Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the sum of series $\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$ I wanted to know how can I start to find the sum of the series:
$$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}=\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}\cdots$$
I am having no clue.
Thanks.
| Not nearly as impressive as the other answer, but elementary:
$$\begin{align}
\frac{(4n)!}{(4n+4)!} &= \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}\\
&= \left(\frac{1}{4n+1} - \frac{1}{4n+2}\right)\left(\frac{1}{4n+3} - \frac{1}{4n+4}\right)\\
&= \frac{1}{6} \frac{1}{4n+1} - \frac{1}{2}\frac{1}{4n+2} + \frac{1}{2}\frac{1}{4n+3} - \frac{1}{6}\frac{1}{4n+4}\\
&= \frac{1}{3}\left(\frac{1}{4n+1} - \frac{1}{4n+2} + \frac{1}{4n+3} - \frac{1}{4n+4}\right) - \frac{1}{6}\left(\frac{1}{4n+2} - \frac{1}{4n+4}\right) - \frac{1}{6} \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right)
\end{align}$$
It is well-known that
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \log 2,$$
and the first parenthesis comprises four consecutive terms of that series, with no overlap, so from that we obtain $\frac{\log 2}{3}$. From the second parentheses, we can pull out a factor of $\frac12$ from both terms, then we obtain
$$\frac{1}{12} \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)$$
which comprises two consecutive terms of the $\log 2$ series, again without overlap, so together these two yield
$$\left(\frac{1}{3} - \frac{1}{12}\right)\log 2 = \frac{\log 2}{4}.$$
Another well-known series is Leibniz series
$$\frac{\pi}{4} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$$
and the last parenthesis comprises two consecutive terms of that, once again without overlap.
Since all parenthesised terms are dominated by $\frac{1}{n^2}$, we can split and rearrange to obtain
$$\begin{align}
\sum_{n = 0}^\infty \frac{(4n)!}{(4n+4)!} &= \frac13 \sum_{n = 0}^\infty \left(\frac{1}{4n+1} - \frac{1}{4n+2} + \frac{1}{4n+3} - \frac{1}{4n+4}\right)\\
&\quad -\frac{1}{12}\sum_{n=0}^\infty \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)\\
&\quad - \frac16 \sum_{n=0}^\infty \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right)\\
&= \frac{\log 2}{3} - \frac{\log 2}{12} - \frac16\frac{\pi}{4} = \frac{\log 2}{4} - \frac{\pi}{24}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
What should be the method for solving and solution of $\frac{x^2+y^2}{x^2-y^2}$? If $\frac{x+y}{x-y}=3$
then $\frac{x^2+y^2}{x^2-y^2} = ?$
| Hint :
$$
z = \frac xy \Rightarrow \frac{z+1}{z-1} = 3
$$
$$
\frac{x^2+y^2}{x^2-y^2} = \frac{z^2+1}{z^2-1} = ?
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How many zeroes are there at the end of the sum $1^1 + 2^2 + 3^3 + \cdots+ 100^{100}$?
Find the number of zeroes at the end of the sum
$$1^1 + 2^2 + 3^3 + \cdots+ 100^{100}$$
I tried a lot and my answer came 4700 but it was not correct.
| Let $A=1^1 +2^2 +3^3 +\cdots+100^{100}.$
$\phi(5^2)=20,(a,5)=1\implies a^{20}\equiv 1\pmod {5^2},a^4\equiv 1\pmod 5.$
$A\equiv (1^1+2^2+3^3+\cdots+20^{20})+(21^1+22^2+\cdots40^{20})+\cdots+(81^1+\cdots+100^{20})\\ \equiv(1+21+41+61+81)+\sum_{k=0}^4\sum_{i=2}^{20}(20k+i)^i
\\\equiv 5+\sum_{k=0}^4\sum_{i=2}^{20}(i^i+i\cdot20k\cdot i^{i-1})
\\\equiv 5+\sum_{k=0}^4(20k+1)\sum_{i=2}^{20}i^i
\\\equiv 5+5\sum_{i=2}^{20}i^i
\\\equiv 5\sum_{i=1}^{20}i^i \pmod {25}$
$\sum_{i=1}^{20}i^i\equiv (1+2^2+3^3+4^0)+(1^2+2^3+3^0+4^1)+(1^3+2^0+3^1+4^2)+(1^0+2^1+3^2+4^3)
\\\equiv 3+4+1+1=4\pmod5
\\\implies A \equiv 20\not \equiv 0\pmod{25},A\equiv 0\pmod 5$
It's easy to find that $A$ is even, hence $10\mid A,100\nmid A.$ Hence there is only one zero at the end of $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
What this sine function equation means? Apostol's book "Calculus" asks to prove that
$$\sin\frac{\pi }{6}=\frac{1}{2}$$
using the fact that
$$\sin 3x=3\sin x-4\sin^3 x$$
and
$$\sin \frac{\pi}{2}=1$$
So, we take $x=\frac{\pi}{6}$ and we have
$$\sin\frac{\pi}{2}=3\sin\frac{\pi}{6}-4\sin^3 \frac{\pi}{6}$$
$$1=3\sin\frac{\pi}{6}-4\sin^3 \frac{\pi}{6}$$
if we take $y=\sin \frac{\pi}{6}$
$$4y^3-3y+1=0$$
$$\left( {y- \frac{1}{2}} \right)^2(y+1)=0$$
and finally $y=\sin\frac{\pi}{6}=\frac{1}{2}$ or $y=\sin\frac{\pi}{6}=-1$. What it means?, it not should only be $\sin\frac{\pi}{6}=\frac{1}{2}$?
| If it is true that $\sin\frac\pi6=\frac12$, then it is true that EITHER $\sin\frac\pi6=\frac12$ OR $\sin\frac\pi6=-1$.
There are two different angles that, when multiplied by $3$, give angles whose sine is $1$. (I am considering two angles the same if they differ by a multiple of $2\pi$.) Those two angles are $\pi/6$ and $-\pi/2$. Three times a $30^\circ$ turn to the left is a $90^\circ$ turn to the left, and three times a $90^\circ$ turn to the right is also a $90^\circ$ turn to the left. That's why you get two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$ Prove that $\tan(6^\circ)\tan(42^\circ) = \tan(12^\circ) \tan(24^\circ)$.
I don't know how to approach this problem. One approach might be to note that $42-6= 24+12$, and then apply the identities for $\tan(A+B)$ and $\tan(A-B)$, but it just makes it more complex:
$$\frac{\tan(42^{\circ}) - \tan(6^{\circ})}{1+\tan(42^{\circ})\tan(6^{\circ})}= \frac{\tan(24^{\circ}) + \tan(12^{\circ})}{1 - \tan(24^{\circ})\tan(12^{\circ})}.$$
Can anyone here please provide a hint?
| We can prove $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$ (Proof below)
Putting $x=6^\circ,$ $$\tan 6^\circ\tan 66^\circ\tan 54^\circ=\tan18^\circ$$
Putting $x=18^\circ,$ $$\tan 18^\circ\tan42^\circ\tan78^\circ=\tan54^\circ$$
On multiplication, $$\tan 6^\circ\tan 66^\circ\tan42^\circ\tan78^\circ=1$$
$$\tan 6^\circ\tan42^\circ=\cot66^\circ\cot78^\circ=\tan(90^\circ-66^\circ)\tan(90^\circ-78^\circ)=\tan24^\circ\tan12^\circ$$
[
Proof $1:$ $$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan x\cdot\frac{\tan60^\circ-\tan x}{1+\tan60^\circ\tan x}\cdot\frac{\tan60^\circ+\tan x}{1-\tan60^\circ\tan x}$$
$$=\tan x\cdot \frac{\sqrt3-\tan x}{1+\sqrt3\tan x}\cdot \frac{\sqrt3+\tan x}{1-\sqrt3\tan x}$$
$$=\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3x$$
Proof $2:$
If $\tan3x=\tan 3y$
$$\frac{3\tan x-\tan^3x}{1-3\tan^2x}=\tan3y$$
$$\implies \tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$
If $\tan x_1,\tan x_2,\tan x_3$ are the three roots of above cubic equation,
$\implies \tan x_1\cdot\tan x_2\cdot\tan x_3=-\tan3y$
Again, as $\tan3x=\tan 3y$
$3x=180^\circ n+3y$ where $n$ is any integer $\implies x=60^\circ n+y$
So, the in-congruent values of $x$ are $y,60^\circ +y,60^\circ\cdot2+y$
So, the corresponding roots are $\tan y,\tan(60^\circ +y), \tan(60^\circ\cdot2+y)=\tan\{180^\circ-(60^\circ-y)\}=-\tan(60^\circ-y)$
$\implies -\tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=-\tan3y$
$\implies \tan(60^\circ-y)\cdot\tan y\cdot\tan(60^\circ +y)=\tan3y$
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/455070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to factorize $x^3 - 7x + 6$? How do you factorize this polynomial:
$${x^3 - 7x + 6}$$
Some online solver doesn't even work saying: using GCF method doesn't work, but sites like Mathway.com gave me the answer, is there a pre-step you need to do before factorizing?
the answer is $(x-1)(x-2)(x+3)$.
This is actually part (b) of a question, it said use the answer for part (a) i.e $x^3-8$ and factorize. I don't get the relationship, what does this hint actually shows?
| The method below will not always work, but it is another way to arrive at the same answer for this cubic.
This is difficult because it is a cubic, but if you consider the quadratic $x^2 - 7x + 6$, that factors nicely as $(x-1)(x-6)$. So we rewrite our given cubic as $$x^3 - 7x + 6 = (x^3 - x^2) + (x^2 - 7x + 6).$$ Factorising each bracket, we have
\begin{align*}
x^3 - 7x + 6 &= (x^3 - x^2) + (x^2 - 7x + 6)\\
&= x^2(x - 1) + (x - 1)(x - 6)\\
&= (x - 1)(x^2 + x - 6)\\
&= (x - 1)(x - 2)(x + 3).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/455427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Alternating Recurrence relation $a_n = b_{n-1} + 5$ and $b_n = na_{n-1}$ I am racking my brain on solving the relation where:
$$a_n = b_{n-1} + 5$$
$$b_n = na_{n-1}$$
where $a_0$ = $b_0$ = 1
I am trying to find the closed form for $a_n$. I have tried to shifting $b_n = na_{n-1}$ to $b_{n-1} = (n-1)a_{n-2}$ by substituting n with n-1. Then I plug $b_{n-1}$ in to $a_n = b_{n-1} + 5$ to get $a_n = (n-1)a_{n-2} + 5$
I am trying to use the method of generating functions to do this. I have that $A(x) = \sum_{n=0}^{\infty} a_{n}x^{n}$. Now I move to the n = 3 term of the summand and get $1+6x+6x^2 + \sum_{n=3}^{\infty} a_{n}x^{n}$ as $a_0 = 1, a_1 = a_2 = 6$ Then we substitute in the relation to get $1+6x+6x^2 + x^2\sum_{n=3}^{\infty} ((n-1)a_{n-2}+5)x^{n-2}$ = $1+6x+6x^2 + x^2\sum_{n=3}^{\infty} (n-1)a_{n-2}x^{n-2}+\sum_{n=3}^{\infty}5x^{n-2}$
I know what to do with the latter term as $\sum_{n=3}^{\infty}5x^{n-2} = 5x/(1-x)$. I however am not sure what to do with the $\sum_{n=3}^{\infty} (n-1)a_{n-2}x^{n-2}$ term as there is an (n-1) in it. I do however see that $(n-1)a_{n-2}x^{n-2}$ looks like it could be integrated with respect to x to become $a_{n-2}x^{n-1}$. If I then pull out a x I could end up with:
$1+6x+6x^2 + x^3/(1-x)+ 5x/(1-x)$
But I am first shaky on how I came to this, but am furthermore wondering how this would be a closed form i.e. we don't even have x's in the sequence.
Thoughts would be very appreciated.
Thanks,
Brian
| Start with $a_n = (n − 1) a_{n − 2} + 5$, and note that this is a first order linear recurrence in $b_n = a_{2 n + 1}$:
$$
b_{n + 1} = 2 n b_n + 5
$$
Summing factor is $\prod_{0 \le k \le n} 2 n = 2^n n!$:
$$
\frac{b_{n + 1}}{2^n n!} - \frac{b_n}{2^{n - 1} (n - 1)!}
= \frac{5}{2^n n!}
$$
Sum for $1 \le k \le n$ to get:
$$
\frac{b_{n + 1}}{2^n n!} - b_1 = 5 \sum_{1 \le k \le n} \frac{2^{-n}}{n!}
$$
The right hand side is (almost) $5 \left(e^{1/2} - 1 \right)$, so $a_{2 n + 1} = b_n \approx a_1 + 5 \cdot 2^{n - 1} (n - 1)! \left(e^{1/2} - 1\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find the equation of a line Find the equation of the line through (12/5 , 1), forming with the axes a triangle area of 5.
There are 4 solutions and how can i get it?
| The line crossing $A(2.4,1)$ and forming a triangle with the axes and crosses x axis at $x=a$ and y axis at $y=b$ is of the form $$\frac{x}{a} + \frac{y}{b} = 1$$
The triangle may be formed on quarter 1,2 and 4.
In your case,$$|a|\times |b|=2\times5=10$$
$$\frac{2.4}{a} + \frac{1}{b} = 1$$
There are two cases;Solving the corresponding systems yields:
$$\bullet b=\frac{10}{a}\implies 0.1\times a^2 -a+2.4=0$$
$$a=\frac{1+0.2}{0.2},a=\frac{1-0.2}{0.2}\implies a=6, b=\frac53,a=4, b=2.5$$
$$\bullet b=-\frac{10}{a}\implies -0.1\times a^2 -a+2.4=0$$
$$a=\frac{1+1.4}{-0.2},a=\frac{1-1.4}{-0.2}\implies a=-12, b=\frac56,a=2, b=-5 \blacktriangle $$
Here are the plots of the 4 cases:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
| Basically, your proof is saying
$$
-0.5=\sqrt{(-0.5)^2}=\sqrt{0.5^2}=0.5
$$
Now find the error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 0
} |
Variable substitution in probability
In modeling the number of claims filed by an individual under an automobile policy during a three-year period, an actuary makes the simplifying assumption that for all integers $n \ge 0$, $p_{n+1}=\frac{1}{5} p_n$, where $p_n$ represents the probability that the policyholder files $n$ claims during the period. Under this assumption, what is the probability that a policyholder files more than one claim during the period?
My first step was the convert the probability function from future values by variable substitution. So I assign $k=n+1 \implies n=k-1$:
$\therefore p_k=\frac{1}{5} p_{k-1} \text{ for } k \ge 1\tag{1}$
Now we can calculate the probabilites for some k values:
$k=0: p_0$
$k=1: p_1=\frac{1}{5}p_0$
$k=2: p_2=\frac{1}{5}p_1=(\frac{1}{5})^2 p_0$
$\vdots$
$k=k: p_k=(\frac{1}{5})^k p_0$
Since the probabilities sum to 1:
We can solve for $p_0: 1 = p_0 ( 1 + (\frac{1}{5}) + (\frac{1}{5})^2 + \cdots )=\cfrac{p_0}{1-1/5}$
$\implies p_0=0.8$
Now this is where I get confused. The question asks us to calculate $\Pr[n>1]$, so:
$\Pr[n>1]=1-\Pr[n \le 1]\tag{2}$
Now when calculating equation(2), I thought I would have to transform $n$ to $k$, so Equation (2) becomes:
$\Pr[n>1]=1-\Pr[k \le 2]=1-\Pr[k=0]-\Pr[k=1]-\Pr[k=2]\tag{3}$
But this is not correct. The solution is just replacing exactly n with k with no substitution. In other words, the solution is:
$\Pr[k>1]=1-\Pr[k \le 1]\tag{4}=1-0.8-0.16$
Can someone please explain why this the case? Why don't I have to transform n to k when solving this question? If $n$ is the number of claims during the period, then k is the number of claims per period PLUS ONE. Thanks in advance! Have a great weekend!
| Let us keep $k$ out of the game, it interferes with the analysis. Let the probability of no claims be $p_0$. Then the probability of $1$ claim is $p_0\cdot\frac{1}{5}$, the probability of $2$ claims is $p_0\cdot\frac{1}{5^2}$, and so on. Thus
$$p_0\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\cdots\right).$$
The sum of the geometric series $1+\frac{1}{5}+\frac{1}{5^2}+\cdots$ is $\frac{5}{4}$, and therefore, as you found, $p_0=\frac{4}{5}$.
The probability of $0$ claims or $1$ claim is $\frac{4}{5}+\frac{4}{5}\cdot\frac{1}{5}$. This is $\frac{24}{25}$, and therefore the probability of $2$ or more claims is $\frac{1}{25}$.
Another way: A coin has probability $\frac{1}{5}$ of landing heads, and $\frac{4}{5}$ of landing tails. Fate decides on the number of claims by tossing the coin until she gets a head. The number of tails she got before the first head is the the number of claims. It is straightforward to verify that this leads to the probability distribution discussed above. So the probability of $2$ or more claims is the probability that the first $2$ tosses result in tails. The probability of this is $\left(\frac{1}{5}\right)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to verify these series expansions? We know that $$ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots (-1)^n \frac{x^{n+1}}{n+1} + \cdots = \log (1+x) $$ whenever $-1<x<1$. What can we say if $x=1$?
That is, does the series $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots (-1)^n \frac{1}{n+1} + \cdots $$ converge to $\log 2$? If so, how to prove this fact rigorously?
And what about the behavior of the series $$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots (-1)^n \frac{x^{2n+1}}{2n+1} + \cdots, $$
which converges to $\arctan x$ for $-1<x<1$, at the points $x = \pm 1$?
| Daniel's comment already has the solution: since
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}n$$
converges (Leibniz series) and the power series
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}n=f(x)$$
converges for $\,|x|<1\,$ to a function $\,f(x)\;$ , then Abel's Theorem gives us
$$\lim_{x\to 1^-}f(x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}n$$
from here, the sum of the above series indeed is
$$\lim_{x\to 1^-}\log(1+x)=\log2\;$$ since, of course, $\,f(x)=\log(1+x)\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$
Find the value of $3^9\cdot 3^3\cdot 3\cdot 3^{1/3}\cdot\cdots$
Doesn't this thing approaches 0 at the end? why does it approaches 1?
| $$3^9 * 3^3 * 3 * 3^{\frac{1}{3}} * ...=$$
$$3^{9\sum_{n=0}^{\infty}3^{-n}}=$$
$$3^{9*1.5}=$$
$$3^{13.5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Finding $\lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}}$ I need help with finding
$\displaystyle \lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}}$
Thanks!
| $$
\lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}} = \lim_{n \to \infty} \frac{(\frac{2}{3})^{n+1} + 1}{(\frac{1}{3})(\frac{2}{3})^{n}+\frac{1}{3}} = \frac{\,\,1\,\,}{\frac{1}{3}} = 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Compute $(-1)^n\sum_{k=1}^n (-1)^k\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$
*
*Compute $(-1)^n\sum_{k=1}^n (-1)^k\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$
*Define $a_{k,m}=\frac{(-1)^{k+m}(n+k-1)!(n+m-1)!}{(k+m-1)[(k-1)!(m-1)!]^2(n-m)!(n-k)!}$
Compute $\sum_{k=1}^n\sum_{m=1}^na_{k,m}$
(Hint: one approach is considering the Hilbert matrix. Other approaches are also welcome. Thanks.)
| For the first one, suppose we are trying to evaluate
$$(-1)^n \sum_{k=1}^n (-1)^k
\frac{(k+n-1)!}{(k-1)!(k-1)!(n-k)!}$$
which is
$$(-1)^n \sum_{k=1}^n (-1)^k
{n\choose k} k {n+k-1\choose k-1}.$$
Introduce
$${n+k-1\choose k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+k-1}}{z^k} \; dz.$$
This yields for the sum
$$\frac{(-1)^n}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n-1}
\sum_{k=1}^n (-1)^k {n\choose k} k \frac{(1+z)^k}{z^k}
\; dz.$$
Now observe that
$$x((1+x)^n)' = \sum_{r=1}^n {n\choose r} r x^r
= nx(1+x)^{n-1}.$$
We thus obtain
$$\frac{n\times (-1)^{n+1}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z}
\left(1-\frac{1+z}{z}\right)^{n-1}
\; dz
\\ = \frac{n\times (-1)^{n+1}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z}
\frac{(-1)^{n-1}}{z^{n-1}}
\; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^n}
\; dz.$$
This evaluates to $$n\times {n\choose n-1} = n^2$$
by inspection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/461646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Existance and uniqueness of solution for a point with fixed distances to three other points I have two sets of known points in $\mathbb{R}^2$: Four points $\mathbf{p_1}, \mathbf{p_2}, \mathbf{p_3}, \mathbf{p_x}$ , and three other points $\mathbf{q_1}, \mathbf{q_2}, \mathbf{q_3}$. I would like to find $\mathbf{q_x}$ such that the ratios of Euclidian distances between $\mathbf{q_x}$ and $\mathbf{q_1}, \mathbf{q_2}, \mathbf{q_3}$, are the same as between $\mathbf{p_x}$ and $\mathbf{p_1}, \mathbf{p_2}, \mathbf{p_3}$. That is, I want to find $\mathbf{q_x}$ such that
$\frac{d(\mathbf{q_1},\mathbf{q_x})}{\sum_i d(\mathbf{q_i},\mathbf{q_x})} = \frac{d(\mathbf{p_1},\mathbf{p_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})}$ and
$\frac{d(\mathbf{q_2},\mathbf{q_x})}{\sum_i d(\mathbf{q_i},\mathbf{q_x})} = \frac{d(\mathbf{p_2},\mathbf{p_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})}$ and $\frac{d(\mathbf{q_3},\mathbf{q_x})}{\sum_i d(\mathbf{q_i},\mathbf{q_x})} = \frac{d(\mathbf{p_3},\mathbf{p_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})},$
where $d(\cdot,\cdot)$ is the Euclidian distance. Or to rewrite it a bit nicer:
$d(\mathbf{q_x},\mathbf{q_k}) = w \cdot d(\mathbf{p_x},\mathbf{q_k}) $ for $ k = 1,2,3$,
where $w$ is the unknown factor $w = \frac{\sum_i d(\mathbf{q_i},\mathbf{q_x})}{\sum_i d(\mathbf{p_i},\mathbf{p_x})}$. Note also the more to the point problem formulation by Blue in the comments.
This looks like three equations with three unknowns (x-, and y- coordinates of $\mathbf{q_x}$ and $w$). However, I have been struggling for quite a while now to find out if there is always a unique solution, and to find this solution, or to find an optimal solution if no solution exists.
