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The perimeter of an isosceles triangle $\triangle ABC$
An isosceles triangle $\triangle ABC$ is given with $\angle ACB=30^\circ$ and leg $BC=16$ $cm$. Find the perimeter of $\triangle ABC$.
We have two cases, right? When 1) $AC=BC=16$ and 2) $AB=BC=16$.
For the first case: let $CH$ be the altitude through $C$. Since the triangle is isosceles, $CH$ is also the angle bisector and $\measuredangle ACH=\measuredangle BCH=15^\circ$. How to approach the problem further? I have not studied trigonometry.
|
The first case.
Let $BK$ be an altitude of $\Delta ABC$.
Thus, $$BK=8,$$
$$CK=\sqrt{BC^2-BK^2}=\sqrt{16^2-8^2}=8\sqrt3$$ and
$$AB=\sqrt{AK^2+BK^2}=\sqrt{(16-8\sqrt3)^2+8^2}=$$
$$=\sqrt{8^2(2-\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}=16\sqrt{2-\sqrt3},$$
which gives the answer: $32+16\sqrt{2-\sqrt3}.$
We can solve the second problem by the similar way.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting three shops, they return home. Find the probability that they have only one umbrella.
My Attempt
$A_i/\bar{A}_i$ : $A$ remembers or forgets umbrella at shop $i$
$B_i/\bar{B}_i$ : $B$ remembers or forgets umbrella at shop $i$
$B_o/\bar{B}_o$ : $B$ remembers or forgets umbrella at shop home
$A_i,,B_i=3/4;\bar{A}_i,\bar{B}_i=1/4$ and $B_o,\bar{B}_o=1/2$
$$
\text{Req. Prob.}=A_1A_2A_3\bar{B}_o+A_1A_2A_3B_o\bar{B}_1+A_1A_2A_3B_oB_1\bar{B}_2+A_1A_2A_3B_oB_1B_2\bar{B}_3+\bar{A}_1B_oB_1B_2B_3+A_1\bar{A}_2B_oB_1B_2B_3+A_1A_2\bar{A}_3B_oB_1B_2B_3\\
=\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)+\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
+\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3+\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
=\frac{27}{128}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}\\
=\frac{3726}{128*64}=\frac{3726}{8192}
$$
But my reference gives the solution $\frac{7278}{8192}$, so what am I missing in my attempt ?
Note: Please check $b$ part of link for a similar attempt to solve the problem to obtain the solution as in y reference.
|
I think your answer is correct. Each of them can forget the umbrella independently.
So:
Pr{A forgets}= $\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+\frac{3}{4}\frac{3}{4}\frac{1}{4}= \frac{74}{128}$
Pr{B forgets}=$\frac{101}{128}$
Consequently: Pr{one umbrella} = Pr{A forgets}.(1-Pr{B forgets})+(1-Pr{A forgets}).Pr{B forgets}= $\frac{3726}{8192}$
|
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|
$a+b+c+d=1, a,b,c,d ≠ 0$, then prove that $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 + (c + \frac{1}{c})^2 + (d + \frac{1}{d})^2 \ge \frac{289}{4} $ If $a+b+c+d=1$, $a,b,c,d ≠ 0$, prove that $$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 + \left(c + \frac{1}{c}\right)^2 + \left(d + \frac{1}{d}\right)^2 \ge \frac{289}{4} $$
I tried expanding the entire LHS and I got
$a^2 + b^2 + c^2 + d^2 + \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + 8$ on the LHS. Then I tried applying AM-GM to the first 4 terms and the last 4 terms and I got
$4 \sqrt{abcd} + \frac{4}{\sqrt{abcd}}$ but that didnt seem useful.
Any hints?
Thank you in advance.
|
Let $$f(x)=x^2+\frac{1}{x^2}.$$
Thus, $f$ is a convex function on $(0,+\infty)$ and $f$ is a convex function on $(-\infty,0)$.
We need to consider the following cases.
*
*$a\geq b\geq c\geq d>0$.
The case $0>a\geq b\geq c\geq d$ is impossible because $a+b+c+d=1.$
Now, our inequality follows from the Jensen's inequality:
$$8+\sum_{\text{cyc}}\left(a^2+\frac{1}{a^2}\right)\geq8+4\left(\left(\frac{a+b+c+d}{4}\right)^2+\frac{1}{\left(\frac{a+b+c+d}{4}\right)^2}\right)=\frac{289}{4}.$$
*$a\geq b\geq c>0>d$ (the case $a>0>b\geq c\geq d$ is a similar).
*$a\geq b>0>c\geq d$.
In the last case we obtain a counterexample: $$(a,b,c,d)=\left(-1,-1,\frac{3}{2},\frac{3}{2}\right).$$
In this case $$\sum_{\text{cyc}}\left(a+\frac{1}{a}\right)^2=\frac{313}{18}<\frac{289}{4},$$ which says that your inequality is wrong!
|
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|
If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate:
$$z^2 + \frac{4}{z^2} = -3$$
$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?
|
Since $z \neq 0$, you have
$$\begin{equation}\begin{aligned}
& z^2 + z + 2 = 0 \\
& z + 1 + \frac{2}{z} = 0 \\
& z + \frac{2}{z} = -1 \\
& \left(z + \frac{2}{z}\right)^2 = (-1)^2 \\
& z^2 + 4 + \frac{4}{z^2} = 1 \\
& z^2 + \frac{4}{z^2} = -3
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
|
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|
The polynomial $x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$ has no real roots.
Prove that the polynomial $$x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$$ has no real roots.
Here is the solution
Transcribed from this image
*For $x \leq 0$ we have obviously $p(x)>0$. Let $x>0$. We transform the polynomial in the same way as a geometric series:
$$
\begin{aligned}
p(x) &=x^{2 n}-2 x^{2 n-1}+3 x^{2 n-2}-\cdots-2 n x+2 n+1 \\
x p(x) &=x^{2 n+1}-2 x^{2 n}+3 x^{2 n-1}-4 x^{2 n-2}+\cdots+(2 n+1) x.
\end{aligned}
$$
Adding, we get
$$
\begin{aligned}
x p(x)+p(x) &=x^{2 n+1}-x^{2 n}+x^{2 n-1}-x^{2 n-2}+\cdots,+x+2 n+1 \\
(1+x) p(x) &=x \cdot \frac{1+x^{2 n+1}}{1+x}+2 n+1.
\end{aligned}
$$
From here we see that $p(x)>0$ for $x>0$.
Can anyone explain how the last line of the solution implies that the polynomial has no real roots?
|
Since other users already explained the last part of the given proof, I propose another way where there is no need to distinguish between the positive and the negative case, and it does not involve any geometric sum.
By splitting the terms of even degree in a convenient way, the polynomial can be written as
$$(0+1)x^{2n} - 2x^{2n-1} + (1+2)x^{2n-2}- \cdots ((n-1)+n)x^2-2nx + (n + (n+1))$$
that is
$$\left(x^n-x^{n-1}\right)^2+2\left(x^{n-1}-x^{n-2}\right)^2+\dots+n\left(x-1\right)^2+\left(n+1\right)$$
which implies that its minimum value over $\mathbb{R}$ is $n+1>0$ (attained for $x=1$).
Therefore the polynomial has no real roots.
|
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|
Inequality $\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=3$. Then prove that $$\sum_{cyc}\frac{a}{2a^2+a+1}\leq \frac{3}{4}$$
I tried to use tangent line method. Let $$f(x)=\frac{x}{2x^2+x+1}$$
Then $$f'(x)=\frac{1-2x^2}{2x^2+x+1}$$ and since we know that equality occurs if $a=b=c=1$, tangent line is: $$y=f(1)+(x-1)f'(1)=\frac{5-x}{16}$$
But $$\frac{x}{2x^2+x+1}\leq\frac{5-x}{16}\implies x\in(-\infty,\frac{5}{2}]$$
it means this is false for $x\in(\frac{5}{2},3]$. Then how to prove original inequality for $max(a,b,c)\geq\frac{5}{2}$? Can anyone help me?
|
This inequality is true for any reals $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, if $abc=0$ so the inequality is obvious.
But for $abc\neq0$, by AM-GM and your work (after assuming that $a\geq\frac{5}{2}$ and $x\neq0$ ) we obtain:
$$\sum_{cyc}\frac{a}{2a^2+a+1}\leq\frac{\frac{5}{2}}{2\left(\frac{5}{2}\right)^2+\frac{5}{2}+1}+\frac{2|x|}{2x^2+|x|+1}=$$
$$=\frac{5}{32}+\frac{2}{2|x|+\frac{1}{|x|}+1}\leq\frac{5}{32}+\frac{2}{2\sqrt2+1}<\frac{3}{4}.$$
|
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|
Double Counting(combinatorial proof) for $1^2\binom{n}{1} + 2^2\binom{n}{2}+3^2\binom{n}{3}+...+n^2\binom{n}{n}$ = $n(n+1)2^{n-2}$ can anyone help me with double-counting proof for this equation:
$1^2\binom{n}{1} + 2^2\binom{n}{2}+3^2\binom{n}{3}+...+n^2\binom{n}{n}$ = $n(n+1)2^{n-2}$
I tried this example: we have n+1 bits and at least two "1" bits. it gives just the right side. and also looking at right side in $\binom{n+1}{2}2^{n-1}$ may help.
|
I only know how to prove it analytically. Let $f(x)=(1+x)^{n}=\sum_{k=0}^{n}{n \choose k}x^{k}.$
Differentiating, we have
$$
f'(x)=\sum_{k=1}^{n}kx^{k-1}{n \choose k}.
$$
Multiply both sides by $x$ and differentiate them again, then we have
\begin{eqnarray*}
\frac{d}{dx}\{xf'(x)\} & = & \frac{d}{dx}\sum_{k=1}^{n}kx^{k}{n \choose k}\\
f'(x)+xf''(x) & = & \sum_{k=1}^{n}k^{2}x^{k-1}{n \choose k}.
\end{eqnarray*}
Put $x=1$, then we have
\begin{eqnarray*}
& & \sum_{k=1}^{n}k^{2}{n \choose k}\\
& = & f'(1)+f''(1)\\
& = & n\cdot2^{n-1}+n(n-1)2^{n-2}\\
& = & 2^{n-2}n(n+1),
\end{eqnarray*}
provided that $n\geq2$.
|
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|
Is there a way to find an upper bound for $n^2+an+b$? I was solving the Project Euler: Problem 27.
Considering quadratics of the form $n^2 + an + b$, where $|a| \lt 1000$ and $|b| \le 1000$
Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.
In order to optimize the algorithm to solve the problem, we can make use of the following fact:
*
*$b$ must always be prime
*Since $n^2+an+b>0$, Determinant must be negative i.e $D=a^2-4b<0$. This bounds the value of $a$.
I realized that if we predetermine a sieve of Eratosthenes then we can further speed up the process for checking whether $n^2+an+b$ is a prime of not. But, this requires us to find an upper bound to $n^2+an+b$ which is required to generate the sieve.
Substituting $a=\pm2\sqrt{b}$, gives $n^2+an+b = n^2 \pm2\sqrt{b}n+b=(n\pm\sqrt{b})^2 $. However this leads nowhere. Is there a way to find the upper bound for $n^2+an+b$ based on the given conditions for $a$ and $b$?
EDIT: Some additional info.
*
*If we keep $b$ as primes below 1000 and $|a|<1000$, maximum observed value of $n^2+an+b$ is 12989 at $a=889,b=347,n=14$.
*If we keep $b$ as primes below 1000 and $|a|<\sqrt{4b}$, maximum observed value of $n^2+an+b$ is 1681 at $a=-1,b=41,n=41$.
|
I would like to extend the answer by @Ingix.
The given quadratic is $f(n)=n^2+an+b$. Let $n=ax+by$ where $x,y \in \mathbb{Z}$.
\begin{equation}
f(ax+by)=a^2(x^2+x)+aby(2x+1)+b^2y^2+b
\end{equation}
If $x^2+x=0$, then $b|f(n)$. This gives $x=0,-1$. For $x=0$, we get $n=by$ and $y=1$ gives the best answer ($y$ cannot be zero otherwise $f(n)$ will be prime).
For $x=-1$, we need $n>0$ i.e. $by-a>0$ i.e. $y> a/b$. Therefore we can say that
\begin{equation}
n = b\left( \left\lfloor \frac{a}{b} \right\rfloor +1 \right)-a
\end{equation}
Note that we cannot use $y=\lceil a/b \rceil$ because if $a$ is divisible by $b$, then $n$ will become zero.
However this bound doesn't always work when $a$ is negative. A simple example would be $f_1(n)=n^2-n+3$. Here $f_1(0)=3,f_1(1)=3,f_1(2)=5,f_1(3)=9$. But the bound shows $3\left(\left\lfloor\frac{-1}{3}\right\rfloor+1\right)+1=3(-1+1)+1=1$. So for negative $a$ values, we can use $x=0$.
To sum up, for any pair $(a,b)$ where $b$ is prime and $a$ is odd,
an upper bound $n$ is
\begin{cases}
n<b, & a<0 \ \mathrm{and} \ a\ne -b \\
n<2b, & a<0 \ \mathrm{and} \ a= -b \\
n<b\left(\left\lfloor\frac{a}{b}\right\rfloor+1\right)-a, & a>0
\end{cases}
In order to keep $f(n)$ always positive, we need
\begin{equation}
n\notin \left[(-a-\sqrt{a^2-4b})/2,(-a+\sqrt{a^2-4b})/2 \right]
\end{equation}
It easy to show that this interval lies on the positive axis only when $a\le-\sqrt{4b}$.
Consider the case where $a=-b$. The lower bound for the interval becomes $(b-\sqrt{b^2-4b})/2$. The lower bound decreases while the interval width increases when $b$ increases, and the lower bound's maximum value is at $b=5$ where the interval itself is $[1.38,3.61]$. It is clear that $n<2$ for any value of $b$ when $a=-b$. Hence, we can ignore the case $a=-b$.
In general a pair $(a,b)$ where $a<-\sqrt{4b}$ can be ignored if it satisfies all the following condition:
*
*$(-a-\sqrt{a^2-4b})/2$ is less than the current max value of number of consecutive primes.
*The interval for which $f(n)$ is negative contains at least one integer point. This is true only when $\lfloor(-a-\sqrt{a^2-4b})/2\rfloor\ne \lfloor (-a+\sqrt{a^2-4b})/2 \rfloor$.
For the case when $a=-\sqrt{4b}$, the interval is reduced to a point $n=-a/2$ which is not an integer since $a$ is odd. However $a$ will be an integer only when $b$ is a square number which is not possible since $b$ is prime. So we will never encounter this case.
Desmos
Finally plugging in optimum values for $a$ and $b$ under given constraints, we get the sieve sizes as
\begin{equation}
f(n)=\begin{cases}
994009 & n=b,\ a<0,\ a=-1,\ b=997 \\
994009 & b>a>0,\ a=1,\ b=997 \\
1985015 & a>b,\ a=999,\ b=997
\end{cases}
\end{equation}
|
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Calculate $\int_0^1 \Big( \int _0^x \sqrt{y^2+\frac{y^2}{x^2}+\frac{y^4}{x^4}} dy \Big) dx$
Calculate $\int_0^1 \Big( \int _0^x \sqrt{y^2+\frac{y^2}{x^2}+\frac{y^4}{x^4}} dy \Big) dx$
I have a problem with this task because I have no idea what parametrization should be used to calculate it. I tried $x=r\sin \alpha, y=r \cos \alpha$ or $s=y, t=\frac yx$ or $s=y, t=(\frac yx)^2$ but in each case I got too complicated calculations.
Does anyone have any clever ideas?
|
What do you mean by too complicated? Upon substitution $y=x\sqrt{z}$ the integral becomes $$\int_0^1 {\rm d}x \frac{x}{2} \int_0^1 \sqrt{x^2+z+1} \, {\rm d}z = \int_0^1 {\rm d}x \frac{x}{3} \left[ \left( x^2 + 2 \right)^{3/2} - \left(x^2+1\right)^{3/2} \right] \\
\stackrel{t=x^2}{=} \frac{1}{6} \int_0^1 {\rm d}t \left[ (t+2)^{3/2} - (t+1)^{3/2} \right] = \frac{1}{15} - \frac{8\sqrt{2}}{15} + \frac{3\sqrt{3}}{5}\, .$$
|
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Find all ordered pairs $(x,y)$ of positive integers such that the expression $x^2+y^2+xy$ is a perfect square. My approach so far:
Let $x^2+y^2+xy=n^2,$ where $n\in\mathbb Z$.
$\implies (x+y) ^2-xy=n^2$
$\implies (x+y) ^2-n^2=xy$
$\implies (x+y+n) (x+y-n) =xy$
Only one case is possible:
When $x+y+n=xy$ and $x+y-n=1$. On adding, we get: $2(x+y) =xy+1$. Since LHS is an even number so is RHS. That's why, $xy$ must be odd$\implies x$ and $y$ both are odds. So I checked for few odd numbers and found out that for $x=3$ and $y=5,\;x^2+y^2+xy$ becomes $49$ which is a perfect square.. I didn't dare to check for more odds as there could be many...
Recently I found out that for $x=5$ and $y=16$(an even number), $x^2+y^2+xy=361=19^2.$ (Surprising!!)
So now I can say, I am stuck very badly.. All of my observations miserably failed...
Please suggest something!
|
Firstly, we can ssume Multiplying by $4$, we get $(2x+y)^2+3y^2$ is a perfect square. Thus we want to find $a,b$ such that $a^2+3b^2$ is a perfect square with $gcd(a,b)=1$
What follows is a well-known theory of finding out rational points on conic sections.
Now this amounts to finding rational points on the ellipse $E: X^2+3Y^2=1$.
We know $(1,0)$ is a rational point. If $P$ is another such point, then the slope of the line joining $P$ and $(1,0)$ is rational $m$. Say $L$ intersects $E$ at $(u,v)$. Then we get $$ \frac{v}{u-1}=m
$$ and $$u^2+3v^2=1$$
This gives us $u^2+3m^2(u-1)^2=1$
Since $1$ ia anyway a root of the quadratic, the other root is $\frac{3m^2-1}{3m^2+1}$ and this gives $u=\frac{3m^2-1}{3m^2+1},v=\frac{-2m}{3m^2+1}$
Thus all rational points are parametrized by $\left (\frac{3m^2-1}{3m^2+1},\frac{-2m}{3m^2+1} \right ) \ ; \ m\in \mathbb Q$ or $\left (\frac{3m^2-1}{3m^2+1},\frac{2m}{3m^2+1} \right ) \ ; \ m\in \mathbb Q$
So we get if $a^2+3b^2=c^2$, then $\frac{a}{c}=\frac{3m^2-1}{3m^2+1}$ $\frac{b}{c}=\frac{2m}{3m^2+1}$ for some $m\in \mathbb Q$
writing everything in terms of integers we get $$a=\pm(3p^2-q^2),b=\pm 2pq, c= \pm (3p^2+q^2)$$
Thus we get $y=6kpq$ where $p,q,k>0$ and $2x+y=k|3p^2-q^2|$
If $p,q$ have opposite parity, this forces $k$ to be even and we get $k=2l$
So we finally get $y=4lpq$ and $x=l|3p^2-q^2|-2lpq$
If, on the other hand both $p,q$ are odd then we get $y=2kpq, x=\frac{k}{2}|3p^2-q^2|-kpq$
|
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|
Evaluation of a tricky binomial sum The Question:
$$
\mbox{To prove that:}\quad
\frac{3!}{2(n+3)} = \sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}
$$
My Attempt:
I start off by writing $\sum_{r=0}^{n}{(-1)^r\frac{\binom{n}{r}}{\binom{r+3}{r}}}$ as $\sum_{r=0}^{n}{(-1)^r\frac{n!3!}{(n-r)!(r+3)!}}$.
Now, since there is a term of $3!$ in it, I thought it would be a good idea to convert the term inside the expression to a $\binom{n+3}{r+3}$ term (by multiplying and dividing by $(n+1)(n+2)(n+3)$ i.e.
$\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}{(-1)^r\frac{(n+3)!}{(n-r)!(r+3)!}}$.
This becomes,
$\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}{(-1)^r\binom{n+3}{r+3}}$
Beyond this, I absolutely have no clue. I have worked at this for hours but I still can't seem to get an alternative to this method so, I tried sticking to it but unfortunately, I couldn't come up with anything. I just can't seem to figure out what I can do to further simplify this expression!!Any hint on how to progress will be greatly appreciated.
Thanks!
|
Start from
$$
\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=0}^{n}(-1)^r \binom{n+3}{r+3}
$$Reindex:
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r-3} \binom{n+3}{r}
$$
$$
=\frac{-3!}{(n+1)(n+2)(n+3)}\sum_{r=3}^{n+3}(-1)^{r} \binom{n+3}{r}
$$Now add and subtract the terms $0\leq r\leq 2$:
$$
=\frac{-3!}{(n+1)(n+2)(n+3)}\left(\sum_{r=0}^{n+3}(-1)^{r} \binom{n+3}{r}-\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right)
$$The binomial theorem graciously takes care of the first series for us, and then the result falls out:
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\left(0+\sum_{r=0}^{2}(-1)^{r} \binom{n+3}{r}\right)
$$
$$
=\frac{3!}{(n+1)(n+2)(n+3)}\cdot \frac{(n+1)(n+2)}{2} = \frac{3}{n+3}
$$
|
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|
Finding all real functions satisfying $f(x) + f \left( \frac{x - 1}{x} \right) = \frac{5x^2 - x - 5}{x}$
Find all functions $f:\mathbb R \to \mathbb R$ that satisfy
$$f(x) + f \left( \frac{x - 1}{x} \right) = \frac{5x^2 - x - 5}{x}$$
for all nonzero $x$.
|
We set up a system of $3$ functional equations and one of the variables will be $f(x)$ and we will solve for it. First replace $x$ by $\frac{1}{1-x}$ and we get: $f(\frac{1}{1-x}) + f(x) = P(x)$, with $P(x)$ is the right hand side evaluated at $\frac{1}{1-x}$. Next replace $x$ by $\dfrac{x-1}{x}\implies f(\frac{x-1}{x})+f(\frac{1}{1-x}) = Q(x)$, with $Q(x)$ is the right hand side evaluated at $\frac{x-1}{x}$. Combine with the original equation: $f(x) + f(\frac{x-1}{x}) = \dfrac{5x^2-x-5}{x}$, you do have a system of "linear" equations in $3$ variables each of which is a function and your $f(x)$ is one of them, and you can solve for $f(x)$ by elimination method in a standard college algebra course.
|
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|
Solving a nonlinear periodic ODE Is it possible to solve the below ODE for arbitrary real values of $c$?
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0,$$
where $\theta=\theta(x)$. Also, $\theta$ itself does not have meaning, but ($\cos(2\theta),\sin(2\theta)$) are important and need to be continuous. There is no boundary condition since $x$ is not constrained by a boundary but goes to infinity. It means that the solution must cover the range $(-\infty,\infty)$.
|
The given equation can be easily integrated:
$$-\dfrac12\left((\cos 2\theta+c\sin2\theta)\theta'\right)'=0\Rightarrow
(\cos 2\theta+c\sin2\theta)\theta' = \dfrac12C_1,\\
(\cos 2\theta+c\sin2\theta)\,\mathrm d(2\theta) = C_1\,\mathrm dx\Rightarrow
\sin 2\theta-c\cos2\theta = C_1 x+C_2 = y,$$
wherein the constants $C_1,C_2\ $ can be defined from the initial conditions.
