$$
Solving this linear system for $b$ and $c$ gives
$$
b = -q-d + \frac{2 d q r}{(p + q)(r - s)} \qquad\qquad
c = -p-d + \frac{2 d p r}{(p + q)(r - s)} \tag3
$$
Since $\triangle ABC$ is right, we also know
$$\begin{align}
2r = |AC|+|AB|-|BC| &= b+c-(p+q) \\
&= 2\,\frac{ ds- (p+q)(r-s)}{r - s} \\
\to \qquad (p+q)(r-s) &= ds -r(r-s)\tag4
\end{align}$$
By Stewart's Theorem, we have
$$b^2p+c^2q=(p+q)(d^2+pq) \quad\underset{(3)}{\to}\quad
(p + q) s (r - s)= d r (2s-r) \tag5$$
Combining $(4)$ and $(5)$ to eliminate $p+q$ gives
$$r s (r - s) = d (r - s)^2 \quad\to\quad r s = d (r - s) \quad\to\quad
\frac1d=\frac1s-\frac1r \tag{$\star$}$$
Substituting $d=2$ gives the specific result for the question as stated. $\square$
There's almost-certainly a quicker path to the target relation. Note that
$$s(b+c+2d+p+q)=2|\triangle ABC| = r(b+c+p+q) \qquad\to\qquad \frac1s-\frac1r=\frac{d}{|\triangle ABC|}$$
So, really, "all we have to do" is show $d^2=|\triangle ABC|$. I'm not seeing a particularly good way to do that. Even so, looking at this as $2d^2=bc$ gives an easy way to construct an accurate figure in, say, GeoGebra, for further investigation.
| {
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Polynomials and Girard If the polynomial $P(x)=x^3 - 3x^2 -7x -1$ has roots $a,b,c$, find the value of $(\frac{1}{a-b} + \frac{1}{b-c} + \frac{1}{c-a})^2$.
My attempt:
I developed the expression and by Girard I was able to simplify the numerator by finding an integer. But the denominator couldn't
|
by Girard I was able to simplify the numerator by finding an integer
This leaves the denominator to calculate, which, as noted in a comment, is actually the discriminant of the cubic (for verification, its value is $1300$ per WA). The following is a shortcut to calculate the denominator without using the cubic discriminant formula (or WA).
Completing the cube, $\,P(x) = x^3 - 3x^2 \color{red}{+ 3 x - 3 x} -7x -1= (x-1)^3 - 10(x-1) - 10\,$.
Let $\,Q(t) = P(t+1) = t^3 - 10 t - 10\,$, then the roots of $\,Q(t)\,$ are $\,\alpha=a-1\,$,$\,\beta=b-1\,$,$\,\gamma=c-1\,$, so the denominator $\,(a-b)^2(b-c)^2(c-a)^2 = (\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2\,$.
Subtracting $\,\alpha^3 - 10 \alpha - 10 = 0\,$ and $\,\beta^3 - 10 \beta- 10 = 0\,$ gives:
$$
0 = \alpha^3-\beta^3 - 10(\alpha-\beta) = (\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2 -10)
$$
The first factor is non-zero since $\,Q(t)\,$ has no multiple roots, so $\,\alpha \ne \beta\,$, then:
$$
\alpha^2+\alpha\beta+\beta^2-10 = 0 \quad\iff\quad (\alpha-\beta)^2 = 10 - 3\alpha\beta = 10 - \frac{30}{\gamma}
$$
Then, using that $\,\alpha+\beta+\gamma=0\,$, $\,\alpha\beta+\alpha\gamma+\beta\gamma=-10\,$, $\,\alpha\beta\gamma = 10\,$ by Vieta's relations:
$$
\require{cancel}
\begin{align}
(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2 &= \left(10 - \frac{30}{\alpha}\right)\left(10 - \frac{30}{\beta}\right)\left(10 - \frac{30}{\gamma}\right)
\\ &= 1000 - 3000\left(\frac{1}{\alpha}+ \frac{1}{\beta} + \frac{1}{\gamma}\right) \\ &\quad\quad + \cancel{9000\left(\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}\right)} - \frac{27000}{\alpha\beta\gamma}
\\ &= 1000 - 3000 \cdot \frac{-10}{10} - \frac{27000}{10}
\\ &= 1300
\end{align}
$$
| {
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$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{6-2\epsilon } \right \rfloor +1
$$
I choose epsilon to be 0.01, and N is 1, which is incorrect.
Then I manipulated the inequality again by using the triangle inequality one more time:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right |
$$
$$
\leqslant \left | \sqrt{n^2+3n-3} \right | +n+ \frac{3}{2} =\epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{2\epsilon } \right \rfloor +1
$$
and this time, when epsilon is 0.01, N is 2624, which is correct
I would like to know why the first approach is wrong and the second one is right, Thank you.
| You cannot “choose” $\varepsilon$, forget it. You want to show that, for every $\varepsilon>0$, the inequality
$$
\Bigl|{\textstyle\sqrt{n^{2} +3n-3}} -n - \frac{3}{2}\Bigr|<\varepsilon
$$
is satisfied for all $n$ greater that some integer $N$ (depending on $\varepsilon$).
Your idea of applying the triangle inequality might be good: indeed, if you are able to see that
$$
\bigl|{\textstyle\sqrt{n^{2} +3n-3}} -n\bigr|+\frac{3}{2}<\varepsilon
$$
is satisfied for $n>N$, then also the wanted inequality would be satisfied. Unfortunately this turns out not to be a good strategy, because when $\varepsilon<3/2$, the inequality is satisfied for no value of $n$.
The required inequality is equivalent to
$$
\textstyle\bigl|\sqrt{4n^2+12n-12}-2n-3\bigr|<2\varepsilon
$$
and under the square root you can complete the square:
$$
4n^2+12n-12=4n^2+12n+9-21=(2n+3)^2-21
$$
and you can set temporarily $2n+3=m$, so you have
$$
\bigl|\textstyle\sqrt{m^2-21}-m\bigr|<2\varepsilon
$$
Since $\sqrt{m^2-21}<m$, this can be simplified to
$$
\textstyle m-\sqrt{m^2-21}<2\varepsilon
$$
hence to
$$
\textstyle m<\sqrt{m^2-21}+2\varepsilon
$$
that we can square to find
$$
\textstyle 0<-21+4\varepsilon^2+4\varepsilon\sqrt{m^2-21}
$$
so
$$
\textstyle 4\varepsilon\sqrt{m^2-21}>21-4\varepsilon^2
$$
which is certainly valid when
$$
m^2-21>\frac{(21-4\varepsilon)^2}{16\varepsilon^2}
$$
hence for
$$
m>\sqrt{21+\frac{(21-4\varepsilon)^2}{16\varepsilon^2}}
$$
Getting back to $n$, we find
$$
n>\frac{1}{2}\biggl(\sqrt{21+\frac{(21-4\varepsilon)^2}{16\varepsilon^2}}-3\biggr)
$$
and we're done.
| {
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Show the function defined on $[0,1] \times [0,1]$ via $\frac{x^2-y^2}{(x^2+y^2)^2}$ if $(x,y) \neq 0$ and $0$ otherwise is not integrable. Let $f: [0,1] \times [0,1] \rightarrow \Bbb{R}$ be defined via
$$f(x,y)=
\begin{cases}
\frac{x^2-y^2}{(x^2+y^2)^2} &(x,y) \neq (0,0) \\
(0,0) & \text{otherwise}
\end{cases}$$
So I know if I integrate with respect to $x$ I get
$$\frac{-x}{x^2+y^2}$$
and if I integrate this with respect to $y$ I get
$$-\tan^{-1}(\frac{y}{x})$$
which is bounded. How is this function non-integrable? also when integrating w.r.t. $x$ was I supposed to plug in end points $0$ and $1$ before moving onto integrating $y$?
| I claim that
$$\int_0^1 \int_0^1 f(x,y) dx dy \neq \int_0^1 \int_0^1 f(x,y) dy dx.$$
We compute the LHS:
\begin{align*}
\int_0^1 \int_0^1 f(x,y) dx dy &= \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dx dy\\
&= \int_0^1 -\frac{x}{x^2+y^2} \bigg\vert_0^1 dy\\
&=-\int_0^1 \frac{1}{1+y^2}dy\\
&=- \tan^{-1}(y) \bigg \vert_0^1\\
&= -\frac{\pi}{4}
\end{align*}
Now we compute the RHS:
\begin{align*}
\int_0^t \int_0^1 f(x,y) dydx &= \int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} dy dx\\
&=\int_0^1 \frac{y}{x^2+y^2} \bigg \vert_0^1 dx\\
&= \int_0^1 \frac{1}{x^2+1} dx\\
&= \tan^{-1}(x) \bigg \vert_0^1 \\
&= \frac{\pi}{4}
\end{align*}
thus $f(x,y)$ is not integrable.
| {
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Prove that for all natural numbers $n \geq 2$, $n^2 \geq n + 2$ Prove that for all natural numbers $n \geq 2$, $n^2 \geq n + 2$.
In solving this proof, I tried two methods, first working backward and then using the contradiction approach. i wish to solve this proof directly or using a contradiction however I am stuck on how to verify it for $n\geq 2$.
| Perhaps the quickest way:
If $n \ge 2$ then $n^2 = n \times n \ge n\times 2 = n + n \ge n + 2$.
If that was too slick:
We now that $n^2 > n$ in general. But by how much and when? $n^2 - n = n(n-1)$. As $n \ge 2$ and $n-1 \ge 1$ then $n^2 - n \ge 2\times 1$ and $n^2 \ge n + 2$.
A cute thing happens when we try a proof by contradiction.
Suppose $n^2 < n + 2$ and $n\ge 2$. If we divide both sides by $n > 0$ we get $n < 1 + \frac 2n$. But as $n \ge 2$ we have $\frac 2n \le 1$ so $n < 1 + \frac 2n \le 1 + 1 =2$ so $n < 2$ which is a contradiction.
A creative way: $n^2 = n^2$ and $n^2 - 1 = (n+1)(n-1)$. And $n-1 \ge 1$ we have $n^2 -1 = (n+1)(n-1) \ge (n+1)\times 1 = (n+1)$ so $n^2 = n + 2$.
And there's always figuring out necessary and sufficient and/or equivalent statements.
$n^2 \ge n + 2 \iff n^2 - n \ge 2 \iff n(n-1) \ge 2$ and is that true? As $n \ge 2$ and $n-1\ge 1$ it is.
Then there is always induction:
If $n = 2$ then $n^2 =4$ and $n + 2 = 4$ so $n^2 \ge n+2$.
If we assume that $n^2 \ge n+2$ then we have $(n+1)^2 = n^2 + 2n + 1$ and as $n^2 \ge n+2$ we have $(n+1)^2 = n^2 + 2n + 1 \ge (n+2) + 2n+ 1 = 3n + 3$ but is $3n +3 \ge 2(n+1) + 2$? Well, $3n + 3 = 2(n+1) + (n+1)$ and as $n \ge 2$ we have $2(n+1) +(n+1)\ge 2(n+1) + 3 > 2(n+1) + 2$.
[i.e. $(n+1)^2 = n^2 + 2n + 1 \ge (2n+2) + (2n +1)=2(n+1) + (2n+1)> 2(n+1) + 2$]
So our induction step holds.
| {
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Solving the Diophantine system $pqr=a^4$, $p+q+r=b^4$ I am trying to find solutions of the following system of diophantine equations:
$$\left\{\begin{array}{rcl}pqr&=&a^4\\p+q+r&=&b^4\end{array}\right.$$
where $a$, $b$, $p$, $q$ ans $r$ are positive integers such that $\gcd(p,q,r)$ is not divisible by $\theta^4$, $\theta>1$.
I found the following solutions $(p,q,r)$ with a computer program :
$(3\,;6\,;72)$ , $(25\,;60\,;540)$ , $(72\,;576\,;648)$ and $(162\,;448\,;686)$.
The system has infinitely many solutions : take $(p\;q\,;r)=\left(A^4\,;B^4\,;C^4\right)$, where $A^4+B^4+C^4=D^4$ and $A$, $B$, $C$ are coprime (see this article).
But can we prove that there are infinitely many solutions using more elementary ways ?
Thank you for your help !
| We have the below Identity:
$(b^2+c^2-a^2)^2+(2ab)^2+(2ac)^2=(b^2+c^2+a^2)^2$
Let:
$p=(b^2+c^2-a^2)^2$
$q=(2ab)^2$
$r=(2ac)^2$
$p+q+r=(a^2+b^2+c^2)^2$
To make (RHS) a fourth power we take:
$b^2+c^2=8a^2$ ---$(1)$
Hence,
$p+q+r=(9a^2)^2=(3a)^4$
$(pqr)=((b^2+c^2-a^2)(2ab)(2ac))^2$
=$(2a)^4(bc(b^2+c^2-a^2))^2$
Since, $(b^2+c^2=8a^2)$ we have $(b^2+c^2-a^2)=(7a^2)$
and to make, $(bc(b^2+c^2-a^2))$, a square we take,
$b=7c$ ---$(2)$
Hence,
$pqr=(2a)^4(7ac)^4=(14a^2c)^4$
Now eqn (1) & (2) above is satisfied by $(a,b,c)=(5,14,2)$
Hence:
$(p+q+r)=(15)^4$
$(p,q,r)=(700)^4$
And,
$(p,q,r)=(30625,19600,400)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Why can't I take the 3rd root of both sides to simplify the implicit differentiation of $\tan^3\left(xy^2+y^{\:}\right)=x$? The task is to find the implicit derivative of $$\tan^3\left(xy^2+y^{\:}\right)=x$$
I can calculate
$$ \frac{dy}{dx}\mathrm{\:of\:}\tan ^3\left(xy^2+y\right)=x $$
as
$$\frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
which agrees with the textbook, Calculus: Early Transcendentals (2016) by Anton, Bivens and Davis.
