Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integer it has to be $k=101$ cause $gcd(k,k-101)=1\lor101$ and both $k,k-101$ can't be both squares of an integer,so $k=101t$ and $t(t-1)$ is never an square except for $t=1,0$ and $t=0$ is not possible hence $k=101$ is only possible integer solution
EDIT: So if $\gcd(k,k-101)=1$ then $k=h^2,k-101=(h-s)^2$ then $h(2h-s)=101$ which can be $s=1,h=51$ or $s=101,h=51$.$y=51^2+51\cdot50=51\cdot 101,x=51$
And since $x=2k-y$ is integer then $k=\frac{h}{2}$,if $h=2q$ then $k$ is integer otherwise if $h=2q+1$ then $$\frac{2q+1}{2}+\frac{1}{2}\sqrt{4q^2-400q-201}=\frac{2q+1}{2}+\frac{1}{2}\sqrt{(2q-10)^2-301}\\(2q-10)^2-301=r^2\\2q-10=z\\z^2-301=(z-c)^2\\c(2z-c)=301,c=1,z=151,c=7,z=25,c=43,z=25$$
The $z=151$ is impossible cause $2q-10$ is even,and $z=25$ is also impossible because $2q-10$ is even.Hence the only solutions are $(x,y)=(101,101),(51,5151),(5151,51)$ The last one clearly from symmetry
| A start: Rewrite as $2xy-101x-101y=0$ and then as $4xy-202x-202y=0$ and then as $(2x-101)(2y-101)=101^2$. There are not many ways to factor $101^2$.
Remark: The approach in the OP is fine, a little more complicated. If you go through the path outlined above, you will find that a couple of solutions were missed.
| {
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"url": "https://math.stackexchange.com/questions/1650875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How many $3$-tuples $(a, b, c) \in M^3$ are there such $a+b+c$ is even? The task is the following:
$M= \left \{ 1,2, ... 99,100 \right \}$
How many $3$-tuples $(a, b, c) \in M^3$ are there such $a+b+c$ is even?
I tried to solve it this way:
There are only two possibilities that $a+b+c$ is even:
First I look at this as a unordered set:
Order isn't important and you can put something back gives me:
one element is even and the other two are not:
$$\binom{50+1-1}{1}\binom{50 +2 - 1}{2}$$
or all three elements are even which gives me:
$$\binom{50+3-1}{3}$$
Now I add both and multiply it by $3!$ because there are $6$ possibilities for an unordered set to be ordered and order is important in tuples.
I get
$$3! \cdot \left[\binom{50 +2}{3} + 50 \binom{50 +1}{2}\right] = 515. 100$$
I think the solution is $500.000$ isn't it? Can't find my mistake...
| Your assumption that there are $3!$ of arranging each ordered triple $(a, b, c)$ is only valid if $a$, $b$, and $c$ are distinct. However, $(a, b, c) \in M^3$, so $a$, $b$, and $c$ need not be distinct.
Let's consider cases.
Three even numbers are selected:
*
*There are $50$ ordered triples of the form $(a, a, a)$.
*Ordered triples $(a, b, c)$ in which exactly two of the numbers are equal. We have $50$ choices for the repeated number, $\binom{3}{2}$ choices for their places in the triples, and $49$ choices for the other even number, so there are $$50 \cdot 49 \cdot \binom{3}{2}$$ such triples.
*Ordered triples $(a, b, c)$ in which each even number is distinct. There are $50$ choices for $a$, $49$ choices for $b$, and $48$ choices for $c$. Hence, there are $50 \cdot 49 \cdot 48$ of these.
One even number and two odd numbers are selected:
*Ordered triples with one even number and a repeated odd number. There are $50$ choices for the even number, $3$ choices for its location, $50$ choices for the repeated odd number, and one way of placing the repeated odd number in the open locations. Hence, there are $50^2 \cdot 3$ of these.
*Ordered triples with one even number and two distinct odd numbers. There are $50$ choices for the even number, $\binom{50}{2}$ choices for the two odd numbers, and $3!$ permutations of the three distinct numbers. Hence, there are $$50 \cdot \binom{50}{2} \cdot 3!$$ of these.