Let
$$t = \tan\theta\Rightarrow \cos2\theta = \dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta} = \dfrac{1-t^2}{1+t^2},\quad \sin2\theta = \dfrac{2\sin\theta\cos^2\theta}{\cos^2\theta+\sin^2\theta} = \dfrac{2t}{1+t^2},$$
$$2t-c(1-t^2) = (1+t^2)y,\quad (c-y)t^2 + 2t - (c+y) = 0,$$
with the common solutions
$$\small t_{1,2} = \dfrac{1\pm\sqrt{1+c^2-y^2}}{c-y},$$
$$\cos2\theta = \dfrac{-cy\mp\sqrt{1+c^2-y^2}}{c^2+1},\ \sin2\theta = \dfrac{-y\pm c\sqrt{1+c^2-y^2}}{c^2+1}, \tag1$$
and the special case
$$y=c\Rightarrow t=c,\quad \cos2\theta =\dfrac{1-c^2}{1+c^2},\quad \sin2\theta =\dfrac{2c}{1+c^2}.\tag2$$
If $y^2 > 1+c^2,\ $ then the solutions $(1)$ are complex. This fact points to the bounded values of the variable $x.$
|
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|
I don’t understand how to reduce this fraction to the stated solution: The fraction is as follows:
$$
\frac{9 \cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44 }{
22 \cdot 27 + 44 \cdot 54 + 66 \cdot 81 + 88 \cdot 108}
$$
That’s all fine. What I don’t get is that my textbook says this reduces to the following:
$$\frac{9\times 11 + (1^2 + 2^2 + 3^2 + 4^2)}{22 \times 27 \times (1^2 + 2^2 + 3^2 +4^2)}$$
I don’t understand how the sum of consecutive squares can be deduced from that fraction, or why the denominator contains $22\times 27 \times\dots $ as opposed to the numerator which is $9 \times 11 + \dots$”
Any insight would be really appreciated!
|
First correct the error in your expression. The first addition sign on top should be a multiplication sign.
Observe for instance that on the numerator, you have the first term $9\cdot11= (9\cdot 1)(11\cdot 1) = 9\cdot11(1^2) $ and the second term is $18\cdot 22 = (9\cdot 2)(11\cdot 2) = 9\cdot11 (2^2)$.
Using exactly the same reasoning, you can convert the remaining terms on top to very similar forms, allowing you to express the numerator as $9\cdot11(1^2 + 2^2 + 3^2 + 4^2)$.
Apply analogous reasoning to the denominator (here you're dealing with terms like $22\cdot27(1^2),22\cdot27(2^2)\dots$ etc.
|
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|
Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following equation for $x$:
$$1-x=(1-u)^2$$ $$x=u(2-u)$$
and replacing $u$ by and $x$, and $=$ by $\to$ , which leads to
$$\frac{1}{(1-x)^2}=1+x(2-x)+x^2(2-x)^2+x^3(2-x)^3+... \tag{2}$$
Succumbing to the same 'substitution'approach with $\frac{x}{1+x^2}$, I put
$$\frac{1}{1-x}=\frac{u}{1+u^2}$$ $$x=1-\frac{1}{u}-u$$
so with $x \to 1-\frac{1}{x}-x$
I should have:
$$\frac{x}{1+x^2}=1+\left(1-\frac{1}{x}-x\right)+\left(1-\frac{1}{x}-x\right)^2+\left(1-\frac{1}{x}-x\right)^3+...\tag{3}$$
but this does not seem correct (at least according to my Desmos graph for $|x|<1$).
Can someone please explain what my conceptual errors are?
|
Hint:
Replace $x$ with $-y^2$ to find
$$\dfrac1{1-(-y^2)}=\sum_{r=0}^\infty(-y^2)^r$$
|
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|
Limits: Epsilon and Delta questions on proving limits Suppose you have f(x)= $ \frac1{x^2+1}$ . We want to show that $\lim_{x\to {-1}} \frac1{x^2+1} = \frac12 $.
This is how I approached this issue
Suppose $ \lvert x+1 \rvert \lt \delta $ and $x \ne -1 $.
Then, $ \lvert f(x)-\frac12 \rvert = \lvert \frac1{x^2+1} -\frac12\rvert = \lvert\frac{-x^2 +1}{2(x^2 +1)}\rvert$.
This can be simplified to:
$ \lvert\frac{(1-x)(1+x)}{2(x^2 +1)}\rvert $.
Therefore, let's assume that $ \lvert x+1 \rvert \lt 1 $.
So, $ -2 \lt x \lt 0 $.
Hence, $ 0 \lt -x \lt 2 \iff 1 \lt -x+1 \lt 3 \iff \lvert -x+1 \rvert \lt 3$, and similarly, $ -2 \lt x \lt 0 \iff 0 \lt x^2 \lt 4 \iff 1 \lt x^2 +1 \lt 5 \iff \frac15 \lt \frac{1}{x^2 +1} \lt 1 \iff \lvert \frac1{x^2 +1} \rvert \lt 1. $
Hence, $ \lvert f(x)-\frac12 \rvert = \lvert\frac{(1-x)(1+x)}{2(x^2 +1)}\rvert \lt \frac{3. \lvert x+1 \rvert}{2.1} = \frac32 \lvert x+1 \rvert. $
To conclude, given any $ \epsilon \gt 0 $, let $ \delta$ = min (1, $ \frac{2 \epsilon}3) $.
Any pointers, mistakes or constructive criticism for this proof?
|
Looks good to me. But one can simplify it further by noticing that
$\displaystyle\left|\frac{1}{x^2+1}\right|\leq 1$ for any $x$.
|
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|
Proving That $\sum^{n}_{k=0} \bigl(\frac{4}{5}\bigr)^k < 5$
Using induction, prove that $$\sum_{k=0}^n \biggl(\frac 4 5 \biggr)^k = 1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n<5$$ for all natural numbers $n.$
What I have tried is as follows.
Consider the statement $$P(n):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^n<5.$$
For $n=1$, we have that $\displaystyle P(1):1+\frac{4}{5}<5$ is true.
For $n=k$, we assume that $$\displaystyle P(k):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^k<5.$$
$\displaystyle P(k+1):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^k+\bigg(\frac{4}{5}\bigg)^{k+1}<5+\bigg(\frac{4}{5}\bigg)^{k+1}$
How can I prove that the sum on the left is $< 5?$ Help me please. Thanks.
|
Assume
$$
1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^k < 5
$$
for some positive integer $k$.$\;$Then
\begin{align*}
&
\frac{4}{5}
{\,\cdot}
\left(
1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^k
\right) < \frac{4}{5}{\,\cdot\,}5
\\[4pt]
\implies\;&
\frac{4}{5}+\left(\frac{4}{5}\right)^2+\left(\frac{4}{5}\right)^3+\cdots +\left(\frac{4}{5}\right)^{k+1} < 4
\\[4pt]
\implies\;&
1+\frac{4}{5}+\left(\frac{4}{5}\right)^2+\cdots +\left(\frac{4}{5}\right)^{k+1} < 5
\\[4pt]
\end{align*}
which completes the induction.
|
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|
Smallest positive integer that has $x_{1},x_{2}, \dots, x_{t}$ with $x_1^3 + x_2^3 + \dots + x_t^3 = 2002^{2002}?$ What is the smallest positive integer $t$ such that there exist integers $x_{1},x_{2}, \dots, x_{t}$ with $x_1^3 + x_2^3 + \dots + x_t^3 = 2002^{2002}?$
I don't really get what the question means by "such that there exist integers $x_{1},x_{2}, \dots, x_{t}$ " and I don't know how to start.
|
Evaluating $2002^{2002}$ mod $9$, we get
\begin{align*}
2002^{2002}
\!&\equiv 4^{2002}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4{\,\cdot\,}4^{2001}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4{\,\cdot\,}(4^3)^{667}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4{\,\cdot\,}1^{667}\;(\text{mod}\;9)\\[4pt]
\!&\equiv 4\;(\text{mod}\;9)\\[4pt]
\end{align*}
Noting that $\{x^3\;\text{mod}\;9\mid 0\le x \le 8\}=\{0,1,8\}$, it's easily verified that
$$x_1^3+x_2^3+x_3^3\equiv 4\;(\text{mod}\;9)$$
has no integer solutions.
It follows that the equation
$$x_1^3+x_2^3+x_3^3=2002^{2002}$$
has no integer solutions.
Hence the least qualifying value of $t$ must be greater than $3$.
But identically we have
\begin{align*}
2002^{2002}
&=
2002^{2001}{\,\cdot\,}\,2002\\[4pt]
&=
2002^{2001}{\,\cdot\,}(1000+1000+1+1)\\[4pt]
&=
\left(2002^{667}\right)^3{\,\cdot\,}(10^3+10^3+1^3+1^3)\\[4pt]
\end{align*}
which is a sum of $4$ cubes.
Therefore $t=4$ is the least qualifying value of $t$.
|
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|
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$ I'm having trouble proving that for any $x,y,z>0$ such that $x+y+z=1$ the following inequality is true:
$\frac{3x+1}{x+1}+\frac{3y+1}{y+1}+\frac{3z+1}{z+1} \le \frac{9}{2}$
It seems to me that Jensen's inequality could do the trick, but I'm having trouble finding the right function and the right arguments. Any help is appreciated.
|
Hint:
$$\left(\frac1{x+1}+\frac1{y+1}+\frac1{z+1}\right)(x+1+y+1+z+1) \ge 9$$
And
$$\frac{3x+1}{x+1} = 3 - \frac{2}{x+1}.$$
|
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|
Convergence of $s_{n+1}=\sqrt{1+s_n}$ Does the sequence $s_{n+1}=\sqrt{1+s_n}$ always converge, no matter what the initial value of $s_1$ is?
Is this sequence always increasing and bounded? I think so, but what's throwing me off is that to find what the sequence converges to, we just solve $s^2=1+s$ to get $s=\frac{1+\sqrt{5}}{2}$.
How can the sequence converge to this number if its inital value is $s_1=3$ for example?
|
For the case $s_1 = 3$, then $s_2 = 2 < 3 = s_1$ and $f(x) = \sqrt{1+x}\implies f'(x)=\dfrac{1}{2\sqrt{1+x}}> 0\implies f$ is an increasing function $\implies s_n$ is strictly decreasing sequence and is bounded below by $0$ as $s_n > 0, \forall n\ge 1$, hence is convergent to $L$ which is the solution of $L = \sqrt{1+L}\implies L = \dfrac{1+\sqrt{5}}{2}$ as claimed. In general, $s_1 \ge -1$ to begin with. Now if $\dfrac{1-\sqrt{5}}{2} < s_1 < \dfrac{1+\sqrt{5}}{2}\implies s_2 > s_1$ and the sequence is strictly increasing and is bounded above by $\dfrac{1+\sqrt{5}}{2}$ which can be shown by induction on $n \ge 1$. Thus it converges to $L =\dfrac{1+\sqrt{5}}{2}$ again. If $s_1 = \dfrac{1+\sqrt{5}}{2}$, then $s_n = \dfrac{1+\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L =\dfrac{1+\sqrt{5}}{2}$ also. If $s_1 > \dfrac{1+\sqrt{5}}{2}\implies s_n$ is a decreasing sequence as before and is bounded below by $0$ so is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ because $L \ge 0$. If $s_1 = \dfrac{1-\sqrt{5}}{2}\implies s_n = \dfrac{1-\sqrt{5}}{2}, \forall n \ge 1\implies s_n \to L = \dfrac{1-\sqrt{5}}{2}$ as $n \to \infty$. Finally, if $-1 \le s_1 < \dfrac{1-\sqrt{5}}{2}\implies s_1 >s_2\implies s_n$ is a strictly decreasing sequence, and is bounded below by $0$ hence is convergent to $L = \dfrac{1+\sqrt{5}}{2}$ since $L \ge 0$. This completes the analysis regarding the possible values of the initial term $s_1$.
|
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|
Given a tetrahedron, whose sides are $AB=3,AC=4,BC=5,AD=6,BD=7,CD=8$ . Find the volume of the tetrahedral $ABCD$ .
Given a tetrahedron, whose sides are $AB= 3, AC= 4, BC= 5, AD= 6, BD= 7, CD= 8$ . Find the volume of the tetrahedral $ABCD$ .
Assume that the tetrahedral $ABCD$ has its height $DH$ , whose length I will find by using vectors and the following lemma:
Given three real numbers $f, t, u$ so that $f+ t+ u= 1$ and $H$ on the plane $BCD$ . We'll have
$$\overrightarrow{DH}= f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC}$$
From hypothesis
$$\overrightarrow{DH}\cdot \overrightarrow{AB}= 0\Rightarrow \left ( f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC} \right )\left ( \overrightarrow{DA}- \overrightarrow{DB} \right )= 0$$
$$\Rightarrow f\overrightarrow{DA}\cdot \overrightarrow{DB}+ 49t+ u\overrightarrow{DC}\cdot \overrightarrow{DB}- 36f- t\overrightarrow{DB}\cdot \overrightarrow{DA}- u\overrightarrow{DC}\cdot \overrightarrow{DA}= 0$$
$$49t- 36f+ (f- t)\overrightarrow{DA}\cdot \overrightarrow{DB}+ u\overrightarrow{DC}\cdot \overrightarrow{DB}- u\overrightarrow{DC}\cdot \overrightarrow{DA}= 0$$
On the other hand
$$\cos DCA= \frac{64+ 36- 16}{2\cdot 8\cdot 6}= \frac{7}{8}$$
$$\cos ADB= \frac{36+ 49- 9}{2\cdot 6\cdot 7}= \frac{19}{21}$$
$$\cos DBC= \frac{64+ 49- 25}{2\cdot 8\cdot 7}= \frac{11}{14}$$
Therefore
$$\overrightarrow{DA}\cdot \overrightarrow{DB}= 6\cdot 7\cdot \frac{19}{21}= 38$$
$$\overrightarrow{DC}\cdot \overrightarrow{DB}= 8\cdot 7\cdot \frac{11}{14}= 44$$
$$\overrightarrow{DC}\cdot \overrightarrow{DA}= 8\cdot 6\cdot \frac{7}{8}= 42$$
$$\therefore 49t- 36f+ 38(f- t)+ 44u- 42u= 0\Rightarrow 2f+ 11t+ 2u= 0$$
Similarly
$$\overrightarrow{DH}\cdot \overrightarrow{BC}= 0\Rightarrow \left ( f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC} \right )\left ( \overrightarrow{DC}- \overrightarrow{DB} \right )= 0$$
$$\Rightarrow f\left ( \overrightarrow{DA}\cdot \overrightarrow{DC}- \overrightarrow{DA}\cdot \overrightarrow{DB} \right )+ t\overrightarrow{DB}\cdot \overrightarrow{DC}- t\left | \overrightarrow{DB} \right |^{2}+ u\left | \overrightarrow{DC} \right |^{2}- u\overrightarrow{DC}\cdot \overrightarrow{DB}= 0$$
$$\Rightarrow 4f+ 44t- 49t+ 64u- 44u= 0\Rightarrow 4f- 5t+ 20u= 0$$
A solution to the system of linear equations given by
$$f= \frac{115}{72}, t= -\frac{2}{9}, u= -\frac{3}{8}$$
$$\Rightarrow \overrightarrow{DH}= \frac{115}{72}\overrightarrow{DA}- \frac{2}{9}\overrightarrow{DB}- \frac{3}{8}\overrightarrow{DC}$$
$$\Rightarrow \left | \overrightarrow{DH} \right |^{2}= \frac{115^{2}}{75^{2}}\cdot 36+ \frac{4}{81}\cdot 49+ \frac{9}{64}\cdot 64- \frac{115\cdot 4}{72\cdot 9}\cdot 38+ \frac{2\cdot 6}{9\cdot 8}\cdot 44- \frac{115\cdot 6}{72\cdot 8}\cdot 42= \frac{1199}{36}$$
Is there any way to find the length of $DH$ ? Thanks a real lot.
|
Let the coordinate of $D$ be ($x,y,z)$ and $\triangle ABC$ be on $xy$- plane, where $A$ is the origin, and $B$ and $C$ are on the coordinate axes.
Distance of $D$ from $B,A,C$ are
$$\begin{align}
(x-3)^2+y^2+z^2 &=49\\
x^2+y^2+z^2 &=36\\
x^2+(y-4)^2+z^2&=64
\end{align}$$ respectively.
Solving these equations we get the value of $z$ which is the height.
$$V=\frac{1}{3}(\text{area of $\triangle ABC$})\times z$$
|
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|
Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $ Find maximum $k \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$
for all $a, b, c$ that are distinct positive real numbers ( $a \neq b$, $b \neq c$, $a \neq c$)
Usually when I see this kind of cyclic, symmetrical inequality, the extreme values are taken at $a = b = c$， which is obviously not the case here. So I am not sure how to approach this one..
|
Let $c\rightarrow0^+$.
Thus, $$\frac{a^3}{b^2}+\frac{b^3}{a^2}\geq k(a+b),$$ which gives that $k\leq1.$
We'll prove that $1$ is a maximal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a^3}{(b-c)^2}\geq a+b+c.$$
Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$(u^2-uv+v^2)^2a^3+3(u^3+v^3)(u-v)^2a^2+3(u^4-u^2v^2+v^4)(u-v)^2a+$$
$$+(u+v)(u^2+uv+v^2)(u-v)^4\geq0$$ and we are done!
|
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|
A problem of definite integral inequality? $$
\text { Minimum odd value of $a$ such that }\left.\left|\int_{10}^{19} \frac{\sin x d x}{\left(1+x^{a}\right)}\right|<\frac{1}{9} \text { is (where } a \in N\right)
$$
I proceed this way
As $|\sin x|<1$
Integration $|\frac{\sin x}{1+x^a}|<|\frac{1}{1+x^a}|$
And using trial and error method to get $a=3$
Any other way ??
|
$$\left|\int_{10}^{19} \frac{\sin x \,dx}{1+x^{a}}\right|\le \int_{10}^{19}\left|\frac{\sin x }{1+x^{a}}\right|\,dx\le \int_{10}^{19}\frac{1 }{1+x^{a}}\,dx<\int_{10}^{19}\frac{1}{x^a}\,dx=
\frac{19^{1-a}-10^{1-a}}{1-a}.$$
Since the denominator contains $1-a$, we will assume that $a >1$. For $a=2$,
$$ \frac{19^{-1}-10^{-1}}{-1}=\frac{9}{190}<\frac19.$$
For $a =3$,
$$\frac{19^{-2}-10^{-2}}{-2}=\frac{261}{72200}<\frac19.$$
Therefore, the smallest odd value of $a$ which satisfies the inequality is $a=3$.
|
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Solving a second-order recurrence relation with complex characteristic roots in polar form. I am self-studying this topic from a textbook and am stuck with trying work through one example.
Suppose we are solving the recurrence equation, $u_n = 2u_{n-1} - 2u_{n-2}$.
This has the characteristic equation $r^2 - 2r + 2 = 0$, which has two characteristic complex roots $1\pm i$. (assume this is correct)
The complex roots can be written in three forms (modulus = $\sqrt{2}$, arguments = $\pm\frac{\pi}{4}$).
\begin{align*}
\text{Rectangular: }&1 \pm i \\
\text{Polar: }&\sqrt{2}\left(\cos\frac{\pi}{4} \pm i\sin\frac{\pi}{4}\right) \\
\text{Exponential: }&\sqrt{2}e^{\pm\frac{\pi i}{4}}
\end{align*}
The general solution to this recurrence relation with two roots is: $u_n = A(\text{root}_1)^n + B(\text{root}_2)^n$, where $A$ and $B$ are arbitrary constants (assume this is correct).
I imagine the complex roots can be used in any of their three forms in the general solution, but am having particular trouble with the polar form.
\begin{align*}
u_n &= A\left(\sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\right)^n + B\left(\sqrt{2}\left(\cos \frac{\pi}{4} - i\sin \frac{\pi}{4}\right)\right)^n && \\
&= A (\sqrt{2})^n\left( \cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4} \right) + B (\sqrt{2})^n\left( \cos\frac{n\pi}{4} - i\sin\frac{n\pi}{4} \right) && \text{Using De Moivre's theorem} \\
&= (\sqrt{2})^n\left( (A+B)\cos\frac{n\pi}{4} + (A-B)i\sin\frac{n\pi}{4} \right) && \text{Switching $A\pm B$ for other arbitrary constants} \\
&= (\sqrt{2})^n\left( C\cos\frac{n\pi}{4} + D\begingroup\color{red}i\endgroup\sin\frac{n\pi}{4} \right)
\end{align*}
I am not sure where I went wrong, but this result does not agree with the result in the textbook I am studying from, in that the imaginary number (colored in red) is absent from the textbook result. I might have thought that is a texbook error, but it explicitly draws attention to the fact that using the general solution in polar form includes only real numbers!
The textbook example does not show the derivation in steps and seems to assume this is straightforward. Also it does not explicitly show that complex roots in exponential form can be used (can they?).
|
The sequence you have found is a generalization of the Fibonacci sequence. Referring to a previous answer I have posted here, the general solution can be expressed as
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}$$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$ and $f_{0,1}$ are the initial conditions. Here, I find that $\alpha,\beta=1\pm i$. Notice that when $\alpha,\beta$ are complex conjugates
$$\begin{align}
& \frac{{{\alpha }^{n}}-{{\beta }^{n}}}{\alpha -\beta }=\frac{{{r}^{n}}\left( {{e}^{in\theta }}-{{e}^{in\theta }} \right)}{r\left( {{e}^{i\theta }}-{{e}^{i\theta }} \right)}=\frac{{{r}^{n}}\sin n\theta }{r\sin \theta } \\
& \frac{{{\alpha }^{n}}+{{\beta }^{n}}}{\alpha +\beta }=\frac{{{r}^{n}}\left( {{e}^{in\theta }}+{{e}^{in\theta }} \right)}{r\left( {{e}^{i\theta }}+{{e}^{i\theta }} \right)}=\frac{{{r}^{n}}\cos n\theta }{r\cos \theta } \\
\end{align}
$$
where $\alpha,\beta=re^{\pm i\theta}$. Following through with the substitutions you end up with
$$f_n=\sqrt{2}^n\bigg[(f_1-f_0)\sin\frac{n\pi}{4}+f_0\cos\frac{n\pi}{4}\bigg]$$
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|
An interesting limit
Let $x\in\mathbb{R}.$ For all $i,j\in\mathbb{N},$ define $a_{i0} = \frac{x}{2^i}, a_{ij} = a_{i,j-1}^2 + 2a_{i,j-1}.$ Find, with proof, $\lim\limits_{n\to\infty} a_{nn}.$
Below is my attempt.
Let for each $n, p_n(x) = a_{nn}$. Then observe that $a_{n+1,n} = p_n(\frac{x}2).$ As well, $p_{n+1}(x)+1 = (p_n(\frac{x}2)+1)^2.$ Iterating gives $p_2(x) = ((p_0(\frac{x}4)+1)^2+1)^2, p_3(x) = (((p_0(\frac{x}8)+1)^2+1)^2+1)^2, p_n(x) = ((\dots (p_0(\frac{x}{2^n})+1)^2\dots)^2+1)^2,$ but I'm not sure how this can be converted to a more useful form such as $(1+\frac{x}{2^n})^{2^n}.$
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For $n\in \mathbb{N}$ ,
$a_{n,0}=\frac{x}{2^n}$
$a_{n,1}+1=(a_{n,0}+1)^2=(\frac{x}{2^n}+1)^2$
$a_{n,2}+1=(a_{n,1}+1)^2=(\frac{x}{2^n}+1)^4$
Continuing this way,
For $1\le j\le n$,
$a_{n,j}+1=(\frac x{2^n}+1)^{2^j}$
Thus for $n=j$,
$a_{n,n}+1=(\frac x{2^n}+1)^{2^n}$
Taking limit as $n\to \infty$, we get
$lim_{n\to \infty} a_{n,n}+1=e^x$ ( Since the right limit is a subsequence of the sequence $(1+\frac xn)^n$ converging to $e^x$)
This gives required limit as $e^x-1$.
|
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|
Group of order $1+2+3+ \cdots + n$ with class equation $1+2+3+ \cdots + n$ It is known that the symmetric group $S_3$ of order $1+2+3$ has the class equation $1+2+3$. But a non-abelian group of order $10 = 1+2+3+4$(which is just $D_5$) cannot have the class equation $1+2+3+4$. My question is the following: "are there finite groups $G$ (other than $S_3$) of order $\frac{n(n+1)}{2}(n \in \Bbb N$) with class equation $1+2+3+\dots+n$ "
I've no idea how to find an example of this/disprove this claim. Kindly shed some light on this matter.