However, my initial idea was to take each side to the power of $\frac{1}{3}$. Yet the resulting calculation gives a different result:
$$
\frac{dy}{dx}\mathrm{\:of\:}\tan ^{\frac{3}{3}}\left(xy^2+y\right)=x^{\frac{1}{3}}
$$
now gives
$$\frac{1-3y^2x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)}{3x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
I was surprised by this result. I have the naive idea that, so long as I perform an operation on both sides of an equation, any manipulation is fair game.
I can see why I have this naive belief. After all, adding $2$ to both sides does not change the result:
$$\frac{dy}{dx}\mathrm{\:of\:}\tan ^3\left(xy^2+y\right)+2=x+2$$
is
$$ \frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}$$
I can even continue with my manipulation by multiplying both sides by 2, still giving the same result:
$$\frac{dy}{dx}\mathrm{\:of\:}2\left(\tan ^3\left(xy^2+y\right)+2\right)=2\left(x+2\right)$$
is
$$\frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}$$
But it seems my naive view of things has to change.
Some additional thoughts, although not very sophisticated, as I try to challenge my naive viewpoint, is that taking both sides to an even power or the reciprocal of an even power could be an issue.
I'm thinking that, after taking both sides to an even power, if I punted that equation on to someone else and told them only what I did and not what I started from, they wouldn't know whether the roots were originally positive or negative and could only guess at how to go back to the original equation.
And if I take both sides to the reciprocal of an even power, I could have negative values inside my radicals for some values of $x,y$, which is problematic for me (I only know of but don't use imaginary numbers).
Is that reasoning correct? Even if it is, I don't think I can apply the same reasoning to taking both sides to the reciprocal of an odd power. So I'm still mathematically naive about what is going on here.
| It does yield the same result
$$
\frac{1-3y^2x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)}{3x^{\frac{2}{3}}\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
we also have
$$
x^{1/3} = \tan\left(xy^2+y^{\:}\right)
$$
or
$$
x^{2/3} = \tan^2\left(xy^2+y^{\:}\right)
$$
insert into the above
$$
\frac{1-3y^2\tan^2\left(xy^2+y^{\:}\right)\sec ^2\left(xy^2+y\right)}{3\tan^2\left(xy^2+y^{\:}\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
which compared with your initial calculation
$$
\frac{1-3y^2\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)}{3\tan ^2\left(xy^2+y\right)\sec ^2\left(xy^2+y\right)\left(2xy+1\right)}
$$
is consistent
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Expected value in coin flipping process You flip a coin, and if the result is tails, you lose. If the result is heads, you get to play again. What is the expected value of throws before you lose?
My Approach
The expected value is the sum of all the outcomes multiplied by their respective probabilities:$$\sum_{i=1}^{n}V_iP_i$$So for this problem:$$\sum_{i=1}^{\infty}i(\frac{1}{2^i})=0.5+0.5+0.375+…$$I can’t figure out how to find the sum, even though I know it converges.
| This is an Arithmetic-Geometric Series. Let $S=\displaystyle \sum_{i=1}^{N}\frac{i}{2^i}$. Then,
$$S=\displaystyle\frac {1}{2^1}+\frac {2}{2^2}+\frac{3}{2^3}+…+\frac{N}{2^N}.\tag{1}$$
Now, divide the whole equation by $2$ and arrange it by shifting the terms by a step: $$\frac{S}{2}=\frac {1}{2^2}+\frac {2}{2^3}+\frac{3}{2^4}+…+\frac{N-1}{2^{N}}+\frac{N}{2^{N+1}}.\tag{2}$$ $(1)-(2)$:$$\frac{S}{2}= \frac {1}{2^1}+\frac {1}{2^2}+\frac{1}{2^3}+…+\frac{1}{2^N}+\frac{1}{2^{N+1}}.$$This is a G.P., so sum is $$\frac{S}{2}=\frac 12\left(\frac{1-\left(\frac12\right)^{N+1}}{1-\frac12}\right)$$$$\implies S =\left(\frac{1-\left(\frac12\right)^{N+1}}{1-\frac12}\right)$$ Taking an infinite sum gives us $S_{\infty}=2$, which is the required answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4497470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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} |
For an arbitrary $n$, evaluate $\left(z^n+\frac{1}{z^{n}}\right)\left(z^n +\frac{1}{z^n}+1\right)$ Let me put you in context, I have the following problem:
Let $z$ be a complex number such that $$\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=1.$$
For an arbitrary $n$, evaluate
$$\left(z^n +\frac{1}{z^n}\right)\left(z^n+\frac{1}{z^n}+1\right)$$
I've been trying to get the answer, however I am not sure if my proof is completely rigorous. My solution:
Given the hypothesis, we have:
\begin{align*}
\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right) & = 1\\
z^2+1+z+1+\frac{1}{z^2} + \frac{1}{z} &= 1 \\
z^2+1+z+\frac{1}{z^2} + \frac{1}{z} &= 0 \\
\frac{z^4+z^2+z^3+1+z}{z^2}&= 0\\
\frac{(z^4+z^2+z^3+1+z)(z-1)}{z^2(z-1)}&= 0\\
\frac{(z^5-1)}{z^2(z-1)}&= 0,z\neq 1
\end{align*}
Also, we have the expression to evaluate:
$$\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right)$$
If $n=0$:
$$\left(z^{0}+\frac{1}{z^{0}}\right)\left(z^{0}+\frac{1}{z^{0}}+1\right)= (1+1)(1+1+1) =6$$
If $n\neq 0$:
\begin{align*}
\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right) & = 1\\
z^{2n}+1+z^n+1+\frac{1}{z^{2n}} + \frac{1}{z^n} &= 1 \\
z^{2n}+1+z^n+\frac{1}{z^{2n}} + \frac{1}{z^n} &= 0 \\
\frac{z^{4n}+z^{2n}+z^{3n}+1+z^n}{z^{2n}}&= 0\\
\frac{(z^{4n}+z^{2n}+z^{3n}+1+z^n)(z^{n}-1)}{z^{2n}(z^{n}-1)}&= 0\\
\frac{(z^{5n}-1)}{z^{2n}(z^{n}-1)}&= 0, z^n\neq 0
\end{align*}
Therefore,
$$\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right)=\begin{cases}
6, & n=0\\
1, & n\neq 0
\end{cases}$$
What do you think?
| Your answer has something wrong, especially the case where $n$ is divided by $5$. Let me explain why.
From $\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=1$ we have $z^5=1$ and $z\neq 1$. So $z$ is a root of unity $5$, so $z=\zeta^m$, where $\zeta=e^{\frac{2\pi i}5}$ and $m\in\{1,2,3,4\}$. Now it is easy to check that $z^n=1$ if and only if $n$ is divided by $5$, or equivalently, $n=5k$ for some $k\in \mathbb Z$ (the essence is that $5$ is a prime number.)
Now,
$$\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right)=\frac{z^{4n}+z^{3n}+z^{2n}+z^n+1}{z^{2n}}+1.$$
If $z^n\neq 1$, i.e. $5\not\mid n$, then we can apply the summation formula for geometric progression:
$$\frac{z^{4n}+z^{3n}+z^{2n}+z^n+1}{z^{2n}}+1=\frac{(z^{5n}-1)}{z^{2n}(z^{n}-1)}+1=0+1=1,$$
where we use $z^5=1\Longrightarrow z^{5n}=1$. If $z^n=1$, i.e., $5\mid n$, then
$$\frac{z^{4n}+z^{3n}+z^{2n}+z^n+1}{z^{2n}}+1=\frac{1+1+1+1+1}1+1=6.$$
Therefore, the answer is
$$\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right)=\begin{cases}
6, & 5\mid n,\\
1, & \text{otherwise}.
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How can we prove $\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}} = 2\cos\left(\frac {\pi }{2^{n+1}}\right)$ without induction I wanted to know the proof without induction / substitution method for the equation
$$\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{n-times}}= 2\cos\left(\frac {\pi }{2^{n+1}}\right)$$
Proof with induction:
\begin{align}
& n=1: \\
& \sqrt{2}=2\cos\left( \dfrac{\pi}{4} \right). \\
\ \\
& \text{Assume that the equation is valid when }n=k. \\
\ \\
&n=k+1; \\
& \underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{k+1-times}}=\sqrt{2+\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{k-times}}} \\
& = \sqrt{2+2\cos\left( \dfrac{\pi}{2^{k+1}} \right)} = \sqrt{2}\cdot\sqrt{2\cos^2\left( \dfrac{\pi}{2^{k+2}} \right)} = 2\cos\left(\dfrac{\pi}{2^{k+2}}\right). \blacksquare
\end{align}
| Let's assume the problem. Then, let $x_n=2\cos \left( \dfrac {\pi}{2^{n+1}} \right).$ We can easily get $x_{n+1}=\sqrt{2+x_n}, x_1=\sqrt{2}$. Then, we can change the problem:
Prove that if $x_{n+1}=\sqrt{2+x_n}, x_1=\sqrt{2}$, then $x_n=2\cos \left( \dfrac {\pi} {2^{n+1}} \right)$.
We also know that $\displaystyle \lim_{n \to \infty}x_n=2$, because $x_n=\sqrt{2+x_n}.$ Also, $x_n$ is an increasing sequence. So, we have that $0<x_i\leq2$ for all $i$.
So, let $x_i=2\cos(a_i).(0<a_i<\frac{\pi}{2}.)$ Then, we have to prove that $a_n=\dfrac{\pi}{2^{n+1}}$.
$2\cos(a_{n+1})=x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(a_n)}=\sqrt{2(1+\cos(a_n)}=\sqrt{2 \cdot 2\cos^2\left(\frac{a_n}{2}\right)}=2\cos\left( \frac{a_n}{2} \right).$
Since $0<a_i<\frac {\pi}{2}$, $\cos$ is injective. Therefore, $a_{n+1}=\frac{a_n}{2}$.
Then, we get:
$$a_{n+1}=\dfrac{a_1}{2^n}.$$
$x_1=\sqrt{2}=2\cos\left( \frac {\pi}{4} \right), a_1=\frac{\pi}{4}.$
Finally, we get:
$$a_n=\dfrac{\pi}{2^{n+1}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as well but not the generalized form. Here, I conjecture a generalization -
$$ I = (-1)^{n+1} \binom{-1/2}{n-1} \cdot \frac{\pi}{2 \, a^{2n-1}} $$
This can be further simplified by
$$ \binom{-1/2}{n} = \left(\frac{-1}{4}\right)^n \binom{2n}{n} $$
Is this conjecture true?
| $$J(b)=\int_0^\infty \frac{1}{x^2+b}dx=\frac{\pi}{2\sqrt{b}}$$
$$I=\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{db^{n-1}}J(b)|_{b=a^2}$$
$$\frac{d^{n-1}}{db^{n-1}}J(b)=\frac{\pi}{2}(-1)^{n-1}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{b^{\frac{2n-1}{2}}}$$
$$I=\frac{\pi}{2}\frac{1}{(n-1)!}\frac{(2n-3)!!}{2^{n-1}}\frac{1}{a^{2n-1}}$$
where
$$(2n-3)!!=\frac{(2n-2)!}{2^{n-1}(n-1)!}$$
$$I=\frac{\pi}{2}\frac{1}{(n-1)!}\frac{(2n-2)!}{(n-1)!~4^{n-1}}\frac{1}{a^{2n-1}}$$
Final result:
$$I=\frac{\pi}{2}\binom{2n-2}{n-1}\frac{1}{4^{n-1}}\frac{1}{a^{2n-1}},~~~n=1,2,3...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Show that $a^3+5a$ is an integer I've been given the following task. Let
$$a = \sqrt[3]{1+\sqrt{\frac{152}{27}}}-\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$$
Show that $a^3+5a$ is an integer.
I tried calculating it by hand but the small page of my copybook is not large enough for the very long calculations. Is there a trick I could use here instead of calculating it by hand?
I tried factoring $a^3+5a$ into $a(a^2+5)$ to make it more simple to calculate but it still gets a tad bit complicated when multiplying with $a$ again.
Thank you in advance.
| Let $x=1+\sqrt{\frac{152}{27}},~~y=-1+\sqrt{\frac{152}{27}}$
$$x^{\frac{1}3}y^{\frac{1}3}=\frac{5}{3}$$
$$a=x^{\frac{1}3}-y^{\frac{1}3}$$
$$\begin{align}
a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=\left(x^{\frac{1}3}-y^{\frac{1}3}\right)\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)\\
\\
a\left(x^{\frac{2}3}+x^{\frac{1}3}y^{\frac{1}3}+y^{\frac{2}3}\right)&=x-y=2\\
\\
a\left(\left(x^{\frac{1}3}-y^{\frac{1}3}\right)^2+3x^{\frac{1}3}y^{\frac{1}3}\right)&=2\\
\\
a\left(a^2+5\right)&=2\\
\\
a^3+5a&=2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does $a^y = x+\sqrt{x^2+1}$ imply $a^{-y} = -x+ \sqrt{x^2+1}$ and then $x=\sinh(y\ln a)$? I want to understand the steps in an argument given for the following question.