Since the cases are mutually exclusive, the total number of ways in which the numbers can be selected is
$$50 + 50 \cdot 49 \cdot \binom{3}{2} + 50 \cdot 49 \cdot 48 + 50^2 \cdot 3 + 50 \cdot \binom{50}{2} \cdot 3! = 500,000$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
I understand that when factored out, you have $(x+3)^3 = x^3+9x^2+27x+27$ as opposed to the other side which is $x^3+27$, so even if you plug in $0$ the left side will always be greater. Is this the right thinking process for determining the answer? Or am I missing something?
| ${(x+3)^3 = x^3 + 3^3}$
Expand
${x^3 + 9x^2 + 27x + 27 = x^3 + 27}$
Cancel
${9x^2 + 27x = 0}$
Factor
${x(9x + 27) = 0}$
${9x(x + 3) = 0}$
Therefore ${x = 0, x = -3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$,
If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that
$$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$
My solution:
From the given,
$$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$
Now,
L.H.S$$=\cos3A+\cos3B+\cos3C$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C-3\cos C$$
$$=12\cos A \cos B \cos C.$$
My solution ends here. How should I complete?
NOTE: I am not allowed to use complex numbers at my level. So, please help me solve this problem without using complex numbers.
| $\cos (A+B+C)=\cos (A+B) \cos C- \sin (A+B) \sin C \\ = (\cos A \cos B -\sin A \sin B) \cos C -( \sin A \cos B +\cos A \sin B) \sin C \\ =\cos A \cos B \cos C- \sum_{cyc} \sin A \sin B \cos C $
$( \sin A+ \sin B)^2= (-\sin C)^2 , (\cos A +\cos B )^2=(-\cos C)^2 \\ \implies 2+ 2\sin A \sin B +2 \cos A \cos B =1 \\ \implies \sin A \sin B = -\cos A \cos B -\dfrac{1}{2} \\ \implies \sum_{cyc} \sin A \sin B \cos C =\sum_{cyc} (-\cos A \cos B -\dfrac{1}{2})\cos C \\ =-3 \cos A \cos B \cos C -\dfrac{1}{2} (\cos A + \cos B + \cos C)=-3 \cos A \cos B \cos C$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting the terms one place:
$$
\begin{align}
1+2+3+4+\cdots & \\ 1+2+3+\cdots & \\
=1+3+5+7+\cdots & =2S
\end{align}
$$
This is the sum of the odd numbers
If we were to now sum the odd numbers and the even numbers like below:
$$ 2+4+6+8+\cdots \\[6pt] 1+3+5+7+\cdots \\[6pt] \text{if we add the terms in a certain order we can get } 1+2+3+4+5+6+7+\cdots$$
This supposedly tells us that:
$$4S = S\\[6pt] 4 \left(\frac{-1}{12}\right)=\frac{-1}{12} \\[6pt] \frac{-1}{3} = \frac{-1}{12} $$
What is faulty with this proof.
| Lets try the same in a more general way: Instead of $-\frac{1}{12}$ I will use $C\in\mathbb{R}$.
So our infinite sum:
$$1+2+3+4+\cdots = C$$
Using the same technique you get these two equalities:
$$
2+4+6+8+\cdots = 2C
\\
1+3+5+7+\cdots = 2C
$$
And when you add them together:
\begin{align}
1+2+3+4+\cdots &= 4C
\\
C &= 4C
\end{align}
So for every nonzero real number $C$ you get $1 = 4$ which is obviously not true. So to answer your question: Why is the proof faulty for C = $-\frac{1}{12}$?? Because it's faulty for every nonzero real number. The mistake is, as it's been already said, in the first line. You can't prove something that is wrong. Learn more about this infinite series here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of ways to get 17 by rolling dice 4 times (with combinatorial argument)
A dice is rolled four times. In how many ways we can get total of 17?
We can solve this by finding coefficient of $x^{17}$ in the ordinary enumerator $(x+x^2+x^3+x^4+x^5+x^6)^4$. But I feel this requires little bit more effort.
We can directly solve similar problem like:
In how many ways we can select four positive non zero integers having total of 17
as $\binom{17-1}{4-1}$ (since we have to select 3 positions out of 16 separating 17 objects for them to be divided in 4 groups). This way we dont have to go for enumerator. How solution to the original problem can be obtained with such combinatorial argument instead of using generating function?