Any help that you can offer is greatly appreciated!
| I'll work from the re-formulation I gave in a comment, but with a minor notational changes.
Given non-negative $u$, $v$, $w$ such that $u+v+w=1$, and given a non-degenerate $\triangle ABC$, we seek a point $P$ such that
$$\frac{|PA|}{u} = \frac{|PB|}{v} = \frac{|PC|}{w} = |PA|+|PB|+|PC| =: r \qquad (1)$$
For convenience, introduce these coordinates:
$$A = (0,0) \qquad B = (c, 0) \qquad C = ( b \cos A, b \sin A ) \qquad P=(x,y)$$
where $b$, $c$ (and, below, $a$) are lengths of the sides opposite angles $B$, $C$ (and $A$) in $\triangle ABC$. Then, from $(1)$, write
$$\begin{align}
(x-0)^2 + ( y - 0 )^2 = r^2 u^2 \quad &\to \quad x^2 + y^2 = r^2 u^2 \\
(x-c)^2 + ( y - 0 )^2 = r^2 v^2 \quad &\to \quad x^2+y^2-2xc + c^2 = r^2 v^2 \\
(x-b \cos A)^2 + ( y - b \sin A )^2 = r^2 w^2 \quad &\to \quad x^2+y^2-2xb\cos A - 2 y b \sin A+ b^2 = r^2 w^2 \\
\end{align}$$
We can use the first equation to eliminate $x^2+y^2$ from the other two equations. (In effect, we're determining the radical axis of each pair of circles.) The result is a linear system
$$\begin{align}
2 x c &= c^2 + r^2 \left( u^2 - v^2 \right) \\
2 x b \cos A + 2 y b \sin A &= b^2 + r^2 \left( u^2 - w^2 \right)
\end{align}$$
whose solution is (after some manipulation that probably needs to be double-checked)
$$\begin{align}
x &= \frac{1}{2c}\left( c^2 + r^2 \left( u^2 - v^2 \right) \right) \\
y &= \frac{1}{2bc\sin A}\left( a b c \cos C + r^2 \left( u^2 a \cos B + v^2 b \cos A - w^2 c \right) \right)
\end{align}$$
Substituting these coordinates into the first circle equation gives a quadratic equation in $r^2$:
$$\begin{align}
0 &= r^4 \; \left( \;
a^2 ( v^2 - u^2 ) ( u^2 - w^2 )
+ b^2 ( w^2 - v^2 ) ( v^2 - u^2 )
+ c^2 ( u^2 - w^2 ) ( w^2 - v^2 )
\; \right) \\
&+ 2 r^2 a b c \; \left( \; u^2 a \cos A + v^2 b \cos B + w^2 c \cos C \; \right) \\
&- a^2 b^2 c^2
\end{align}$$
If the coefficient of $r^4$ vanishes ---for instance, when (and only when?) $u=v=w$--- we have
$$r^2 = \frac{a b c}{2\; \left( \; u^2 a \cos A + v^2 b \cos B + w^2 c \cos C \; \right)} \qquad (2) $$
Otherwise, we invoke the Quadratic Formula, noting that the discriminant factors nicely as
$$\begin{align}
&\phantom{\cdot} \left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right) \\
&\cdot\left(ua+vb+wc\right)\left(-ua+vb+wc\right)\left(ua-vb+wc\right)\left(ua+vb-wc\right)
\end{align}$$
The first set of factors constitute Heron's Formula for $16T^2$, where $T$ is the area of $\triangle ABC$. The second set of factors can be interpreted as $16{T_\star}^2$, where $T_\star$ is the area of an auxiliary triangle whose sides are $ua$, $vb$, $wc$ (if such a triangle exists!). We conclude that
$$r^2 = \frac{\pm 8 T T_\star - a b c \; \left( \; u^2 a \cos A + v^2 b \cos B + w^2 c \cos C \; \right)}{a^2 ( v^2 - u^2 ) ( u^2 - w^2 )
+ b^2 ( w^2 - v^2 ) ( v^2 - u^2 )
+ c^2 ( u^2 - w^2 ) ( w^2 - v^2 )} \qquad (3)$$
With an appropriate choice for "$\pm$", we can substitute-back into the formulas for $x$ and $y$ to find $P$. That messy business is left as an exercise for the reader.
I'm not prepared to assert that $u=v=w$ is the only circumstance under which the coefficient of $r^4$ vanishes, putting equation $(2)$ into play. Nevertheless, note that when the coefficient doesn't vanish, solution to the problem requires that
$$\left(ua+vb+wc\right)\left(-ua+vb+wc\right)\left(ua-vb+wc\right)\left(ua+vb-wc\right) \ge 0$$
(so that $r^2$ is real). This equation is equivalent the the existence of the auxiliary triangle with sides $ua$, $vb$, $wc$, which is therefore necessary for the existence of $P$. Confirmation (or refutation) that it is also sufficient is another exercise for the reader.
As for uniqueness of the solution (when it exists): The independence of the linear system would seem to guarantee that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:
$$ a^2 < 2 \implies a < \frac{2}{a}\\
\text{Let}\hspace{1cm} B=\frac{a+\frac{2}{a}}{2}=\frac{a^2+2}{2a}
$$
$B$ is greater than $a$ because:
$$ \begin{array} {aa} B>a & \implies \frac{a^2+2}{2a}>a \\
& \implies a^2 + 2 > 2a^2 \\
& \implies 2 > a^2 \\
& \implies a^2 < 2
\end{array}$$
If $B^2$ is less than $2$, then $B^2-2<0$, but:
$$\begin{array} {aa} B^2-2 < 0 & \implies \left( \frac{a^2+2}{2a} \right)^2 - 2 < 0 \\
& \implies \frac{a^4+4a^2+4}{4a} - \frac{8a^2}{4a^2} < 0 \\
& \implies \frac{(a^2-2)^2}{(2a)^2} < 0
\end{array}$$
Which is a contradiction since the left hand side of the inequality will be positive for all values of $a$.
But I think we must be able to find such a $B$ since based on my understanding of this answer, we can find a another rational number whose distance from $a$ is less than the distance between $a$ and $\sqrt{2}$
Therefore, I have 2 questions to ask:
*
*Why does this approach work in the case of $a^2>2$ but not when $a^2<2$?
*How should I approach these kind of questions since it seems that there are a few ways to construct a $B$ that satisfies a given set of restrictions? For example, see here (the proof is immediately before the section "13. The Completeness Axiom".
| I too, remembered that this was explained in Baby Rudin:
Associate to each rational $p>0$ , the number:
$p':=\frac{2p+2}{p+2}$
then, if p is in $S_A$:={$x:x^2<2 $}, then $p'^2-2=[(2p+2)/(p+2)]^2-2= \frac{2(p^2-2)}{(p+2)^2}<0$ , so $\frac{2p+2}{p+2}$ is in $S_A$.
Similarly, if p is in $S_B$:={$x:x^2>2$} , then we have, since $p^2>2$ , that p'is in $S_B$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the sum $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$ I would like to check I have this correct
Find the sum
$$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$$
Hint: rationalise the denominators to get a 'telescoping' sum: a sum of terms in which many pairs add up to zero.
I rationalised the denominators to get a series like this:
$$\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} +...+\frac{\sqrt{99}-\sqrt{100}}{-1} $$
Which can be written:
$$\sqrt{2}-\sqrt{1} + \sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}...+ \sqrt{99}-\sqrt{98} +\sqrt{100}-\sqrt{99}$$
Which is the telescoping sum the question talks about.
Most of the terms drop out to leave
$$-\sqrt{1} +\sqrt{100} = 9$$
Have I got this correct?
| you could also do it by induction
conjecture: $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} = \sqrt{n}-1$ For $n \in [2,3,4 ...]$
For $n=2$:
$$\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2}-1$$
For $n+1$: $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$
$$= \sqrt{n} - 1 + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$
$$= \sqrt{n} - 1 + \sqrt{n+1} - \sqrt{n}$$
$$= \sqrt{n+1}-1$$
$$Q.E.D.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Does this inequality involving differences between powers hold on a particular range? Let
$$f(x)=\left(1-\frac{2x}{x+c}\right)^{-n}-\left(1+\frac{x}{c}\right)^{2n}$$ and
$$g(x)=\left(1-\frac{2x}{c}\right)^{-n}-\left(1-\frac{x}{c}\right)^{-2n}$$
where $c>0$ and $n>0$ are constants.
I am wondering if $f(x)\leq g(x)$ for $0\leq x <c/2$ (I am actually interested in small positive $x$).
Clearly, this holds with equality when $x=0$ and I think the inequality holds for $0< x <c/2$. This is based on the numerical evaluations as well as the series expansion in Mathematica of $h(x)=g(x)-f(x)$ which yields:
$$h(x)=\frac{2 n x^3}{c^3}+\frac{3 \left(2 n^2+n\right) x^4}{c^4}+\frac{2 \left(4 n^3+5 n^2+3 n\right) x^5}{c^5}+\ldots$$
However, I am having trouble actually proving this. Any help?
| Let's normalise things a bit first.
$$\left(1 - \frac{2x}{x+c}\right)^{-n} = \left(\frac{c-x}{c+x}\right)^{-n} = \left(\frac{1+\frac{x}{c}}{1-\frac{x}{c}}\right)^n.$$
So we have everything clearly as a function of $\frac{x}{c}$, whence we can without loss of generality assume $c = 1$. Then the inequality we want to show becomes
$$\begin{align}
\left(\frac{1+x}{1-x}\right)^n - (1+x)^{2n} &\leqslant \frac{1}{(1-2x)^n} - \frac{1}{(1-x)^{2n}}\\
\iff (1-x^2)^n - (1-x^2)^{2n} &\leqslant \left(\frac{(1-x)^2}{1-2x}\right)^n - 1\\
\iff (1-x^2)^n\left(1 - (1-x^2)^{n}\right) &\leqslant \left(1+\frac{x^2}{1-2x}\right)^n - 1
\end{align}$$
Now, for $0 \leqslant x < \frac12$, we have $x^2 \leqslant \frac{x^2}{1-2x}$, and by Bernoulli's inequality, we have
$$(1+x^2)^n - 1 \geqslant n\cdot x^2.$$
On the other hand, also by Bernoulli's inequality, we have
$$1 - (1-x^2)^n \leqslant n\cdot x^2,$$
and since that is multiplied with $(1-x^2)^n \leqslant 1$, we conclude
$$\begin{align}(1-x^2)^n\left(1 - (1-x^2)^{n}\right) &\leqslant (1-x^2)^n\cdot n\cdot x^2\\
&\leqslant n\cdot x^2\\
&\leqslant (1 + x^2)^n - 1\\
&\leqslant \left(1+\frac{x^2}{1-2x}\right)^n - 1.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof on an equilateral triangle with a cevian extended to its circumcircle Consider the following figure with equilateral triangle $ABC$ and a cevian $AQ$ extended to $P$ on its circumcircle.
We are required to prove that:
$\frac{1}{PB} + \frac{1}{PC} = \frac{1}{PQ}$
Let $\angle PAC = \alpha$ and let length of $AB = s$
By the Law of Sines,
$\frac{PC}{\sin\alpha} = \frac{s}{\sin\angle CPA} = \frac{s}{\sin60^{\circ}} \implies PC = \frac{s\sin\alpha}{\sin60^{\circ}}$
Similarly, $PB = \frac{s\sin(60^{\circ} - \alpha)}{\sin60^{\circ}}$
$\frac{1}{PB} + \frac{1}{PC} = \frac{\sin60}{s}\left(\frac{\sin(60 - \alpha) + \sin\alpha}{\sin\alpha\sin(60 - \alpha)}\right)$
It remains to be proven that:
$PQ = \frac{\sin60\sin(60 - \alpha)\sin\alpha}{\sin60\sin(60 - \alpha) + \sin60\sin\alpha}$
I'm utterly lost from here.
| Observe that the cuadrilateral $ABPC$ is cyclic, so by similarity we have $\triangle PQB \sim \triangle CQA$ y $\triangle PQC \sim \triangle BQA$. Therefore:
\begin{equation*}
\dfrac{PQ}{PB} = \dfrac{QC}{AC} \qquad \text{y} \qquad \dfrac{PQ}{PC} = \dfrac{BQ}{AB}.
\end{equation*}
Thus, taking into account that $AB = AC = BC$, we have:
\begin{equation*}
\dfrac{PQ}{PB} + \dfrac{PQ}{PC} = \dfrac{QC}{AC} + \dfrac{BQ}{AB} = \dfrac{QC + BQ}{AB} = \dfrac{BC}{AB} = 1.
\end{equation*}
Dividing by $PQ$, we conclude:
\begin{equation*}
\dfrac{1}{PB} + \dfrac{1}{PC} = \dfrac{1}{PQ}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How prove this $\tan{x}>\frac{3x}{2+\sqrt{1-x^2}}$ let $0<x<1$,prove that
$$\tan{x}>\dfrac{3x}{2+\sqrt{1-x^2}}$$
This problem have nice solution?
my idea:
let $$f(x)=\tan{x}-\dfrac{3x}{2+\sqrt{1-x^2}}=\tan{x}-3x\dfrac{2-\sqrt{1-x^2}}{3+x^2}$$
and other idea:
let $x=\cos{t}$,then
$$\tan{(\cos{x})}>\dfrac{3\cos{t}}{2+\sin{t}}$$
other (2) idea:
$$\tan{x}>x+\dfrac{1}{3}x^3$$
| why dont you go for the calculus approach.
assume $ f(x) = tanx-3x/2+\sqrt{1+x^2} $.
differentiate w.r.t to x
we get $ f'(x) = sec^{2}x - 3/(2+\sqrt{1+x^2} + 3(x^2)/(2+\sqrt{1+x^2})^2 * 1/\sqrt{1+x^2} $ ,
under the given interval , $ f'(x) > 0 $ , implying it is an increasing function.
we know that $f'(x) > 0 <=> f(x) > 0 .$.
thus $ tanx > 3x/1+\sqrt{+x^2} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/470172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find inverse for the closed-form expression of linear recurrence relation I am trying to find an inverse of the following formula:
$$
a_n=\frac{2+\sqrt{6}}{4}(1+\sqrt{6})^n+\frac{2-\sqrt{6}}{4}(1-\sqrt{6})^n
$$
This formula is a closed-form expression of a linear recurrence relation and I'd like to find a closed-form expression for the inverse in order to test whether and where (index) a given number occurs in the linear recurrence relation.
I do not know whether this is at all possible (if not, why not?), but a similar problem for the linear recurrence relation describing the Fibonacci numbers has been solved:
$$
\begin{align}
F_n&=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^n-\frac{1}{\sqrt{5}}(\frac{1-\sqrt{5}}{2})^n\\
&=\frac{\varphi^n-(-\varphi)^{-n}}{\varphi+\varphi^{-1}} = \frac{\varphi^n- (-\varphi)^{-n}}{\sqrt{5}}\\
\text{where } \varphi &= \frac{1 + \sqrt{5}}{2}\\
\text{and }n&=\log_\varphi(\frac{F_n\sqrt{5}+\sqrt{5F_n^2\pm4}}{2})
\end{align}
$$
If it is possible to find an inverse for the mentioned formula, I'm interested in a more general method for approaching the problem, in particular how to manipulate the form:
$$
c_0(a+\sqrt{b})^n+c_1(a-\sqrt{b})^n
$$
Is this form a special form (is it named and/or does it have special properties)? I tried searching for it using terms like "sum of conjugate binomial powers" but failed finding anything that describes this form.
| EDIT : Dah, doesn't work as the first commentor pointed out. I guess I had to write it all out to see where it failed! I leave the answer there for the curious.
The quadratic approach works, it just requires some good thinking (at least in this case it works!).
So you begin with
$$
a_n = \frac{2+\sqrt 6}4 \left( 1 + \sqrt 6 \right)^n + \frac{2-\sqrt 6}4 \left( 1 - \sqrt 6 \right)^n.
$$
I decided to write $\varphi_6 \overset{def}= 1 + \sqrt 6$ and $\overline{\varphi}_6 \overset{def}= 1 - \sqrt 6$, so that the above equation can be re-written as
$$
4 a_n = (1 + \varphi_6) \varphi_6^n + (1+\overline{\varphi}_6)\overline{\varphi}_6^n.
$$
Note that $\varphi_6 \overline{\varphi}_6 = -5$, hence
$$
4 a_n \varphi_6^n = (1+\varphi_6)\varphi_6^{2n} + (1+\overline{\varphi}_6)(-5)^n.
$$
The quadratic equation
$$
(1+\varphi_6) y^2 - 4a_n y + (-5)^n(1+\overline{\varphi}_6) = 0
$$
has the solution $\varphi_6^n$, hence by noticing that
$$
(1+\varphi_6)(1+\overline{\varphi}_6) = (2+\sqrt 6)(2-\sqrt 6) = -2,
$$
we obtain
$$
\varphi_6^n = \frac{4a_n \pm \sqrt{ 16a_n^2 -4(1+\varphi_6)(1+\overline{\varphi}_6)(-5)^n } }{2(1 + \varphi_6)} = \frac{4a_n \pm \sqrt{ 16a_n^2 + 8(-5)^n } }{2(1 + \varphi_6)} ;
$$
but we need to determine this sign, as we only know $\varphi_6^n$ belongs to one of the two possible signs. If $n$ is even, then
$$
4a_n = \sqrt{16a_n^2} < \sqrt{16 a_n^2 + 8 (-5)^n},
$$
hence since $\varphi_6 > 0$, we must take the $+$ sign. If $n$ is odd, denote by $\psi_n$ the other root of the quadratic that $\varphi_6^n$ solves. Then for a quadratic with given coefficients $a,b,c$ and roots $r_1, r_2$, we see that
$$
ay^2 + by + c = a(y^2 + \frac ba y + \frac ca) = a (y - r_1)(y - r_2) = a(y^2 - (r_1+r_2)y + r_1 r_2) \quad \Longrightarrow \quad r_1 r_2 = \frac ca.
$$
Using this identity with the approximations $\varphi_6 \approx 3.45$ and $\frac 5{\varphi_6} \approx 1.45$, we can see that
$$
\varphi_6^n \psi_n = \frac{1+\overline{\varphi}_6}{1+\varphi_6} (-5)^n = 5^n \frac{-(1+\overline{\varphi}_6)}{1+\varphi_6} \approx 5^n \cdot 0.1 \quad \Longrightarrow \quad \psi_n \approx \frac{(1.45)^n}{10} < (3.45)^n = \varphi_6^n,
$$
so that $\varphi_6^n$ corresponds again to the $+$ sign since the two roots are $\varphi_6^n$ and $\psi_n$ but $\varphi_6^n > \psi_n$.
(Note that the discriminant of the quadratic for $\varphi_6^n$ must be positive since $\varphi_6$ is a real number ; in other words, $a_n^2 > \frac{5^n}2$ for $n$ odd, we don't have to make the check.)
So we have
$$
\varphi_6^n = \frac{4a_n + \sqrt{ 16a_n^2 + 8(-5)^n } }{2(1 + \varphi_6)},
$$
hence
$$
n = \frac{ \log \left( \frac{4a_n + \overset{\, \, \, \quad \qquad \displaystyle{\downarrow}}{\sqrt{ 16a_n^2 + 8(-5)^n } }}{2(1 + \varphi_6)} \right) }{\log(\varphi_6)}.
$$
(Arrow points to why it failed.)
Hope that helps,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/470714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Quadratic equation - $\alpha$ and $\beta$ Roots If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 8x - 5 = 0$, find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
My working out so far:
I know that $\alpha + \beta = -8$ and $\alpha \beta = -5$ (from the roots) and thenIi go on to work out that
$\alpha= -8-\beta$ and $\beta= -8-\alpha$, then I substitute into what the question asks me.
$\frac{-8-\beta}{-8-\alpha}$ and $\frac{-8-\alpha}{-8-\beta}$ however I do not know how to proceed further. I might be doing this completely wrong and my apologies for that.
Another solution came to me that if $\alpha$ and $\beta$ are roots of the other unknown equation. I can somehow manipulate that to find the answer. But I don't think that will work. All help is appreciated thank you.
| If $α$ and $β$ are the roots of the quadratic equation $ax^2+bx+c=0$ ,that is, $x^2+\frac{b}{a}x+\frac{c}{a}=0$ ($a$ is not equal to zero), then the equation can be written as $x^2-(α+β)x+αβ=0$.
The equation whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ is: $x^2-(\frac{α}{β}+\frac{β}{α})x+(\frac{α}{β})(\frac{β}{α})=0$.
$\implies x^2-(\frac{α^2+b^2}{αβ})x+1=0$
$\implies x^2-(\frac{(α+b)^2-2αb}{αβ})x+1=0$
$α$ and $β$ are the roots of the equation $x² + 8x - 5 = 0$
$α+β=-8$
$αβ=-5$
Therefore the required equation is $x^2-(\frac{(-8)^2-2(-5)}{-5})x+1=0$
$\implies x^2-(\frac{74}{-5})x+1=0$
$\implies x^2+\frac{74}{5}x+1=0$
$\implies 5x^2+74x+5=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
} |
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$ This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $6$(d)
Prove that if $x^n = y^n$ and $n$ is even, then $x = y$ or $x = -y$.
Proof. Suppose $x^n = y^n$ and $n$ is even. We consider the following cases.
Case 1. $x \geq 0$ and $y \geq 0$. Now suppose, for the sake of contradiction, $x \neq y$. Then $x > y$ or $x < y$. If $x > y$, then $0 \leq y < x$, so $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < y$ leads to a contradiction. Thus $x = y$.
Case 2. $x \geq 0$ and $y < 0$. Now suppose, for the sake of contradiction, $x \neq -y$. Then $x > -y$ or $x < -y$. If $x > -y$, then $0 < -y < x$, so $(-y)^n < x^n$. Since $n$ is even, it follows that $y^n < x^n$. But this contradicts our assumption that $x^n = y^n$. Similarly, $x < -y$ leads to a contradiction. Thus $x = -y$.
Case 3. $x < 0$ and $y \geq 0$. Applying case 2 with $x$ and $y$ interchanged, we get $y = -x$. Therefore $x = -y$.
Case 4. $x <0$ and $y < 0$. Then $-x > 0$ and $-y > 0$. Applying case 1 to $-x$ and $-y$, we get $-x = -y$. Therefore $x = y$.
