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How to determine the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$ I have a question
Find the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$
There ARE questions like this on stack exchange already I know, but I'm not able to formulate a pattern or know how to apply that thing here... I've tried making combinations of $1$'s and $x^{10}$'s, $x$'s and $x^9$'s etc but I am unable to solve it. Please help.
PS. How to do it using combinations exclusively.
|
Try:
$\begin{align*}
1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10} \\
&= \frac{1 - x^{11}}{1- x} \\
(1 + x + &x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10})^4 \\
&= \frac{(1 - x^{11})^4}{(1- x)^4} \\
&= (1 - 4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
\cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
[x^{10}] (1 - &4 x^{11} + 6 x^{22} - 4 x^{33} + x^{44})
\cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
&= [x^{10}] \sum_{k \ge 0} (-1)^k \binom{-4}{k} x^k \\
&= (-1)^{10} \binom{-4}{10} \\
&= \binom{10 + 4 - 1}{4 - 1} \\
&= 286
\end{align*}$
|
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|
Using ring R=$\mathbb{Z}[X]/(X^n+1)$ Currently studying Ring-LWE,here, and having trouble grasping a concrete definition of what this means, as a ring. My understanding is that everything in this ring is an integer polynomial, and that it must be reduced $X^n+1$, is this correct? Essentially working modulo $X^n+1$.
It then goes on to work in quotient group $R_q=R/qR$. When in this quotient group, would all elements have already been reduced $X^n+1$ and finally then reduced mod $q$?
|
The polynomials are just multiplied as usual but any term $X^N$ is replaced by $-1$, $X^{n+1} \equiv -X$ etc. , as we consider $X^n+1$ to be equivalent to $0$.
All coefficients of those polynomials lie in $\mathbb{Z}_q$, so whenever they go above $q$, reduce them.
So a simple example: $n=3$, $q=3$, then $$(1+X+X^2)(1+2X+2X^2)= 1 + 2X + 2X^2 + X + 2X^2 + 2X^3 + X^2 + 2X^3 + 2X^4=1 + 3X + 5X^2 +4X^3 + 2X^4 \equiv 1+ 2X^2 + 4(-1) + 2(-X) \equiv (1-4) - 2X +2X^2 \equiv X +2X^2$$
for example (modulo calculation errors, working from screen)
|
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|
$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$ I want to prove that
$$\frac{(x-y)(x-z)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-x)(z-y)}{z^2}\geq 0$$
for positive numbers $x,y,z$.
I don't know how to even begin. I must say I'm not 100% certain the inequality always holds.
I tried the sort of factoring involved in proving schur's inequality, but it doesn't seem to work here. I also tried to distribute the denominators to obtain terms of form (1-y/x)(1-z/x) and then maybe substituting x/y=a, y/z=b, z/x=a etc
|
There is no loss of generality in assuming $0 < x \leq y \leq z$. Rewrite the inequality as
$$\frac{(x-y)^2+(y-z)(x-y)}{x^2}+\frac{(y-x)(y-z)}{y^2}+\frac{(z-y)^2+(y-x)(z-y)}{z^2} \geq 0$$
and rearrange the terms as follows:
$$\frac{(x-y)^2}{x^2}+\frac{(z-y)^2}{z^2}+(z-y)(y-x)\left(\frac{1}{x^2}+\frac{1}{z^2}-\frac{1}{y^2}\right)\geq 0.$$
Since $(z-y)(y-x)\geq 0$ by assumption, it suffices to prove
$$\frac{1}{x^2}+\frac{1}{z^2}-\frac{1}{y^2} \geq 0,$$
which is equivalent to
$$\left(\frac{y}{x}\right)^2+\left(\frac{y}{z}\right)^2-1\geq0.$$
But the latter is obviously true as $y/x \geq 1$.
|
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|
In how many ways can $5$ boys and $5$ girls stand in a queue such that exactly four of the girls stand consecutively in the queue? I have two combinatorics questions.
Let $n$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then, the value of $m/n$ is?
Let $n_1<n_2<n_3<n_4<n_5$ be positive integers such that $n_1+n_2+n_3+n_4+n_5 = 20$. The number of distinct arrangements of ($n_1, n_2, n_3, n_4, n_5$) is?
So for the first question, I was able to find out $ n = 6! × 5!$ but while finding $m$, I could do $4! × 7!$ But that also include all the fives girls accidentally coming together.. So how do I eliminate those cases? I can't proceed further.
The second one, I thought was one of those stars and bars problems, so I did $C(24, 4)$ but then I realized that the condition on the values of numbers is not that simple. I tried converting it into simpler problem, as they do with stars and bars problems but I couldn't achieve something useful. I'm stuck :/
Can you help me get through these questions?
P.S.-These questions are sometimes meant to be done using a trick, so if you think you know some trick to make that easier, please be sure to tell it. And otherwise a true solution would be as helpful :)
|
In how many ways can $5$ boys and $5$ girls stand in a queue if all five girls stand consecutively in the queue?
You are correct that there are $6!5!$ ways for all five girls to stand consecutively in the queue.
Method 1: We treat the block of five girls as a single object. We then have six objects to arrange, the block of girls and the five boys. The objects can be arranged in $6!$ ways. The five girls can be arranged within the block in $5!$ ways. Thus, there are $6!5!$ ways for five boys and five girls to stand in a queue if all five girls stand consecutively in the queue.
Method 2: Line up the five boys, which can be done in $5!$ ways. This creates six spaces in which to place the block of five girls, four between successive boys and two at the ends of the row.
$$\square b_1 \square b_2 \square b_3 \square b_4 \square b_5 \square$$
Choose one of these six spaces in which to place the block of girls, then arrange the five girls within the block. This can be done in $6 \cdot 5!$ ways. Hence, the number of admissible arrangements is $6!5!$.
In how many ways can $5$ boys and $5$ girls stand in a queue if exactly four girls stand consecutively in the queue?
We modify the second method above.
Line up the five boys in $5!$ ways. This creates six spaces in which to place the girls. Choose which four of the five girls stand consecutively, which can be done in $\binom{5}{4}$ ways. Choose which of the six spaces the block of four girls fills. Arrange the four girls in that space in $4!$ ways. That leaves five spaces in which to place the remaining girl. Hence, the number of ways five boys and five girls can stand in a queue if exactly four girls stand consecutively is
$$5!\binom{5}{4} 6 \cdot 4! \cdot 5 = 5! \cdot 5 \cdot 6 \cdot 5 \cdot 4! = 5 \cdot 6!5!$$
In how many ways can $20$ be expressed as the sum of five distinct increasing positive integers?
Since $20$ is a small number, we can simply write down all the possibilities:
\begin{align*}
20 & = 1 + 2 + 3 + 4 + 10\\
& = 1 + 2 + 3 + 5 + 9\\
& = 1 + 2 + 3 + 6 + 8\\
& = 1 + 2 + 4 + 5 + 8\\
& = 1 + 2 + 4 + 6 + 7\\
& = 1 + 3 + 4 + 5 + 7\\
& = 2 + 3 + 4 + 5 + 6
\end{align*}
Notice that any sum of five distinct positive integers is at least $1 + 2 + 3 + 4 + 5 = 15$. We then have to distribute five more ones in such a way that we preserve the increasing sequence. Since $5$ can be partitioned into at most five positive integers in the following seven ways,
\begin{align*}
5 & = 5\\
& = 4 + 1\\
& = 3 + 2\\
& = 3 + 1 + 1\\
& = 2 + 2 + 1\\
& = 2 + 1 + 1 + 1\\
& = 1 + 1 + 1 + 1 + 1
\end{align*}
we can do so in the following ways:
\begin{align*}
(0, 0, 0, 0, 5)\\
(0, 0, 0, 1, 4)\\
(0, 0, 0, 2, 3)\\
(0, 0, 1, 1, 3)\\
(0, 0, 1, 2, 2)\\
(0, 1, 1, 1, 2)\\
(1, 1, 1, 1, 1)\\
\end{align*}
Adding these, respectively, to the vector $(1, 2, 3, 4, 5)$ yields the solutions
\begin{align*}
(1, 2, 3, 4, 10)\\
(1, 2, 3, 5, 9)\\
(1, 2, 3, 6, 8)\\
(1, 2, 4, 5, 8)\\
(1, 2, 4, 6, 7)\\
(1, 3, 4, 5, 7)\\
(2, 3, 4, 5, 6)
\end{align*}
that correspond to the seven sums we wrote above.
|
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|
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?
|
Let $x=t^2,$ where $t>0$.
Thus, $$t^3-9t-\sqrt8=0$$ or
$$t^3+\sqrt8t^2-\sqrt8t^2-8t-t-\sqrt8=0$$ or
$$(t+\sqrt8)(t^2-\sqrt8t-1)=0$$ or
$$x-\sqrt{8x}=1.$$
|
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Proving non-trivial integer solutions don't exist for $z^2=10x^2-5y^2$ Given $z^2=10x^2-5y^2$.
I am wanting to show this equation has integer solutions only $0=x=y=z$.
I attempted by considering modulo $2$, however this didn't give me a contradiction. Any help or hints would be appreciated.
Solution:
Consider modulo $5$. Then we have that $z^2 \equiv 0 \mod5$. This is impossible as:
$0^2=0 \mod5$
$1^2=1 \mod5$
$2^2=4 \mod5$
$3^2=4 \mod5$
$4^2=1 \mod5$
|
As Hoseyn Heydari's comment stated, you should consider checking modulo $5$ instead for your equation of
$$z^2 = 10x^2 - 5y^2 \tag{1}\label{eq1}$$
As the RHS is a multiple of $5$, this means $z = 5z_1$ for some integer $z_1$. Thus, \eqref{eq1} becomes
$$5z^2_1 = 2x^2 - y^2 \tag{2}\label{eq2}$$
Since the possible congruences of squared values modulo $5$ are $0, 1$ and $4$, then $2x^2$'s values modulo $5$ must be $0, 2$ or $3$ (as $2 \times 4 \equiv 3 \pmod 5$). Thus, since the LHS is a multiple of $5$ and the only match between the $2$ sets of $\{0,2,3\}$ and $\{0,1,4\}$ is $0$, this means $2x^2 - y^2$ can only be congruent to $0$ modulo $5$ if both $x$ and $y$ are as well. Let $x = 5x_1$ and $y = 5y_1$ to get
$$z^2_1 = 10x^2_1 - 5y^2_1 \tag{3}\label{eq3}$$
Note this is of the same form as \eqref{eq1}, but with each value being $5$ times smaller. As the procedure can be repeated indefinitely, this is only possible if each of the values were $0$ initially, giving that the sole solution is $x = y = z = 0$.
|
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|
Define the domain of $f(x)=\sqrt{x^2+4}$ I am to define the domain of $f(x)=\sqrt{x^2+4}$.
I arrived at $[-2,\infty)$ whereas the textbook solution is $(-\infty,\infty)$.
To arrive at my solution I set the radicand to be greater than or equal to zero:
$x^2+4\ge0$
$x+2\ge0$ # square root of each side
$x\ge-2$
Thus I get the domain as $[-2,\infty)$.
Why is the domain actually $(-\infty, \infty)$?
[edit I had a typo the radicand is $x^2+4$]
|
As you said, the radicand must be greater than or equal to zero. Since the square of any real number is at least zero, $x^2 + 4 \geq 4 > 0$ for every real number $x$, which implies that the domain is indeed $(-\infty, \infty)$.
In your calculation, you should have had
\begin{align*}
x^2 + 4 & \geq 0\\
x^2 & \geq -4
\end{align*}
which is true for every real number $x$.
It is not true that $\sqrt{x^2 + 4} = x + 2$. Notice that if $x = 1$, then $\sqrt{x^2 + 4} = \sqrt{1 + 4} = \sqrt{5}$ while $x + 2 = 1 + 2 = 3$. Squaring $x + 2$ yields
\begin{align*}
(x + 2)^2 & = (x + 2)(x + 2)\\
& = x(x + 2) + 2(x + 2)\\
& = x^2 + 2x + 2x + 4\\
& = x^2 + 4x + 4
\end{align*}
Hence, $\sqrt{x^2 + 4} = x + 2$ is only true when $4x = 0 \implies x = 0$.
|
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|
An exercise from Apostol's book I am trying to solve following problem from Apostol's Mathematical Analysis. The problem could be very trivial, but I am not getting clue for it.
Let $\{a_n\}$ be a sequence of real numbers in $[-2,2]$ such that
$$|a_{n+2}-a_{n+1}|\le \frac{1}{8} |a_{n+1}^2-a_n^2| \,\,\,\, \mbox{ for all } n\ge 1.$$
Prove that $\{a_n\}$ is convergent.
Q. Any hint for solving this? (I was not getting the restrictions of interval and the factor $\frac{1}{8}$).
My try: since $a_i\in [-2,2]$ so $a_i^2\in [0,4]$. Thus, $|a_{n+1}^2-a_n^2|\le 4$ and so $|a_{n+2}-a_{n+1}|\le \frac{1}{2}$. After this, I couldn't proceed.
Any HINT is sufficient.
|
Let $A_{m+1}^-:=\left|a_{m+1}-a_m\right|$ and $A_{m+1}^+:=\left|a_{m+1}+a_m\right|$ for convenience. Then for a positive integer $k$, \begin{align}A_{n+k+1}^-&\le\frac{1}{8}\left|a_{n+k}^2-a_{n+k-1}^2\right|=\frac{1}{8}A_{n+k}^+A_{n+k}^-\\&\le\frac{1}{8}A_{n+k}^+\cdot\frac18\left|a_{n+k-1}^2-a_{n+k-2}^2\right|=\frac1{8^2}A_{n+k}^+A_{n+k-1}^+A_{n+k-1}^-\\&\le\frac1{8^3}A_{n+k}^+A_{n+k-1}^+A_{n+k-2}^+A_{n+k-2}^-\\&\le\cdots\\&\le\frac{A_{n+k-(n+k-1)+1}^-}{8^{n+k-1}}\prod_{i=1}^{n+k-1}A_{n-k-i+1}^+\\&\le\frac{A_2^-}{8^{n+k-1}}\prod_{i=1}^{n+k-1}|2+2|\quad(\text{since}\, \left|a_j\right|\le2\quad\forall j\ge1)\\&=\frac{A_2^-}{2^{n+k-1}}\end{align} Therefore, as $k\to\infty$, we have $\left|a_{n+k+1}-a_{n+k}\right|\to0$ and convergence is shown.
|
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Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that
$$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$
has only real solutions
This is what I got till now.
$$\left(\frac{1+i(a+ib)}{1-i(a+ib)}\right)^n = a+ib$$
$$\left(\frac{(1-b)+ia}{(1+b)-ia}\right)^n = a+ib$$
$$\frac{(1-b)+ia}{(1+b)-ia}.\frac{(1+b)+ia}{(1+b)+ia} = \frac{1-b^2+2ia-a^2}{1+2b+b^2+a^2}$$
then
$$\left(\frac{1-b^2-a^2}{1+2b+b^2+a^2}+\frac{2ia}{1+2b+b^2+a^2}\right)^n = a+ib$$
where
$$a=0 \text{; & } 1+2b+b^2\neq0$$
|
Set
$$
w=\frac{1+iz}{1-iz}
$$
so you can solve for $z$, getting
$$
z=i\frac{1-w}{1+w}
$$
This is real if and only if
$$
i\frac{1-w}{1+w}=-i\frac{1-\bar{w}}{1+\bar{w}}
$$
that becomes
$$
1+\bar{w}-w-w\bar{w}=-1+\bar{w}-w+w\bar{w}
$$
that is, $w\bar{w}=1$.
Therefore $|a+bi|=1$ is a necessary condition. Now if we write $a+bi$, the equation becomes
$$
\frac{1+iz}{1-iz}=u
$$
where $u$ is any $n$-th root of $a+bi$ and $|u|=1$. By the same argument as before, the solution is real.
|
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|
Exercises to see if certain series converge.
*
*Does the series $\sum \frac{2^k+1}{3^k}$ converge?
*Does the series $\sum (-1)^k\frac{1}{k^2}$ converge?
*Does the series $\sum \frac{2^k+2^{2k}}{4^k}$ converge?
*Does the series $\sum \frac{k!}{(2k)!}$ converge?
For the first question, do we use the comparison test comparing it to $\left(\frac{2}{3}\right)^k$???
For the second question, this series does converge. To prove this observe that this converges absolutely because
\begin{equation*}
\sum_{k=1}^{\infty} \left|(-1)^k\frac{1}{k^2}\right| = \sum_{k=1}^{\infty} \frac{1}{k^2}
\end{equation*}
converges. To see that this does converge, we can use the integral test.
The function $f:[1,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x^2}$ is continuous, non-negative and decreasing by examining the derivative $f'(x) = -\frac{2}{x^3} < 0$ for all $x\geq 1$. So we have
\begin{equation*}
\begin{split}
\int_{1}^{\infty} \frac{1}{x^2} \; dx &= \left[-\frac{1}{x}\right]_{1}^{\infty} \\
&= \lim_{a\to\infty} \left(-\frac{1}{a}+1\right) \\
&= 1
\end{split}
\end{equation*}
This integral converges so by the integral test $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges as well and hence, $\sum_{k=1}^{\infty} (-1)^k\frac{1}{k^2}$ converges.
For the third question, this series does NOT converge. To show that this series diverges, we just need to show that
\begin{equation*}
\lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} \neq 0
\end{equation*}
or doesn't exist. Well, for $k\geq 0$,
\begin{equation*}
\frac{2^k+2^{2k}}{4^k} = \frac{2^k+4^k}{4^k} = \left(\frac{1}{2}\right)^k+1.
\end{equation*}
Thus,
\begin{equation*}
\begin{split}
\lim_{k\to\infty} \frac{2^k+2^{2k}}{4^k} &= \lim_{k\to\infty} \left(\frac{1}{2}\right)^k+1 \\
&= 0+1 \\
&= 1 \neq 0.
\end{split}
\end{equation*}
For the fourth question, if we let $a_k = \frac{k!}{(2k)!}$ then using the ratio test we get
\begin{equation*}
\begin{split}
\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right| &= \lim_{k\to\infty} \left|\frac{\frac{(k+1)!}{(2k+1)!}}{\frac{k!}{(2k)!}}\right| \\
&= \lim_{k\to\infty} \left|\frac{(k+1)!}{(2k+1)!}\times \frac{(2k)!}{k!}\right| \\
&= \lim_{k\to\infty} \left|\frac{(k+1)\times k\times (k-1)\times\ldots\times (2k)\times (2k-1)\times (2k-2)\times\ldots}{(2k+1)\times 2k\times (2k-1)\times\ldots\times k\times (k-1)\times (k-2)\times\ldots}\right| \\
&= \lim_{k\to\infty} \frac{k+1}{2k+1} \\
&= \lim_{k\to\infty} \frac{1+\frac{1}{k}}{2+\frac{1}{k}} \\
&= \frac{1}{2} < 1.
\end{split}
\end{equation*}
Hence, $\sum_{k=0}^{\infty} \frac{k!}{(2k)!}$ converges.
|
For 1) use the fact that $\sum (\frac 2 3)^{k}$ and $\sum \frac1 {3^{k}}$ both converge.
Your answer for 2) and 3) are OK.
In 4) $a_{k+1}$ is not $\frac {(k+1)!} {(2k+1)!}$. It is $\frac {(k+1)!} {(2k+2)!}$. [You have to replace $k$ by $k+1$ in the formula for $a_k$]. But your approach is good and you will be able to complete it after making this correction.
|
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what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
|
Let $a = 1$. Then $4^b = 3 \Rightarrow b = \log_4 3 = \frac{\log 3}{\log 4}$.
Therefore:
$$9^{a/b} + 16^{b/a}$$
$$=9^{\log 4 / \log 3} + 16^{\log3 / \log 4}$$
$$=3^{2 \log 4 / \log 3} + 4^{2 \log 4 / \log 3}$$
$$=3^{2 \log_3 {4} } + 4^{2 \log_4 {3}}$$
$$= \left( 3^{\log_3 4} \right )^2 + \left( 4^{\log_4 3} \right)^2 $$
|
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Generating function on number of ways to purchase a bouquet with exactly 50 dollars There are two kinds of flowers in a shop. Roses cost 3 dollars each while carnations cost 2 dollars each. How many different kinds of bouquets can be bought with exactly 50 dollars?
My solution:
We can purchase $1,2,3,...$ roses, each cost $3$ dollars which gives the choices $(1+x^3+x^6+...)$. Likewise we can choose $(1+x^2+x^4+...)$ carnations. We define our generating function as:
$G(x) = (1+x^2+x^4+...)(1+x^3+x^6+...)=(-1/(x^3-1))(-1/(x^2-1))=(x^2-1)^{-1}(x^3-1)^{-1}$
We are to count the number of combinations where the total cost is 50 dollars, hence we're looking for $[x^{50}]$.
How can one compute $[x^{50}]G(x)?$
|
Use trusty partial fractions...
$\begin{align*}
[x^{50}] \frac{1}{(1 - x^2) (1 - x^3)}
&= [x^{50}] \left(
\frac{1}{3 (1 + x + x^2)}
+ \frac{1}{4 (1 + x)}
+ \frac{1}{4 (1 - x)}
+ \frac{1}{6 (1 - x)^2}
\right) \\
&= [x^{50}] \frac{1 - x}{3 (1 - x^3)}
+ \frac{1}{4} \cdot (-1)^{50}
+ \frac{1}{4}
+ \frac{1}{6} \cdot \binom{-2}{50} \\
&= \frac{1}{3} \cdot [x^{50}] \frac{1}{1 - x^3}
- \frac{1}{3} \cdot [x^{49}] \frac{1}{1 - x^3}
+ \frac{1}{2}
+ \frac{1}{6} \cdot \binom{50 + 2 - 1}{2 - 1} \\
&= 9
\end{align*}$
The powers of $x$ that appear in $(1 - x^3)^{-1}$ are all multiples of 3, and neither 50 nor 49 qualify.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Result from $X^2+Y^2=p^{n-1}$ with $p^{n-1} \equiv 1 \pmod{8}$ I'm reading an article and they have the equation $X^2+Y^2=p^{n-1}$ where $p$ is a prime and $p^{n-1} \equiv 1 \pmod{8}$ and $p \equiv 1 \pmod{4}$.
They conclude that since $p^{n-1} \equiv 1 \pmod{8}$ it must follow that
$$\begin{cases}
X \equiv 0 \pmod{4} \\ Y \equiv \pm 1 \pmod{4}
\end{cases}
\textrm{ or }
\begin{cases}
X \equiv \pm 1 \pmod{4} \\ Y \equiv 0 \pmod{4}
\end{cases}.$$
Where does this come from?
|
Note $p^{n-1} \equiv 1 \pmod 8$ means both sides of $X^2 + Y^2 = p^{n-1}$ is odd. Thus, one of $X$ and one of $Y$ must be even and the other odd. Consider $X$ to be even and $Y$ to be odd. As $Y$ is odd (so $Y \equiv \pm 1 \pmod 4$), when squared, it's $Y^2 \equiv 1 \pmod 8$. With $X$ being even, if it's $X \equiv 2 \pmod 8$, then it's square is $X^2 \equiv 4 \pmod 8$, so $X^2 + Y^2 \equiv 5 \pmod 8$. However, since $X^2 + Y^2 \equiv 1 \pmod 8$, this means $X \equiv 0 \pmod 4$.
The two sets of congruences you ask about just list this for $X$ and $Y$ as I describe above, and since the equation is symmetric in $X$ and $Y$, it also lists the possibilities for the conditions on $X$ and $Y$ to be switched around.
|
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|
Find function $f(x)$ knowing that $[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$ is satisfied for $\forall x \in \mathbb R$.
Find function $f(x)$ knowing that $$\large [f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$$ is satisfied for $\forall x \in \mathbb R$ $(f(x) \ne c$ with $c$ being a constant$)$.
Okay, where to start. $f(x) = x$ doesn't work as $$(x^2 - x + 1)(x^3 + 1) - (x^4 - 2x^3 + x + 1)(x + 1) = 3(x - 1)x(x + 1)$$. Furthermore, if $f(x) = x^2 - x$ then $f(x^2 - x) = x^4 - 2x^3 + x$. However, at this case,
$$[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1) = (x - 1)x(x + 1)(x^6 - 3x^4 + 2x^3 + 1)$$
|
If we assume a solution of the form $f(x)=\dfrac{x}{a}$ then
$$ \left[\frac{1}{a}(x^2-x)+1\right]\cdot\left[\frac{1}{a}(x^3+1)\right]
=\left[\frac{1}{a}(x^4-2x^3+x)+1\right]\cdot\left[\frac{1}{a}(x+1)\right] $$
$$ (x^2-x+a)\cdot(x^3+1)=(x^4-2x^3+x+a)\cdot(x+1) $$
So either $x=-1$ or
$$ (x^2-x+a)\cdot(x^2-x+1)=x^4-2x^3+x+a $$
$$ x^4-2x^3+(a+2)x^2-(a+1)x+a=x^4-2x^3+x+a $$
$$ (a+2)x^2-(a+1)x=x $$
So $a=-2$ is the only solution.