Find the inverse of the following function:
$$f(x) = \log_a(x+\sqrt{x^2+1})$$
We find:
$$\begin{align}
a^y &= \phantom{-}x+\sqrt{x^2+1} \tag1 \\[4pt]
a^{-y} &= -x+ \sqrt{x^2+1} \tag2
\end{align}$$
whence:
$$x = \frac{1}{2} (a^y-a^{-y}) = \sinh(y\ln a) \tag3$$
I understand how he converts it to an exponential $(1)$, but then $(2)$ doesn’t make sense to me, as well as part $(3)$.
| Right, so what you should do is to write these things out explicitly. So:
$$a^y = \sqrt{x^2+1}+x$$
$$a^{-y} = \frac{1}{a^y} = \frac{1}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}+x} \cdot \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x} = \frac{\sqrt{x^2+1}-x}{(x^2+1)-x^2} = \sqrt{x^2+1}-x$$
Then, notice that:
$$a^y-a^{-y} = (\sqrt{x^2+1}+x)-(\sqrt{x^2+1}-x) = 2x$$
It follows that $x = \frac{1}{2}(a^y-a^{-y})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\int_{0}^{\pi} \frac{d x}{a-\cos x} \right)\right|_{a=3} \\
&=\left.\frac{2(-1)^{n} \pi}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\frac{1}{\sqrt{a^{2}-1}}\right)\right|_{a=3}
\end{aligned}
$$
Multiplying both the numerator and denominator $\sec^4x$ yields
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\left(\sec ^{2} x+\tan ^{2} x\right)^{2}} d x \tag*{} $
Letting $ t=\tan x$ gives
$\displaystyle \begin{aligned}I&= \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+2 t^{2}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{1+\frac{1}{t^{2}}}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{\frac{3}{4}\left(2+\frac{1}{t^{2}}\right)-\frac{1}{4}\left(2-\frac{1}{t^{2}}\right)}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\frac{3}{4} \int_{0}^{\infty} \frac{d\left(2 t-\frac{1}{t}\right)}{\left(2 t-\frac{1}{t}\right)^{2}+8}-\frac{1}{4} \int_{0}^{\infty} \frac{d\left(2 t+\frac{1}{t}\right)}{\left(2 t+\frac{1}{t}\right)^{2}}\\&=\left[\frac{3}{8 \sqrt{3}} \tan ^{-1}\left(\frac{2 t-\frac{1}{t}}{2 \sqrt{2}}\right)+\frac{1}{4\left(2 t+\frac{1}{t}\right)}\right]_{0}^{\infty}\\&=\frac{3 \pi}{8 \sqrt{2}}\end{aligned}\tag*{} $
Is there any method other than tangent half-angle substitution?
| Start as J.G., then using differentiation yields
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\frac{1-\cos 2 x}{2}\right)^{2}} &=4 \int_{0}^{\frac{\pi}{2}} \frac{d x}{(3-\cos 2 x)^{2}} \\
&=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{2}} \\
&=-\left.2 \frac{\partial}{\partial a} \int_{0}^{\pi} \frac{d x}{a-\cos x}\right|_{a=3} \\
&=-\left.2 \frac{\partial}{\partial a}\left(\frac{\pi}{\sqrt{a^{2}-1}}\right)\right|_{a=3} \\
&=\frac{3 \pi}{8 \sqrt{2}}
\end{aligned}
$$
where the last second line comes from my post$\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta} =\frac{\pi}{\sqrt{a^{2}-b^{2}}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Do you get another method for $\int \frac{\sin x}{\sin 5 x}dx$? First of all, we use De Moivres’ Theorem to express $\sin 5x $ in terms of $\cos x$ and $\sin x$.
$$
\begin{aligned}
&\cos 5 x+i \sin 5 x\\=&(\cos x+i \sin x)^{5} \\
=& \cos^{5} x+5 i \cos ^{4} x \sin x-10 \cos^{3} x \sin ^{2} x-10 i \cos ^{2} x \sin ^{3} x+5 \cos x \sin ^{4} x+i \sin ^{5} x
\end{aligned}
$$
Comparing the imaginary parts on both sides yields
$$
\sin 5 x=5 \cos 4 \sin x-10 \cos ^{2} x \sin ^{3} x+\sin ^{5} x
$$
$$
\begin{aligned}
\because \int\frac{\sin x}{\sin 5 x} d x &=\int \frac{1}{5 \cos ^{4} x-10 \cos ^{2} x \sin ^{2} x+\sin ^{4} x} d x \\
&=\int \frac{\sec ^{4} x d x}{5-10 \tan ^{2} x+\tan ^{4} x} \\
&=\int \frac{1+t^{2}}{t^{4}-10 t^{2}+5}dt, \quad \textrm{ where }t=\tan x.
\end{aligned}
$$
Playing a small trick on the integrand yields
$$
\begin{aligned}
\int \frac{\sin x}{\sin 5 x}dx&=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{5}{t^{2}}-10} d t\\
&= \int \frac{\frac{\sqrt{5}+1}{2}\left(1+\frac{\sqrt{5}}{t^{2}}\right)+\frac{\sqrt{5}-1}{2}\left(1-\frac{\sqrt{5}}{t^{2}}\right)}{t^{2}+\frac{5}{t}-10} d t\\
&=\frac{\sqrt{5}+1}{2} \int \frac{d\left(t-\frac{\sqrt{5}}{t}\right)}{\left(t-\frac{\sqrt{5}}{t}\right)^{2}-(10-2 \sqrt{5})}+\frac{\sqrt{5}-1}{2} \int \frac{d\left(t+\frac{\sqrt{5}}{t}\right)}{\left(t+\frac{\sqrt{5}}{t}\right)^{2}-(10+2 \sqrt{5})}\\&=\frac{\sqrt{5}+1}{2 \sqrt{10-2 \sqrt{5}}} \tan ^{-1}\left(\frac{t-\frac{\sqrt{5}}{t}}{\sqrt{10-2 \sqrt{5}}}\right)+\frac{\sqrt{5}-1}{4 \sqrt{10+2 \sqrt{5}}} \ln \left|\frac{t+\frac{\sqrt{5}}{t}-\sqrt{10+2 \sqrt{5}}}{t+\frac{\sqrt{5}}{t}+\sqrt{10+2 \sqrt{5}}}\right|+C\\&=\frac{\sqrt{5}+1}{2 \sqrt{10-2 \sqrt{5}}} \tan ^{-1}\left(\frac{\tan^2 {x}-\sqrt{5} }{\tan {x}\sqrt{10-2 \sqrt{5}}}\right)\\&+\frac{\sqrt{5}-1}{4 \sqrt{10+2 \sqrt{5}}} \ln \left| \frac{\tan ^{2} x-\sqrt{10+2 \sqrt{5}} \tan x+ \sqrt{5}}{\tan ^{2} x+\sqrt{10+2 \sqrt{5}} \tan x+\sqrt{5}}\right|+C
\end{aligned}
$$
Your comments and alternative solutions are highly appreciated.
| In terms of the Gaussian hypergeometric function : if
$$I_n=\int \frac{\sin(x)}{\sin(nx)}\,dx$$then
$$I_n=\frac{i e^{i (n+1) x} \, _2F_1\left(1,\frac{n+1}{2 n};\frac{1}{2}
\left(3+\frac{1}{n}\right);e^{2 i n x}\right)}{n+1}-\frac{i e^{i (n-1) x} \, _2F_1\left(1,\frac{n-1}{2 n};\frac{1}{2}
\left(3-\frac{1}{n}\right);e^{2 i n x}\right)}{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$a,b,c \in \mathbb{R}^+, \dfrac{a}{a+7b}+\dfrac{b}{b+7c}+\dfrac{c}{c+7a} \geq \dfrac{3}{8}$ How to prove
$a,b,c \in \mathbb{R}^+$
$$\dfrac{a}{a+7b}+\dfrac{b}{b+7c}+\dfrac{c}{c+7a} \geq \dfrac{3}{8}$$
my solution
this equivalent to
$$\dfrac{7(13a^2b+13b^2c+13c^2a+35a^2b^2+35b^2c^2+35b^2c^2-144abc)}{8(a+7b)(b+7c)(c+7a)} \geq 0$$
from AM-GM, we can get
$$13a^2b+13b^2c+13c^2a \geq 39abc$$
$$35b^2c^2+35b^2c^2+35b^2c^2 \geq 105abc$$
done
I want to know better way to prove this
| Your solution is correct.
Let $A$ be your expression and $B=a(a+7b) + b(b+7c) +c(c+7a)$. Then it is enought, by Cauchy Shwarz inequality, to prove $$8(a+b+c)^2\geq 3(a^2+b^2+c^2+7ab+7bc+7ca)$$
i.e.
$$5(a^2+b^2+c^2)\geq 5(ab+bc+ca)$$ which is true by again CS or AG.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find postive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square
Find positive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square.
My try: Let $\dfrac{x^6-1}{x-1}=y^2$, so $(x^4+x^2+1)(x+1)=y^2$.
Let $d= \gcd(x+1,x^4+x^2+1)$.
And
$d \vert x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, because $d\vert x+1$ so $d\vert 2$.
Case 1: $d=1$. If $d=1$ so $(x^4+x^2+1)$,$(x+1)$ is perfect square and $(x^2+\dfrac{1}{2})^2\le x^4+x^2+1\le(x^2+1)^2$ so no satisfactory solution.
Case 2: $d=2$. If $d=2$ so $\dfrac{x^4+x^2+1}{2}$,$\dfrac{x+1}{2}$ is perfect square.
But now I stuck. Please give me a hint. Thank you.
| There's a simpler solution if you consider $\frac{x^6 - 1}{x - 1} = (x^2 + x + 1)(x^3 + 1)$. Let $d = \gcd(x^2 + x + 1, x^3 + 1)$, then $d \mid x^2 + x + 1 \mid x^3 - 1$ and $d \mid x^3 + 1$. So $d \mid 2$ but the number $x^2 + x + 1 = x(x + 1) + 1$ is odd. Therefore $d = 1$ and the numbers $x^2 + x + 1$ and $x^3 + 1$ are perfect squares. But for $x > 1$ we have $x^2 < x^2 + x + 1 < (x + 1)^2$. So there's no such $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Integration by substitution and by partial fractions lead to different results When integrating $\int \frac{3x-2}{x+1}dx$ we can take two paths.
$\alpha)$ Let $u=x+1$, so
$$\int \frac{3x-2}{x+1}dx = \int \frac{3u-5}{u}du = 3u- 5\ln|u|=3(x+1)-5\ln|x+1|$$
$\beta)$ See that $3x-2=3(x+1)-5 \implies \frac{3x-2}{x+1}=3-\frac{5}{x+1}$. Then $$\int \frac{3x-2}{x+1}dx=\int (3-\frac{5}{x+1})dx = 3x-5\ln|x+1|$$
Clearly, the two results are very similar. They only differ in that $x \neq x+1$. Of course I must be making a mistake in some of the approaches (or both), but I can not seem to find where the error is. Anybody care to point it out for me? Thanks in advance.
| Both of them are correct, because you forget the $+C$ part.
$$\begin{align}\int \frac{3x-2}{x+1}dx&=3(x+1)-5\ln|x+1|+C_1\\
\\
\int \frac{3x-2}{x+1}dx&=3x-5\ln|x+1|+C_2\end{align}$$
Since $C_1$ and $C_2$ are arbitrary constants, if you define $C_2=3+C_1$, they are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
find $\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1} dx$
Find (with proof) $\displaystyle\int_0^\infty \frac{|\cos (\pi x)|}{4x^2 - 1}dx$
It's actually not even clear that the integral converges. If there were only sines/cosines in the integral, a standard technique would be to use the trigonometric identity $\sin \theta = \frac{2t}{1+t^2}$, $\cos \theta = \frac{1-t^2}{1+t^2}$, $t = \tan(\theta/2)$. I know that $\frac{2}{4x^2-1} = \frac{1}{2x-1} -\frac{1}{2x+1}$, so maybe one could plug this into the given integral to obtain an integral that's easier to evaluate?
Edit: from a comment, I think it would be useful to note that $\cos (\pi x) \leq 0$ iff $(2k+1/2)\pi \leq \pi x \leq (2k + 3/2)\pi\iff(2k + 1/2)\leq x\leq 2k+3/2$ (for some integer $k$) and $\cos(\pi x) \ge 0$ iff $(2k - 1/2) \leq x \leq (2k+1/2)$ (for some integer $k$). So we can split the integral according to these ranges. Then it might be useful to apply integration by parts. Let $a\in \mathbb{R}$. Then
$$\begin{align}\int \frac{\cos(ax)}{4x^2-1}dx &= \frac{1}2\left(\int \frac{\cos(a x)}{2x-1}dx - \int \frac{\cos(a x)}{2x+1}dx\right)\\
\\
&= \frac{1}2\left(\frac{1}2 \ln(2x-1)\cos (ax) +\frac{1}2a \int \sin(ax)\ln(2x-1)dx\right) \\
\\&-\frac{1}2\left(\frac{1}2 \ln(2x+1)\cos (ax) +\frac{1}2a \int \sin(ax)\ln(2x+1)dx\right)\end{align}$$
but I'm not sure how to simplify the result.
For the bounty, I'm looking for formal proofs. In particular, I'd like to see justifications for interchanging an integral and a sum. One can freely interchange an integral and a sum if the terms are nonnegative (as $\sum \int f_n = \int \sum f_n$ for nonnegative Lebesgue measurable functions $f_n$ and the Lebesgue integral equals the Riemann integral for Riemann integrable functions).
Also I'd like to see justifications for why $$\int_0^\infty \frac{\cos^2(\pi (2m+1) x)}{4x^2 - 1}dx = 0 = \int_0^\infty \frac{\sin^2(\pi (2m+1)x)}{4x^2-1} dx,$$ where $m$ is any nonnegative integer.
| Since others addressed the first part of the question, I will look at the second part
$$\int_0^\infty \dfrac{\cos^2(\pi (2m+1) x)}{4x^2 - 1}dx = 0 = \int_0^\infty \dfrac{\sin^2(\pi (2m+1)x)}{4x^2-1} dx$$
To confirm or deny this statement let is consider the following integrals
$$\int_0^\infty \dfrac{\cos^2(b x)}{1+a^2x^2}dx=\frac{\pi}{4a}\left(1+e^{-\frac{2b}{a}}\right)$$
$$\int_0^\infty \dfrac{\sin^2(b x)}{1+a^2x^2}dx=\frac{\pi}{4a}\left(1-e^{-\frac{2b}{a}}\right)$$
$a$ and $b$ are parameters
Now, replace $a$ with $ia$ where $i$ is the imaginary unit.