Is my proof correct? Is there a shorter way to prove this?
| The argument for Case 1 handles all four cases if you note that $x^n = |x|^n$ (etc.) when $n$ is even. (Just answerifying an earlier comment.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
General solution for ordinary differential equation $y'' + 9y = 18x + 36 \sin 3x$ Find a (real) general solution for
$$y'' + 9y = 18x + 36 \sin 3x$$
I need some help in finding the solution, have been struggling for long. My answer does not match with the final answer which is:
$$y_h = A\cos 3x + B\sin 3x $$
$$y_p = 2x - 6x\cos 3x$$
| Given:
$$\tag 1 y'' + 9y = 18x + 36 \sin 3x$$
Homogeneous
$$m^2 +9 = 0 \rightarrow m_1 = 3i, m_2 = -3i$$
So,
$$y_h = A \cos 3x + B \sin 3x$$
Particular
Since we have a $\sin 3x$ term in the homogeneous solution, we will choose a particular solution:
*
*$y_p = a x + b x \cos 3x + c x \sin 3x$
*$y'_p = a + b \cos 3x + 3 c \cos 3x + c \sin 3x -3b \sin 3x$
*$y''_p = 6 c \cos 3x - 9b x \cos 3x-6 b \sin 3x - 9c x \sin 3x$
Substituting into $(1)$ yields:
$\tag 2 y''_p + 9y_p = 9~a x - 6 b ~\sin 3x + 6 c ~\cos 3x = 18 x + 36 \sin 3x$
Equating terms yields:
$$a = 2, b = -6, c = 0$$
Our final solution:
$$y(x)= y_h + y_p = A \cos 3x + B \sin 3x + 2x -6 x \cos 3x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $a^2b$ . If a variable tangent to the curve $x^2y = c^3$ makes intercepts $a$ and $b$ on the $x$ and $y$ axis respectively then find the value of $a^2b$.
| The tangent in the point $(x_0,\frac{c^3}{x_0^2})$ is $y'=-\frac{2c^3}{x^3}$ so the tangent lines crossing each point of the function are:
$$
y-\frac{c^3}{x_0^2} = -\frac{2c^3}{x_0^3}(x-x_0)
$$
$a$ is where tangent line crosses $x$-axis i.e. $y=0$ , and $b$ is where $x=0$.
$$
\left.
\begin{array}{cc}
y=0 \Rightarrow a &= \frac32 x_0\\
x=0 \Rightarrow b &= 3\frac{c^3}{x_0^2}
\end{array}
\right\}
\Rightarrow a^2b = \frac{27}4c^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The Diophantine equation $x^2 + 2 = y^3$ How to solve the Diophantine equation $x^2 + 2 = y^3$ with $x,y>0$ ?
($x,y$ are integers.)
| Lemma.
Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $n=t^2+2u^2$ divides $bt-au$.
Proof.
Assume the theorem is false, and let $m$ be a minimal counterexample. Evidently $m > 1$ since the theorem is trivially true for $m=1$.
Note that $b$ is coprime to $m$. Let $A$ be an integer such that $Ab \equiv a\!\pmod{m}$, chosen so that $\tfrac{-m}{2} < A \le \tfrac{m}{2}$. Then $A^2+2 = lm$ for some positive integer $l < m$. Clearly $l$ cannot be a smaller counterexample than $m$, and so there exist coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$.
Let $t = \tfrac{ar+2bs}{m}$ and $u=\tfrac{br-as}{m}$. Direct calculation confirms the equations for $a$, $b$, and $n$. From $n=t^2+2u^2$, we deduce that $t$ is an integer because $u$ is an integer, and $t$ and $u$ are coprime because $\gcd(t,u)$ divides both $a$ and $b$. Finally, note that $n$ divides $bt-au=sn$.
Hence $m$ is not a counterexample, contradicting the original assumption. $\blacksquare$
Corollary.
Let $a$ and $b$ be coprime integers with $m$ an integer such that $m^3=a^2+2b^2$. Then there are coprime integers $r$ and $s$ such that $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$.
Proof.
Evidently $m$ is odd since $a^2+2b^2$ is at most singly even. And $a$ and $m$ must be coprime. Using the theorem, we have $m=r^2+2s^2$ and $m^2=t^2+2u^2$. Then $m$ divides $a(ur-ts)=t(br-as)-r(bt-au)$, and therefore $m \mid (ur-ts)$. The lemma can then be reapplied with $a$ and $b$ replaced by $t$ and $u$. Repeating the process, we eventually obtain integers $p$ and $q$ such that $p^2+2q^2=1$. The only solution is $q=0$ and $p=\pm1$. Ascending the path back to $a$ and $b$ (reversing signs along the way, if necessary) yields $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$, as claimed. $\blacksquare$
Theorem.
The Diophantine equation $X^3 = Y^2+2$ has only one integer solution, namely $(x,y) = (3, \pm 5)$.
Proof.
Evidently $y$ and $2$ are coprime. By the corollary, we must have $b=1=s(3r^2-2s^2)$ for integers $r$ and $s$. The only solutions are $(r,s)=(\pm 1,1)$. Hence $a=y=r(r^2-6s^2)=\pm 5$, so $(x,y)=(3,\pm 5)$. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Trigonometry : verify that $\cos \frac{A}{2} +\cos 2A = 0$ for $A=\frac{2\pi}{5}$ Question in trigonometry: verify that $\cos\frac{A}{2} + \cos 2A = 0$.
Let $A = \frac{2\pi}{5}$.
*
*Verify that $\cos \frac{A}{2} + \cos 2A = 0$.
*Show that $x=\cos A$ satisifies the equation $x=2(4x^4-4x^2+1)-1$.
*Factor $8x^4-8x^2-x+1$ over $\mathbb Z$, and deduce that $\cos A$ is a zero of a quadratic polynomial over $\mathbb Z$.
*Determine $\cos A.$
My solution is as follows :
We know $\cos(A/2)=\sqrt{(1+\cos A)/2}$
and $\cos2A=2\cos^2A-1$.
$\cos2A=2\cos^2A-1$
It is sufficient if we prove that $\cos2A=-\cos(A/2)$ for the first question.
$2\cos^2A-1=\pm\sqrt{(1+\cos A)/2}$
squaring on both sides and simplifying:
$8\cos4A+2-8\cos^2A=1+\cos A$
subsitituting $\cos A$ with $x$ :
$8x^4- 8x^2+2-1 = x$
$2(4x^4-4x^2+1)-1=x$
I believe that if I do the same thing backwards, I will get the answer for question (2)
I have factorized the equation and ended up with this:
$(x-1)[8x^3+8x-1]$
Now, the third question asks me to deduce that cosA is a zero of quadratic polynomial, but I have only linear and cubic factors, how do I proceed?
Regarding the first question, I’m clueless, I have simplified it as much as possible, but its not going anywhere.
| $1$.
$\cos \frac A2=\cos \frac \pi5$ and
as $\cos(\pi-x)=-\cos x,$
$\cos2A=\cos\frac{4\pi}5=\cos\left(\pi-\frac\pi5\right)=-\cos\frac\pi5 $
$2$.
As $5A=2\pi,A=2\pi-4A$
So using $\cos2x=2\cos^2x-1$ and $\cos(2\pi-x)=\cos x$,
$\cos A=\cos(2\pi-4A)=\cos 4A=2(\cos2A)^2-1=2(2\cos^2A-1)^2-1=8\cos^4A-8\cos^2A+1$
$3$.
But $\cos4A=\cos A$
$\implies 4A=2n\pi\pm A$ where $n$ is any integer
Taking $'+'$ sign, $A=\frac{2n\pi}5$ where $n=0,1,2,3,4$
So, the roots are $\cos0=1,$
$\cos\frac{2\pi}5, \cos\frac{4\pi}5=\cos\left(\pi-\frac{\pi}5\right)=-\cos\frac\pi5<0,$
$\cos\frac{6\pi}5=\cos\left(\pi+\frac{\pi}5\right)=-\cos\frac\pi5,$
$\cos\frac{8\pi}5=\cos\left(2\pi-\frac{8\pi}5\right)=\cos\frac{2\pi}5$
Taking $'-'$ sign, $A=\frac{2n\pi}3$ where $n=0,1,2$
So, the roots are $\cos0=1,$
$\cos\frac{2\pi}3=\cos\left(\pi-\frac{\pi}3\right)=-\cos\frac\pi3=-\frac12,$
$\cos\frac{4\pi}3=\cos\left(\pi+\frac{\pi}3\right)=-\cos\frac\pi3=-\frac12,$
So, $8x^4-8x^2-x+1=(x-1)\{x-(-\frac12)\}\left(x-\cos\frac{2\pi}5\right)\{x-(-\cos\frac{\pi}5)\}$
So, $\cos\frac{2\pi}5>0,-\cos\frac{\pi}5<0$ are the roots of the qaudratic eqaution $$\frac{8x^4-8x^2-x+1}{(x-1)(x+\frac12)}=0\ \ \ \ (1)$$
$4$.
As $A=\frac{2\pi}5,0<\frac{2\pi}5<\frac\pi2,$ the value of $\cos\frac{2\pi}5$ will be the positive root of $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/475807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computig the series $\sum\limits_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)$ So I have this problem for midterm reviews:
$$\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right)=\text{ ?}$$
I know that you can find the series form of a natural log, as shown here:
$$\ln\left(1-\frac{1}{n^2}\right)=-\sum_{k=2}^\infty \left(\frac{1}{n^{4k}}\right)\left(\frac{1}{2k}\right) $$
But the above doesn't seem to help very much since it results in two summation notations mushed together. Is there a somewhat nontedious way to go about this? Thanks! All help appreciated.
| $$ \sum_{n=2}^{\infty} \ln \left( 1 - \frac{1}{n^2} \right) = \ln \left( \prod_{n=2}^{\infty} \left(1 - \frac{1}{n^2} \right) \right) $$
$$ = \ln \left( \prod_{n=2}^{\infty} \frac{n-1}{n}\frac{n+1}{n} \right) $$
$$ \ln \left( \frac{1}{2} \times \frac{3}{2} \times \frac{2}{3} \times \frac{4}{3} \times \frac{3}{4} \times ... \right) $$
$$ = \ln \left( \frac{1}{2} \right) = -\ln 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| $A+B=180-C$
$\tan(A+B)=\tan(180-C)$
$[\tan(A)+\tan(B)]/[1-\tan(A)\tan(B)]=-\tan(C)$
$\tan(A)+\tan(B)=-\tan(C)+\tan(A)\tan(B)\tan(C)$
$\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 4
} |
Given that $z=2-i$ and $z^2=3-4i$ find the roots of the equation $(z+i)^2=3-4i$ Given that $z=2-i$ and $z^2=3-4i$ find the roots of the equation $(z+i)^2=3-4i$
How do you use the given properties to find the roots? I can only obtain them the long way by working through
$$\begin{align}
&(z+i)(z+i)&=3-4i\\
&z^2+2zi+i^2&=3-4i\\
&z^2+2zi-1-3+4i&=0\\
&z^2+2zi+4i-4&=0\\
&z^2+2z-2z+2iz+4i-4&=0\\
&(z-(2-2i))(z+2)&=0\\
\end{align}$$
$$z=2-2i, z=-2$$
| First, you can find one root: $z+i=2-i$, so $z_1=2-2i$.
You can also get from the coefficients that $z_1+z_2=-2i$, then $z_2=-2$.
You can refer to Vieta's formulas.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of $\Gamma(n+a) / \Gamma(n+b)$ If $a$ and $b$ are positive real numbers, such that $b > a + 1$, can we find the sum $$\sum_{n=0}^{\infty} \frac{\Gamma(n+a)}{\Gamma(n+b)}?$$
For example I have found that $$\sum_{n=0}^{\infty} \frac{\Gamma(n+3/2)}{\Gamma(n+3)} = \sqrt{\pi} = \Gamma(1/2)$$ and $$\sum_{n=0}^{\infty} \frac{\Gamma(n+4/3)}{\Gamma(n+4)} = \frac{3}{10}\Gamma(4/3)$$ but no general rule.
| in general
$$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}\Gamma{(b+k)}}{k!\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(b)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$$
note that
$$I=\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-x)^{-a}dx=B(b,c-a-b)=\dfrac{\Gamma{(b)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}}$$
and we have
$$(1-x)^{-a}=\sum_{k=0}^{\infty}\binom{a-1+k}{k}x^k$$
then
\begin{align*}I&=\sum_{k=0}^{\infty}\binom{a-1+k}{k}\int_{0}^{1}x^{b+k-1}(1-x)^{c-b-1}dx=\sum_{k=0}^{\infty}\dfrac{(a-1+k)!}{(a-1)!k!}B(b+k,c-b)\\
&=\sum_{k=0}^{\infty}
\dfrac{\Gamma{(a+k)}}{\Gamma{(a)}k!}\cdot\dfrac{\Gamma{(b+k)}\Gamma{(c-b)}}{\Gamma{(c+k)}}
\end{align*}
so
$$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}\Gamma{(b+k)}}{k!\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(b)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$$
so let $b=1$
then
$$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}}{\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(c-a-1)}}{\Gamma{(c-1)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Maximize the expression $\frac{1}{x}+\frac{1}{y}$, given that $\frac{1}{x^n}+\frac{1}{y^n}<1$, for natural numbers $x,y,n$ $x$, $y$ are natural numbers such that $\dfrac{1}{x^n}+\dfrac{1}{y^n}<1$. Maximise $\dfrac{1}{x}+\dfrac{1}{y}$.
Here $n$ is a natural number also.
| You can try using Lagrangian multiplies to find stationary points (saddle points, maxima, minima). As Macavity suggested we gonna split the problem into 2 sub-problems.
Case 1: $n \geq 2$
The function is: $f(x,y) = \frac{1}{x^n} + \frac{1}{y^n} = x^{-n} + y^{-n}$. And we have two constraints those are: $x,y \geq 2$, because otherwise the function will violate the condition $\frac{1}{x^n} + \frac{1}{y^n} < 1$
$$f(x,y) = x^{-n} + y^{-n}$$
$$g(x) = x \geq 2$$
$$h(y) = y \geq 2$$
$$F(x,y,\lambda,\lambda_1) = x^{-n} + y^{-n} + \lambda(x-2) + \lambda_1(y-2)$$
Now we take partial derivatives:
$$F_x = -nx^{-n-1} + \lambda = 0$$
$$F_y = -ny^{-n-1} + \lambda_1 = 0$$
Beacuse the two last terms has to be equal to zero we have 4 cases.
Cases $\lambda = \lambda_1 = 0$ and $\lambda_1 = 0,x = 2$ and $\lambda = 0, y = 2$ wouldn't work, beacuse that would imply that one of the variables is 0, which isn't possible. That leaves us with only one option: $x = y = 2$
Now checking into the initial function for $(x,y) = (2,2)$ and for a random n, we get that this point is maxima of the function.
So the maximum of the function occurs at $(2,2)$ and it's $f(2,2) = \frac{1}{2^{n-1}}$
Case 2: $n = 1$
The function now looks like this:
$f(x,y) = x^{-1} + y^{-1}$ and we have constraints the same two constraints and an additional one, which is: $\frac{x+y}{xy} < 1$
$$f(x,y) = x^{-1} + y^{-1}$$
$$g(x) = x \geq 2$$
$$h(y) = y \geq 2$$
$$j(x,y) = x+y \leq xy - 1$$
$$F(x,y,\lambda,\lambda_1, \lambda_2) = x^{-1} + y^{-1} + \lambda(x-2) + \lambda_1(y-2) + \lambda_2(x+y+1-xy)$$
Now we take partial derivatives:
$$F_x = -x^{-2} + \lambda + \lambda_2 - \lambda_2y = 0$$
$$F_y = -y^{-2} + \lambda_1 + \lambda_2 - \lambda_2x = 0$$
Because the last 3 terms have to be equal to 0, we have 8 distinct cases:
Sub-case 1: $\lambda = \lambda_1 = \lambda_2 = 0$
This method doesn't give a solution, because it implies $x=y=0$, which is impossible.
Sub-case 2: $\lambda = \lambda_1 = x+y-xy+1 = 0$
From $x+y-xy+1 = 0$ we obtain:
$$y+1 = xy-x \implies x(y-1) = y+1 \implies x = \frac{y+1}{y-1}$$
Beacuse x is an integer, this implies $y=2$ and $x=3$.
Also we have:
$$x+1 = xy-y \implies y(x-1) = x+1 \implies y = \frac{x+1}{x-1}$$
Beacuse y is an integer, this implies $x=2$ and $y=3$.
So in this case we obtained two solutions: $$(x,y) = (2,3), (3,2)$$
Sub-case 3: $\lambda = y-2 = \lambda_2 = 0$
This implies $(0,2)$ as solution which is imposible.
Sub-case 4: $\lambda = y-2 = x+y-xy+1 = 0$
This implies a solution that we've already obtained, which is $(3,2)$.
Sub-case 5: $x-2 = \lambda_1 = \lambda_2 = 0$
This implies $(2,0)$ as solution which is imposible.
Sub-case 6: $x-2 = \lambda_1 = x+y-xy+1 = 0$
This implies a solution that we've already obtained, which is $(2,3)$.
Sub-case 7: $x-2 = y-2 = \lambda_2 = 0$
This implies a solution $(2,2)$, which violate the third constrain and is impossible.
Sub-case 8: $x-2 = y-2 = x+y-xy+1 = 0$
This is a contradiction, because it's impossible these 3 constraints to be equal to 0.
Because we exhausted all the sub-cases, we check the two solution we obtained and we get that they are global maxima of the function.
Conclusion
If $n \geq 2$ then the maximum occurs at point $(2,2)$, otherwise it occurs at points $(3,2)$ and $(2,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute the integral $\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$
Compute the indefinite integral
$$
\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}
$$
My Attempt:
$$
\int\frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}=\int\frac{1}{(x^2-x+1)^{3/2}.\sqrt{\dfrac{x^2+x+1}{x^2-x+1}}}\,dx
$$
Now define $t$ such that $t^2=\dfrac{x^2+x+1}{x^2-x+1}$ to get
$$
\begin{align}
2t\,dt &= \frac{(x^2-x+1)(2x+1)-(x^2+x+1)\cdot (2x-1)}{(x^2-x+1)^2}\,dx\\
2tdt &= \frac{-4x^2+2x+2}{(x^2-x+1)^2}dx
\end{align}
$$
I don't know how to proceed from here.
| Instead of the Euler substitution we can use the following technique adapted
from a general procedure explained in 2.252 of Table of Integrals,
Series, and Products by I.S. Gradshteyn and I.M. Ryzhik, 7th. ed.
The substitution
\begin{equation*}
x=\frac{1-t}{t+1},\quad dx=-\frac{2}{\left( t+1\right) ^{2}}dt,\quad t=-%
\frac{x-1}{x+1}
\end{equation*}
reduces the given integral
\begin{equation*}
I=\int \frac{dx}{(x^{2}-x+1)\sqrt{x^{2}+x+1}}
\end{equation*}
to the sum of the two integrals
\begin{equation*}
I=-2\int \frac{t}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}dt-2\int \frac{1}{
\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}dt.
\end{equation*}
The first integral can be evaluated by the substitution
\begin{equation*}
u=\sqrt{t^{2}+3},
\end{equation*}
while the second one is integrable by the substitution
\begin{equation*}
v=\frac{t}{\sqrt{t^{2}+3}}.
\end{equation*}
Both substitutions transform the integrands into simple rational fractions as follows
\begin{eqnarray*}
I_{1} &=&-2\int \frac{t}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}dt=-2\int
\frac{1}{-8+3u^{2}}\,du,\qquad u=\sqrt{t^{2}+3} \\
&=&\frac{\sqrt{6}}{6}\operatorname{arctanh}(\frac{\sqrt{6}}{4}u) \\
I_{2} &=&-2\int \frac{1}{\left( 3t^{2}+1\right) \sqrt{t^{2}+3}}dt=-2\int
\frac{1}{8v^{2}+1}\,dv,\qquad v=\frac{t}{\sqrt{t^{2}+3}} \\
&=&-\frac{\sqrt{2}}{2}\arctan (2\sqrt{2}v) \\
I &=&I_{1}+I_{2}=\frac{\sqrt{6}}{6}\operatorname{arctanh}\frac{\sqrt{6}\sqrt{t^{2}+3}
}{4}-\frac{\sqrt{2}}{2}\arctan \frac{2\sqrt{2}t}{\sqrt{t^{2}+3}}+C.
\end{eqnarray*}
We finally get
\begin{equation*}
I=\frac{\sqrt{6}}{6}\operatorname{arctanh}\frac{\sqrt{6}\sqrt{x^{2}+x+1}}{2\left(
x+1\right) }+\frac{\sqrt{2}}{2}\arctan \frac{\sqrt{2}\left( x-1\right) }{
\sqrt{x^{2}+x+1}}+C.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Ancient Babylonian problem: solve the system $x + y = 50$, $x^2 + y^2 + (x - y)^2 = 1400$ $x + y = 50, x^2 + y^2 + (x - y)^2 = 1400$. [Hint: Subtract the square of the first equation from twice the second equation to get a quadratic in $x - y$.] I have gotten it reduced to $x^2 + y^2 -3xy = 550$ but am not sure how to get it in terms of $x - y$ to solve it.
| The hint is saying that you should take two times $x^2 + y^2 + (x - y)^2$ (the second equation) and subtract the square of the first, $(x + y)^2 = x^2 + 2xy + y^2$.
So the result is
$$
2(x^2 + y^2 + (x-y)^2) - (x^2 + 2xy + y^2)
$$
You should be able to simplify this into something simple in terms of just $(x-y)$. But then, we also know that the above is equal to
$$
2(1400) - (50)^2
$$
So from here you can solve for $x-y$. Then once you know $x - y$, since you already know $x + y$ you can find out what $x$ and $y$ are individually.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding point of contact and value of b If a line $y = 2x - b$ connects with a curve $y = 3x^2 + 2$, how do I approach finding the value of b? To find the coordinates, if I'm not mistaken, I need to set the derivative of $y = 3x^2 + 2$ equal to the equation of the line. Is this correct? Many thanks in advance!
| By differentiation, we have
$$\dfrac{dy}{dx} = 6x$$
Since we know that the slope of a line is $2$,
$$\dfrac{dy}{dx} = 2 = 6x$$
$$x = \frac{1}{3}$$
Therefore,
$$y = 3(\frac{1}{3})^2 + 2$$
$$y = \frac{7}{3}$$
Thus,
$$y - \frac{7}{3} = 2(x - \frac{1}{3})$$
$$y = 2x + \frac{5}{3}$$
So $b = \frac{5}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Irreducible but not prime
In $\mathbb Z[\sqrt{5}]$, $2$ and $1+\sqrt{5}$ are irreducible but not prime.