Or may be I can rephrase the problem as what is the direct formula for such question where we have upper limit on number of objects of particular type to be selected (6 in case of dice) to get particular sum. When their is no upper limit the formula turns out to be $\binom{n-1}{r-1}$.
Note:
Second problem can be solved by finding coefficient of $x^{17}$ in $(x+x^2+x^3+x^4+...)^4$, the approach which I dont want to follow for the first problem.
| Let $x_k$ be the outcome of the $k$th roll. Then we want the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 17 \tag{1}$$
in the positive integers with the restriction that $x_k \leq 6$ for $1 \leq k \leq 4$.
Let $y_k = 6 - x_k$ for $1 \leq k \leq 4$. Then each $y_k$ is a non-negative integer satisfying $0 \leq y_k \leq 5$. Substituting $6 - y_k$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
6 - y_1 + 6 - y_2 + 6 - y_3 + 6 - y_4 & = 17\\
-y_1 - y_2 - y_3 - y_4 & = -7\\
y_1 + y_2 + y_3 + y_4 & = 7 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. A particular solution of equation 2 corresponds to the placement of three addition signs in a row of seven ones. For example,
$$1 1 + 1 1 1 + 1 1 +$$
corresponds to the solution $y_1 = 2$, $y_2 = 3$, $y_3 = 2$, and $y_4 = 0$, while
$$1 + 1 1 + 1 1 1 + 1$$
corresponds to the solution $y_1 = 1$, $y_2 = 2$, $y_3 = 3$, and $y_4 = 1$. Thus, the number of solutions of equation 2 is the number of ways three addition signs can be inserted into a row of seven ones, which is
$$\binom{7 + 3}{3} = \binom{10}{3}$$
since we must choose which three of the ten symbols (seven ones and three addition signs) will be addition signs.
However, we must exclude those solutions of equation 2 in which one or more of the $y_k$'s exceeds $5$. Since $2 \cdot 6 = 12 > 7$, at most one of the $y_k$'s can exceed $5$.
Suppose $y_1 > 5$. Let $z_1 = y_1 - 6$. Then $z_1$ is a non-negative integer. Substituting $z_1 + 6$ for $y_1$ in equation 2 yields
\begin{align*}
z_1 + 6 + y_2 + y_3 + y_4 & = 7\\
z_1 + y_2 + y_3 + y_4 & = 1 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with $4$ solutions. By symmetry, any of the four variables could have exceeded $5$ in equation 2. Thus, the number of solutions of equation 2 in which one of the variables exceeds $5$ is
$$\binom{4}{1} \cdot \binom{4}{1}$$
Hence, the number of ways to obtain a sum of $17$ with four rolls of a six-sided die is
$$\binom{10}{3} - \binom{4}{1}^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $ \sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =5\,$? How to find $x$ in:
$$
\sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =5
$$
| Hint: $~x+1=\sqrt{x^2+2x+1}=\sqrt{x^2+\sqrt{4x^2+4x+1}}=\ldots~$ Can you take it from here ? :-$)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show $\lim\limits_{n\rightarrow \infty}\left(\raise{5pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $? How to show this is true?
$$\lim_{n\rightarrow \infty}\left(\raise{3pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $$
| Writing $I_n=\int_0^{\frac{\pi}{2}}\sin^n x \, dx$, one can establish:
*
*$I_0=\frac{\pi}{2}, I_1=1$
*$I_n=\frac{n-1}{n}I_{n-2}$
*$I_{2n}=\frac{\binom{2n}{n}}{2^{2n}}\frac{\pi}{2}, I_{2n-1}=\frac{1}{2n}\frac{2^{2n}}{\binom{2n}{n}},I_{2n+1}=\frac{1}{2n+1}\frac{2^{2n}}{\binom{2n}{n}}$
*$I_{n+1}\le I_n$ (as $0\le \sin x \le 1 \implies \sin^{n+1} x \le \sin^n x$)
*$I_n \le \frac{n+1}{n}I_{n+1}$ (as $I_n\le I_{n-1}=\frac{n+1}{n}I_{n+1}$)
So, $1\le \frac{I_n}{I_{n+1}} \le 1+\frac{1}{n}$, whence $\frac{I_n}{I_{n+1}}\to1$ by the squeeze theorem. Now:
$$\frac{I_{2n+1}}{I_{2n}}=\frac{2^{4n}}{\binom{2n}{n}^2}\frac{2}{\pi(2n+1)}\to 1$$
$$\frac{I_{2n}}{I_{2n-1}}=\frac{\binom{2n}{n}^2}{2^{4n}}\frac{n\pi}{1}\to 1$$
each of which verify your claimed limit.