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Note that one necessary condition for this to occur is that $\binom{n+1}{2} = \frac{(n+1)n}{2}$ must be divisible by $1, 2, \ldots, n-1, n$, since all the terms in the class equation are indices of centralizers, hence are divisors of the order of the group. Since $n$ and $n-1$ are relatively prime, this means that $n(n-1) \mid \frac{(n+1)n}{2}$, so $n(n-1) \leq \frac{(n+1)n}{2}$. But this implies
\begin{align*}
n-1 \leq \frac{n+1}{2} \implies 2n - 2 \leq n+1 \implies n \leq 3 \, .
\end{align*}
Since $n=2$ corresponds to $\mathbb{Z}/3\mathbb{Z}$ which has class equation $1+1+1$, then $S_3$ and the trivial group are the only examples with class equation $1 + \cdots + n$.
|
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|
Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.
$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}+1}{6}\geqslant
\sqrt[6]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[6]{\frac{1}{108}}\Rightarrow x^3+\frac{1}{x^2}\geqslant 6\sqrt[6]{\frac{1}{108}}-1 $
Sadly $\ 6\sqrt[6]{\frac{1}{108}}-1$ is not correct answer, it is not the minimum.
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A bit late answer but I thought it might be worth noting.
I was wondering whether we could squeeze out the minimum directly from Young's inequality:
$$ab\leq \frac{a^p}{p} + \frac{b^q}{q};\: a,b \geq 0;\: p,q>1 \text{ and } \frac 1p + \frac 1q = 1$$
And, yes indeed, this works well, too. Since we need the powers of $x$ to cancel out, I start constructing the exponents with $x^6$:
\begin{eqnarray*} x^3 + \frac 1{x^2}
& = & \left(x^6\right)^{\frac 12} + \left(\frac 1{x^6}\right)^{\frac 13}\\
& = & \left(\sqrt[5]{x^6}\right)^{\frac 52} + \left(\frac 1{\sqrt[5]{x^6}}\right)^{\frac 53}\\
& = & \frac 25\cdot \left(\sqrt[5]{\frac{5^2}{2^2}}\sqrt[5]{x^6}\right)^{\frac 52} + \frac 35\cdot\left(\sqrt[5]{\frac{5^3}{3^3}}\frac 1{\sqrt[5]{x^6}}\right)^{\frac 53} \\
& \stackrel{Young}{\geq} & \sqrt[5]{\frac{5^2}{2^2}\cdot \frac{5^3}{3^3}} = \frac 5{\sqrt[5]{2^2\cdot 3^3}}
\end{eqnarray*}
Equality holds for $a^p = b^q$ which means
$$\frac 52 x^3 = \frac 53 \frac 1{x^2} \Leftrightarrow x=\sqrt[5]{\frac 23}$$
|
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|
Find the pedal equation of the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ Find the pedal equation of the ellipse $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
My Attempt:
Given equation of ellipse is
$$\frac {x^2}{a^2} + \frac {y^2}{b^2}=1$$
Differentiating both sides wrt $x$
$$\frac {2x}{a^2} + \frac {2y}{b^2} \cdot \frac {dy}{dx} = 0$$
$$\frac {dy}{dx}=-\frac {b^2x}{a^2y}$$
Equation of tangent at point $(x,y)$ is
$$Y-y=-\frac {b^2x}{a^2y} (X-x)$$
$$b^2x X+ a^2y Y-a^2y^2-b^2x^2=0$$
Now, length of perpendicular from $(0,0)$ to the tangent is
$$p=\frac {-(a^2y^2+b^2x^2)}{\sqrt {b^4x^2+a^4y^2}}$$
How to solve further?
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The equation of the tangent to given ellipse at the point $~(X,Y),~$$~\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1~,\tag1$ is$~\dfrac {xX}{a^2} + \dfrac {yX}{b^2} = 1~.\tag2$
Compared the equation of the tangent of the ellipse with $~AX + BY + C = 0 ~,$ we have $~A=\frac {x}{a^2},~B=\frac {y}{b^2}~$ and $~C=-1~.$
Now perpendicular distance of the tangent of the ellipse form $~(0,0)~$ is
$$p=\left|\frac {C}{\sqrt{A^2+B^2}}\right|=\left|\frac {-1}{\sqrt{\left(\frac {x}{a^2}\right)^2+\left(\frac {y}{b^2}\right)^2}}\right|\implies \frac 1{p^2}=\frac {x^2}{a^4} + \frac {y^2}{b^4}~.\tag3$$
We know that $$~r^2=x^2+y^2\implies r^2-b^2=x^2+(y^2-b^2)$$
$$\implies\frac{r^2-b^2}{a^2-b^2}=\frac{x^2+(y^2-b^2)}{a^2-b^2}=\frac{x^2-\frac{b^2}{a^2}x^2}{a^2-b^2}=\dfrac {x^2}{a^2}~.\qquad[\text{using equation $(1)$}]$$
Hence $~\dfrac {x^2}{a^2}=\dfrac{r^2-b^2}{a^2-b^2}~$.
Similarly, $~\dfrac {y^2}{b^2}=\dfrac{a^2-r^2}{a^2-b^2}~$.
Putting these two values in equation $(3)$, we have
$$\frac 1{p^2}=\frac {1}{a^2}\left(\frac{r^2-b^2}{a^2-b^2}\right) + \frac {1}{b^2}\left(\frac{a^2-r^2}{a^2-b^2}\right)$$
$$\implies \frac {a^2b^2}{p^2}=\frac {(r^2b^2-b^4)+(a^4-a^2r^2)}{a^2-b^2}=(a^2+b^2-r^2)$$
Hence the pedal equation of the ellipse $(1)$ is $$ \frac {a^2b^2}{p^2}=(a^2+b^2-r^2)~.$$
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For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that this is divisible by 10 for all odd n, but I don't know how to find the other solutions, if any.
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This can be simplified using the patterns for exponents. $1^n$ always ends in $1$. $2^n$ repeats a pattern where its last digit ends in $2, 4, 8, 6$. $3^n$ repeats a pattern where its last digit ends in $3, 9, 7, 1$. $4^n$ repeats a pattern where its last digit ends in $4, 6$. So adding up our final digits for the four unique cases, we get:
CASE 1: $1+2+3+4 = 10$
CASE 2: $1+4+9+6 = 20$
CASE 3: $1+8+7+4 = 20$
CASE 4: $1+6+1+6 = 14$
Each case corresponding to one of four equivalence classes that the sum can be partitioned into using the equivalence relation that groups the sum by the value each number is being raised to, n, modulus 4.
The cases where the sum of these four least significant digits of the members of the sum are 0 are solutions to our problem. Each of these cases happens 25 times throughout the numbers 1...100 so there are 75 solutions.
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|
Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.
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Following @AdityaDwivedi's suggestion, $dt=4x^{-5}(x^2-1)dx$ so the integral is $\int\frac{dt}{4\sqrt{2+t}}=\tfrac12\sqrt{2+t}+C$. As to how you'd come up with this idea, note the original integral is $\int\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$, which suggests the proposed substitution, or better still $u=2-2x^{-2}+x^{-4}$.
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$f(xy + x +y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $f : \mathbb R \to \mathbb R$ that satisfies both 2 conditions ,
$f(xy + x + y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$ $\forall x,y \in \mathbb R$.
Determine all such $f$.
My solution.
Let $P(x,y) $ be $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $Q(x,y)$ be $f(xy + x + y) = f(xy) + f(x) + f(y)$.
First , I show that $f$ is decreasing $\forall x \in \mathbb R$
Proof : Let $x \gt y$ , from $P(x,y)$ we get that $f(x) \leq f(y)$. (Because $(x - y) \leq 0)$
So, $f$ in decreasing.
$Q(0,0)$ $\to$ $f(0) = 3f(0)$ $\to$ $f(0) = 0$.
$Q(x,-x)$ $\to$ $f(-x^2) = f(-x^2) + f(x) + f(-x)$. $\to$ $f(x) = -f(-x)$.
$P(x,2x)$ $\to$ $xf(x) \geq xf(2x)$. $\to$ $f(x) \geq f(2x)$.
Since $f(x) \geq f(2x)$ , take $x \lt 0$ , we get that $f$ is increasing $\forall x \lt 0$. This implies $f(x) = c$ and $f(x) = -c$ ; $c \in \mathbb R$.
$P(x,y)$ when $x,y \gt 0$ implies $-c = -3c$ , $c = 0$.
So, $f(x) = 0$ $\forall x \in \mathbb R$. Q.E.D.
Is my proof correct? (This is from 2016 Thailand POSN Camp 2)
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Putting $y = 1$,
$f(x + x + 1) = f(x) + f(x) + f(1)$
$f(2x + 1) = 2f(x) + f(1)$
Under the assumption that $f$ is continuously differntiable, you can say the following:
$2f'(2x+1) = 2f'(x) \implies f'(2x+1) = f'(x)$
This gives us,
$f'(x) = f'(2(\frac{x-1}{2}) + 1) = f'(\frac{x-1}{2})$
$f'(\frac{x-1}{2}) = f'(2(\frac{x}{4} - \frac{1}{4} - \frac{1}{2}) + 1) = f'((\frac{x}{4} - \frac{1}{4} - \frac{1}{2}))$
In general,
$f'(x) = f'(\frac{x}{2^n} - \frac{1}{2^n} - \frac{1}{2^{n-1}} - ...-\frac{1}{2})$
That is,
$f'(x) = f'(\frac{x}{2^n} - \frac{2^{n-1} - 1}{2^n})$
Because of continuity of $f'$, we have
$\lim_{n \to \infty}f'(x) = \lim_{n \to \infty}f'(\frac{x}{2^n} - \frac{2^{n} - 1}{2^n})$
$f'(x) = f'(0 - 1 + \frac{1}{2^n}) = f'(-1)$
So, $f'(x) = f'(-1) \implies f(x) = xf'(-1) + c$
Recall, $f(2x + 1) = 2f(x) + f(1)$, putting $x = 0$, we have,
$f(1) = 2f(0) + f(1) \implies f(0) = 0$, so $c = 0$, in above.
$f'(x) = f'(-1)x$, by second condition you have already shown that $f'$ is -ve. So, I wonder if we can say $f'(x) = -cx, c > 0$.
Now, I know that the assumption of continuity and differentiability is a big one and it is quite possible that the given conditions could be used to get there or I could be completely wrong altogether. In any case, I thought I would update what I had so far.
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Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like $\cos^2{x}=1-\sin^2{x}$ after getting a common denominator.
$$4\int_0^{\frac{\pi}{4}} \frac{\sin^2{(5x)\cos^2{x}-\cos^2{(5x)}\sin^2{x}}}{\sin^2{(2x)}} \mathop{dx}$$
Where should I go from here? Any help is appreciated!
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De Moivre's law says that:
$(\cos x + i\sin x)^5 = \cos 5x + i\sin 5x$
To find $\cos 5x, \sin 5x$ we just need to separate the real and imaginary parts of the left hand side.
$\cos^5 x + 5i\cos^4x\sin x - 10\cos^3x \sin^2x - 10i\cos^2x\sin^3 x+ 5\cos x\sin^4x + i\sin^5 x$
$\cos 5x = \cos^5x - 10\cos^3x\sin^2x + 5\cos x\sin^4x\\
\sin 5x = 5\cos^4x\sin x - 10\cos^2x\sin^3x + \sin^5x$
$\frac {\cos^2 5x}{\cos^2 x} = (\cos^4 x - 10\cos^2 x\sin^2 x + 5\sin^4 x)^2\\
\frac {\sin^2 5x}{\sin^2 x} = (\sin^4 x - 10\cos^2 x\sin^2 x + 5\cos^4 x)^2$
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Evaluate integral: $\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x +b^2\sin^2x)dx$? For $a,b>0$, show that
$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$
I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.
For $a>b$ then, we can write the $$ a^2\cos^2x +b^2\sin^2x =b^2\cos^2x+b^2\sin^2x + k\cos^2x = b^2 +k\cos^2x $$ where $ k= a^2-b^2$ and similarly, for $ b>a$ we can write $a^2\cos^2x +b^2\sin^2x = a^2+l\sin^2x$ where $l= b^2-a^2$. Hence we have $$ \int_0^{\frac{\pi}{2}}\ln(b^2+k\cos^2x )dx=\\
\int_0^{\frac{\pi}{2}}\left(\ln(b^2) + \ln\left(1+\frac{k}{b^2}\cos^2x\right)\right)dx\cdots(1)$$ and $$\int_0^{\frac{\pi}{2}}\ln(a^2+l\sin^2x )dx \\= \int_0^{\frac{\pi}{2}}\left(\ln(a^2) + \ln\left(1+\frac{l}{a^2}\sin^2x\right)\right)dx, \; \; b>a$$ Since $\left| \frac{k}{b^2}\cos^2 x\right|\leq 1$ so we use the Machlaurin series of $\ln(1+x)$ for $|x|<1$ giving us $$ \pi \ln(b)+\sum_{m\geq 1} \frac{(-1)^{m-1}}{m}\left(\frac{k}{4b^2}\right)^m\int_0^{\frac{\pi}{2}}\cos^{2m}xdx$$ The latter integral is well know result know as Wallis integral thus gives us $$ \frac{\pi}{2}\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\cdots(2)$$ Recalling the generating function of Central binomial coefficients for $|y| <\frac{1}{4}$, ie $$\sum_{m\geq 1} {2m\choose m} y^{m}=\frac{1}{\sqrt{1-4y}}-1.$$ Dividing by $y$ and an integrating from $0$ to $z$ we have $$ \sum_{m\geq 1}\frac{1}{m}{2m\choose m} z^{m} =-2\ln\left(\frac{\sqrt{1-4z}+1}{2}\right) $$ multiplying thoroughly by $-1$ and set $z=-\frac{k}{4b^2}=-\frac{a^2-b^2}{4b^2}$ gives us $$\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\\ = 2\ln\left(\frac{{a+b}}{2}\right)-2\ln b $$ From $(1)$ and $(2)$ we have $$\pi\ln(b)+\pi\ln\left(\frac{a+b}{2}\right)-\pi\ln b \\= \pi\ln\left(\frac{a+b}{2}\right)$$ Similarly, for the case of $b>a$ we replace $ \cos^2x$ by $\sin^2x$, $k$ by $ l$. We obtained the same desired result. Moreover, we can also have the following result.
$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\frac{1}{2}\int_0^{\pi}\ln(a^2\cos^2x+b^2\sin^2x)dx=\pi\ln\left(\frac{a+b}{2}\right)$$
Now I'm wishing to know other different approaches/proofs for the aforementioned integral.
Thank you.
|
I thought it might be instructive to present an approach that relies on contour integration in the complex plane. To that end, we proceed.
MODIFYING THE INTEGRAL
Let $I(a,b)$ be defined by the integral
$$I(a,b)=\int_0^{\pi/2}\log(a^2\cos^2(x)+b^2\sin^2(x))\,dx\tag1$$
Using the double angle formulas, $\sin^2(x)=\frac{1-\cos(2x)}{2}$ and $\cos^2(x)=\frac{1+\cos(2x)}{2}$, in $(1)$ we obtain
$$\begin{align}
I(a,b)&=\int_0^{\pi/2}\log\left(\frac{a^2+b^2}2+\frac{a^2-b^2}{2}\cos(2x)\right)\,dx\tag2\\\\
&=\frac\pi2 \log\left(\frac{a^2+b^2}{2}\right)+\frac14 \int_{-\pi}^{\pi} \log\left(1+\alpha\cos(x)\right)\,dx\tag3
\end{align}$$
In going from $(2)$ to $(3)$ we enforced the substitution $x\mapsto x/2$ and exploited the evenness of the integrand.
From here, we could use Feynman's trick, differentiate under the integral (tangent half-angle or contour integration), evaluate the integral of the derivative, and finish by integrating. Instead, we will apply contour integration to directly evaluate the integral.
MOVING TO THE COMPLEX PLANE
Next, we transform the integration on the right-hand side of $(3)$ to the complex plane through the substitution $z=e^{ix}$ to obtain
$$\begin{align}
\int_{-\pi}^{\pi} \log\left(1+\alpha\cos(x)\right)\,dx&=\oint_{|z|=1}\log\left(\frac{\alpha(z^2+2z/\alpha+1)}{2z}\right)\,\frac1{iz}\,dz\tag4
\end{align}$$
where $-1<\alpha =\frac{a^2-b^2}{a^2+b^2}<1$.
We select the principal branch of the logarithm with branch cuts emanating from branch points, extending along the negative real axis, and terminating at $z=-\infty$.
Denote the roots of $z^2+2z/\alpha+1=0$ by $r_1$ and $r_2$.
We will assume that $a<b$ so that $-1<r_1=\frac{a-b}{a+b}<0$ lies inside the unit circle $|z|=1$ while $r_2=\frac{a+b}{a-b}<-1$ lies outside the unit circle.
Then, we have from $(4)$
$$\begin{align}
\oint_{|z|=1}\log\left(\frac{\alpha(z^2+2z/\alpha+1)}{2z}\right)\,\frac1{iz}\,dz&=\oint_{|z|=1} \frac{\log\left(-\alpha/2\right)}{iz}\,dz-\oint_{|z|=1} \frac{\log\left(z\right)}{iz}\,dz\\\\
&+\oint_{|z|=1} \frac{\log\left(z-r_1\right)}{iz}\,dz+\oint_{|z|=1} \frac{\log\left(z-r_2\right)}{iz}\,dz\tag5
\end{align}$$
where the negative sign on $\alpha$ is required to ensure that the sum of the arguments of the logarithms is $0$ on the unit circle.
EVALUATING THE INTEGRALS IN $(5)$
From the residue theorem, we see that
$$\oint_{|z|=1} \frac{\log\left(\alpha/2\right)}{iz}\,dz=2\pi \log\left(\alpha/2\right)\tag6$$
From Cauchy's Integral Theorem, we have
$$\begin{align}\oint_{|z|=1}\frac{\log(z)}{z}\,dz&=\lim_{\varepsilon\to0}\left(\int_{-1}^{-\varepsilon}\frac{(\log(-x)-i\pi)-(\log(-x)+i\pi)}{ix}\,dx+\int_{-\pi}^\pi \frac{\log(\varepsilon e^{i\phi})}{i\varepsilon e^{i\phi}}\,d\phi\right)\\\\
&=0\tag7
\end{align}$$
From the residue theorem, we find that
$$\begin{align}
\oint_{|z|=1}\frac{\log(z-r_2)}{iz}\,dz&=2\pi \log(-r_2)\tag8
\end{align}$$
Finally, from Cauchy's Integral Theorem, we have
$$\begin{align}
\oint_{|z|=1}\frac{\log(z-r_1)}{iz}\,dz&=2\pi \log(-r_1)\\\\
&+\lim_{\varepsilon\to0}\left(\int_{-1}^{r_1-\varepsilon}\frac{(\log(-(x-r_1)) -i\pi)-(\log(-(x-r_1))+i\pi)}{ix}\,dx\\\\+\int_{-\pi}^{\pi} \frac{\log(\varepsilon e^{i\phi})}{i(r_1+\varepsilon e^{i\phi})}\,i\varepsilon e^{i\phi}\,d\phi\right)\\\\
&=0\tag9
\end{align}$$
PUTTING IT ALL TOGETHER
Using $(6)-(9)$, we find that
$$\begin{align}
I(a,b)&=\frac\pi2 \log\left(\frac{a^2+b^2}{2}\right)+\frac\pi2 \log(-\alpha/2)+\frac\pi2\log(-r_2)\\\\
&=\pi\log\left(\frac{a+b}{2}\right)
\end{align}$$
which agrees with the expected result!
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|
Calculating $\binom {2016}0 - \binom {2016}3 + \binom {2016}6 - \binom {2016}9 + ... +\binom {2016}{2016}$ Hello everyone how can I calculate this expression
$\binom {2016}0 - \binom {2016}3 + \binom {2016}6 - \binom {2016}9 + ... +\binom {2016}{2016}$?
I tried to mark $\omega = \frac{\sqrt{3}i-1}{2}$ and $\omega^3 = 1$,
But I don't know how to continue.
|
Hint: Instead, let $\omega$ be such that $\omega^6 = 1$ (so $\omega = \frac 12 + i\frac{\sqrt{3}}2$). Let $f(x) = (1 + x)^{2016}$. Consider the sum
$$
f(\omega) + f(\omega^3) + f(\omega^5).
$$
We can calculate the above quantity as
$$
(1 + \omega)^{2016} + (1 + \omega^{3})^{2016} + (1 + \omega^5)^{2016} = \\
(1 + \omega)^{2016} + (1 + \omega^5)^{2016} + (1 - 1)^{2016} = \\
2 \operatorname{Re}[(1 + \omega)^{2016}].
$$
Note that
$$
1 + \omega = \frac 32 + i \frac{\sqrt 3}{2} = \sqrt{3} \left[\frac{\sqrt{3}}{2} + \frac 12i\right] = \sqrt{3}\alpha,
$$
where we note that $\alpha^{12} = 1$. Noting that $2016 = 12 \cdot 168$, we find that
$$
(1 + \omega)^{2016} = 3^{2016/2}\alpha^{2016} = 3^{1008}.
$$
|
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|
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
Is this correct?
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)'=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)=\int \frac{3x^2}{{(2x+1)}^{\frac{3}{2}}} \mathop{dx}$$
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hint
For the last integral, put
$$2x+1=u^2$$
$$x=\frac{u^2-1}{2}$$
$$dx=udu$$
$$3x^2=\frac 34(u^4+1-2u^2)$$
it becomes
$$\frac 34\int \frac{u^4-2u^2+1}{u^3}udu$$
|
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|
$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\frac{2}{(x^2+1)^2}dx$$ and then proceed by using partial fraction decomposition on the first integral. The second integral could be dealt with by substituting $x=\tan \theta$.
Is there a way to evaluate this integral without using partial fractions, and preferably without splitting it into two integrals as I did here?
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To avoid partially fractionizing the integrand, proceed with the substitution below instead
\begin{align}
\int\frac{x^3+3x+2}{(x^2+1)^2(x+1)}\overset{x=\frac{1-y}{1+y}}{dx}
=&\ \frac12\int \frac{y^3-3y^2-3y-3}{(1+ y^2)^2}dy\\
=& \ \frac12\int \frac{(y^2-3)y}{(1+ y^2)^2}-\frac3{1+ y^2}\ dy\\
=& \ \frac1{1+ y^2}+\frac14\ln(1+ y^2)-\frac32\tan^{-1}y
\end{align}
|
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|
Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\sin(x)=b$, $(a^2+b^2=1)$, the expression becomes
$$\frac{4(a^2-b^2)^2-4a^2+3b^2}{4b^2-4a^2b^2}=\frac{4a^4-8a^2b^2+4b^4-4a^2+3b^2}{4b^4}=\bigg(\frac{a^2}{b^2}-1\bigg)^2-\frac{4a^2-3b^2}{4b^4}$$But I don't think I got anything useful... Any help is appreciated.
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Let us express everything in term of $s:=\sin x$:
$$\frac{4(1-2s^2)^2-4(1-s^2)+3s^2}{4s^2-4s^2(1-s^2)}=\frac{16s^4-9s^2}{4s^4}=4-\frac9{4s^2}.$$
|
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|
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 17 \text{ (mod $k$)}$.
Knowing the sum of cubes formula we get that ($\frac{(k-5)(k-4)}{2})^2\equiv 17 \text{ (mod $k$)}$.
From here I'm not sure how I should continue. What would be my options?