Thus, as Luca Bressan found, not only is $f(x)=-\frac{1}{2}x$ a solution of the form $f(x)=cx$ it the only solution of that form.
ADDENDUM: By a similar analysis it can be shown that if $f(x)=\dfrac{1}{a}x+b$ then it must be the case that $a=-2$ and $b=0$.
|
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|
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Here is my attempt:
$x = 0$ and $y = 0$, these both line go through the $x$ and $y$ axes. And also the circle touches those two lines. So the center will be $C(p,p)$. That means $r = k = h = p$. The circle also touches the $x = a$ line. That means $r = a/2$. From that, I determined the center as $C(a/2, a/2)$.
The equation:
$$\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$$
$$x^2 + y^2 - 2ax - 2ay + \frac{a^2}{4} = 0$$
which is not the correct answer.
Now can anyone tell me what's wrong with this attempt?
|
If the simple error in your calculation [coefficient in middle $2xy$ term in expansion of $(x+y)^2$ ] is removed the correct equation is
$$ (x^2+y^2)-a (x+y)+ \frac{a^2}{4}=0 $$
The origin/corner touching circle is in the first or fourth quadrant according as odd term $y$ is positive or negative.
$$ (x^2+y^2)-a (x-y)+ \frac{a^2}{4}=0 $$
Everything else is quite OK.
|
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|
Find $\lim_{n\rightarrow \infty}\int_0^\infty \frac{x^{n-2}\cos(n\pi x)}{1+x^n} dx$ I feel it goes to zero I separate the integral $\int_0^1 \frac{x^{n-2}\cos(n\pi x)}{1+x^n}$ the sequence goes to zero and it bounded by $0.5$ so by bounded convergence theorem the limit is zero. Now for the $\int_1^\infty \frac{x^{n-2}\cos(n\pi x)}{1+x^n} $ first $\frac{x^{n-2}}{1+x^n}\leq \frac{1}{x^2}$ which integrable. Another thought as $n$ goes large enough the $\int_1^\infty\cos(n\pi x)$ will be zero and the other part will be almost constant. Any help with full details.
Trying Partial derivative as @EuklidAlexandria suggested
$$\int_1^\infty \frac{x^{n-2} \cos(n\pi x) }{1+x^n} dx = 0 -\int_1^\infty \underbrace{\frac{x^{n-3}(n-2-2x^n)\sin(n\pi x)}{n\pi(1+x^n)^2}}_{f_n}$$
then the sequence
$$\left| f_n \right| \leq \left|\frac{x^{n-3}(n-2x^n)}{n\pi x^{2n}}\right|\leq \frac{n+2x^n}{n x^{n+3}}\leq \frac{1}{x^n} + \frac{1}{x^3}\leq \frac{2}{x^3}$$ since we are concern about limit we can consider $n\geq 3$
Now by Lebesgue Dominated convergence theorem we have $f_n \rightarrow 0$ p.w and $|f_n| \leq \frac{2}{x^3}$ which is integrable on $[1,\infty)$ Hence
$$\lim_{n\rightarrow \infty} \int_{1}^{\infty} \frac{x^{n-2}\cos(n\pi x)}{1+x^n} = \lim_{n\rightarrow \infty}- \int_1^\infty \frac{x^{n-3}(n-2-2x^n)\sin(n\pi x)}{n\pi(1+x^n)^2} =\int 0 = 0$$
Is that correct?
|
Your solution seems fine. For a bit of further simplification, we may do as follows. Write $I_n$ for the integral. Then
\begin{align*}
&\left| I_n - \int_{1}^{\infty} \frac{\cos(n\pi x)}{x^{2}} \, \mathrm{d}x \right| \\
&\leq \int_{0}^{1} \left| \frac{x^{n-2}\cos(n\pi x)}{1+x^n}\right| \, \mathrm{d}x + \int_{1}^{\infty} \left| \frac{x^{n-2}}{1+x^n} - \frac{1}{x^2} \right|\left| \cos(n\pi x)\right| \, \mathrm{d}x \\
&\leq \int_{0}^{1} x^{n-2} \, \mathrm{d}x + \int_{1}^{\infty} \frac{1}{x^2(1 + x^n)} \, \mathrm{d}x.
\end{align*}
It is easy to check that this bound converges to $0$ as $n\to\infty$. Now, to show that $I_n \to 0$, it suffices to check that
$$ \int_{1}^{\infty} \frac{\cos(n\pi x)}{x^{2}} \, \mathrm{d}x \xrightarrow[n\to\infty]{} 0, $$
which is an immediate consequence of the Riemann-Lebesgue lemma. (Or, of course, integration by parts works perfectly.)
|
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|
From a set of N elements how many reflexive relations are out of the anti symmetric relations As I know in a set of N elements there are $2^{n^2}$ (two to the power of $n$ squared) relations, in which there are $3^{\frac12(n^2-n)}$ anti-symmetric relations.
How can I find out of those anti-symmetric relations the amount that is also reflexive?
|
A binary relation on a set $S$ with $N$ elements can be represented by an $(N \times N)$-table in which the cell $(a, b)$ is marked if and only if $(a, b)$ belongs to the relation (i.e., $a$ is related to $b$).
For example, if $S = \{ 1, 2, 3, 4 \}$, a simple example of binary relation on $S$ is:
\begin{array}{c||c|c|c|c|}
& 1 & 2 & 3 & 4 \\
\hline
1 & & \times & \times & \\
\hline
2 & & & & \\
\hline
3 & \times & & & \\
\hline
4 & & & & \times \\
\hline
\end{array}
This table represents the relation $\{(1, 2), (1, 3), (3, 1), (4, 4) \}$.
Since the table has $N^2$ cells and each cell can be either marked or not ($2$ options), there are $2^{N^2}$ binary relations on $S$.
A binary relation on $S$ is reflexive if $(a, a)$ belongs to the relation for any $a \in S$. This means that all the cells on the diagonal are marked (but other cells can be marked as well). For example, we can turn the relation above into a reflexive one as follows (the diagonal is highlighted):
\begin{array}{c||c|c|c|c|}
& 1 & 2 & 3 & 4 \\
\hline
1 & \color{red}\times & \times & \times & \\
\hline
2 & & \color{red}\times & & \\
\hline
3 & \times & & \color{red}\times & \\
\hline
4 & & & & \color{red}\times \\
\hline
\end{array}
Again, the table has $N^2$ cells, of which the $N$ cells on the diagonal must be marked (only $1$ option), while the remaining $N^2 - N$ cells not on the diagonal can either be marked or not ($2$ options). Therefore there are $1^N \cdot 2^{N^2 - N} = 2^{N^2 - N}$ reflexive relations on $S$.
A reflexive relation on $S$ is also antisymmetric if at most one of the pairs $(a, b)$ and $(b, a)$ belong to the relation for any distinct $a, b \in S$. This means all the cells on the diagonal are marked as before, and for $a \neq b$ there are exactly $3$ options:
*
*Neither $(a, b)$ nor $(b, a)$ are marked;
*$(a, b)$ is marked and $(b, a)$ is not marked;
*$(b, a)$ is marked and $(a, b)$ is not marked.
The reflexive relation above isn't antisymmetric because both $(1, 3)$ and $(3, 1)$ are marked, but we can turn it into an antisymmetric one by removing e.g. the mark on the cell $(1, 3)$:
\begin{array}{c||c|c|c|c|}
& 1 & 2 & 3 & 4 \\
\hline
1 & \times & \times & & \\
\hline
2 & & \times & & \\
\hline
3 & \times & & \times & \\
\hline
4 & & & & \times \\
\hline
\end{array}
There are $N$ elements on the diagonal which must be marked ($1$ option) and there are $\binom N 2 = \frac {N (N - 1)} 2$ ways of choosing two elements $a \neq b$ for which there are the $3$ options above. Therefore there are $1^N \cdot 3^{N(N-1)/2} = 3^{N(N-1)/2}$ reflexive antisymmetric relations on $S$.
|
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|
Find the maximum angle $X$ in the range $0^\circ \leq x < 360^\circ$ which satisfies the equation $\cos^2(2x)+\sqrt{3}\sin(2x)-\frac{7}{4}=0$
Find the maximum angle $X$ in the range $0^\circ \leq x < 360^\circ$ which satisfies the equation
$\cos^2(2x)+\sqrt{3}\sin(2x)-\frac{7}{4}=0$
$\cos^2(2x)=\sin^2(2x)-1$, so we can substitute $t$ for $\sin(2x)$, and we have the equation $t^2+\sqrt{3}t-\frac{7}{4}-1=0$.
Solving for $t$, $t$ is either $\frac{-\sqrt{3}+\sqrt{14}}{2}$ or $\frac{-\sqrt{3}-\sqrt{14}}{2}$
$x= \frac{\arcsin(t)}{2}$
How can I find the maximum angle $X$?
Edit: $\cos^2(2x) = -\sin^2(2x)+1$ , not $\cos^2(2x)=\sin^2(2x)-1$
|
Note you have $\cos^22x=1-\sin^22x$; with $\sin2x=t$ you get
$$
1-t^2+t\sqrt{3}-\frac{7}{4}=0
$$
so the equation can be rewritten
$$
4t^2-4t\sqrt{3}+3=0
$$
This is actually $(2t-\sqrt{3})^2=0$, so the only root is $t=\sqrt{3}/2$. Hence $2x=\pi/3+2k\pi$ or $2x=2\pi/3+2k\pi$. Hence
$$
x=\frac{\pi}{6}+k\pi\qquad\text{or}\qquad x=\frac{\pi}{3}+k\pi
$$
If you prefer degrees,
$$
x=30^\circ+k180^\circ\qquad\text{or}\qquad x=60^\circ+k180^\circ
$$
It shouldn't be difficult to find the maximum solution.
|
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|
Find all $(x,y)$ of positive integers s.t. $x^3-y^3=xy+61$ Given
$x^3-y^3=xy+61$.
Find all pairs (x,y) of positive integers which satisfies the given equation.
I tried solving this by following method-
$(x-y)(x^2+y^2+xy)-xy=61$
$(x-y)[(x-y)^2+3xy]-xy=61$
$(x-y)^3+3xy(x-y)-xy=61$
I don't know if doing this was in any means fruitful or not.
Remember $x>y$, only then a solution is possible.
|
$x\geq y+1$ so:
$61 = x^3-y^3-xy=x(x^2-y)-y^3\geq (y+1)(y^2+y+1)-y^3=2y^2+2y+1$
$y^2+y\leq 30$, so $1\leq y\leq 5$ - just check this numbers
|
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|
Number of particular paths of even length Let $W_n=\{ w=a_1a_2\cdots a_{2n}: a_i\in\{1,-1\},\ a_1+\cdots+a_j\geq 0,\ \forall\ 1\leq j\leq 2n\}$. Let $W_n^k=\{w\in W_n:w\ \text{does not cross the line}\ y=k \}$. Is there any way to find $\# W_n^k$?
Example: if $n=2$, then $\# W_2=6$ and $\# W_2^1=1$ and $\#W_2^2=4$.
|
Fix $k$ and let $a_{n,i}$ be the number of paths of length $n$ with step sizes $\pm 1$ that never leave the strip $\{0, 1, \ldots, k\}$ and end in state $i$, $0 \leq i \leq k$. If $0 < i < k$, so $i$ is not on the boundary, there are two ways that we could have arrived at the state $i$: either from $i+1$ or $i-1$. So $a_{n,i} = a_{n-1, i-1} + a_{n-1, i+1}$. Furthermore, $a_{n, 0} = a_{n-1, 1}$ and $a_{n, k} = a_{n-1, k-1}$ since if we are on the bottom / top we must have come from just above / below. Thus we have a linear recurrence. If we write the recurrence as a matrix equation, we get (e.g., for k = 3)
$$ \left( \begin{array}{cc}
a_{n, 0} \\
a_{n, 1} \\
a_{n, 2} \\
a_{n, 3}
\end{array} \right) =
\quad
\left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{array} \right)\left( \begin{array}{cc}
a_{n-1, 0} \\
a_{n-1, 1} \\
a_{n-1, 2} \\
a_{n-1, 3}
\end{array} \right)
$$
So inductively,
$$ \left( \begin{array}{cc}
a_{n, 0} \\
a_{n, 1} \\
a_{n, 2} \\
a_{n, 3}
\end{array} \right) =
\quad
\left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{array} \right)^n\left( \begin{array}{cc}
1\\
0 \\
0 \\
0
\end{array} \right)
$$
since the path starts in state $0$.
Then you may quickly compute $a_{n,i}$, and $$\#W_n^k = \sum_{i=0}^k a_{2n, k}.$$ For $k = 3$ this gives OEIS sequence A001519. If you want an explicit formula, you could diagonalize the matrix; OEIS gives the formula $$(\phi^{2n-1} + \phi^{1-2n})/\sqrt{5},$$ $\phi=(1+\sqrt{5})/2$ the golden ratio. I don't know whether there is a formula for general $k$.
|
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|
I need to prove that $\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$ I need to prove the following limit
$$\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$$
Using the Squeeze Theorem
$$0\le|\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}|\leq\frac{|x|\log(1+y)}{y}\to0$$
For $(x,y)\to(0,0)$
Here I have used the fact that
$$\sqrt{x^2+y^2}\ge\sqrt{y^2}=y$$
Such that
$$\frac{1}{\sqrt{x^2+y^2}}\le\frac{1}{y}$$
And the fact that
$$\lim_{y\to0}\frac{\log(1+y)}{y} =0$$
So I can conclude that
$$0\le\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}\le\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{y}\le0$$
And thus by the Squeeze Theorem
$$\lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}}=0$$
Is my proof correct?
|
$\begin{aligned}
& \lim_{(x, y)\to(0,0)}\frac{|x|\log(1+y)}{\sqrt{x^2+y^2}} \xrightarrow{\begin{cases}x=r \cdot \cos\theta
\\ y=r \cdot \sin\theta
\end{cases}}\lim_{r\to 0}{\frac{|r \cdot \cos\theta| \ln{(1+r \cdot \sin\theta)} }{ \sqrt{(r \cdot \cos\theta)^2+(r \cdot \sin\theta)^2} } }
\\& \le \lim_{r\to 0}\frac{|r \cdot \cos\theta|(r\cdot \sin\theta)}{ \sqrt{(r \cdot \cos\theta)^2+(r \cdot \sin\theta)^2} }= \lim_{r\to 0}\frac{r ^2\cdot| \cos\theta|(\sin\theta)}{r}=\lim_{r\to 0} r\cdot| \cos\theta|(\sin\theta)
\\& \le \lim_{r\to 0} r\cdot1 =0
\end{aligned}$
|
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|
The sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$
*
*Show that the sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$ is monotone decreasing and bounded below.
Let $a_n = \frac{n^2}{9^n}$. For $n\geq 1$ we have
\begin{equation*}
a_n-a_{n+1} = \frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}.
\end{equation*}
To show this sequence is decreasing we want to show that this quantity is positive. Now we know that $n^2 > 0$, and $(n+1)^2 > 0$ for $n\geq 1$ so $a_n-a_{n+1} > 0$, i.e. $a_n > a_{n+1}$. Since it is decreasing it must be monotone decreasing as required.
A sequence is bounded below if there is a number $m$ such that $m\leq a_n$ for all $n$. Since all the terms in this sequence are all positive, the sequence is bounded below by $0$, i.e. $a_n\geq 0$ as required.
*Use the result in (i) to prove that $\lim_{n\to\infty} \frac{n^2}{9^n} = 0$.
Since this sequence is monotone decreasing and bounded below, it converges by the monotone convergence theorem. Lets re-write this a little. We can do this by writing
\begin{equation*}
\frac{n^2}{9^n} = \frac{n^2}{\left(e^{\ln{(9)}}\right)^n} = \frac{n^2}{e^{n\ln{(9)}}}
\end{equation*}
Now we have a limit of the form $\frac{\infty}{\infty}$ so we can apply L'Hopital's rule which gives
\begin{equation*}
\begin{split}
\lim_{n\to\infty} \frac{n^2}{9^n} &= \lim_{n\to\infty} \frac{n^2}{e^{n\ln{(9)}}} \\
&= \lim_{n\to\infty} \frac{2n}{\ln{(9)}e^{n\ln{(9)}}} \\
&= \lim_{n\to\infty} \frac{2}{\left(\ln{(9)}\right)^2e^{n\ln{(9)}}} \\
&= 0
\end{split}
\end{equation*}
as required.
Is this good? Thanks!
|
Notice that
\begin{align*}
a_n-a_{n+1}&=\frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}\\
&=\frac{1}{9^n}\left(\frac{9n^2-n^2-1-2n}{9}\right)\\
&=\frac{8n^2-1-2n}{9^{n+1}}.
\end{align*}
Now, $8n^2-1-2n=(4n+1)(2n-1)\gt 0$ because $n\geq 1$.
|
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|
The proof of van der Corput inequality $\newcommand{\lrp}[1]{\left(#1\right)}$
$\newcommand{\lrmod}[1]{\left|#1\right|}$
I am trying to understand the proof of the van der Corput inequality given in Lemma 1 of this blog entry due to Tao. We will use the Big-O notation, whose definition can be found here.
Lemma. (Van der Corput Inequality).
Let $a_1 a_2, a_3 , \ldots$ be a sequence of complex numbers bounded by magnitude $1$.
Then for any $1\leq H\leq N$ we have
$$
\lrmod{
\frac{1}{N} \sum_{n=1}^N a_n
}
\leq
\lrp{
\frac{1}{H}\sum_{h=0}^{H-1} \lrmod{ \frac{1}{N} \sum_{n=1}^N a_{n+h}\bar a_n}
}^{1/2} + O\lrp{ \frac{H}{N}}
$$
The proof Tao has provided proceeds as follows. It is easy to see that
$$
\lrmod{
\frac{1}{N} \sum_{n=1}^N a_n - \frac{1}{N} \sum_{n=1}^N a_{n+h}
}
=
O\lrp{\frac{H}{N}}
$$
Therefore
$$
\frac{1}{H}\sum_{h=0}^{H-1}\lrmod{
\frac{1}{N} \sum_{n=1}^N a_n - \frac{1}{N} \sum_{n=1}^N a_{n+h}
}
=
O\lrp{\frac{H}{N}}
$$
and hence by triangle inequality we have that
$$
\lrmod{\frac{1}{N} \sum_{n=1}^N a_n - \frac{1}{N}\sum_{n=1}^N \frac{1}{H}\sum_{h=0}^{H-1} a_{n+h}}
=
O\lrp{\frac{H}{N}}
$$
Write
$$
x= \frac{1}{N} \sum_{n=1}^N a_n \text{ and } y_n= \frac{1}{H} \sum_{h=0}^{H-1} a_{n+h}
$$
So the last equation becomes
$$
\lrmod{x- \frac{1}{N}\sum_{n=1}^N y_n} = O\lrp{ \frac{H}{N}}
$$
Again by triangle inequality we have that
$$
|x| \leq \lrmod{ \frac{1}{N} \sum_{n=1}^N y_n} + O\lrp{ \frac{H}{N}} \leq \frac{1}{N} \sum_{n=1}^N |y_n| + O\lrp{ \frac{H}{N}}
$$
By Jensen's inequality we have
$$
\lrp{\frac{1}{N} \sum_{n=1}^N |y_n|}^2 \leq \frac{1}{N}\sum_{n=1}^N |y_n|^2
$$
Thus we get
$$
|x| \leq \lrp{ \frac{1}{N}\sum_{n=1}^N |y_n|^2}^{1/2} + O\lrp{ \frac{H}{N}}
$$
which is nothing but
$$
\lrmod{ \frac{1}{N}\sum_{n=1}^N a_n} \leq \lrp{ \frac{1}{N} \sum_{n=1}^N \lrmod{\frac{1}{H} \sum_{h=0}^{H-1} a_{n+h}}^2}^{1/2} + O\lrp{ \frac{H}{N}}
$$
At this point Tao writes that ``Expanding out the square and rearranging a bit, we soon obtain the desired upper bound."
I am unable to follow this step. Can somebody please give some details here. Also, if you can comment on what this inequality is trying to capture then please feel free to share.
|
You are mistaking the inequality. Tao is saying that there is a constant $C > 0$ with $|\frac{1}{N}\sum_{n \le N} a_n| \le C(\frac{1}{H}\sum_{h \le H} |\frac{1}{N}\sum_{n \le N} a_{n+h}\overline{a_n}|)^{1/2} + C\frac{H}{N}$.
Actually, I think there should be a $C\frac{1}{H^{1/2}}$ on the right hand side as well (see the link I gave).
Based on where you got to, and using $\sqrt{x+y} \le \sqrt{x}+\sqrt{y}$, it suffices to show $$\frac{1}{N}\sum_{n \le N} |\frac{1}{H}\sum_{h \le H} a_{n+h}|^2 \le \frac{2}{H}\sum_{h \le H} |\frac{1}{N}\sum_{n \le N} a_{n+h}\overline{a_n}|+O(\frac{1}{H})+O(\frac{H}{N}).$$
Merely by expanding the square, the left hand side is $$\frac{1}{H^2}\sum_{h_1,h_2 \le H} \frac{1}{N}\sum_{n \le N} a_{n+h_1}\overline{a_{n+h_2}}.$$ The term $h_1 = h_2$ has contribution $$\frac{1}{H^2}\sum_{h \le H} \frac{1}{N}\sum_{n \le N} |a_{n+h}|^2 = O(\frac{1}{H}),$$ so we can ignore it. We are left with
\begin{align*}
\frac{2}{H^2}\sum_{1 \le h_1 < h_2 \le H} \frac{1}{N}\sum_{n \le N} a_{n+h_1}\overline{a_{n+h_2}} &= \frac{2}{H^2}\sum_{1 \le h_1 < h_2 \le H} \frac{1}{N}\sum_{n=h_1+1}^{N+h_1} a_n\overline{a_{n+h_2-h_1}} \\ &= \frac{2}{H^2}\sum_{1 \le h_1 < h_2 \le H} \left[\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h_2-h_1}} + O(\frac{H}{N})\right].
\end{align*}
The $O(\frac{H}{N})$ term merely comes out of the triple sum (since everything is averaged), so we can ignore it. We are left with $$\frac{2}{H^2} \sum_{1 \le h_1 < h_2 \le H} \frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h_2-h_1}} = \frac{2}{H^2}\sum_{h \le H}\sum_{\substack{1 \le h_1 < h_2 \le H \\ h_2-h_2 = h}} \frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}}.$$
Now, we can pull out the sum in $n$ and note that the number of $1 \le h_1 < h_2 \le H$ with $h_2-h_1 = h$ is $H-h$ to get $$\frac{2}{H^2}\sum_{h \le H} (H-h)\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}},$$ which is upper bounded by $$\frac{2}{H^2}\sum_{h \le H} (H-h)\left|\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}}\right|,$$ which is of course upper bounded by $$\frac{2}{H}\sum_{h \le H} \left|\frac{1}{N}\sum_{n=1}^N a_n\overline{a_{n+h}}\right|,$$ as desired.
In regards to what this inequality is trying to capture. First, note that it holds in any Hilbert space, i.e. $||\frac{1}{N}\sum_{n \le N} v_n|| \le C(\frac{1}{H}\sum_{h \le H} ||\frac{1}{N}\sum_{n \le N} \langle v_{n+h},v_n \rangle||)^{1/2}+C\frac{1}{H^{1/2}}+C\frac{H}{N}$ (just go through the proof). Now suppose the $v_n$'s are mutually orthogonal. Then the left hand side is $\frac{1}{\sqrt{N}}$, while the right hand side is $O(\frac{1}{H^{1/2}})+O(\frac{H}{N})$ (which actually shows the need for the $O(\frac{1}{H^{1/2}})$ term), where the $O(\frac{1}{H^{1/2}})$ term came from $\frac{1}{N}\sum_{n \le N} \langle v_n,v_n \rangle$, the "diagonal term" in the proof above. The point is that the Van der Corput inequality allows one to get good bounds on $||\frac{1}{N}\sum_{n \le N} v_n||$ when the $v_n$'s are "almost orthogonal" (the preceding sentence shows it specializes to actual orthogonality).
|
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|
Prove inequality for a sum to $2^{n+1}$ Problem:I realize I made a mistake on one of the last questions I posted.Here is the problem:
Prove for all $n \in \mathbb{N}$
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$
Attempt:
I'm having trouble recognizing the pattern for the general term of the inequality, can someone help me?
trying cases
for $n=1$ then $\sum\limits_{k=1}^{2^{1+1}} \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}$
for $n=2$ then $\sum\limits_{k=1}^{2^{2+1}}\frac
{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+ \frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}$
Can someone tell me if this is the correct pattern to use in the proof?Would this be a proof by induction or a direct proof?