To show irreducible
I tried that there exists $\alpha$ and $\beta$ such that
$$2=\alpha\cdot \beta $$
$$N(2)=N(\alpha)\cdot N(\beta)$$
$$4=N(\alpha)\cdot N(\beta)$$
then there are 3 possibilities (namely $1.4 , 4.1 , 2.2$)
Now,I have to show that $N(\alpha)=2$ is not possible
that is $|a^2-5b^2|=2$
$a^2-5b^2=\pm 2$
How can I show that this is not possible? and how can i show that $2$ is not prime in $Z[\sqrt{5}]$?
| To show that $2$ is not prime, observe that $2$ divides $(\sqrt{5}+1)^2$ but $2$ does not divide $\sqrt{5}+1$. Or else we can use $(\sqrt{5}-1)(\sqrt{5}+1)$.
Similarly, we can show that $\sqrt{5}+1$ is not prime. For $(\sqrt{5}+1)(\sqrt{5}-1)=4$. So $\sqrt{5}+1$ divides $(20(2)$, but $\sqrt{5}+1$ does not divide $2$.
To show that $a^2-5b^2$ cannot be equal to $\pm 2$, observe that any square is congruent to $0$, $1$, or $-1$ modulo $5$. Since $a^2-5b^2\equiv a^2\pmod{5}$, we cannot have $a^2-5b^2$ equal to anything congruent to $\pm 2\pmod{5}$. In particular, it cannot be equal to $2$ or $-2$.
Remark: It turns out that this flaw can be fixed. If we consider the numbers of the form $a+b\sqrt{5}$, where $a$ and $b$ are integers, together with numbers of the form $\frac{a+b\sqrt{5}}{2}$, where $a$ and $b$ are odd integers, we get a structure in which every non-unit irreducible is prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/483620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$, given $abc=1$ Let a, b, c be positive numbers such that $abc=1$. Prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$
The usual methods do not seem to work, including a substitution $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and trying to apply Muirhead's inequality.
| We have $(ab+bc+ca)^2 \geq 3 (a+b+c)abc $, which is equivalent to $ (ab-bc)^2 + (bc-ca)^2 + (ca-ab)^2 \geq 0$. Hence,
$$ 1 + \frac{3}{a+b+c} \geq 2 \sqrt{ \frac{3}{a+b+c} } \geq \frac{6}{ab+bc+ca}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/484263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric equation in two ways gives different answer I have been given the equation $\sin^2 x+ \cos x +1 = 0.$ I tried to solve it in two ways. First, $1 - \cos^2(x) + \cos(x) + 1 = 0,$ $\cos^2(x) - \cos(x) - 2 = 0,$ $\cos x=2$ or $\cos x=-1$ thus $x=\pi+n2\pi$ for all $n\in \mathbb{Z}.$
Second approach. Let $\tan \frac{x}{2}=t.$ Then $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}.$ Thus the equation is
$$\left (\frac{2t}{1+t^2}\right )^2+\frac{1-t^2}{1+t^2}+1=0.$$ Therefore
$$4t^2+(1-t^2)(1+t^2)+(1+t^2)^2=0.$$ But $$4t^2+(1-t^2)(1+t^2)+(1+t^2)^2=4t^2+1-t^4+1+2t^2+t^4=6t^2+2>0.$$
and there is no solutions.
What is my mistake in the second reasoning?
| You are defining $t = \tan\cfrac{x}{2} = \cfrac{\sin \cfrac{x}2}{\cos \cfrac{x}2}$, which disqualifies any root for which $\cos \cfrac{x}2 = 0$.
What happens when $x = \pi + 2\pi n$?
The second method is perfectly fine, but you have to take care of undefined values of $t$ separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derive power series for $\frac{1}{1-x^2}$ from $\frac{1}{1-x}$? The series $\sum_{k=0}^{\infty} x^k$ is known and equals $\frac{1}{1-x}$ for $|x|<1$.
Can I use this fact and derive the power series for $\frac{1}{1-x^2}$ from it, eg using $\frac{1}{1-x^2}=\frac{1}{1-x}\frac{1}{1+x}, x\neq\pm1$. Or must I derive it from the definition $\sum_{k=0}^{\infty} \frac{f^k(0)}{k!}x^k$
Thanks in advance!
Alexander
| First way:
$$\frac1{1-x^2}=\frac1{1-(x^2)}=1+(x^2)+(x^2)^2+\ldots=1+x^2+x^4+\ldots$$
Second way: using that
$$\frac1{1+x}=\frac1{1-(-x)}=1+(-x)+(-x)^2+(-x)^3+\ldots=1-x+x^2-x^3+\ldots$$
we decompose by partial fractions
$$\frac1{1-x^2}=\frac12\left(\frac1{1-x}+\frac1{1+x}\right)=
\frac12\left(1+x+x^2+\ldots+1-x+x^2-x^3+\ldots\right)=$$
$$=\frac12\left(2+2x^2+2x^4+\ldots\right)=1+x^2+x^4+\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
The maximum and minimum of a function How can I find the maximum and minimum of $$L=\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2},$$
given that $x^2+y^2+z^2=1$?
I worked on some basic algebra, but it doesn't get me anywhere, so I'm stuck on it.
| By Cauchy-Schwarz inequality
$$\sum_{cyc}\sqrt{1-x^2}\le\sqrt{3\cdot(1-x^2+1-y^2-z^2)}=\sqrt{6}$$
By the other hand,we have
\begin{align*}
&\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2}=\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}\\
&\ge\sqrt{0+y^2}+\sqrt{0+z^2}+\sqrt{y^2+z^2}\\
&=|y|+|z|+\sqrt{y^2+z^2}
\end{align*}
this problem since $y^2+z^2=1$,find the $|y|+|z|+1$ minum
let
$$|y|=a\ge 0,|z|=b\ge0, a^2+b^2=1$$
then
$$|y|+|z|+1=a+b+1=\cos{t}+\sin{t}+1=\sqrt{2}\sin{(t+\dfrac{\pi}{4})}+1\ge 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factorize : $x^6 − 10x^3 + 27$ I want to factorize $$x^6 − 10x^3 + 27$$
I tried two methods , first I let $y=x^3 $ and converted it into a quadratic but the solutions are not real .
The second method I tried was getting it to the form $a^3+b^3+c^3-3abc$ but I could only get close ( $(x^2)^3+(x)^3+3^3-3(x^2)(x)(3)$ but that is actually $x^6-8 x^3+27$ , close but not exact )
I'm run out of ideas , so any help is appreciated .
Note : I know that the answer is $$(x^2+2 x+3) (x^4-2 x^3+x^2-6 x+9)$$ , but I want to know how !
| Seems to me in your attempted form $a^3+b^3 + c^3 - 3abc$ intuitively you came quite close, but quit too soon.
$x^6 − 10x^3 + 27 = (x^2)^3 + (2x)^3 +(3)^3 - 3\cdot (x^2) (2x)(3) $
The rest I presume you can factorise.
Of course such tricks work only on a few polynomials like the one in your question(s).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
A nonlinear system of equations I want to solve the system of equations ,
$$\left\{ \begin{array}{rcl}
y^3 − 9x^2 + 27x − 27 &=& 0\\
z^3 − 9y^2 + 27y − 27 &=& 0\\
x^3 − 9z^2 + 27z − 27 &=& 0\\
\end{array}
\right.$$
Since I've never solved a non linear system of equations in more than 2 variables and containing different degree terms in each equation ( Here , each equation contains degree 3 , degree 2 and degree 1 terms ) , I don't know where to begin. I'm only concerned about real roots .
Actually I checked wolfram alpha and it seems that all real roots are integers , the solution is $(3,3,3)$ . However I want to know how to solve this system ( by hand )
Please help.
| For integer solutions:
You can rewrite the equations using
$$y^3 - 9x^2 +27x - 27 = y^3 - x^3 + (x-3)^3$$
and similarly for the two other equations. Adding the three equations then yields
$$(x-3)^3 + (y-3)^3 + (z-3)^3 = 0.$$
It is long known that in all integer solutions of the equation $a^3 + b^3 = c^3$, one of the integers must be $0$. Our equation is (modulo rearranging) exactly that, so we must have $x = 3$ or $y = 3$ or $z = 3$.
Suppose $x=3$. Then the first equation reduces to $y^3 - x^3 = 0$, hence $y^3 = 27$, hence $y = 3$. In the same way, it follows that $z = 3$. Starting from $y = 3$ or $z = 3$ yields the same. So $(3,3,3)$ is the only integer solution to the system.
Now, for any real solution, we still have by adding the equations
$$(x-3)^3 + (y-3)^3 + (z-3)^3 = 0,$$
so if one of $x,y,z$ were different from $3$, one of the three at least would be larger than $3$. Without loss of generality, let $x = 3 +\delta > 3$. Then, from the first equation, we obtain
$$\begin{align}
y^3 &= x^3 - (x-3)^3\\
&= (3+\delta)^3 - \delta^3\\
&= 3^3 + 27\delta + 9\delta^2\\
&> 3^3,
\end{align}$$
so also $y > 3$. The same reasoning for the second equation then yields $z > 3$, and so
$$(x-3)^3 + (y-3)^3 + (z-3)^3 > 0$$
contradicting the assumption that $(x,y,z)$ is a solution of the system.
Hence $(3,3,3)$ is the only real solution of the system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question on definite integral How to compute the following:
$$ \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x(1+x^2)}dx $$
If you could, please, solve it completely, for, I am afraid, a hint won't do for me.
| Set $$\vartheta(a)=\int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x(1+x^2)}dx$$
Differentiate and use partial fractions to integrate.
Spoiler
$$\vartheta '(a) = \int_0^\infty {\frac{1}{{1 + {a^2}{x^2}}}\frac{1}{{1 + {x^2}}}} dx$$
and
$$\frac{1}{{1 + {a^2}{x^2}}}\frac{1}{{1 + {x^2}}} = \frac{1}{{1 - {a^2}}}\left( {\frac{1}{{1 + {x^2}}} - \frac{{{a^2}}}{{1 + {a^2}{x^2}}}} \right)$$
Then use that
$$\int_0^\infty {\frac{{dx}}{{1 + {x^2}}}} = \frac{\pi}2$$
to obtain
$$\vartheta '(a) = \frac{1}{{1 - {a^2}}}\left( {\frac{\pi }{2} - a\frac{\pi }{2}} \right) = \frac{{1 - a}}{{1 - {a^2}}}\frac{\pi }{2} = \frac{\pi }{2}\frac{1}{{1 + a}}$$
And I guess you can take it from there.
ADD Note I missed a detail on the integral $$\int_0^\infty {\frac{{{a^2}}}{{1 + {a^2}{x^2}}}dx} $$ If $a>0$, we have that $$a\int_0^\infty {\frac{1}{{1 + {{\left( {ax} \right)}^2}}}d\left( {ax} \right)} = a\int_0^\infty {\frac{1}{{1 + {t^2}}}dt} = \frac{{\pi a}}{2}$$ but if $a<0$ $$a\int_0^\infty {\frac{1}{{1 + {{\left( {ax} \right)}^2}}}d\left( {ax} \right)} = a\int_0^{ - \infty } {\frac{1}{{1 + {t^2}}}dt} = - \frac{{\pi a}}{2}$$ Thus in any case $$\vartheta'(a)=\frac{1}{{1 - {{\left| a \right|}^2}}}\left( {\frac{\pi }{2} - \left| a \right|\frac{\pi }{2}} \right) = \frac{\pi }{2}\frac{1}{{1 + \left| a \right|}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Divisibility of ceiling function of surd Show that:
The integer next greater than $(\sqrt{7}+\sqrt{3})^{2n}$ is divisible by $4^n$
| Since
$$
(\sqrt7+\sqrt3)^2+(\sqrt7-\sqrt3)^2=20\tag{1}
$$
and
$$
(\sqrt7+\sqrt3)^2(\sqrt7-\sqrt3)^2=16\tag{2}
$$
$(\sqrt7+\sqrt3)^2$ and $(\sqrt7-\sqrt3)^2$ satisfy the equation
$$
x^2-20x+16=0\tag{3}
$$
The recursion
$$
a_n=20a_{n-1}-16a_{n-2}\tag{4}
$$
where $a_0=2$ and $a_1=20$ is therefore satisfied by
$$
a_n=(\sqrt7+\sqrt3)^{2n}+(\sqrt7-\sqrt3)^{2n}\tag{5}
$$
Since $(\sqrt7-\sqrt3)^2\lt1$, we have
$$
a_n=\left\lceil(\sqrt7+\sqrt3)^{2n}\right\rceil\tag{6}
$$
Divide $(4)$ by $4^n$ to get
$$
\frac{a_n}{4^n}=5\frac{a_{n-1}}{4^{n-1}}-\frac{a_{n-2}}{4^{n-2}}\tag{7}
$$
Since $\dfrac{a_0}{4^0}=2$ and $\dfrac{a_1}{4^1}=5$, what does $(7)$ tell you about $a_n$? Combine with $(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$ I am stuck with the following problem which one of friends gave me :
If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$
The options are $\,\,k^5+4k^3+4k, \,k^5+5k^3+6k,\,k^5+5k^3+5k,\,k^5+5k^3+4k $.
We see that $x-\frac 1 x=k \implies x=\frac{k \pm \sqrt{k^2+4}}{2}$. Now putting this value to $\,\,x^5-\frac {1}{x^2}$ makes the calculation complicated.
Can anyone help? Thanks and regards to all.
EDIT: The problem contained a typo and thanks to @noam for pointing that out. Now using binomial expansion of $x^5-\frac{1}{x^5}$, we see that option 3 is the correct choice.
| HINT:
Applying the binomial expansion
$$
k^5=\left(x-\frac1x\right)^5-5\left(x^3-\frac1{x^3}\right)+10k.
$$
Now expand $k^3=\left(x-\frac1x\right)^3$ to express the cubic factor in terms of $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Algebra confusion: $A^2 + B^4$? So i am currently trying to solve a practice question and i have hit a stump.
The first term of a GP is $1$ and the sum of the third and fifth term is $90$. Find the common ratio.
Using $T_n = a \cdot r^{n-1}$ I got that $a = 1$
Hence
$ar^2 + ar^4 = 90 \implies r^2 + r^4 = 90$
But this is where i got stumped. The answer is given as $ \pm \;3$
Any help is highly appreciated.
| All of your calculations are right. We find ourselves needing to solve $r^2+r^4=90$.
Rearranging gives $r^4+r^2-90=0$. This can be thought of as a quadratic in $r^2$. Let $y=r^2$ then $y^2=r^4$ and so $r^4+r^2-90=0$ becomes $y^2+y-90=0$.
It turns out that $y^2+y-90$ factorises to gives $(y-9)(y+10)$.
Since $y$ was shorthand for $r^2$ we see that $r^4+r^2-90 \equiv (r^2-9)(r^2+10)$.
The solutions to $(r^2-9)(r^2+10) =0$ are $r^2=9$ and $r^2 = -10$. The equation $r^2=9$ has solutions $r=\pm 3$, while the equation $r^2=-10$ has no real solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum Value of a complex expression Find Minimum value of $\dfrac {\mathrm{Im}~ z^5} {(\mathrm{Im}~ z)^5}$ for $z = x+iy$.
Here I started with $z = r \exp(i \theta)$
solve the expression finally i got the answer in terms of $\tan$ which i find little difficult to get along with
| If we set $z=x+iy$
Using Binomial Theorem,
$z^5=(x+iy)^5=x^5+\binom51x^4(iy)+\binom52x^3(iy)^2+\binom53x^3(iy)^3+\binom53x(iy)^4+(iy)^5$
$=\cdots+i\{\binom51x^4y-\binom53x^2y^3+y^4\}$
So, img$(z^5)=\binom51x^4y-\binom53x^2y^3+y^4=5x^4y-10x^2y^3+y^4$
$$\implies \frac{\text{img}(z^5)}{(\text{img}(z))^5}=5\left(\frac xy\right)^4-10\left(\frac xy\right)^2+1=5\left(\left(\frac xy\right)^2-1\right)^2+1-5\ge-4$$
The equability occurs when $\left(\frac xy\right)^2-1=0\implies \frac xy=\pm1\iff x=\pm y$
If we start using polar coordinate,
as $\displaystyle \frac xy=\frac{r\cos\theta}{r\sin\theta}=\cot\theta$
we shall reach at $\displaystyle 5\cot^4\theta-10\cot^2\theta+1=5(\cot^2\theta-1)^2+1-5\ge-4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Factorising a 3 x 3 determinant - What Am I doing Wrong? $$\begin{vmatrix}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3 \\
\end{vmatrix}$$
subtracting the top row from the middle and bottom rows
$$ =
\begin{vmatrix}
1 & a & a^3 \\
0 & (b - a) & (b^3 - a^3)\\
0 & (c - a) & (c^3 - a^3)\\
\end{vmatrix}
$$
expanding the determinant
$$
= \begin{vmatrix}
(b - a) & (b^3 - a^3)\\
(c - a) & (c^3 - a^3)\\
\end{vmatrix}
$$
pulling out factors
$$
= (b-a)(c-a)
\begin{vmatrix}
1 & (b^2 + a^2)\\
1 & (c^2 + a^2)\\
\end{vmatrix}
$$
and evaluating gives
$$
= (b-a)(c-a)(c^2 - b^2)
$$
yet the answer in the book gives
$$(a+b+c)(a-b)(b-c)(c-a)$$
I am self studying so thanks in advance
| A way to compute the determinant via polynomials, à la Vandermonde.
Consider the polynomial
$$
f(x)
=
\begin{vmatrix}
1 & x & x^{3} \\
1 & b & b^{3} \\
1 & c & c^{3} \\
\end{vmatrix}.
$$
Clearly two roots of $f(x)$ are $b$ and $c$, because $f(b)$ is the determinant of a matrix with two equal rows, and similarly for $f(c)$.
Also, the coefficient of $x^{3}$ is
$$\begin{vmatrix}
1 & b\\
1 & c
\end{vmatrix}
=
c - b
$$
and the coefficient of $x^{2}$ is $0$. Since the coefficient of $x^{2}$ is the negative of the sum of the roots, the three roots of $f(x)$ are $$b, \quad c, \quad - b - c,$$ and thus
$$
f(x) = (c - b) (x - b) (x - c) (x + b + c).
$$
Now set $x = a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to get maximum value of a function with three variables satisfying a condition? The maximum value of the function
$$f(x, y, z) = \left(x -\frac{1}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 + \left(z - \frac{1}{3}\right)^2$$
subject to the constraints-
$x + y + z = 1; x \ge 0; y \ge 0; z \ge 0$
what is the actual method to find that ? it may be simple but i dont know , please help me to understand .
| First expand the function:
$$f(x, y, z) = \left(x -\frac{1}{3}\right)^2 + \left(y - \frac{1}{3}\right)^2 + \left(z - \frac{1}{3}\right)^2$$
$$f(x, y, z) = x^2 - \frac 23 x + \frac 19 + y^2 -\frac 23 y + \frac 19 + z^2 - \frac 23 z + \frac 19$$
$$f(x, y, z) = x^2 + y^2 + z^2 - \frac 23 x - \frac 23 y - \frac 23z + \frac 13$$
To minimize/maximize a function with a given constain it's best to apply Lagrange multiplier. So after applying it the function will look like this:
$$F(x, y, z, \lambda) = x^2 + y^2 + z^2 - \frac 23 x - \frac 23 y - \frac 23z + \frac 13 + \lambda (x+y+z-1)$$
Now we take partial derivatives and we set them to $0$.
$$F_x = 2x - \frac 23 + \lambda = 0$$
$$F_y = 2y - \frac 23 + \lambda = 0$$
$$F_z = 2z - \frac 23 + \lambda = 0$$
$$F_{\lambda} = x + y + z -1 = 0$$
Now we get three relation:
$$x = y = z = \frac{\frac {2}{3} - \lambda}{2} = \frac{2-3\lambda}{6}$$
Plugging this into the fourth partial derivative relation we have:
$$x+y+z = 1$$
$$\frac{2-3\lambda}{6} + \frac{2-3\lambda}{6} + \frac{2-3\lambda}{6} = 1$$
$$3(2-3\lambda) = 6$$
$$6-3\lambda = 6$$
$$-3\lambda = 0 \implies \lambda = 0$$
This would imply that:
$$x = y = z = \frac 13$$
But this is the minima, because we have:
$$f(\frac 13, \frac 13, \frac 13) = 0$$
So because this is the only stationary point, the maxima must be on the boundary. And indeed the maxima occurs at these 3 point $$(x,y,z) = (1,0,0), (0,1,0), (0,0,1)$$
$$f(1,0,0) = f(0,1,0) = f(0,0,1) = \frac 23$$
Note that because all these terms are from the same form and we don't have terms of the for $xy$ in the expansion it means that the partial derivatives will have same forms, implying that at the stationary point we'll find the variables will all be the same.
When you check all statinary point in a certain region and you don't get what you want like in this case, then check the boundaries. In this case the region is bounded between $0$ and $1$ so we set one variable at a time to this boundaries and this becomes a 2 variable function which is much easier to maximize.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving of Boundedness and Monotonicity of a sequence $e(n) = (1 + \frac 1n)^n$ is convergent. How do we prove it?(proofs other than binomial expansion are welcome)
| By Bernoulli's inequality, we have $$\left(1-\frac 1{(n+1)^2}\right)^{n+1}\ge 1-\frac 1{n+1}=\frac n{n+1}$$ Now note that $$1-\frac 1{(n+1)^2}={n^2+2n\over n^2+2n+1}={n(n+2)\over (n+1)^2}=\frac {n+2}{n+1}\frac n{n+1},$$ so the above inequality reads $$\left(\frac {n+2}{n+1}\right)^{n+1}\left(\frac n{n+1}\right)^{n+1}\ge \frac n{n+1}$$ Multiplying both sides with $\displaystyle\left(\frac {n+1}n\right)^{n+1}$ gives you the monotonicity of the sequence.