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As $\sum\limits_{i=1}^ni^3=\frac14 n^2(n+1)^2$ we consider cases $n=2k$ and $n=2k-1$.
*
*$n=2k$, $\ \frac14 n^2(n+1)^2=(n+5)\left(2 k^3 - 3 k^2 + 8 k + \frac{100}{2 k + 5} - 20\right)$ so $(2k+5)|(100-17)$, but $100-17=83$ is a prime, thus $2k+5=83$ and $n=78$.
2a. $n=4k+1$, $\ \frac14 n^2(n+1)^2-17=(n+5)\left(16 k^3 + 13 k + \frac{83}{2 (2 k + 3)} - \frac{33}{2}\right)$ thus $\frac{83}{2 (2 k + 3)} - \frac{1}{2}$ should be a whole number, thus $\frac{83}{ (2 k + 3)}$ should be odd integer, $(2 k + 3)|83$, $k=40$ and $n=4k+1=161$
2b. $n=4k+3$, $\ \frac14 n^2(n+1)^2-17=(n+5)\left(16 k^3 + 24 k^2 + 25 k + \frac{83}{4 (k + 2)} - 8\right)$ thus $\frac{83}{4 (k + 2)}$ should be a whole number, however, it's not the case as $4\not|83$.
So the answer is $78+161$.
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|
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
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Problem: Let $a, b, c > 0$ with $a^2+b^2+c^2+abc=4$. Prove that $a^3+b^3+c^3+(abc)^3\ge 4$.
Solution: It suffices to prove that, for $a, b, c > 0$,
$$a^3+b^3+c^3+(abc)^3 - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{1}$$
It is verified by Mathematica.
By Vasc's Equal Variable Theorem [1, Corollary 1.9], we only need to prove the case when
$a=b$. The proof is ugly. Omitted. If I find nice proofs, I will update the post.
I hope to see simple SOS (Sum of Squares) solutions of (1).
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
% https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
Remark: The stronger version is also true: for $a, b, c > 0$,
$$a^3+b^3+c^3+2(abc)^2 - abc - 4 - 2(a^2+b^2+c^2+abc - 4) \ge 0. \tag{2}$$
(Note: $(abc)^3 + abc \ge 2(abc)^2$.)
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|
A square is cut into three equal area regions by two parallel lines find area of square. A square is cut into three equal area regions by two parallel lines that are 1 cm apart, each one passing through exactly one of two diagonally opposed vertices. What is the area of the square ?
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Let $AE=x$. By symmetry (equal area etc.) $FC=x$. Let the side of square be $a$. Then $DE=BF=\sqrt{a^2+x^2}$.
Area of the parallelogram $FBED$ = $\sqrt{a^2+x^2} \cdot 1=\sqrt{a^2+x^2}$.
Area of $\triangle DAE$ = Area of $\triangle FCB$ = $\frac{ax}{2}$.
$$\frac{ax}{2}=\sqrt{a^2+x^2} \implies \color{red}{x^2=\frac{4a^2}{a^2-4}}.$$
And area of square is thrice the area of the triangle gives us
$$a^2=\frac{3ax}{2} \implies 2a=3x.$$
Solve these to get $a^2=13$ (area of the square).
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$2^x\leq x+1$ for $x\in [0,1]$ I tried using mean value theorem but couldn't show $2^{2^c} < e$ for $c ∈ (0,x)$. Writing taylor expansion of $2^x$ also don't work because we need to show $2^x$ is smaller than something, not greater
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Since other people have explained the typical way to do this, I'll mention that there's a very tricky way to show $2^x \leq 1 + x$ for $0 \leq x \leq 1$ with an infinite series expansion. But not a Taylor series - an infinite series of binomial coefficients.
If we define the binomial coefficients $\binom{x}{k}$ for non-integer arguments $x$ as $$\binom{x}{0} := 1 \text{ and } \binom{x}{k} := \frac{x(x-1)...(x-k+1)}{k!} \text{ for } k \geq 1,$$
then $$2^x = \sum_{k = 0}^\infty \binom{x}{k}$$ is an absolutely convergent series for $x > 0$, and a conditionally convergent series for $0 > x > -1$†, as we have the asymptotic formula $$\binom{x}{k} \approx \frac{(-1)^k}{\Gamma(-x)k^{x+1}}$$ as $k \rightarrow \infty$. Writing out the first few terms explicitly:
$$2^x = \sum_{k = 0}^\infty \binom{x}{k} = 1 + x + \binom{x}{2} + \binom{x}{3} + ...$$
We are now ready to get our inequality $2^x \leq 1 + x$ for $0 \leq x \leq 1$. In fact, we can prove something stronger: $2^x < 1 + x$ for $0 < x < 1$ (i.e. the inequality is strict everywhere except the endpoints).
For $0 < x < 1$ and $k \geq 1$, the terms $\binom{x}{k}$ are nonzero, alternating in sign, and their absolute values are monotone decreasing towards zero, since $$\binom{x}{k+1} = \frac{x-k}{k+1} \binom{x}{k} = (-1)\frac{k-x}{k+1} \binom{x}{k},$$
and $0 < \frac{k-x}{k+1} < 1$ when $k \geq 1$ and $0 < x < 1$.
Therefore, in the range $0 < x < 1$, $$1 + x + \binom{x}{2} + \binom{x}{3} + ...$$ satisfies the conditions of the alternating series test from the second term onwards. Because in an alternating series, the odd partial sums are upper bounds and the even partial sums are lower bounds when the first term (in this case, $1+x$) is positive, we get a whole family of inequalities for $2^x$, among them the one in the original post:
\begin{align*}
2^x & \leq 1 + x \\
2^x & \geq 1 + x + \binom{x}{2} \\
2^x & \leq 1 + x + \binom{x}{2}+\binom{x}{3} \\
2^x & \geq 1 + x + \binom{x}{2}+\binom{x}{3}+ \binom{x}{4} \\
& ... \\
2^x & \leq 1 + x + \binom{x}{2}+...+\binom{x}{2n-1} \\
2^x & \geq 1 + x + \binom{x}{2}+...+\binom{x}{2n} \\
&... \
\end{align*}
for all $0 \leq x \leq 1$ and all $n \geq 1$, with equality only if $x = 0$ or $x = 1$.
† When $x$ equals $0$, it is a finite series: $\binom{0}{0} = 1$ and $\binom{0}{k} = 0$ for all other $k$, so $2^0 = \sum_{k = 0}^\infty \binom{0}{k} = 1 + 0 + 0 + ... = 1.$ This is true more generally: for $x$ any nonnegative integer, the series will be finite, with all but its first $x+1$ terms equal to $0$.
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|
$\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule I'm trying to calculate $\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule.
The trigonometrical identity $\sin^2(x) = \frac{1-\cos(2x)}{2}$ doesn't seem to lead anywhere. I also attempted to calculate using $\cos^2(x) + \sin^2(x) = 1$ without success.
Any ideas?
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The way you tried to solve:
$(i) \ \displaystyle \sin^2(x) = \displaystyle \dfrac{1-\cos(2x)}{2}$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{1}{2}\left(\dfrac{1-\cos (2 x)}{1-\cos (x)}\right)$$
$$=\dfrac{1}{2} \displaystyle \lim _{x \rightarrow 0} \frac{\cos (0)-\cos (2 x)}{\cos(0)-\cos (x)}=\frac{1}{2} \displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{\sin ^{2} \frac{x}{2}}$$
(Using the identity $\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$ )
Using $(ii) \ \sin^2(\theta)+ \cos^2(\theta)=1$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{1-\sin \left(\frac{\pi}{2}-x\right)}$$
Let $\frac{\pi}{2}-x=2\theta$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{\cos ^{2} (\theta)+\sin ^{2}( \theta)-2 \sin (\theta) \cos (\theta)} =\displaystyle \lim _{x \rightarrow 0} \frac{\sin ^{2} x}{(\cos \theta-\sin \theta)^{2}} $$
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin^2(x)}{\left[\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)-\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]^{2}}$$
Expand the sine and cosine terms with respective formula to again get:
$$\displaystyle \lim _{x \rightarrow 0} \dfrac{\sin ^{2} x}{2\sin ^{2} \frac{x}{2}}$$
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|
Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}}\right)^2\\\\
&=(2x+b+k)(2x+b-k).
\end{align}
Thus,
\begin{align}
I&=4\int_0^{\infty}\frac{\ln x}{(2x+b+k)(2x+b-k)}\,dx\\\\
&=\frac4{2k}\int_0^{\infty}\left(\frac1{2x+b-k}-\frac1{2x+b+k}\right)\ln x\, dx\\\\
&=\frac2k\left[\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b-k}\,dx}_{=I_1}-\underbrace{\int_0^{\infty}\frac{\ln x}{2x+b+k}\,dx}_{=I_2}\right]
\end{align}
For $I_1:$ Letting $2x+b-k=t$ yields
\begin{align}
I_1&=\int_{b-k}^{\infty} \frac{\ln(t-b+k)-\ln 2}t\,dt\\\\
&=\int_{b-k}^{\infty}\frac{\ln(t-b+k)}t\, dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t
\end{align}
Now $\ln(t-b+k)=\ln(k-b)+\ln\left(\frac{t}{k-b}+1\right).$ So,$$I_1=\ln(k-b)\int_{b-k}^{\infty}\frac{dt}t+\int_{b-k}^{\infty} \frac{\ln\left(\frac t{k-b}+1\right)}t\,dt-\ln 2\int_{b-k}^{\infty}\frac{dt}t.$$
If we let: $\frac t{k-b}=-u.$ Then $$I_1=\ln\left(\frac{k-b}2\right)\int_{b-k}^{\infty} \frac{dt}t+\int_1^{\infty}\frac{\ln(1-u)}u\,du.$$
It's getting messier and messier. Also it seems that $I_1$ and hence $I$ diverges. Please tell me, am I heading towards something useful? It has become difficult for me to handle this integral. Please show me a proper way.
|
I will just show how to solve it in the case $b=0$, since the idea is the same, and the calculation becomes easier. Substituting $x=c\tan\theta$, we have:
\begin{equation}
\begin{split}
\int_0^\infty\frac{\ln x}{x^2+c^2}dx&=c^{-1}\int_0^{\pi/2}\ln(c\tan\theta)d\theta\\
&=c^{-1}\int_0^{\pi/2}\ln(c)d\theta+\int_0^{\pi/2}\ln{\sin\theta}d\theta-\int_0^{\pi/2}\ln{\cos\theta}d\theta\\
&=c^{-1}\int_0^{\pi/2}\ln(c)d\theta\\
&=c^{-1}\frac{\pi}2\ln(c)\\
\end{split}
\end{equation}
|
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|
Find the minimum of the set $A=\left\{\int_0^1(t^2 - at-b)^2 dt\, : \,a,b \in \mathbb{R}\right\}$.
Let $$A=\left\{\int_0^1(t^2 - at-b)^2 dt\, : \,a,b \in \mathbb{R}\right\}\,.$$ Find the minimum of $A$.
$\textbf{My attempt:}$
Well, we have
$ 0 \leq\int_0^1(t^2 - at-b)^2 dt = \frac{1}{5} - \frac{a}{2} + \frac{a^2-b}{3} + ab+b^2$.
Ok, I can see that like a funtion $f: \mathbb{R}^2 \to \mathbb{R}$ given by $f(a,b) = \frac{1}{5} - \frac{a}{2} + \frac{a^2-b}{3} + ab+b^2$, then I need to find $(a,b)$ s.t $f(a,b)$ is minimun.
But I don't know how can I do that...Can you give me a hint?
|
Your calculation is incorrect. For $a,b\in\mathbb{R}$, if $f(a,b):=\displaystyle\int_0^1\,(t^2-at-b)^2\,\text{d}t$, then
$$f(a,b)=\frac{a^2}{3}+ab+b^2-\frac{a}{2}-\frac{{\color{red}2}b}{3}+\frac{1}{5}\,.$$
Thus,
$$f(a,b)=\frac{1}{3}\,\left(a+\frac{3(2b-1)}{4}\right)^2+\frac{1}{4}\left(b+\frac{1}{6}\right)^2+\frac{1}{180}$$
for all $a,b\in\mathbb{R}$. This shows that $f(a,b)\geq \dfrac1{180}$ for each pair $(a,b)\in\mathbb{R}^2$. The inequality becomes an equality if and only if
$$a+\frac{3(2b-1)}{4}=0\text{ and }b+\frac{1}{6}=0\,,$$
which is equivalent to
$$(a,b)=\left(1,-\frac16\right)\,.$$
|
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|
Let $a_2,a_3,\cdots,a_n$ be positive real numbers and $s=a_2+a_3+\cdots+a_n$. Show that $\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}I tried this problem in this way:-
Let the real numbers are $b_2,b_3,\cdots,b_n$. such that $(b_2,b_3,\cdots,b_n)$is a permutation of the numbers given $(a_2,a_3,\cdots,a_n)$. Hence $s=b_2+b_3+\cdots+b_n$
Now denote $a_2=b_2,\ a_3=b_3,\cdots ,\ a_n=b_n$
Hence$$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_2^{1-\frac{1}{2}} + b_3^{1-\frac{1}{3}} + \cdots +b_n^{1-\frac{1}{n}}$$
Now denote $a_2=b_n,\ a_3=b_2,\ a_4=b_3,\cdots ,\ a_n=b_{n-1}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_n^{1-\frac{1}{2}} + b_2^{1-\frac{1}{3}} + b_3^{1-\frac{1}{4}}+ \cdots +b_{n-1}^{1-\frac{1}{n}}$$
Now denote $a_2=b_{n-1},\ a_3=b_n,\ a_4=b_2,\cdots ,\ a_n=b_{n-2}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_{n-1}^{1-\frac{1}{2}} + b_n^{1-\frac{1}{3}} + b_2^{1-\frac{1}{4}}+ \cdots +b_{n-2}^{1-\frac{1}{n}}$$
$$\vdots$$ $$\vdots$$
Now denote $a_2=b_{3},\ a_3=b_4,\ a_4=b_5,\cdots ,\ a_{n-1}=b_n,\ a_n=b_{2}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_{3}^{1-\frac{1}{2}} + b_4^{1-\frac{1}{3}} + \cdots +b_{n}^{1-\frac{1}{n-1}}+b_{2}^{1-\frac{1}{n}}$$
Add all these and we get:- $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}} \right)$$
Hence we need to prove $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}} \right)<(n-1)\left(s+\sqrt{s}\right)$$
Now $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}}\right)=\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(b_k^{-\frac{1}{i}}\right)\right)$$
Now let a positive real number $m$ and a positve integer $p$. then $$m^{\frac{1}{p}}\leq \frac{\frac{1}{m}+\overbrace{1+1+\cdots}^{(p-1)\ \text{times}}}{p}$$
Hence $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(b_k^{-\frac{1}{i}}\right)\right)\leq\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(\frac{\frac{1}{b_k}+\overbrace{1+1+\cdots+1}^{(i-1)\ \text{times}}}{i}\right)\right)= \sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n \left(\frac{1}{i}\right) + b_k\sum\limits_{i=2}^n\left(\frac{i-1}{i}\right)\right)=
(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right) + \sum\limits_{k=2}^n \left(b_k\sum\limits_{i=2}^n\left(1-\frac{1}{i}\right)\right)=
(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right) + (n-1)\sum\limits_{k=2}^n b_k -\sum\limits_{k=2}^n\left(b_k\sum\limits_{i=2}^n\left(\frac{1}{i}\right)\right)=(n-1)s+(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right)-\sum\limits_{k=2}^n\left(b_k\sum\limits_{i=2}^n\left(\frac{1}{i}\right)\right)$$
Now i am stuck. Can anyone help me? If you have any other process please mention that also. This problem was in the Old and New Inequalities Volume 2. So please try to be limited in AM-GM and Cauchy-Schwarz.
This problem was proposed by George Tsintsifas in American Mathematical Monthly. So if anyone gives the original solution to this problem i will gladly welcome that.
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We can prove that $a_k^{1-1/k} < a_k + \frac{2}{k} \sqrt{a_k}$.
Indeed, if $a_k \ge 1$, it is obvious;
and if $0 < a_k < 1$, by Bernoulli inequality $(1+x)^r \le 1 + rx$ for $0 < r \le 1$ and $x > -1$, we have
$a_k^{1-1/k} = a_k (a_k^{-1/2})^{2/k} = a_k(1 + a_k^{-1/2} - 1)^{2/k}
\le a_k [1 + (a_k^{-1/2} - 1)\frac{2}{k}]
< a_k + \frac{2}{k}\sqrt{a_k}$.
Thus, by Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
\sum_{k=2}^n a_k^{1-1/k} &< \sum_{k=2}^n a_k + \sum_{k=2}^n
\frac{2}{k} \sqrt{a_k}\\
&= \sum_{k=2}^n a_k + \sqrt{\sum_{k=2}^n \frac{4}{k^2}}\sqrt{\sum_{k=2}^n a_k}\\
&= s + 2\sqrt{\sum_{k=1}^n \frac{1}{k^2} - 1}\ \sqrt{s}\\
&< s + 2\sqrt{s}
\end{align}
where we have used $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$
to get $\sqrt{\sum_{k=1}^n \frac{1}{k^2} - 1} < \sqrt{\frac{\pi^2}{6} - 1} < 1$.
(Q. E. D.)
|
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|
Question about finding roots of a polynomial and studying the nature The number of real roots of the equation $1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{2}}{3}+\cdots+\frac{x^{7}}{7}=0$
(without factorial) is
My work
Let,$\mathrm{f}(\mathrm{x})=1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{x^{6}}{6}$
[Let, f has a minimum at $x=x_{0},$ where then $f^{\prime}\left(x_{0}\right)=0$]
$\Rightarrow 1+x_{0}+x_{0}^{2}+x_{0}^{3}+x_{0}^{4}+x_{0}^{5}=0$
$\Rightarrow \frac{x_{0}^{6}-1}{x_{0}-1}=0$
$\Rightarrow \frac{\left(x_{0}^{3}-1\right)\left(x_{0}^{3}+1\right)}{x_{0}-1}=0$
$\Rightarrow\left(x_{0}^{2}+x_{0}+1\right)\left(x_{0}^{2}-x_{0}+1\right)\left(x_{0}+1\right)=0$
Which has a real root $x_{0}=-1$
But, $f(-1)=1-1+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{6}>0$
The $f(x)>0$ and hence $f$ has no real zeros.
Now let, $g(x)=1+\frac{x}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots+\frac{x^{7}}{7}$
An odd degree polynomial has at least one real root.
If our polynomial g has more than one zero, say $x_{1}, x_{2}$
Then by Role's theorem in $\left(x_{1}, x_{2}\right)$ we have $^{\prime} x_{3}$ ' such that $\mathrm{g}^{\prime}\left(x_{3}\right)=0$
$\Rightarrow 1+x_{3}+x_{3}^{2}+\cdots+x_{3}^{6}=0$
But this has no real zeros. Hence the given polynomial has exactly one real zero.
correct me if i Am wrong
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Define
$$f(x)=1+\sum_{n=1}^7 \frac{x^n}{n}$$
to be the given function. Then we have
\begin{align}
f'(x)
&=\sum_{n=1}^7x^{n-1}\\
&=\cases{\frac{x^7-1}{x-1}&$x\ne1$\\7&$x=1$}\\
\end{align}
But $x^7=1$ has a single real root namely $x=1$ because it's derivative is $7x^6\ge0$. Thus $f'(x)$ has no real roots. So $f(x)$ is strictly increasing and thus has at most one real root. Using the fact that $f(x)\sim x^7/7$ for $|x|\to\infty$ we can conclude that $f(x)$ has exactly one real root (a sign change must occur).
|
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Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ = \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$= \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$
Therefore,
$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$
That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?
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HINT:
Using Euler's formula, we have $\cos(2x)=\frac12(e^{i2x}+e^{-i2x})$.
Now make the substitution $z=e^{i2x}$ with $dx=\frac1{i2z}\,dz$
|
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|
Simplify $\sum^{20}_{k=10} k\binom{k-1}{9}$.
Simplify $$\sum^{20}_{k=10} k\binom{k-1}{9}$$ as much as possible.
I feel like I could utilize the hockey stick identity, but have not found a way to do so with that extra $k$.
Any help would be appreciated.
|
If you couldn't think about how to get rid of the '$k$':
What we need to evaluate is
\begin{align*}
\sum^{20}_{k=10} k\binom{k-1}{9}&=20{19\choose 10}+19{18\choose 9}+\cdots+10{9\choose 9}\\
\end{align*}
Using Hockey-stick identity, we can evaluate
\begin{align*}
\sum^{20}_{k=10} k\binom{k-1}{9}&=10{9\choose 9}+\cdots+2{17\choose 9}+{18\choose 10}\\
\sum^{20}_{k=10} k\binom{k-1}{9}&=\underbrace{\underbrace{{9\choose 9}+{10\choose 9}+\cdots+{18\choose 9}}_{{19\choose 10}+}+\underbrace{{9\choose 9}+{10\choose 9}+\cdots+{17\choose 9}+\cdots}_{{18\choose 10}+\cdots}+\underbrace{{9\choose 9}}_{{10\choose 10}}}_{\color{red}{20\choose 11}}\\
\end{align*}
But we seek $20\displaystyle\sum^{20}_{k=10} \binom{k-1}{9}-\color{red}{20\choose 11}=20{20\choose10}-{20\choose 11}=10{21\choose 11}$.
|
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|
Use linearisation of a certain function to approximate $\sqrt[3]{30}$ Background
Find the linearisation of the function
$$f(x)=\sqrt[3]{{{x^2}}}$$
at
$$a = 27.$$
Then, use the linearisation to find
$$\sqrt[3]{30}$$
My work so far
Applying the formula
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) + f^\prime\left( a \right)\left( {x – a} \right),}$$
where
$${f\left( a \right) = f\left( {27} \right) }={ \sqrt[3]{{{{27}^2}}} }={ 9.}$$
Then, the derivative using the power rule:
$${f^\prime\left( x \right) = \left( {\sqrt[3]{{{x^2}}}} \right)^\prime }={ \left( {{x^{\frac{2}{3}}}} \right)^\prime }={ \frac{2}{3}{x^{\frac{2}{3} – 1}} }={ \frac{2}{3}{x^{ – \frac{1}{3}}} }={ \frac{2}{{3\sqrt[3]{x}}}.}$$
then
$${f^\prime\left( a \right) = f^\prime\left( {27} \right) }={ \frac{2}{{3\sqrt[3]{{27}}}} }={ \frac{2}{9}.}$$
Substitute this in the equation for $L(x)$:
$${L\left( x \right) = 9 + \frac{2}{9}\left( {x – 27} \right) }={ 9 + \frac{2}{9}x – 6 }={ \frac{2}{9}x + 3.}$$
Then, to use this linearisation to find
$$\sqrt[3]{30}$$
I perform the following $\Delta x = x – a = 30 – 27 = 3$ as the condition is $x =30$ and the staring point is $a=27$
As the the derivative of this particular function is given by $f\left( x \right) = \sqrt[\large 3\normalsize]{x}$
$${f’\left( x \right) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\frac{2}{3}\normalsize}} } = {\frac{1}{{3\sqrt[\large 3\normalsize]{{{x^2}}}}},}$$
and its value at point $a$ is equal to
$${f’\left( {a} \right) = \frac{1}{{3\sqrt[\large 3\normalsize]{{{{27}^2}}}}} } = {\frac{1}{{3 \cdot {3^2}}} = \frac{1}{{27}}.}$$
Thus, getting the solution
$${f\left( x \right) \approx f\left( {a} \right) + f’\left( {a} \right)\Delta x,\;\;}\Rightarrow {\sqrt[\large 3\normalsize]{{30}} \approx \sqrt[\large 3\normalsize]{{27}} + \frac{1}{{27}} \cdot 3 } = {3 + \frac{1}{9} } = {\frac{{28}}{9} \approx 3,111.}$$
Is my process correct so far? Or, did I go wrong in the second part? Also, as $a=27$ is from the original linearisation, this would be brought into the linearisation approximation for $\sqrt[3]{30}$?
|
There is something wrong : your definition of $f$ is not consistant : sometimes you use $f(x)=\sqrt[3]{x^2}$ and other times $f(x)=\sqrt[3]{x}$... You should only use the second one I think. Get that right and you got the good idea !
|
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|
How to solve this ODE: $x^3dx+(y+2)^2dy=0$? I am trying to solve $$ x^3dx+(y+2)^2dy=0 \quad( 1)$$
Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$ (1) \to (y+2)^2y'=-x^3 $$
Hence,
$$ \int (y(x)+2)^2y'(x) dy = \int -x^3dx = - \frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integration by parts:
$$ \int (y(x)+2)^2y'(x) dy = y(x)(y(x)+2)^2 - \int y^3dy- \int 4y^2dy + \int4ydy \iff$$
$$ \iff - \frac{y^4}{4} + -\frac13 y^3 + 4y^2 + 8y + c_2$$
But then solving the ODE for $y(x)$ is a struggle:
$$ \iff - \frac{y^4}{4} + -\frac13 y^3 + 4y^2 + 8y = - \frac{3}{4}x + C$$
Any ideas on how I can solve this?
|
Substitute $Y=y+2$ initially. Solving $x^3dx+Y^2dY=0$ yields $Y=\left(C^\prime-\frac{3x^4}4\right)^{\frac13}=y+2$.
|
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|
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
What is the smallest integer $n>1$ for which the mean of the square numbers $1^2,2^2 \dots,n^2 $ is a perfect square?