| {
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"url": "https://math.stackexchange.com/questions/1659664",
"timestamp": "2023-03-29T00:00:00",
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Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
| Yes, indeed,
$$(a+b+c)^3$$
$$=a^3+b^3+c^3+3a^2b+3a^2c+3b^2a+3b^2c+3c^2a+3c^2b+6abc$$
$$=a^3+b^3+c^3+3a^2b+3a^2c+3abc+3b^2a+3b^2c+3abc+3c^2a+3c^2b+3abc-3abc$$
$$=a^3+b^3+c^3+3a(ac+ab+bc)+3b(ac+ab+bc)+3c(ac+ab+bc)-3abc$$
$$=a^3+b^3+c^3+3(a+b+c)(ac+ab+bc)-3abc$$
So
$$\color{red}{a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ac+ab+bc)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What's $\alpha+\beta$ if we have: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ ($\alpha$ and $\beta$ are Real) What's $\alpha+\beta$ if we have $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$? Here $\alpha$ and $\beta$ are real.
Firstly, I subtracted the two equations and got the following:
$$\alpha^3-\beta^3-6(\alpha^2-\beta^2)+13(\alpha-\beta)=-18$$ Then I tried to factorize the left hand side as: $$(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2-6\alpha-6\beta+13)=-18$$
At this point it seems we can't go on!
Then I tried adding the two equations as: $$(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2-6\alpha-6\beta+13)+12\alpha\beta=20$$
unfortunately, again I can't continue!
Is there any special creativity needed for solving this question?
| In order to get rid of the quadratic terms we write $\alpha:=a+2$, $\beta:=b+2$. Then
$$1=\alpha^3-6\alpha^2+13\alpha=(a+2)^3-6(a+2)^2+13(a+2)=a^3+a+10\ ,$$
and similarly
$$19=\beta^3-6\beta^2+13\beta=b^3+b+10\ .$$
It follows that $a$ and $b$ satisfy the equations
$$a^3+a=-9,\quad b^3+b=9\ .$$
Now (an unexpected) symmetry comes to our rescue:
As $t\mapsto t^3+t$ is monotonically increasing there is exactly one $b=:b_*>0$ satisfying the second of these equations, and it is then obvious that $a:=-b_*$ is the unique solution of the first of these equations. It follows that necessarily
$$\alpha+\beta=-b_*+2 +b_*+2=4\ .$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Derivative of $4x^5 \tan(\frac{-1}{x})$ I can't seem to be able to get the correct signs for this derivative.
$\frac{d}{dx}(4x^5\tan(\frac{-1}{x}))$
Here's my work:
$= (4x^5)'\tan(\frac{-1}{x}) + 4x^5(\tan(\frac{-1}{x}))'$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\tan(\frac{-1}{x}))'$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\frac{-1}{x})'\sec^2(\frac{-1}{x})$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\frac{-1}{x})'\sec^2(\frac{-1}{x})$
And here is where I think I make my mistake:
$(\frac{-1}{x})' = (-1x^{-1})'$
$(\frac{-1}{x})' = (1x^{-2})$
$(\frac{-1}{x})' = \frac{1}{x^2}$
So plugging that derivative back into the step I left off on in the last equation:
$= 20x^4tan(\frac{-1}{x}) + 4x^5\frac{1}{x^2}sec^2(\frac{-1}{x})$
$= 20x^4tan(\frac{-1}{x}) + 4x^3sec^2(\frac{-1}{x})$
However the answer key and wolfram alpha display the answer to be
$= 20x^4\tan(\frac{-1}{x}) - 4x^3\sec^2(\frac{-1}{x})$
I can't figure out the step where I went wrong... Please help!
| Hint $sec(-\theta)=sec(\theta)$ thats the only mistake.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a possible f(x) Do you think that it is possible to find a $f$ such that given a floating point constant $c \gt 0$ and an integer constant $n \gt 0$, then $\forall x_i \gt 1, i=1, 2,...,n$ :
$$f\left({1\over x_1}+{1\over x_2}+...+{1\over x_n}\right) = {1\over x_1 + c}+{1\over x_2 + c}+...+{1\over x_n + c}$$
| Update: Generalized for $n\geq3, c>0$, not just the case $n=3,c=1$.