After separating real and imaginary parts we get formal equalities
$$\int_0^\infty \dfrac{\cos^2(b x)}{1-a^2x^2}dx=\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)-i\frac{\pi}{4a}\left[\cos \left( \frac{2b}{a}\right)+1\right]$$
$$\int_0^\infty \dfrac{\sin^2(b x)}{1-a^2x^2}dx=-\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)+i\frac{\pi}{4a}\left[\cos \left( \frac{2b}{a}\right)-1\right]$$
Obviously, if the imaginary part is different from zero then the integrals diverge or are undefined.
But they have so called the Cauchy Principal Value wich is given by the real part
$$PV\int_0^\infty \dfrac{\cos^2(b x)}{1-a^2x^2}dx=\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)$$
$$PV\int_0^\infty \dfrac{\sin^2(b x)}{1-a^2x^2}dx=-\frac{\pi}{4a}\sin \left( \frac{2b}{a}\right)$$
where $PV$ indicates the Cauchy Principal Value.
If we take $a=2$ then in order for these expressions to be zero $b$ must take values $$b=\pi m$$
where $m$ is an integer, positive or negative.
So the equality on the top of this post is correct. These integrals converge to zero.
If the imaginary part in an expression above is zero then we have a convergent integral. For example if we take $\frac{2b}{a}=\pi$ then
$$\int_0^\infty \dfrac{\cos^2(\frac{a \pi}{2}x)}{1-a^2x^2}dx=0$$
This is really convergent integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by a plus sign, and it yields $\pi^2$.
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx \text{
— (I)} \\
\implies \mathcal{I} = \int_0^\pi (\pi - x) \cdot \frac{\cos{\frac{x}{2} + \sin{\frac{x}{2}}}}{\sqrt{\sin{x}}} dx \text{
— (II)}
$$
On (I) + (II), we have,
$$
\mathcal{I} = \frac{\pi}{2}\int_0^\pi \frac{\sin{\frac{x}{2}} + \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx = \frac{\pi}{2} \int_0^\pi \frac{\sin{\frac{x}{2}}+\cos{\frac{x}{2}}}{\sqrt{1 - (\sin{\frac{x}{2}-\cos{\frac{x}{2}}})^2}} dx
$$
On substitution,
$$
\sin{\frac{x}{2}} - \cos{\frac{x}{2}} = u \implies \left(\sin{\frac{x}{2}} + \cos{\frac{x} {2}}\right) dx = 2 \cdot du
$$
The upper and lower limits changes to 1 and -1. Now, we have
$$
\mathcal{I} = \frac{\pi}{2} \int_{-1}^1 \frac{2 \cdot du}{\sqrt{1 - u^2}} du = \pi \cdot \left[\arcsin{u}\right]_{-1}^1 = \pi^2
$$
But...
*
*The sign was not supposed to be changed. We get $(2x - \pi)$ instead of $\pi$ in the nominator when adding both integrals. It complicates the problem.
*Using integral-calculator.com or a scientific calculator is helpless.
*The answer to the original problem should be $2\pi \cdot \ln{2}$ (approx 4.35.)
| Evaluate \begin{align}
I=& \int_0^{\pi/2} \frac{x(\sin x- \cos x)}{\sqrt{\sin{2x}}} dx
=\int_0^{\pi/2} x\ d\bigg(- \tanh^{-1}\frac{\sqrt{2\tan x}}{1+\tan x}\bigg)\\
\overset{ibp}=&\int_0^{\pi/2} \tanh^{-1}\frac{\sqrt{2\tan x}}{1+\tan x}
\overset{t= \tan x}{dx}
=\int_0^\infty \frac1{1+t^2}\tanh^{-1}\frac{\sqrt{2t}}{1+t^2}dt=J(1)
\end{align}
where $J(a)= \int_0^\infty \frac1{1+t^2}\tanh^{-1}\frac{a\sqrt{2t}}{1+t}dt$
\begin{align}
J’(a)=&\int_0^\infty\frac{\sqrt{2t}(t+1)}{(1+t^2)(t^2+2(1-a^2)t+1)}\overset{t\to t^2}{dt}\\
=& \ \frac{\sqrt2}{1-a^2}
\int_0^\infty \frac{t^2+1}{t^4+1}- \frac{t^2+1}{t^4+2(1-a^2)t^2+1}\ dt\\
= &\ \frac\pi{1-a^2}\bigg(1-\frac1{\sqrt{2-a^2}}\bigg)
\end{align}
Then
\begin{align}
\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx=&\ 4I=4J(1)=4\int_0^1 J’(a)da\\
=&\int_0^1 \frac{4\pi}{1-a^2}\bigg(1-\frac1{\sqrt{2-a^2}}\bigg)da=2\pi\ln 2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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How to compute $\lim\limits_{x\to 0} (-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$? Consider the function
$$f(x)=(-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
How do we compute $\lim\limits_{x\to 0} f(x)$?
Here is my attempt
If $t\geq 0$ then $\frac{t^{2n+2}}{1+t^2} \leq t^{2n+2}$ and if $t\leq 0$ then $t^{2n+2}\leq\frac{t^{2n+2}}{1+t^2}$. That is, for any $t$
$$-|t|^{2n+2}\leq\frac{t^{2n+2}}{1+t^2}\leq |t|^{2n+2}$$
Thus
$$-\int_0^x|t|^{2n+2}\leq\int_0^x\frac{t^{2n+2}}{1+t^2}\leq \int_0^x|t|^{2n+2}$$
$$-\frac{|x|^{2n+3}}{2n+3}\leq\int_0^x\frac{t^{2n+2}}{1+t^2}\leq \frac{|x|^{2n+3}}{2n+3}$$
$$-\lim\limits_{x\to 0}\frac{|x|^{2n+3}}{2n+3}\leq\lim\limits_{x\to 0}\int_0^x\frac{t^{2n+2}}{1+t^2}\leq \lim\limits_{x\to 0}\frac{|x|^{2n+3}}{2n+3}$$
$$0\leq\lim\limits_{x\to 0}\int_0^x\frac{t^{2n+2}}{1+t^2}\leq 0$$
$$\lim\limits_{x\to 0}\int_0^x\frac{t^{2n+2}}{1+t^2}=0$$
With this result in hand, we go back to the original limit.
$$\lim\limits_{x\to 0}(-1)^{n+1} \frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
$$=(-1)^{n+1}\lim\limits_{x\to 0}\frac{\int_0^x \frac{t^{2n+2}}{1+t^2}dt}{x^{2n+1}}$$
$$=\frac{0}{0}$$
By L'Hopital's Rule
$$=(-1)^{n+1}\lim\limits_{x\to 0}\frac{\frac{x^{2n+2}}{1+x^2}}{(2n+1)x^{2n}}$$
$$=(-1)^{n+1}\lim\limits_{x\to 0} \frac{x^2}{(2n+1)(1+x^2)}$$
$$=0$$
is this calculation correct? I know the result is correct, but I am interested in the steps.
For context, $f(x)$ is the remainder in a Taylor Polynomial of the $\arctan$ function. The book I am reading (Spivak's Calculus) skipped over the details of this calculation.
| Too much advanced but this is just for your curiosity.
Sonner or later, you will learns that
$$\int \frac{t^{2n+2}}{1+t^2}\,dt=\frac{t^{2 n+1} }{2 n+1}\left(1-\,
_2F_1\left(1,n+\frac{1}{2};n+\frac{3}{2};-t^2\right)\right)$$ where appears the Gaussian hypergeometric function.
$$\int_0^x \frac{t^{2n+2}}{1+t^2}\,dt=\frac{x^{2 n+1} }{2 n+1}\left(\,
_2F_1\left(1,n+\frac{1}{2};n+\frac{3}{2};-x^2\right)-1\right)$$
$$R=\frac{\int_0^x \frac{t^{2n+2}}{1+t^2}\,dt }{x^{2n+1}}=\frac{1 }{2 n+1}\left(\,
_2F_1\left(1,n+\frac{1}{2};n+\frac{3}{2};-x^2\right)-1\right)$$ Developed as a series around $x=0$ gives
$$R=\sum_{p=1}^\infty (-1)^p\frac{ x^{2 p}}{2 n+2 p+1}=-\frac {x^2}{2n+3}+O(x^4)$$
Not truncated, $R$ is another special function.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equation as
$$\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(\frac{\theta}{2}\right).$$
Equating the argument of the $\sin$ functions, we see that
\begin{align}
\frac{\pi}{2}-2\theta&=\frac{\theta}{2} \\
\frac{5}{2}\theta&=\frac{\pi}{2} \\
\theta&=\frac{\pi}{5}.
\end{align}
While this yields the correct answer, I do not know how we can justify equating the arguments of the $\sin$ functions.
| That guarantees that that is one of the answers.
It's justified because if $x = y$ then $f(x) = f(y)$ for any possible function.
So IF $\frac \theta 2 = \frac \pi 2 - 2\theta$ then $\sin \frac \theta 2 = \sin (\frac \pi 2 -2\theta) = \cos 2\theta$, then any solution to $\frac \theta 2 = \frac \pi 2 - 2\theta$ (there is only one such solution) will be a solution to $\sin \frac \theta 2 = \cos 2\theta$.
But it doesn't mean it is the only solution.
It is true that if $x = y \implies f(x) = f(y)$ but it is not true that $f(x) = f(y) \implies x=y$. $x=y$ will be one pair of solutions (always) but there may be others.
But we can take this further.
As $\cos 2\theta = \sin (\frac \pi 2 - 2\theta)$ always, it will suffice to find all solutions to $\sin\frac {\theta} 2 = \sin (\frac \pi 2 - 2\theta)$.
So we have to ask ourselves how do we find all solutions to $\sin x = \sin y$.
Well $x = y$ is one but so is $x = y + 2k\pi$ for any integer $k$. And $x = \pi - y$ is another but so is $x = \pi- y +2j\pi=-y + (2j+1)\pi$ for any integer $j$. And that is all of them.
So if $\sin\frac {\theta} 2= \sin (\frac \pi 2- 2\theta)$ then we have either
$\frac \theta 2 = \frac \pi 2 - 2\theta + 2k\pi$ for some integer $k$.
Or we have
$\frac \theta 2 = 2\theta - \frac \pi 2 + (2j+1)\pi$ for some integer $j$.
....
Now we just solve those.
If $\frac \theta 2 = \frac \pi 2 - 2\theta + 2k\pi$ then
$\frac 52\theta = \frac \pi 2 + 2k \pi$ so
$\theta = \frac \pi 5 + \frac 45k\pi$.
That is $\frac \pi 5$ is a solution but so are $\pi, \frac 95\pi, \frac{13}5\pi, \frac {17}5\pi$ etc.
And if $\frac \theta 2 = 2\theta - \frac \pi 2 + (2j+1)\pi$ then we have
$\frac 32\theta = \frac \pi 2 - (2j+1)\pi$.
(For simplicity sake we can replace $-(2j+1)$ with any other odd integer so lets replace it with a positive $(2m + 1)$ to get... )
$\frac 32\theta =\frac \pi 2 + (2m+1)\pi$
$\theta = \frac \pi 3 + \frac {4m+2}3\pi$.
So $\theta = -\frac{\pi} 3 \pi, \frac {7\pi}3, \frac {11\pi}3$ are all solutions.
| {
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"source": "stackexchange",
"question_score": "1",
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If $\frac{x^2+y^2+x+y-1}{xy-1}$ is an integer for positive integers $x$ and $y$, then its value is $7$. I saw this on quora
and haven't been able to solve it.
If $\dfrac{x^2+y^2+x+y-1}{xy-1}$
is an integer for positive integers
$x$ and $y$,
then its value is $7$.
If
$y=1$ this is
$\dfrac{x^2+x+1}{x-1}
= x+2+\dfrac{3}{x-1}$
which is an integer only when
$x=2$ or $x=4$ and has value $7$.
Looking at the values from 2 to 20
for $x$ and $y$,
this is an integer only for
$x=2, y=12$ (and $x=12, y=2$).
This value is 7.
So it looks like this might be correct,
and I don't know how to show it.
| This can be shown using Vieta Jumping. Consider the polynomial
$$P(x,y)=x^2+y^2+x+y-1-kxy+k$$
We will show that for $x,y,k\in\mathbb{N}$, $P(x,y)=0$ implies $k=7$.
$1)$ If $(x,y)$ is a solution, then $(y,x)$ is a solution
$2)$ If $y=1$ then clearly $k=7$.
$3)$ $k\geq 3$. At $k=1$ solving for $x$ in $P(x,y)=0$ gives a discriminant of
$$\Delta = -3y^2-6y+1$$
At $k=2$ we get a discriminant of
$$\Delta = -8y-3$$
$4)$ $(x,x)$ cannot be a solution.
$$k=1\Rightarrow x\in\{-2,0\}$$
$$k=2\Rightarrow x=-\frac{1}{2}$$
For $k\geq 3$ we have the discriminant of $P(x,x)$
$$\Delta=k^2-(3k-3)$$
The difference between $k^2$ and $(k-1)^2$ is
$$k^2-(k-1)^2=2k-1<3k-3$$
The difference between $k^2$ and $(k-2)^2$ is
$$k^2-(k-2)^2=4k-4>3k-3$$
Thus, $\sqrt{\Delta}\not\in \mathbb{Q}$.
$5)$ Suppose we have a solution $(x,y)$ with $x> y>1$. Then
$$x\in\left\{\frac{ky-1\pm\sqrt{(k y-1)^2-4 \left(k+y^2+y-1\right)}}{2}\right\}$$
Since $x\in\mathbb{N}$ and both solutions are positive, we know that both are also natural numbers. Now, can $x$ be the lesser solution? Note that
$$\frac{ky-1-\sqrt{(k y-1)^2-4 \left(k+y^2+y-1\right)}}{2}<\frac{ky-1-\sqrt{(k y-1-2y)^2}}{2}=y$$
Here, this inequality follows since
$$0<1 - k - 2 y - 2 y^2 + k y^2$$
holds for all $(y,k)\in \{2,3,...\}\times \{3,4,...\}$ except for $(2,3)$. We need only consider these pairs from points $2)$ and $3)$ above. We can ignore the final pair though since this gives the polynomial
$$x^2-5x+8$$
with complex solutions. For all other acceptable $(y,k)$ though we have
$$0<1 - k - 2 y - 2 y^2 + k y^2$$
$$=(k y-1)^2-4 \left(k+y^2+y-1\right)-(k y-1-2 y)^2$$
Thus, $x$ is the greater solution and from our original solution $(x,y)$ where $x>y$ we can get a new solution $(z,y)$ where $y>z$.