Direct proof attempt
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k}>\frac{n2^{n}}{2^{n+1}}=\frac{n}{2}$
|
Induction will work. Although, the observation made by José shows why induction works.
Suppose that $n\in\mathbb N$. For the base case when $n=1$ we have
$$\sum\limits_{k=1}^{2^2} \frac{1}{k}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{1}{2}$$
then for the induction hypothesis we will assume that
$$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$$
for all $k\in\mathbb N$. So, for the induction step we need to show that
$$\sum\limits_{k=1}^{2^{n+2}} \frac{1}{k} > \frac{n+1}{2}$$
Starting from the LHS
\begin{align}\sum\limits_{k=1}^{2^{n+2}} \frac{1}{k}&=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots + \frac{1}{2^{n+1}}+\frac{1}{2^{n+1}+1}+\dots+\frac{1}{2^{n+2}}\\&=
\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k}+\frac{1}{2^{n+1}+1}+\dots+\frac{1}{2^{n+2}}
\\& >\frac{n}{2}+\frac{1}{2^{n+1}+1}+\dots+\frac{1}{2^{n+2}}\\&>\frac{n}{2}+\overbrace{\frac1{2^{n+2}}+\frac1{2^{n+2}}+\cdots+\frac1{2^{n+2}}}^{2^{n+1}\text{ times}}\\&=\frac{n}{2}+\frac{2^{n+1}}{2^{n+2}}\\&=\frac{n}{2} + \frac{1}{2}\\&=\frac{n+1}{2}
\end{align}
|
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|
In simplifying the formula that I've derived for finding the square root of a complex number to the standard formula. So by easy means, I derived
$\sqrt{a+ib} = \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$
But then I checked for the actual formula it is this;
$\sqrt{a+ib} = \sqrt{\frac{\sqrt{a^2 + b^2}+a}{2}}± i(\sqrt{\frac{\sqrt{a^2 + b^2}-a}{2}})$
So how do you simplify that to this?
Btw here is the derivation;
Let $\sqrt{a+ib} = x+iy$
$a+ib=x^2 - y^2 + i2xy$
We know, $(x^2-y^2)^2 + (2xy)^2 = x^2 + y^2$
or, $x^2 + y^2 = a^2 + b^2$
Hence we get $x = \sqrt{\frac{a+a^2+b^2}{2}}$ and
$y=\frac{b}{\sqrt{2}\sqrt{a^2+a+b^2}}$
Therefore;
$\sqrt{a+ib} =
> \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$
|
You made a mistake in your derivation. You should have $(x^2+y^2)^\color{red}2=a^2+b^2$.
This follows from $a=x^2-y^2$ and $b=2xy$ (or from known properties of complex modulus).
Thus, $a=x^2-\dfrac {b^2}{4x^2}$;
solving this quadratic equation in $x^2$ yields $x^2=\dfrac{a+\sqrt{a^2+b^2}}2$ as the correct answer.
|
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|
PMF for the maximum of two four-sided dices
Suppose we roll two four-sided dice, that is, each die has four sides, numbered $1, 2, 3, 4$. Let $X_1$ and $X_2$ be the numbers that appear on the first and second die respectively, and let $Z = \max\{X_1, X_2\}$, that is $Z$ is the larger of the two numbers rolled. Find the probability mass function of $Z$.
The tricky part for me is that it takes the $\max$ of two numbers. Therefore, my initial approach was listing all the possible combinations where there could be a $\max$ number, e.g.: $(1,2, 2,1, 2,2)$ for side $2$. But I assumed that, because there is no $0$ side, $1$ can't be max. I'm struggling to get the probabilities $f_z(z) = 12/16$, $f_z(z) = 4/16$, $f_z(z) = 3/16$. $f_z(z) = 5/16$. What $z$ values would they have to be to get to these probabilities? Note that I used $f_z(z)$, for $Z$. I initially thought the $z$ would be an interval of values but I was completely wrong.
|
Assuming the die is fair, there are $4^2 = 16$ equally likely outcomes.
The maximum of a set, if it exists, is the largest element of the set.
\begin{array}{c c}
\text{outcome} & \text{maximum}\\ \hline
(1, 1) & 1\\
(1, 2) & 2\\
(1, 3) & 3\\
(1, 4) & 4\\
(2, 1) & 2\\
(2, 2) & 2\\
(2, 3) & 3\\
(2, 4) & 4\\
(3, 1) & 3\\
(3, 2) & 3\\
(3, 3) & 3\\
(3, 4) & 4\\
(4, 1) & 4\\
(4, 2) & 4\\
(4, 3) & 4\\
(4, 4) & 4
\end{array}
By inspection, a maximum of $1$ occurs once, a maximum of $2$ occurs three times, a maximum of $3$ occurs five times, and a maximum of $4$ occurs $7$ times. Hence, we obtain the probability mass function
\begin{align*}
\Pr(Z = 1) & = \frac{1}{16}\\
\Pr(Z = 2) & = \frac{3}{16}\\
\Pr(Z = 3) & = \frac{5}{16}\\
\Pr(Z = 4) & = \frac{7}{16}
\end{align*}
As a sanity check, notice that we have accounted for all possible outcomes and that $$\Pr(Z = 1) + \Pr(Z = 2) + \Pr(Z = 3) + \Pr(Z = 4) = \frac{1}{16} + \frac{3}{16} + \frac{5}{16} + \frac{7}{16} = 1$$
as required
|
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|
How do I derive the following expression for the sum of orthogonal matrices? In Johansen's book 'Likelihood-based inference in cointegrated vector autoregressive models', in order to get the expression for the Granger's representation theorem he claims that:
$$\beta_\bot(\alpha'_\bot \beta_\bot )^{-1} \alpha'_\bot + \alpha (\beta' \alpha)^{-1} \beta' = I \tag{1}$$
Where:
$\alpha$ is $N\times R$, $\text{rank}(\alpha) =R$
$\beta$ is $N\times R$, $\text{rank}(\beta) =R$
$\beta_\bot $ is $N\times (N-R)$, $\text{rank}(\beta_\bot) =N-R$
$\alpha_\bot $ is $N\times (N-R)$, $\text{rank}(\alpha_\bot) =N-R$
$\alpha' \alpha_\bot =0$
$\beta' \beta_\bot =0$
I am not able to prove (1). Can you help me, please?
|
The stated identity need not be valid under the assumptions provided. Consider for example $\alpha=\beta_{\bot}=\binom{1}{0}$ and $\beta=\alpha_{\bot}=\binom{0}{1}$ corresponding to $N=2$, $R=1$. These vectors all have rank $R=N-R=1$, and they satisfy $\alpha'\alpha_\bot = \beta'\beta_\bot = 0$. But $\alpha'_\bot \beta_\bot=\beta'\alpha=0$, so $(\alpha'_\bot \beta_\bot)^{-1}$, $(\beta'\alpha)^{-1}$ do not exist.
The identity is true, however, if we may assume that the 2-by-1 block matrices $(\alpha,\beta_\bot)$, $(\beta,\alpha_\bot)$ are nonsingular. In that case we have the following derivation:
\begin{align}
I
&=\begin{pmatrix} \alpha & \beta_\bot \end{pmatrix} \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}^{-1} \begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}^{-1}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix} \\
&= \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}
\left[\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix} \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix} \right]^{-1}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}\\
&= \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}
\begin{pmatrix} \beta'\alpha & \beta'\beta_\bot \\ \alpha_\bot'\alpha & \alpha'_\bot \beta_\bot \end{pmatrix}^{-1}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}\\
&= \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}
\begin{pmatrix} \beta'\alpha & 0 \\ 0 & \alpha'_\bot \beta_\bot \end{pmatrix}^{-1}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}\\
&= \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}
\begin{pmatrix} (\beta'\alpha)^{-1} & 0 \\ 0 & (\alpha'_\bot \beta_\bot)^{-1} \end{pmatrix}\begin{pmatrix} \beta' \\ \alpha_\bot' \end{pmatrix}\\
&= \begin{pmatrix} \alpha & \beta_\bot \end{pmatrix}
\begin{pmatrix} (\beta'\alpha)^{-1}\beta' \\ (\alpha'_\bot \beta_\bot)^{-1}\alpha_\bot' \end{pmatrix}\\
&=\alpha(\beta'\alpha)^{-1}\beta' + \beta_\bot(\alpha'_\bot \beta_\bot)^{-1}\alpha_\bot'.
\end{align}
Thus the identity follows algebraically if we grant that the above block-matrix inverses exist. It may be possible to weaken the premises further.
|
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|
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
|
Also, we can use C-S here:
$$19.5=\sqrt{6.5^2\cdot9}=\sqrt{(2.5^2+6^2)((2x-1)^2+(y+1)^2)}\geq$$
$$\geq2.5(2x-1)+6(y+1)=5x+6y+3.5,$$
which gives $$5x+6y\leq16.$$
The equality occurs for $$(2.5,6)||(2x-1,y+1),$$ which says that $16$ is a maximal value.
|
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|
Solve for $\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x$ It is known that $$\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x=\pi \ln \left|2 \cos \frac{a}{2}\right|+a \ln \left|\tan \frac{a}{2}\right|+2 \sum_{k=0}^{+\infty} \frac{\sin (2 k+1) a}{(2 k+1)^{2}}$$
In an attempt to derive the equation, I utilized
$$\sum_{n=0}^{\infty}\left(\frac{e^{i x}}{a}\right)^{n}=\frac{1}{1-\frac{e^{i x}}{a}}=\frac{a(a-\cos x)}{a^{2}-2 a \cos x+1}+i \frac{a \sin x}{a^{2}-2 a \cos x+1}$$, where a>1
and $$\sum_{n=0}^{\infty}\left(ae^{i x}\right)^{n}=\frac{1}{1-ae^{i x}}=\frac{a(a-\cos x)}{a^{2}-2 a \cos x+1}+i \frac{a \sin x}{a^{2}-2 a \cos x+1}$$, where a<1
The series expansion for $\ln \left(x^{2}+2 \sin a \cdot x+1\right)$ can be obtained by integrating the imaginary part of the equation and plugging in $x=\frac{\pi}{2}-a$, $a=x$
However, I still failed to get the desired result by using this series expansion. I wonder whether I can obtain the equation through such Fourier series or I just missed some important point. Any help would be appreciated.
|
Let's follow the advice by @Sonal_sqrt. For $-\pi/2<a<3\pi/2$,
$$\int_0^\infty\frac{\cos a\ dx}{1+2x\sin a+x^2}=\tan^{-1}\left.\frac{x+\sin a}{\cos a}\right|_{x=0}^{x=\infty}=\frac{\pi}{2}-a,$$
so, if $I(a)$ is the given integral, then (for the same range of $a$)
\begin{align}
I'(a)&=\int_0^\infty\frac{2x\cos a\ dx}{(1+x^2)(1+2x\sin a+x^2)}
\\&=\cot a\int_0^\infty\left(\frac{1}{1+x^2}-\frac{1}{1+2x\sin a+x^2}\right)dx
\\&=\cot a\left(\frac{\pi}{2}-\frac{1}{\cos a}\Big(\frac{\pi}{2}-a\Big)\right)=\frac{a}{\sin a}-\frac{\pi}{2}\tan\frac{a}{2}.
\end{align}
Knowing $I(0)=-2\int\limits_0^{\pi/2}\ln\cos t\,dt=\pi\ln 2$, and integrating $a/\sin a$ by parts, we have
$$I(a)=\pi\ln\Big|2\cos\frac{a}{2}\Big|+a\ln\Big|\tan\frac{a}{2}\Big|-\int_0^a\ln\Big|\tan\frac{x}{2}\Big|\,dx.$$
Finally, note that
$$-\ln\Big|\tan\frac{x}{2}\Big|=\Re\ln\frac{1+e^{ix}}{1-e^{ix}}=2\sum_{n=0}^{\infty}\frac{\cos(2n+1)x}{2n+1},$$
and integrability of this series is shown using Dirichlet's test.
|
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|
Finding the length of the arc for function $8y = x^4 +2x^{-2}$
Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$
I first isolated for $y$ and derived:
$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$
$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$
Then found the arc length:
$$L = \int ^2 _1 \sqrt{1+[f'(x)]^2} dx
\\ = \int ^2 _1 \sqrt{1+({1\over 2}x^3 -{1 \over 2}x^{-3})^2} dx
\\ = \int ^2 _1 \sqrt{{1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}} dx$$
Let $u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}$
$du = {{3 \over 2} x^5 - {3\over 2}x^{-7} dx}$
However, I cannot seem to integrate this
How can I integrate this equation?
|
Hint. Note that
$${1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}=\frac{(x^3+x^{-3})^2}{4}.$$
so it remains to evaluate
$$L=\frac{1}{2}\int ^2 _1(x^3+x^{-3})\,dx.$$
|
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|
Evaluating the following integral: $\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx$ How can we compute this integral for all $\operatorname{Re}(a)>0$ and $\operatorname{Re}(b)>0$?
$$\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx$$
Is there a way to compute it using methods from real analysis?
|
**my attempt **
$$I=\int_{0}^{\infty }\frac{x\ cos(ax)}{e^{bx}-1}dx=\sum_{n=1}^{\infty }\int_{0}^{\infty }x\ e^{-bnx}cos(ax)dx\\
\\
\\
=\frac{1}{2}\sum_{n=1}^{\infty }\int_{0}^{\infty }x\ e^{-bnx}\ (e^{iax}-e^{-iax})dx=\frac{1}{2}\sum_{n=1}^{\infty }[\int_{0}^{\infty }x\ e^{-(bn-ia)}dx+\int_{0}^{\infty }x\ e^{-(bn+ia)}dx]\\
\\
\\
=\frac{1}{2}\sum_{n=1}^{\infty }(\frac{\Gamma (2)}{(bn-ai)^2}+\frac{\Gamma (2)}{(bn+ia)^2})=\frac{1}{2b^2}\sum_{n=0}^{\infty }\frac{1}{(n-\frac{ai}{b})^2}+\frac{1}{2b^2}\sum_{n=0}^{\infty }\frac{1}{(n+\frac{ai}{b})^2}+\frac{1}{a^2}\\
\\$$
$$=\frac{1}{a^2}+\frac{1}{2b^2}(\Psi ^{1}(\frac{ai}{b})+\Psi ^{1}(\frac{-ai}{b}))\\\\\
\\
but\ we\ know\ \Psi ^{(1)}(\frac{-ai}{b})=\Psi ^{(1)}(1-\frac{ai}{b})-\frac{b^2}{a^2}\\
\\
\\
\therefore \ I=\frac{1}{a^2}+\frac{1}{2b^2}\left ( \Psi ^{(1)}(1-\frac{ai}{b}) +\Psi ^{(1)}(\frac{ai}{b})-\frac{b^2}{a^2}\right )\\
\\
\\
by\ using\ the\ reflection\ formula\ :\ \Psi ^{(1)}(1-\frac{ai}{b})+\Psi ^{(1)}(\frac{ai}{b})=\frac{\pi ^2}{sin^2(\frac{i\pi a}{b})}\\
\\$$
so we have
$$\therefore I=\frac{1}{2a^2}+\frac{1}{2b^2}\left ( \frac{-\pi ^2}{sinh^2(\frac{\pi a}{b})} \right )\\
\\
\\
=\frac{1}{2a^2}-\frac{\pi ^2}{2b^2sinh^2(\frac{\pi a}{b})}\ \ \ \ \ \ , b>0$$
note that :
$$\frac{\pi ^2}{sin^2(\frac{i\pi a}{b})}=-\frac{\pi ^2}{sinh^2(\frac{\pi a}{b})}$$
|
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|
Finding polynomial functions that satisfy a given condition
Find all polynomial functions $P(x)$ which satisfy $P(x)^2=P(P(x))$.
Attempt at reaching answer:
Let $d$ be the degree of $p(x) $.
We have $d+d=d*d \implies d=0 \vee d=2$.
If $d=0$ we have that $p(x)$ is constant, and this is indeed a solution.
Otherwise, we have $d=2 \implies p(x)=ax^2+bx+c$.
Plugging this into the original equation, we get
$(ax^2+bx+c)^2=a(ax^2+bx+c)^2+b(ax^2+bx+c)+c$.
Checking the coefficient of $x^4$ gives trivially $a=1$.
Rewriting and simplifying gives
$bx^2+b^2x+bc+c=0 \forall x \implies b=0 $ , $ c=0$.
Hence our solutions are
$p(x) = x $ and $p(x)= c$, with $c \in \mathbb{R} $.
Apparently, the first function (p(x)=x) is correct, but the second one isn't because it's not a polynomial function and it doesn't work for $x^2+1$... What would the correct answer then be?
|
You may proceed as follows:
Differentiating gives
$$2P(x)P'(x) = P'(P(x))P'(x)$$
Obviously the identity is satisfied, if $P'(x) = 0 \Rightarrow P(x) = c \stackrel{c^2=c}{\Longrightarrow} c=0$ or $c=1$ .
Now, if $P'(x) \neq 0$, then setting $y=P(x)$ gives us on a non-empty open interval the polynomial equation
$$2y = \frac{dP}{dy} \Rightarrow P(y) = y^2+d$$
Plugging this into the original identity gives
$$(y^2+d)^2 \stackrel{!}{=}(y^2+d)^2 + d \Leftrightarrow d=0$$
So, the possible polynomials are $P(x) =x^2$, $P(x) =1$, and $P(x) =0$.
|
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|
Fully simplifying $\sqrt{13+2\left(\sqrt{2}-\sqrt{10}-\sqrt{20}\right)}$ I have this statement:
Simplify this: $\sqrt{13+2\left(\sqrt{2}-\sqrt{10}-\sqrt{20}\right)}$
I know how to simplify roots like $\sqrt{a \pm k\sqrt{b}}$, But i don't know how to simplify this.
Any hint for an elegant solution is appreciated.
|
Notice that $$(-\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z-2\sqrt{xy}+2\sqrt{yz}-2\sqrt{xz}$$
So $$\sqrt{x+y+z-2\sqrt{xy}+2\sqrt{yz}-2\sqrt{xz}}=-\sqrt{x}+\sqrt{y}+\sqrt{z}$$
In your case on comparision you have $x+y+z=13$, $xy=20, zx=10, yz=2.$ So you get $xyz=20$, then $x=10,y=2,z=1$, Hence the answer is $\sqrt{10} -\sqrt{2}- \sqrt{1}.$
|
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|
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\tan\theta=-5$
Ref$(\theta)$=$\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}$
$$
\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-25}{1+25}=\frac{-24}{26}=\frac{-12}{13}\\
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{-10}{26}=\frac{-5}{13}\\
Ref(\theta)\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta)
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}=\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-13\end{bmatrix}\\
=\frac{1}{13}\begin{bmatrix}-48+65\\-20-156\end{bmatrix}=\frac{1}{13}\begin{bmatrix}17\\-176\end{bmatrix}
$$
Why am I not getting the required solution in Method two using matrix method ?
Thanx @ganeshie8 for the remarks, so in that case how do I find the operator for reflection of a point over the line not passing through the origin ?
|
As ganeshie8 suggested, your matrix formula is not working because the line does not pass through the origin.
When you translate everything up by $6$ units, the line now passes through the origin and you can continue as follows:
$$\begin{bmatrix}
\dfrac{-12}{13} & \dfrac{-5}{13} \\ \dfrac{-5}{13} & \dfrac{12}{13}
\end{bmatrix}\begin{bmatrix}4\\-7\end{bmatrix}\\$$
$$=\frac{1}{13} \begin{bmatrix}
-48+35 \\ -20-84
\end{bmatrix}\\$$
$$= (-1, -8)$$
And now translate down by $6$ units to find that the original coordinate is at $(-1,-14)$.
|
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|
An interesting identity involving the abundancy index of divisors of odd perfect numbers Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
A number $y$ is said to be perfect if $\sigma(y)=2y$.
Denote the abundancy index of $z$ by $I(z)=\sigma(z)/z$.
Euler proved that an odd perfect number $N$, if one exists, must necessarily have the form $N = q^k n^2$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
While considering the difference
$$I(n^2) - I(q^k)$$
for $k=1$, I came across the interesting identity
$$\frac{d}{dq}\bigg(I(n^2)-I(q)\bigg)=\frac{d}{dq}\bigg(\frac{q^2 - 2q - 1}{q(q+1)}\bigg)=\frac{3q^2 + 2q + 1}{q^2 (q+1)^2}.$$
This is interesting because of
$$I(n^2)+I(q)=\frac{2q}{q+1}+\frac{q+1}{q}=\frac{3q^2 + 2q + 1}{q(q+1)}=q(q+1)\bigg(\frac{3q^2 + 2q + 1}{q^2 (q+1)^2}\bigg)$$
so that we have the identity (or differential equation (?))
$$q(q+1)\frac{d}{dq}\bigg(I(n^2)-I(q)\bigg)=I(n^2)+I(q).$$
Two questions:
[1] Is there a simple explanation for why the identity (or differential equation (?)) holds?
[2] Are there any other identities that could be derived in a similar fashion?
Update (July 25, 2020 - 10:15 AM Manila time)
I tried computing the derivative
$$\frac{d}{dq}\bigg(I(n^2)+I(q)\bigg)$$
and I got
$$q(q+1)\frac{d}{dq}\bigg(I(n^2)+I(q)\bigg)=q(q+1)\frac{d}{dq}\bigg(\frac{3q^2 + 2q + 1}{q(q+1)}\bigg)=q(q+1)\bigg(\frac{q^2 - 2q - 1}{q^2 (q+1)^2}\bigg)=I(n^2)-I(q).$$
|
There is nothing too special about the identities, as we shall soon see.
Hereinafter, we let $q^k n^2$ be an odd perfect number with special prime $q$, and we assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds true. Also, we denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)$ is the classical sum of divisors of $x$.
We compute:
$$\frac{d}{dq}(I(n^2) + I(q))=\frac{d}{dq}\bigg(\frac{3q^2 + 2q + 1}{q(q+1)}\bigg)=\frac{d}{dq}\bigg(3 - \frac{q - 1}{q(q + 1)}\bigg)=\frac{d}{dq}\bigg(\frac{-1}{q+1}+\frac{1}{q(q+1)}\bigg)$$
$$=\frac{d}{dq}\bigg(\frac{-2}{q+1}+\frac{1}{q}\bigg)=\frac{2}{(q+1)^2}-\frac{1}{q^2}= \frac{2q^2 - (q+1)^2}{(q(q+1))^2}$$
But
$$I(n^2) - I(q) = \frac{2q}{q+1} - \frac{q+1}{q} = \frac{2q^2 - (q+1)^2}{q(q+1)}.$$
Similarly, we obtain:
$$\frac{d}{dq}(I(n^2) - I(q))=\frac{d}{dq}\bigg(\frac{q^2 - 2q - 1}{q(q+1)}\bigg)=\frac{d}{dq}(I(n^2) + I(q) - 2I(q))$$
$$=\frac{d}{dq}\bigg(3 - \frac{q - 1}{q(q + 1)} - \frac{2(q+1)}{q}\bigg)=\frac{d}{dq}\bigg(\frac{-1}{q+1}+\frac{1}{q(q+1)}+\frac{-2}{q}-2\bigg)$$
$$=\frac{d}{dq}\bigg(\frac{-2}{q+1}+\frac{1}{q}+\frac{-2}{q}-2\bigg)=\frac{d}{dq}\bigg(\frac{-2}{q+1}-\frac{1}{q}-2\bigg)=\frac{2}{(q+1)^2}+\frac{1}{q^2}$$
$$=\frac{2q^2 + (q+1)^2}{(q(q+1))^2}.$$
But
$$I(n^2)+I(q)=\frac{2q}{q+1}+\frac{q+1}{q}=\frac{2q^2 + (q+1)^2}{q(q+1)}.$$
In the above computations, we have used the identity
$$\frac{1}{q(q+1)}=\frac{1}{q}-\frac{1}{q+1}.$$
|
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|
How to show $13\left(x^{2}+1\right)+\sqrt{2\left(x^{4}+1\right)}+62 x-\frac{45\left(x^{2}-1\right)}{\ln x}>0$ for $x\in(0,1)$ I am doing a problem that can be reduced to proving the following inequality:
$$13\left(x^{2}+1\right)+\sqrt{2\left(x^{4}+1\right)}+62 x-\frac{45\left(x^{2}-1\right)}{\ln x}>0, x \in(0,1)
$$
But I don't know how to handle it.
I tried to separate $\ln x$ from the function，because it is easier to differentiate. But I cannot handle the radical expression.
Any ideas?
|
Here's my sketch proof:
For the following assume $x\in(0,1)$. First use the following lemma:
$$\left(x^2+1\right)+\sqrt{2}\sqrt{x^4+1}>2x+2\left(\frac{x^4+1}{x^2+1}\right)$$
Proof:
$${\frac{d}{dx}\left(2(x^4+1)-\left\{2x+2\left(\frac{x^4+1}{x^2+1}\right)-(x^2+1)\right\}^2\right)\\=\frac{4(x-1)^5(x^4+2x^3+4x^2+4x+1)}{(x^2+1)^3}<0}$$
So plugging in $x=1$ to the original gives $4=4$ and we're done.