To prove boundedness, use the binomial theorem to show $\left(1+\frac 1n\right)^n\le 3$ noting that $${n\choose k}\left(\frac 1n\right)^k=\frac 1{k!}{n(n-1)...(n-k+1)\over n^k}=\frac 1{k!} \frac{n}n\frac{n-1}n...\frac{n-k+1}n\le \frac 1{k!}\le \left(\frac 12\right)^{k-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How find this integral $I={\iint}_{D}\frac{(x+y)\ln(1+y/x)}{\sqrt{1-x-y}}\,dx\,dy$ find the value
$$I={\Large\iint}_{D}\dfrac{(x+y)\ln\left(1+\dfrac{y}{x}\right)}{\sqrt{1-x-y}}\,dx\,dy,$$
where $$D=\{(x,y)\mid x+y\le 1,x\ge 0,y\ge 0\}.$$
| The integral is
$$\int_0^1 \int_0^{1-x} \frac{ (x+y) \log(1 + \frac y x)}{ \sqrt{1 - x - y}} dy dx$$
It can be evaluated by parts.
$$
\begin{align*}
I_1 &= \int_0^{1-x} \frac{ (x+y) \log(1 + \frac y x)}{ \sqrt{1 - x - y}} dy \\
&= \left[\log (1 + \frac yx)\frac{-2}{3} \sqrt{-x-y+1} (x+y+2) \right ]_0^{1-x} \\ & - \int_0^{1-x} \frac 1 {x+y} \frac{-2}{3} \sqrt{-x-y+1} (x+y+2) dy \\
&= \frac 2 3 \int_0^{1-x} \frac{(x+y+2)\sqrt{1-x-y}}{x+y}dy \\
&= \frac 2 3 \left( -\frac{2}{3} \left(\sqrt{1-x} (x+5)-6 \sinh ^{-1}\left(\sqrt{\frac{1}{x}-1}\right)\right. \right )\\
\end{align*}$$
\begin{align*}
I_2 &= \int_0^1 \frac 2 3 \left( -\frac{2}{3} \left(\sqrt{1-x} (x+5)-6 \sinh ^{-1}\left(\sqrt{\frac{1}{x}-1}\right)\right. \right ) dx \\
&= - \frac 4 9 \left( \int_0^1 \sqrt{1-x} (x+5)dx - 6 \int_0^1 \sinh^{-1} \sqrt{ \frac 1 x - 1 }dx\right )\\
&= - \frac 4 9 \left( \left[ -\frac 2 5 (1 - x)^{3/2} (9 + x) \right ]_0^1 - 6 \int_0^1 \log \left( \sqrt{\frac 1 x -1} + \sqrt{ \left(\frac 1 x -1 \right ) + 1}\right )dx \right )\\
&= - \frac 4 9 \left( \frac {18}5 - 6 \left[ x \log \left(\sqrt{\frac{1}{x}-1}+\sqrt{\frac{1}{x}}\right)-\sqrt{1-x} \right ]_0^1 \right ) \\
&= - \frac 4 9 \left( \frac{18}{5} - 6 \right ) = \frac {16}{15}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/496133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How prove this inequality $ \frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c} \ge 2(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)$ let $a,b,c$ are positive numbers, show that
$$ \frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c} \ge 2(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)\cdots (1)$$
my try:
$$\Longleftrightarrow \frac{b^3+c^3}{a}+a^2+\frac{c^3+a^3}{b}+b^2+\frac{a^3+b^3}{c}+c^2 \ge 3(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)$$
$$\left(a^3+b^3+c^3\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 9(a^2+b^2+c^2)-6(ab+bc+ac)$$
$$(a^3+b^3+c^3)(ab+bc+ac)\ge abc[9(a^2+b^2+c^2)-6(ab+bc+ac)]$$
some days ago,I have ask this same problem：How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$
then I can't prove it,Thank you
maybe $(1)$ have other nice methods?
| We need to prove that
$$\sum_{cyc}\left(\frac{b^3+c^3}{a}-2a^2-3(b-c)^2\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{b^3+c^3}{a}-8a^2+6ab\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{a^3}{b}+\frac{b^3}{a}-4a^2-4b^2+6ab\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^4}{ab}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integration of $\int_{}^{}\frac{1}{x(1+x)^3}dx $ we got this integration problem
$$\int_{}^{}\frac{1}{x(1+x)^3}dx $$
it seems a fairly simple problem but what i am struggling with it is doing its partital fractions
$\int_{}^{}\frac{1}{x(1+x)^3}dx $ =$\int_{}^{}\frac{a}{x}dx$ +$\int_{}^{}\frac{b}{(1+x)^3}dx$+
$\int_{}^{}\frac{c}{(1+x)^2}dx$ + $\int_{}^{}\frac{d}{1+x}dx$ now how to get values of $a,b,c,d$ ? it seems confusion by doing it by $a(1+x)^3+$ $bx+$ $cx(1+x)$ $+d(x(1+x)^2) =1$ is there any other way to break it into partial fractions and solve it ?
| You need to begin with the good partial fraction decomposition! That is,
$$
\frac 1{x(1+x)^3} = \frac ax + \frac b{(1+x)} + \frac c{(1+x)^2} + \frac d{(1+x)^3}.
$$
This is the correct way because the partial fraction decomposition works when you put irreducible factors individually in their separate fractions, together with $k$ terms if the irreducible factor appears to the $k^{\text{th}}$ power, each term being a different power of that irreducible factor. For instance, since $(1+x)^3$ is the irreducible factor $(1+x)$ (to the third power) of $x(1+x)^3$, it must appear three times, once for each power of $(1+x)$.
This gives you the equation $a(1+x)^3 + bx(1+x)^2 + cx(1+x) + dx = 1$. Expand the factors. You'll get
$$
a + (3a + b + c + d)x + (3a + 2b + c)x^2 + (a + b) x^3 = 1.
$$
By comparing the two polynomials' coefficients, it is straightforward that $a=1$ and $b = -a = -1$. Therefore $0 = 3a + 2b + c = 3 - 2 + c = 1+c$, hence $c=-1$ and $0 = 3a + b + c + d = 3 - 1 - 1 + d = 1+d$, hence $d = -1$. This gives you
$$
\frac 1{x(1+x)^3} = \frac 1x + \frac {-1}{(1+x)} + \frac {-1}{(1+x)^2} + \frac {-1}{(1+x)^3}.
$$
Hope that helps,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Sum of Series $ \frac{F_0}{6}+\frac{F_1}{6^1} +\frac{F_2}{6^2}+\frac{F_3}{6^3}+\frac{F_4}{6^4} + \ldots$ I want to find of the following series where $ F_0 = 1 $ , $ F_1 = 1 $ and $ F_n=F_{n-1}+F_{n -2} $ .
$$ S = \frac{F_0}{6}+\frac{F_1}{6^1} +\frac{F_2}{6^2}+\frac{F_3}{6^3}+\frac{F_4}{6^4} + \ldots$$
I have tried to proceed with the thought of finding a close formula for this problem but cant be able to derive formula . Can you please help me to find the formula for this problem ?
| First, the series converges because $0 \leq F_n \leq 2^n$ by induction.
Then, you can compute
$$ S = \sum_{n=0}^{+\infty} \frac{F_n}{6^n}
= F_0 + \frac{F_1}{6} + \sum_{n=2}^{+\infty} \frac{F_n}{6^n}
= F_0 + \frac{F_1}{6} + \sum_{n=2}^{+\infty} \frac{F_{n-1}}{6^n} + \sum_{n=2}^{+\infty} \frac{F_{n-2}}{6^n}. $$
Reordering should get you to
$$ S = F_0 + \frac{F_1}{6} + \frac{S-F_0}{6} + \frac{S}{6^2}. $$
Then it's easy to find $S$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find this limit $I=\lim_{x\to 0}\left(\frac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)^{\cot^2{x}}$ Find the limit of the value
$$I=\lim_{x\to 0}\left(\dfrac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)^{\cot^2{x}},(0<a\neq1,0<b\neq 1,a\neq b)$$
Following is My ugly methods
$$I=e^{\displaystyle\lim_{x\to 0}\cot^2{x}\ln{\left(\dfrac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)}}$$
so we only find following this limit
$$A=\lim_{x\to 0}\cot^2{x}\ln{\left(\dfrac{a^x-x\ln{a}}{b^x-x\ln{b}}\right)}=\lim_{x\to 0}\dfrac{\ln{(a^x-x\ln{a})}-\ln{(b^x-x\ln{b})}}{x^2}$$
By the L'Hospital
$$A=\lim_{x\to 0}\dfrac{\dfrac{a^x\ln{a}-\ln{a}}{a^x-x\ln{a}}-\dfrac{b^x\ln{b}-\ln{b}}{b^x-x\ln{b}}}{2x}$$
so
\begin{align*}A&=\lim_{x\to 0}\dfrac{a^xb^x(\ln{a}-\ln{b})+\ln{a}\ln{b}x(b^x-a^x)+a^x\ln{b}-b^x\ln{a}}{(a^x-x\ln{a})(b^x-x\ln{b})(2x)}\\
&=\lim_{x\to 0}\dfrac{(\ln{a}-\ln{b})+\ln{a}\ln{b}x(b^x-a^x)+(\ln{b}-\ln{a})}{2x}\\
&=\lim_{x\to 0}\dfrac{\ln{a}\ln{b} x(b^x-a^x)}{2x}\\
&=0
\end{align*}
so $$I=e^{0}=1$$
Have other nice methods and my result is true? Thank you
| Since $\ln (a^x-x\ln{a}) = \ln (1+\frac{x^2}{2}(\ln{a})^2 + o(x^2)) = \frac{x^2}{2}(\ln{a})^2 + o(x^2)$, $\ln (b^x-x\ln{b}) = \frac{x^2}{2}(\ln{b})^2 + o(x^2)$, we have
$$\cot^2{x} \left(\ln (a^x-x\ln{a}) - \ln (b^x-x\ln{b}\right) = \frac{(\ln{a})^2-(\ln{b})^2}{2} + o(1).$$
Hence
$$I = \sqrt{e^{(\ln{a})^2-(\ln{b})^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help with an induction proof Here's the thing I'm trying to prove
$2^k - (2^{k-1} + 2^{k-2} + ... + 2^2 + 2^1) = 2$
It's obviously easy for the k = 1 case, but I'm stuck on the k + 1 case.
| Hint: Note that
\begin{align}
2^{k + 1} - \left(2^k + 2^{k - 1} + ... + 2^2 + 2^1\right) &= 2^{k} + 2^k - \left(2^k + 2^{k - 1} + ... + 2^2 + 2^1\right) \\ &= 2^k - \left(2^{k - 1} + ... + 2^1\right)\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of series ${n\choose 2a}{a\choose 0}+ {n\choose {2a+2}}{{a+1}\choose 1} + {n\choose {2a+4}}{{a+2}\choose 2} + \ldots$ I wanted to check the rationality of the cosine function for some rational multiples of $\pi$. And I found out that, $\cos(n \cdot\arccos x)$ generates a polynomial in $x$ whose co-efficients have the form:
$${n\choose 2a}{a\choose 0}+ {n\choose {2a+2}}{{a+1}\choose 1} + {n\choose {2a+4}}{{a+2}\choose 2} + \ldots$$
For $a = 0$ the answer is simple, but what for other cases? Please help.
| Suppose we seek to evaluate
$$\sum_{k=0}^{n} {n\choose 2a+2k} {a+k\choose k}$$
where $n\ge a\ge 0.$
Introduce
$${n\choose 2a+2k} = {n\choose n-2a-2k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a-2k+1}}
(1+z)^{n} \; dz.$$
Observe that this is zero when $2k\gt n-2a$ which is the correct
value. Therefore we may extend the range of $k$ to infinity to obtain
for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}}
(1+z)^{n}
\sum_{k\ge 0} {a+k\choose a} z^{2k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}}
(1+z)^{n} \frac{1}{(1-z^2)^{a+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}}
(1+z)^{n-a-1} \frac{1}{(1-z)^{a+1}} \; dz.$$
Extracting the residue we get
$$\sum_{q=0}^{n-2a} {n-a-1\choose q}
{n-2a-q+a\choose a} =
\sum_{q=0}^{n-2a} {n-a-1\choose q}
{n-a-q\choose a}
\\ = \sum_{q=0}^{n-2a} {n-a-1\choose q}
{n-a-q\choose n-2a-q}.$$
This time introduce
$${n-a-q\choose n-2a-q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a-q+1}}
(1+z)^{n-a-q} \; dz$$
which is zero when $q\gt n-2a$ so that we may extend $q$ to $n-a-1$ to
get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}}
(1+z)^{n-a}
\sum_{q=0}^{n-a-1} {n-a-1\choose q} \frac{z^q}{(1+z)^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2a+1}}
(1+z)^{n-a}
\left(1+\frac{z}{1+z}\right)^{n-a-1}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1+z}{z^{n-2a+1}}
\left(1+2z\right)^{n-a-1}
\; dz.$$
This yields the closed form
$$2^{n-2a} {n-a-1\choose n-2a} +
2^{n-2a-1} {n-a-1\choose n-2a-1}
\\ = 2^{n-2a} \frac{a}{n-2a} {n-a-1\choose n-2a-1} +
2^{n-2a-1} {n-a-1\choose n-2a-1}
\\ = \left(2 \frac{a}{n-2a} + 1\right)
2^{n-2a-1} {n-a-1\choose n-2a-1}
= \frac{n}{n-2a} 2^{n-2a-1} {n-a-1\choose n-2a-1}.$$
This holds when $2a\lt n.$ When $2a=n$ the initial closed form yields
$1$ because the second binomial coefficient vanishes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving $\arccos(11/14)+\arcsin(-1/7)=x$ Solve $\arccos(11/14)+\arcsin(-1/7).$
I'm using $\cos(a+b)=\cos(a)\cos(b)+\sin(a)\sin(b).$
Attempt: sheet1 and sheet2
| Hint:
$$
\begin{align}
&\sin\left(\cos^{-1}\left(\frac{11}{14}\right)-\sin^{-1}\left(\frac{1}{7}\right)\right)\\
&=\sin\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\cos\left(\sin^{-1}\left(\frac{1}{7}\right)\right)
-\cos\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\sin\left(\sin^{-1}\left(\frac{1}{7}\right)\right)\\
&=\frac{5\sqrt3}{14}\frac{4\sqrt3}{7}-\frac{11}{14}\frac17\\
&=\frac{49}{98}
\end{align}
$$
Another Hint:
$$
\begin{align}
&\cos\left(\cos^{-1}\left(\frac{11}{14}\right)-\sin^{-1}\left(\frac{1}{7}\right)\right)\\
&=\cos\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\cos\left(\sin^{-1}\left(\frac{1}{7}\right)\right)
+\sin\left(\cos^{-1}\left(\frac{11}{14}\right)\right)\sin\left(\sin^{-1}\left(\frac{1}{7}\right)\right)\\
&=\frac{11}{14}\frac{4\sqrt3}{7}+\frac{5\sqrt3}{14}\frac17\\
&=\frac{49\sqrt3}{98}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How prove this inequality $\frac{2^x-1}{3^x-2^x}\le3\left(\frac{1}{x-1}-\frac{1}{x}\right)$ show that
$$\dfrac{2^x-1}{3^x-2^x}\le3\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\cdots(1)$$
This problem (1) is from when I solve following
$$1+\dfrac{3}{5}+\dfrac{7}{19}+\cdots+\dfrac{2^n-1}{3^n-2^n}\le 4-\dfrac{3}{n}\cdots (2)$$
My try if we have prove this
$$\dfrac{2^n-1}{3^n-2^n}\le 3(\dfrac{1}{n-1}-\dfrac{1}{n}),n\ge 2$$
then $(2)$ is prove it.
By the way,I have use other methods solve it $(2)$
because I have try
$$\dfrac{2^n-1}{3^n-2^n}\le\left(\dfrac{2}{3}\right)^{n-1}$$
$$\Longleftrightarrow 6^n+2\cdot 3^n\ge 3\cdot 4^n$$
since $AM-GM$
$$\dfrac{1}{3}\cdot 6^n+\dfrac{2}{3}\cdot 3^n\ge\left(\dfrac{1}{3}\cdot 6+\dfrac{2}{3}\cdot 3\right)^n=4^n$$
so
$$\sum_{k=1}^{n}\dfrac{2^k-1}{3^k-2^k}\le\sum_{k=1}^{n}\left(\dfrac{2}{3}\right)^{k-1}=3-3\cdot\left(\dfrac{2}{3}\right)^n$$
and it easy prove
$$4-\dfrac{3}{n}\ge 3-3\cdot(\dfrac{2}{3})^n$$
so the $(2)$ prove by done.
Now my question How can prove by $(1)$,and I think $(1)$ have lot of nice methods.Thank you
| We can show the stronger inequality (i.e. with increased left side)
$$\dfrac{2^x}{3^x-2^x}\le3\left(\dfrac{1}{x-1}-\dfrac{1}{x}\right)\tag{1}
$$
holds for $x\ge 6.$ Division of the numerator and denominator on the left by $2^x$ and algebra gives the equivalent form
$$(3/2)^x \ge \frac{x^2-x+3}{3}. \tag{2}$$
This holds when $x=6$, and one can show the derivative of the left side exceeds that of the right side for $x\ge 6.$ If we increase the right side derivative to $2x/3$, then one could get exact values [using Lambert's W function] for when $\ln(3/2)\cdot(3/2)^x$ crosses through the line $y=2x/3$, and these are approximately at $0.863$ and $5.373$, so that for $x\ge 6$ we have the derivative of the left side of $(2)$ exceeds that of the right side, while also $(2)$ holds at $x=6$, from which we can conclude $(2)$, and so also $(1)$, hold for $x\ge 6$
Note: the inequality $(1)$ fails to hold only on a short interval $(\alpha,\beta)$ contained in the interval $[4,6]$. The above approach doesn't fill that gap, and it seems (to me) that without dropping the subtraction of $1$ in the given inequality makes the manipulations more complicated, making an approach via derivatives difficult.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
When computing $\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$, I don't get -2. This limit:
$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}$
is supposed to be equal to -2. My textbook and Wolfram Alpha both state that. However, I can't seem to get same exact result.
Here's what I tried to do:
$$\lim_{x\to -\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} = \lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}} \cdot \frac{1 \over x}{1 \over x} = \lim_{x\to -\infty}\frac{\frac{x}{x}\cdot(2+\frac7x)}{\sqrt{\frac{x^2}{x^2}\cdot(1+\frac{2}{x}-\frac{1}{x^2})}} = \frac{1\cdot(2+0)}{\sqrt{1\cdot(1+0-0)}} =\frac{2}{\sqrt{1}}=2$$
Where did I make a mistake?
| The mistake at the second $=$ where you use $\frac{1}{x}=\sqrt{\frac{1}{x^2}}$ which is false when $x$ is negative. The correct formula is $\sqrt{x^2}=|x|$ which gives $\left|\frac{1}{x}\right|=\sqrt{\frac{1}{x^2}}$ in your case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
no. of onto functions from A to A Let
$$A =\{1,2,3,4,5,6,7\}\quad and \quad f:\mathbb{A}\rightarrow \mathbb{A}.$$
What is the no. of onto functions in which at least $3$ elements of $A$ have self Images?
I did not understand the Language of Question
| Here is a very tedious approach:
Let $\eta_n$ be the number of bijections of $\{1,...,n\}$ that have no fixed point (that is, $f(x) \neq x$ for all $x$). We note that $\eta_1 = 0$, and by inspection $\eta_2 = 1, \eta_3 = 2$. Computing $\eta_4$ is a little more work, enumerating gives $\eta_4 = 9$ (the possibilities are $(2,1,4,3), (2,3,4,1)$, $(2,4,1,3)$, $(3,1,4,2)$, $(3,4,1,2)$, $(3,4,2,1)$, $(4,1,2,3)$, $(4,3,1,2)$, $(4,3,2,1)$).
Consider the number of bijections that have exactly $k$ fixed points. We need to consider $k=3,...,7$.
Note that there are 6 fixed points iff there are 7 fixed points, and there is exactly one of these. So we just need to consider $k=3,4,5$ and add one.
First choose $k \in \{3,4,5\}$ points, there are $\binom{7}{k}$ of these. We have $\eta_{7-k}$ possibilities for the remaining $7-n$ numbers. Hence there are $\binom{7}{k} \eta_{7-k}$ of these.
So the number we want is $1+\sum_{k=3}^5 \binom{7}{k} \eta_{7-k} = 1 + \binom{7}{3}9 + \binom{7}{4}2 + \binom{7}{5}1 = 407$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove, that $\sin x- a^3\cos x\leq \frac 1 3 \sqrt{1+a^6}$ Let a and $x$ be natural numbers with the property that $\sin x\leq a\cos x$.
Prove that $\sin x- a^3 \cos x\leq \frac 1 3 \sqrt{1+a^6}$.
Again, I'm looking for a second solution. I don't know how to use LaTex and my solution is hard to write here. It would be very nice to see a different solution from mine.
| Consider the problem
$$\max_{\mathbf{y}\in \mathbb{R}^2} \mathbf{y} \cdot \mathbf{c}$$ subject to $\mathbf{b}\cdot \mathbf{y} \le 0$ and $\|\mathbf{y}\|=1$, where $\mathbf{b}=(-a,1)$, $\mathbf{c}=(-a^3,1)$. Since $\mathbf{b} \cdot \mathbf{c} \ge 0$ it's not too hard to see that the optimal argument is given by $\mathbf{y}^*=(\cos x,\sin x) = (-1,-a)/\sqrt{1+a^2}$. (Drawing a picture of the feasible region helps.) So then we want to check that:
$$\mathbf{y}^*\cdot \mathbf{c} = \frac{a^3-a}{\sqrt{1+a^2}} \le \frac{1}{3}\sqrt{1+a^6}$$
Since $a \ge 1$ this is equivalent to:
$$9 a^2(a^2-1)^2 \le (1+a^2)(1+a^6).$$
Which then by collecting terms and factoring, is equivalent to:
$$a^8-8a^6+18a^4-8a^2+1=(a^4-4a^2+1)^2\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that an Integral is Zero does someone have any idea about how to prove that
\begin{equation}
\int_{0}^{\infty} e^{-x^{4/3}} \cos x^{4/3} x^{4n+1} dx = 0,
\end{equation}
for $n=0,1,2,...$?
I met this integral while I was solving a PDE.
Thank you very much in advance for your invaulable help.