Let us rehash what we have showed: if we have a solution $(x_0,y_0)$ with $x_0>y_0>0$, then we can make a new solution $(x_1,y_1)$ with $0<x_1<y_1=y_0$. Since the solutions are symmetric, this gives us another solution $(x_2,y_2)$ with $x_2>y_2>0$. Since all of these solutions $(x_n,y_n)$ are natural numbers, this sequence must continue until it hits some minimum. But this minimum can only be $1$ since otherwise we can always find a smaller solution, implying that $k=7$ from the beginning.
EDIT: Since people are interested, this process can be run in reverse to generate all of an infinite number of solutions. Consider the sequences defined
$$a_1=1$$
$$a_2=2$$
$$a_n=\frac{\sqrt{45 a_{n-1}^2-18 a_{n-1}-23}+7 a_{n-1}-1}{2}$$
and
$$b_1=1$$
$$b_2=4$$
$$b_n=\frac{\sqrt{45 b_{n-1}^2-18 b_{n-1}-23}+7 b_{n-1}-1}{2}$$
Then all solutions are of the form $(a_n,a_{n-1})$ or $(b_n,b_{n-1})$ (up to symmetry) and it is easy to generate an infinite amount of them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Prove $abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}$ for $a+b+c+d=1$
Let $a,b,c$, and $d$ be four positive reals satisfying $a+b+c+d=1$. Show that
$$abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}.$$
I tried the inequality between $27abc$ and $(a+b+c)^3$ but it didn't help me
| Inspired by https://math.stackexchange.com/a/1748125/823641.
Define $f(a,b,c,d) = abc+abd+acd+bcd - \frac{1}{27} - \frac{176abcd}{27}$.
We have $f(1/4,1/4,1/4,1/4) = 0$.
We shall show that $f(1/4,1/4,1/4,1/4)$ is the maximum among all $a,b,c,d$ satisfying the conditions.
Suppose not, and assume that another value of $(a,b,c,d) = (x,y,z,w) \neq (1/4,1/4,1/4,1/4)$ corresponds to the maximum $f(x,y,z,w)$.
WLOG, we assume that $x \neq y$.
We have
\begin{align}
&f(\frac{x+y}{2},\frac{x+y}{2},z,w) - f(x,y,z,w) \\
=~&(x-y)^2(z+w)/4 - \frac{176zw}{27 * 4}(x-y)^2 \\
=~& \frac{(x-y)^2}{4}(z+w-\frac{176zw}{27}),
\end{align}
where
$$\frac{(x-y)^2}{4} > 0.$$
If $z+w-\frac{176zw}{27} > 0$ then we are done because in that case we have $f(\frac{x+y}{2},\frac{x+y}{2},z,w) > f(x,y,z,w)$, completing the proof by contradicting the assumption of maximality.
Now, we assume that
$z+w-\frac{176zw}{27} \leq 0$,
then we have
$$f(x,y,z,w) = xy(z+w - 176zw/27) + (x+y)zw - 1/27$$
$$\leq (x+y)zw - 1/27 \leq (1 - t)t^2/4 - 1/27,$$
where we have used the Cauchy–Schwarz inequality and $t = z+w$.
Let $g(t) = (1 - t)t^2/4 - 1/27$.
It is easy to check (by $g'(t)$) that $g(t)$ achieves its maximum at $t = 2/3$ with $g(t) = 0$.
Therefore, we have $f(x,y,z,w) \leq 0$, completing the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Calculate $\frac{4-5\sin\alpha}{2+3\cos\alpha}$
Calculate $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}$$ if $\cot\dfrac{\alpha}{2}=-\dfrac32$.
My first approach was to somehow write the given expression only in terms of the given $\cot\frac{\alpha}{2}$ and just put in the value $\left(-\dfrac{3}{2}\right)$. Now I don't think that's possible because we have constants (4 and 2). My try, though: $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}\right)}=\dfrac{4-10\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\cos^2\frac{\alpha}{2}-3\sin^2\frac{\alpha}{2}}$$ My second idea was to find the value of the trig functions of $\alpha$. I don't know if this is the most straight-forward approach, but $$\cot\alpha=\dfrac{\cot^2\frac{\alpha}{2}-1}{2\cot\frac{\alpha}{2}}=\dfrac{\frac94-1}{-2.\frac32}=-\dfrac{5}{12}.$$ Am I now supposed just to find the values of $\sin\alpha$ and $\cos\alpha$? Nothing more elegant? We would have $\dfrac{\cos\alpha}{\sin\alpha}=-\dfrac{5}{12}\Rightarrow\cos\alpha=-\dfrac{5}{12}\sin\alpha$ and putting into $\sin^2\alpha+\cos^2\alpha=1$ we'd get $\cos\alpha=\pm\dfrac{12}{13}$.
| We have $$\cot{\frac{\alpha}{2}}=\frac{\sin \alpha}{1-\cos \alpha}=-\sqrt{\frac{1+\cos \alpha}{1-\cos \alpha}}=-\frac{3}{2} \implies \sin \alpha=1.5\cos \alpha -1.5;$$ $$3\sqrt{1-\cos \alpha}=2\sqrt{1+\cos \alpha}$$
Thus, $9(1-\cos \alpha)=4(1+\cos \alpha) \implies \cos \alpha= \frac{5}{13}$
$$\frac{4-5\sin \alpha}{2+3\cos \alpha}=\frac{4-7.5\cos \alpha+7.5}{2+3\cos \alpha}=\frac{112}{41}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).
The given answer is $\dfrac{79}{128}$.
| using the hint: $x^5-y^5 = (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$
Let's create the parts:
since $x^2+y^2=1$ we have:
$(x^2+y^2)^2 = x^4+2x^2y^2+y^4 = 1$ (2)
We have a big part of it but we are missing $xy$ so
we know $x-y = 1/2$ squaring both side: $x^2 + y^2 -2xy = 1/4$ since $x^2+y^2=1$ then $xy=3/8$ (1).
using (2)
$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4) = \frac{1}{2} (1 - x^2y^2 + xy(x^2+y^2)) = \frac{1}{2} (1 - \frac{9}{64} + \frac{3}{8}\times1) = \frac{79}{128}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triangle ABC with $\angle{ACB} = 3\angle{ABC }$ and $AB = \frac{10}{3}BC$. Find $\cos{A}\cos{B}\cos{C}$ assume that $\angle{ABC}=y, \angle{ACB}=3y, BC=3x,$ and $AB=10x$ then by sine rule, i obtain following
$ \frac{10x}{\sin{3y}}=\frac{AC}{\sin{y}}=\frac{3x}{\sin{4y}}$
by cosine rule in try to figure out AC
$AC=\sqrt{109x^2-60x^2\cos{y}}$
i have no idea how to combine these two informations in order to solve the problem
| From the first equation:
$$10\sin(4y)-3\sin(3y)=0$$
Let's expand everything in terms of $\sin y$ and $\cos y$:
$$\begin{align}\sin(4y)&=\sin(2(2y))\\&=2\sin2y\cos2y\\&=4\sin y\cos y(1-2\sin^2y)\\&=4\sin y\cos y-8\sin^3 y\cos y\\\sin(3y)&=\sin(y+2y)\\&=\sin y\cos2y+\cos y\sin2y\\&=\sin y(1-2\sin^2y)+2\sin y\cos^2y\\&=\sin y-2\sin^3y+2\sin y-2\sin^3y\\&=3\sin y-4\sin^3y\end{align}$$
We then get
$$40\sin y\cos y-80\sin^3y\cos y=9\sin y-12\sin^3y$$
$\sin y=0$ is not a good solution for the problem, so we can divide by it:
$$40\cos y-80(1-\cos^2y)\cos y-9+12(1-\cos^2y)=0$$
Let $\cos y=z$. Then$$40z-80z+80z^3+3-12z^2=0\\80z^3-12z^2-40z+3=0$$
It is not trivial, but from the rational root theorem you get that one root is $z=\frac34$
So the equation becomes $$(4z-3)(20z^2+12z-1)=0$$
Note that $y$ has to be less than $\frac\pi4$, since you have angles $y$, $3y$, and the sum of them has to be less than $\pi$. You can see that $\cos y=\frac34$ is the only acceptable solution. Then the rest should be easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying the determinant of a matrix. Suppose $$A = \begin{pmatrix} 1+a_{1}+a_{1}b_{1}+b_{2} & 1+a_{1} & 1 & 0\\ a_{2}+a_{2}b_{1}+b_{3} & 1+a_{2} & 1 & 1\\ a_{3}+a_{3}b_{1} + b_{4} & a_{3} & 1 & 1\\ a_{4} + a_{4}b_{1} & a_{4} & 0 & 1\end{pmatrix}$$ Show that $$\det(A) = \det\begin{pmatrix} b_{1}-b_{2}+b_{4} & 1+a_{1}-a_{3}+a_{4}\\ -1-b_{2}+b_{3} & a_{1}-a_{2}+a_{4}\end{pmatrix}$$
I am not sure how to show this. I tried to perform some row and column operations but could simplify matrix $A$.
Note that for a general matrix of size $(3k+1)\times (3k+1)$, the determinant of $A$ can be written in a similar form. Say for a $7 \times 7$ matrix,
$$A_{7 \times 7} = \begin{pmatrix} 1+a_1+a_1b_1+b_2 & 1+a_1 & 1& 0 &0 &0 &0\\ a_2 + a_2b_1 + b_3 & 1+a_2 & 1 & 1&0&0&0\\ a_3+a_3b_1+b_4 & a_3 & 1 & 1 & 1& 0 &0\\ a_4+a_4b_1+b_5 & a_4 & 0 & 1 & 1&1 &0\\ a_5+a_5b_1+b_6 & a_5 & 0 & 0 & 1&1 &1\\ a_6+a_6b_1+b_7 & a_6 & 0 & 0 & 0&1 &1\\ a_7 + a_7b_1 & a_7 & 0 & 0 & 0&0 &1
\end{pmatrix}$$
$$\det(A) = \det\begin{pmatrix} b_{1}-b_{2}+b_{4}-b_{5}+b_{7} & 1+a_{1}-a_{3}+a_{4}-a_{6}+a_{7}\\ -1-b_{2}+b_{3}-b_{5}+b_{6} & a_{1}-a_{2}+a_{4}-a_{5}+a_{7}\end{pmatrix}$$
Any thoughts on how to show this in general?
| First, for any integer $n\geq 1$ and any numbers $A_1,...,A_n,B_1,...,B_n$ we have
$$
\det\left(I_n+\begin{pmatrix}A_1 & B_1 & 0 & \cdots &0 \\
A_2 & B_2 & 0 & \cdots &0 \\
\vdots & \vdots & \vdots & \ddots &\vdots \\
A_n & B_n & 0 & \cdots &0 \\\end{pmatrix}\right)=\det\begin{pmatrix}1+A_1 & B_1 \\
A_2 & 1+B_2\end{pmatrix}.
$$
The determinant under consideration (for all $n$) is
$$
\det(M+N)=\det M\cdot\det(I_n+M^{-1}N)
$$
where
$$
M=\begin{pmatrix} 1 & 1 & 1 & 0 & \cdots &\cdots\\
0 & 1 & 1 & 1 & \cdots&\cdots \\
0 & 0 & 1 & 1 & \cdots&\cdots \\
0 & 0 & 0 & 1 & \cdots&\cdots \\
\vdots & \vdots& \vdots& \vdots& \ddots&\vdots\\
0&0&0&0&\cdots&1\end{pmatrix},\quad
N=
\begin{pmatrix}
a_1(1+b_1)+b_2 & a_1 & 0 & \cdots &0 \\
a_2(1+b_1)+b_3 & a_2 & 0 & \cdots &0\\
a_3(1+b_1)+b_4 & a_3 & 0 & \cdots &0\\
\vdots & \vdots & \vdots & \ddots &\vdots\\
a_n(1+b_1)+b_{n+1} & a_n & 0 & \cdots &0
\end{pmatrix}
$$
We have $\det M=1$ and $M^{-1}$ is the matrix whose
first row is $(1,-1,0,1,-1,0,1,-1,0,...)$ and the $(k+1)$th row is obtained from the first by adding $k$ zeros to the left and truncating it to the right:
$$
M^{-1}=\begin{pmatrix}
1&-1&0&1&-1&0&\cdots\\
0&1&-1&0&1&-1&\cdots\\
0&0&1&-1&0&1&\cdots\\
0&0&0&1&-1&0&\cdots\\
0&0&0&0&1&-1&\cdots\\
0&0&0&0&0&1&\cdots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\
0&0&0&0&0&0&\cdots&1
\end{pmatrix}
$$
By inspection, $M^{-1}N$ has the 3rd, 4th,... columns equal to the zero vector (as $N$ does so) and therefore we can apply the formula stated in the beginning to obtain
$$
\det(M+N)=\det(I_n+M^{-1}N)=\det\begin{pmatrix}1+A_1 & B_1 \\
A_2 & 1+B_2\end{pmatrix}
$$
with
\begin{align}
A_1&=\sum_{j=1}^n i_j(a_j(1+b_1)+b_{j+1}),&
B_1&=\sum_{j=1}^n i_ja_j,\\
A_2&=\sum_{j=1}^n i_{j-1}(a_j(1+b_1)+b_{j+1}),&
B_2&=\sum_{j=1}^n i_{j-1}a_j,
\end{align}
where we set $i_0:=0,i_1:=1,i_2:=-1$ and $i_{j+3k}:=i_j$ for all integers $k$. (To match the matrix in the original question, $b_{n+1}=0$, but the formula is valid for any value of $b_{n+1}$.)