Then it remains to show that
$$12\left(x^2+1\right)+2\left(\frac{x^4+1}{x^2+1}\right)+64x>45\left(\frac{x^2-1}{\ln
x}\right)$$
Proof: By algebraic manipulation, this is equivalent to
$$\frac{2}{45}\ln x<\frac{x^4-1}{7x^4+32x^3+12x^2+32x+7}$$
So
$$\frac{d}{dx}\left(\frac{2}{45}\ln x-\frac{x^4-1}{7x^4+32x^3+12x^2+32x+7}\right)\\[5px]=\frac{2(x-1)^6(49x^2+22x+49)}{45x(7x^4+32x^3+12x^2+32x+7)^2}>0$$
and plugging in $x=1$ to the equivalent form gives $0=0$ and we're done.
|
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|
We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$
Can someone help me with the second one?
I forgot to tell that solving the equation is not an option in my case.
|
Take the step below,
$$(x_1-x_2)^2= x_1^2+x_2^2 - 2x_1x_2=\dfrac{7}{4}-2(-\frac 12) = \frac{11}{4}
$$
Thus,
$$|x_1-x_2|=\frac{\sqrt{11}}{2}$$
|
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"language": "en",
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|
computing an integral without making any substitution We are trying to evaluate
$$ \int\limits_0^1 ( \sqrt{2-x^2} - \sqrt{2x-x^2} ~) ~ dx $$
without any substitution (well, this is how this problem is supposed to be solved)
Idea:
We notice that if $y=f(x)$ is the integrand, then $f(1) = \sqrt{2}$ and $f(1)=0$ as is evident. So, my idea would be to evaluate
$$ \int\limits_0^{\sqrt{2}} f^{-1}(y) ~ dy $$
But, this would make it harder since we would need to solve $y = \sqrt{2-x^2} - \sqrt{2x-x^2}$ for $x$... Any ideas how would we tackle this problem?
|
One possible approach to solve this problem consists in interpreting integrals as areas. Indeed, consider the functions
\begin{align*}
\begin{cases}
f(x) = \sqrt{2 - x^{2}}\\
g(x) = \sqrt{2x - x^{2}} = \sqrt{1 - (x-1)^{2}}
\end{cases}
\end{align*}
The graph of the function $f$ is the upper semicircle centered at the origin with radius $r_{1} = \sqrt{2}$. Thus the area corresponding to the first integral is given by
\begin{align*}
\int_{0}^{1}f(x)\mathrm{d}x = \frac{\pi r_{1}^{2}}{4} - \left(\frac{\pi r_{1}^{2}}{8} - \frac{1}{2}\right) = \frac{\pi r_{1}^{2}}{8} + \frac{1}{2} = \frac{\pi}{4} + \frac{1}{2}
\end{align*}
On the other hand, the graph of the function $g$ is the upper semicircle centered at $(1,0)$ whose radius equals $r_{2} = 1$. Therefore we have
\begin{align*}
\int_{0}^{1}g(x)\mathrm{d}x = \frac{\pi r_{2}^{2}}{4} = \frac{\pi}{4}
\end{align*}
Finally, we have
\begin{align*}
\int_{0}^{1}[f(x) - g(x)]\mathrm{d}x = \frac{1}{2}
\end{align*}
|
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|
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?
|
You may use
*
*$(x^2-1)|(x^{2k} -1)$ by applying $(x^2)^k - 1 = (x^2-1)(x^{2(k-1)} + \cdots + x^2 + 1)$
\begin{eqnarray*} x^{81} + x^{49} + x^{25} + x^{9} + x
& = & x\left((x^2)^{40} -1 + (x^2)^{24} -1 + (x^2)^{12} -1 + (x^2)^{4} -1 + 4+1\right) \\
& = & x((x^2-1)p(x) + 5)\\
& = & x(x^2-1)p(x) + 5x
\end{eqnarray*}
So, the remainder is $5x$.
|
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|
Cant get partial fraction decomposition I would like to prove that: $$ L = (\frac{1+z^{-1}}{1+0.5z^{-1}} ) \cdot (\frac{1}{1-z^{-1}}) = \frac{0.166}{z + 0.5} + \frac{1.33}{z - 1} + 1$$
How do I get from left to right? The same problem have other solution that I succeed to get $L = \frac43\frac{z}{z-1} - \frac13\frac{z}{z + 0.5}$(and Its good enough for solving the engineering problem) but I want to understand how do I get to the first
|
First, write the expression as a ratio of expanded polynomials in lowest terms, that is,
$$
\frac{2z^2+2z}{2z^2-z-1}.
$$
Next, use long division to find the quotient and remainder of $\frac{2z^2+2z}{2z^2-z-1}$ and then rewrite it as the quotient plus the remainder over the denominator, as follows
$$
1+\frac{3z+1}{2z^2-z-1}.
$$
Hence, we only have to find the partial fraction decomposition of $\frac{3z+1}{2z^2-z-1}$. To do so, factor the denominator into linear and irreducible quadratic terms and equal it to the partial fraction expansion form
$$
\frac{3z+1}{(z-1)(2z+1)}=\frac{\theta_1}{z-1}+\frac{\theta_2}{2z+1}.
$$
Multiplying both sides by $(z-1)(2z+1)$ and rewriting the previous identity yields
$$
3z+1=\theta_1-\theta_2+z(2\theta_1+\theta_2).
$$
Finally, equating the coefficients on both sides yields the following system
$$
\begin{cases}
1=\theta_1-\theta_1\\
3=2\theta_1+\theta_2
\end{cases}
$$
which yields $\theta_1=\frac43$ and $\theta_2=\frac13$. Hence,
$$
\left(\frac{1+z^{-1}}{1+0.5z^{-1}} \right)\left(\frac{1}{1-z^{-1}}\right)=1+\frac{4}{3(z-1)}+\frac{1}{3(2z+1)}=1+\frac{1.(3)}{z-1}+\frac{0.1(6)}{z+0.5}.
$$
Note that $\frac43=1.333...=1.(3)$ and $\frac{1}{6}=0.1666...=0.1(6)$.
|
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|
Calculate the maximum value of $\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} $ where $a, b, c \in \mathbb R^+$ satisfying $abc = 1$.
Calculate the maximum value of $$\large \frac{bc}{(b + c)^3(a^2 + 1)} + \frac{ca}{(c + a)^3(b^2 + 1)} + \frac{ab}{(a + b)^3(c^2 + 1)}$$ where $a, b, c$ are positives satisfying $abc = 1$.
We have that $$\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} \le \frac{1}{2} \cdot \sum_{cyc}\frac{1}{(b + c)(a^2 + 1)} \le \sum_{cyc}\frac{1}{(b + c)(a + 1)^2}$$
$$ = \sum_{cyc}\frac{1}{a(ab + ca + b + c)(bc + 1)} \le \dfrac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt{bc}(ab + ca + 2\sqrt{bc})}$$
$$ = \frac{1}{2} \cdot \sum_{cyc}\frac{1}{a\sqrt a(b + c) + 2}$$
That's all I got, not because I can't go for more, but since I went overboard with this, there's no use trying to push for more.
|
Let $a=b=c=1$.
Thus, we obtain a value $\frac{3}{16}.$
We'll prove that it's a maximal value.
Indeed, by AM-GM $$\sum_{cyc}\frac{bc}{(b+c)^3(a^2+1)}\leq\sum_{cyc}\frac{bc}{8\sqrt{b^3c^3}(a^2+1)}=\sum_{cyc}\frac{\sqrt{a}}{8(a^2+1)}.$$
Id est, it's enough to prove that
$$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{3}{2},$$ where $a$, $b$ and $c$ are positives such that $abc=1$.
We have $$\frac{3}{2}-\sum_{cyc}\frac{a}{a^4+1}=\sum_{cyc}\left(\frac{1}{2}-\frac{a}{a^4+1}-\frac{1}{2}\ln{a}\right).$$
Let $f(x)=\frac{1}{2}-\frac{x}{x^4+1}-\frac{1}{2}\ln{x}.$
Thus, easy to see that $f(x)\geq0$ for all $0<x<2$
and from here our inequality is proven for $\max\{a,b,c\}<2$.
Let $a\geq2$.
Thus, by AM-GM:
$$\sum_{cyc}\frac{a}{a^4+1}\leq\frac{2}{2^4+1}+2\cdot\frac{x}{4\sqrt[4]{x^4\cdot\left(\frac{1}{3}\right)^3}}=\frac{2}{17}+\frac{\sqrt[4]{27}}{2}<\frac{3}{2}$$ and we are done!
|
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|
$a_{n+1}\ge a_n$ for $a_n=\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$ Let $a_n = {1 \ \over n+1} +{1 \ \over n+2}+ \ldots\ +{1 \ \over 2n}. $
Prove that for $ n \ge\ 3 $ one has $ a_{n+1}\ge\ a_n $. and based on this conclude that $ {a_{2019}}>{3 \over 5}$
I try
$ a_{n+1}- a_n = {1 \ \over n+2}+{1 \ \over n+3} +{1 \ \over n+4}+\ldots\ + {1 \ \over 2(n+1)} - {1 \ \over n+1} -{1 \ \over n+2}-{1 \ \over n+3}-{1 \ \over n+4}- \ldots\ -{1 \ \over 2n} $
$$ = {1 \ \over 2(n+1)}-{1 \ \over n+1}={1 \ \over (n+1)} \gt\ 0 $$
|
Supppose $n \geq 3$.\begin{align*} a_{n+1} &= \frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n+1} + \frac{1}{2n+2} \\ & =\left( \frac{1}{n+1} + \frac{1}{n+2}+ \cdots + \frac{1}{2n-1} + \frac{1}{2n} \right) + \left( \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} \right) \\ & = a_n + \left( \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} \right) \end{align*}
Observe $$ \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} > \frac{1}{2n+2} + \frac{1}{2n+2} - \frac{1}{n+1} =0$$
Thus $a_{n+1} \geq a_n$.
To prove $a_{2019} > \frac{3}{5}$, I recommended you to compute first few values of $\{a_n\}_{n \geq 3}$. Fortunately, at $n=3$ we have
$$a_{2019} \geq a_3 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{37}{60} >\frac{3}{5}$$
|
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|
Is the series $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$ convergent or divergent $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$
$\begin{align}
\sum_{n=1}^{\infty} \frac{4+3^n}{2^n} &= \sum_{n=1}^{\infty} \frac{4}{2^n} + \sum_{n=1}^{\infty} \frac{3^n}{2^n} \\
&= \sum_{n=1}^{\infty} \frac{4}{2 \cdot 2^{n-1}} + \sum_{n=1}^{\infty} \bigg(\frac{3}{2}\bigg)^n \\
&= 2\sum_{n=1}^{\infty}\frac{1}{2^{n-1}} + \sum_{n=1}^{\infty}\frac{3}{2} \bigg(\frac{3}{2}\bigg)^{n-1}
\end{align}$
Now I observe that the two terms are both geometric series and although the first one converges because $\frac{1}{2} < 1$, the second one doesn't because $\frac{3}{2} > 1$. Then can I assume that the series diverges and that is it?
|
Yes. Whenever you have a convergent series $\sum_{n=0}^\infty a_n$ and a divergent series $\sum_{n=0}^\infty b_n$, the series $\sum_{n=0}^\infty(a_n+b_n)$ diverges.
Or you can apply the ratio test to reach the same conclusion.
|
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|
Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/smarter way to do this?
|
Observe:
$(x - \frac{1}{x})^2 = x^2 - 2 + \frac {1}{x^2}\\
(x + \frac{1}{x})^2 = x^2 + 2 + \frac {1}{x^2}\\
(x - \frac{1}{x})^2= (x + \frac 1x)^2 - 4$
We make this substitution in the original expression
$(x + \frac 1x)^4(x - \frac{1}{x})^2)\\
(x + \frac 1x)^4((x + \frac{1}{x})^2 - 4)\\
(x + \frac 1x)^6 - 4(x + \frac{1}{x})^4$
Note, we are doing something not entirely unlike $\sin^2 x = 1 - \cos^2 x$ here.
And use the binomial expansion on each term above.
$(x + \frac 1x)^6 = (x^6 + \frac1{x^6}) + 6 (x^4 + \frac1{x^4}) + 15 (x^2 + \frac1{x^2}) + 20\\
(x + \frac 1x)^4 = (x^4 + \frac1{x^4}) + 4 (x^2 + \frac1{x^2}) + 6$
$(x + \frac 1x)^6 - 4(x + \frac 1x)^4 = (x^6 + \frac1{x^6}) + 2 (x^4 + \frac1{x^4}) - (x^2 + \frac1{x^2}) - 4$
|
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|
Summation or Algebra? I am trying to visualize the concept that should be involved with this problem: What is $x$ in the equation
$$(x-1)-2(x-2)+3(x-3)-4(x-4)+\dots-10(x-10)=0\quad ?$$
What should I do? Express in summation (which I find very hard) or just algebra?
|
$$(x-1)-2(x-2)+3(x-3)-4(x-4)+\dots-10(x-10)=0$$
$$\implies x(1-2+3-4+\cdots-10)-(1-2^2+3^2-\cdots-10^2)=0$$
$$\implies x=\dfrac{1-2^2+3^2-\cdots-10^2}{1-2+3-4+\cdots-10}$$
$$\implies x=11$$
$$1-2^2+3^2-\cdots-10^2=\sum_{n=1}^{10}(-1)^{n-1}n^2=-55$$
and $$1-2+3-4+\cdots-10=\sum_{n=1}^{10}(-1)^{n-1}n=-5$$
|
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|
Definite integral of rational expression involving quadratics I was given the following exercise on my Calculus class:
$$
\int_2^4 \frac{x^2+4x+24}{x^2-4x+8}dx
$$
I studied from a book (author, Stewart) various methods to solve integrals of the form $\int \frac{P(x)}{Q(x)}dx$; a basic idea common to all of this methods is that $P(x)$ should have a smaller degree than $Q(x)$, in order for us to make a useful factorization. Since this is not the case on the integral I was given, I first computed the division of $P(x)=x^2+4x+24$ over $Q(x)=x^2-4x+8$, which took me to the fact that
$P(x)= Q(x) + (8x+32)$. So the integral can be rewritten in the following manner:
$$
\int_2^4 \frac{(x^2-4x+8)+(8x+32)}{x^2-4x+8}dx=\int_2^4 dx+\int_2^4 \frac{8x+32}{x^2-4x+8}dx
$$
Putting aside the first term of the resulting expression, which is a very simple integral, it all comes down to solving the second term. Gladly I have managed to translate the first integral into a new one in which the numerator is of smaller degree than the denominator. But how do I go about solving $\int_2^4 \frac{8x+32}{x^2-4x+8}dx$.
Well, according to Stewart, once we have managed to solve our little degree problem, we have to factorize the denominator, and from there rewrite the whole expression. But the quadratic on the denominator does not seem to have a complete factorization, and so I am clueless on what to do. This is the only method I was taught to solve this kind of integrals. Am I making a mistake somewhere, or how would you go about solving this problem? Thanks in advance.
|
HINT
You may start by noticing that
\begin{align*}
\frac{8x+32}{x^{2}-4x+8} & = 4\times\frac{2x + 8}{x^{2}-4x+8} = 4\times\frac{2x - 4 + 4 + 8}{x^{2}-4x+8}\\\\
& = 4\times\frac{2x - 4}{x^{2}-4x+8} + 4\times\frac{12}{x^{2}-4x+8}
\end{align*}
In order to integrate the first expression, observe that $(x^{2}-4x+8)^{\prime} = 2x-4$. So it remains to solve the integral of the second expression. For this purpose, we can rearrange the denominator as $(x-2)^{2} + 4$. Then you apply the substitution (or any other trigonometric substitution you prefer)
\begin{align*}
\cosh(u) = \frac{x-2}{2}
\end{align*}
Can you take it from here?
|
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|
Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + 1) + 18$
System
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As Steve Kass's question comment suggests, you have
$$\begin{equation}\begin{aligned}
(x + 1) (y + 2) (z + 1) & = (x + 1) (y + 1) (z + 1) + 8 \\
((y + 1) + 1)\left((x + 1) (z + 1)\right) & = (x + 1) (y + 1) (z + 1) + 8 \\
(x + 1) (y + 1) (z + 1) + (x + 1)(z + 1) & = (x + 1) (y + 1) (z + 1) + 8 \\
(x + 1)(z + 1) & = 8
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
$$\begin{equation}\begin{aligned}
(x + 4) (y + 1) (z + 1) & = (x + 1) (y + 1) (z + 1) + 18 \\
((x + 1) + 3)\left((y + 1) (z + 1)\right) & = (x + 1) (y + 1) (z + 1) + 18 \\
(x + 1) (y + 1) (z + 1) + 3(y + 1)(z + 1) & = (x + 1) (y + 1) (z + 1) + 18 \\
3(y + 1)(z + 1) & = 18 \\
(y + 1)(z + 1) & = 6
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since the factors are positive integers and $z + 1$ is a factor in both \eqref{eq1A} and \eqref{eq2A}, then $z + 1$ must divide $\gcd(8,6) = 2$. Thus, $z + 1 = 1$ or $z + 1 = 2$. For the first case, this gives $z = 0$, $x + 1 = 8 \implies x = 7$ and $y + 1 = 6 \implies y = 5$. I'll leave the $z + 1 = 2$ case for you to do.
|
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|
Using integration to compute the area of a region bounded by a curve For the area bounded by curves
$$(x-1)^2 + y^2 =1 $$
and
$$x^2 + y^2 =1$$
I tried it by finding $y$ and integrating within the suitable limits through $y\,\mathrm{d}x$ method. But when I tried the $x\,\mathrm{d}y$ way the answer was different. For the $y\,\mathrm{d}x$ method the answer was $\frac{2\pi}{3} - \frac{\sqrt3}2$ . But the answer for $x\,\mathrm{d}y$ method was not same.
(y.dx means I found the function y in terms of x and integrated in the respective limits. x.dy means I found the function x in terms of y and integrated within respective limits.)
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Through geometry, we can see that the area bound by the curves is made up of $4$ equal areas -- namely the area bound by the upper red semicircle and the $x$-axis on the interval $x\in[0.5,1]$.
If the shaded area above is $A$, then the total area bound by the curves is $4A$.
We can then write the red semicircle as a function $y$ in terms of $x$, and perform the integration.
$$x^2+y^2=1\Rightarrow y=\sqrt{1-x^2}$$
\begin{align}
A&=\int_{0.5}^1\sqrt{1-x^2}\ dx\\
4A&=4\int_{0.5}^1\sqrt{1-x^2}\ dx\\
&=4\int_{\frac{\pi}6}^{\frac{\pi}2}\sqrt{1-\sin^2\theta} \cos\theta\ d\theta&(\textrm{let }x=\sin\theta\Rightarrow dx=\cos\theta\ d\theta)\\
&=4\int_{\frac{\pi}6}^{\frac{\pi}2}\cos^2\theta\ d\theta\\
&=4\int_{\frac{\pi}6}^{\frac{\pi}2}\frac12(1+\cos 2\theta)\ d\theta\\
&=2\int_{\frac{\pi}6}^{\frac{\pi}2}1+\cos 2\theta\ d\theta\\
&=2(\theta+\frac12\sin 2\theta)\bigg]_{\frac{\pi}6}^{\frac{\pi}2}\\
&=2\theta+\sin 2\theta\bigg]_{\frac{\pi}6}^{\frac{\pi}2}\\
&=(\pi-\frac{\pi}3)+(\sin\pi-\sin\frac{\pi}3)\\
&=\frac{2\pi}3-\frac{\sqrt3}2
\end{align}
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|
proving the identity $b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$ for $b \in \mathbb{N}$, $b > 2$, by counting a set in different ways I'm trying to prove that $\forall b \in \mathbb{N}, b>2,$ $b^{3} = 6\binom{b} {3} +6 \binom{b}{2} +b$ without just using algebra. The idea I've thought about is that there are $b^{3}$ ways to choose a triple of numbers from $\{1,\ldots, b\}$, and there are obviously $b$ ways to choose one element, the issue is I'm not sure how the $\binom{b}{3}$ and $\binom{b}{2}$ terms contribute. Any help would be greatly appreciated.
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$$b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$$
Method 1: We count ordered triples with entries drawn from the set $\{1, 2, 3, \ldots, b\}$ with replacement in two ways.
There are $b$ choices for each of the three entries, so there are $b^3$ such triples.
For the RHS, we consider cases, depending on how many different numbers appear in the triple.
Three different numbers appear in the triple: There are $\binom{b}{3}$ ways to select three different numbers to appear in the triple, and $3!$ ways to arrange them within the triple. There are
$$\binom{b}{3}3! = 6\binom{6}{3}$$
such triples.
Exactly two different numbers appear in the triple: There are $\binom{b}{2}$ ways to select two different numbers to appear in the triple, two ways to choose which of those selected numbers will appear twice, and three ways to choose which of the three entries in the triple will be filled by the singleton. Hence, there are
$$\binom{b}{2} \cdot 2 \cdot 3 = 6\binom{b}{2}$$
such triples.
Exactly one number appears in the triple: As you observed, there are $b$ ways to fill each entry in the triple with the same number.
Hence, the number of ordered triples with entries drawn from the set $\{1, 2, 3, \ldots, b\}$ with replacement is
$$6\binom{b}{3} + 6\binom{b}{2} + b$$
Since we have counted the same set of triples in two different ways, the counts must be the same. Hence,
$$b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$$
Method 2: We distribute three different prizes to $b$ people, where each person may receive up to three prizes.
There are $b$ ways to select the recipient of each of the three prizes, so the prizes can be distributed in $b^3$ ways.
For the RHS, we consider cases, depending on how many different recipients receive the prizes.
Three different recipients of the prizes: There are $\binom{b}{3}$ ways to select three recipients for the prizes and $3!$ ways to distribute the prizes among them, giving
$$\binom{b}{3}3! = 6\binom{b}{3}$$
such distributions.
Two different recipients of the prizes: For this to occur, one person must receive two prizes and a different person must receive the other prize. There are $\binom{b}{2}$ ways to select two recipients for the prizes, two ways to choose which of them receives two prizes, and $\binom{3}{2}$ ways to decide which two of the three prizes that person receives. The other person must receive the remaining prize, so there are
$$\binom{b}{2}\binom{2}{1}\binom{3}{2} = \binom{b}{2} \cdot 2 \cdot 3 = 6\binom{b}{2}$$
such distributions.
One recipient receives all three prizes: There are $b$ ways to select which person receives all three prizes.
Since the three cases are mutually exclusive and exhaustive, the prizes can be distributed in
$$6\binom{b}{3} + 6\binom{b}{2} + b$$
ways.
Since we have counted the same set of prize distributions in two different ways, our counts must be equal. Hence,
$$b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$$
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|
What are the number of ordered pairs $(x,y)$ where both $x$ and $y$ divide $20^{19}$, however $xy$ doesn't? What are the number of ordered pairs $(x,y)$ where both $x$ and $y$ divide $20^{19}$, however $xy$ doesn't?
I started by taking the prime factorization of $20^{19}$ to get : $2^{38}5^{19}$.
I then noticed, that the only way for $xy$ to not divide this number was if the powers of $x$ and $y$ add to greater than $38$ or $19$. For example, if $x=2^{38}$ and $y = 2^{2}$, then $xy$ would be $2^{40}$ which would not divide $2^{38}5^{19}$ evenly.
My question is, what would be the total number of ordered pairs?
|
Well, If $x = 2^a5^c$ and $y = 2^b5^d$
thenn we have $a+b > 38$ or $c+d > 19$.
To have $a+b > 38$ while $a \le 38$ and $b \le 38$ if we have $a = k$ then $b$ can be as small as $39-k$ and can be as large as $38$. Those are $k$ options. $(38+1) - (39-k) = k$. So there are $\sum_{k=1}^{38}k = \frac {38*39}2 = 741$ such pairs.
And likes there are $\sum_{k=1}^{19} k = 190$ such possible pairs.
If $(a,b)$ is such a pairc $c,d$ can be anything from $0$ to $19$. Thatn is $20^2 = 400$ options.
So there are $741*400$ pairs where $a+b > 38$.
Like wise there are $190*39^2$ pairs were $c + d > 19$.
But we counted all the pairs where $a+b > 38$ and $c+d >19$ twice! So we must subtract that number of pairs. There are $190*741$ so pairs.
So the total number of pairs where $a+b>38$ or $c+d > 19$ or both is
$741*400+190*39^2 - 190*741$.
|
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|
Find the maximal value of $c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$ in terms of $c^3 + a^3 = m$ where $a, b$ and $c$ are positives.
Let $a, b$ and $c$ be positive real numbers such that $c^3 + a^3 = m$ and $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0.$$ Find the maximal value of $$c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$$ in terms of $m$.