Initially, this seemed like one could work it out with $AM-GM$, but it doesn't seem so.
From $AM-GM$ one gets that $$\frac{1^2+2^2+ \dots+n^2}{n} \geqslant \sqrt[\leftroot{-1}\uproot{2}n]{1^2\cdot2^2\dots\cdot n^2}
$$
is this of any help here?
Remark. Thanks to Favst, the source of the problem is Problem 1 of 1994 British Mathematical Olympiad Round 2
|
Following a hint by Mostafa Ayaz, we have $(n+1)(2n+1)=6k^2$ for some integer $k$. That is,
$$(4n+3)^2-3(4k)^2=1\,.$$
Hence, $(4n+3)+(4k)\sqrt{3}=(2+\sqrt{3})^m$ for some nonnegative integer $m$. Therefore,
$$4n+3=\sum_{r=0}^{\left\lfloor\frac{m}{2}\right\rfloor}\,\binom{m}{2r}\,2^{m-2r}\,3^r\,.$$
If $m$ is odd, then
$$4n+3\equiv 2m\cdot 3^{\frac{m-1}{2}}\pmod{4}\,.$$
If $m$ is even, then
$$4n+3\equiv 3^{\frac{m}{2}}\pmod{4}\,.$$
Since $$4n+3\equiv 3\pmod{4},$$ we need $$m\equiv 2\pmod{4}\,.$$
That is, $m=4s+2$ for some nonnegative integer $s$
$$4n+3+(4k)\sqrt{3}=(7+4\sqrt{3})\,(97+56\sqrt{3})^s\,.$$
That is, $n=a_s$ and $k=b_s$, where
$$a_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s+(7-4\sqrt{3})\,(97-56\sqrt{3})^s-6}{8}$$
and
$$b_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s-(7-4\sqrt{3})\,(97-56\sqrt{3})^s}{8\sqrt{3}}\,.$$
Note that $a_0=1$, $a_1=337$, and
$$a_s=194\,a_{s-1}-a_{s-2}+144\text{ for }s=2,3,4,\ldots\,.$$
Furthermore, $b_0=1$, $b_1=195$, and
$$b_s=194\,b_{s-1}-b_{s-2}\text{ for }s=2,3,4,\ldots\,.$$
Therefore, the next smallest pair $(n,k)$ apart from the one given by Robert Israel is
$$(n,k)=(a_2,b_2)=(65521,37829)\,.$$
|
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|
Let $\Phi$ be standard Gaussian CDF and $u > 0$. What is good u-bound for $\int_0^1\Phi(u/r - ur)dr$ as a function of $u$? Let the function $\Phi(x) := (1/\sqrt{2\pi})\int_{-\infty}^x e^{-t^2/2}dt$ be the standard Gaussian CDF. For $u > 0$, define $I(u) := \int_0^1 \Phi(u/r-ur)dr$.
Question. What are good upper-bounds for $I(u)$ in terms of $u$ ?
|
A good simple upper bound for $u > 1$:
Using integration by parts and then the substitution $w = \frac{u}{r} - ur$, we have
\begin{align}
I(u) &= \frac{1}{2} + \frac{1}{2u\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{1}{2}w^2}
(\sqrt{w^2+4u^2} - w)\mathrm{d} w\\
&\le \frac{1}{2} + \frac{1}{2u\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{1}{2}w^2}
\left(2u + \frac{w^2}{4u} - w\right)\mathrm{d} w\\
&= 1 - \frac{1}{2u\sqrt{2\pi}} + \frac{1}{16u^2}.
\end{align}
The upper bound $I_1(u) = 1 - \frac{1}{2u\sqrt{2\pi}} + \frac{1}{16u^2}$ is good if $u > 1$.
For example, $\frac{I_1(1)-I(1)}{I(1)} \approx 0.009176113058$,
$\frac{I_1(2)-I(2)}{I(2)} \approx 0.0007002868670$, and $\frac{I_1(10)-I(10)}{I(10)} \approx 0.000001187547307$.
|
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|
Show that the elements of the sequence are divisible by $2^n$ I am trying to prove the following:
Consider the sequence defined by $A_{n+2}=6A_{n+1}+2A_n, A_0=2, A_1=6$. Show that $2^n|A_{2n-1}$ but $2^{n+1}\nmid A_{2n-1}$.
The first terms of this sequence are 2, 6, 40, 252, 1592, 10056, 63520.
In fact, the maximal exponent of $2$ that divides $A_n$ seems to follow a pattern with a period of 4:
1, 1, 3, 2, 3, 3, 5, 4, 5, 5, 7, 6...
But I haven't been able to prove this.
|
We can't do induction on just the odd terms without generalizing something about the even terms.
Let's try a few terms and see why this seems to occur...and to see what else we will need to assure this will always be true:
$2^1|a_{2*1-1} =a_1 = 6$ because ... it does and $2^2\not \mid a_1 = 6$ because ... it doesn't.
$a_{2*2 -1} = a_3 = 6a_1 + 2a_2$. Now $2^2|6a_1$ because $2^1|a_1$ and $2^3\not \mid 2a_1$ because $2^2\not \mid a_1$.
So $2^2|6a_1 + 2a_2$ if and only if $4|2a_2$ or if $2|a_2$ . But if $2|a_2$ but $4\not \mid a_2$ we would have $a_1 = 2odd_1$ and $a_2=2odd_2$ for twe odd integers $odd_1$ and $odd_2$, and $a_3=6a_1 + 2a_2=3*4odd_1 + 4odd_2 = 4(3odd_1+odd_2)$ which would be divisible by $8$ because $3odd_1 + odd_2$ would be even.
So for this to be true we need $2^2|a_2$. And as $a_2 = 6a_0 + 2a_1 = 6*2+2*6=24$ we do have $4|a_2$.
But for our induction step to work we will always need $2^{k+1}|a_{2k}$.
So that's what we are going to need to have induction work:
$P(n) =2^n\mid a_{2n-1}$ but $2^{n+1}\not \mid a_{2n-1}$ and $2^{n+1}\mid a_{2n}$.
Base case: $n = 1$.
This is true for $k=1$ and $2|a_1=6$ but $4\not \mid a_1=6$ and $4|a_2 = 24$.
Induction case: Assume true for $n = k$. Show is true for $n=k+1$
Assuming that is true for some $k$ so that $2^k|a_{2k-1}$ but $2^{k+1}\not \mid|a_{2k-1}$ and $2^{k+1} |a_{2k}$ so there exist an odd integer $b$ and and integer $c$ so that $a_{2k-1} = 2^kb$ and $a_{2k}=2^{k+1}c$ then
$a_{2(k+1)-1}=a_{2k+1} = 6a_{2k-1} + 2 a_{2k}=$
$6*2^{k}b + 2*2^{k+1}c=$
$2^{k+1}(3b + 2c)$
So $2^{k+1}|a_{2(k+1)-1}$ but $2^{k+2}\not \mid a_{2(k+1)-1}$ as $3b+2c$ is odd.
And $a_{2(k+1)}=a_{2k+2}= = 6a_{2k} + 2a_{2k+1}=$
$6*2^{k+1}c + 2(2^{k+1}(3b + 2c))=$
$3*2^{k+2}c + 2^{k+2}(3b+2c)$ .
So $2^{k+2}|a_{2(k+1)}$
So we are done.
|
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|
Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^\circ}..{\sin ^2}{89^\circ}$
$\left( {\frac{{1 - \cos {2^\circ}}}{2}} \right) = {\sin ^2}{1^\circ}$
$T = \left( {\frac{{1 - \cos {2^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {6^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {{10}^\circ}}}{2}} \right)....\left( {\frac{{1 - \cos {{178^\circ}}}}{2}} \right)$
$2 + \left( {n - 1} \right)4 = 178 \Rightarrow n = 45$
$T = \frac{1}{{{2^{45}}}}\left( {1 - \cos {2^\circ}} \right)\left( {1 - \cos {6^\circ}} \right)\left( {1 - \cos {{10}^\circ}} \right)....\left( {1 - \cos {{178^\circ}}} \right)$
$\left( {1 - \cos {{178}^\circ}} \right) = \left( {1 + \cos {2^\circ}} \right)$
$T = \frac{1}{{{2^{45}}}}\left( {1 - {{\cos }^2}{2^\circ}} \right)\left( {1 - {{\cos }^2}{6^\circ}} \right)\left( {1 - {{\cos }^2}{10^\circ}} \right)....\left( {1 - {{\cos }^2}{{86}^\circ}} \right)\left( {1 - \cos {{90}^\circ}} \right)$
$T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}}$
Not able to proceed from here
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The connection of these kind of products with Chybshev polynomials in my opinion is the most important. Because, these polynomials show up everywhere like a unification of many things. Some nice properties of them are listed here. I know so few of them!: https://en.wikipedia.org/wiki/Chebyshev_polynomials
$T_n(x)=2^{n-1}\prod_{k=0}^{n-1}(x-\cos(\frac{2k+1}{2n}\pi))$ and $T_n(0)=(-1)^{\frac{n}{2}}$ when $n$ is even. So we get $2^{n-1}\prod_{k=0}^{n-1}\cos(\frac{2k+1}{2n}\pi)=(-1)^{\frac{n}{2}}$ when $n$ is even. Now let $n=90$. Then $2^{89}\prod_{k=0}^{89}\cos(\frac{2k+1}{180}\pi)=-1$ and thus $2^{89}\prod_{k=0}^{44}\cos^2(\frac{2k+1}{180}\pi)=2^{89}\prod_{k=0}^{44}\sin^2(\frac{2k+1}{180}\pi)=2^{89}\prod_{k=0}^{89}\sin(\frac{2k+1}{180}\pi)=1$.
|
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|
How can this integral be convergent? According to ${\tt Mathematica}$, the following integral converges if
$\beta < 1$.
$$
\int_{0}^{1 - \beta}\mathrm{d}x_{1}
\int_{1 -x_{\large 1}}^{1 - \beta}\mathrm{d}x_{2}\, \frac{x_{1}^{2} + x_{2}^{2}}{\left(1 - x_{1}\right)\left(1 - x_{2}\right)}
$$
How is this possible ?. For $x_{1} = 0$ the integration over
$x_{2}$ hits $1$ on the boundary, so the denominator vanishes and hence the whole expression should diverge.
How can this integral be convergent ?.
|
Changing a little notations (and trying to show the steps), considering
$$I=\int^{1-\beta}_{0}dx \int^{1-\beta}_{1-x} \frac{x^2+y^2}{(1-x)(1-y)}\,dy$$
$$\int \frac{x^2+y^2}{(1-x)(1-y)}\,dy=\frac{\left(x^2+1\right) \log (y-1)+\frac{1}{2} (y-1)^2+2 (y-1)}{x-1}$$
$$J(x)=\int^{1-\beta}_{1-x} \frac{x^2+y^2}{(1-x)(1-y)}\,dy$$ $$J(x)=-\frac{-2 \left(x^2+1\right) \log (-\beta )+2 \left(x^2+1\right) \log (-x)+(x-\beta
) (\beta +x-4)}{2 (x-1)}$$ Integrating this last one with respect to $x$ gives
$$2 \int J(x)\,dx=-4 \text{Li}_2(x)+\log (1-x) \left(\beta ^2-4 \beta +4 \log (-\beta )-4 \log
(-x)+3\right)-x (-(x+2) \log (-\beta )+(x+2) \log (-x)-5)$$
$$2\int^{1-\beta}_{0} J(x)\,dx= -\log (\beta -1) \left(\beta ^2-4 \beta +4 \log (\beta )+3\right)+\log (-\beta )
\left(\beta ^2-4 \beta +4 \log (\beta )+3\right)+(\beta -1) ((\beta -3) \log
(\beta )-5)-4 \text{Li}_2(1-\beta )$$ which does not exist (at least as a real if $\beta >1$.
Now, assuming $\beta < 1$, this reduces to
$$2I=\left(\beta ^2-4 \beta +4 \log (\beta )+3\right) \log \left(\frac{\beta }{1-\beta
}\right)+(\beta -1) ((\beta -3) \log (\beta )-5)-4 \text{Li}_2(1-\beta )$$
|
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|
How to integrate $\int {2\over (x^2+2)\sqrt{x^2+4}}dx$?
Solve the following indefinite integral:
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx$$
My approach:
I used the substitution: $x=2\tan t$, $dx=2\sec^2t dt$
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx=\int \frac{2}{(4\tan^2t+2)\sqrt{4\tan^2t+4}}\cdot 2\sec^2t\ dt$$
$$=\int \frac{4\sec^2t }{2(2\tan^2t+1)2\sec t} dt$$
$$=\int \frac{\sec t}{2\tan^2t+1}dt$$
In numerator I have $\sec t$ but not $\sec^2t$ therefore I can't see a way to take it further. Please help me solve this integral. Thanks in advance.
|
The problem's already solved but just another method.
Substitute $x = \frac{1}{t}$
$$=\int \frac{2}{\big(\frac{1}{t^2} + 2\big) \sqrt{\frac{1}{t^2} + 4}} \frac{-dt}{t^2}$$
$$=\int \frac{2t^3}{(1+2t^2) \sqrt{1+4t^2}} \frac{-dt}{t^2}$$
Now substitute $1+4t^2 = u^2$.
Integral will simplify to
$$\int - \frac{du}{u^2 + 1}$$
$$=- \tan^{-1} u = -\tan^{-1}(\sqrt{1+4t^2}) = -\tan^{-1}\sqrt{1 + \frac{4}{x^2}}$$
$$ = \tan^{-1}\frac{x}{\sqrt{x^2+4}} + C$$
|
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|
Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hence,
$$ (1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = - \sum _{n=1}^{\infty } \frac1{2^{n+1}} = - \sum _{n=2}^{\infty } \frac1{2^{n}} = -\left( \sum _{n=0}^{\infty } \left(\frac12\right)^n -1-\frac12\right) = -2+\frac32 =-\frac12 $$
Therefore,
$$ \boxed{\sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} =-\frac12} $$
I can't spot any algebraic mistakes, thus I assume the sum is correct. But I don't understand the result. How can a real valued sum converge to a negative number?
|
I reckon your original sum is
$$\sum _{n=1}^{\infty }\left(\frac{1}{2^{2n-1}}-\frac{1}{2^{2n}}\right).$$
Of course, this is a GP.
|
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|
Definite integration $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx$ How do I integrate $$\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx\quad ?$$
The actual integral that I encountered is:
$$\int_{-\infty}^\infty dx \left(\frac{N}{\cosh(\frac{\pi }{c}(x-1))}+\frac{1}{\cosh(\frac{\pi}{c}x)} \right) 2 \tan^{-1}\left(\frac{2x-2}{c} \right)$$ where c is a constant with $$\Re c>0$$ Not sure if these two terms makes it easier.
I was trying to solve just the last term, but I couldn't make any progress. Numerical integration gives $\int _{-\infty}^\infty \frac{\tan^{-1}(2x-2)}{\cosh(\pi x)}dx= -1.01334 $. Any hint on how to do it analytically?
|
For $a>0$ and $b\in \mathbb{R}$,
$$\tag{*}\color{blue}{\int_{ - \infty }^\infty {\frac{{\arctan (ax + b)}}{{\cosh \pi x}}dx} = 2\Im\left[ \log\Gamma(\frac{3}{4}+\frac{i (b-i)}{2 a})- \log\Gamma(\frac{1}{4}+\frac{i (b-i)}{2 a})\right]}$$
Here, $\log\Gamma$ is the log gamma function.
To begin, assume $\Im(c)>0, \Re(s)<0, \xi\in \mathbb{R}$, we have the following Fourier transform (hold pointwise except possibly for $\xi=0$):
$$\int_{ - \infty }^\infty {{{(x + c)}^s}{e^{ - 2\pi ix\xi }}dx} = \frac{{{e^{\pi is/2}}}}{{\Gamma ( - s){{(2\pi )}^s}}}{\xi ^{ - s - 1}}{e^{2\pi ic\xi }}{\chi _{(0,\infty )}}(\xi )$$
this can be proved by shifting the path of integration using a parallelogram, then use the result for $\int_0^\infty x^s \exp(-2\pi i x\xi) dx$. Here $\chi_A$ is the characteristic function for set $A$.
Fourier transform of $\text{sech } \pi x$ is itself, Plancherel theorem implies
$$\int_{ - \infty }^\infty {\frac{{{{(x + c)}^s}}}{{\cosh \pi x}}dx} = \frac{{{e^{\pi is/2}}}}{{\Gamma ( - s){{(2\pi )}^s}}}\int_0^\infty {\frac{{{x^{ - s - 1}}{e^{2\pi icx}}}}{{\cosh \pi x}}dx} $$
this holds only for $\Re(s)<0$, a minor modification will make it holds for $\Re(s)<2$:
$$\int_{ - \infty }^\infty {\frac{{{{(x + c)}^s}}}{{\cosh \pi x}}dx} = \frac{{{e^{\pi is/2}}}}{{\Gamma ( - s){{(2\pi )}^s}}}\int_0^\infty {{x^{ - s - 1}}{e^{2\pi icx}}(\frac{1}{{\cosh \pi x}} - 1)dx} +(-ic)^s e^{\pi i s/2} $$
Differentiate both sides with respect to $s$, then put $s=0$ yields
$$\tag{1}\int_{ - \infty }^\infty {\frac{{\log (x + c)}}{{\cosh \pi x}}dx} = - \int_0^\infty {\frac{{{e^{2cix}}}}{x}(\frac{1}{{\cosh x}} - 1)dx} + \log c \qquad \Im(c)>0$$
We claim that
$$\tag{2}\int_0^\infty {\frac{{{e^{ - 2cx}}}}{x}(\frac{1}{{\cosh x}} - 1)dx} = \log \frac{c}{2} + 2\log \Gamma (\frac{1}{4} + \frac{c}{2}) - 2\log \Gamma (\frac{3}{4} + \frac{c}{2}) \qquad c>0$$
It is not difficult to show the Laplace transform of $\text{sech }x$ is $\frac{1}{2} (\psi(\frac{s+3}{4})-\psi(\frac{s+1}{4}))$, therefore, by a property of Laplace transform,
$$\int_0^\infty {\frac{{{e^{ - 2cx}}}}{x}(\frac{1}{{\cosh x}} - 1)dx} = \int_{2c}^\infty {\left[ { - \frac{1}{s} + \frac{1}{2}\left( { - \psi (\frac{{1 + s}}{4}) + \psi (\frac{{3 + s}}{4})} \right)} \right]ds} $$
because $\int \psi(x)dx = \log\Gamma(x)$, $$\small \int_{2c}^R {\left[ { - \frac{1}{s} + \frac{1}{2}\left( { - \psi (\frac{{1 + s}}{4}) + \psi (\frac{{3 + s}}{4})} \right)} \right]ds} = \log (2c) - \log R + 2\log \frac{{\Gamma (\frac{1}{4} + \frac{c}{2})\Gamma (\frac{{3 + R}}{4})}}{{\Gamma (\frac{3}{4} + \frac{c}{2})\Gamma (\frac{1+R}{4})}}$$
making $R\to \infty$ proves $(2)$.
Combining $(1), (2)$ and analytic continuation shows, $$\int_{ - \infty }^\infty {\frac{{\log (x{c^{ - 1}} + 1)}}{{\cosh \pi x}}dx} = 2\log \Gamma (\frac{3}{4} - \frac{{ci}}{2}) - 2\log \Gamma (\frac{1}{4} - \frac{{ci}}{2}) - \log \frac{c}{2} + \frac{\pi }{2}i\qquad \Im(c)>0$$
Taking complex conjugation both sides
$$\int_{ - \infty }^\infty {\frac{{\log (x{c^{ - 1}} + 1)}}{{\cosh \pi x}}dx} = 2\log \Gamma (\frac{3}{4} + \frac{{ci}}{2}) - 2\log \Gamma (\frac{1}{4} + \frac{{ci}}{2}) - \log \frac{c}{2} - \frac{\pi }{2}i\qquad \Im(c)<0$$
WLOG, assume $a>0$, then
$$\int_{ - \infty }^\infty {\frac{{\arctan (ax + b)}}{{\cosh \pi x}}dx} = \Im \left[ {\int_{ - \infty }^\infty {\frac{{\log (\frac{{ia}}{{1 + bi}}x + 1)}}{{\cosh \pi x}}dx} + \log (1 + bi)} \right]$$
(some arguments are needed to justify the separation of $\log$), so after some simplification we have $(*)$.
|
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|
Finding the limit of $f(f(x))$ type problem Let $f$ be a differentiable function and equation
of normal to the graph of $y = f(x)$ at $x = 3$ is
$3y = x + 18$. If $L = \mathop {\lim }\limits_{x \to 1} \frac{{f\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) - f\left( {{{\left( {3 + f\left( 3 \right) - x - 6} \right)}^2}} \right)}}{{{{\sin }^2}\left( {x - 1} \right)}}$ then find the value of L.
My approach is as follow
$y = \frac{{x + 18}}{3} \Rightarrow f\left( x \right) = \frac{{x + 18}}{3}$
$f'\left( x \right) = \frac{1}{3}$
Confused after applyying L.Hosp rule
$L = \mathop {\lim }\limits_{x \to 1} \frac{{f'\left( {3 + {{\left( {4{{\tan }^{ - 1}}x - \pi } \right)}^2}} \right) \times 2\left( {4{{\tan }^{ - 1}}x - \pi } \right) \times \frac{4}{{1 + {x^2}}} - f'\left( {{{\left( {3 + f\left( 3 \right) - x - 6} \right)}^2}} \right) \times 2\left( {3 + f\left( 3 \right) - x - 6} \right) \times - 1}}{{\sin 2\left( {x - 1} \right)}}$
|
Hint:
Normal $3y=x+18$ gives us $y=7$ for $x=3$ so it intersects the curve at $(3,7)$ so $f(3)=7$ and the slope of the normal is $\frac{1}{3}$, so $f'(3)=-3$ (normal being perpendicular to the tangent etc.).
|
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|
Integrating $\int \sqrt{a^2+x^2} \ \mathrm{d} x$ with trig. substitution I am trying to come up with all the formulas I have myself and I stumbled upon a roadblock again.
Integrating $\int \sqrt{a^2+x^2} \ \mathrm{d} x$ with Trig Substitution.
So I imagined a triangle where the hypotenuse is $1$, $\sin(y) = x$ opposite over hypotenuse and $\cos(y) = a$ adjacent over hypotenuse.