Best Regards,
Maurizio Barbato
| I guess you already found out that this integral is not equal to zero. I hope
Actually you can cope with this integral and I hope those derivations will be of any help.
$$
\begin{eqnarray}
\int_{0}^{\infty} e^{-x^{4/3}} \cos x^{4/3} x^{4n+1}\;\mathrm dx &=& \int_{0}^{\infty} e^{-x^{4/3}} \left(\frac{e^{\mathrm ix^{4/3}}+e^{-\mathrm ix^{4/3}}}{2}\right) x^{4n+1}\;\mathrm dx=\\
&=&\frac{1}{2}\int_{0}^{\infty} e^{-x^{4/3}(1+\mathrm i)} x^{4n+1}\;\mathrm dx+\frac{1}{2}\int_{0}^{\infty} e^{-x^{4/3}(1+\mathrm i)} x^{4n+1}\;\mathrm dx=\\
\end{eqnarray}
$$
Multiplying anf dividing by $\frac{4}{3}x^{1/3}$, changing the variable to $t=x^{4/3}$:
$$
\begin{eqnarray}
\int_{0}^{\infty} e^{-x^{4/3}(1\pm\mathrm i)} x^{4n+1}\;\mathrm dx&=&\int_{0}^{\infty} e^{-x^{4/3}(1\pm\mathrm i)} x^{4n+1}\left(\frac{3}{4}x^{-1/3}\right)\;\mathrm dx^{4/3}=\\
&=&\frac{3}{4}\int_{0}^{\infty} e^{-t(1\pm\mathrm i)} t^{3n+\frac{1}{2}}\;\mathrm dt
\end{eqnarray}
$$
Changing the variable to $z=(1\pm \mathrm i)t$:
$$
\begin{eqnarray}
\frac{3}{4}\int_{0}^{\infty} e^{-t(1\pm\mathrm i)} t^{3n+\frac{1}{2}}\;\mathrm dt&=&
\frac{3}{4}(1\pm\mathrm i)^{-3n-\frac{3}{2}}\int_{0}^{\infty} e^{-z} z^{3n+\frac{1}{2}}\;\mathrm dt=\\
&=&\frac{3}{4}(1\pm\mathrm i)^{-3n-\frac{3}{2}}\Gamma\left(3n+\frac{3}{2}\right)
\end{eqnarray}
$$
So
$$
\begin{eqnarray}
\int_{0}^{\infty} e^{-x^{4/3}} \cos x^{4/3} x^{4n+1}\;\mathrm dx &=&
\frac{3}{8}\Gamma\left(3n+\frac{3}{2}\right)\left((1+\mathrm i)^{-3n-\frac{3}{2}}+(1-\mathrm i)^{-3n-\frac{3}{2}}\right)=\\
&=&\frac{3}{8}\Gamma\left(3n+\frac{3}{2}\right)2^{-\frac{3n}{2}-\frac{3}{4}}\left(e^{\mathrm i \frac{\pi}{4}(-3n-\frac{3}{2})}+e^{-\mathrm i \frac{\pi}{4}(-3n-\frac{3}{2})}\right)=\\
&=&\frac{3}{8}\Gamma\left(3n+\frac{3}{2}\right)2^{-\frac{3n}{2}+\frac{1}{4}}\cos\left(\frac{3\pi}{4}\left(n+\frac{1}{2}\right)\right)
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$? Problem :
If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the
value of $\cos(\theta -\frac{\pi}{4})$?
My approach :
Solution: $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$
$\Rightarrow \tan(\pi \cos\theta) = \tan \{ \frac{\pi}{2} - (\pi \sin\theta) \} $
$\Rightarrow \pi \cos\theta = \frac{\pi}{2} - (\pi \sin\theta)$
$\Rightarrow \frac{1}{2} =\frac{1}{\sqrt{2}}[\sin\frac{\pi}{4} \cos\theta + \cos\frac{\pi}{4} \sin\theta] $
$\Rightarrow \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4} + \theta)$
$\Rightarrow \frac{\pi}{4} = \frac{\pi}{4} + \theta$
$\Rightarrow \theta = 0$
$\therefore \cos(\theta - \frac{\pi}{4})$ = $\frac{1}{\sqrt{2}}$ But this is wrong answer.. please suggest where I am wrong... thanks.
| $$\tan (\pi \cos x)=\cot(\pi \sin x)$$
or,
$$\tan (\pi \cos x)=\tan (\frac{\pi}{2} - \pi \sin x)$$ [∵, tan(π/2+Ф)=cotФ]
or, πcosx=π/2-πsinx
or, π(sinx+cosx)=π/2
or, sinx+cosx=1/2
or, (sinx+cosx)1/√2=1/2√2
or, cosx(1/√2)+sinx(1/√2)=1/2√2
or, cosxcosπ/4+sinxsinπ/4=1/2√2
or, cos(x-π/4)=1/2√2
∴, a)1/2√2 is the right answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Coefficient power series problem #49 What is the coefficient of $x^n$ in the power series form of $(1-2x)^{1/3}$? This problem is taken from bona chapter 4, third edition.
| If $\sum_{n=0}^\infty a_n x^n$ is the Taylor expansion of $f$ about zero, then $a_n = \frac{f^{(n)} (0)}{n!}$.
Here $f(x) = (1-2x)^{\frac{1}{3}}$. A little work shows that $f^{(n)}(x) = (-2)^n (1-2x)^{( \frac{1}{3}-n )} \left( (\frac{1}{3}) (\frac{1}{3}-1) \cdots (\frac{1}{3}-(n-1)) \right)$, and so
$a_n = \frac{f^{(n)} (0)}{n!} = (-2)^n \frac{ (\frac{1}{3}) (\frac{1}{3}-1) \cdots (\frac{1}{3}-(n-1)) }{n!}$.
Alternatively (although basically the same thing), we have $(1+x)^{\frac{1}{3}} = \sum_{n=0}^\infty \binom{\frac{1}{3}}{n}x^n$, and so $(1-2x)^{\frac{1}{3}} = \sum_{n=0}^\infty \binom{\frac{1}{3}}{n}(-2)^nx^n$, hence $a_n = (-2)^n \binom{\frac{1}{3}}{n}$. The term $\binom{\alpha}{n}$ is defined in http://en.wikipedia.org/wiki/Binomial_series, for example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $\sqrt{1+{4\over x}+{4\over x^2} }$? If $$abx^2 = (a-b)^2(x+1)$$ then what is $$\sqrt{1+{4\over x}+{4\over x^2} }$$
(A) $a+b \over a-b$ (B)$a-b\over a+b$ (C) $a^2+ab$ (d) None
EDIT: What I've done is this:
$$abx^2=(a^2+b^2-2ab)(x+1)$$
=> $$abx^2 - (a^2+b^2+2ab)x - (a-b)^2=0$$
=> $$abx^2 - (a-b)^2x - (a-b)^2=0$$
but this is getting me nowhere.
| $\textbf{Shortest Solution}$:
$$\frac{ab}{(a-b)^2}=\frac{(x+1)}{x^2}$$
$$\frac{ab}{(a-b)^2}=\frac{1}{x^2}+\frac{1}{x}$$
Multiplying $4$ both sides,
$$\frac{4ab}{(a-b)^2}=\frac{4}{x^2}+\frac{4}{x}$$
Adding $1$ both sides,
$$1+\frac{4ab}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$
$$\frac{(a-b)^2+4ab}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$
$$\frac{(a+b)^2}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$
Taking Square root both sides,
$$\sqrt{\frac{(a+b)^2}{(a-b)^2}}=\sqrt{1+\frac{4}{x^2}+\frac{4}{x}}$$
$$\frac{(a+b)}{(a-b)}=\sqrt{1+\frac{4}{x^2}+\frac{4}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
What is the value of $f(0)+f(8)$? Suppose $f$ is a polynomial of degree $7$ which satisfies $f(1) =2$, $f(2)=5$, $f(3)=10$, $f(4)=17$, $f(5)=26$, $f(6)=37$ and $f(7)=50$. What is the value of $f(0)+f(8)$?
| Note that the differences between the given values form an arithmetic sequence:
$$ +3,\ +5,\ +7,\ +9,\ +11,\ +13\,.$$
So that, we have an easy solution $f$, having $f(0)=1$ and $f(8)=65$.
(Anyway, $f(x)=x^2+1$.)
Note that $7$ values determine uniquely a polynomial of degree $\le \bf 6$.
So, any other polynomial that satisfies the conditions is of the form
$$f(x)= x^2-1\ + \ g(x)\cdot(x-1)(x-2)\dots(x-7)\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Operation of Sine and Cosine Use condition $\displaystyle\sin\theta-\cos\theta=\sqrt{2}$,
Please find the value of $\displaystyle\frac1{\sin^{10}\theta}+\frac1{\cos^{10}\theta}$.
| If we set $a=\sin\theta,b=\cos\theta$
we have $a-b=\sqrt2$ and
as $a^2+b^2=1,(a-b)^2=2\iff -2ab=1$
$\displaystyle (a^2-b^2)^2=(a^2+b^2)^2-4a^2b^2=1^2-(2ab)^2=0$
$\displaystyle\implies a^2-b^2=0$ and we have $a^2+b^2=1$
Can you solve for $a^2,b^2?$
We need to find $\displaystyle\frac1{a^{10}}+\frac1{b^{10}}=\frac1{(a^2)^5}+\frac1{(b^2)^5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the marginal distribution of $(X,Y)$ Given the next robability density function$\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty}\displaystyle\int_{-\infty}^{\infty} \frac{1}{\pi^2(1+x^2+y^2+z^2)^2}\, dx dy dz $
a)Find the marginal distribution of $(X,Y)$
The marginal distribution of (X,Y) is $\displaystyle\int_{-\infty}^{\infty} \frac{1}{\pi^2(1+x^2+y^2+z^2)^2}\,dz$
How can I solve this integral not using spherical coordinates ??
This integral looks like a Cauchy Distribution, then I know that $\displaystyle\int_{-\infty}^{\infty} \frac{a}{\pi((x-b)^2+a^2)}\,dx = 1$
Is the marginal of $(X,Y)$ a Cauchy distribution??
Thanks for your help :D
| Since
$$
\int_{-\infty}^\infty \frac{1}{A^2+z^2} dz=\frac{z}{A^2+z^2}\Big|_{-\infty}^\infty+\int_{-\infty}^\infty \frac{2z^2}{(A^2+z^2)^2} dz=\int_{-\infty}^\infty \frac{2A^2+2z^2-2A^2}{(A^2+z^2)^2} dz
$$
Then
$$
\int_{-\infty}^\infty \frac{2A^2}{(A^2+z^2)^2} dz=\int_{-\infty}^\infty \frac{1}{A^2+z^2} dz=\frac{1}{A}\arctan(\frac{z}{A})\Big|_{-\infty}^\infty=\frac{\pi}{A}
$$
Therefore
$$
\int_{-\infty}^\infty \frac{1}{(A^2+z^2)^2} dz=\frac{\pi}{2A^3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find points on curve $y = (x^2 - 1)^3 / (x^2 + 1)^3$ when slope of tangent line is zero? As soon as a get to the quotient rule I can't simplify to solve for $x$.
This is what I have tried:
Since the derivative much be equal to zero:
$$0 = [6x(x^2 -1)^2 (x^2+1)^3] - [6x(x^2 + 1)^2(x^2-1)^3]$$
How do I simplify further and solve for $x$?
| HINT:
$$\displaystyle \frac{d\left(\frac{x^2-1}{x^2+1}\right)^3}{dx} =\frac{d\left(\frac{x^2-1}{x^2+1}\right)^3}{d\left(\frac{x^2-1}{x^2+1}\right)}\cdot\frac{d\left(\frac{x^2-1}{x^2+1}\right)}{dx}$$
Now, $$\frac{d\left(\frac{x^2-1}{x^2+1}\right)^3}{d\left(\frac{x^2-1}{x^2+1}\right)}=3\left(\frac{x^2-1}{x^2+1}\right)^2$$
and $$\frac{d\left(\frac{x^2-1}{x^2+1}\right)}{dx}=\frac{d\left(1-\frac2{x^2+1}\right)}{dx}=\frac{2\cdot2x}{(x^2+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can someone help me solve this limits question? $$\begin{align}\lim x → ∞\end{align}$$
$$\begin{align}
f(x) = {\frac{2^{x+1}+{3^{x+1}}}{2^x + 3^x}} \\
\end{align}$$
I tried using L Hopitable but that gives the same expression. Also tried using substitution but I didn't get anywhere. Help would be appreciated.
| Here is another idea. Try writing,
\begin{align*}
\frac{2^{x+1}+3^{x+1}}{2^x+3^x} &= \frac{2 \cdot 2^x+3 \cdot 3^x}{2^x+3^x} \\
&=\frac{3\cdot 2^x - 2^x+3 \cdot 3^x}{2^x+3^x} \\
&=\frac{3\left( 2^x+3^x \right)-2^x}{2^x+3^x} \\
&=3-\frac{2^x}{2^x+3^x} \\
&=3-\frac{\frac{2^x}{2^x}}{\frac{2^x}{2^x}+\frac{3^x}{2^x}} \\
&=3-\frac{1}{1+\left( \frac{3}{2} \right)^x}.
\end{align*}
Now most people would just look at this and say
$$\lim_{x \rightarrow \infty} \left( 3-\frac{1}{1+\left( \frac{3}{2} \right)^x} \right)=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
no. of quadrilateral in 12 sided polygon Find the number of quadrilaterals that can be made using the vertices of a polygon of 12 sides as their vertices and having
(1) exactly 1 sides common with the polygon.
(2) exactly 2 sides common with the polygon.
$\underline{\bf{My \; Try}}::$ Let $A_{1},A_{2},A_{3},................A_{12}$ points of a polygon of side $=12$.
(1) part:: Let We Select adjacents pairs $A_{1}A_{2}$, Then other two vertices are from $A_{4},A_{5},.........A_{11}$.
Here $A_{12}$ is not included because it is Left consecutive point corrosponding to $A_{1}$
So this can be done by $\displaystyle \binom{7}{2}$ similarly we can take another consecutive pairs $A_{2}A_{3}$.
So there are Total $12$ adjacents pairs in Anticlock-wise sence.
So Total no. of Quadrilateral in which one side common with $12$ sided polygon is
$\displaystyle = \binom{7}{2}\times 12 = 21\times 12 = 252$
(2) part :: If $2$ selected sides are consecutive:
Let we select $A_{1}A_{2}$ and $A_{2}A_{3}$. Then we select one points from the vertices $A_{4},A_{5},A_{6},.........A_{12}$
This can be done by $\displaystyle {9}{1}$ ways.
Now we select consecutive adjacents sides in Anti-clockwise sence by $(11)$ ways.
So Total ways in above case(for two adjacents sides) is $\displaystyle = \binom{9}{1}\times 11 = 99$
If $2$ selected sides are not consecutive:
Now I did not understand How can i calculate in that case
Help required
Thanks
| Let us count the number of quadrilaterals $A_1A_2A_{2+i}A_{2+i+j}$ where $i >1, j > 1, 2+i+j < 12$. Such vertices count the quadrilaterals with exactly $A_1A_2$ as common side. This is same as the number of solutions to $i+j <10, i >1, j > 1 $. Putting $x = i-1, y = j-1$, we need the solutions $x+y < 8$ where $x >0, y > 0$. By stars and bars method this is $\binom{1}{1}+\binom{2}{1}+\binom{3}{1}+\binom{4}{1}+\binom{5}{1} +\binom{6}{1}= 21$. Thus the answer for part 1 is $21\times 12=252$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove lower bound $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}\ge\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\frac{\sqrt{2}}{2n}$
Consider $$S_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}$$ and show that, for every positive integer $n$,
$$S_n\ge\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\dfrac{\sqrt{2}}{2n}$$
I can prove a related upper bound:
$$S_n=\sum_{k=1}^{n}\dfrac{1}{n\sqrt{1+\left(\dfrac{k}{n}\right)^2}}\le\int_{0}^{1}\dfrac{1}{\sqrt{1+x^2}}dx=\ln{(1+\sqrt{2})}$$
but I can't prove the lower bound.
| The inequality to be proved is
$$\sum\limits_{k=1}^{n}\frac{1}{\sqrt{n^2+k^2}}>\left(1-\frac{1}{n}\right)\ln{(1+\sqrt{2})}+\frac{\sqrt{2}}{2n}$$
equivalent to
$$\frac{1}{n-1}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{1+(\frac{k}{n})^2}}>\ln{(1+\sqrt{2})}=\int_0^1\frac{1}{\sqrt{1+x^2}}.$$
LHS is the Riemann Sum of integral $\displaystyle\int_0^1\frac{1}{\sqrt{1+x^2}}$. When $n\to\infty$, LHS=RHS.
So what we have to prove is that LHS is monotonically decreasing with $n$.
Consider a funtion $f(x)=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+(1-x)^2}}$.
Easy to know its 2nd derivative in $[0,1]$ is negative.
So we have
$$f\Big(\frac{k}{n}\Big) > \Big(1-{k\over n}\Big)f\Big({k\over n+1}\Big)+{k\over n}f\Big({k+1\over n+1}\Big).$$
Sum over $n$, we get
$$\sum_{k=1}^{n-1} f\Big(\frac{k}{n}\Big)>\sum_{k=1}^{n-1}\Big(1-\frac{k}{n}\Big)f\Big(\frac{k}{n+1}\Big)+\sum_{k=1}^{n-1}\frac{k}{n}f\Big(\frac{k+1}{n+1}\Big)=\frac{n-1}{n}\sum_{k=1}^nf\Big(\frac{k}{n+1}\Big).$$
So $$\frac{1}{n-1}\sum_{k=1}^{n-1}\frac{1}{\sqrt{1+\Big(\dfrac{k}{n}\Big)^2}}=\frac{1}{2(n-1)}\sum_{k=1}^{n-1} f\Big(\frac{k}{n}\Big)$$ is monotonically decreasing with $n$. Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
How prove this inequality $6+5|x+y+z|+2|yz+xz+xy|\ge 3(|x|+|y|+|z|)$ let $x,y,z$ be real numbers,show that
$$6+5|x+y+z|+2|yz+xz+xy|\ge 3(|x|+|y|+|z|)\cdots\cdots\cdots (1)$$
I think this inequality is nice,and I have see this
$$1+|x+y+z|+|xy+yz+xz|+|xyz|\ge\dfrac{\sqrt[3]{2}}{2}(|x|+|y|+|z|)\cdots\cdots(2)$$
for any complex numbers $x,y,z$
(2) This solution:
let
$1\ge|x|\ge |y|\ge |z|,y=bx,z=cx,1\ge|b|\ge |c|$,and let $\theta=|x|\le 1$
then
$$LHS=1+\theta|1+b+c|+\theta^2|b+c+bc|+\theta^3|bc|-\dfrac{\sqrt[3]{2}}{2}\theta(1+|b|+|c|)$$
then
$$LHS\ge 1+\theta^3(|1+b+c|+|b+c+bc|+|bc|)-\dfrac{\sqrt[3]{2}}{2}\theta(1+|b|+|c|)\ge1+\theta^3-\dfrac{3\sqrt[3]{2}}{2}\theta\ge 0$$
when $|x|,|y|,|z|\ge 1$ is obvious.Done!
But this $(1)$ I can't prove it.Thank you
| By switching all the signs, we can clearly assume that at least 2 variables are non-negative.
If $x,y,z\geq 0$, there is nothing to prove, so we only need to prove that:
$$\forall x,y,z\geq 0,\quad LHS=6+5|x+y-z|+2|xy-xz-yz|\geq 3(x+y+z)=RHS. $$
Now there are 3 cases. If $z\leq\frac{xy}{x+y}=\frac{1}{2}HM(x,y)$, then:
$$ LHS-RHS \geq 6+5(x+y-z)-3(x+y+z)\geq\frac{2}{x+y}\left((x+y)^2-4xy\right)\geq 0.$$
If $z\geq x+y = 2 AM(x,y)$, then:
$$ LHS-RHS \geq 6+2x+2y+2z(x+y-4)-2xy \geq \frac{3}{2}(x+y-2)^2 \geq 0.$$
If $\frac{1}{2}HM(x,y)=\frac{xy}{x+y}\leq z \leq x+y = 2 AM(x,y)$, then:
$$ LHS-RHS = 2\left((x+y-4)z+(x+y+3-xy)\right), $$
so we have to prove that $\frac{xy-x-y-3}{x+y-4}$ cannot belong to the interval $\left(\frac{xy}{x+y},x+y\right)$ for any couple $(x,y)$ of non-negative real numbers. If $x+y\geq 4$, proving $\frac{xy-x-y-3}{x+y-4}<\frac{xy}{x+y}$ is equivalent to prove $-3(x+y)-(x+y)^2+4xy<0$, that is trivial since the LHS is $\leq -12$. If $x+y<4$, proving $\frac{xy-x-y-3}{x+y-4}>x+y$ is equivalent to prove $xy+3(x+y)-3<(x+y)^2$, that is a consequence of $(x+y-2)^2>0$ since $xy\leq\frac{1}{4}(x+y)^2$. The proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How prove this $(x,y)H\binom{x}{y}\ge p(x^2+y^2)$ let $H_{2\times 2}$ is positive definite matrix,
show that:
There exist $p>0$, such
$$(x,y)H\binom{x}{y}\ge p(x^2+y^2)$$
My try: let
$$H=\begin{bmatrix}
a_{1}&a_{2}\\
b_{1}&b_{2}
\end{bmatrix}$$
and such $$|H|=a_{1}b_{2}-a_{2}b_{1}>0$$
so
$$(x,y)H\binom{x}{y}=(a_{1}x+b_{1}y,a_{2}x+b_{2}y)\binom{x}{y}=a_{1}x^2+b_{2}y^2+(a_{2}+b_{1})xy$$
then How find this $p$,such
$$a_{1}x^2+b_{2}y^2+(a_{2}+b_{1})xy\ge p(x^2+y^2)$$
so I let $t=\dfrac{x}{y}$,then
$$p\le\dfrac{a_{1}t^2+b_{2}+(a_{2}+b_{1})t}{t^2+1}=y$$
then we have
$$(a_{1}-y)t^2+(a_{2}+b_{1})t+b_{2}-y=0$$
so
$$\Delta=(a_{2}+b_{1})^2-4(a_{1}-y)(b_{2}-y)\ge 0$$
$$\Longrightarrow \dfrac{a_{1}+b_{2}}{2}-\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}}\le y\le \dfrac{a_{1}+b_{2}}{2}+\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}}$$
so,There are exist $p \in[I_{1},I_{2}]$
where $$I_{1}= \dfrac{a_{1}+b_{2}}{2}-\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}},I_{2}= \dfrac{a_{1}+b_{2}}{2}+\sqrt{\dfrac{(a_{2}+b_{1})^2}{4}+\dfrac{(a_{1}-b_{2})^2}{4}}$$
| Let $v=[x,y]^T$ and $H$ has the eigen decomposition $H=P^TDP$ where $D$ is the $2\times 2$ diagonal matrix with eigenvalues $\lambda_1,\lambda_2$ as its diagonal entries. Let $\lambda_1\geq \lambda_2$. Define $y=Pv=[y_1,y_2]^T$. Then
\begin{align}
v^THv=v^TP^TDPv=y^TDy=\lambda_1y_1^2+\lambda_2y_2^2\geq \lambda_2(y_1^2+y_2^2)=\lambda_2(x^2+y^2)
\end{align}
Convince yourself that $y_1^2+y_2^2=x^2+y^2$. Also check if you derive something in the same lines for any $N\times N$ positive definite matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding derivative by using L'Hospital's rule $$ f(x)= \begin{cases} \frac{1}{x\ln 2} - \frac{1}{2^x -1} \text { , if } x \neq 0 \\ \\ \frac{1}{2} \text{ , if }x=0 \end{cases}$$
I'm supposed to find derivative at point 0. I thought I should start by finding limits of both directions, showing that they equal by applying L'Hospital as many times as needed.
| I give it a try without any powerful theorems of calculus, because I think the question belongs to an introductory course on calculus.