We have that $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0$$
$$\iff (b^3 - c^3)(a^3 - b^3) \ge 0 \implies b^3(c^3 + a^3 - b^3) \ge (ca)^3 \iff m - b^3 \ge \left(\frac{ca}{b}\right)^3$$
It is evident that $$c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a} = \frac{m - b^3 + 3b + 2}{c + a} \ge \frac{(ca)^3 + 3b^4 + 2b^3}{b^3(c + a)}$$
How should I continue?
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Claim: The maximal value is $${\frak m}=\frac12\left(\frac m2\right)^{2/3}+\left(\frac m2\right)^{-1/3}+\frac32\tag1$$ attained at $a=b=c$.
Let $$d=c^2 - ca + a^2 - \dfrac{(b + 1)^2(b - 2)}{c + a}=\frac{m-b^3+3b+2}{a+c}\tag2$$ with $a^3+c^3=m$. Supposing that $$(b^3 - abc) + ab(a + b) - c(a^2 + b^2)=r\tag3$$$$(abc - b^3) - bc(b + c) + a(b^2 + c^2)=s\tag4$$ with $rs\le0$, subtracting $(3)$ from $(4)$ yields \begin{align}r-s&=-2(abc-b^3)+a^2b-ab^2-a^2c-b^2c+b^2c+bc^2-ab^2-ac^2\\&=2b^3+b(a-c)^2-ac(a+c)\tag5.\end{align} This forms the equality constraint for Lagrange multipliers; in particular, $${\cal L}(a,b,c,\lambda)=\frac{m-b^3+3b+2}{a+c}+\lambda(2b^3+b(a-c)^2-ac(a+c)-r+s)\tag6$$ with the partial derivatives $$\begin{bmatrix}\dfrac{\partial{\cal L}}{\partial a}\\\dfrac{\partial{\cal L}}{\partial b}\\\dfrac{\partial{\cal L}}{\partial c}\end{bmatrix}=\begin{bmatrix}-\dfrac{m-b^3+3b+2}{(a+c)^2}+\lambda(2b(a-c)-c(2a+c))\\\dfrac{-3b^2+3}{a+c}+\lambda(6b^2+(a-c)^2)\\-\dfrac{m-b^3+3b+2}{(a+c)^2}+\lambda(-2b(a-c)-a(a+2c))\end{bmatrix}\tag7$$ which give, respectively, $$m+b^3-3b-2+\lambda(a+c)^2(2ab-2ac-2bc-c^2)=0\tag8$$$$-3b^2+3+\lambda(a+c)(6b^2+(a-c)^2)=0\tag9$$$$m+b^3-3b-2+\lambda(a+c)^2(-2ab-2ac+2bc-a^2)=0\tag{10}.$$ Subtracting $(8)$ from $(10)$ yields $$\lambda(a+c)^2(4ab-4bc+a^2-c^2)=\lambda(a+c)^2(a-c)(4b+a+c)=0\tag{11}$$ and since $a,b,c>0$, either $\lambda=0$ or $a=c$. If the former, $(8)$ implies that $m+b^3-3b-2=0$, so $d=0$. This is clearly not maximal as $a=b=c=1$ yields $d=3>0$. Therefore $$a=c\implies a^3+a^3=m\implies a^3=\frac m2.\tag{12}$$ Replacing $c$ by $a$ in $(3)$ and $(4)$ gives $$r=b^3-a^2b+a^2b+ab^2-a^3-ab^2=b^3-a^3\tag{13}$$$$s=a^2b-b^3-a^2b-ab^2+ab^2+a^3=a^3-b^3\tag{12}$$ so that $$(b^3-a^3)(a^3-b^3)=-(b^3-a^3)^2\le0\tag{14}.$$ This forces equality to hold at $a=b$, and together with $(12)$, the maximal value is $${\frak m}=\frac{m-a^3+3a+2}{a+a}=\frac{m-\frac m2+3\left(\frac m2\right)^{1/3}+2}{2\left(\frac m2\right)^{1/3}}\tag{15}$$ which proves $(1)$ and hence the claim.
|
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Oresme's proof of the harmonic series' divergence I've been trying to understand Oresme's proof that the harmonic series diverges since it's greater than the series of halves, which diverges.
I'm struggling to capture an aspect of the relationship which I think can be expressed as: "the series of halves is not surjective on the harmonic series".
My best idea so far is that the series of halves goes up to $\log_2$ of the $n$ for the harmonic series. Does that seem right?
$$ \begin{alignat*}{10}
\sum_{k=1}^{n}\frac{1}{k} &=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\dots\\\\
&=\frac{1}{1}+\frac{1}{2}+[\frac{1}{3}+\frac{1}{4}]+[\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}]\dots\\\\
&> \begin{cases}
\begin{aligned}
&\frac{1}{1}+\frac{1}{2}+[\frac{1}{4}+\frac{1}{4}]+[\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}]\dots\\\\
=&1+\begin{cases}
\begin{aligned}
&\frac{1}{2}+\frac{2}{4}+\frac{4}{8}\dots\\\\
= &\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\dots\\\\
= &\sum_{k=1}^{\log_2(n)}\frac{1}{2}\\\\
= &\frac{\log_2(n)}{2}\\
= &\log_2(n)-1
\end{aligned}
\end{cases}\\
\end{aligned}
\end{cases}
\end{alignat*}$$
|
Another way to write this.
Consider the first $m$
groups of $2^k$ terms
for $k = 0$ to $m-1$.
The $k$-th group is
$g(k)
=\sum_{j=2^k}^{2^{k+1}-1} \dfrac1{j}
$.
Since $j$ for each term
is less than $2^{k+1}$,
and there are
$2^k$ terms,
the sum satisfies
$g(k)
\gt\sum_{j=2^k}^{2^{k+1}-1} \dfrac1{2^{k+1}}
=\dfrac{2^k}{2^{k+1}}
=\dfrac12$.
Therefore,
the sum of $m$ of these groups
is greater than
$\dfrac{m}{2}$,
and this can be made arbitrarily large.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\cos \frac{165}{2}$ What are the steps to solve $$\frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4}$$ into $$\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
Please explain. I got them from derivation of $$\cos(82.5^\circ)$$ in 2 different ways.
$$\cos(60^\circ+22.5^\circ)$$ and $$\cos\frac{(90^\circ+75^\circ)}{2}$$
Thanks in advance
|
Note that $\sqrt{2\sqrt{2}} = \sqrt{4/\sqrt{2}}$ so you have
$$
\begin{split}
f &= \frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4} \\
&= \frac12 \sqrt{\frac{2\sqrt{2}-1-\sqrt3}{\sqrt2}} \\
&= \frac12 \sqrt{2 - \frac{1+\sqrt3}{\sqrt2}}
\end{split}
$$
so remains to prove that $$ \frac{1 + \sqrt{3}}{\sqrt2} = \sqrt{2+\sqrt{3}}$$
To do that,multiply by $\sqrt2$ and square both sides to get
$$
1^2 + 3 + 2\sqrt{3} = 2(2+\sqrt3)
$$
which are obviously equal to $4+2\sqrt3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3378575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
"Altered" Alternating Series Diverges or Converges? Consider the series
$$
1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
which alternates between a block of positives and a block of negatives, with the block sizes,
$$
1, 1, 2, 3, 4, 5, 6, \dots
$$
and so on.
Does this series converge or diverge?
|
Write $H_n = \sum_{k=1}^{n} \frac{1}{k} $ for the $n$-th harmonic number. It is well-known that $H_n$ has the asymptotic form
$$ H_n = \log n + \gamma + \mathcal{O}\left(\frac{1}{n}\right) \quad \text{as} \quad n\to\infty. $$
Now returning to OP's question, let $s_n$ denote the $n$-th partial sum. Also, write
$$m_k = \frac{k(k+1)}{2} + 1 \qquad \text{and} \qquad K_n = \max\{ k \geq 1 : m_k \leq n \}.$$
Using this, we can write $s_n$ in the the following block form
$$ s_n = 1 + \Bigg( \sum_{k = 1}^{K_n} (-1)^k (H_{m_k} - H_{m_{k-1}}) \Bigg) + (-1)^{K_n+1} (H_n - H_{m_{K_n}} ). $$
Then by using the asymptotic form of $H_n$, we easily find that
$$ H_{m_k} - H_{m_{k-1}} = \frac{2}{k} + \mathcal{O}\left(\frac{1}{k^2}\right), $$
and so,
$$ s_n = 1 + \sum_{k=1}^{K_n} \left( (-1)^{k} \frac{2}{k} + \mathcal{O}\left(\frac{1}{k^2}\right) \right) + \mathcal{O}\left(\frac{1}{K_n}\right). $$
Therefore it follows that $s_n$ converges as $n\to\infty$
|
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|
Recursive sequence. Value is resetting to 0
Solve
$$\begin{cases}
{a_n} = 5a_{n-1} - 6a_{n-2} + 6 \cdot 3^{n}\\
{a_0} = 1, a_1=50
\end{cases}$$
The problem I've got here is the C resetting.
It happens to me sometimes and I don't know why.
Hope someone can clear it up for me.
Thanks.
|
$a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n$, with $a_0=1$, $a_1=50$.
First we find solution of homogenous recurrence relation
$$a_n-5a_{n-1}+6a_{n-2}=0.$$
$r^2-5r+6=0$, we have $r=2$ with multiplicity 1 and $r=3$ with multiplicity 1. So,
$$a_n^h=A\cdot 2^n+B\cdot 3^n.$$
Because in the right-hand side $$a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n,$$
Now we have $r=2$ with multiplicity 1 and $r=3$ with multiplicity 2. Let the particular solution
$$a_n^p=C\cdot 2^n+D\cdot 3^n + E\cdot n\cdot 3^n.$$
Now delete the terms which appear similar in the $a_n^h$,
$$a_n^p=E\cdot n\cdot 3^n.$$
Substituting new $a_n^p$ into $$a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n,$$
we have
\begin{eqnarray}
a_n-5a_{n-1}+6a_{n-2}&=&6\cdot 3^n\\
E\cdot n\cdot 3^n-5E\cdot (n-1)\cdot 3^{n-1}+6E\cdot (n-2)\cdot 3^{n-2}&=&6\cdot 3^n\\
E\cdot n\cdot 3^n-\dfrac{5}{3}E\cdot (n-1)\cdot 3^{n}+\dfrac{2}{3}E\cdot (n-2)\cdot 3^{n}&=&6\cdot 3^n\\
E\cdot n\cdot 3^n-\dfrac{5}{3}E\cdot n\cdot 3^{n}+\dfrac{5}{3}E\cdot 3^{n}+\dfrac{2}{3}E\cdot n\cdot 3^{n}-\dfrac{4}{3}E\cdot 3^{n}&=&6\cdot 3^n\\
\dfrac{1}{3}E\cdot 3^{n}&=&6\cdot 3^n\\
\end{eqnarray}
Now, we have $E=18$. So, the particular solution will be
$$a_n^p=18\cdot n\cdot 3^n.$$
Now,
$$a_n=a_n^h+a_n^p.$$
$$a_n=A\cdot 2^n+B\cdot 3^n+18\cdot n\cdot 3^n.$$
To find $A$ and $B$, we substitute initial value.
$$a_0=A\cdot 2^0+B\cdot 3^0+18\cdot 0\cdot 3^0=A+B=1.$$
$$a_1=A\cdot 2^1+B\cdot 3^1+18\cdot 1\cdot 3^1=2A+3B+54=50\iff 2A+3B=-4.$$
Solving last two equations, we get $A=7$ and $B=-6$. So we get the solution is
$$a_n=7\cdot 2^n-6\cdot 3^n+18\cdot n\cdot 3^n.$$
|
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|
antidervative of $\ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $ Question
How can the antiderivative of
$$
\ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $$ be obtained?
|
A hint from trigonometry
Consider the Figure below, where $\triangle ABC$ is right-angled, $\overline{AB} = |x|$ and $\overline{AC}=1$, so that, by definition
$$\alpha = \angle CAB = \arccos |x|.$$
Draw the bisector of $A$, that intersects $BC$ in $D$. From $D$ draw the perpendicular to $AD$. Let, on it, $E$ be a point such that $ED \cong AD$. Finally draw from $E$ the perpendicular to $AB$, that intersects $AB$ in $H$.
Clearly we have
\begin{eqnarray}
\angle EAB &=& \frac{\alpha}2+\frac{\pi}4=\\
&=&\frac{\arccos |x|}2 + \frac{\pi}4,
\end{eqnarray}
and
\begin{eqnarray}
\tan \angle EAB &=& \tan \left(\frac{\arccos |x|}2 + \frac{\pi}4\right)=\\
&=& \frac{\overline{EH}}{\overline{AH}}.
\end{eqnarray}
So our first aim is to write this ratio in terms of $x$ "directly".
First we calculate
$$\overline{AD} = \frac{\overline{AB}}{\cos\frac{\alpha}2}=\frac{\sqrt 2 |x|}{\sqrt{1+|x|}},$$
by the bisection formula, and
$$\overline{BD} = |x|\sqrt{\frac{1-|x|}{1+|x|}},$$
by Pythagorean Theorem on $\triangle ABD$.
Let then $F$ be the intersection between $EH$ and $AD$. We have $\triangle EFD\sim\triangle ABD$, and, therefore
$$\overline{EF} = \frac{2|x|}{1+|x|}.$$
Pythagorean Theorem on $\triangle AFD$ gives
$$\overline{FD} = \frac{|x|\sqrt{2(1-|x|)}}{1+|x|}.$$
As a consequence
$$\overline{AF} = \frac{\sqrt 2 |x| (\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}.$$
Similarity $\triangle AFH\sim\triangle ABD$ yields
$$\overline{FH} = \frac{|x|\sqrt{1-|x|}(\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}$$
and
$$\overline{AH} = \frac{|x|(\sqrt{1+|x|}-\sqrt{1-|x|})}{\sqrt{1+|x|}}.$$
The tangent we are looking for is thus
\begin{eqnarray}
\tan \angle EAB &=& \frac{\frac{2|x|}{1+|x|}+\frac{|x|\sqrt{1-|x|}(\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}}{\frac{|x|(\sqrt{1+|x|}-\sqrt{1-|x|})}{\sqrt{1+|x|}}} =\\
&=&\frac{1+|x|+\sqrt{1-x^2}}{1+|x|-\sqrt{1-x^2}}=\\
&=&\frac{\left(1+|x|+\sqrt{1-x^2}\right)^2}{(1+|x|)^2-1+x^2}=\\
&=&\frac{1+\sqrt{1-x^2}}{|x|}.
\end{eqnarray}
Therefore, integrating by parts leads to the result
\begin{eqnarray}
\mathcal I &=& \int\log\left|\tan\left(\frac{\arccos x}2+\frac{\pi}4\right)\right| dx=\\
&=& \int \log\left|\frac{1+\sqrt{1-x^2}}{|x|}\right| dx=\\
&=& \int \log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right) dx=\\
&=& \boxed{x\log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right) + \arcsin x + C}.
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Interesting sequence with a beautiful term : $U_{n+1}=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}$ Question :
Find the term of $U_{n}$ such that :
$U_{0}=2$ and $U_{1}=\frac{5}{2}$
$U_{n+1}=U_{n}(U_{n-1}^{2}-2)-\frac{5}{2}$
My try :
We have ; $U_{2}=\frac{5}{2}$ $U_{3}=2^{3}+2^{-3}$
After calculus ... I think the term is :
$U_{n}=2^{a_n}+2^{-a_n}$
Where $a_n=\frac{2^{n}-(-1)^{n}}{3}$ but I don't
need use induction ?
If any one have another method please tell me ?
|
Given any $v > 1$, construct sequence $\displaystyle\;v_n = \begin{cases} 1, & n = 0\\ v, & n = 1\\ v_{n-1}v_{n-2}^2 & n > 1\end{cases}$
Notice
$$v_{n+1} = v_n v_{n-1}^2 \iff
\frac{v_{n+1}}{v_n^2} = \frac{v_{n-1}^2}{v_n}
\implies
\frac{v_{n+1}}{v_n^2} + \frac{v_n^2}{v_{n+1}} =
\frac{v_{n-1}^2}{v_n} + \frac{v_n}{v_{n-1}^2}
$$
The value of expression $\displaystyle\;\frac{v_{n+1}}{v_n^2} + \frac{v_n^2}{v_{n+1}}$ is independent of $n$ and equals to $v + \frac1v$.
Let $\displaystyle\;u_n = v_n + \frac1{v_n}$, we obtain
$$\begin{align} u_{n+1} &= v_{n+1} + \frac1{v_{n+1}}
= v_n v_{n-1}^2 + \frac{1}{v_n v_{n-1}^2}\\
&= v_n v_{n-1}^2 + \frac{1}{v_n v_{n-1}^2} +
\left[\left(\frac{v_n}{v_{n-1}^2} + \frac{v_{n-1}^2}{v_n}\right) - \left(v + \frac1v\right)\right]\\
&= \left( v_n + \frac1{v_n}\right)\left(v_{n-1}^2 + \frac{1}{v_{n-1}^2}\right)
- \left( v + \frac1v \right)\\
&= u_n \left( u_{n-1}^2 - 2 \right) - \left(v + \frac1v\right)
\end{align}
$$
Compare the recurrence relations between $U_n$ and $u_n$, we find $U_n$ is a special case of $u_n$ for $v = 2$.
To derive an explicit expression for $v_n$ and hence for $U_n$, we can use the fact
$$v_{n+1}v_n = (v_n v_{n-1})^2, \forall n \ge 1
\quad\implies\quad
v_n v_{n-1} = (v_1 v_0)^{2^{n-1}} = v^{2^{n-1}}, \forall n \ge 1$$
This leads to
$$v_n = (v_{n-1} v_{n-2}) v_{n-2} = v^{2^{n-2}} v_{n-2}, \forall n \ge 2$$
Together with $v_0 = 1, v_1 = v$, we find
$$\begin{align}
v_{2k} &= v^{ 2^{2k-2} + 2^{2k-4} + \cdots + 2^0}(1) = v^{\frac{2^{2k} - 1}{3}}\\
v_{2k+1} &= v^{ 2^{2k-1} + 2^{2k-3} + \cdots + 2^1}(v) = v^{\frac{2^{2k+1} + 1}{3}}
\end{align}
\quad\implies\quad v_n = v^{\frac{2^n - (-1)^n}{3}}
$$
As a special case for $v = 2$, we obtain
$$U_n = 2^{a_n} + 2^{-a_n}\quad\text{ where }\quad a_n = \frac{2^n - (-1)^n}{3}$$
|
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|
Show that $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $z=-\frac12(1\pm i\cot(k\pi/8))$; and follow-up questions
*
*(a) Show that the equation $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $-\frac12\left(1\pm i\cot\frac{k\pi}{8}\right)$, where $k=1,2,3$.
*(b) Hence show that
$$(z+1)^8-z^8=\tfrac18(2z+1)(2z^2+2z+1)\left(4z^2+4z+\csc^2\tfrac{\pi}{8}\right)\left(4z^2+4z+\csc^2\tfrac{3\pi}{8}\right)$$
*(c) By making a suitable substitution into this identity, deduce that
$$\cos^{16}\theta - \sin^{16}\theta = \tfrac1{16}\cos 2\theta(\cos^22\theta+1)\left(\cos^22\theta+\cot^2\tfrac{\pi}{8}\right)\left(\cos^22\theta+\cot^2\tfrac{3\pi}{8}\right) $$
original problem image
How do you reformat the polynomial to reach the answer required?
I have tried to turn the LHS into $[(z+1)/z]^8 = 0$ and solve the roots of unity, but I keep getting $\operatorname{cis}(2k\pi/8)$ instead of $k\pi/8$ which is required.
Thanks in advance
|
$$
(1+z)^8=z^8\tag1
$$
$(1)$ implies that
$$
1+z=e^{\pi ik/4}z\tag2
$$
That is
$$
\begin{align}
z
&=\frac1{e^{\pi ik/4}-1}\\
&=\frac1{\cos(\pi k/4)-1+i\sin(\pi k/4)}\\
&=\frac{\cos(\pi k/4)-1-i\sin(\pi k/4)}{(\cos(\pi k/4)-1)^2+\sin^2(\pi k/4)}\\
&=\frac{\cos(\pi k/4)-1-i\sin(\pi k/4)}{2-2\cos(\pi k/4)}\\
&=\frac{-\sin^2(\pi k/4)-i\sin(\pi k/4)(1+\cos(\pi k/4)}{2\sin^2(\pi k/4))}\\
&=-\frac12-\frac i2\cot(\pi k/8)\tag3
\end{align}
$$
where we have used this identity
$$
\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}\tag4
$$
|
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|
Complex Numbers - Converting to Polar form I understand the basics of converting to polar form but I have just come across a question that I haven't seen before.
Usually the complex number is expressed as $z=a+bi$, but this time the complex number I was given is $z^3=-4+4{\sqrt3}i$.
Do I need to somehow remove the 3 power?
Do I just simply use my usually formulas and ignore the power?
Thanks.
EDIT: I thought I should add additional information. I need to convert to polar form, as I will then use Moivre's rule to calculate the roots of the complex number.
I found the polar form for the RHS, $z=8(\cos2.09+i\sin1.05)$. Do I need to cube root this to find the polar form of the original complex number?
Moivre expression is in the picture attached Moivre Expression
where $n=3$ and $k=0,1,2$ I am to find the roots $z0, z1, z2$
|
Now $z^3=8(\cos \frac{2\pi}{3}+i \sin \frac{2 \pi}{3})$: using DeMoivre's formula we get
$$
z_0=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 0 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 0 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{2\pi}{9}+i \sin \frac{2 \pi}{9}\right)\\
z_1=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 1 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 1 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{8\pi}{9}+i \sin \frac{8 \pi}{9}\right)\\
z_2=\sqrt[3]{8}\left(\cos \left(\frac{1}{3}\cdot\frac{2\pi}{3}+\frac{2 \cdot 2 \cdot \pi}{3}\right)+i \sin \left(\frac{2 \pi}{9}+\frac{2 \cdot 2 \cdot \pi}{3}\right)\right)=2\left(\cos \frac{14\pi}{9}+i \sin \frac{14 \pi}{9}\right)
$$
|
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|
Inverse element of $x^2 + x + 1$ in $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ I know that $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field since $x^3 + x + 1$ is irreducible in $\mathbb Z_2$. But I still can't seem to find any inverse element for $x^2 + x + 1$. I want to find an element $g(x) \in \mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ such that $(x^2 + x + 1)g(x) = 1$. I've tried multiplying every single element of $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ by $x^2 + x + 1$, but I never seem to get any of them to be equal to 1 or equivalent. How is this possible if $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field?
Thanks in advance.
|
Write $(x^2+x+1)(ax^2+bx+c)=1\implies ax^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c=a(-x^2-x)+(a+b)(-x-1)+(a+b+c)x^2+(b+c)x+c=(b+c)x^2+cx+c=(b+c)x^2+cx+c-a=1\implies a=1,b=0,c=0$, where I have used the fact that $x^3+x+1=0$ to substitute for $x^4$ and $x^3$.
So $x^2$ is your inverse.
|
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|
$1^3+2^3+...+n^3= $? How I can find formula $1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$ . We can verify with ınduction and I know how I can prove it. How we find what is the formula...
I tried like this
$1^3+2^3+...+n^3= An^4+ Bn^3+Cn^2+Dn+E$
after a long process i found it this way
$A=-299/54,$
$B=-4/9,$
$C=107/54,$
$D=4.$
And then I tried how can show that
But I didn't find Maybe The value of a,b,c,d is wrong
Can I get proof of this formula in this way or another way
if we didn't know the formula, how would we get this formula
|
You can expand
$$\left(\frac {n(n+1)}2\right)^2=\frac 14n^2(n^2+2n+1)=\frac 14n^4+\frac 12n^3+\frac 14n^2$$
and read off $A=\frac 14,B=\frac 12,C=\frac 14,D=0,E=0$
You can also solve simultaneous equations
$$1^3=1=A1^4+B^3+C1^2+D1+E=A+B+C+D+E\\
1^3+2^3=9=A2^4+B2^3+C2^2+D2+E=16A+8B+4C+2D+E$$
and three more similar. You should get the same result as above.
|
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|
Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
Hence,
\begin{equation*}
\lim_{n\to\infty} \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1} = \frac{0+\infty}{0+1} = \infty.
\end{equation*}
Thus, $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ diverges.
Is this correct? Thanks.
|
Your argument is correct. However, you could come to the same conclusion through a different method. Let
$$a_n=\frac{2^n+5^n}{3^n+4^n}$$
and consider what happens as $n\to\infty$. In the numerator, $5^n$ will grow faster than $2^n$ and in the denominator $4^n$ will grow faster than $3^n$. Therefore
$$\lim_{n\to\infty}a_n=\lim_{n\to\infty} \frac{5^n}{4^n}=\lim_{n\to\infty}\left(\frac{5}{4}\right)^n=\infty$$
which implies by the term test that the series diverges.
|
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"url": "https://math.stackexchange.com/questions/3397130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does the series $\sum_{n = 1}^{\infty}\frac{\cos n}{\sqrt{n}}(1 + \frac{1}{1!} + \frac{1}{2!} + ... + \frac{1}{n!})$ converge? I want to check if this series converge or diverge:
$$\sum_{n = 1}^{\infty}\frac{\cos n}{\sqrt{n}}(1 + \frac{1}{1!} + \frac{1}{2!} + ... + \frac{1}{n!})$$
Which technique can i use there ?