Then $dx = \cos(y)dy$ and factoring out the a out of the square root,
I get $ \int a $ $\sqrt{1+\frac{x^2}{a^2}} \ \mathrm{d} x$ or $ \int \sqrt{1+\tan y^2}\cos y^2 \ \mathrm{d} y$
which I could simplify to $\int \cos y \ \mathrm{d} y$. This gives me the wrong answer.
Can you pinpoint or hint at where I am going wrong with my substitution once more?
Apologies, I appreciate it.
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Let $x=a\tan\theta\implies dx=a\sec^2\theta\ d\theta$
$$\int \sqrt{a^2+x^2}dx=\int \sqrt{a^2+a^2\tan^2\theta}\ a\sec^2\theta \ d\theta$$
$$=\int a\sec\theta\ a \sec^2\theta \ d\theta$$
$$=a^2\int \sec^3\theta \ d\theta$$
Using integration by parts
$$=a^2\left(\frac{1}{2}\sec\theta \tan\theta+\frac12\ln|\sec\theta+\tan\theta|\right)+C$$
substituting back to $x$,
$$=\frac{1}{2}\left(x\sqrt{a^2+x^2}+a^2\ln|x+\sqrt{a^2+x^2}|\right)+C$$
|
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|
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
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$P(x)=(x-1)^{100}+(x-2)^{200}=Q(x)×(x^2-3x+2)+ax+b$
$P(1)=(-1)^{200}=a+b, a+b=1$
$P(2)=(1)^{100}=2a+b, 2a+b=1$
$a=0 , b=1$
Hence remainder :- $R(x)=1$
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|
How to make a perfect square from a number given in a surd form $a+b\sqrt{c}$? Is there a way of checking that a number can be written as a perfect square and hence finding it if the number is given in the surd form?
For example, if I expand and simplify $$(1+\sqrt{2})^{2}=3+2\sqrt{2}$$.
Is there a way of finding that perfect square (assuming it has one or after checking it has one) from $3+2\sqrt{2}$?
|
$$(a + b \sqrt{c})^2 = a^2 + b^2 c + 2 a b \sqrt{c}$$
So if you are given $s + t \sqrt{c}$ (with $s, t, c$ rational, $t \ne 0$ and $c$ not a square)
and want to write it in this form, you want to find rationals $a$ and $b$ to solve the equations
$$ \eqalign{a^2 + b^2 c &= s\cr
2 a b &= t\cr}$$
Since $a=0$ won't work, we can write $b = t/(2a)$ and the equation becomes
$$ a^2 + \frac{c t^2}{4a^2} = s $$
which we can solve for $a^2$ and then for $a$:
$$ a = \pm \frac{\sqrt{ 2\,s+2\,\sqrt {s^2-c{t}^{2}}}}{2} $$
That is, $s^2 - c t^2$ must be the square of a rational, and then $2s + 2 \sqrt{s^2 - ct^2}$ must be the square of a rational, and if so we get a solution with this $a$ and $b = t/(2a)$.
|
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|
If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
|
The Alexey Burdin's hint gives the following substitution.
Let $x=4a$, $y=2b$ and $z=4c$.
Thus, $abc=1$ and by AM-GM we obtain:
$$x^2+4xy+4y^2+2z^2=16a^2+32ab+16b^2+32c^2=$$
$$=16(a^2+2ab+b^2+2c^2)\geq16\cdot6\sqrt[6]{a^4b^4c^4}=96.$$
The equality occurs for $a=b=c=1,$ which says that we got a minimal value.
|
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|
How to compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$?
How to compute the following integral?
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$$
So what I did is to change $\sin(x)$ to $\cos(x)$ with cofunction identity, which is $\sin(\frac{\pi}{2} -x) = \cos(x)$. The integral changes into easier:
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^3(x)+ \cos^3(x)} \, \mathrm{d}x$$
And then I divided by $\cos^3(x)$. It will turn everything to
$$\int_{0}^{\frac{\pi}{2}} \frac{\sec^2(x)}{\tan^3(x)+1} \, \mathrm{d}x.$$
And I used $u$-substitution setting $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$ and the bounds $u = \tan(\frac{\pi}{2}) = \infty$ and $u =\tan(0) = 0$ and the integral changed into integral to
$$ \int_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3} $$
This is where I used partial fraction decomposition and my answer is divergent and my answer is wrong:
\begin{align*}
\int_{0}^{\infty} \frac{1}{3(u+1)} + \frac{-u+2}{3(u^2-u+1)} \, \mathrm{d}u
\end{align*}
by this it looks like it will divergent.
The correct answer is $\frac{2\pi}{3\sqrt{3}}$. So, what do I do, next?
Then I did the another method which gives me divergent again,
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sin^3(x)+\cos^3(x)} \, \mathrm{d}x
&= \int_{0}^{\frac{\pi}{2}} \frac{\tan(x)\sec^2(x)}{1+\tan^3(x)} \, \mathrm{d}x
\end{align*}
and $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$, the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{u}{1+u^3} \, \mathrm{d}u$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{-1}{3(u+1)} + \frac{u+1}{3(u^2-u+1)} \, \mathrm{d}u
\end{align*}
and did partial fraction and this gives divergent.
I have no idea what to do next for the first method of work, or the second method of work.
|
Note
\begin{align}
\int_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3}
&=\frac13
\int_{0}^{\infty} \left(\frac{1}{1+u}+ \frac{2-u}{u^2-u+1}\right)du\\
&= \frac13\int_{0}^{\infty} \left(\frac{1}{1+u}-\frac12 \frac{2u-1}{u^2-u+1}+ \frac32\frac1{(u-\frac12)^2+\frac34} \right)du\\
&=\frac13 \left( \ln\frac{u+1}{\sqrt{u^2-u+1}}+\frac1{\sqrt3}\tan^{-1}\frac{2u-1}{\sqrt3}\right)\bigg|_0^\infty=\frac{2\pi}{3\sqrt3}
\end{align}
|
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|
Finding all possible values of $z$ given three conditions. Suppose that $x,y,z$ are positive integers satisfying $x \le y \le z$, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of $z$?
I have only found $2$ sets of positive integers solutions satisfying all conditions, which are $(x, y, z)=(1, 3, 8), (1, 4, 5).$ Would this mean our final solution is $8+5=13?$ I'm not sure if there are any more pairs, or if I did something wrong.
|
$\frac{1}{2} = \frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} \leq \frac{3}{x^2}$.
Therefore, $x^2 \leq 6$ which only leaves $x=1$ and $x=2$
For $x=1$, you get $(x,y,z) = (1,3,8)$ and $(1,4,5)$. I'll leave it to you to prove that these are the only possibilities.
Hint: Use the same technique as above to get an upper bound on $y$ and then try all possible values.
Hint2: $yz = 2 (1+y+z) \Rightarrow \frac{1}{2} = \frac{1}{yz} + \frac{1}{z} + \frac{1}{y} \leq \frac{3}{y}$
For $x=2$, you get $(x,y,z) = (2,2,4)$. I'll let you prove that this is the only possibility
So the sum of all $z$ would be 17.
|
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|
If $f(x)$ is differentiable for all real numbers, then what is the value of $\frac{a+b+c}{2}$?
If $f(x)=\begin {cases} a^2 + e^x & -\infty <x<0 \\ x+2 & 0\le x \le 3 \\ c -\frac{b^2}{x} & 3<x<\infty \end{cases}$, where $a,b,c$ are positive quantities. If $f(x)$ is differentiable for all real numbers, then value of $\frac{a+b+c}{2}$ is
Left hand derivative at $x=0$
$$Lf’(0) =\lim_{h\to 0} \frac{2 - (a^2 +e^{-h})}{h}$$
For limit to exist, $2-a^2=0 \implies a=\pm \sqrt 2$
$$L f’(0)=1$$
Right hand derivative at $x=0$
$$R f’(0) =\lim_{h\to 0} \frac{ h+2 -2}{h} =1$$
Left hand derivative at $x=3$
$$Lf’(3) =\lim_{h\to 0} \frac{5- (3-h+2)}{h}=1$$
And
$$Rf’(3) =\lim_{h\to 0} \frac{ c -\frac{b^2}{3+h}-5}{h}$$
For limit to exist, $c=h$
$$Rf’(3) =\lim_{h\to 0} \frac{b^2}{(3+h)(h)}=\infty$$
Where am I going wrong?
|
First see my comment, and Peter Foreman's comments.
As $x \to 3^-$ (that is, as $x$ approaches 3 from below)
$f'(x) = 1.$
Also, $f(3) = 5.$
As $x \to 3^+$:
$f(x) \to c - \frac{b^2}{3}.$
$f'(x) \to \frac{b^2}{3^2}.$
Thus, $\frac{b^2}{3^2} = 1 \Rightarrow b^2 = 9.$
Further, $c - \frac{b^2}{3} = 5 \Rightarrow c - 3 = 5.$
Addendum: sneaking in intuition expansion through the back door
Taking a bird's-eye view, neither $a$ nor $c$ will have any effect on any LHS or RHS derivatives. Therefore, the derivative constraint will only affect $b.$
Consequently, $a$ and $c$ can be interpreted as "free-form" variables whose only purpose is to raise or lower the corresponding segment of the overall function to ensure continuity.
|
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|
How can I show the quotient of the $k$th partial sums of $\sum\limits_{n=1}^{k} n$ and $\sum\limits_{n=1}^{k} n^2$ is $\frac{3}{2k+1}$? I've found using pen and paper that any trivial case of the sum of a sequence of integers from $1$ to $k$ divided by the sum of the squares of these integers is equal to $\frac{3}{2k+1}$
for example,
$$
\begin{split}
\frac{1+2+3}{1^{2}+2^{2}+3^{2}} &= \frac{3}{7}\\
\frac{1+2+3\ +\ 4}{1^{2}+2^{2}+3^{2}+4^{2}} &= \frac{3}{9}\\
\frac{1+2+3\ +\ 4\ +5}{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}} &= \frac{3}{11}\\
\frac{1+2+3\ +\ 4\ +5+6}{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}} &= \frac{3}{13}
\end{split}
$$
Using a computer I can verify this is true for very large values of $k$ and so I intuitively assume it is true for all values of $k$ but how do I prove this?
I've come up with the following equation to represent this as the quotient of two partial sums but don't know how to get from the left side of the equation to the right.
$$\frac{\displaystyle \sum_{n=1}^k n}{\displaystyle \sum_{n=1}^k n^2} = \frac{3}{2k+1}$$
|
As we know or we can easily prove, $\sum_{n=1}^kn = \dfrac{k(k+1)}{2}$
Now expanding, $(n-1)^3 = n^3 - 3n^2 + 3n - 1$
$n^3 - (n-1)^3 = 3n^2 - 3n + 1$
$ \sum_{n=1}^k n^3 - (n-1)^3 = 3\sum_{n=1}^kn^2 - 3\sum_{n=1}^kn + \sum_{n=1}^k1$
$ k^3 = 3\sum_{n=1}^kn^2 - 3\dfrac{k(k+1)}{2} + k$
$ 3\sum_{n=1}^kn^2 = k^3 + 3\dfrac{k(k+1)}{2} - k$
$ \sum_{n=1}^kn^2 = \dfrac{k(k+1)(2k+1)}{6}$
|
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|
Problem with factoring $x^4-x^3+x^2-x+1$ I want to calculate following integral Using partial fraction: $$\int{1\over x^5+1}dx$$So I decompose the denominator:
$$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$$
For the next step I searched on internet and find out I should decompose$x^4-x^3+x^2-x+1$ like this:
$$x^4-x^3+x^2-x+1=(x^2-ax+1)(x^2-bx+1)$$
And then $a,b$ can be found easily.
My question is Why the coefficients of $x^2,x^0$ are $1$?
Because I can rewrite:
$$x^4-x^3+x^2-x+1=(ax^2+bx+c)(dx^2+ex+f)$$
And only thing I can see in the first look is $ad=1,cf=1$ and I have no clue that why $a=d=c=f=1$
You can see his answer below:
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In general, the two polynomials are given up to multiplication of a constant (you can multiply one by $k$ and other by $1/k$), so you can arrange it in a way that $a=d=1$ is guaranteed. For example $x^2+4x+4$ can be factored as $(x+2)(x+2)$ but also as $(2x+4)(\frac{1}{2}x+1)$. So we are free to fix one of the coefficient to make the answer unique. However if you do this, then you don't have the choice for others, so a correct start here is something like
$$x^4-x^3+x^2-x+1=(x^2−ax+b)(x^2−cx+d).$$
Sure you can do some calculation further to get more information about the constant coefficients, but not before that.
Also following slightly modified example shows that assuming both leading and constant coefficients to be $1$ from the start is wrong:
$$
x^4-x^3+x^2+x+1=\\(x^2+0.86676039+0.46431261)(x^2-1.86676039x+2.15372137)
$$
However, as pointed in the linked other question, in this case it was probably used (but not explained) that the polynomial is palindromic (self-reciprocal), which implies its roots come in pairs $\alpha, \frac{1}{\alpha}$ (it's a result of $x^4f(1/x)=f(x)$). This allows you to expect the factors in a form $$(x-\alpha)(x-\frac{1}{\alpha})=x^2-(\alpha+\frac{1}{\alpha})x+1,$$ or more generic $x^2-ax+1$.
|
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|
Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those forms then, it implies that D is an orthogonal matrix. But how can I prove this? If D is orthogonal, then it must be of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Update: a and b must satisfy that $a^2+b^2=1$
|
To Prove: If A is a 2x2 orthogonal matrix, A is of the form given in the question.
Proof:
For any general non-singular matrix, $A= \begin{pmatrix}a & b \\
c & d\end{pmatrix}$,
$\begin{equation}\det(A) = ad-bc \end{equation}$ and
$A^T= \begin{pmatrix}a & c \\
b & d\end{pmatrix}$.
Using $ \begin{align}A A^T = I \end{align} $, we get $\begin{equation}a^2 + b^2 = c^2 + d^2 = 1\end{equation}$ and $\begin{equation}ac=-bd\end{equation}$.
Using $\begin{align} A^T = A^{-1} \end{align}$, we get $\begin{equation}ad-bc = \frac{d}{a} = \frac{a}{d} = \frac{c}{-b} = \frac{-b}{c}\end{equation}$.
On solving $\begin{equation}d^2 = a^2 \end{equation}$ and $\begin{equation} b^2 = c^2 \end{equation}$ (obtained from above equation) and satisfying in $\begin{equation}ac=-bd\end{equation}$.
You get the required matrices.
|
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|
Evaluate $\int e^{2x}(7+e^x)^{1/2}\,dx$ $\int e^{2x}(7+e^x)^{1/2}\,dx$
Let $u=7+e^x \rightarrow du=e^xdx$
So the integral becomes:
$\int u^{\frac{3}{2}}-7u^{\frac{1}{2}}du$ and so our answer is
$\frac{2}{5}(7+e^x)^{\frac{5}{2}}-\frac{14}{3}(7+e^x)^{\frac{3}{2}}+C$
But this is not what wolfram says. Did I make a mistake?
|
The integral becomes
\begin{align}
\int (u-7)u^{\frac{1}{2}}du
&=\frac{2}{5}(7+e^{x})^{\frac{5}{2}}-\frac{14}{3}(7+e^{x})^{\frac{3}{2}}+C\\
&=\frac{2}{15}(7+e^{x})^{\frac{1}{2}}(3(7+e^{x})^{2}-35(7+e^{x}))+C\\
&=\frac{2}{15}\sqrt{7+e^{x}}(7e^{x}+3e^{2x}-98)+C
\end{align}
as required.
|
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|
Can $\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$ where $R$ and $x$ are positive constants, be solved using substitution? While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$$
where $R$ and $x$ are positive constants.
Can this integral be solved using substitution?
|
Once even I tried to derive this result and got stuck on the same step. After searching many articles and abstracts on internet, I concluded that there is no simple analytical solution of this integral rather a lot of heavy computation is required.
The potential equation was like $$V=\int_0^{2\pi}k\cdot\frac{Q\ d\theta}{2\pi}\cdot\frac{1}{\sqrt{R^2+x^2-2Rx\cos\theta}}=\frac{k\ Q}{\pi\ R}\int_0^{\pi}\frac{d\theta}{\sqrt{a-b\cos\theta}}$$
where $\displaystyle a=1+\frac{x^2}{R^2}$ and $\displaystyle b=\frac{2x}{R}.$
1. One possibility is to express the integrand as a power series in $\cosθ$, and then integrate term by term.
Integrate this series, using $$\int_0^{\pi} \cos^nθ\ dθ=\begin{cases}\frac{(n−1)!!π}{n!!} &, \text{if $n$ is even}\\ 0 &, \text{if $n$ is odd}\end{cases}$$
$$V=\frac{k\ Q}{R}(1+\frac{3}{16}c^2+\frac{105}{1024}c^4+\frac{1155}{16384}c^6+\frac{25025}{4194304}c^8+...)$$
where $\displaystyle c=\frac{b}{a}=\frac{2(x/R)}{(x/R)^2+1}.$
2. We can obtain a power series in $x/R$ .
Consider the expression $\displaystyle \frac{1}{\sqrt{R^2+x^2−2Rx\cosθ}}=\frac{1}{R\sqrt{1+(x/R)^2−2(x/R)\cosθ}}$
Expanding this trinomial, we get a Legendre Polynomial whose co-efficients can be calculated. After integrating, we get,
$$V=\frac{k\ Q}{R}\left(1+\frac{1}{4}(\frac{x}{R})^2+\frac{9}{64}(\frac{x}{R})^4+\frac{25}{256}(\frac{x}{R})^6+\frac{1225}{16384}(\frac{x}{R})^8...)\right)$$
|
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|
Pointwise and uniform convergence of ${\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}x^n}$ I want to check the pointwise and uniform convergence of the below power series.
*
*$\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}x^n}$
We have that \begin{equation*}\sqrt[n]{\left (1+\frac{1}{2n}\right )^{4n^2}}=\left (1+\frac{1}{2n}\right )^{4n}=\left [\left (1+\frac{1}{2n}\right )^{2n} \right ]^2\to e^2=\frac{1}{R} \Rightarrow R=\frac{1}{e^2}\end{equation*}
So we have pointwise convergence for $|x|<\frac{1}{e^2}$.
We check also the boundaries $x=-\frac{1}{e^2}$ and $x=\frac{1}{e^2}$.
At $x=-\frac{1}{e^2}$ we have $\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}\left (-\frac{1}{e^2}\right )^n}$.
At $x=\frac{1}{e^2}$ we have $\displaystyle{\sum_{n=1}^{+\infty}\left (1+\frac{1}{2n}\right )^{4n^2}\left (\frac{1}{e^2}\right )^n}$.
How do we check these series?
|
You have
$$\ln \left[\left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n \right] = 4n^2 \ln \left( 1 + \frac{1}{2n}\right) - 2n = 4n^2 \left( \frac{1}{2n} - \frac{1}{8n^2} + o\left( \frac{1}{n^2}\right)\right) - 2n$$
$$= -\frac{1}{2} + o \left( 1\right)$$
So
$$\lim_{n \rightarrow +\infty} \left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n = \frac{1}{\sqrt{e}}$$
so both $$\sum \left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{-1}{e^2}\right)^n \quad \text{ and } \quad \sum \left( 1 + \frac{1}{2n}\right)^{4n^2} \left( \frac{1}{e^2}\right)^n$$ diverge.
|
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|
How does $3\cos x + 4\sin x$ become $5\cos(x - \arctan\frac{4}{3})$? I'm not sure how any rule is being applied to manipulate the $4\sin x$
None of the double angle/compound angle formulas have the trig functions in this layout
|
First observe that $\sqrt{3^2 + 4^2} = 5,$ and so $\sqrt{\left( \frac 3 5 \right)^2 + \left( \frac 4 5 \right)^2} = 1.$ So there is some angle $\varphi$ for which $\cos\varphi=\frac 3 5$ and $\sin\varphi=\frac 4 5.$ Since the sine and cosine of that angle are both positive, it is in the first quadrant. Its tangent is its sine divide by its cosine, so that is $4/3.$ Hence $\varphi = \arctan \frac 43.$
So
\begin{align}
& 3\cos x + 4\sin x = 5\left( \tfrac 3 5 \cos x + \tfrac 4 5 \sin x \right) \\[8pt]
= {} & 5\left( \cos\varphi \cos x + \sin\varphi \sin x \right) \\[8pt]
= {} & 5\cos(x-\varphi) = 5\cos(x - \arctan \tfrac 4 3).
\end{align}
|
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|
A summation of a series based on the Fibonacci sequence. A sequence is defined as follows: $$a_n=a_{n-1}+a_{n-2}\ \forall \ n\geq3\ ,\ n\in Z. $$
If $a_1=a$ and $a_2=b$, find $$S=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{a_{2i-1}a_{2i+1}}$$ in terms of a and b.
My Approach:
I first found the n-th term as $a_n=F_{n-1}a_2+F_{n-2}a_1$ where $F_i$ is the i-th term of the well-known Fibonacci sequence $1,1,2,3,5,8,13...$
I also noticed that by setting $a=b$ the resulting sum comes to $\frac{1}{a^{2}}\left(\frac{1}{F_1F_3}+\frac{1}{F_3F_5}+\frac{1}{F_5F_7}+...\right)$. And by this theorem,(https://link.springer.com/article/10.1186/s13662-016-0860-0)
The final sum should be, $\frac{1}{a^{2}}*\frac{1}{F_2}=\frac{1}{a^{2}}$.
I also tried expanding the i-th term of $S$ but it did not simplify well.
Another result which may be of use is $\frac{1}{a_{2i-1}a_{2i+1}}=\frac{1}{a_{2i}}\left(\frac{1}{a_{2i-1}}-\frac{1}{a_{2i+1}}\right)$
A solution or any hint/insight on how to solve this problem will be much appreciated.
Thank you
|
Telescopic summation works
Let a=b=1, for simplicity. Use the property of Fibonacci numbers $F_m$:
$$F_{m+1} F_{m+2} -F_m F_{m+3}=(-1)^m.$$ Here $A_m=F_m.$
Let $m=2(n-1)$, we get
$$A_{2n-1}A_{2n} - A_{2n-2} A_{2n+1}=1.$$
Hence $$\frac{1}{A_{2n-1} A_{2n+1}}= \frac{A_{2n-1}A_{2n} - A_{2n-2} A_{2n+1}}
{A_{2n-1} A_{2n+1}}=\frac{A_{2n}}{A_{2n+1}}-\frac{A_{2n-2}}{A_{2n-1}}=B_n-B_{n-1}$$
Telescopic summation:
$$S=\sum_{k=1}^{n} \frac{1}{A_{2n-1} A_{2n+1}}=[(B_1-B_0)+(B_2-B_1)+(B_3-B_2)+...+(B_{n}-B_{n-1}]$$
$$\implies S=B_n-B_0=\frac{A_{2n}}{A_{2n+1}}-0=\frac{A_{2n}}{A_{2n+1}},$$
as $B_0=\frac{F_0}{f_2}=0$
|
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|
Solve $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$ I tried to compute $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$
\begin{equation}\begin{aligned}
&\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x&\\
&=\int_{0}^{\pi} \frac{2(a-1)+4 \sin ^{2} \frac{x}{2}}{(a-1)^{2}+4 a \cdot \sin ^{2} \frac{x}{2}} d x\\
&=2 \int_{0}^{\frac{\pi}{2}} \frac{2(a-1)+4 \sin ^{2} x}{(a-1)^{2}+4 \operatorname{asin}^{2} x} d x\\
&=\frac{2}{a} \int_{0}^{\frac{\pi}{2}}\left(1+\frac{2 a(a-1)-(a-1)^{2}}{(a-1)^{2}+4 a \sin ^{2} x}\right) d x\\
&=\frac{\pi}{a}+\frac{2}{a} \int_{0}^{\frac{\pi}{2}} \frac{a^{2}-1}{(a-1)^{2}+4 a \sin ^{2} x} d x\\
&=\frac{\pi}{a}+\frac{2}{a} \int_{0}^{\frac{\pi}{2}} \frac{\frac{a+1}{a-1} \sec ^{2} x}{1+\left(\frac{a+1}{a-1}\right)^{2} \tan ^{2} x} d x\\
&=\frac{\pi}{a}+\frac{2}{a} \int_{0}^{\infty} \frac{1}{1+u^{2}}d u\\
&=\frac{2 \pi}{a}
\end{aligned}\end{equation}
But I can't find my mistake.
|
Well,your mistake is in the last step, I am assuming that you made the substitution $\frac{a+1 }{a-1}\tan x=u$ and when $-1<a<1$ the limits would be $-\infty$ and $0$.