For $n = 3, c>0, x_1 = x_2 = x_3 = 4c$:
\begin{gather}
\frac1{4c} + \frac1{4c} + \frac1{4c} = \frac3{4c}, \\
\frac1{4c+c} + \frac1{4c+c} + \frac1{4c+c} = \frac3{5c}.
\end{gather}
For $n = 3, c>0, x_1 = x_2 = 8c, x_3 = 2c$:
\begin{gather}
\frac1{8c} + \frac1{8c} + \frac1{2c} = \frac3{4c}, \\
\frac1{8c+c} + \frac1{8c+c} + \frac1{2c+c} = \frac5{9c}.
\end{gather}
So for any possible function $f$,
$$
f\left( \frac1{4c} + \frac1{4c} + \frac1{4c} \right)
= f\left( \frac3{4c} \right)
= f\left( \frac1{8c} + \frac1{8c} + \frac1{2c} \right)
$$
but
$$
\frac1{4c+c} + \frac1{4c+c} + \frac1{4c+c}
\neq \frac1{8c+c} + \frac1{8c+c} + \frac1{2c+c}.
$$
Hence for $n=3$ it is impossible that
$f\left({1\over x_1}+{1\over x_2}+...+{1\over x_n}\right)$
can be equal to ${1\over x_1 + c}+{1\over x_2 + c}+...+{1\over x_n + c}$
in all cases; in at least one of these two cases the function
is not equal to the latter expression.
To extend this to $n>3$,
let $x_4 = \cdots = x_n = 2$, and modify the sums above
by adding $\frac12 + \cdots + \frac12$ to the sums of $\frac{1}{x_i}$
and $\frac1{2+c} + \cdots + \frac1{2+c}$ to the sums of $\frac{1}{x_i + c}$;
we again have two equal sums of $\frac{1}{x_i}$
but unequal sums of $\frac{1}{x_i + c}$.
| {
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Is there any formula for computing the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2k)$ I need to compute the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2n)$ in order to use the Stirling's formula to calculate, but I am stuck. Thanks!
I am a little bit confused now... should the last term of $w(n/2+1/2)$ be $(n+1-2n)$ or $(n+1-2k)$?
| While @alex.gordan beat me to a full proof that's simpler than mine functionally, I figured I would share the way I approached this problem!
First note that $(a)_k = a(a+1)(a+2)\cdots(a+k-1)$. Thus,
$$\begin{align}\left(\frac{1-n}{2}\right)_k & = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+1\right)\cdots\left(\frac{1-n}{2}+k\right) \\
& = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+\frac{2}{2}\right)\left(\frac{1-n}{2}+\frac{4}{2}\right)\cdots\left(\frac{1-n}{2}+\frac{2k-2}{2}\right)\\
& = \left(\frac{1-n}{2}\right)\left(\frac{3-n}{2}\right)\left(\frac{5-n}{2}\right)\cdots\left(\frac{2k-1-n}{2}\right) \\& = \frac{(1-n)(3-n)(5-n)\cdots(2k-1-n)}{2^k}\end{align}$$
Let's now adjust this to fit your product. First let's remove the denominator
$$2^k\left(\frac{1-n}{2}\right)_k =(1-n)(3-n)(5-n)\cdots(2k-1-n)$$
Now let's pull out $k$ negative signs to flip each term
$$2^k(-1)^k\left(\frac{1-n}{2}\right)_k =(n-1)(n-3)(n-5)\cdots(n+1-2k)$$
Now we simplify the powers and multiply by $(n+1)$
$$(-2)^k(n+1)\left(\frac{1-n}{2}\right)_k =(n+1)(n-1)(n-3)(n-5)\cdots(n+1-2k)$$
We now use the definition $(a)_k = \frac{\Gamma(a+k)}{\Gamma(a)}$, let $n = 2m$, and transform this into
$$\frac{(-2)^k (n+1) \Gamma(\frac{2k+1-n}{2})}{\Gamma(\frac{1-n}{2})}$$
$$=(-2)^k (n+1)\frac{\Gamma(k-m+1/2)}{\Gamma(1/2-m)}$$
Now using the fact that $\Gamma(z) \sim \sqrt{\frac{2\pi}{z}} \left(\frac{z}{e}\right)^z$ (known as Stirling's approximation)
$$\sim (-2)^k (n+1)\frac{\sqrt{\frac{2\pi}{k-m+1/2}} \left(\frac{k-m+1/2}{e}\right)^{k-m+1/2}}{\sqrt{\frac{2\pi}{1/2-m}} \left(\frac{1/2-m}{e}\right)^{1/2-m}}$$