Finally, the constraint on the remainder mod $3$ of the size is not necessary, the formula holds for all $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximum length of points of tangency of an ellipse and a variable circle Let ellipse $4x^2+16y^2=64$ and circle $x^2+y^2=r^2$ have a common tangent touching at A and B respectively. The maximum length of AB can be?
(A) 4
(B) 3
(C) 2
(D) 5
I tried solving the question using parametric coordinates, using slope form of tangents but there are a lot more variables and much-complicated equations. Can't think of some intuitive or out-of-the-box solution. Please help.
|
The dual conics are $16X^2+4Y^2=1$ and $r^2X^2+r^2Y^2=1,$ and they intersect when r is between $2$ and $4$ making the common tangents $$\pm x \sqrt{r^2-4}/(2 \sqrt3 r) \pm y \sqrt{16-r^2}/(2\sqrt3 r) + 1=0$$ and $$A: (\pm(r \sqrt{r^2-4})/(2 \sqrt3),\pm(r \sqrt{16-r^2})/(2 \sqrt3))$$ $$B: ((\pm 8\sqrt{r^2-4})/(\sqrt3 r),\pm (2 \sqrt{16-r^2})/(\sqrt3 r)),$$
so the squared distance is $((4-r)(4+r)(r-2)(r+2))/r^2$ which maximizes at
$r=\sqrt8$ when the distance is $$2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving using Mathematical Induction from my discrete math class This is a practice exercise from our class about proving inequalities using mathematical induction. I've been stuck on the last step for quite a while now.
This is the Question. "Prove that $\sum_{k=1}^n\frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n}$ whenever $n$ is a positive integer."
This is my attempt to prove it.
Step 1: Base case $(n=1)$
$\sum_{k=1}^1\frac{1}{(1)^2}\le2-\frac{1}{1}$
$1\le1$
Step 2: Induction hypothesis
Assume that $\sum_{k=1}^n\frac{1}{k^2}\le 2-\frac{1}{n}$ is true for $n=c$
Then, $\sum_{k=1}^c\frac{1}{k^2}\le 2-\frac{1}{c}$
Step 3: prove that it is true for $n=c+1$
$\sum_{k=1}^{c+1}\frac{1}{k^2}$ $\le$ $2-\frac{1}{c}+\frac{1}{(c+1)^2}$
$\le 2+\frac{-(c+1)^2+c}{c(c+1)^2}$
$\le 2+\frac{-(c^2+2c+1)+c}{c(c+1)^2}$
$\le 2+\frac{-c^2-2c-1+c}{c(c+1)^2}$
$\le 2-\frac{c^2+c+1}{c(c+1)^2}$
$\le 2-\frac{c^2+c}{c(c+1)^2}-\frac{1}{c(c+1)^2}$
$\le 2-\frac{c(c+1)}{c(c+1)^2}-\frac{1}{c(c+1)^2}$
$\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$
I'm Stuck on this step.
| In step3, you added a new term $\frac{1}{c+1}$ (while $n=c+1$) to the both sides of the induction hypothesis ($\sum_{k=1}^{c}\frac{1}{k^2} \le 2-\frac{1}{c}$) and ended with:
$\sum_{k=1}^{c+1}\frac{1}{k^2} \le 2-\frac{1}{c}+\frac{1}{(c+1)^2} \le ...
\le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2}$
Because $n=c+1$ and $n>1$, then $c>0$, then $\frac{1}{c(c+1)^2}>0$. Now we have:
$\sum_{k=1}^{c+1}\frac{1}{k^2} \le ... \le 2-\frac{1}{c+1}-\frac{1}{c(c+1)^2} \le 2-\frac{1}{c+1}$
And you are done with the deduction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532680",
"timestamp": "2023-03-29T00:00:00",
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Find the all real roots of the polynomial $x^6+3 x^5+3 x-1=0$ in closed form
Find the all real roots of the polynomial
$$x^6+3 x^5+3 x-1=0$$
in exact form.
WolframAlpha gives only numerical results. I've asked a few similar questions before. The source of the problem comes from the algebra precalculus workbook (but not homework workbook).
The rational root theorem doesn't work. Because, $x=1$ and $x=-1$ are not roots.
Wolfram says , we have only $2$ real roots. What kind of factorization should I try?Can a factorization of the form $P(x)=(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$ work?
I also tried the trick $\frac {P(x)}{x^n}$, but I failed.
| This is such a cursed question.
The only elementary method I can think of is to assume the existence of a factorization of the form $$P(x) = x^6 + 3x^5 + 3x - 1 = (x^2 + kx \pm 1)(x^4 + ax^3 + bx^2 + cx \mp 1) \tag{1}$$ and then after tediously expanding and equating coefficients, we require
$$(a,b,c) = (3-k, 1-3k+k^2, 3-2k+3k^2-k^3) \tag{2}$$ for the case where the quadratic factor is $x^2 + kx - 1$. For the second case where the quadratic is $x^2 + kx + 1$, the solution for $(a,b,c)$ leads to a contradiction $c = k$ and the linear term becomes $0$; thus is inadmissible.
After substituting $(2)$ into $(1)$ we find
$$P(x) = x^6 + 3x^5 - k g(k) x^2 + (g(k) + 3)x - 1, \tag{3}$$ where $$g(k) = k^3 - 3k^2 + 3k - 6. \tag{4}$$ Then if $g(k) = 0$, we have a valid factorization. This tells us we are guaranteed a solution in radicals. We note
$$\begin{align}
g(k) &= k^3 - 3k^2 + 3k - 6 \\
&= (k^3 - 3k^2 + 3k - 1) - 5 \\
&= (k - 1)^3 - 5. \tag{5}
\end{align}$$
Therefore, $g(k) = 0$ has the unique real root $k = 1 + \sqrt[3]{5}$, from which we can refer to $(1)$ to recover the two real roots
$$x = \frac{-k \pm \sqrt{k^2 + 4}}{2} = \frac{-1 - 5^{1/3} \pm \sqrt{5^{2/3} + 2 \cdot 5^{1/3} + 5}}{2}. \tag{6}$$ The quartic does not have any real solutions, but by this point I'm already exhausted by all the tedious math and don't want to prove it by hand. The problem with this approach is that it is highly speculative--there is no guarantee of it working, and we basically got lucky: the coefficients in $(3)$ turn out "just right" in which the solution of the cubic $(4)$ makes both the linear and quadratic terms correct. Moreover, the solution to $(4)$ is "easy" and does not require the cubic formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1$
Prove that $\sin^n (2x) + (\sin^n x - \cos^n x)^2 \leq 1$.
Let $a = \sin x, b = \cos x.$ Then we want to show $2^n a^n b^n +a^{2n}-2a^n b^n + b^{2n}\leq 1,$ which follows once we show that $(2^n-2)a^n b^n \leq \sum_{j=1}^{n-1}{n\choose j} a^{2(n-j)}b^{2j}$. I think the AM-GM inequality is useful, but I'm not sure how to apply it in this case. I don't think that for all j we have $a^{2(n-j)} b^{2j}\ge a^n b^n$; for instance if $a<b,$ then $a^{2(n-j)}b^{2j} < a^nb^n$.
| Assume that $n$ is a positive integer.
Let $a = \sin x$ and $b = \cos x$.
Then $a^2 + b^2 = 1$ and $|2ab| \le a^2 + b^2 = 1$.
Let $$f(n) := (2^n-2) a^n b^n + a^{2n} + b^{2n}.$$
We have $f(1) = a^2 + b^2 = 1$ and $f(2) = 2a^2b^2 + a^4 + b^4 = (a^2 + b^2)^2 = 1$.
First, we have
\begin{align*}
&f(2n) - f(2n+1)\\
=\,& (2^{2n}-2) a^{2n} b^{2n} + a^{4n} + b^{4n}
- [(2^{2n+1}-2) a^{2n+1} b^{2n+1} + a^{4n+2} + b^{4n+2}]\\
=\,& (2^{2n}-2)(1 - 2ab)a^{2n}b^{2n} - 2a^{2n+1}b^{2n+1} + a^{4n}(1 - a^2) + b^{4n}(1 - b^2)\\
\ge\,& - 2a^{2n+1}b^{2n+1} + a^{4n}b^2 + b^{4n}a^2\\
\ge\,& - 2a^{2n+1}b^{2n+1} + 2\sqrt{a^{4n}b^2 \cdot b^{4n}a^2}\\
\ge\,& 0.
\end{align*}
Second, we have
\begin{align*}
&f(2n) - f(2n+2)\\
=\,& (2^{2n}-2) a^{2n} b^{2n} + a^{4n} + b^{4n} - [(2^{2n+2}-2) a^{2n+2} b^{2n+2} + a^{4n+4} + b^{4n+4}]\\
=\,& (2^{2n}-2)(1-4a^2b^2)a^{2n}b^{2n} - 6a^{2n+2}b^{2n+2} + a^{4n}(1-a^4) + b^{4n}(1-b^4)\\
\ge\,& - 6a^{2n+2}b^{2n+2} + a^{4n}b^2(1+a^2) + b^{4n}a^2(1+b^2)\\
=\,& a^2b^2(a^{4n} + b^{4n} - 2a^{2n}b^{2n}) + a^{4n}b^2 + b^{4n}a^2 - 4a^{2n+2}b^{2n+2}\\
\ge\,& 2\sqrt{a^{4n} b^2 b^{4n} a^2} - 2ab \cdot 2a^{2n+1}b^{2n+1}\\
\ge\,& 0.
\end{align*}
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$
My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm 1$, but this implies $a=0$ and this is not always correct. Then I realized that,
$$x^4-2x^2+1=(x^2-1)^2$$
is a perfect square. So, I tried to write the original equation as
$$ax^4-x^3+x+\bigg(x-\frac 1x\bigg)^2=0$$
$$x^2\bigg(ax^2-x+\frac 1x\bigg)+\bigg(x-\frac 1x\bigg)^2=0$$
But I failed again. I couldn't spot the palindromic property.
| You are on the right track.
Let,
$$P(x)=ax^6-x^5+x^4+x^3-2x^2+1$$
We observe that, $0$ is not a possible root of $P(x)$. Therefore, we can divide all terms of the polynomial by $x^2\, (x\neq 0)\,:$
$$
\begin{align}\frac {P(x)}{x^2}&=ax^4-x^3+x+\left(x-\frac {1}{x}\right)^2\\
&=x^2\left(ax^2-\left(x-\frac 1x\right)\right)+\left(x-\frac {1}{x}\right)^2\\
&=ax^4-x^2\left(x-\frac 1x\right)+\left(x-\frac {1}{x}\right)^2\\
&=\color{red}{\left(x-\frac {1}{x}\right)^2}-\color{blue}{x^2}\color{red}{\left(x-\frac 1x\right)}+\color{blue}{ax^4}=0.
\end{align}$$
Finally, substitute $x-\frac {1}{x}=u$, then you have:
$$
\begin{align}
&u^2-ux^2+ax^4=0\\
\implies &\Delta_u=x^4(1-4a)\\
\implies &u_{1,2}=\frac {x^2\pm x^2\sqrt {1-4a}}{2}\\
\implies &u_{1,2}=x^2\left(\frac {1\pm \sqrt {1-4a}}{2}\right)\\
\implies &x-\frac {1}{x}=x^2\left(\frac {1\pm \sqrt {1-4a}}{2}\right)\\
\implies &\left(\frac {1\pm \sqrt {1-4a}}{2}\right)x^3-x^2+1=0.
\end{align}
$$
Note that, the last line is the result you want to achieve.
| {
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"url": "https://math.stackexchange.com/questions/4538322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Finding the limit in terms of a without using L'Hospitals Find $\;\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}\;$ in terms of a without using L'Hospitals Rule.
I first graphed the function and noticed that that limit tends to go towards $-a^2$
I also tried this approach:
$\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}=\lim_\limits{x \to 0} \dfrac{2 \cdot \frac{1-\cos(ax)}{2}}{1-\sqrt{1+x^2}} = \lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1 - \sqrt{1+x^2}} $
However, I'm not sure how to proceed. I tried rationalizing the denominator but it only makes it more tangled.
| Thank to @Wang YeFei for the hint.
$\lim_\limits{x \to 0} \dfrac{1-\cos(ax)}{1-\sqrt{1+x^2}}=\lim_\limits{x \to 0} \dfrac{2 \cdot \frac{1-\cos(ax)}{2}}{1-\sqrt{1+x^2}} = \lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1 - \sqrt{1+x^2}} $
Then we proceed to multiply the numerator and denominator by $1 - \sqrt{1+x^2}$:
$\lim_\limits{x \to 0} \dfrac{2\sin^2(\frac{ax}{2})}{1-\sqrt{1+x^2}} \cdot \dfrac{1 - \sqrt{1+x^2}}{1 - \sqrt{1+x^2}} = \lim_\limits{x \to 0} -\dfrac{[2\sin^2(\frac{ax}{2})][1 - \sqrt{1+x^2}]}{x^2}$
Then, to be able to use the fact that $\lim_ \limits{x \to 0} \dfrac{\sin x}{x}=1$, we multiply the numerator and denominator by $\dfrac{a^2}{4}$:
$\lim_\limits{x \to 0} -\dfrac{[2\sin^2(\frac{ax}{2})][1 - \sqrt{1+x^2}]}{x^2} \cdot \dfrac{\frac{a^2}{4}}{\frac{a^2}{4}} = \lim_\limits{x \to 0} \dfrac{[\sin^2(\frac{ax}{2})][-\frac{1}{2}a^2][1 - \sqrt{1+x^2}]}{\dfrac{a^2x^2}{4}}$
$=\lim_\limits{x \to 0} \dfrac{\sin^2(\frac{ax}{2})}{\dfrac{a^2x^2}{4}} \cdot \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})$
$= \left( \lim_\limits{x \to 0} \dfrac{\sin(\frac{ax}{2})}{\dfrac{ax}{2}}\right)^2 \cdot \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})$
$= \lim_\limits{x \to 0} \hspace{0.2cm} (-\frac{1}{2}a^2)(1 - \sqrt{1+x^2})=-a^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the value of given integral The given integral is :$$\displaystyle\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta$$
I tried solving it with some trigonometric identities but comes out it does not work here.