And your integral would vanish.
|
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|
Find the inequality with the best possible $k= constant$ (with the condition $x^{2}+ y^{2}\leq k$).
Find the inequality with the best possible $constant$
*
*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{7}$. Prove that
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}\leq \frac{3}{1+ \left ( \frac{x+ y}{2} \right )^{2}}$$
where $constant= \frac{2}{7}$ is the best possible.
*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{5}$. Prove that
$$\frac{1}{\sqrt{1+ x^{2}}}+ \frac{1}{\sqrt{1+ y^{2}}}+ \frac{1}{\sqrt{1+ xy}}\leq \frac{3}{\sqrt{1+ \left ( \frac{x+ y}{2} \right )^{2}}}$$
where $constant= \frac{2}{5}$ is the best possible.
They are my two examples. I'm looking forward to seeing more many inequalities alike. Thanks for all your nice comments.
|
Hint.
After rotating CCW the inequality
$$
\frac{3}{\frac{1}{4} (a+b)^2+1}-\frac{1}{a^2+1}-\frac{1}{a b+1}-\frac{1}{b^2+1}\ge 0
$$
we have
$$
\frac{6}{y^2+2}-\frac{2}{-x^2+y^2+2}-\frac{2}{(x+y)^2+2}-\frac{2}{(x-y)^2+2}\ge 0\ \ \ \ \ \ (1)
$$
this has at the equality, the trace in blue shown in the figure below.
Thus to guarantee $x^2+y^2 = k$ with maximum $k$, the circle should be internally tangent to this curve. This could be easily calculated by doing in $(1)$, $x=0$ but this curve has immersed a double zero so we should proceed with
$$
\lim_{x\to 0}\frac{\frac{6}{y^2+2}-\frac{2}{-x^2+y^2+2}-\frac{2}{(x+y)^2+2}-\frac{2}{(x-y)^2+2}}{x^2} = \frac{4-14 y^2}{\left(y^2+2\right)^3}
$$
and thus solving $4-14 y^2=0$ we obtain the value for $k = \frac 27$
Applying the same procedure in the case of
$$
\frac{3}{\sqrt{\frac{1}{4} (a+b)^2+1}}-\frac{1}{\sqrt{a^2+1}}-\frac{1}{\sqrt{a b+1}}-\frac{1}{\sqrt{b^2+1}}\ge 0
$$
we obtain
$$
ineq=\frac{3}{\sqrt{y^2+2}}-\frac{1}{\sqrt{-x^2+y^2+2}}-\frac{1}{\sqrt{(x+y)^2+2}}-\frac{1}{\sqrt{(x-y)^2+2}}\ge 0
$$
and
$$
\lim_{x\to 0}\frac{ineq}{x^2} = \frac{2-5 y^2}{\sqrt{2} \left(y^2+2\right)^{5/2}}
$$
thus obtaining $k = \frac 25$ and also in the case of
$$
\frac{3}{\sqrt[3]{\frac{1}{4} (a+b)^2+1}}-\frac{1}{\sqrt[3]{a^2+1}}-\frac{1}{\sqrt[3]{a b+1}}-\frac{1}{\sqrt[3]{b^2+1}}\ge 0
$$
analogously we obtain
$$
k = \frac{6}{13}
$$
etc.
NOTE
for the $\sqrt[n]{\cdot}$ case we have
$$
k_n = \frac{2n}{3n+4}
$$
|
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|
Morrie's law with sines While it's trivial to prove $\prod_{k=0}^{n-1}\cos(2^kx)=\frac{\sin(2^nx)}{2^n\sin x}$, Wikipedia refers to a "similar" identity $\sin\tfrac{\pi}{9}\sin\tfrac{2\pi}{9}\sin\tfrac{4\pi}{9}=\frac{\sqrt{3}}{8}$. How does this generalize to a result for $\prod_{k=0}^{n-1}\sin(2^kx)$? Failing that, how do we prove this special case?
|
I can show you a proof of the similar identity, but I don't know a general formula for $\prod_{k=0}^{n-1}\sin(2^kx)$.
According to Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$,
we have that
$$\prod _{k=1}^{n-1}\,\sin \left({\frac {k\pi }{n}} \right)=\frac{n}{2^{n-1}} .$$
For $n=9$, by symmetry, we have that
$$\left(\sin \left({\frac {\pi }{9}}\right) \sin \left({\frac {2\pi }{9}}\right)\sin \left({\frac {4\pi }{9}}\right)\right)^2\underbrace{\sin^2\left({\frac {\pi }{3}}\right)}_{3/4}=\prod _{k=1}^{8}\,\sin \left({\frac {k\pi }{9}} \right)=\frac{9}{256} $$
and it follows that
$$\sin \left({\frac {\pi }{9}}\right) \sin \left({\frac {2\pi }{9}}\right)\sin \left({\frac {4\pi }{9}}\right)=\sqrt{\frac{3}{64}}=\frac{\sqrt{3}}{8}.$$
In a similar way, for $n=7$, we show that
$$\sin \left({\frac {\pi }{7}}\right) \sin \left({\frac {2\pi }{7}}\right)\sin \left({\frac {4\pi }{7}}\right)=\frac{\sqrt7}{8}.$$
|
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|
Have I followed through on this proof by induction correctly? We are asked to prove that $1^3+2^3+\cdots+n^3 = (1+2+\cdots + n)^2$ by induction.
Basis: $n=1 \\ 1^3 = 1^2 \quad\checkmark \\ n=2 \\ 1^3+2^3 = (1+2)^2 = 9 \quad\checkmark
$
We rewrite both series using the following summation notation:
$$\sum_{i=1}^{n}i^3 = \left(\sum_{i=1}^{n}i\right)^2$$
Inductive Hypothesis:
We assume that this equality holds for an arbitrary value $k$ such that:
$$\sum_{i=1}^{k}i^3 = \left(\sum_{i=1}^{k}i\right)^2$$
Inductive Step:
We next want to show that the equality holds for $k+1$ by rewriting the sum as:
$$1^3+2^3+\cdots+k^3+(k+1)^3 = \sum_{i=1}^{k}i^3+(k+1)^3=\left(\sum_{i=1}^{k}i\right)^2+(k+1)^3=\left(\sum_{i=1}^{k+1}i\right)^2$$
$$\square$$
|
Your last step is not justified, unless you just want to say that the result is trivial and be a bada%# mathematician. If not, then we need to recall that $$\sum_{i=1}^{k}i=\frac{k(k+1)}{2} $$
Continuing from the last line, $$\begin{align}\bigg(\sum_{i=0}^{k} i \bigg)^2 +(k+1)^3&= \frac{k^2(k+1)^2}{4} +(k+1)^3\\&=\frac{(k+1)^2}{4}\bigg[k^2+4(k+1)\bigg]\\&=\frac{(k+1)^2}{4}\bigg[ k^2+4k+4\bigg]\\&=\frac{(k+1)^2}{4}(k+1+1)^2\\&=\bigg[\frac{(k+1)(k+1+1)}{2}\bigg]^2\\&=\bigg(\sum_{i=0}^{k+1}i\bigg)^2\end{align}$$
|
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|
Fast way to solve $4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$ The question is this:
$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}$
For some reason, I keep on getting 289/3, even though it is the wrong answer. This is from a timed test, and my way is wrong and extremely slow.
|
Use this formula $$\boxed{(a-b)^3 = a^3-3ab(a-b)-b^3}$$
$$4 = \sqrt[3] {x+10}-\sqrt[3] {x-10}\;\;\;/^3$$
$$64 = x+10 -3\sqrt [3]{(x+10)(x-10)}(\underbrace{\sqrt[3] {x+10}-\sqrt[3] {x-10}}_{4})-x+10$$
So $$11 = -3\sqrt [3]{x^2-100}\implies x = \pm\sqrt{{-11^3\over 27}+100}$$
|
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|
Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
taken from here, to solve this problem.
My attempt
I can write the numerator of the expression as
$$
3x +4 = 3(x-1)+7 = \sum_{n\ge0}a_n (x-1)^n
$$
where
$$
a_n=\begin{cases}
7 & n=0 \\
3 & n=1 \\
0 & n>1
\end{cases}
$$
On the other hand, given $|x-1| <2$ we get
\begin{align*}
\frac{1}{x+1} = \frac{1}{2} \cdot \frac{1}{1-\left(\frac{1-x}{2}\right)} = \frac{1}{2} \sum_{k\ge0} \left(\frac{1-x}{2}\right)^n = \sum_{k\ge0}\underbrace{\frac{-1}{(-2)^{k+1}}}_{\color{blue}{b_k}} (x-1)^k
\end{align*}
And then, using equation $(1)$ we get
\begin{align*}
\frac{3x+4}{x+1} &= \left(\sum_{n\ge0}a_n (x-1)^n\right)\left(\sum_{k\ge0}b_k (x-1)^k\right)\\
&= \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(a_0 b_n + a_1 b_{n-1} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(7 \frac{-1}{(-2)^{n+1}} + 3 \frac{-1}{(-2)^{n}} \right)(x-1)^n\\
&= \sum_{n\ge0}\left( \frac{-7 +3(-1)(-2)}{(-2)^{n+1}} \right)(x-1)^n\\
&= \sum_{n\ge0} \frac{-1}{(-2)^{n+1}} (x-1)^n\\
\end{align*}
And this seems to imply that $\frac{3x+4}{x+1} = \frac{1}{x+1}$, at least for $|x-1| <2$, which is clearly not true.
I've gone over the steps, but I don't see where my mistake is. Could anyone tell me where my solution went wrong? Thank you!
|
To make life simpler, let $x=y+1$ $$A=\frac{3x+4}{x+1}=\frac{3y+7}{y+2}=3+\frac 1 {y+2}$$ Let $y=2t$
$$A=3+\frac 12 \frac 1 {1+t}$$
Make the expansion around $t=0$ and go back to $y$ and then $x$
|
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|
$3x^2+y^2 \ge -ax(x+y)$, solve for $a$ so this inequality is true for all $x,y$ I started trying to reorganize the equation
$$3x^2+y^2\ge -ax(x+y)$$
$$3x^2+y^2+ax^2+axy\ge0$$
notice:$$\left(\frac {ax}2+y\right)^2=axy+\frac14x^2a^2+y^2$$
now i should be able to simply add $\frac14x^2a^2$ as this number is always bigger or equal to zero so it wont affect the inequality?
$$x^2(3+a)+\left(\frac {ax}2+y\right)^2\ge0$$
this inequality should hold for all $x,y$ when $3+a \ge 0, a\ge-3$
but this is wrong, where is the mistake?
|
We have that
$$3x^2+y^2\ge-ax(x+y) \iff(3+a)x^2+axy+y^2 \ge 0$$
which requires $a\ge -3$ for $y=0$ and for $y\neq 0$ by $\frac x y =t$ we have
$$\iff y^2\left((3+a)t^2+at+1\right) \ge 0 \iff a^2-4(3+a) \le 0 \iff a^2-4a-12 \le 0$$
which requires $-2\le a \le 6$.
Note that adding the factor $\frac14x^2a^2$ affects the inequality, for $x\neq 0$, for that reason your solution can't be right.
|
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|
Let $r, s, t$ be integers. Suppose $\gcd(a, b) = 1$. Let $d = \gcd(a − b, ra^2 + sab + tb^2 )$. Show that $d|(r + s + t)$. Let $r, s, t$ be integers. Suppose $\gcd(a, b) = 1$. Let $d = \gcd(a − b, ra^2 + sab + tb^2)$. Show that $d|(r + s + t)$.
Any hints for this question? I am stuck with it for days.
|
Since $d$ divides $a - b$, there exists an integer $c$ such that $a = b +cd$. Since we also have $ra^2 + sab + tb^2 \equiv 0 \bmod d$, one gets
$$
ra^2 + sab + tb^2 = r(b +cd)^2 + s(b +cd)b + tb^2 \equiv 0 \bmod d
$$
whence
$$
(r+s+t)b^2 = rb^2 + sb^2 + tb^2 \equiv 0 \bmod d.
$$
Since $\gcd(a,b) = 1$, there exist by Bézout's identity two integers $n$ and $m$ such that $na + mb = 1$. Since $a \equiv b \bmod d$, it follows that
$$
1 = na + mb \equiv nb + mb \equiv (n+m)b \bmod d.
$$
Thus $b^2(n+m)^2 \equiv 1 \bmod d$ and
$$
(r+s+t)b^2(n+m)^2 \equiv (r+s+t) \equiv 0 \bmod d.
$$
Therefore $d$ divides $r+s+t$.
|
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|
A tricky inequality: $n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.$ Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$ be the $n-$th harmonic number,
it is not difficult to prove that:
(1) $n(n+1)^{\frac{1}{n}}<n+H_n,$ for $n>1$; (use AM-GM inequality)
(2) $(n-2)n^{\frac{1}{n-2}}>n-H_n$, for $n>2$. (Hint: $n=3$ is obvious,
when $n>3$, $n-(n-2)n^{\frac{1}{n-2}}<2<H_n$)
So from (1), we know
$$H_n>n(n+1)^{\frac{1}{n}}-n, n>1;\ (3)$$
from (2), we know
$$H_n>n-(n-2)n^{\frac{1}{n-2}},n>2.\ (4)$$
It seems inequality $(3)$ is better than $(4)$, that is to say,
$$\boxed{n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.}$$
My concern is whether there are tricky proofs for above inequality without complicated derivative calculations.
Any helps, hints and comments will welcome.
|
By Bernoulli
$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}=\frac{n}{\left(1+\frac{1}{n+1}-1\right)^{\frac{1}{n}}}+\frac{n-2}{\left(1+\frac{1}{n}-1\right)^{\frac{1}{n-2}}}\geq$$
$$\geq\frac{n}{1-\frac{1}{n+1}}+\frac{n-2}{1+\frac{1}{n-2}\cdot\frac{1-n}{n}}=n+1+\frac{n(n-2)^2}{n^2-3n+1}=$$
$$=2n+\frac{1}{n^2-3n+1}>2n$$
|
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|
Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$
I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?
|
Let $z:=\sqrt{3x-4}\ge0$. We have
$$z+\sqrt[3]{1-z^2}=1$$
or by cubing,
$$(z-1)^3+(1-z^2)=z^3-4z^2+3z=z(z-1)(z-3)=0.$$
This yields three solutions in $x$,
$$\frac43,\frac53,\frac{13}3.$$
|
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|
Proving $\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$
Prove this trigonometric identity:
$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$
I've simplified it until
$$\frac{2\cos^2\theta}{1-\sin\theta}$$
but couldn't get $2+2\tan\theta$ from it.
|
We have that
$$\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}=\frac{\cos\theta(1+\cos\theta)}{1-\sin\theta}+\frac{\cos\theta(\cos\theta-1)}{1+\sin\theta}=$$
$$=\frac{\cos\theta(1+\cos\theta)(1+\sin\theta)+\cos\theta(\cos\theta-1)(1-\sin\theta)}{1-\sin^2\theta}=$$
$$=\frac{(1+\cos\theta)(1+\sin\theta)+(\cos\theta-1)(1-\sin\theta)}{\cos\theta}=$$
$$=\frac{2\cos \theta+2\sin \theta}{\cos\theta}=2+2\tan\theta$$
|
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|
Minimum of a function without calculus. $a=\frac{{(1+t^2)}^3}{t^4}$
Find the the minimum value of $a$. .$$a=\frac{{(1+t^2)}^3}{t^4}$$.
Instead of calculus i tried using the AM-GM inequality.,as follows:we have $$3+\frac{1}{t^4}+\frac{3}{t^2}+3\left( \frac{t^2}{3}\right)\ge 3+{\left(\frac{1}{9}\right)}^{1/5}$$ which is not the minimum i got by using calculus.What mistake am i making using this inequality? And also it would be very helpul if anyone could arrive at the minima using basic inequalities.(withoutv calculus)
using calculus minima occured at ,$t=\sqrt{2}$.
|
A very short proof that the minimal value is $\frac{27}4$ can be given by noting that $$(1+t^2)^3-\frac{27}4 t^4=\frac{(\sqrt 2 - t)^2 (\sqrt 2 + t)^2 (1 + 4 t^2)}4\geq0$$
with equality if and only if $t=\sqrt 2$ or $t=-\sqrt 2$.
EDIT: Systematically we can proceed as follows: For any $a>0$, we have by weighted AM-GM, $$1+t^2=1+ a\frac{t^2}a\geq(1+a)\left(\frac{t^2}a\right)^\frac{a}{1+a}.$$
We want the power of $t$ on the right-hand side to be $\frac43$, so that if we raise it to the third power, it is of the same power as $t^4$. For this, we have to have $\frac{a}{1+a}=\frac23$ which is equivalent to $a=2$. If we use the inequality for $a=2$, we get $$1+t^2\geq\frac{3}{2^\frac23} t^\frac43.$$
It follows that $$(1+t^2)^3\geq \frac{3^3}{2^2} t^4=\frac{27}4 t^4.$$
(Equality iff $1=\frac{t^2}2$.)
|
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|
Finding cube roots of a unity - proper explanation is needed SQUARE ROOT
Let's say that I have:
\begin{align*}
x^2 &= 9
\end{align*}
And I $\sqrt{\phantom{x}}$ an entire equation to get:
\begin{align*}
x_{1,2} &= \sqrt{9}\\
x_{1} &= +3\\
x_{2} &= -3
\end{align*}
Experience and "fundamental theorem of algebra" thought me that:
There are always two square roots ($x_1$, $x_2$)!
Experience thought me that:
Product of mixed roots doesn't return back the original:
\begin{align*}
x_1 \cdot x_2 &= -3 \cdot 3 = -9\\
x_2 \cdot x_1 &= 3 \cdot (-3) = -9
\end{align*}
Experience also thought me that:
Only if roots are multiplied by itself they return back the result:
\begin{align*}
x_1 \cdot x_1 &= 3 \cdot 3 = 9\\
x_2 \cdot x_2 &= (-3) \cdot (-3) = 9
\end{align*}
CUBE ROOT
Then there comes this:
\begin{align*}
x^3 &= 1\\
\end{align*}
If I $\sqrt{\phantom{x}}$ an entire equation I get:
\begin{align*}
x_{1,2,3} &= \sqrt[3]{1}\\
x_1 &= 1\\
x_2 &= \text{???}\\
x_3 &= \text{???}
\end{align*}
Here I have no experience whatsoever but "fundamental theorem of algebra" says that:
There are always three cube roots ($x_1$, $x_2$, $x_3$)!
Now pay attention! Like for square root it should be true that:
If roots $x_1$, $x_2$, $x_3$ are multiplied by itself they return $1$. Their mixed products will probably return something else.
\begin{align*}
x_1 \cdot x_1 \cdot x_1 = 1\\
x_2 \cdot x_2 \cdot x_2 = 1\\
x_3 \cdot x_3 \cdot x_3 = 1
\end{align*}
So at this point everybody's already rushing to their paper and drawing this complex plane:
And in complex plane:
If we multiply complex numbers, length of their position vectors multiply and their angles sum up.
So everybody will now draw $x_1$ on the $\mathbb{R}$ axis and tell me that this is the first root. I agree with that. Then they say:
Length of a position vector of other two roots ($x_2$, $x_2$) must be $1$ as well.
And I agree with that. But what they say now, I can't agree with no matter how I look at it. They say:
Because according to "fundamental theorem of algebra" there are three cube roots, and we will multiply them in order to get to $1$, angle between them must be $120^\circ$. We now start at $1$ and if we do a full circle we end at $1$ again. Woila! Problem solved...
I can't agree with this explanation as is because this only holds for $x_1$. If I take $x_1$ and make a full circle I end up getting $1$. But with $x_2$ and $x_3$ my starting point is not $1$. Starting points are actually:
\begin{align*}
\text{starting point for $x_2$:} \hspace{5mm} -\frac{1}{2} - \frac{\sqrt{3}}{2}i \\
\text{starting point for $x_3$:} \hspace{5mm} -\frac{1}{2} + \frac{\sqrt{3}}{2}i
\end{align*}
And in case of $x_2$ and $x_3$ when we do a full circle we don't end up getting $1$, but we geometricaly looking end at positions:
\begin{align*}
-1/2 - \sqrt{3}/2 i \\
-1/2 + \sqrt{3}/2 i
\end{align*}
If the starting positions are as I state they are, angle that we need in order to get to the $1$ should be $\pm240/3 = \pm80^\circ$.
But then why does this explanation that they use, work? At the end of the day e.g. $x_2$ solves the problem:
\begin{align*}
\left(x_2\right)^3 &= 1\\
\left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right)^3 &= 1\\
\left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right) \left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right) \left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right)&= 1\\
\left(\frac{1}{4} - \frac{\sqrt{3}}{4}i - \frac{\sqrt{3}}{4}i + \frac{3}{4}i^2\right) \left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right) &= 1\\
\left(\frac{1}{4} - \frac{2\sqrt{3}}{4}i - \frac{3}{4}\right) \left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right) &= 1\\
\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \left( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \right) &= 1\\
\frac{1}{4} - \frac{\sqrt{3}}{4} i + \frac{\sqrt{3}}{4} i - \frac{3}{4} i^2 &= 1\\
\frac{1}{4} + \frac{3}{4} &= 1
\end{align*}
I am sure there is something wrong with the explanation everybody are using but it is correct to assume roots are $120^\circ$ appart...
Probably the point here is not about making a full circle at all! Like most of the people say. It must be about adding conjugate pairs so that artifficialy we create enough roots to satisfy fundamental theorem of algebra! Thats all! So odd number root $3$ was historicaly first one that forced us to get out of the horizontal axis in complex plane and invent complex roots...
Can anyone present a proper explanation of why we choose $120^\circ$ for cube roots?
|
There is really no "first" root, "second" root, etc.
In the case of $x^2=1$, there are two roots, $-1$ and $1$. It is your choice to give these two roots whatever names you like.
In the case of $x^3=1$, there are three distinct roots. Again, it does not matter which one you call $x_1$.
In general, (complex) solutions to the equation $x^n=1$ together with complex multiplication form a group: https://en.wikipedia.org/wiki/Root_of_unity
This group tells you what happens when you multiply two solutions to the equations $x^n=1$.
I don't know where the "everybody explanation" is from; I would not do that anyway.
One way to find the three roots for $x^3=1$ is quite straightforward. Write $x=re^{i\theta}$ where $r\ge 0$ and $\theta\in[0,2\pi)$. Then
$$
(re^{i\theta})^3=1
$$
Taking the absolute value on both sides you get $r=1$. On the other hand,
$$
e^{i(3\theta)}=1
$$
which implies that $3\theta=2k\pi$ where $k$ is some integer. Now write
$$
\theta=\frac{2k\pi}{3},\quad k\in\mathbb{Z}
$$
and pick the ones that are in $[0,2\pi)$. By periodicity, this is not a unique way to represent the roots. You could, for example, restrict $\theta$ to $[-\pi,\pi)$ at the beginning of the calculation.
|
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|
There are triads of perfect squares that are consecutive terms of arithmetic progression? Prove that there exist infinitely many triples of positive integers $ x , y , z $ for which the numbers $ x(x+1) , y(y+1) , z(z+1) $ form an increasing arithmetic progression.