We have
$\displaystyle \begin{aligned}f'(0) &= \lim_{x \to 0}\frac{f(x) - f(0)}{x}\\
&= \lim_{x \to 0}\dfrac{\dfrac{1}{x\log 2} - \dfrac{1}{2^{x} - 1} - \dfrac{1}{2}}{x}\\
&= \lim_{x \to 0}\dfrac{2(2^{x} - 1) - 2x\log 2 - x(2^{x} - 1)\log 2}{2x^{2}(2^{x} - 1)\log 2}\\
&= \lim_{x \to 0}\dfrac{2(2^{x} - 1) - 2x\log 2 - x(2^{x} - 1)\log 2}{2x^{3}\log 2}\dfrac{x}{2^{x} - 1}\\
&= \lim_{x \to 0}\dfrac{2^{x + 1} - x\log 2 - 2^{x}x\log 2 - 2}{2x^{3}(\log 2)^{2}}\\
&(\text{apply L'Hospital rule})\\
&= \lim_{x \to 0}\dfrac{2^{x + 1}\log 2 - \log 2 - (2^{x} + 2^{x}x\log 2)\log 2}{6x^{2}(\log 2)^{2}}\\
&= \lim_{x \to 0}\dfrac{2^{x}\log 2 - \log 2 - 2^{x}x(\log 2)^{2}}{6x^{2}(\log 2)^{2}}\\
&(\text{apply L'Hospital rule})\\
&= \lim_{x \to 0}\dfrac{2^{x}(\log 2)^{2} - (2^{x} + 2^{x}x\log 2)(\log 2)^{2}}{12x(\log 2)^{2}}\\
&= \lim_{x \to 0}\dfrac{ -2^{x}x(\log 2)^{3}}{12x(\log 2)^{2}}\\
&= -\lim_{x \to 0}2^{x}\cdot\frac{\log 2}{12}\\
&= -\frac{\log 2}{12}\end{aligned}$
I have used the fact that $$\lim_{x \to 0}\frac{2^{x} - 1}{x} = \log 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A formula for the power sums: $1^n+2^n+\dotsc +k^n=\,$? Is there explicit formula for the expression $1^n + 2^n + \dotsc + k^n\,$?
I know that for $n=1$ the explicit formula becomes $S=k(k+1)/2$ and for $n=3$ the formula becomes $S^2$. But what about general $n$?
I know there is a way using the Taylor expansion of $f(x)=1/(1-x)=1+x+x^2+\dotsc\;$, by differentiating it and then multiplying by $x$ and then differentiating again. Repeating this $n$ times, we get
$$\frac{d}{dx}(x\frac{d}{dx}(\dots x\frac{d}{dx}f(x))\dots )=1+2^nx^n+3^nx^n\dots.$$
Now do the same process but with the function $g(x)=x^{k+1}f(x)$. Then subtract them and we get $1+2^nx^n+\dots k^nx^n$. Because we have the explicit formulas $f(x)$ and $g(x)$ we can find the explicit formula by this process for arbitrary $n$. A big problem is that as $n$ grows, it is going take a lot of time finding the explicit formula. My question is therefore: are there other ways?
| There is a result which does not depend on the Bernouli numbers or the Eulerian numbers. I found it in my first year at university, so you can be sure it involves nothing too complicated.
The method is that of discrete calculus, first I define $ \Delta f(x) = f(x+1)-f(x) $ and $ \Sigma f(x) = \sum\limits_{k=0}^{x-1} f(k) $ as discrete equivalents of differentiation and integration respectively.
The result will follow from a discrete equivalent of integration by parts, which I first proove:
$ f(x)g(x) = \Sigma f(x+1)g(x+1) -\Sigma f(x)g(x) $
$ \hspace{30 pt} = \Sigma f(x+1)g(x+1) -\Sigma f(x)g(x) + \Sigma f(x)g(x+1) - \Sigma f(x)g(x+1) $
$ \hspace{30 pt} = \Sigma (f(x+1)-f(x))g(x+1) + \Sigma f(x)(g(x+1)-g(x)) $
$ \hspace{30 pt} = \Sigma (\Delta f(x))g(x+1) + \Sigma f(x) \Delta g(x) $
$ \therefore \space \Sigma f(x) \Delta g(x) = f(x)g(x) - \Sigma (\Delta f(x))g(x+1) $
The expression we want a formula for will be a polynomial one degree higher than the power n, so we choose f(x) and g(x) with consideration that $ \deg(f(x)g(x)) $ should be $ n+1 $.
For a dash of simplicity, we can get $ f(x) \Delta g(x) = x^{n} $ by choosing $ f(x) = x^{n-1} $ and $ g(x) = \frac{1}{2} x(x-1) $.
For the other terms in the equivalence we need:
$ f(x)g(x) = \frac{1}{2} (x-1) x^n $
$ (\Delta f(x))g(x+1) = ((x+1)^{n-1}-x^{n-1})(\frac{1}{2} x(x+1)) = \frac{1}{2} x(x+1) \sum\limits_{m = 0}^{n-2} \binom{n-1}{m} x^m $
$ \hspace{57 pt} = \frac{1}{2} x(x+1)((n-1) x^{n-2} + \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} x^m) $
$ \hspace{57 pt} = \frac{1}{2}(n-1) x^n + \frac{1}{2}(n-1) x^{n-1} + \frac{1}{2} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (x+1) x^{m+1} $
Finally, plugging these in and rearranging:
$ \sum\limits_{k=0}^{x-1} k^n = \frac{1}{2} (x-1) x^n - \sum\limits_{k=0}^{x-1} \left( \frac{1}{2}(n-1) k^n + \frac{1}{2}(n-1) k^{n-1} + \frac{1}{2} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (k+1) k^{m+1} \right) $
$ 2\sum\limits_{k=0}^{x-1} k^n = (x-1) x^n - (n-1) \sum\limits_{k=0}^{x-1} k^n - (n-1) \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (k+1) k^{m+1} $
$ (n+1) \sum\limits_{k=0}^{x-1} k^n = (x-1) x^n + (1-n) \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} (k+1) k^{m+1} $
$ \sum\limits_{k=0}^{x-1} k^n = \frac{x-1}{n+1} x^n + \frac{1-n}{n+1} \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} \frac{k+1}{n+1} k^{m+1} $
$ \therefore \sum\limits_{k=1}^{x} k^n = x^n + \sum\limits_{k=0}^{x-1} k^n = x^n + \frac{x-1}{n+1} x^n + \frac{1-n}{n+1} \sum\limits_{k=0}^{x-1} k^{n-1} - \sum\limits_{k=0}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} \frac{k+1}{n+1} k^{m+1} $
$ \hspace{31 pt} = \frac{x+n}{n+1} x^n + \frac{1-n}{n+1} \sum\limits_{k=1}^{x-1} k^{n-1} - \sum\limits_{k=1}^{x-1} \sum\limits_{m = 0}^{n-3} \binom{n-1}{m} \frac{k+1}{n+1} k^{m+1} $
Note that the power of k in the sums on the RHS does not exceed $ n-1 $.
Example with $ n = 3 $; the formula simplifies down to:
$$ \sum\limits_{k=1}^{x} k^3 = \frac{1}{4} (x^4 + 3x^3 -3 \sum\limits_{k=1}^{x-1} k^2 - \sum\limits_{k=1}^{x-1} k) $$
Which further simplifies to the correct polynomial. Only the partial sums for $k^2$ and $k$ need be known, and this formula will provide that of any higher degree.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/533772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 6,
"answer_id": 4
} |
Decomposition of an ideal as a product of two ideals How to show $$5\mathbb{Z}[\sqrt[3]{2}] = (5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1).$$
Firstly, I think that I can say that $$(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1)= (25,5(\sqrt[3]{2}+2),5((\sqrt[3]{2})^2+3\sqrt[3]{2}-1),5((\sqrt[3]{2})^2+\sqrt[3]{2})).$$ Since 5 divides each term, I have that $(5, \sqrt[3]{2}+2)(5, (\sqrt[3]{2})^2+3\sqrt[3]{2}-1) \subseteq (5)$. But now how do I show the other direction? Any help would be appreciated. Thanks in advance.
| Working in $\Bbb Z[x]/(x^3-2)$ I get
$$(5,x+2)(5,x^2+3x-1)=\big(25,5(x^2+3x-1),5(x+2),(x+2)(x^2+3x-1)\big).$$
Then $(x+2)(x^2+3x-1)=x^3+5x^2+5x-2=5x(x+1)$. Hence the above ideal is
$$=(5)(5,x^2+3x-1,x+2,x^2+x).$$
Use the Euclidean division algorithm (which applies here) to simplify the right-hand factor. For example $x^2+3x-1\equiv x-1$ mod $x+2$ so we may replace $x^2+3x-1$ with $x-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/536611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\left(x^{2010}+1\right)\left(1+x^2+x^4+x^6+.......+x^{2008}\right)=2010x^{2009}$ Solve for $x$
$\left(x^{2010}+1\right)\left(1+x^2+x^4+x^6+.......+x^{2008}\right)=2010x^{2009}$
solution should be by hand
| First it is clear that $x \geq 0$.
By AM-GM inequality we have:
$$x^{2010}+1 \geq 2 x^{1005}$$
$$1+x^2+x^4+x^6+.......+x^{2008} \geq 1005 \sqrt[1005]{1x^2...x^{2008}}=1005 x^{\frac{2+4+6+...+2008}{1005}}=1005x^{1004}$$
Multiplying you get
$$\left(x^{2010}+1\right)\left(1+x^2+x^4+x^6+.......+x^{2008}\right) \geq2010x^{2009}$$
You get equality if and only if you have equality in the above. Thus
$$x^{2010} =1$$
and
$$1=x^2=x^4=...=x^{2008}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the next step in the parametric arc length problem? Find the length of the curve $x = \frac{1}{3}(2t^2 +1)^\frac{3}{2}$, $y = \frac{1}{2}t^2$ from 0 to 4.
I believe the arc length formula for a parametric equation is $\int_a^b \sqrt{ (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } dt $
So, I'll need to get the derivative of both x:
$$ \frac{dx}{dt} = \frac{3}{2} (\frac{1}{3} )(2t^2 +1)^\frac{1}{2} $$
Simplified:
$$ \frac{1}{2} (2t^2 + 1)^\frac{1}{2} * (4t) $$
Square it (did I square it correctly?):
$$ 4t^2 (2t^2 + 1) $$
Multiply through:
$$ 8t^4 + 4t^2 $$
Derivative of y:
$$ y = t $$
Square it:
$$ y = t^2 $$
Plug it in:
$$ \int_0^4 \sqrt{ 8t^4 + 4t^2 + t^2 } $$
Add like terms
$$ \int_0^4 \sqrt{8t^4 + 5t^2} $$
Pull out $t^2$
$$ t \int_0^4 \sqrt{8t^2 + 5} $$
If everything is correct so far, what do I do next?
| Using the method of $u$-substitution let $u=8t^2+5$, then ${1\over 16}du=tdt$. When $t=0$, $u=5$ and when $t=4$, $u=133$. Thus $$\int_0^4 t\sqrt{8t^2+5}dt={1\over 16}\int_5^{133}\sqrt{u}du.$$ Integrating the right-hand side we obtain
$$\left.\left({1\over 24}u^{3\over 2}\right)\right|_5^{133}.$$ Which becomes $${1\over {24}}(133\sqrt{133}-5\sqrt5)\approx63.44.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/542202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\
x^2=b^2+(2b-1)y+y^2\cdots\cdots
\end{cases}$$
$(1)+(2)$
then
$$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$
so
$$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$
we take $(3)$ in $(2)$,we have
$$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so
$$(2ab-a+b)=0$$
or
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
if
$$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$
and if
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
I don't prove
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
| This is not a solution, but brute force can be used to remove the radicals. Let $A=y^2-x$ and $B=x^2+y$. We have
$$\sqrt{AB}-y\sqrt{A}-x\sqrt{B}+xy=y$$
Isolating $\sqrt{AB}$ and squaring both sides:
$$\sqrt{AB}=y\sqrt{A}+x\sqrt{B}+y(1-x)\quad(1)$$
$$AB=y^2A+x^2B+y^2(1-x)^2+2xy\sqrt{AB}+2y^2(1-x)\sqrt{A}+2xy(1-x)\sqrt{B}$$
(1) allows us to remove $\sqrt{AB}$. We do this and also recall what $A$ and $B$ equal.
$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy\left(y\sqrt{A}+x\sqrt{B}+y(1-x)\right)+2y^2(1-x)\sqrt{A}+2xy(1-x)\sqrt{B}$$
Group $\sqrt{A}$ and $\sqrt{B}$ terms, then rearrange a bit:
$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy^2(1-x)+2y^2\sqrt{A}+2xy\sqrt{B}$$
$$x^2y^2+y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2xy^2+x^2y^2+2xy^2-2x^2y^2+2y\left(y\sqrt{A}+x\sqrt{B}\right)$$
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y\left(y\sqrt{A}+x\sqrt{B}\right)$$
(1) allows us to sub out the quantity in parentheses:
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y\left(y(x-1)+\sqrt{AB}\right)$$
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y^2(x-1)+2y\sqrt{AB}$$
$$y^3-x^3-xy=y^4+xy^2+x^4+x^2y-y^2-2x^2y^2+2y\sqrt{AB}$$
$$y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2=2y\sqrt{AB}$$
Squaring both sides, we've reached a goal of no longer having radicals.
$$(y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2)^2=4y^2(y^2-x)(x^2+y)$$
I had a CAS expand this, move it all to one side, and then, as expected, $(x+y)$ factors out of it (twice).
$$(x+y)^2 p(x,y)=0$$
where $$p(x,y)=x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2$$
is a monster. It would be sufficient to show that $p(x,y)$ is never $0$ in the region of the plane where both $\sqrt{A}$ and $\sqrt{B}$ are defined aside from points along $x+y=0$ (like $(0,0)$). This is a pretty messy polynomial, but at least it's a polynomial.
EDIT: This approach seems to be useless; a CAS plot of the zero set of $p$ has several components, all of which are in the region where $\sqrt{A}$ and $\sqrt{B}$ are defined. They must be extraneous solutions from the squaring that was done twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/543607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 7,
"answer_id": 4
} |
How prove this inequality $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{3^n}\right)\ge\frac{14}{25}$
show that
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$
My try:
I only prove following not strong inequality:
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{1}{2}$$
proof: use Bernoulli inequality
$$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge 1+x_{1}+x_{2}+\cdots +x_{n},$$where $x_{i}\ge -1$
so
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge1-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^n}\right)$$
so
$$LHS\ge1-\sum_{n=1}^{\infty}\dfrac{1}{3^n}=1-\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}$$
But for $(1)$,I can't prove it,Thank you
| You can use the same idea, but first pull out a couple of factors from the product:
$$(2/3) (8/9) (26/27) \left(1-\sum_{j=4}^\infty 1/3^j\right)=11024/19683>14/25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
How find this sum $\sum_{n=1}^{\infty}\frac{1}{n^2-n+a}$ Today Question
if $a>\dfrac{1}{4}$, show that
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2-n+a}=\dfrac{\pi}{\sqrt{4a-1}}\cdot\dfrac{e^{\pi\sqrt{4a-1}}-1}{e^{\pi\sqrt{4a-1}}+1}\tag{1}$$
I have konw that solution:
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{\pi}{2}\dfrac{e^{2\pi}+1}{e^{2\pi}-1}-\dfrac{1}{2}$$
solution:
note
$$e^{ax}=\dfrac{\pi^{2\pi a}-1}{\pi}\left(\dfrac{1}{2a}+\sum_{n=1}^{\infty}\dfrac{a\cos{nx}-n\sin{x}}{n^2+a^2}\right)$$
let $a=1,x=0$,then we have
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{\pi}{2}\dfrac{e^{2\pi}+1}{e^{2\pi}-1}-\dfrac{1}{2}$$
But for $(1)$ I have
$$\sum_{n=1}^{\infty}\dfrac{1}{(n-\frac{1}{2})^2+a-\frac{1}{4}}$$
Then I can't ,Thank you for your help.
| First thing you can try is
$$
\sum_{n=1}^{\infty} \frac{1}{n^2+4a-1}=\sum_{n=1}^{\infty} \frac{1}{(2n)^2+4a-1}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2+4a-1}
$$
You can find
$$
\sum_{n=1}^{\infty} \frac{1}{n^2+4a-1},
$$
and
$$
\sum_{n=1}^{\infty} \frac{1}{(2n)^2+4a-1}$$
using the identity.
By the way, the formula you wrote for Fourier series does not look right, maybe check for typos.
The right formula that you should obtain is
$$
\sum_{n=1}^{\infty} \frac{1}{n^2+a^2}=\frac{1}{2a}\left(\pi\frac{e^{2\pi a}+1}{e^{2\pi a}-1}-\frac{1}{a}\right)
$$
The first sum is:
$$
\frac{1}{2\sqrt{4a-1}}\left(\pi\frac{e^{2\pi \sqrt{4a-1}}+1}{e^{2\pi \sqrt{4a-1}}-1}-\frac{1}{\sqrt{4a-1}}\right)
$$
The second sum is:
$$
\frac{1}{4}\frac{1}{\sqrt{4a-1}}\left(\pi\frac{e^{\pi \sqrt{4a-1}}+1}{e^{\pi \sqrt{4a-1}}-1}-\frac{2}{\sqrt{4a-1}}\right)
$$
Hence, the final result is:
$$
\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2+4a-1}=\frac{1}{4}\frac{1}{\sqrt{4a-1}}\left(\pi\frac{e^{\pi \sqrt{4a-1}}-1}{e^{\pi \sqrt{4a-1}}+1}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Possible Class equation for a group Determine the possible class equation for a group of order 21?
Until now I have found the following:
$1+3+3+7+7$
$1+1+1+3+3+3+9$
$1+1+1+1+1+1+1+7+7$
$1+1+1+1+1+1+1+1+1+3+3+3+3$
$1+1+1+\cdots +1 \ (21 \ \text{times})$
Is there any way to eliminate the choices from this equation? More importantly, how would we know that this is a complete list (Until now my attempt has just been guess and check after I found possible occurences of 1's) Is there any easier way to determine the class equation?
| Given that |G|=21.Obviously,second,third and forth can be not possible class equations . Because from 1+1+1+1+1+1+1+7+7 ,we see that |Z(G)|=7 and |G/Z(G)|=3, which is prime.So G is abelien.Then we must have G=Z(G),which implies that |G|=|Z(G)| but |Z(G)|=7.
Again from 1+1+1+3+3+3+9,we see that |Z(G)|=3 and |G/Z(G)|=7, which is prime.So G is abelian.So we must have G=Z(G) and |G|=|Z(G)| but |Z(G)|=3.finally from 1+1+1+1+1+1+1+1+1+3+3+3+3,we see that |Z(G)|=9, also we know Z(G) is a subgroup of G and |G| is divisible by |Z(G)| but |Z(G)|=9 ,|G| is not divisible by |Z(G)|
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/549593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How prove this inequality $\sum_{i=1}^{n}\frac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1}$ show that
$$\sum_{i=1}^{n}\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1}$$
My try:
$$x\in(n-1,n)\Longrightarrow \sqrt{x}>\sqrt{n-1}$$
so
$$\int_{n-1}^{n}\sqrt{x}dx>\sqrt{n-1}$$
so
$$\dfrac{2}{3}n^{\frac{3}{2}}=\int_{0}^{n}\sqrt{x}dx>\sqrt{n-1}+\sqrt{n-2}+\cdots+1$$
so
$$1+\sqrt{2}+\cdots+\sqrt{n}<\sqrt{n}+\dfrac{2}{3}n^{\frac{3}{2}}$$
But this not use.Thank you
| Try induction. $P(1)$ is obvious.
$P(n) \Rightarrow P(n_+1)$.
$$\sum_{i=1}^{n+1}\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{i}}{i^2}\le\sqrt{2n-1} +\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2}$$
To complete the proof we need to show
$$\sqrt{2n-1} +\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2} \leq \sqrt{2n+1} \Leftrightarrow \\
\dfrac{\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1}}{(n+1)^2} \leq \sqrt{2n+1}-\sqrt{2n-1} \Leftrightarrow \\
\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1} \leq (n+1)^2[ \sqrt{2n+1}-\sqrt{2n-1}] = \frac{2(n+1)^2}{\sqrt{2n-1}+\sqrt{2n+1}}
$$
Now, by your inequality we have
$$\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n+1} \leq \sqrt{n+1}+\frac{2}{3}(n+1)^{\frac{3}{2}}$$
and the inequality
$$\sqrt{n+1}+\frac{2}{3}(n+1)^{\frac{3}{2}} \leq \frac{2(n+1)^2}{2\sqrt{2n+1}} \leq \frac{2(n+1)^2}{\sqrt{2n-1}+\sqrt{2n+1}} $$
is easy to prove and completes the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/551666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convolution of a Function I am calculating $f*f$, where $f(x) = \frac{1}{a^2 + x^2}$. So I need to calculate
$$ (f*f)(x) = \int^\infty_{-\infty} \frac{1}{(a^2+y^2)(a^2+(x-y)^2)} dy$$
I guess the answer is $\frac{2\pi}{a(4a^2+x^2)}$. Also, we can assume $a > 0$. I have tried really hard on this problem, and I am at a complete loss. I appreciate your help tremendously!
| Things can get easier a lot if you can use Fourier transform. Note that:
$\mathcal{F}(\frac{1}{x^2+a^2})=\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|\omega|}$
$\mathcal{F}(\frac{1}{x^2+a^2}*\frac{1}{x^2+a^2})=(\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|\omega|})(\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|\omega|})=\frac{1}{a^2}\frac{\pi}{2}e^{-2a|\omega|}$
$\frac{1}{x^2+a^2}*\frac{1}{x^2+a^2} = \mathcal{F^{-1}}(\frac{1}{a^2}\frac{\pi}{2}e^{-2a|\omega|})=\frac{\sqrt{2\pi}}{a(4a^2+x^2)}$
Even if you cannot use Fourier transform in your solution may be looking at the proofs related to it help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$ Is it true that for $x\in[0,2\pi]$ we have
$$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}$$
How can I prove it?