However solution given starts with these equations mentioned down below and later they substituted $z= \cos{\theta}+ i\sin(\theta)$
$$\begin{array}{r}
z^{10}-1=\left(z^2-1\right)\left[z^2-2 \cos \left(\frac{\pi}{5}\right) z+1\right]\left[z^2-2\left(\cos \frac{2 \pi}{5}\right) z+1\right] \\
\quad \times\left(z^2-2 \cos \frac{4 \pi}{5} z+1\right)\left(z^2-2 \cos \frac{6 \pi}{5} z+1\right)
\end{array}$$
$$\begin{aligned}
z^5-\frac{1}{z^5}=\left(z-\frac{1}{z}\right) &\left(z-2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right) \\
& \times\left(z-2 \cos \frac{4 \pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{6 \pi}{5}+\frac{1}{z}\right)
\end{aligned}$$
Can anyone help me understand how I can derive these equations and use them here in this integral?
Thank you for your help.
| Note
$$ \sin(5\theta)=\sin \theta(1+2\cos2\theta+2\cos4\theta), \cos^4\theta=\frac18(3+4\cos2\theta+\cos4\theta) $$
from here.
Using
$$ \cos^2{\frac{\pi}{5}}+\cos^2{\frac{2\pi}{5}}=\frac{3}{4}, \cos^2{\frac{\pi}{5}}\cos^2{\frac{2\pi}{5}}=\frac{1}{16} $$
one has
\begin{eqnarray}
&&(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})\\
&=&\cos^4\theta-(\cos^2{\frac{\pi}{5}}+\cos^2{\frac{2\pi}{5}})\cos^2\theta+\cos^2{\frac{\pi}{5}}\cos^2{\frac{2\pi}{5}}\\
&=&\cos^4\theta-\frac34\cos^2\theta+\frac1{16}\\
&=&\frac18(3+4\cos2\theta+\cos4\theta)-\frac34\frac{1+\cos2\theta}{2}+\frac1{16}\\
&=&\frac1{16}(1+2\cos2\theta+2\cos4\theta)
\end{eqnarray}
and hence
$$\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta
= \int_0^1 \frac1{16}d\theta = \frac1{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$
We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^\frac13+\dfrac23x\left(x^2-1\right)^{-\frac23}(x^2-1)(x^2+1)\\=\dfrac23x\left(x^2-1\right)^\frac13\left(6x^2+x^2+1\right)\\=\dfrac23x\left(x^2-1\right)^\frac13(7x^2+1)$$ The given answer is $$y'=\dfrac{2x(7x^4-6x^2-1)}{3\sqrt[3]{\left(x^2-1\right)^2}}$$ I don't see my mistake...
| The two formulas are equivalent.
$$\begin{align}
&\frac{2}{3}x\left(x^{2}-1\right)^{\frac{1}{3}}\left(7x^{2}+1\right)-\frac{2x\left(7x^{4}-6x^{2}-1\right)}{3\sqrt[3]{\left(x^{2}-1\right)^{2}}}
\\&=\frac{2x}{3\left(x^{2}-1\right)^{\frac{2}{3}}}\left(\left(7x^{2}+1\right)\left(x^{2}-1\right)^{\frac{1}{3}}\cdot\left(x^{2}-1\right)^{\frac{2}{3}}-\left(7x^{4}-6x^{2}-1\right)\right)
\\&=\frac{2x}{\ldots}\left(\left(7x^{2}+1\right)\left(x^{2}-1\right)-\left(7x^{4}-6x^{2}-1\right)\right)
\\&=0
\end{align}$$
It simply depends on whether you choose to bring together the sum given by the product rule in terms of $\left(x^{2}-1\right)^{\frac{1}{3}}$ or $\left(x^{2}-1\right)^{\frac{2}{3}}$. Any reasonable examiner should find both formulas correct, especially if they follow your correct working.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what comes to mind is finding $(a + b + c + d)^3$ and subtracting whatever we don't need to get $a^3 + b^3 + c^3 + d^3$. However,
\begin{equation*}
(a+b+c+d)^3 = a^3+3a^2b+3a^2c+3a^2d+3ab^2+6abc+6abd+3ac^2+6acd+3ad^2+b^3+3b^2c+3b^2d+3bc^2+6bcd+3bd^2 + c^3 + 3c^2d + 3cd^2 + d^3
\end{equation*}
There is simply no good way to calculate $6abc + 6abd + 6acd + 6bcd $ without expanding everything.
| You're not using the most important parts of the question: namely that $\xi$ is a fifth root of unity, and also that conveniently, all the coefficients are either $20$ and $13$.
Calculating $a^3 + b^3 + c^3 + d^3$ directly and collecting like coefficients, we have:
$$a^3 + b^3 + c^3 + d^3 = 20^3(\xi^6 + \xi^{12} + \xi^9 + \xi^3) + 3 \cdot 20^2 \cdot 13(\xi^5 + \xi^{10} + \xi^{10} + \xi^5)$$
$$+ 3 \cdot 20 \cdot 13^2 (\xi^4 + \xi^8 + \xi^{11} + \xi^{7}) + 13^3 (\xi^3 + \xi^6 + \xi^{12} + \xi^9)$$
and since $\xi^5 = 1$:
$$= 20^3 \left(\frac{1 - \xi^{15}}{1 - \xi^3} - 1 \right) + 3 \cdot 20^2 \cdot 13 \cdot4 +3 \cdot 20 \cdot 13^2 (\xi^4 + \xi^3 + \xi^1 + \xi^2) + 13^3 \left(\frac{1 - \xi^{15}}{1 - \xi^3} - 1 \right)$$
Can you continue from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show a function is continuous on $\mathbb{R}$ by any method Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.
Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|<\delta$. So, $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|=|K||\frac{x(K^2+y^2)-y(K^2+x^2)}{(K^2+y^2)(K^2+x^2)}|\leq |\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}|\leq|\frac{xK^2-yK^2+xy^2}{(K^2+y^2)(K^2+x^2)}|$ I'm a bit stuck here on how to relate this inequality to $|x-y|$. Can someone help out? Thanks!
| Since you are asking a solution by any method, you can try to use the limit definition of continuity. Recall that a function $f$ is continuous at a point $c$ of its domain if $$\lim_{x\rightarrow c} f(x) = f(c)$$
So for any $c\in \mathbb{R}$ we have $\lim_{x\rightarrow c}Kx = Kc$ and $\lim_{x\rightarrow c} K^2+x^2 = K^2+c^2 \neq 0$. So we have:$$
f(c) = \frac{Kc}{K^2+c^2} = \frac{\lim_{x\rightarrow c}Kx}{\lim_{x\rightarrow c}K^2+c^2} = \lim_{x\rightarrow c}\frac{Kx}{K^2+x^2} = \lim_{x\rightarrow c} f(x)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Sum of the areas of all n-sided polygons inscribed within each other Imagine you take an $n$-sided regular polygon with side length $x$, and connect the midpoints of each of its sides to construct a smaller but identical $n$-sided regular polygon within it.
Now perform the same process with the smaller polygon, and then with the next one, and the next and so on.
If you continue this, endlessly inscribing smaller and smaller regular $n$-sided polygons, what is the sum of all of their individual areas in terms of $x$ and $n$?
Below is an image describing this question when n = 6:
I know there must be a common ratio between each of the areas, so i'm thinking about turning this into a geometric series of the form $\frac{a}{1-r}$ (since $r$ < 1), where $a$ = the area of the first polygon, however I do not know how to find the specific values for $a$ and $r$ in terms of $x$ and $n$.
Any help with this question would be greatly appreciated!
| Let the side length of the outermost polygon and side length of the first constructed polygon be $x_0$ and $x_1$ respectively. The external angles of the polygons will each be $\frac{2\pi}{n}$, so the internal angles will each be $\pi - \frac{2\pi}{n}$. Consider the triangle shown in the diagram below:
Applying the cosine law, we get
$$\begin{align}
(x_1)^2 &= \left( \frac{x_0}{2} \right)^2 + \left( \frac{x_0}{2} \right)^2 - 2\left( \frac{x_0}{2} \right)\left( \frac{x_0}{2} \right) \cos \left( \pi - \frac{2\pi}{n} \right) \\
\\
&= \frac{(x_0)^2}{2} + \frac{(x_0)^2}{2} \cos \left( \frac{2\pi}{n} \right) \\
\\
&= \frac{(x_0)^2}{2} \left[ 1 + \cos \left( \frac{2\pi}{n} \right) \right] \\
\\
&= \frac{(x_0)^2}{2} \left[ 2 \cos^2 \left( \frac{\pi}{n} \right) \right] \\
\\
&= (x_0)^2 \cos^2 \left( \frac{\pi}{n} \right) \\
\\
x_1 &= x_0 \cos\left( \frac{\pi}{n} \right)
\end{align}$$
Thus the common ratio of the sides is $\cos \left( \frac{\pi}{n} \right)$, so the common ratio of the areas will be $\boxed{\cos^2 \left( \frac{\pi}{n} \right)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)$$
How does one prove this step-by-step?
| \begin{align}
&\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln x}\,dx \\
=&\int_0^1 \frac{x^{m -1}-x^{n -1}}{\ln x}\sum_{k\ge 0} (-x^p)^k \,dx\\
=&\ \sum_{k\ge 0}(-1)^k
\int_0^1 \frac{x^{m +pk-1}-x^{n +pk -1}}{\ln x}
\> \ \overset{u=-\ln x}{dx}\\
=&\ \sum_{k\ge 0} (-1)^k
\int_0^\infty \frac{e^{-(n +pk)u}-e^{-(m +pk)u}}{u}du\\
=&\ \sum_{k\ge 0} (-1)^k\ln \frac{m+kp}{n+kp}
= \ln \frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Spivak, Calculus, Ch. 22: How do we compute the limit $\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$? My question is simply how do we compute the limit
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$$
I know the limit is $1$, both from the context below and because I checked in Maple.
Here is the context in which this limit arose.
In Chapter 22 of Spivak's Calculus, Problem 13 asks us to show first that if $f$ is increasing on $[1,\infty)$ then
$$f(1)+...+f(n-1)<\int_1^n f(x)dx<f(2)+...+f(n)$$
When we apply this result to the function $f(x)=\log{(x)}$ we easily obtain the relationship
$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^n}\tag{1}$$
Spivak concludes for us that given this result we can say that
$$\lim\limits_{n\to\infty} \frac{\sqrt[n]{n!}}{n}=\frac{1}{e}\tag{2}$$
I am interested in going from $(1)$ to $(2)$.
Starting at $(1)$, if we take the n-th root and divide by $n$ we have
$$0<\frac{1}{e^{\frac{n-1}{n}}}<\frac{\sqrt[n]{n!}}{n}<\frac{(n+1)^{\frac{n+1}{n}}}{n}\cdot \frac{1}{e}$$
Now,
$$\lim\limits_{n\to\infty} \frac{1}{e^{\frac{n-1}{n}}}=\frac{1}{e}$$
I would like to compute the limit
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}\cdot \frac{1}{e}$$
Which necessitates computing the limit that gave rise to the current question
$$\lim\limits_{n\to\infty} \frac{(n+1)^{\frac{n+1}{n}}}{n}$$
| Another way:
Using $$ \lim_{n \to \infty} \frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right) = 0$$ and $\frac{\ln n}{n} < 1$ then
$$ \frac{1}{n} \, (n+1)^{\frac{n+1}{n}} = \frac{n+1}{n} \, e^{\left( \ln n + \ln\left(1 + \frac{1}{n}\right) \right)/n} = \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)} \, e^{\ln n/n} $$
which gives the limit
\begin{align}
L &= \lim_{n \to \infty} \, \frac{1}{n} \, (n+1)^{\frac{n+1}{n}} \\
&= \lim_{n \to \infty} \, \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)} \, e^{\ln n/n} \\
&= \lim_{n \to \infty} \, \left(1 + \frac{1}{n}\right) \, e^{\frac{1}{n} \, \ln\left(1 + \frac{1}{n}\right)} \, \left(1 + \frac{\ln n}{n} + \frac{\ln^2 n}{2 \, n^2} + \mathcal{O}\left(\frac{\ln^3 n}{n^3}\right) \right) \\
&= 1
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \frac n4$ How do you prove that $\displaystyle\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \dfrac{n}{4}$, where $x\in\mathbb{R}$?
I tried mathematical induction, but it doesn't work. I also have some ideas on complex number solution, such as setting $z=\cos\left(x\right)+i\sin\left(x\right)$ and working on $Re\left(z\right)$, but neither does that work.
| By periodicity $\left|\cos(jx)\right|=|\cos (j(\pi+x))|=|\cos (j(\pi-x))|$ and the fact that $\cos$ even, it's enough to assume $0 <x \le \pi/2$ as the inequality is obvious for $x=0$.