UPD: I thought using Abel's theorem here as following. Series
$$\sum_{n = 1}^{\infty}a_n = \sum_{n = 1}^{\infty} \frac{\cos n }{\sqrt{n}}$$
converges. But is sequence
$$\{(1 + \frac{1}{1!} + \frac{1}{2!} + ... + \frac{1}{n!})\}_{n = 1}^{\infty}$$
bounded ?
|
Recall that
$$
e= \lim_{n \to \infty} 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!}
$$
Now consider $S_N= \cos n (1+1/1!+\cdots+1/N!)$. We have $\sum_{n=1}^M S_n$ is bounded for every integer $M$; to see this, consider see this post and use the fact that you know the sum of the reciprocal factorials tends to $e$, and the sequence $\{1/\sqrt{n}\}$ tends to $0$. Therefore, by Dirichlet's Test, $\displaystyle \sum_{n=1}^\infty \dfrac{\cos n}{\sqrt{n}}\left( 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots + \dfrac{1}{n!}\right)$ converges.
|
{
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"url": "https://math.stackexchange.com/questions/3398073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find matrix A of linear transformation of two bases. If we have set E = {v1, v2, v3} with
$v1 = \begin{pmatrix}1\\1\\1\end{pmatrix}$ ,
$v2 = \begin{pmatrix}1\\0\\-2\end{pmatrix}$ ,
$v3 = \begin{pmatrix}1\\1\\0\end{pmatrix}$
and
$L(v1) = \begin{pmatrix}1\\2\\3\\0\end{pmatrix}$ ,
$L(v2) = \begin{pmatrix}3\\0\\1\\0\end{pmatrix}$ ,
$L(v3) = \begin{pmatrix}-1\\-2\\-3\\0\end{pmatrix}$
How do I find a matrix A of L in basis E and standard basis R^4 and a matrix B of L in standard basis R^3 and R^4?
I'm not sure I understand how to find a matrix of L for two bases?
|
You can write
$LV=U$, where $L_{4 \times 3}$ is the unknowm matrix $V_{3 \times 3}$ is the matricx made from the columns $v_i$ and $U_{4 \times 3}$ is made from the columns $u_i$
$V=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & -2 & 0 \end{bmatrix}$ $~~U=\begin{bmatrix} 1 & 3 & -1 \\ 2 & 0 &-2 \\ 3 & 1 & -3 \\ 0 & 0 & 0 \end{bmatrix}.$ Then $L=U V^{-1}= \begin{bmatrix} 7 & -8 & 2 \\ 8 & -10 & 4\\-13 & -16 & 6\\ 0 & 0& 0 \end{bmatrix}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Where is the error in this apparent contradiction arising from the function $f(\pm n)=\left(\frac{n}{n+1}\right)^{\pm 1}$? We define a function $ f \left( \pm n \right) =\left( \frac{ n }{ n + 1 } \right) ^ { \pm 1 } $.
That is, $$ f \left( m \right) = \begin{cases}\frac{ m }{ m + 1 } &\text{ if } m \ge 0 \\ \frac{ -m + 1 }{ -m } & \text{ if }m \lt 0 \end{cases}. $$
From this, $ f \left( - a \right) = \frac{ a + 1 }{ a } = \frac{ a }{ a } + \frac{ 1 }{ a }= 1 + \frac{ 1 }{ a } $
However, we can also do this:
$ f \left( \frac{ \alpha }{ \beta } > 0 \right) = \frac{ \alpha }{ \beta } \div \left( \frac{ \alpha }{ \beta } + 1 \right) = \frac{ \alpha }{ \beta } \div \left( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \beta } \right) = \\ = \frac{ \alpha }{ \beta } \div \frac{ \alpha + \beta }{ \beta } = \frac{ \alpha \beta }{ \alpha \beta + \beta ^ 2 } $
Since $ \frac{ \alpha \beta + \beta ^ 2 }{ \alpha \beta } = 1 + \frac{ \beta ^ 2 }{ \alpha \beta } = 1 + \frac{ \beta }{ \alpha } = \frac{ \alpha + \beta }{ \alpha } $, $ \frac{ \alpha \beta }{ \alpha \beta + \beta ^ 2 } = \frac{ \alpha }{ \alpha + \beta } $.
Since $ f \left( k \right) = \frac{ 1 }{ f \left( - k \right)} $, $ f \left( - \frac{ \gamma }{ \delta } \right) = \frac{ \gamma }{ \gamma + \delta } $.
But this conflicts with the previous definition!
It says that $f \left( -2 \right) = \frac{ 1 }{ 2 }$ AND $= 1 + \frac{ 1 }{ 2 }$!
Where is the problem?
|
you have , where you used alpha and beta
$\frac{a}{b}\div \frac{a + b}{b} = \frac{a}{a +b}$
but you've ended up with $b^2$ etc which is right, but you could cancel b
then you say 'since' and you don't cancel b out properly
|
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|
How to solve this limit, and what did I do wrong? I have to solve:
$$ \lim_{n\to\infty}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1} $$
Here's my attempt:
$$ \lim_{n\to\infty}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1} = \lim_{n\to\infty}\frac{(1+\frac{2}{n+1})^{1/6}-1}{\frac{2}{n+1}}\cdot\frac{2}{(n+1)(\sqrt[n]{2}-1)}=\frac{1}{6}\lim_{n\to\infty}\frac{2}{\frac{2^{1/n}-1}{1/n}+\sqrt[n]{2}-1}=\frac{2}{6(\ln2-1)}=\frac{1}{3\ln2-3}$$
But the result is supposed to be $\frac{1}{3\ln2}$, what did I do wrong?
PS: I used the following limits in my attempt to solve it:
$$\lim_{n\to\infty}\frac{(1+x_n)^r-1}{x_n}=r$$
$$\lim_{n\to\infty}\frac{a^{x_n}-1}{x_n}=\ln a$$
|
From here
$$\ldots=\frac{1}{6}\lim_{n\to\infty}\frac{2}{\frac{2^{1/n}-1}{1/n}+\sqrt[n]{2}-1}=\frac{1}{6}\frac{2}{\ln 2+1-1}=\ldots$$
As an alternative
$$\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\sqrt[n]{2}-1}=\frac{\frac1n}{\sqrt[n]{2}-1}\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\frac1n}\to \frac{1}{3\ln2}$$
indeed
*
*$\frac{\frac1n}{\sqrt[n]{2}-1}\to \frac{1}{\ln2}$
and by binomial approximation
*
*$\frac{\sqrt[6]{\frac{n+3}{n+1}}-1}{\frac1n}=n\left(\left(1+\frac{2}{n+1}\right)^\frac16-1\right) \approx n\left(1+\frac{2}{6n+6}-1\right)=\frac{n}{3n+3}\to \frac13$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!
*
*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because $\alpha=3 > 1$. Is this true?
*To calculate the integral, using the integration by parts where $u = \ln x$ and $dv = \frac{x \ln x}{(x^2+1)^2} dx$.
So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx = \frac{- \ln x}{2(x^2+1)}- \frac{1}{4x^2} +\frac{\ln |x|}{2} ~\Big|_{0}^{\infty} $$
and this undefined while it should converge to $0$ ! what I missed?
I found an error in the calculation so the integral = So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx = \frac{- \ln x}{2(x^2+1)}+ \frac{1}{8} \Big( \ln x^2 - \ln (x^2+1) \Big) ~\Big|_{0}^{\infty} $$
and it's still undefined!
|
The limit of $\frac{x \log{x}}{(x^2+1)^2}$ has the same behavior as $\frac{\log{x}}{x^3}$.
You can use L'Hôpital's rule and see that it converges to $\frac{1}{3x^3}$, which goes to $0$ as $x$ approaches infinity.
Using integration by parts with $u=\log{x}$ and $\frac{dv}{dx} = \frac{x}{(x^2+1)^2}$. We know that $\frac{du}{dx} = \frac{1}{x} $, and that $ v = \int_0^x \frac{x}{(x^2+1)^2} dx$. Using the substitution $x=\tan{y}$, we obtain $ v = \int_0^{\tan^{-1}{x}} \frac{\tan{y}}{(\tan^2{y}+1)^2} dx
= \int_{{\pi}/{2}}^{\tan^{-1}{x}}\sin{y}\cdot \cos^3{y} \space dx = -\frac{1}{4}\cos^4{y} \space|_{\pi/2}^{\tan^{-1}{x}} = -\frac{1}{4(x^2+1)^2}
$.
The integral becomes $\int_0^\infty \frac{x \log{x}}{(x^2+1)^2} dx= uv -\int_0^\infty v \frac{du}{dx}dx
=-\frac{\log{x}}{4(x^2+1)^2}|_0^\infty + \int_0^\infty \frac{1}{4x(x^2+1)^2}
=\frac{1}{4}-\frac{1}{4}=0
$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Arrangement of letters with one pair of letters always between two letters
How many ways are there to arrange $11$ letters, $A,B,C,D,E,F,x,x,x,y,y$ so that every $y$ lies between two $x$ (not necessarily adjacent)$?
Attempt:I fixed the letters $ x \quad y \quad y \quad x$. Totally there are 5 gaps(all distinct) for the remaining 7 letters(these are all distinct) So there are $5^7$ ways to distribute these letters. But I know that this isn't done yet as I haven't accounted for the relative positions of the letters once placed in the gap(We need to permute that also).
So I wrote down the various distributions possible of number of letters in each gap.
$(3+1+1+1+1)$ type $\rightarrow 3! \cdot 5!/4!$ permutations
$(4+1+0+1+1)$ type $\rightarrow 4 ! \cdot 5!/3!$ permutations.
$5+1+1+0+0$ type $\rightarrow 5! \cdot5!/(2!\cdot2!)$ permutations.
$(6+1+0+0+0)$ type $\rightarrow 6 ! \cdot 5!/3!$ permutations.
$(7+0+0+0+0)$ type $\rightarrow 7 ! \cdot 5!/4!$ permutations.
$2+2+3+0+0$ type $\rightarrow 2 ! \cdot2!\cdot3!\cdot 5!/(2!\cdot2!)$ permutations.
$2+3+1+1+0$ type $\rightarrow 2 ! \cdot3!\cdot 5!/(2!)$ permutations.
$2+4+1+0+0$ type $\rightarrow 2 ! \cdot4!\cdot5!/(2!)$ permutations.
$2+5+0+0+0$ type $\rightarrow 2 ! \cdot5!\cdot 5!/(3!)$ permutations.
$3+3+1+0+0$ type $\rightarrow 3! \cdot3!\cdot 5!/(2!\cdot2!)$ permutations.
$3+4+0+0+0$ type $\rightarrow 4!\cdot3!\cdot 5!/(3!)$ permutations.
$2+2+2+1+0$ type $\rightarrow 2!\cdot2!\cdot2!\cdot 5!/(3!)$ permutations.
$2+2+3+0+0$ type $\rightarrow 2!\cdot2!\cdot5!/(2!\cdot2!)$ permutations.
Adding all these up and multiplying the sum with $5^7$ I get some huge answer which is wrong :(
Obviously there was to be another way to solve this but I couldn't think of any.
|
The capital letters can be arranged in $6!=720$ ways. The lowercase letters can be arranged in $\binom{3}{1}=3$ ways, since we are forced to have x's at either end.
With that done, the capital letters and lowercase letters can be merged in $\binom{11}{5}=462$ ways -- all we need to do is choose which five spots out of the 11 will have lowercase letters since we have already determined the order in which they appear.
$720\cdot3\cdot462=997920$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
using epsilon delta definition to proof limits
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
let $$f(x)=\frac{2x+3}{x-1}$$ then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_f \left(
0 < \left|x-1 \right|=x-1<\delta\Longrightarrow \large\left|\frac{2x+3}{x-1}\right|\right)>M$
$$M<\left|\frac{2x+3}{x-1}\right|=\frac{\left|2x+3\right|}{x-1}$$
take $\delta\le1$ implies:$$M<\frac{\left|2x+3\right|}{x-1}<\frac{7}{x-1}$$$$x-1<\frac{7}{M}$$
hence $$\delta\le\min\left\{1,\left(\frac{7}{M}\right)\right\}$$
is it true?
$$\lim_{\large x \to 1} \frac{\left(-1\right)^{\left[x\right]}}{x-1}=-∞$$
let $$g(x)=\frac{\left(-1\right)^{\left[x\right]}}{x-1}$$ ($\left[x\right]$ is floor function )
then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_g \left(
0 < \left|x-1 \right|<\delta\Longrightarrow \large\frac{\large \left(-1\right)^{\left[\large x\right]}}{x-1} <-M \right)$
take $\delta\le1$ implies:$0<x<2$, here is the problem, what is exactly $\left[x\right]$?, I tried another $\delta$ less than $1$, but still I have the problem.
|
FIRST:--- Choose $M>0$. Let $\delta>0$ be such that,$$0<\delta<\frac{5}{M+2}$$ $$\implies\frac{5}{\delta}-2=\frac{5-2\delta}{\delta}>M.$$ Hence, for any $x\in \Bbb R$ with $1<x<1+\delta$ we have, $$\frac{2x+3}{x-1}=\frac{2(x-1)+5}{x-1}>\frac{5-2(x-1)}{x-1}>\frac{5-2\delta}{\delta}>M.$$ So, $$\lim_{\large x \to 1+} \frac{2x+3}{x-1}=∞.$$
SECOND:--- Choose $M>0$. Then for any $0<\delta<\frac{1}{M}$ we have, $$1-\delta<x<1\implies\frac{(-1)^{\lfloor x\rfloor}}{x-1}=\frac{(-1)^0}{x-1}=\frac{1}{x-1}<-\frac{1}{\delta}<-M$$$$\text{and}$$$$1<x<1+\delta\implies\frac{(-1)^{\lfloor x\rfloor}}{x-1}=\frac{(-1)^1}{x-1}=\frac{-1}{x-1}<-\frac{1}{\delta}<-M.$$ So $$\lim_{x\to 1}\frac{(-1)^{\lfloor x\rfloor}}{x-1}=-\infty.$$
|
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|
how to find $\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$ how to find
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$$ without L'hopital or taylor/Laurent series
I tried but did not get any answer:
$$\frac{\tan^{2}\left(x\right)\tan\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\sin\left(x\right)\cos^{2}\left(x\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2\left(\sin^{3}\left(x\right)-3\left(\sin\left(x\right)\left(1-\sin^{2}\left(x\right)\right)\right)\right)}{\cos^{3}\left(x\right)\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}=\frac{2}{1}\frac{-3\left(\sin(x)-\sin^{2}\left(x\right)\right)+\sin^{3}\left(x\right)}{\left(1-\sin^{2}\left(x\right)\right)(\cos(x))\left(\sqrt{3}\cos\left(x\right)-\sin\left(x\right)\right)}$$
|
Let $y=x-\pi/3$, so we are to compute
\begin{align*}
\lim_{y\rightarrow 0}\dfrac{\tan^{3}(y+\pi/3)-3\tan(y+\pi/3)}{\cos(y+\pi/2)}.
\end{align*}
Note that
\begin{align*}
\tan^{3}(y+\pi/3)-3\tan(y+\pi/3)&=\tan(3(y+\pi/3))(3\tan^{2}(y+\pi/3)-1)\\
&=(\tan(3y))(3\tan^{2}(y+\pi/3)-1),
\end{align*}
and the term reduces to
\begin{align*}
\dfrac{(\tan(3y))(3\tan^{2}(y+\pi/3)-1)}{-\sin y}.
\end{align*}
But
\begin{align*}
\lim_{y\rightarrow 0}\dfrac{\tan 3y}{\sin y}=3\lim_{y\rightarrow 0}\dfrac{\sin 3y}{3y}\dfrac{1}{\cos 3y}\dfrac{y}{\sin y}=3.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int {\dfrac{x^2+\left(n-1\right)n}{\left(x\sin x+n\cos x\right)^2}}dx $
$$\int {\dfrac{x^2+\left(n-1\right)n}{\left(x\sin\left(x\right)+n\cos\left(x\right)\right)^2}}dx $$
My Try:
I multiple $x^{2n-2}$ to both N and D, then took D as $u$ and then solved to get $\dfrac{n\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+n\cos\left(x\right)} + C$ as answer.
My teacher told that this would have been much easier if we had applied linearity and written question as
$={\displaystyle\int}\dfrac{x\sin\left(x\right)+\left(n-1\right)\cos\left(x\right)}{x\sin\left(x\right)+n\cos\left(x\right)}\,\mathrm{d}x-{\displaystyle\int}\dfrac{\left(\left(1-n\right)\sin\left(x\right)+x\cos\left(x\right)\right)\left(n\sin\left(x\right)-x\cos\left(x\right)\right)}{\left(x\sin\left(x\right)+n\cos\left(x\right)\right)^2}\,\mathrm{d}x$
I didn't get it, how did we write the above statement? I mean, please explain the method or steps for reaching to this part.
|
Use shorthands $s= \sin x$ and $c=\cos x$ below. Decompose the integrand with $s^2+c^2 = 1$
\begin{align}
& {\dfrac{x^2+(n-1)n}{(x\sin x+n\cos x)^2}} \\
=&\frac{x^2+(n-1)n}{(xs+nc)^2}
= \frac{x^2(s^2+c^2)+(n-1)n(s^2+c^2)}{(xs+nc)^2} \\
=& \frac{[xs+(n-1)c](xs+nc)-[(1-n)xs+xc](ns-xc) }{(xs+nc)^2}\\
= & \frac{xs+(n-1)c}{xs+nc}-
\dfrac{[(1-n)xs+xc](ns-xc)}{\left(xs+nc\right)^2} \\
= &\dfrac{x\sin x+(n-1)\cos x} {x\sin x+n\cos x}-\dfrac{[(1-n)\sin x+x\cos x](n\sin x-x\cos x)}{(x\sin x+n\cos x)^2} \\
= & \frac{d}{dx}\left( \dfrac{n\sin x-x\cos x}{x\sin x+n\cos x} \right) \\
\end{align}
Thus
$$\int {\dfrac{x^2+(n-1)n}{(x\sin x+n\cos x)^2}}
= \dfrac{n\sin x-x\cos x}{x\sin x+n\cos x} +C$$
|
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|
Integrate $\int_{0}^{\pi/2}{\frac{x\cos\left(x\right)-\sin\left(x\right)}{\sin\left(x\right)+x^2}}dx$
$$\int_{0}^{\pi/2}{\dfrac{x\cos\left(x\right)-\sin\left(x\right)}{\sin\left(x\right)+x^2}}dx$$
I am unable to exploit the properties of definite integral, neither it seems that indefinite integration is possible.
|
I think that power 2 of $\sin x$ is missed and solve the other one as below. At the first sight, the numerator of the integrand seems to be an antiderivative of $x\sin x$. However, the negative sign in between reminds me that it should be an antiderivative of $\dfrac{\sin x}{x}$. So I decided to divide both numerator and denominator of the integrand by $x^2$ and it works wonderfully.
$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{x \cos x-\sin x}{\sin ^{2} x+x^{2}} d x =& \int_{0}^{\frac{\pi}{2}} \frac{\frac{x \cos x-\sin x}{x^{2}}}{\frac{\sin ^{2} x}{x^{2}}+1} d x \\=& \int_{0}^{\frac{\pi}{2}} \frac{d\left(\frac{\sin x}{x}\right)}{\left(\frac{\sin x}{x}\right)^{2}+1} \\=&\left[\tan ^{-1}\left(\frac{\sin x}{x}\right)\right]_{0}^{\frac{\pi}{2}} \\=& \tan ^{-1}\left(\frac{2}{\pi}\right)-\frac{\pi}{4} \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right) \end{aligned}$
|
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|
Find the harmonic conjugate Let
$$u(x,y)=x^3+ax^2y+bxy^2+2y^3$$
Be a harmonic function and $v(x,y)$ be its harmonic conjugate .If $v(0,0)=1$ then find the
$$|a+b+v(1,1)|$$
The solution i tried-since $u$ is harmonic so by laplace equation we get
$$6x+2ay+2bx+12y=0$$
By comparing coefficients we get $a=-6 $ and $b=-3$ so we get
$$u(x,y)=x^3-6x^2y-3xy^2+2y^3$$
Next i am not getting how to proceed please give me a hint
Thankyou.
|
$$u(x,y) =x^3 -6x^2 y -3xy^2 +2y^3 =(x^3 -3xy^2 ) +2 (y^3 -3x^2 y) =Re( (x+iy)^3 ) - Im(2 (x+iy)^3 )= Re z^3 -2Im z^3 == Re [(1-2i)z^3 ]$$
Hence the conjugate is equal to
$$v(x,y)=Im [(1-2i) z^3 ]$$
$$|a+b +v(1,1) |=|-9 +Im [(1-2i) (1+i)^3]|=|-9 +6|=3$$
|
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|
Evaluate $\displaystyle\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}$ I have been trying to evaluate
\begin{equation*}
\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}.
\end{equation*}
We have
\begin{equation*}
\frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x} = \frac{x^2 + x}{\ln(1 + x + x^2) - x} \cdot \frac{1}{\sqrt{x^2 + x + 4} + 2} \quad \text{for all}\ x \in \mathbb{R},
\end{equation*}
so I think my exercise boils down to
\begin{equation*}
\lim_{x \to 0^{-}} \frac{x^2 + x}{\ln(1 + x + x^2) - x},
\end{equation*}
and this limit equals $-\infty$ by De L'Hôpital's Theorem.
Can I evaluate this limit without De L'Hôpital's Theorem?
|
Hint:
$$ (1\pm f(x))^n \approx 1\pm nf(x) $$
$$ \ln(1\pm f(x)) \approx \pm f(x)$$
for $ f(x) \approx 0$ as $x\to 0$.
|
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|
Prove $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1+2^N \bmod 2^{N+1}$ I want to prove $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1+2^N \bmod 2^{N+1}$ for $N\geq 4$.
First, let's prove $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1 \bmod 2^{N}$.
$$1+i-i^2\equiv1+j-j^2$$
$$\Leftrightarrow (i-j)(i+j-1)\equiv0$$
Since the parity of $i-j$ and $i+j-1$ are different, $j\equiv i \lor j=2^N-i+1$.
Thus, $1+i-i^2$ takes all odd number below $2^{N-1}$ once. From Gauss-Wilson's theorem, $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1 \bmod 2^N$ holds
I found $\prod_{i=1}^{2^{N-1}} (1+i-i^2) \equiv 1+2^N \bmod 2^{N+1}$ up to $N=27$ by numerical calculation. How can we prove this?
|
Let $a_i=1+i-i^2$ and
$$p_n=\prod_{i=1}^{2^n}a_i$$
so that we have to show
$$p_{n-1}\equiv 1+2^n\pmod{2^{n+1}}\tag 1$$
Note that $(1)$ implies $p_{n-1}^2\equiv 1+2^{n+1}\pmod{2^{n+2}}$.
By induction on $n\geq 4$, we have:
\begin{align}
p_n
&=p_{n-1}\prod_{i=2^{n-1}+1}^{2^n}a_i\\
&=p_{n-1}\prod_{i=1}^{2^{n-1}}a_{2^{n-1}+i}\\
&=p_{n-1}\prod_{i=1}^{2^{n-1}}(1+2^{n-1}+i-(2^{n-1}+i)^2)\\
&=p_{n-1}\prod_{i=1}^{2^{n-1}}(1+2^{n-1}+i-2^{2n-2}-2^ni-i^2)\\
&=p_{n-1}\prod_{i=1}^{2^{n-1}}(a_i+2^{n-1}(1-2^{n-1}-2i))\\
&\equiv p_{n-1}\left(\prod_{i=1}^{2^{n-1}}a_i+\sum_{j=1}^{2^{n-1}}2^{n-1}(1-2^{n-1}-2j)\prod_{i\neq j}a_i\right)\\
&\equiv p_{n-1}^2\left(1+2^{n-1}\sum_{j=1}^{2^{n-1}}\frac{1-2^{n-1}-2j}{a_j}\right)\\
&\equiv (1+2^{n+1})\left(1+2^{n-1}\sum_{j=1}^{2^{n-1}}\frac{1-2^{n-1}-2j}{a_j}\right)\\
&\equiv 1+2^{n+1}+2^{n-1}\sum_{j=1}^{2^{n-1}}\frac{1-2^{n-1}-2j}{a_j}\pmod{2^{n+2}}
\end{align}
hence the assertion reduces to proving
$$\sum_{j=1}^{2^{n-1}}\frac{1-2^{n-1}-2j}{a_j}\equiv 0\pmod{8}$$
and
\begin{align}
\sum_{j=1}^{2^{n-1}}\frac{1-2^{n-1}-2j}{a_j}
&\equiv\frac{2^{n-1}}8\sum_{j=1}^{8}\frac{1-2^{n-1}-2j}{a_j}\\
&\equiv{2^{n-4}}\sum_{j=1}^{8}\frac{1-2j}{a_j}\pmod{8}
\end{align}
which is clearly $\equiv 0\pmod 8$ if $n\geq 7$
|
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|
Parabola equation in form of quadratic $ax^2+bx+c=y$ where $a+b+c$ is an integer Suppose a parabola has vertex $(\frac{1}{4},\frac{-9}{8})$ and equation $ax^2+bx+c=y$ where $a>0$ and $a+b+c $ is an integer. Find the minimum possible value of $a$ under the given condition.