$$= \frac{(-2)^k (n+1)(k-m+1/2)}{e^k}\sqrt{\frac{1/2-m}{k-m+1/2}} \left(\frac{(k-m+1/2)}{(1/2-m)}\right)^{1/2-m}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac{1-v^2}{(1+v^2)(-v^2+v+1)} \, dv$$
Using partial fractions
$$-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}=\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{2v}{v^2+1}-\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{4}{v^2+1}+\frac{1}{5}\int^{\frac{\pi}{8}}_0 \frac{-2v+1}{-v^2+v+1}$$
$$=\frac{1}{5}ln|v^2+1|-\frac{4}{5}\arctan(v)+\frac{1}{5}ln|-v^2+v+1|$$ from $\frac{\pi}{8}$ to $0$
$0.02-0+0.299-0+0.04-0=0.359$
But it should come out 0.32
| Before doing any substitutions are algebra, you can see that the value of this integral will be a finite positive number because you're integrating a positive function over a bounded interval.
Multiplying the numerator and denominator by $(1-v^2)(1+v^2)$, we get:
$$
\frac{\frac 2 {1+v^2}}{2+\frac{2v}{1-v^2}} = \frac{2(1-v^2)}{2(1+v^2)(1-v^2)+2v(1+v^2)} = \frac{1-v^2}{(1 - v^4) + v+v^3}.
$$
Since the denominator is not $0$ when $v$ is $1$ or $-1$, nothing cancels. You may need numerical methods to factor this; I'm not sure.
Notice that as $x$ goes from $0$ to $\pi/4$, then $v$ goes from $0$ to $\tan\dfrac\pi8$, so your bounds of integration need to take that into account.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
| We'll prove that our inequality is true for all reals $x$, $y$ and $z$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since $\prod\limits_{cyc}(x^2+xy+y^2)=27(3u^2v^4-u^3w^3-v^6)$,
we see that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a linear function.
Thus, $f$ get's a minimal value for an extremal value of $w^3$.
Since $x$, $y$ and $z$ are real roots of the equation $(X-x)(X-y)(X-z)=0$ or
$X^3-3uX^2+3v^2X-w^3=0$ or $X^3-3uX^2+3v^2X=w^3$,
we see that a line $Y=w^3$ and a graph $Y=X^3-3uX^2+3v^2X$ should have three common points and the extremal value of $w^3$ happens for the equality case of two variables.
Since our inequality is homogeneous and even degree, it remains to check one case only: $y=z=1$,
which gives $(x-1)^2(5x^2+8x+5)\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Inequality $ab+bc+cd \geq 3abcd$ If $a,b,c,d$ are real numbers such that $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$, then is this inequality true: $$ab+bc+cd\geq 3abcd$$
| Let $$a=\frac{5}{2},b=\frac{1}{2}\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right),c=2,,d=\frac{1}{2}\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)$$ we find $$a+b+c+d=6$$$$a^2+b^2+c^2+d^2=12$$ and $$ab+bc+cd=\frac{1}{8}\left(39-5\sqrt{5}\right)\approx 3.477$$ and $$3abcd=\frac{15}{4}=3.75$$ with $$ab+bc+cd<3abcd$$ so the inequality is not always true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
| Integrate by parts
\begin{align}\int \sqrt{1 + \frac{1}{x^2}}\ dx
=x\sqrt{1 + \frac{1}{x^2}}+\int \frac{dx}{x^2 \sqrt{1 + \frac{1}{x^2}}}
=x\sqrt{1 + \frac{1}{x^2}}-\sinh^{-1}\frac1x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
System of Equations , Finding value of coefficient so equations are inconsistent
Question; Calculate a value for the coefficient $'a'$ of $x$ so that the solutions to the three equations are inconsistent. Demonstrate the resulting system of equations are then inconsistent:
$$ x+2y+2z=6$$
$$ x-y+z=-4$$
$$ ax+y-9z=-1$$