$ \bigg( $ It is equivalent to find all triples of $ 4x(x+1)+1=(2x+1)^{2} , 4y(y+1)+1=(2y+1)^{2} , 4z(z+1)+1=(2z+1)^{2} $ $ \bigg) $
Note : I know $ \big( 1^{2} , 5^{2} , 7^{2} \big) $ , $ \big( 7^{2} , 13^{2} , 17^{2} \big) $ , $ \big( 7^{2} , 17^{2} , 23^{2} \big) $ , $ \big( 17^{2} , 25^{2} , 31^{2} \big) $ , but how i can found all triples ?
|
Looking at your examples,
$$ 2(5^2)=1^2+7^2=(4-3)^2+(4+3)^2 \rightarrow (3,4,5) $$
$$ 2(13^2)=7^2+17^2=(12-5)^2+(12+5)^2 \rightarrow (5,12,13) $$
$$ 2(17^2)=7^2+23^2=(15-8)^2+(15+8)^2 \rightarrow (8,15,17) $$
$$ 2(25^2)=17^2+31^2=(24-7)^2+(24+7)^2 \rightarrow (7,24,25) $$
Use any Pythagorean triplet such that $a^2+b^2=c^2$, then the identity at work is :
$$ (a-b)^2+(a+b)^2=2(a^2+b^2)=2(c^2) $$
So start with the general form of Pythagorean triplet and you may arrive at your desired construction.
Addendum :
$m^2+n^2=2(p)^2$ satisfying triplets make automedian triangles.
|
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|
To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\frac{1}{2}\sin C \\c=2\sin C \Rightarrow \frac{1}{c}=\frac{1}{2}*\sin C $$
so, $$\frac{1}{c}=ab \Rightarrow abc=1 \Rightarrow \sqrt{abc}=1$$
now the problem becomes
$$ab+bc+ac > \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}},$$
with $0<a\leq b\leq c\leq 2$, and $a+b>c$.
But even so I don't know how to prove it.
Any help or hint is appreciated. Thank you.:)
|
$ \sum_{cyc}\frac{1}{a} = \frac{1}{2} \sum_{cyc}\frac{1}{a} + \frac{1}{b} \ge \sum_{cyc}\frac{1}{\sqrt {ab}} = \sum_{cyc}\sqrt {a} \,$ (using AM-GM and $abc = 1)$
EDIT: I think I missed an important point earlier that when $a = b = c = 1$, R = $\frac{1}{2\sin 60^0} = \frac{1}{\sqrt3}$ which does not meet the condition. Hence the equality is not possible.
|
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|
How prove this limit via epsilon delta? $$\lim_{(x,y) \to (4,1)}{\frac{y}{2x-y}}=\frac{1}{7}$$
I know so far that $|x-4|<\delta\quad \&\quad |y-1|<\delta$, and that I can use
$$\bigg|\frac{y-1}{2x-y} + \frac{1}{2x-y} - \frac{1}{7}\bigg|$$ and I get one delta, but how to continue from here, or is it even correct?
|
To simplify we can change coordinates $u=x-4$ and $v=y-1$ such that
$$\lim_{(x,y) \to (4,1)}{\frac{y}{2x-y}}=\lim_{(u,v) \to (0,0)}{\frac{v+1}{2u-v+7}}=\frac{1}{7}$$
and assuming wlog $|u|,|v|\le 1$
$$\left|\frac{y-1}{2x-y} - \frac{1}{7}\right|=\left|\frac{v+1}{2u-v+7} - \frac{1}{7}\right|=\left|\frac{8v-2u}{7(2u-v+7)}\right|\le$$
$$\le\frac87\left|\frac{v}{2u-v+7}\right|+\frac27\left|\frac{u}{2u-v+7}\right|\le$$
$$\le \frac87\left|\frac{v}{4}\right|+\frac27\left|\frac{u}{4}\right| \le \frac87\frac{\delta}{4}+\frac27\frac{\delta}{4}=\frac{5}{14} \delta$$
|
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|
Let $\{a_n\}$ be a sequence of real numbers such that $a_1=2$, $a_{n+1} = a_n^2 -a_n+1$, for $n=1,2,3..$. [Cont]
Let $\{a_n\}$ be a sequence of real numbers such that $a_1=2$, $a_{n+1} = a_n^2 -a_n+1$, for $n=1,2,3..$. Let $S=\frac{1}{a_1}+\frac{1}{a_2} ....+\frac{1}{a_{2018}}$, then prove that
*
*$S>1-\frac{1}{2018^{2018}}$
*$S<1$
*$S>1-\frac{1}{2017^{2017}}$
$$a_{n+1}-1 = a_n^2-1-a_n+1$$
$$a_{n+1} -1 = (a_n-1)(a_n)$$
$$\frac{1}{a_{n+1}-1} =\frac{1}{a_n-1} -\frac{1}{a_n}$$
$$\frac{1}{a_n}=\frac{1}{a_n-1} -\frac{1}{a_{n+1}-1}$$
So $$S=1-\frac{1}{a_{2019}-1}$$
How do I proceed from here?
|
You have already shown that
$S = 1-\frac{1}{a_{2019}-1}$
So, $S \gt 1-\frac{1}{2018^{2018}}$ if $\, a_{2019} \gt {2018}^{2018} + 1$
By induction -
As $a_{2} = {a_1}^2 - a_1 + 1 = 3 \gt 1^1 + 1$
and $a_{3} = {a_2}^2 - a_2 + 1 = 7 \gt 2^2 + 1$
For a value of $n \ge 3$,
If $a_{n+1} \gt n^n + 1$, we need to show that $a_{n+2} \gt (n+1)^{n+1} + 1$
$a_{n+2} \ge (n^n + 1)^2 - (n^n+1) + 1 = n^{2n} + n^n + 1 \gt (n+1)^{n+1} + 1$
which is true if we can show that $n^{2n} \gt (n+1)^{n+1}$
or if we can show that $\frac {n^n} {n+1} \gt (\frac{n+1}{n})^n$ for $n \ge 3$
For which please see the below links -
The first answer in the first link beautifully shows that
$(\frac{n+1}{n})^n \lt n$ for $n \ge 3$
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
Prove $(\frac{n+1}{n})^{n+1}$ is decreasing
This shows for prove $(\frac{n+1}{n})^n < (\frac{n}{n-1})^n$ if $n\geq2$
|
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|
Rewriting this equation without the square roots? Given the following system of equations:
$\sqrt{(x-x_1)^2+(y-y_1)^2}+s(t_2-t_1) = \sqrt{(x-x_2)^2 + (y-y_2)^2}$
$\sqrt{(x-x_2)^2+(y-y_2)^2}+s(t_3-t_2) = \sqrt{(x-x_3)^2 + (y-y_3)^2}$
$\sqrt{(x-x_3)^2+(y-y_3)^2}+s(t_3-t_1) = \sqrt{(x-x_1)^2 + (y-y_1)^2}$
How could I write this without square roots? Here, the goal is to solve the system to determine the unknowns $x$ and $y$. How do I solve this (the goal is to do so algorithmically)?
|
$$
\sqrt{a^2}+b = \sqrt{c^2} \\
\implies a^2 + b^2 + 2b\sqrt{a^2} = c^2 \\
\implies a^2 + b^2 + 2b(\sqrt{c^2}-b) = c^2 \\
\implies \sqrt{c^2} = \frac{c^2 -a^2 - b^2}{2b} + b \\
\implies c^2 = \left(\frac{c^2 -a^2 - b^2}{2b} + b \right)^2
$$
$$
\iff c = \pm \left(\frac{c^2 -a^2 - b^2}{2b} + b \right)
$$
|
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|
Troubles finding the Fourier series of a sawtooth function plus a cosine function I have a function $y(x) = y_1(x) + y_2(x)$ consisting of two other waveforms, where
$ y_1(x) = \cos{\left(\dfrac{16 \pi}{5} x \right)}; \, y_2(x) = \displaystyle \sum_{k=-\infty}^{\infty} y_3(x - k); \, y_3(x) = \begin{cases} x & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases} \tag*{} $
In other words, $y_2(x)$ is a sawtooth of period 1, amplitude 1, that starts rising at $x = 0$. The period of $y(x)$ is $T = \text{LCM} (\frac{5}{8}, 1) = 5 $. In the following image, $y(x)$ is shown in blue, $y_1(x)$ in yellow, and $y_2(x)$ in green:
Both by hand and using Mathematica I found the Fourier coefficients of $y(x)$, obtaining the trigonometric form:
$ y(x) = \dfrac{1}{2} - \displaystyle \sum_{n=1}^\infty \left[ \dfrac{1 + \cos{(\frac{2 \pi n}{5})} + \cos{(\frac{4 \pi n}{5})} + \cos{(\frac{6 \pi n}{5})} + \cos{(\frac{8 \pi n}{5})}}{\pi n} \sin{\left( \frac{2 \pi n}{5} x \right)} \right] \tag*{} $
or the complex/exponential form:
$ y(x) = \displaystyle \sum_{n=-\infty}^\infty \left[ i \dfrac{(-1)^n \left( (-1)^n + \left( 2 \cos{\frac{\pi n}{5}} + \cos{\frac{3 \pi n}{5}} \right) \right)}{2 \pi n} \exp{\left(i \frac{2 \pi n}{5} x \right)} \right] \tag*{} $
These computations are shown in Mathematica in the following image:
But when I plot any of the two previous expressions, they don't look like the original $y(x)$. In the following image, the original expression for $y(x)$ is shown in blue, while its trigonometric Fourier series approximation (up to the 30th harmonic) is shown in yellow:
For some reason, the Fourier series looks like the sawtooth. What did I do wrong?
Edit: Computation of $a_n$
Following Olivier's answer, I got:
$\begin{align}
a_n &= \dfrac{2}{T} \displaystyle\int_0^T y(x) \cos{(n \omega_0 x)} \, \mathrm dx \\
&= \dfrac{2}{5} \displaystyle\int_0^5 \left( y_1(x) + y_2(x) \right) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx \\
&= \dfrac{2}{5} \displaystyle\int_0^5 \left( \cos{\left(\dfrac{16 \pi}{5} x \right)} \cos{(\dfrac{2 \pi n}{5} x)} + \cos{\left(\dfrac{2 \pi n}{5} x\right)} \displaystyle \sum_{k=-\infty}^{\infty} y_3(x - k) \right) \, \mathrm dx \\
&= \dfrac{2}{5} \left( \underbrace{\displaystyle\int_0^5 \cos{\left(\dfrac{16 \pi}{5} x \right)} \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx}_{I_1} + \cdots \right. \\
& \left. \cdots + \underbrace{\displaystyle\int_0^5 (y_3(x) + y_3(x - 1) + y_3(x - 2) + y_3(x - 3) + y_3(x - 4)) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx}_{I_2} \right)
\end{align}$
Using a table of common integrals, for $I_1$ I got:
$ I_1 = \begin{cases} 0 & \text{if } n \ne 8 \\ \dfrac{5}{2} & \text{if } n = 8 \end{cases} $
For $I_2$:
$I_2 = \displaystyle\int_0^1 x \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx + \displaystyle\int_1^2 (x - 1) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx + \displaystyle\int_2^3 (x - 2) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx + \cdots$
$\cdots + \displaystyle\int_3^4 (x - 3) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx + \displaystyle\int_4^5 (x - 4) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx$
The five integrals of $I_2$ have the general form (where $m$ is an integer):
$\displaystyle\int_m^{m+1} (x - m) \cos{\left(\dfrac{2 \pi n}{5} x\right)} \, \mathrm dx = \dfrac{25}{4 \pi^2 n^2} \cos{\left(\dfrac{2 \pi n}{5} [m+1]\right)} + \cdots $
$\cdots + \dfrac{5}{2 \pi n} [m+1] \sin{\left(\dfrac{2 \pi n}{5} [m+1]\right)} - \dfrac{25}{4 \pi^2 n^2} \cos{\left(\dfrac{2 \pi n}{5} [m]\right)} - \dfrac{5}{2 \pi n} [m] \sin{\left(\dfrac{2 \pi n}{5} [m]\right)} - \cdots $
$\cdots - \dfrac{5}{2 \pi n} [m+1] \sin{\left(\dfrac{2 \pi n}{5} [m+1]\right)} + \dfrac{5}{2 \pi n} [m] \sin{\left(\dfrac{2 \pi n}{5} [m]\right)} $
Subtituting this integral for $m=0,1,2,3,4$ in $I_2$ and simplifying:
$I_2 = 0 $
Subtituting $I_1$ and $I_2$ in $a_n$ and simplifying:
$ a_n = \begin{cases}
0 & \text{if } n \ne 8 \\
1 & \text{if } n = 8
\end{cases} $
I updated the trigonometric Fourier series (shown in yellow) to include this and plotted it, and now it looks like the original $y(x)$ (shown in blue):
I don't understand why Mathematica failed initially to compute $a_n$. And I'm ashamed of needing help for this.
|
I find it suspicious that $a_n=0$ since the function $y-\dfrac{1}{2}$ is not odd. According to your calculations $y-\dfrac{1}{2}$ is a pure sinus wave.
You should check the calculations of $a_n$ when $n=8$, the integral now involves
$$\cos^2\left(\dfrac{16\pi t}{5}\right)$$
and is unlikely to vanish.
|
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing
|
Another way.
For $x\leq0$ it's obvious.
But for $x>0$ by AM-GM we obtain:
$$x^{12}-x^9+x^4-x+1=x^{12}-x^9-x^4+x+2x^4-2x+1=$$
$$=x(x^8-1)(x^3-1)+2x^4+3\cdot\frac{1}{3}-2x\geq$$
$$\geq4\sqrt[4]{2x^4\cdot\left(\frac{1}{3}\right)^3}-2x=2\left(\sqrt[4]{\frac{32}{27}}-1\right)x>0.$$
|
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|
use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I prove that this isn't applicable for any number less than $1$? Also, where do I go after this? Thank you!
|
Just note that
$$
f(n+1)-f(n)=(n+1)^3 + 5(n+1)-(n^3 + 5n)=3 (n^2 + n + 2)
$$
I suspect this is the intended solution.
Actually, no induction is needed:
$$
n^3 + 5n = 6 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
$$
and so is always a multiple of $6$.
|
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How to solve such a system of quadratic equations: $x^2+y^2-xy=a^2, x^2+z^2-xz=b^2, y^2+z^2-yz=c^2$ I don't know how to solve this this system:
$$x^2+y^2-xy=a^2\\x^2+z^2-xz=b^2\\y^2+z^2-yz=c^2$$ The system of quadratic equation in symmetry form has many geometric meaning,
this seems to be a triangular pyramid, three adjacent angles with a common point are $\frac{\pi}{3}$, and opposite to this point is a triangle with $a,b,c$.
But this geometric meaning makes no sense to me, it can't simplify the problem. I don't have other idea about it.
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Solve the first and second equations as quadratic in $y$ and $z$ to get four pairs of solutions
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
\left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x-\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
\left(x+\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x+\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
\left(x-\sqrt{4 b^2-3 x^2}\right)$$
$$y=\frac{1}{2} \left(x+\sqrt{4 a^2-3 x^2}\right)\qquad z=\frac{1}{2}
\left(x+\sqrt{4 b^2-3 x^2}\right)$$
Suppose that the last is the "good" one, replace in the third equation which is
$$(a^2+b^2-c^2)-\frac{5 x^2}{4}+\frac x 4\left(\sqrt{4 a^2-3 x^2}+\sqrt{4 b^2-3 x^2} \right)-\frac{1}{4} \sqrt{4 a^2-3 x^2} \sqrt{4 b^2-3 x^2}=0$$
I suppose that multiple squaring steps would lead to an octic equation in $x$ which, I hope, could be factorized.
However, for given values of $(a,b,c)$, I would prefer to graph the function to locate more or less the solution and after to use Newton method for the solution in $x$ and then $y$ and $z$.
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"url": "https://math.stackexchange.com/questions/3849602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Where is the mistake when finding $\int x \sqrt{4+5 x} \ dx$? We have to find the integral of $\frac{dy}{dx}=x \sqrt{4+5x}$. This is from Morris Kline's book, Chapter $7$, exercise $5$, question $1$m.
I try to solve like following:
Let $u=4+5x$. Then $x=\frac{u-4}{5}$. Hence $\frac{d y}{d x}=\left(\frac{u-4}{5}\right) \sqrt{u}$
Now $$\int(x \sqrt{4+5 x}) \cdot d x=\int\left(\frac{u-4}{5}\right)
\sqrt{u} \cdot d x$$
$$ = \int\left(\frac{u^{3 / 2}-4 u^{1 / 2} }{5}\right)
\cdot d x$$
$$=\frac{\int u^{3 / 2}-4 \int u^{1 / 2}}{5} \cdot d x$$
$$=\frac{\frac{u^{5 / 2}}{5 / 2} -\frac{4 u^{3 / 2}}{3 /
2}}{5} + C$$
$$=\frac{2 u^{5 / 2}}{25}-\frac{8 u^{3 / 2}}{15}+C$$
$$\Rightarrow y=\frac{2(4+5 x)^{5 / 2}}{25} - \frac{8(4+5 x)^{3 /
2}}{15}+C$$
Where $C$ is a constant.
But the answer given in the book is: $$y=\left(\frac{2(4+5 x)^{5 /
2}}{125}\right)-\left(\frac{8(4+5 x)^{3 / 2}}{75}\right)+C.$$
Where is the mistake?
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Remember change of variables requires substituting the differentiable:
$x = \frac{u - 4}{5} \rightarrow dx = \frac{1}{5} du$
So really after substitution you should have:
$\int{\frac{u-4}{5}(\sqrt{u})(\frac{1}{5}) du}$
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"timestamp": "2023-03-29T00:00:00",
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|
Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answer one question.
PROBLEM:
Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.
CHOICES:
A) $a>b$
B) $a<b$
C) $a=b$
D) Given information is not enough
Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.
Any suggestion to solve this problem in a short time? THANKS.
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My intuition is to notice that
$$\frac{2+\frac{1}{2}}{2+\frac{1}{5}}\ ?\ \frac{5+\frac{1}{2}}{5+\frac{1}{5}}$$
(I have used $?$ since I do not know how these expressions relate)
$$\frac{5}{2}\frac{5}{11}\ ?\ \frac{11}{2}\frac{5}{25}$$
$$\frac{25}{22}\ ?\ \frac{55}{50}$$
$$\frac{50}{44}\ ?\ \frac{55}{50}$$
Since $50-44=6>5=55-50$, we know that $?$ is actually $>$. Then
$$\left(\frac{2+\frac{1}{2}}{2+\frac{1}{5}}\right)(3+\frac{1}{3})(4+\frac{1}{4})>(2+\frac{1}{5})(3+\frac{1}{4})\left(\frac{5+\frac{1}{2}}{5+\frac{1}{5}}\right)$$
and the expression simplifies, giving answer $A$.
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"url": "https://math.stackexchange.com/questions/3852464",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
What I tried...
The equation $x^2=y-6$ is of a parabola. To find the slope of the tangent to the parabola at the point $(1,7)$,
$$\frac{dy}{dx}\Bigg|_{(1,7)}=2\tag{Slope of the line tangent to the parabola}$$
So the equation of the line is $2x-y+5=0\implies y=2x+5$
Substituting this in the equation of circle to find the point of intersection of the line with the circle, we get,
$$x^2+(2x+5)^2+16x+12(2x+5)+c=0$$
Solving this, I get a complicated equation and then the answer comes out in terms of $c$ but the actual answer does not contain $c$ at all.
I would prefer a more analytical/geometrical approach if possible
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Because the line "touches" the circle, there is only one point of intersection with the circle. Therefore, the equation,
$$
x^2+(2x+5)^2+16x+12(2x+5)+c=0 \implies 5 x^2 + 60 x + 85 + c= 0
$$
has one solution.
So the determinant $\Delta = 3600 - 4\cdot5\cdot(85+c)=0$. Ergo, $c=95$.
And solving the equation $5x^2 + 60x +180=0$ gives the solution $(-6,-7)$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3853199",
"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$
For any reals $a$, $b$, $c$ and $d$ prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$$
C-S in the IMO 2001 stile does not help here:
\begin{align}
&\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2} \\
\leq&\sqrt{4\left(\tfrac{a^2}{(1+a^2)^2}+\tfrac{b^2}{(1+a^2+b^2)^2}+\tfrac{c^2}{(1+a^2+b^2+c^2)^2}+\tfrac{d^2}{(1+a^2+b^2+c^2+d^2)^2}\right)}\\
\leq&2\sqrt{\tfrac{a^2}{1+a^2}+\tfrac{b^2}{(1+a^2)(1+a^2+b^2)}+\tfrac{c^2}{(1+a^2+b^2)(1+a^2+b^2+c^2)}+\tfrac{d^2}{(1+a^2+b^2+c^2)(1+a^2+b^2+c^2+d^2)}} \\
=&2\sqrt{1-\tfrac1{1+a^2}+\tfrac1{1+a^2}-\tfrac1{1+a^2+b^2}+\tfrac1{1+a^2+b^2}-\tfrac1{1+a^2+b^2+c^2}+\tfrac1{1+a^2+b^2+c^2}-\tfrac1{1+a^2+b^2+c^2+d^2}} \\
<&2
\end{align}
We can assume that our variables are non-negative, of course.
For two variables we can get a best estimation here:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\leq\sqrt{\frac{207+33\sqrt{33}}{512}}\approx0.88...$$
There is also the following Ji Chen's estimation:
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\dotsb+\frac{x_n}{1+x_1^2+x_2^2+\dotsb+x_n^2}<\sqrt{n}-\dfrac{\ln{n}}{2\sqrt{n}},$$
but it does not help.
Thank you!
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A proof for $n=2$.
We need to prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\leq\sqrt{\frac{207+33\sqrt{33}}{512}}.$$
Since $x\leq|x|$ and for $ab=0$ it's obvious, it's enough to prove this inequality for positive variables.
Now, by AM-GM
$$\frac{b}{1+a^2+b^2}=\frac{1}{\frac{1+a^2}{b}+b}\leq\frac{1}{2\sqrt{1+a^2}}$$ and it's enough to prove that $f(a)\leq\sqrt{\frac{207+33\sqrt{33}}{512}},$ where
$$f(a)=\frac{a}{1+a^2}+\frac{1}{2\sqrt{1+a^2}}=\frac{2a+\sqrt{1+a^2}}{2(1+a^2)}.$$
Now, $$f'(a)=\tfrac{\left(2+\frac{a}{\sqrt{1+a^2}}\right)(1+a^2)-\left(2a+\sqrt{1+a^2}\right)\cdot2a}{2(1+a^2)^2}=\tfrac{2-2a^2-a\sqrt{1+a^2}}{2(1+a^2)^2}.$$
Now, $f'(a)=0$ gives $$2(1-a^2)=a\sqrt{1+a^2}$$ and we see that should be $1-a^2>0$.
Thus, $$4(1-2a^2+a^4)=a^2+a^4$$ or
$$3a^4-9a^2+4=0$$ or
$$a^2=\frac{9-\sqrt{33}}{6}$$ or
$$a=\sqrt{\frac{9-\sqrt{33}}{6}}.$$
Now, easy to see that for this value $f$ gets a maximal value and
$$f\left(\sqrt{\tfrac{9-\sqrt{33}}{6}}\right)=\tfrac{2\sqrt{\tfrac{9-\sqrt{33}}{6}}+\sqrt{1+\frac{9-\sqrt{33}}{6}}}{2\left(1+\tfrac{9-\sqrt{33}}{6}\right)}=\tfrac{2\sqrt{9-\sqrt{33}}+\sqrt{15-\sqrt{33}}}{\sqrt2(5\sqrt3-\sqrt{11})}=\sqrt{\tfrac{207+33\sqrt{33}}{512}}.$$
The last equality it's nice.
We can get a proof by the following Carl Schildkraut's beautiful idea:
Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true?
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.