My approach
$(x-\frac{1}{4})^2=4a'(y+\frac{9}{8})$
$\frac{x^2-\frac{x}{2}+\frac{1}{16}}{4a'}-\frac{9}{8}=y$
$\frac{1}{4a'}-\frac{1}{4a'}+\frac{1}{64a'}-\frac{9}{8}=Z$, where $Z$ is an integer
$\frac{9}{64a'}-\frac{9}{8}=Z$, from here I am confused
|
We know that the vertex x coordinates is $\frac{-b}{2a} = \frac{1}{4}$ solve for $b = \frac{-a}{2}$ now the point $\frac{1}{4} , \frac{-9}{8}$ satisfies the parabola so
$\frac{a}{16} - \frac{a}{8} + c = \frac{-9}{8}$
$a-2a + 16c = -18 $
$c = \frac{-18 +a}{16}$ finally
$a+b+c = a + \frac{-a}{2} +\frac{-18+a}{16} \in \mathbb{Z}$ find the solution and pick the smallest one
we will have $16 \mid 9a-18 $
|
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|
Is my Proof for the following problem correct? Prove that $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$ If $a,b,c$ are positive real numbers prove that $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
I state the following:
Multiplying both sides by $abc$ yields $$(abc)(a+b+c)?{a^4+b^4+c^4}$$
By AM-GM for $abc$ we have $$abc\leq\left(\frac{a+b+c}{3}\right)^3\Rightarrow\left(\frac{a+b+c}{3}\right)^3(a+b+c)?{a^4+b^4+c^4}$$
This can be rewritten as the following:
$$\left(\frac{a+b+c}{3}\right)^4\leq\frac{a^4+b^4+c^4}{3}$$
which holds due to the power mean inequality.
I was also thinking that it may be proven by a double application of the Chebyshev inequality as well
To begin, hold that ${a}\ge{b}\ge{c}$ observe that $$\left[\begin{matrix} a^3 & b^3 & c^3 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{matrix}\right]=a^2+b^2+c^2\leq\left[\begin{matrix} a^3 & b^3 & c^3 \\ \frac{1}{b} & \frac{1}{c} & \frac{1}{a}\end{matrix}\right]=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}$$
Thus we can hold confidently even if we assume equality that $$\left[\begin{matrix} {a^2} & {b^2} & c^2 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{matrix}\right]=a+b+c\leq\left[\begin{matrix} \frac{a^3}
{b} & \frac{b^3}{c} & \frac{a^3}{a} \\ \frac{1}{c} & \frac{1}{a} & \frac{1}{b}\end{matrix}\right]=\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
Please provide feedback if you can.
|
Your idea is correct, but the proof is a bit confusing because it is not always clear if you assume the desired inequality or not. I would write it as follows:
Using the AM-GM inequality $(1)$ and the power-mean inequality $(2)$ we have
$$
abc(a+b+c)\underset{(1)}{\leq}(a+b+c)\left(\frac{a+b+c}{3}\right)^3 = 3\left(\frac{a+b+c}{3}\right)^4 \underset{(2)}{\leq} a^4 + b^4 + c^4
$$
and dividing by $abc$ gives
$$
a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab} \, .
$$
|
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|
Proving $364 \mid n^{91} - n^7$ [Generalization of Euler & Fermat Theorems] I am trying to prove this statement which is equivalent to show $n^{91} = n^7 \pmod{ 364}$. By splitting modulus theorem $n^{91} = n^7 \bmod 91$ and $\bmod 4$. Then I don’t know what to do next... Can anyone help me?
|
Use Euler's theorem:
Mod $4$: either $n\equiv0 $ or $2$, so $n^{91}\equiv n^7\equiv0$, or $\color{blue}{n^2\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv (n^2)^{42}n^7\equiv n^7$
Mod $7$: either $n\equiv0$, so $n^{91}\equiv n^7\equiv 0$, or $\color{blue}{n^6\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv(n^6)^{14}n^7\equiv n^7$
Mod $13$: either $n\equiv0$, so $n^{91}\equiv n^7\equiv 0$, or $\color{blue}{n^{12}\equiv 1}$ so $n^{91}\equiv n^{84}n^7\equiv(n^{12})^{7}n^7\equiv n^7$
|
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|
What 's wrong with this integral substitution? What's wrong with the following solution?
Let $t=\sin x,$ then $x=\arcsin t.$ By substitution method we have the following:
$\int_0^{2\pi}\sin^2x \rm{dx}=\int_0^0 \frac{t^2}{\sqrt{1-t^2}}\rm{dt}=0$.
However, we know that $\int_0^{2\pi}\sin^2x \rm{dx}=\int_0^{2\pi}\frac{1-\cos 2x}{2} \rm{dx}=\pi.$
I can't find the problem.
Thanks for any suggestions.
|
The reason why the lower and upper limits turn into the same value $0$ after changing variable is due to the lack of bijectivity of the substitution, even though the calculation is incorrect. Bijectivity in changing the variable is not a strong requirement, but it needs to be careful in calculation when a non-bijective substitution is applied.
Correct calculation using the same substitution is as follows: From $1-t^2 = x$, $t^2 = 1-x \ge 0$ and so
$$t = \begin{cases}
\sqrt{1-x} & \text{for $t \ge 0$,} \\
-\sqrt{1-x} & \text{for $t < 0$.}
\end{cases}$$
with
$$dt = \begin{cases}
-\frac{1}{2\sqrt{1-x}}\,dx & \text{for $t > 0$} \\
\frac{1}{2\sqrt{1-x}}\,dx & \text{for $t < 0$.}
\end{cases}$$
Hence
\begin{align*}
\int_{-1}^{1}(1-t^2)\,dt
& = \int_{-1}^{0} (1-t^2)\,dt + \int_0^1 (1-t^2)\,dt \\
& = \int_0^1 x\left(\frac{1}{2\sqrt{1-x}}\right)\,dx
+ \int_1^0 x\left(\color{red}{-}\frac{1}{2\sqrt{1-x}}\right)\,dx\\
& = \int_0^1 \frac{x}{2\sqrt{1-x}}\,dx
+ \int_\color{red}{0}^\color{red}{1} \frac{x}{2\sqrt{1-x}}\,dx \\
& = \int_0^1 \frac{x}{\sqrt{1-x}}\,dx
= \lim_{\varepsilon \rightarrow 1^-} \int_0^{\varepsilon} \frac{x}{\sqrt{1-x}}\,dx\\
& = \lim_{\varepsilon \rightarrow 1^-} \left[\left.-\frac{2}{3}\sqrt{1-x}(x+2)\right|_0^\varepsilon\right] \\
& = \lim_{\varepsilon \rightarrow 1^-} \left[\left(-\frac{2}{3}\sqrt{1-\varepsilon}(\varepsilon+2)\right) - \left(-\frac{4}{3}\right)\right] \\
& = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3}.
\end{align*}
|
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|
given $ ∫ y dx$ between x=3 and x=1, show that it equals A + B√3 where A and B are integers to be found Where $$∫y dx =2x^{3/2} - 8x^{1/2}$$
I keep getting $6 + 6√3$, which is incorrect. The correct answer should be $6 - 2√3$
But I can't figure out how to correctly derive that answer.
Cheers
|
It must be a calculation error. Indeed, given the antiderivative expression $2x^{\frac 32} - 8x^{\frac 12}$, there is no need to attempt to find $y$, because the integral between $1$ and $3$ is given by the value of this function at $3$ minus its value at $1$, by the fundamental theorem of calculus.
The value at $3$ :
$$
2 \times 3^{\frac 32} - 8 \times 3^{\frac 12} = 2 \times 3 \times 3^{\frac 12} - 8 \times 3^{\frac 12} = (6-8) \times 3^{\frac 12}=-2 \times 3^{\frac 12} = -2 \sqrt3
$$
and the value at $1$ :
$$
2 \times 1^{\frac 32} - 8 \times 1 ^{\frac 12} = 2-8 = -6
$$
giving the answer $-2\sqrt 3 -(-6) = 6 - 2 \sqrt 3$.
|
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|
If $x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$ What is the value of $5x^2-5x-1?$ If $$x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$$ What is the value of $5x^2-5x-1?$
Efforts:
After rationalization, I got $x={\sqrt{5}+1\over 2}$ and $x^2={\sqrt{5}+1\over \sqrt{5}-1}$. Going by this method is very tedious and boring.
Is there a more clever and imaginative way to approach this problem?
Thanks.
|
Rationalizing the denominator of $x$, we get
$$x=\sqrt{\frac{\sqrt5+1}{\sqrt5-1}}=\sqrt{\frac{\sqrt5+1}{\sqrt5-1}\cdot\frac{\sqrt5+1}{\sqrt5+1}}=\sqrt{\frac{(\sqrt5+1)^2}{4}}=\frac{1+\sqrt5}2$$
For the expression to be evaluated, we have
\begin{align}
5x^2-5x-1&=5(x^2-x-\frac15)\\
&=5(x^2-x-\frac15-\frac45+\frac45)\\
&=5(x^2-x-1)+4
\end{align}
Since $x=\dfrac{1+\sqrt5}2$ is a root of $x^2-x-1=0$, we have the expression to be equal to $\boxed4$.
|
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|
Problem with equation in complex numbers
I am supposed to calculate:
$x^{2}=5+i$
I used formula:
$\left | \cos \frac{\alpha }{2} \right |=\sqrt{\frac{1+\cos \alpha }{2}}$
$\left | \sin \frac{\alpha }{2} \right |=\sqrt{\frac{1-\cos \alpha }{2}}$
and I came to the point where:
x= $\sqrt[4]{26}\left ( \frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}\sqrt[4]{26}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}\sqrt[4]{26}} \right)$= $\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
I know that k=0,1 and also that $\sin x, \cos x $ are positive in the first quadrant, so my solution will simple be :
$x_{0}=\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
$x_{1}=-\frac{\sqrt{\sqrt{26}+5}}{\sqrt{2}} + i \frac{\sqrt{\sqrt{26}-5}}{\sqrt{2}} $
It does not seem like the correct solution. Can anyone please tell me, where I made the mistake?
|
$5 + i = \rho (\cos \theta + i\sin theta)\\
\sqrt{\rho (\cos \theta + i\sin \theta)} = \sqrt {\rho}(\cos \frac \theta2 + i\sin \frac \theta2)\\
\rho = \sqrt {5^2 + 1} = \sqrt {26}\\
\theta = \arctan \frac {1}{5}\\
\cos \theta = \frac {5}{\sqrt {26}}\\
\sin\theta = \frac {1}{\sqrt {26}}\\
\tan \frac{\theta}{2} = \frac {1-\cos\theta}{\sin\theta}= \frac {1 - \frac {5}{\sqrt{26}}}{\frac {1}{\sqrt {26}}} = \sqrt {26} - 5\\
$
|
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|
Determinant of a $3 \times 3$ Vandermonde matrix
Let $$A = \begin{bmatrix} p^2 & p & 1\\ q^2 & q & 1\\ r^2 & r & 1\end{bmatrix}$$ Prove that $$\det(A) = (r-q)(r-p)(p-q)$$
|
\begin{align}
\begin{vmatrix}p^2 & p & 1\\ q^2 & q & 1\\ r^2 & r & 1\end{vmatrix}
&=\begin{vmatrix}q^2&q\\r^2&r\end{vmatrix}-\begin{vmatrix}p^2&p\\r^2&r\end{vmatrix}+\begin{vmatrix}p^2&p\\q^2&q\end{vmatrix}\\
&=(q^2r-qr^2)-(p^2r-pr^2)+(p^2q-pq^2)\\
&=q^2r-qr^2-p^2r+pr^2+p^2q-pq^2\\
&=\frac{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2}{(r−q)(r−p)(p−q)}(r−q)(r−p)(p−q)\\
&=\frac{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2}{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2}(r−q)(r−p)(p−q)\\
&=(1)(r−q)(r−p)(p−q)=(r−q)(r−p)(p−q)\\
\end{align}
|
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|
Prove that the equation has only one real root. Prove that $(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only one real root.
It's easy to show that the equation has a real root using Rolle's theorem. But how to show that the real root is unique? By Descartes' rule of sign, it can be shown that it has 3 or 1 real root.
But it doesn't guarantee that the real root is unique. If we calculate the root then it can be shown that it has only one real root.
|
Let $y=\dfrac{x-1+x-2+x-3+x-4}4$
$x=y+2.5$
$$(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=(y+1.5)^3+(y+.5)^3+(y-.5)^3+(y-1.5)^3$$
Now use $(a-b)^3+(a+b)^3=2(a^3+3ab^2)$
$$0=4y^3+6y((1.5)^2+(.5)^2)$$
|
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|
How did we find the generalized eigenvectors corresponding to the eigenvalue 2 here pg.305 in Golan. The example is given below:
How did we find the generalized eigenvectors corresponding to the eigenvalue 2 here?
I watched this video https://www.youtube.com/watch?v=msFp3vOYIoA and I followed the procedure mentioned in it, which say take the eigenvector you calculated first and it is now no longer homogenous equation that you want to find its solution but AX = the eigenvector that you found in the first step. but this procedure did not give me the vectors mentioned in the picture. Could anyone help me please?
|
If you're good with generalized eigenvectors as a concept, skip this. Recall for an eigenvector $Av=\lambda v$ so $(A-\lambda I)v=0$. For generalized eigenvectors we satisfy $(A-\lambda I)^kv=0$, so we can see that an eigenvector can be generalized by solving for $(A-\lambda I)u=v$ where $v$ is a generalized eigenvector of degree $k$ and $u$ of degree $k+1$. Then $(A-\lambda I)^{k+1}u=(A-\lambda I)^k(A-\lambda I)u=(A-\lambda I)^kv=0$.
To the computation. Did you check to subtract $2I$ from the original matrix?
$$(A-2I)v=
\begin{bmatrix}
0 & -2 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}v=0$$
gives us our first eigenvector of $$
\begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}$$ by inspection. A lone zero column does us justice.
Then we want to solve $(A-\lambda I)u=v$ or: $$\begin{bmatrix}
0 & -2 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \begin{bmatrix}
a \\
b \\
c \\
d \\
\end{bmatrix} = \begin{bmatrix}
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}$$
Note to get the third zero, we need that $\begin{bmatrix}
0 & 0 & 0 & 1 \\
\end{bmatrix}\begin{bmatrix}
a \\
b \\
c \\
d \\
\end{bmatrix}=\begin{bmatrix}
0 \\
\end{bmatrix}$ so it must be that $d=0$. Then the second row must satisfy $\begin{bmatrix}
0 & -1 & 1 & 0 \\
\end{bmatrix}\begin{bmatrix}
a \\
b \\
c \\
0 \\
\end{bmatrix}=\begin{bmatrix}
0 \\
\end{bmatrix}$ giving $-b+c=0$ so $b=c$. Then we have from the first row $\begin{bmatrix}
0 & -2 & 1 & 1 \\
\end{bmatrix}\begin{bmatrix}
a \\
b \\
b \\
0 \\
\end{bmatrix}=\begin{bmatrix}
1 \\
\end{bmatrix}$ so $-b=1$ gives $b=c=-1$ so $\begin{bmatrix}
a \\
b \\
c \\
d \\
\end{bmatrix}=\begin{bmatrix}
0 \\
-1 \\
-1 \\
0 \\
\end{bmatrix}$
One more time:
$$\begin{bmatrix}
0 & -2 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \begin{bmatrix}
e \\
f \\
g \\
h \\
\end{bmatrix} = \begin{bmatrix}
0 \\
-1 \\
-1 \\
0 \\
\end{bmatrix}$$
Start with the third row, so $\begin{bmatrix}
0 & 0 & 0 & 1 \\
\end{bmatrix}\begin{bmatrix}
e \\
f \\
g \\
h \\
\end{bmatrix}=\begin{bmatrix}
-1 \\
\end{bmatrix}$ so $h=-1$. Then from the 2nd row, $\begin{bmatrix}
0 & -1 & 1 & 1 \\
\end{bmatrix}\begin{bmatrix}
e \\
f \\
g \\
-1 \\
\end{bmatrix}=\begin{bmatrix}
-1 \\
\end{bmatrix}$ so $-f+g+h=-f+g-1=-1$, so $-f+g=0$ or $f=g$.
By the first row, this implies that $\begin{bmatrix}
0 & -2 & 1 & 1 \\
\end{bmatrix}\begin{bmatrix}
e \\
f \\
f \\
-1 \\
\end{bmatrix}=\begin{bmatrix}
0 \\
\end{bmatrix}$ so $-2f+f+h=-f-1=0$, or $f=g=-1$.
So we get $\begin{bmatrix}
e \\
f \\
g \\
h \\
\end{bmatrix}=\begin{bmatrix}
0 \\
-1 \\
-1 \\
-1 \\
\end{bmatrix}$.
|
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|
Adjoint of Sum = Sum of Adjoints? Didnt find this anywhere, just verifying.
We know that: $$ (A+B)^T= A^T+B^T $$.
Does it follow that:$$ (A+B)^† = A^† +B
^† $$ for all A and B matrices (the dagger here representing the adjoint)?
If so what is the proof?
|
This statement is not true in general.
Wikipedia tells us that for 3 by 3 matrices the first element of the adjoint obtains as,
$$
a_{1,1}^\dagger = a_{2,2}a_{3,3} - a_{2,3}a_{3,2} \quad \text{ and } \quad b_{1,1}^\dagger = b_{2,2}b_{3,3} - b_{2,3}b_{3,2},
$$
but
$$
\begin{align}
(a_{1,1} + b_{1,1})^\dagger &= (a_{2,2} + b_{2,2})(a_{3,3} + b_{3,3}) - (a_{2,3} + b_{2,3})(a_{3,2} + b_{3,2}) \\ &= a_{2,2}a_{3,3} + b_{2,2}b_{3,3} - a_{2,3}a_{3,2} - b_{2,3}b_{3,2} + a_{2,2}b_{3,3} + b_{2,2}a_{3,3} - a_{2,3}b_{3,2} - b_{2,3}a_{3,2} \\ &=
a_{1,1}^\dagger + b_{1,1}^\dagger + a_{2,2}b_{3,3} + b_{2,2}a_{3,3} - a_{2,3}b_{3,2} - b_{2,3}a_{3,2} \\ &\neq a_{1,1}^\dagger + b_{1,1}^\dagger
\end{align}
$$
You may also check numerically with Matlab:
A = randn(3);
B = randn(3);
adjoint(A+B)
adjoint(A)+adjoint(B)
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3446710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Evaluate integral $\int_t^\infty (1-3x^{-4})e^{-x^2/2}dx$ I am trying to evaluate the integral
$$\int_t^\infty (1-3x^{-4})e^{-x^2/2}\,\mathrm{d}x.$$
This is as far as I could get before getting stuck.
$$
\begin{align}
\int_t^\infty (1-3x^{-4})e^{-x^2/2}\,\mathrm{d}x&=\int_t^\infty e^{-x^2/2}\,\mathrm{d}\left(x+\frac{1}{x^3}\right) \\
&=-\left(t+\frac{1}{t^3}\right)e^{-t^2/2}-\int_t^\infty \left(x+\frac{1}{x^3}\right)\,\mathrm{d}(e^{-x^2/2}) \\
&=-\left(t+\frac{1}{t^3}\right)e^{-t^2/2}+\int_t^\infty \left(x^2+\frac{1}{x^2}\right)e^{-x^2/2}\,\mathrm{d}x \\
\int_t^\infty \left(x^2+\frac{1}{x^2}\right)e^{-x^2/2}\,\mathrm{d}x&=\int_t^\infty e^{-x^2/2}\,\mathrm{d}\left(\frac{1}{3}x^3-\frac{1}{x}\right)
\end{align}
$$
|
$$I=I_1-I_2$$
$$I_1=\int_{t}^{\infty} e^{-x^2/2} dx= \sqrt{2} \int_{t/\sqrt{2}}^{\infty} e^{-z^2} dz=\sqrt{\frac{\pi}{2}}~Erfc(t/\sqrt{2})$$
$$I_2=3\int_{t}^{\infty} x^{-4}~ e^{-x^2/2} dx$$
Let $x^2=2u, xdx=du$, then
$$I_2=\frac{3}{4\sqrt{2}} \int_{t^2/2}^{\infty} u^{-5/2} e^{-u} du$$
Next, twice integration by parts leads to
$$I_2=\frac{3}{4\sqrt{2}}\left[-\frac{2}{3} e^{-u}~ u^{-3/2}+\frac{4}{3} e^{-u} u^{-1/2}+\frac{4}{3}\sqrt{\pi} ~Erf(\sqrt{u})~du \right]_{t^2/2}^{\infty}$$
Usw $Erfc(z)=1-Erf(z),~ Erf(\infty)=1$
$$I=\left(\frac{1}{t}-\frac{1}{t^3} \right) e^{-t^2/2}.$$
|
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"url": "https://math.stackexchange.com/questions/3448635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\left(\dfrac{B}{2}\right)=\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{C}{2}\right)$$
$$2\sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$$
$$2\sqrt{\dfrac{(s-a)(s-c)(s-b)}{s(s-b)^2}}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-a)^2}}+\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-c)^2}}$$
$$\dfrac{2}{s-b}=\dfrac{1}{s-a}+\dfrac{1}{s-c}$$
$$\dfrac{2}{s-b}=\dfrac{s-c+s-a}{(s-a)(s-c)}$$
$$\dfrac{2}{s-b}=\dfrac{b}{(s-a)(s-c)}$$
$$2\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-c\right)=b\left(\dfrac{a+b+c}{2}-b\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$b^2-a^2-c^2+2ac=ba+bc-b^2$$
$$2b^2-a^2-c^2+2ac-ba-bc=0\tag{1}$$
$$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$$
$$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$
$$\cos A+\cos C=\dfrac{ab^2+ac^2-a^3+a^2c+b^2c-c^3}{2abc}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac^2-a^3+a^2c-c^3}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac(a+c)-(a+c)(a^2+c^2-ac)}{b}}{2ac}$$
Using equation $(1)$, $2ac-a^2-c^2=ba+bc-2b^2$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{(a+c)(ba+bc-2b^2)}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+(a+c)(a+c-2b)}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+a^2+c^2+2ac-2ba-2bc}{2ac}$$
$$\cos A+\cos C=\dfrac{a^2+c^2+2ac-ab-bc}{2ac}$$
Using equation $(1)$, $2ac-ba-bc=a^2+c^2-2b^2$
$$\cos A+\cos C=\dfrac{a^2+c^2+a^2+c^2-2b^2}{2ac}$$
$$\cos A+\cos C=\dfrac{2a^2+2c^2-2b^2}{2ac}$$
$$\cos A+\cos C=2\cdot\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A+\cos C=2\cos B$$
Is there any nice way to solve this question, mine goes very long. I tried various methods but this was the only way I was able to prove the required result.
|
Writing $A_2$ for $A/2$, etc, and noting that $A_2+B_2+C_2 = \pi/2$, we have
$$\begin{align}
\tan A_2-\tan B_2 &= \tan B_2 - \tan C_2 \\[6pt]
\frac{\sin A_2 \cos B_2 - \cos A_2 \sin B_2}{\cos A_2 \cos B_2}
&=
\frac{\sin B_2 \cos C_2 - \cos B_2 \sin C_2}{\cos B_2 \cos C_2} \\[6pt]
\sin(A_2-B_2)\cos C_2 &= \sin(B_2-C_2)\cos A_2 \\[6pt]
\sin(A_2-B_2)\sin(A_2+B_2) &= \sin(B_2-C_2)\sin(B_2+C_2) \\[6pt]
\frac12\left(\cos 2B_2 - \cos 2 A_2\right) &= \frac12\left(\cos 2C_2-\cos 2B_2\right) \\[6pt]
\cos B - \cos A &= \cos C - \cos B
\end{align}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3450441",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$ $\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$
I wanted to use L' Hopital's rule so I wrote the term as:
$$\frac{\sin^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)}{\cos^2{x}}$$
and found the first derivative of the denumaerator & denominator:
$\frac{{\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\sin{x}\cos{x}}{2\cos{x}(-\sin{x})}$
$$\lim_{x\to \frac{\pi}{2}}{\left(-2\left({\left(\frac{4\sin{x}\cos{x}+3\cos{x}}{2\sqrt{2\sin^2{x}+3\sin{x}+4}}-\frac{2\sin{x}\cos{x}+6\cos{x}}{2\sqrt{\sin^2{x}+6\sin{x}+2}}\right)}\sin^2{x}\;+ \;2\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)\sec{x}\csc{x}\right)}$$
But I didn't know how to continue.