Note that $|\cos x|+|\cos 2x| \geqslant \dfrac{\sqrt{2}}{2} $ with equality at $x=\pi/4$ (splitting into $0 \leqslant x \leqslant \pi/4, \pi/4 \le x \le \pi/2$ and expliciting one gets a quadratic in $\cos x$ etc) so we can assume $n \ge 3$ since $\sqrt 2/2 \ge n/4$ for $n=1,2$
Now $$\sum^{n+1}_{j=1}|\cos(jx)| \ge \sum^{n+1}_{j=1}\cos^2(jx)=\frac{n+1}{2}+\frac{1}{2}\sum^{n+1}_{j=1}\cos(2jx)$$
Now $\sum^{n+1}_{j=1}\cos(2jx)=(\frac{\sin ((2n+3)x)}{\sin x}-1)/2$ and we know that $\sin x \ge \frac{2x}{\pi}$ so $$|(\frac{\sin ((2n+3)x)}{\sin x}-1)/2| \le \frac{1}{2}(\frac{\pi}{2x}+1)$$
Hence if $\frac{\pi}{2x}+1 \le n+2$ or $x \ge \frac{\pi}{2n+2}$ we have $$\sum^{n+1}_{j=1}|\cos(jx)| \ge \frac{n+1}{2}-\frac{1}{2}|(\frac{\sin ((2n+3)x)}{\sin x}-1)/2| \ge \frac{n+1}{2}-\frac{n+2}{4}=\frac{n}{4}$$
If $x \le \frac{\pi}{2n+3}$ then $(2n+3)x \le \pi$ so $\frac{\sin ((2n+3)x)}{\sin x} \ge 0$ hence $$\sum^{n+1}_{j=1}|\cos(jx)| \ge \frac{n+1}{2}-\frac{1}{4}\geqslant \frac{n}{4}$$
hence assume $\frac{\pi}{2n+3} \le x \le \frac{\pi}{2(n+1)}$ then all $\cos jx \geqslant 0, j=1,..n+1$ since $(n+1)x \leqslant \pi/2$ hence:
$$\sum^{n+1}_{j=1}|\cos(jx)|=\sum^{n+1}_{j=1}\cos(jx)=(\frac{\sin ((n+3/2)x)}{\sin x/2}-1)/2$$
$\frac{\pi}{2}\le (n+3/2)x \le \frac{\pi}{2}+\frac{\pi}{4n+4}$ so $\sin ((n+3/2)x) \ge \cos \frac{\pi}{4n+4} \ge 1- (\frac{\pi}{4n+4})^2/2 $ while $\sin x/2 \le x/2 \le \frac{\pi}{4n+4}$ so $\frac{1}{ \sin x/2} \ge \frac{4n+4}{\pi}$, hence
$$2\sum^{n+1}_{j=1}|\cos(jx)| \ge \frac{4n+4}{\pi}(1- (\frac{\pi}{4n+4})^2/2 ) -1 \ge \frac{4n}{\pi}+4/\pi - \pi/32-1\ge \frac{n}{2}$$ since $n \ge 3$, so we are done!
Now $\int_0^{\pi/2}|\cos jx|dx=1, j \ge 1$ so $\int_0^{\pi/2}\sum^{n+1}_{j=1}|\cos(jx)|dx=n+1$ which suggests that the average sum is about $\frac{2n+2}{\pi}$ hence the lower bound $n/4$ most likely can be improved
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute the line integral of $x^2 + y^2$ on curve $|x|+|y|=1$
Compute the integral
$$\int_{\gamma} f(x,y)ds$$
where $\gamma=\{|x|+|y|=1\}$ and $f(x,y)=x^2+y^2$
Note, we have not covered any major theorems on evaluating line integrals, which has been presented is to parametrize the curve then integrate with respect to the parameter(roughly speaking). However, I have no idea how to paramatrize the curve. My intuition tells me we should probably split the curve into four cases, but then the derivative of the parametrizations will be zero? So how would I proceed? Input would be much appreciated!
| $\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4$ is the obvious sum of line segments: 1) $\gamma_{1}$ joins $(1,0)$ and $(0,1)$. 2) $\gamma_{2}$ joins $(0,1)$ and $(-1,0)$. 3) $\gamma_{3}$ joins $(-1,0)$ and $(0,-1)$. 4) 2) $\gamma_{4}$ joins $(0,-1)$ and $(1,0)$.
Then, $I=\int_{\gamma}(x^2+y^2)ds=\sum_{i=1}^4I_{i}$ where $I_i=\int_{\gamma_{i}}(x^2+y^2)ds$.
(1) Parametrization of $\gamma_1$: $x=t$, $y=1-t$, $0\leq t\leq 1$. Then $dx=dt$, $dy=-dt$, $ds=\sqrt{dx^2+dy^2}=\sqrt{2}dt$. Hence, $I_1=\int_0^1(t^2+(1-t)^2)\sqrt{2}dt=\sqrt{2}\int_0^1(2t^2-2t+1)dt=\frac{2\sqrt{2}}{3}.$
(2) Parametrization of $\gamma_2$: $x=t$, $y=t+1$, $-1\leq t\leq 0$. Then $dx=dt$, $dy=dt$, $ds=\sqrt{dx^2+dy^2}=\sqrt{2}dt$. Hence, $I_1=\int_{-1}^0(t^2+(1+t)^2)\sqrt{2}dt=\sqrt{2}\int_{-1}^0(2t^2+2t+1)dt=\frac{2\sqrt{2}}{3}.$
Here, we noticed that the orientation is not important.(Why?) Let's dicuss this. It is confusing. We could choose a counter-clockwise orientation of $\gamma_2$: $x=-t$, $y=1-t$, $0\leq t\leq 1$. Then $dx=-dt$, $dy=-dt$, $ds=\sqrt{dx^2+dy^2}=\sqrt{2}dt$. Hence, $I_1=\int_{0}^1((-t)^2+(1-t)^2)\sqrt{2}dt=\sqrt{2}\int_{0}^1(2t^2-2t+1)dt=\frac{2\sqrt{2}}{3}.$ Same result.
It seems that (3) and (4) computations will give the same result: $\frac{2\sqrt{2}}{3}$. Therefore, $I=4(\frac{2\sqrt{2}}{3})=\frac{8\sqrt{2}}{3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove or disprove $ab+bc+cd+da\leq1$ if $a+b+c+d=2$ Non-negative real numbers $a,b,c,d$ are such that $a+b+c+d=2$. Prove or disprove that
$$ab+bc+cd+da\leq1$$
I see there are multiple equality cases, where $(a,b,c,d)$ is for example $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$, $(\frac{3}{4},\frac{1}{2},\frac{1}{4},\frac{1}{2})$, $(\frac{3}{4},\frac{5}{8},\frac{1}{4},\frac{3}{8})$.
I suspect it's true, and maybe it can be proven with rearrangement, but I have not found a way.
It's reminiscent of Chebyshev's inequality, since the desired inequality is equivalent to
$$4(ab+bc+cd+da)\leq(a+b+c+d)(b+c+d+a)$$
But we cannot assume that $(a,b,c,d)$ and $(b,c,d,a)$ are oppositely ordered.
| We have: $ab+bc+cd+da = (a+c)(b+d) \le \dfrac{((a+c)+(b+d))^2}{4}= \dfrac{(a+b+c+d)^2}{4}= 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Can we find the integral solutions of $x(3y-5)=y^2+1$?
$x(3y-5)=y^2+1$ has no integral solution. True or False?
Answer False, as $(-1,1)$ satisfies it.
I checked on WolframAlpha that the given hyperbola has four integral solutions. $(-1,1),(-1,-4),(5,13),(5,2)$.
I wonder if we can find that without hit and try.
My Attempt:
$y^2-3xy+1+5x=0$ is a quadratic in $y$. So,
$$y=\frac{3x\pm\sqrt{9x^2-4-20x}}{2}$$
For $y$ to be integer, the discriminant should be a perfect square, that too of an even number. Also, $x$ should be even so that the denominator $2$ can be cancelled out.
But wolfram states only odd $x$.
What's my error? Also, how to proceed next?
I tried equating the discriminant with $(2k)^2$ but couldn't conclude.
| Hint. This is a linear equation in $x$, so
$$ x = \frac{y^2+1}{3y-5} = \frac{1}{3}\frac{3y^2+3}{3y-5} = \frac13\frac{y(3y-5)+(5/3)(3y-5)+3+25/3}{3y-5} $$
Therefore
$$ 9x = 3y+5+\frac{34}{3y-5} \implies \frac{34}{3y-5}=9x-3y-5.$$
The right hand side of the equation is an integer, if $x$ and $y$ are integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.
Show that for any $n$, there are infinitely many cubes of the form $2^na - 9$.
Progress: We use induction on $n.$ For $n=1$ it works. Say it works for $n-1$. We will show for $n$. Note that $2^na-9$ is $$2^{n-1}2a-9.$$ If we have infinite $a$ such that $a$ is even for $n-1$, then we are done. Else, only finite amounts of $a$ are even.
So after a large constant $N$, we have $2^{n-1}a-9=x^3$ for only odd $a$.
So say $$2^{n-1}a_1-2^{n-1}a_2=u^3-w^3\implies 2^{n-1}|u^3-w^3$$
But I couldn't get anything further.
I thought about taking $v_2$, as in, if $2^na=x^3+9$ then we can take $x=y^2$. So we have $$2^na=y^2+3^2.$$
| Suppose we have some positive $b$ for which $2^nb-9=u^3$ for some positive integer $u$. Notice that $u$ must be odd. Note then that
\begin{equation}
2^n(b+3u^2+3u2^n+(2^{n})^2)-9=u^3+3u^22^n+3u(2^n)^2+(2^n)^3=(u+2^n)^3,
\end{equation}
and so $a=b+3u^2+3u2^n+(2^n)^2$ also satisfies the equation $2^na-9=v^3$ for some integer $v$.
However, $3u^2+3u2^n+(2^n)^2$ must be odd, as $u$ is odd. If $b$ was even, then $a$ is odd, and if $b$ was odd, then $a$ is even. Reiterating this argument, we find that if one solution $b$ exists, then we must in fact have an infinite amount of solutions, and that these solutions alternate signs, so that we also have an infinite amount of even solutions.
Your proof by induction then works.
| {
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"url": "https://math.stackexchange.com/questions/4561289",
"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$
where $a>0$, $b>0$ are some constants.
I know how to calculate this limit using the L'Hopital's rule:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)};
$$
$$
\lim\limits_{x\to 0} \ln\left(\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}\right)=
\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)}{x}$$
$$
=
\lim\limits_{x\to 0} \frac{2}{a^{\sin x}+b^{\sin x}}\cdot\frac12\cdot
\left( a^{\sin x}\cos x \ln a+b^{\sin x}\cos x \ln b \right)=
\frac12\left( \ln a+ \ln b \right)
$$
$$
\Rightarrow
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}=
e^{\frac12\left( \ln a+ \ln b \right)}=\sqrt{ab}.
$$
I'm allowed to use the limits $\lim_{x\to0}\frac{\sin x}{x}=1$,
$\lim_{x\to0}\frac{a^x-1}{x}=\ln a$,
$\lim_{x\to0}\frac{\log_a(1+x)}{x}=\log_a e$ and $\lim_{x\to0} (1+x)^{1/x}=e$.
| Let's prove this expression:
Given positive $a$ and $b$:
$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}$$.
In order to show this we are going to use the following:
$$\lim_{n\rightarrow\infty}n(\sqrt[n]{a}-1)=\ln a$$ thus by AM-GM
$$\frac{1}{n}\ln\sqrt{ab}\leq\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})=\ln\bigg(\frac{1}{2}(\sqrt[n]{a}-1)+\frac{1}{2}(\sqrt[n]{b}-1)+1\bigg)<\frac{1}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$
now multiply both sides by $n$, we get
$$\ln\sqrt{ab}\leq n\ln\frac{1}{2}(\sqrt[n]{a}+\sqrt[n]{b})\leq\frac{n}{2}\bigg((\sqrt[n]{a}-1)+(\sqrt[n]{b}-1)\bigg)$$
Taking $n\rightarrow\infty$ and using the squeeze theorem:
We get
$$\lim_{n\rightarrow\infty}\bigg(\frac{\sqrt[n]{a}+\sqrt[n]{b}}{2}\bigg)^n=\sqrt{ab}.$$
Let $n=\frac1x$ and the rest is done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3$-var inequality: $\frac{bc}{\sqrt{a}+3}+\frac{ca}{\sqrt{b}+3}+\frac{ab}{\sqrt{c}+3} \leq \frac{3}{4}$ for $a+b+c=3$. Problem: Let $a,b,c$ be positive numbers satisfied $a+b+c=3$. Prove that $$\dfrac{bc}{\sqrt{a}+3}+\dfrac{ca}{\sqrt{b}+3}+\dfrac{ab}{\sqrt{c}+3} \leq \dfrac{3}{4}$$
I've tried U.C.T method but it doesn't reach the solution. I also thought of $p,q,r$ method, but I think it will end up a ugly solution. The only useful thing I get is $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca$. Anyone?
| Fact 1: It holds that, for all $x \ge 0$,
$$\frac{1}{3 + \sqrt x} \le \frac13 - \frac{13}{96}x + \frac{5}{96}x^2.$$
(Note: Letting $x=y^2$,
we have $\mathrm{RHS} - \mathrm{LHS} = \frac{y(5y^2 + 25y + 32)(y-1)^2}{96(y+3)}\ge 0$.
The RHS comes from the Taylor approximation of the LHS around $x=1$,
that is $\frac{1}{3 + \sqrt x}\sim \frac14 - \frac{1}{32}(x-1) + \frac{3}{256}(x-1)^2$, after adjusting the coefficient of $(x-1)^2$.)
By Fact 1, it suffices to prove that
$$\sum_{\mathrm{cyc}} bc \left(\frac13 - \frac{13}{96}a + \frac{5}{96}a^2\right) \le \frac34$$
or
$$\frac13(ab + bc + ca) - \frac{13}{32}abc + \frac{5}{96}abc(a + b + c) \le \frac34$$
or (using $a + b + c = 3$)
$$\frac13(ab + bc + ca) - \frac{1}{4}abc \le \frac34$$
or
$$12(ab + bc + ca) - 9abc \le 27$$
or (using $a + b + c = 3$)
$$4(a + b + c)(ab + bc + ca) - 9abc \le (a + b + c)^3$$
which is degree three Schur.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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