For other intervals what is the value of above series if is convergent?
| See the results posted here, where I show that
$$\sum_{n=-\infty}^{\infty} \frac{\sin^2{a n}}{n^2} = \pi a$$
when $a \in (0,\pi)$. Now, use the fact that
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
and let $S$ be the sum in question. Then
$$\frac{\pi^2}{6} - S = \sum_{n=1}^{\infty} \frac{1-\cos{n x}}{n^2} = 2 \sum_{n=1}^{\infty} \frac{\sin^2{n x/2}}{n^2}$$
Rewrite the last sum as
$$2 \sum_{n=1}^{\infty} \frac{\sin^2{n x/2}}{n^2} = \sum_{n=-\infty}^{\infty} \frac{\sin^2{n x/2}}{n^2} - \left ( \frac{x}{2} \right )^2 = \pi \frac{x}{2} - \frac{x^2}{4}$$
Then
$$\frac{\pi^2}{6} - S = \pi \frac{x}{2} - \frac{x^2}{4} \implies S = \frac{x^2}{4} - \pi \frac{x}{2} + \frac{\pi^2}{6}$$
as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
What is the maximum value of $4(\sin x)^2 + 3(\cos x)^2$ The question is: What is the maximum value of: $4\sin^2\theta + 3\cos^2\theta$
This is the way I did it:
$4\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3$
The max value of $\sin^2\theta$ is $1$, so the answer must be $4$. However my book says that the answer is $5$. Where did I go wrong?
| Though this has been asked and answered, I'll give an alternate solution That avoids messy derivative calculations. Notice that $4\sin^2{x}+3\cos^2{x}$ is very close to $4(\sin^2{x}+\cos^2{x})=4(1)=4$. So we fix it so we get what we want.
$$
4\sin^2{x}+3\cos^2{x}=4\sin^2{x}+3\cos^2{x}+\big(\cos^2{x}-\cos^2{x}\big)
$$
Which gives us
$$
4(\sin^2{x}+\cos^2{x})-\cos^2{x}=4-\cos^2{x}=4+(-\cos^2{x})
$$
So the maximum of our original function will occur whenever $-\cos^2{x}$ has a minimum (since we have added it to our original function). Since the minimum of $-\cos^2{x}$ is $0$, the maximum value of our original function $4\sin^2{x}+3\cos^2{x}$ is $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding $x+y+z$ If $x+1/x = y$, $y+1/y=z$, $z+1/z=x$, then find $x+y+z$. Is there any way to do so without taking out the values of $x$, $y$, $z$?
Please help,
| Let us first write down the equations:
$$x+\frac{1}{x}=y, \; y + \frac{1}{y}=z,\; z+\frac{1}{z}=x$$
We now add these equations together:
$$x+\frac{1}{x}+ y + \frac{1}{y}+z+\frac{1}{z}=x +y+z$$
So we can see cancelling out $x+y+z$ gives us something, which was our motivation in adding them up:
$$\frac{1}{x} + \frac{1}{y}+\frac{1}{z}=0$$
Expanding, we get $xy+yz+xz=0$.
Here, we can note the identity $(x+y+z)^2 = x^2 +y^2+z^2 + 2(xy+yz+zx)$. Hence, if we can find $x^2+y^2+z^2$, we are done.
This can be done by realizing that by multiplying $x$ on both sides of $x+\frac{1}{x} = y$, gives us $x^2+1=xy$. Doing this with other equations and adding them all, we have:
$$x^2+1+y^2+1+z^2+1=xy+yz+xz$$
Substituing $xy+yz+xz=0$, we have $x^2 +y^2+z^2 = -3$. So,
$$x+y+z=\pm\sqrt{x^2 +y^2+z^2 + 2(xy+yz+zx)} = \pm\sqrt{-3+2(0)} = \pm\sqrt{-3} = \pm\sqrt{3}i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
the max value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$ What is the max value of the term $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$
If $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6$,
$a,b,c,d \in \mathbb{R}$
I tried to get from the equation to the term but it's too complicated.
Any idea?
Thanks.
| Let $w=\frac{a}{b}, x=\frac{b}{c}, y=\frac{c}{d}, z=\frac{d}{a}$, so that $wxyz=1, w+x+y+z=6$, then $$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=wx+xy+yz+zw=(w+y)(x+z) \leq \frac{((w+y)+(x+z))^2}{4}=9$$
(Note $st \leq \frac{(s+t)^2}{4} \Leftrightarrow 0 \leq (s-t)^2$)
Equality is achieved when $w+y=x+z$, so $w+y=x+z=3, wxyz=1$. We can achieve this for example by $y=x=\frac{3-\sqrt{5}}{2}, w=z=\frac{3+\sqrt{5}}{2}$, $(a, b, c, d)=(1, \frac{3-\sqrt{5}}{2}, 1, \frac{3+\sqrt{5}}{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/553595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find a matrix $X$ such that $X+X^2+X^3 = \begin{bmatrix} 1&2005\\ 2006&1 \end{bmatrix}$?
Find a matrix $X \in M_{2}(\mathbb Z)$ such that $$X+X^2+X^3=\begin{bmatrix} 1&2005\\ 2006&1\end{bmatrix}$$
My try:
Let
$$X=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}$$
where $a,b,c,d\in Z$
then
$$X^2=\begin{bmatrix}
a^2+bc&ab+bd\\
ac+cd&bc+d^2
\end{bmatrix}$$
then $$X^3=\begin{bmatrix}
a^3+abc+abc+bcd&a^2b+b^2c+abd+bd^2\\
a^2c+acd+bc^2+cd^2&abc+bcd+bcd+d^3
\end{bmatrix}$$
so
$$X+X^2+X^3=\begin{bmatrix}
a^3+2abc+bcd+a^2+bc+a&a^2b+b^2c+abd+bd^2+ab+bd+b\\
a^2c+acd+bc^2+cd^2+ac+cd+c&abc+2bcd+d^3+bc+d^2+d
\end{bmatrix}$$
then we have
$$\begin{cases}
a^3+2abc+bcd+a^2+bc+a=1\\
a^2b+b^2c+abd+bd^2+ab+bd+b=2005\\
a^2c+acd+bc^2+cd^2+ac+cd+c=2006\\
abc+2bcd+d^3+bc+d^2+d=1
\end{cases}$$
Note $2005=5\cdot 401 $ is prime number,so
$$b(a^2+bc+ad+d^2+a+d+1)=2005$$
we can $b=\pm 1$ ,or $b=\pm 2005$ ,or $b=\pm 5$ or $b=\pm 401$
and note $2006=2\times 1003$
then $c=\pm 2$ ,or $c=\pm 1003$or $c=\pm 1$,or $c=\pm 2006$
so following is very ugly. But I can't.Thank you ,maybe this problem have other nice methods,Thank you
| The Elkies's argument: -more generally- let $K$ be a field, $A\in M_k(K)$ and $P\in K[x]$ of degree $d$; we consider the equation $(*)$ $P(X)=A$. If $(*)$ has a solution $X\in K$, then necessarily, the polynomial of degree $kd$: $\det(A-P(x)I_k)$ has a factor of degree $\leq k$ in $K[x]$.
This condition does not suffice as we can see reading the user1551's argument with $K=GF(2)$.
On the other hand, there is a solution of the initial problem in $GF(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Assume that there are $n+1$ vectors of dimension $n$ and all angles of any two vectors equal, is the cosine of the angle $-\frac{1}{n}$? When $n=2$, the $3$ vectors construct a regular triangle, and the consine of the angle is $-\frac{1}{2}$
When $n=3$, the $4$ vectors construct a regular tetrahedon, and the consine of the angle is $-\frac{1}{3}$
Is it true for any $n$? I guess so, but how to prove?
| You can do this without needing any specific assignments of coordinates except for having one node at the origin. This is how you do it.
Let one node be at the origin (obviously). Now, you have a set of $n$ vectors pointing to each of the other nodes. We call these $\mathbf{v}_i$. As all distances between nodes are of the same length, we have that $|\mathbf{v}_i|=|\mathbf{v}_j-\mathbf{v}_i|=C$ for $j\neq i$. Let the centre be $\mathbf{c}=\frac1{n+1}\sum_{i=1}^n \mathbf{v}_i$.
Now, we consider the centre, the origin node, and one other node - without loss of generality, we use $\mathbf{v}_1$. These three points form a triangle, and from the Law of Cosines, we have
$$
\mathbf{c}\cdot(\mathbf{c}-\mathbf{v}_1) = |\mathbf{c}||\mathbf{c}-\mathbf{v}_1|\cos \theta
$$
Now, noting the symmetry of the system, we have
$$
\mathbf{c}\cdot\mathbf{v}_1 = \frac{C^2+(n-1)\mathbf{v_2}\cdot\mathbf{v_1}}{n+1}
$$
Each face forms an equilateral triangle, so $\mathbf{v}_2\cdot\mathbf{v}_1=\frac{C^2}2$. This allows us to write
$$
\mathbf{c}\cdot\mathbf{v}_1 = \frac{C^2}2
$$
Similarly, we have
$$
|\mathbf{c}|^2 = n\frac{C^2+(n-1)\mathbf{v}_2\cdot\mathbf{v}_1}{(n+1)^2}=\frac{nC^2}{2(n+1)}
$$
Substituting these into our equation, and noting that $|\mathbf{c}-\mathbf{v}_1|=|\mathbf{c}|=\frac{C}{\sqrt{2}}$, we have
$$
\cos\theta = \frac{\frac{nC^2}{2(n+1)}-\frac{C^2}2}{\frac{nC^2}{2(n+1)}}=\frac{\frac{n}{n+1}-1}{\frac{n}{n+1}} = \frac{\ \frac{-1}{n+1}\ }{\frac{n}{n+1}} = -\frac1n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find all real zeros of $f(x)=2x^3+10x^2+5x-12$ Hey guys I'm having a little trouble with one problem:
Find all real zeros of
$$f(x)=2x^3+10x^2+5x-12.$$
I got $x=-4,(2x^2+2x-3)$. I'm just having trouble using the quadratic formula to get the rest.
| $2x^3+10x^2+5x-12 = 0 \implies (x+4)(2x^2 + 2x - 3) = 0$
That's where you got to; you found one of the roots to be $-4$. The next step is simple, all you need to do is solve the quadratic equation to find the other two roots. The easiest way to do it is to plug values into the Quadratic Formula (click for video).
The solution of $ax^2+bx+c=0$ is
$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$
But any monkey's uncle can plug in values to a formula and solve. I'm going to show you a method called Completing the Square:
$$2x^2 + 2x - 3 = 0$$
$$\Rightarrow 2x^2 + 2x = 3$$
$$\Rightarrow x^2 + x = \frac{3}{2}$$
$$\Rightarrow x^2 + x + \left(\frac{1}{2}\right)^2 = \frac{3 \cdot 2}{2 \cdot 2} + \left(\frac{1}{2}\right)^2$$
$$\Rightarrow \left(x + \frac{1}{2}\right)^2 = \frac{6}{4} + \frac{1}{4}$$
$$\Rightarrow x+\frac{1}{2} = \pm\sqrt{\frac{7}{4}}$$
$$\Rightarrow x = \frac{1}{2}(\pm\sqrt 7 - 1)$$
So, the remaining roots are approximately $-1.82287565553$ and $0.82287565553$.
Roots of $f(x)=2x^3+10x^2+5x−12$:
The roots of this cubic are $-4, \frac{1}{2}(\sqrt 7 - 1), \frac{1}{2}(- \sqrt 7 - 1)$
Since your working with cubic equations, you might find it helpful to know that one can find the roots of a cubic equations by using the Cubic Formula!
A direct, uncomplicated yet unpopular form of it is as follows:
The solution of $ax^3+bx^2+cx+d=0$ is
$$x = \sqrt[3]{\left({-b^3\over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)+
\sqrt{\left({-b^3 \over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)^2
+\left({c\over 3a}-{b^2 \over 9a^2}\right)^3}} \\
+\sqrt[3]{\left({-b^3\over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)-
\sqrt{\left({-b^3 \over 27a^3}+{bc\over 6a^2}-{d\over 2a}\right)^2
+\left({c\over 3a}-{b^2 \over 9a^2}\right)^3}}
-{b\over 3a}
$$
This formula was by Gerolamo Cardano, and isn't totally complete. Infact no one recommends using it as it has certain problems with 3 real roots. But it's fun to see it this way.
Better ways are listed on the wikipedia page for Cubic Functions.
But as a rule of thumb, the best method has to be "Factor, Factor, Factor"
At times it may also be useful to know a bit more about the roots of a cubic.
Let $\mathcal A$, $\mathcal B$ and $\mathcal C$ be the roots of a basic cubic function.
$$\Rightarrow\mathcal{ax^3 + bx^2 + cx + d = a(x-A)(x-B)(x-C)} $$
$$\Rightarrow\mathcal{ax^3 + bx^2 + cx + d = ax^3 - a(A+B+C)x^2 + a(AB + BC + AC)x - a(A\cdot B\cdot C)}$$
If you are to compare the coefficients of $x$ on both sides,
You can reach three vital relations:
*
*$\mathcal{A + B + C }= \frac{-b}{a}$
*$\mathcal{A\cdot B\cdot C} = \frac{-d}{a}$
*$\mathcal{AB + BC + AC} = \frac{c}{a}$
Hope this helps :D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/556571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
| Writing
$$\frac{1}{n(n+1)(n+2)} = \frac{1}{n+1}\left(\frac{1}{2n} - \frac{1}{2(n+2)}\right) = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)},$$
we see that the sum telescopes to
$$\frac{1}{2(1)(2)} = \frac{1}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 10,
"answer_id": 2
} |
which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$ Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
| Let $A = (\sqrt{7}-\sqrt{6}) - (\sqrt{6}-\sqrt{5} )$
This comes to comes to $A = (\sqrt{7}+\sqrt{5})-2\sqrt{6}$
The square of the first is $12+2\sqrt{35}$, the square of the second is $12+2\sqrt{36}$
Since $35 \lt 36$, then $A \lt 0$, hence $\sqrt{6}-\sqrt{5}$ is bigger.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 15,
"answer_id": 5
} |
Integral of a complex function over a circle I've got the problem. I was to compute:
$\int_0^{2 \pi} \frac{1- cos(n\phi)}{1 - cos(\phi)} d\phi$
using analytic functions methods. My attempt: I came to this:
On a unit circle, we have $z= e^{i\phi}$, where $\phi \in [0, 2\pi)$. Then, we obtain $log(z) = i \phi \Rightarrow dz = ie^{log(z)} d\phi = iz d\phi \Rightarrow d\phi = \frac{1}{iz} dz$.
So my integral is equal to (C is a unit circle):
$\int_C \frac{1}{iz} \frac{1- cos(n\frac{log,z}{i})}{1 - cos(\frac{logz}{i})} dz = \int_C \frac{1}{iz} \frac{1- \frac{e^{nlogz} + e^{-nlogz}}{2}}{1 - \frac{e^{logz} + e^{-logz}}{2}} dz = \int_C \frac{1}{iz} \frac{1- \frac{z^n +z^{-n} }{2}}{1 - \frac{z + z^{-1}}{2}} dz = \int_C \frac{z^{-n}(z^n -1)^2}{i(z-1)^2} dz$
Here I got stuck. How to compute $\int_C \frac{z^{-n}(z^n -1)^2}{i(z-1)^2} dz$? Can you show me? I will be grateful.
| Hint: You now have the fraction
$$
\frac{z^{-n}(z^n - 1)^2}{i(z-1)^2}
$$
Note, however, that $\frac{z^{n}-1}{z-1} = 1 + z + z^2 + \cdots z^{n-1}$. Use this to simplify the above fraction to one without singularities on the unit disk.
Second hint: you now have
$$
\frac 1i \int_C \frac{(1 + z + z^2 + \cdots z^{n-1})^2}{z^n}
$$
By Cauchy's integration formula, we have
$$
\begin{align}
\frac 1i \int_C \frac{(1 + z + z^2 + \cdots z^{n-1})^2}{z^n} &=
\frac{2 \pi }{(n-1)!} \left[\frac{(n-1)!}{2 \pi i} \int_C \frac{(1 + z + z^2 + \cdots z^{n-1})^2}{z^{(n-1)+1}}\right]\\
&= \frac{2 \pi }{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}\left[(1 + z + z^2 + \cdots z^{n-1})^2\right]_{z=0}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find Weight for minimum Manhattan Distance Let's say, I have three points $(1, 4)$, $(4, 3)$ and $(5, 2)$. I need to find weight $w_1$ and $w_2$ so that the point $(1, 4)$ be the centroid of the points in Manhattan Distance. Actually, the point $(1, 4)$ must have the minimum total distance from others.
How can I weight those X and Y coordinates to minimalize the total distance toward point $(1, 4)$?
| Rescale $x$-coordinates with $(2/9)$ and $y$-coordinates with $(8/5)$, giving:
$$
\left(\frac{2}{9}\times 4,\frac{8}{5} \times 3 \right) = \left( \frac{8}{9} , \frac{24}{5} \right) \qquad ; \qquad \left(\frac{2}{9}\times 5,\frac{8}{5} \times 2 \right) = \left( \frac{10}{9} , \frac{16}{5} \right)
$$
Centroid = $(9/9,20/5)$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/566570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How find this $g(x)=f(x)f(1-x)$ maximum and minimum Question:
Let $f: \mathbb{R}\to\mathbb{R}$ is a function such that $$f( \cot x ) = \cos 2x+\sin 2x $$ for all $0 < x < \pi$.
Define $$g(x) = f(x) f(1-x) , -1 \leq x \leq 1$$ Find the maximum and minimum values of $g$ on the closed interval $[-1, 1].$
My try: since
$$\sin{2x}=\dfrac{2\tan{x}}{1+\tan^2{x}},\cos{2x}=\dfrac{1-\tan^2{x}}{1+\tan^2{x}}$$so
$$\sin{(2x)}+\cos{(2x)}=\dfrac{1-\tan^2{x}+2\tan{x}}{1+\tan^2{x}}$$
so
$$f(x)=\dfrac{x^2+2x-1}{x^2+1}$$
then
$$g(x)=f(x)f(1-x)=\dfrac{x^2+2x-1}{x^2+1}\cdot\dfrac{(1-x)^2+2(1-x)-1}{(1-x)^2+1},-1\le x\le 1$$
Then I can't.Thank you for your help
| let $a=x,b=1-x，f\left(a\right)=1+\dfrac{2a-2}{a^2+1},g\left(x\right)=f\left(a\right)*f\left(b\right)=1+\dfrac{\left(2a-2\right)\left(2b-2\right)}{\left(1+a^2\right)\left(1+b^2\right)}+\dfrac{\left(2a-2\right)}{\left(1+a^2\right)}+\dfrac{\left(2b-2\right)}{\left(1+b^2\right)}=1+\dfrac{4-4\left(a+b\right)+4ab}{a^2b^2+a^2+b^2+1}+\dfrac{-2-2b^2-2-2a^2+2\left(a+b\right)+2ab\left(a+b\right)}{a^2b^2+a^2+b^2+1}=1+\dfrac{8ab-4}{\left(ab\right)^2-2ab+3}=1+\dfrac{4-8\left(1-ab\right)}{\left(1-ab\right)^2+2}=1+4\times\dfrac{1-2u}{u^2+1},u=1-ab=1-a+a^2> 0$
$h\left(u\right)=\dfrac{1-2u}{u^2+1},3\ge u>0 $
$h'\left(u\right)=\dfrac{u^2-u-1}{\left(u^2+1\right)^2}=0, u=\dfrac{1+\sqrt{5}}{2}$, is min,
max is $u$ get min which is $u=\dfrac{3}{4}$
edit: there is more simple method to find max and min of $h\left(u\right)$
$u\ge \dfrac{3}{4} \implies 2u-1 \ge 0, t=2u-1,h\left(u\right)=-\dfrac{4t}{t^2+2t+5}=-\dfrac{4}{t+\dfrac{5}{t}+2} $
let $q\left(t\right)=t+\dfrac{5}{t}, \dfrac{1}{2}\le t \le 5 , q\left(t\right) \ge 2\sqrt{5}$ when $t=\sqrt{5}$ ,it is min.
for max, we only need to check two bounds: $q\left(\dfrac{1}{2}\right)=10\dfrac{1}{2},q\left(5\right)=6$, so max is $q=10\dfrac{1}{2}, t=\dfrac{1}{2}$.
now these are all high school methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/568194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How prove this $S_{n}\neq 0$ let $n$ is postive integer numbers,and let
$$S_{n}=\sin{1}-\sin{4}+\cdots+(-1)^{n-1}\sin{(3n-2)}$$
show that
$$S_{n}\neq 0, \forall n\in N^{+}$$
My try: maybe this problem use
$$2\sin{x}\sin{y}=\cos{(x-y)}-\cos{(x+y)}$$
\begin{align*}2\sin{\dfrac{3}{2}}S_{n}&=2\sin{1}\sin{\dfrac{3}{2}}-2\sin{4}\sin{\dfrac{3}{2}}+\cdots+2(-1)^{n-1}\sin{(3n-2)}\sin{\dfrac{3}{2}}\\
&=\cos{\dfrac{1}{2}}-\cos{\dfrac{5}{2}}-\cos{\dfrac{5}{2}}+\cos{\dfrac{11}{2}}+\cdots
\end{align*}
I think this is nice problem, Thank you
|
go with product to sum
then calculate sum of $n$th of sigma
at then end you have $6n-7=+-1+4k\pi$
so
$6k-8=4k\pi$ divide by 4 -->$k$ is not an integer
$6k-6=4k\pi$ divide by 4 -->$k$ is not an integer
so $s(n)$ is not zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof: Divisible by 15 I have to proof that $16^m - 1$ is divisible by $15$. Is my following proof correct?
$$\begin{align}
16^m - 1=&\frac{16^{m+1}}{16}-1\\
=&\frac{16^{m+1}-16}{16} \\
=&(16^{m+1}-16)\cdot\frac{1}{16} \\
=&\underbrace{(16^{m+1}-16)}\cdot\frac{1}{15(1+1/15)} \\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1+1/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{16/15}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{1}\cdot\frac{15}{16}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underbrace{a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{15}{16}}\cdot\frac{1}{15}\\
=&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot\frac{1}{15}\\
\end{align}$$
$$\therefore \boxed{16^m - 1=\frac{b}{15}}$$
Or is this the wrong way and I have to do it with mathematical induction?
| Hint:
$$16^{n+1}-1=16\cdot 16^n-1=15\cdot 16^n+(16^n-1)\;\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
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.