I drew a graph in Geogebra:
|
Instead of using L'Hospital, multiply the top and bottom by conjugate:
$$\frac{\sin^2{x}\Big((2\sin^2{x}+3\sin{x}+4)-(\sin^2{x}+6\sin{x}+2)\Big)}{\cos^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\Big)}=\\
\require{cancel}\frac{\sin^2{x}(\cancel{1-\sin x})(2-\sin x)}{(\cancel{1-\sin x})(1+\sin x)\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}+\sqrt{\sin^2{x}+6\sin{x}+2}\Big)}.$$
And now plug $x=\frac{\pi}{2}$ to get:
$$\frac{1^2\cdot 1}{(1+1)(3+3)}=\frac1{12}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3457123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Minimize $x + z$ subject to $x^2 + y^2 = 1$ and $y^2+z^2 = 4$ I'm trying to solve this problem by KKT's condition:
$$\begin{align*}
\text{min} & \quad x + z \\
\text{s.t} & \quad x^2 + y^2 = 1 \\
& \quad y^2+z^2 = 4
\end{align*}$$
One of the regularity conditions is linear independence constraint constraint qualification (LICQ), which is
The gradients of the active inequality constraints and the gradients of the equality constraints are linearly independent at $x^{*}$.
Could you please verify if I correctly apply the KKT's theorem? Thank you so much for your help!
$\textbf{My attempt:}$
Put $f(x,y,z) = x + z$, $h_1(x,y,z) = x^2+y^2-1$, $h_2(x,y,z)=y^2 + z^2 -4$, and $\mathcal K= \{(x,y,z) \in \mathbb R^3 \mid h_1(x,y,z) = h_2(x,y,z) = 0\}$.
Because $\mathcal K$ is compact and $f$ is continuous, the problem has a solution.
Moreover, $\nabla f (x,y,z) = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, $\nabla h_1 (x,y,z) = \begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix}$, and $\nabla h_2 (x,y,z) = \begin{pmatrix} 0\\ 2y \\ 2z \\ \end{pmatrix}$.
Next, we check if LICQ is satisfied by considering the system of equations: $$\begin{cases}
\mu_1 \nabla h_1 (x,y,z) + \mu_2 \nabla h_2 (x,y,z)=0 &=0 \\
h_1(x,y,z) &= 0\\
h_1(x,y,z) &=0
\end{cases}
\iff
\begin{cases}
\mu_1 \begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix} + \mu_2 \begin{pmatrix} 0\\ 2y \\ 2z \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix} \\
x^2 + y^2 &= 1\\
y^2+z^2 &= 4
\end{cases}$$
$$\iff \begin{cases}
x \mu_1 &= 0\\
y(\mu_1 + \mu_2) &= 0 \\
z \mu_2 &= 0 \\
x^2 + y^2 &= 1\\
y^2+z^2 &= 4
\end{cases}$$
If $\mu_1 \neq 0$ then $x=0$. It follows that $y^2 =1$ and $z^2 =3$. Hence $\mu_2=0$ and thus $y(\mu_1 + \mu_2) \neq 0$, which is impossible. If $\mu_2 \neq 0$ then $z=0$. It follows that $y^2 =4 >1$, which is impossible. As such, $\mu_1= \mu_2 =0$ and thus LICQ is satisfied. It follows from KKT's theorem that the solution satisfies
$$\begin{cases}
h_1(x,y,z) &= 0\\
h_1(x,y,z) &=0 \\
\nabla f (x,y,z) + \mu_{1} \nabla h_1(x,y,z) + \mu_{2} \nabla h_1(x,y,z)&=0
\end{cases}
\iff
\begin{cases}
x^2 + y^2 &= 1\\
y^2+z^2 &= 4 \\
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} +\mu_1 \begin{pmatrix} 2x \\ 2y \\ 0 \end{pmatrix} + \mu_2 \begin{pmatrix} 0\\ 2y \\ 2z \\ \end{pmatrix} &= \begin{pmatrix} 0\\ 0 \\ 0 \\ \end{pmatrix}
\end{cases}$$
$$\iff
\begin{cases}
x^2 + y^2 &= 1\\
y^2+z^2 &= 4 \\
2 \mu_1 x &= -1\\
y(\mu_1 + \mu_2) &=0 \\
2 \mu_2 z &= -1
\end{cases}
\iff
\begin{cases}
x^2 &= 1\\
z^2 &= 4 \\
y &= 0\\
\end{cases}$$
Hence $(x,y,z) =(\pm1,0,\pm2)$. After comparing the values at these points, we get the solution is $(-1,0,-2)$.
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Yes, your solution is correct, I just corrected a small typo in the notations of $f(x, y, z)$.
|
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|
If $x,y,z>0.$Prove: $(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$ If $x,y,z>0.$Prove:
$$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$
I was not able to solve this problem instead I could solve similar inequality when we have two variable.I assumed $y=tx$ and uesd derivative.
Can this be generalized as:
If ${a_i>0}\quad(i=1,2,...,n)$
$$\sum_{i=1}^n a_{i} \sum_{i=1}^n \frac{1}{a_{i}}\geq n^2\sqrt[]\frac{\sum_{i=1}^n a^2_{i} }{\sum_{i=1}^n a_{i}a_{i+1} }$$
$a_{n+1}=a_{1}$
Question from Jalil Hajimir
|
I have a solution using Buffalo way, but it's ugly! I'm sorry about that!
Solution:
Without loss of generality, assume that $x=\min\{x,y,z\}$.
Let $x=a$, $y=a+u$, $z=a+v$ so $a>0$; $u,v \geq 0$
We need to prove: $$(x+y+z)^2 (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 - 81\frac{(x^2 +y^2 +z^2)}{xy+yz+zx} \geq 0$$
After reduction of many fractions to a common denominator, we need to prove:
$$27a^6(u^2 -uv+v^2)+18a^5 (u+v)^3 +3a^4 (u^4 +13u^3 v+78u^2 v^2+13uv^3 +v^4 )+2a^3(4u^5 -7u^4 v+94u^3 v^2 +94u^2 v^3 -7uv^4 +4v^5)+3a^2 uv(4u^4 +4u^3 v+57u^2 v^2 +4uv^3 +4v^4)+6au^2 v^2(u^3 +4uv(u+v)+v^3)+u^3 v^3 (u+v)^2 \geq 0$$
Because: $u^2 -uv+v^2 \geq 0$; $(u+v)^3 \geq 0$; $u^4 +13u^3 v+78u^2 v^2+13uv^3 +v^4 \geq 0$, $uv(4u^4 +4u^3 v+57u^2 v^2 +4uv^3 +4v^4)$;$(u^3 +4uv(u+v)+v^3)\geq 0$; $u^3 v^3 (u+v)^2 \geq 0$
So it suffices to prove: $4u^5 -7u^4 v+94u^3 v^2 +94u^2 v^3 -7uv^4 +4v^5 \geq 0$
But it's obvious true by AM-GM: $$4u^5+94u^3 v^2 -7u^4 v \geq 2\sqrt{(4u^5).(94u^3 v^2)} - 7u^4 v =(4\sqrt{94}-7)u^4 v >0$$
And $$4v^5 +94u^2 v^3 -7uv^4 \ge 2\sqrt{(4v^5).(94u^2 v^3)} -7uv^4 =(4\sqrt{94}-7)uv^4 >0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the Maclaurin series for $f(x)=x\ln(x+1)$
Find the Maclaurin series for the function
$$f(x)=x\ln(x+1)$$
So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.
$$\begin{align}
f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0 \\
f''(x)&=\frac{1}{(x+1)^2}+\frac{1}{x+1} & f''(0)&=2 \\
f^{(3)}(x)&=\frac{-x-3}{(x+1)^3} & f^{(3)}(0)&=0 \\
f^{(4)}(x)&=\frac{2x+8}{(x+1)^4} & f^{(4)}(0)&=8
\end{align}$$
So now plugging in for the Maclaurin form
$$P_n(x)=0+x^2-\frac{1}{2}x^3+\frac{1}{3}x^4+....$$
Is this correct?
|
You may use
$$\frac1{1-t} = 1 + t + t^2 + \>...$$
to obtain,
$$x\ln(1+x) = -x\int_0^{-x} \frac{dt}{1-t}=-x\int_0^{-x}\left(1+t+t^2+\>...\right)dt
= x^2 - \frac{x^3}2 + \frac{x^4}3+\>...$$
|
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"url": "https://math.stackexchange.com/questions/3461995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Give an asymptotic upper bound of this recurrence relation : $T(n) = 2\cdot T(n^{1/2}) + n$ I unrolled the recursion and got this:
$T(n) = 2^k\cdot T(n^{1/2^k}) + [2^0\cdot n^{1/2^0} + 2^1\cdot n^{1/2^1} + 2^2\cdot n^{1/2^2}+\cdots+2^{k-1}\cdot n^{1/2^{k-1}}]$
I considered every $n^{1/2^i}$ as $n$.
$\begin{align}
T(n)&\leq 2^k\cdot T(n^{1/2^k}) + n(2^0 + 2^1 + 2^2+\cdots +2^{k-1})\\
&= 2^k\cdot T(n^{1/2^k}) + n(2^k -1)
\end{align}$
Now, considering, $T(2)=1$
$n^{1/2^k}=2\implies k=\log_2(\log_2n)$
$\therefore T(n) \leq 2^{\log_2(\log_2n)} + n(\log_2n-1)=\mathcal{O}(n\cdot\log_2n)$
Now my questions are :
1. Is there any tighter bound possible?
2. How to solve it using Master's Method?
If I assume, $n=2^m$, the recurrence becomes,
$$T(2^m) = 2\cdot T(2^{m/2}) + 2^m$$
Let, $T(2^m) = S(m)$
$$S(m) = 2\cdot S(m/2) + 2^m$$
But can we apply Master's theorem here since $2^m$ is not polynomially larger than $m$?
|
We have
$$T(m^2)=2T(m)+m^2.$$
Then
$$T(4)=2T(2)+4,\\
T(16)=4T(2)+8+16,\\
T(256)=8T(2)+16+32+256,\\
T(65536)=16T(2)+32+64+512+65536,\\
\cdots\\
T(2^{2^k})=2^kT(2)+2^k2^2+2^{k-1}2^4+2^{k-2}2^8+\cdots2^{2^k},$$
which is
$$T(n)=\log_2n\ T(2)+\log_2n\,2^2+(\log_2n-1)2^4+(\log_2n-1)2^8+\cdots+n=O(n).$$
|
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|
Show that there are infinitely many natural numbers such that $a^2+b^2=c^2+3 .$ This question was asked in the Crux Mathematicorum , October edition , Pg -$5$ , which can found be Here.
The question states that :
Both $4$ and $52$ can be expressed as the sum of two squares as well as
exceeding another square by $3$ :
$$4 = 0^2+2^2 \quad\,, \, 4-3=1^2 $$
$$52 = 4^2+6^2 \quad\,, \, 52-3=7^2 $$
Show that there are an infinite number of such numbers that have these two
characteristics.
My attempt :
I Found $4,52$ and $292$ to have this characteristics. An interesting feature I noticed was $$4 = \color{red}{0^2}+2^2 \quad\,, \, 4-3=\color{green}{1^2}$$
$$52 =4^2 + \color{red}{6^2} \quad\,, \, 52-3=\color{green}{7^2}$$
$$292 = 6^2 + \color{red}{16^2} \quad\,, \, 292-3=\color{green}{17^2}$$
If a number $y$ satisfies this property , then it might be written as :
$$y= a^2+b^2 = c^2+3$$
And based on the above examples , I conjectured that :
$$(k)^2 + b ^2 = (k+1)^2 + 3$$
where $a=k$ and $c=k+1$.
This expression on simplifying gives us :
$b^2 = 2(k+2)$ . For R.H.S to be a perfect square , $(k+2)$ must be of the form $2x^2$ , which on solving , gives $k = 2x^2-2$ and $b$ comes out to be $2x$.
So a solution is given by $\color{blue}{(a,b,c) = (2x^2-2,2x,2x^2-1)}$
And our number becomes $y = 4(x^4 - x^2 + 1)$ for all $x\in\mathbb{N}$ and hence there are infinite numbers with this characteristics.
Although ' maybe ' this proves the question , it is not quite rigorous method to do this . Also , it does not provide all the possible numbers as we have only taken the special case when $a=k \,\,, c= k+1$.
What is the better way to solve this problem and the general formula for the number?
Bonus Question : Prove that the highest power of two dividing the number is $2.$ Or more generally ,show that :
$$2^c\nmid y \quad \quad \text { For any } c \ge 3.$$
Verified it for all $y\le1.5\times10^5$ and it seems quite likely to be true . For my case , where $y = 4(x^4 - x^2 +1)$ , this is obvious as $x^4-x^2+1$ is always odd and hence the number is only divisible by $4$. But what about the other numbers ?
Edit :
The first $5$ numbers with this characteristics are :
$$4 = 0^2+2^2 \quad\,, \, 4-3={1^2}$$
$$52 =4^2 + {6^2} \quad\,, \, 52-3=7^2$$
$$292 = 6^2 + {16^2} \quad\,, \, 292-3={17^2}$$
$$628 = 12^2 + {22^2} \quad\,, \, 628-3={25^2}$$
$$964 = 8^2 + {30^2} \quad\,, \, 964-3={31^2}$$
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Brainstorming from doing something similar with pythagorean triples.
If $b^2 = 2a + 1 + 3=2a+4$ then $a^2 + b^2 = a^2 + 2a + 1 + 3 = (a+1)^2 + 3$.
If $b$ is any even number, $2k; k > 1$ then if $a = \frac {(2k)^2-4}2= 2k^2-2$
And if $c = 2k^2 -1$ then
$a^2 + b^2 = $
$(2k^2 -2)^2 + 4k^2 =$
$4k^4 - 8k^2 + 4 + 4k^2 =$
$4k^4 - 4k^2 + 1 + 3 =$
$(2k^2 - 1)^2 + 3=$
$c^2 + 3$.
Yep.... that seems to do it.
=====
Perhaps more formally I should have done
$a^2 + b^2 = c^2 + 3 \implies$
$c^2 - a^2 = b^2 -3$
$(c-a)(c+a) = b^2 -3$
If I let $d = c-a$ be an odd number and $e = d+2a = c+a$ be a larger odd number then so long as we have
$b^2 -3 = d*e$ is an odd number then we have a solutions.
So if $b$ is any even number $> 2$ then $b^2 -3$ is odd number. If we let $b^2 -3 = d*e; d< e$ be any factors of $b^2-3$ (if $b^2 -3$ is prime or a prime square we can let $d = 1$ and $e=b^2 -3$). Then we let $a = \frac {e-d}2$ and $c = \frac {e+d}2$ we then have
$b^2 -3 = d*e = (c-a)(c+a) = c^2 - a^2$ and
$a^2 +b^2 = c^2 + 3$.
...
If $b$ is odd and $b^2 -3$ is even then $b= 2k+1$ and $b^2 -3 = 4k^2 + 4k -2$ and is divisible by $2$ but not $4$. $c-a$ and $c+a$ must both be the same parity so this will not be possible.
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"timestamp": "2023-03-29T00:00:00",
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|
If $2R+r=r_1$, then prove that $A=90$
R is the circumradius of the triangle ABC, $r$ is the inradius and $r_1$ is the ex-radius on side a which is opposite to angle A
In the expression
$$2R+4R\sin \frac A2 \sin \frac B2 \sin \frac C2=4R\sin \frac A2 \cos \frac B2 \cos \frac C2$$
R gets canceled and the expression becomes
$$2+\sin \frac A2(\cos \frac{B-C}{2}-\cos \frac{B+C}{2})=\sin \frac A2 (\cos \frac{B+C}{2}+\cos \frac{B-C}{2})$$
$$2-\sin^2\frac A2=\sin ^2\frac A2$$
$$\sin \frac A2=1$$
$$A=180$$ which is obviously wrong. Please let me know my mistake.
|
Hint
$$2R=\cdots=4R\sin A/2\cos(B+C)/2=4R\sin^2A/2=2R(1-\cos A)$$
|
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|
What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
]1
Solution:
Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\times\pi r^2=\dfrac{5\pi r^2}{2}$
The radius $R$ of the big circle is equal to $r$ plus the radius of the circumscribed circle of equailateral triangle ABC, whose side is $2r$.
The radius of the circumscribed circle of an equilateral triangle is the length of the sides divided by $\sqrt{3}$. Since the side here measures $2r$, the radius of the circumscribed circle is $\dfrac{2r}{\sqrt{3}}$.
So we have $R = r+\dfrac{2r}{\sqrt{3}}$
The area of the big circle is $\pi \times R^2$, which here is equal to $(r+\dfrac{2r}{\sqrt{3}})^2$
which, when expanded, gives
Big Circle area = $\dfrac{\pi r^2(7+4\sqrt{3})}{3}$
To obtain the shaded fraction, we need to divide the area of the colored region by the area of the big circle:
Shaded fraction = $\dfrac{\dfrac{5\pi r^2}{2}}{\dfrac{\pi r^2(7+4\sqrt{3})}{3}}$
Shaded fraction = $\dfrac{5\pi r^2}{2} \times \dfrac{3}{\pi r^2(7+4\sqrt{3})}$
Shaded fraction = $\dfrac{15}{2(7+4\sqrt{3})} \simeq 53.847 \% $
I think it's wrong. In the drawing the smaller circles are not tangent to the largest
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Take the upper right point $A$ of the red circle, the center $B$ of the yellow circle and the center $C$ of the big circle as vertices of a triangle. Call the radius of the smaller circles $r$. Then $AB=3r$, $BC=r\sqrt3$ and $\angle CBA=30^\circ$. Law of cosine gives $R=AC$, the radius of the big circle.
My results: $R^2=13r^2/3$ and the ratio is $15/26$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit with criteria $\lim_{n \to \infty}n \cdot \left [ \frac1e \left (1+\frac{1}{n+1} \right )^{n+1}-1 \right ]$ $$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]$$
I was trying to calculate a limit that drove me to this case of Raabe-Duhamel's test, but I don't know how to finish it. Please give me a hint or a piece of advise.
I cannot use any of the solution below, but they are clear and good. I'm trying to prove it using squeeze theorem like this:
$$\lim_{n \to \infty}n \cdot \left [ \frac{\left (1+\frac{1}{n+1} \right )^{n+1}}{e}-1 \right ]=\frac{-1}{e} \cdot\lim_{n \to \infty}n \cdot \left [e- \left (1+\frac{1}{n+1} \right )^{n+1} \right ]$$
I found this:
$$\frac{e}{2n+2}<e- \left (1+\frac{1}{n} \right )^{n}<\frac{e}{2n+1}$$
Is this true? How can I prove this? Thanks for the answers.
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An overkilled method:
\begin{align*}
n\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right]&=\dfrac{n}{n+1}\cdot(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right],
\end{align*}
so we are to look at
\begin{align*}
(n+1)\cdot\left[\dfrac{\left(1+\dfrac{1}{n+1}\right)^{n+1}}{e}-1\right],
\end{align*}
somehow it is the same looking at
\begin{align*}
n\cdot\left[\dfrac{\left(1+\dfrac{1}{n}\right)^{n}}{e}-1\right]=\dfrac{1}{e}\cdot n\cdot\left[\left(1+\dfrac{1}{n}\right)^{n}-e\right].
\end{align*}
We note that
\begin{align*}
\lim_{x\rightarrow 0}(1+x)^{1/x}=e,
\end{align*}
so
\begin{align*}
n\cdot\left[\left(1+\dfrac{1}{n}\right)^{n}-e\right]=n\int_{0}^{1/n}\left((1+x)^{1/x}\right)'dx.
\end{align*}
Taking the derivative of the integrand, we find that
\begin{align*}
n\int_{0}^{1/n}\left((1+x)^{1/x}\right)'dx=n\int_{0}^{1/n}\left((1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)\right)dx.
\end{align*}
By Integral Mean Value Theorem applied to the interval $[0,1/n]$, we get
\begin{align*}
&n\int_{0}^{1/n}\left((1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)\right)dx\\
&=(1+\eta_{x})^{1/\eta_{x}}\left(-\dfrac{1}{\eta_{x}^{2}}\log(1+\eta_{x})+\dfrac{1}{\eta_{x}}\dfrac{1}{\eta_{x}+1}\right).
\end{align*}
However, it is not hard to compute that
\begin{align*}
\lim_{x\rightarrow 0}(1+x)^{1/x}\left(-\dfrac{1}{x^{2}}\log(1+x)+\dfrac{1}{x}\dfrac{1}{x+1}\right)=-e/2,
\end{align*}
so as $x\rightarrow 0$, $\eta_{x}\rightarrow 0$ and the limit is $-1/2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculating definite integral with absolute value I need to evaluate, with $a,c>0$, the integral:
$$\int_{-a}^a\left(1-\frac{|x-y|}c\right)\,dy$$
This is what i tried
if $x>y:$
$$\int_{-a}^a\left(1+\frac{y-x}c\right)\,dy=2 a-\frac{2 a x}{c}$$
if $x<y:$
$$\int_{-a}^a\left(1+\frac{x-y}c\right)\,dy=2 a+\frac{2 a x}{c}$$
is this right? Can I combine the two?
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For $x \in (-a,a) $
$$I \equiv \int_{-a}^a\left(1-\frac{|x-y|}c\right)\,dy \\ = \frac 1c\int_{-a}^x\left( c-x+y \right)\,dy
\\ + \frac 1c\int_{x}^a\left( c+x-y \right)\,dy
\\ = \frac 1c \left[ (c-x)(x+a)+(c+x)(a-x) \right]
\\ = \frac 1c (-2x^2+2ac) =2a-2\frac{x^2}{c} $$
for $x<-a$
$$I = 2a+2\frac{ax}c $$
for $x>a$
$$I = 2a-2\frac{ax}c $$
|
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"url": "https://math.stackexchange.com/questions/3469093",
"timestamp": "2023-03-29T00:00:00",
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|
$xy + ax^2 + bx + c = 0 , x+y$ has relative max or min if... Options :
$b^2-4ac > 0$
$b^2/4ac >0$
$b/(c-1) >0$
$c/(a-1)>0$
$a/(b-1)>0$
Relative max or min is local max or min, means max or min at certain open interval..
$xy + ax^2 + bx + c = 0 $
$ax^2 + (y+b)x + c = 0 $ if its a qudratic fuction.
Then the max or min point will be (-b/2a , f(-b/2a))
$y = (-ax^2 - bx - c)/x$
But what does it mean by $x+y$ ?
When $x+y$ will have relative max or min?
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Get $y$ by itself so that $$y = \frac{-ax^{2}-bx-c}{x}$$
Then substitute this for $y$ in $f(x) = x+y$ so we have
$$f(x) = \frac{-ax^{2}-bx-c}{x} + x$$
and we want to know if $f(x)$ will have a relative minimum or maximum. The cases where it will not have a minimum or maximum are when the derivative has no real zeroes. Lets find the derivative using the product rule:
$$f'(x) = \frac{1}{x}(-2ax-b)-\frac{1}{x^{2}}(-ax^{2}-bx-c)+1$$
Simplify fractions
$$f'(x) = \frac{x(-2ax-b)-(-ax^{2}-bx-c)+x^{2}}{x^{2}}$$
Simplify
$$f'(x) = \frac{(-a+1)x^{2} + c}{x^{2}}$$
We wish to know when the derivative will be $0$, and we can multiply away the denominator to get
$$0 = (-a+1)x^{2} + c$$
$$x^{2} = \frac{-c}{-a+1} = \frac{c}{a-1}$$
Notice that we need real roots in order to have a relative minimum or maximum, and we are taking a square root, thus
$$\frac{c}{a-1}>0$$
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|
Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$
I tried showing the equation, but my attempts did not get the result.
I already showed that
$$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})=\frac18$$ $$\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=-\frac12$$
Using in particular the trigonometric formulas:
$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ , $\sin(\pi+\alpha)=-\sin(\alpha)$ , $\sin(\pi-\alpha)=\sin(\alpha)$ , $\cos(\pi-\alpha)=-\cos(\alpha)$ , $\sin(-\alpha)=-\sin(\alpha)$ , and $2\sin(\alpha)\cos(\beta)=\sin(\alpha+\beta)+\sin(\alpha-\beta)$
Any hint would be helpful. Thanks in advance.
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HINT:
$$cos\theta_1cos\theta_2=\frac{cos(\theta_1 - \theta_2) + cos(\theta_1 + \theta_2)}{2}$$
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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.