Cannot use matrix/matrics
What I have attempted:
Using the first 2 equations to eliminate $z$
$$ x+2y+2z=-6$$
$$ x-y+z=-4$$
Multiplying the second by $2$
$$ x+2y+2z=6$$
$$ 2x-2y+2z=-8$$
Subtracting both to get
$$ -x + 4y = 14 $$
$$ y = \frac{14}{4} + \frac{x}{4} $$
Again using the first 2 equations to eliminate y
$$ x+2y+2z=-6$$
$$ x-y+z=-4$$
Multiplying the second by $-2$
$$ x + 2y + 2z = -6 $$
$$ -2x + 2y - 2z = 8 $$
Subtracting
$$ 3x + 4z = -14 $$
$$ z = \frac{-14}{4} - \frac{3x}{4} $$
Subbing both $z$ and $y$ into $ax+y-9z=-1$
$$ ax + \frac{14}{4} + \frac{x}{4} -9(\frac{-14}{4} - \frac{3x}{4}) = -1 $$
$$ax + \frac{14}{4} + \frac{x}{4} + \frac{63}{2} + \frac{27x}{4} = -1 $$
$$ ax + 7x = -36 $$
$$ x(a+7) = -36 $$
Now I am stuck, how do I find the value of a?
| Here is a computer doing basic algebra. It is tedious to do it by hand.
Leting $a= \frac{-28}{4}$ will reduce the system to $0x+0y+0za=1$ and $0 \neq 1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$x^4 -10x^2 +1 $ is irreducible over $\mathbb Q$ I have seen the thread Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ but this didn't really have a full solution.
Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals? And what exactly is a linear factor.
What I did was the following:
Let $y=x^2$. Then $y^2 - 10y +1=0 \iff y^2 -10y = -1$, add $25$ to both sides:
$$y^2 - 10y + 25 = -1 + 25 \iff (y - 5)^2 = 24 \iff y = 5 \pm 2\sqrt6$$
So $x^2=5 \pm 2 \sqrt6= (\sqrt2 \pm \sqrt3 )^2$
So $x^4 -10x^2 +1 =(x^2 -(\sqrt2 + \sqrt3 )^2)(x^2 - (\sqrt2 - \sqrt3)^2)$
Then difference of two squares:
$ (x-(\sqrt2 + \sqrt3))(x+(\sqrt2 + \sqrt3))(x-(\sqrt2 - \sqrt3))(x+(\sqrt2 - \sqrt3))=0$
Expanding this out gives $x^4 -10x^2 +1 =0$ so we have found all the roots and we can confirm that they are the roots right? None of these roots are rational so it must be irreducible?
Please can you advise me on this method that I am using and not direct me to a theorem or any type of shortcut.
Thanks in advance.
| It's quite easy to find a factorization over $\mathbb{R}$:
\begin{align}
p(x)=x^4-10x^2+1
&=(x^2-1)^2-8x^2 \\[6px]
&=(x^2-2\sqrt{2}\,x-1)(x^2+2\sqrt{2}\,x-1) \\[6px]
&=(x-(\sqrt{2}+\sqrt{3}))(x-(\sqrt{2}-\sqrt{3}))\\
&\qquad \cdot (x-(-\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}-\sqrt{3}))\\
\quad
\end{align}
Our task is to show that $p(x)$ is the minimal polynomial of $\sqrt{2}+\sqrt{3}$, which implies it is irreducible. If not, the degree of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ would be less than $4$. However
$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$ (prove it), which has degree $4$ over $\mathbb{Q}$, because $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Showing that the result of an equation may be a perfect square Consider an equation that, for a few select integers $x$, yields a value that is a perfect square. In the case of perfect square quadratics it is easy. For example $(x-2)^2$ will always yield a perfect square value. But what about for an equation like $(x-2)(x-3)$? Or $x^4 -x^3 + x^2 + 2x + 1$? Perhaps it yields a perfect square for a few inputs, but is there a way to determine all such inputs?
| We first show that $\gcd(x-2,x-3)=1$, which is to say that $x-2$ and $x-3$ have no common prime factor. Assume to the contrary that $x-2=pa$ and $x-3=pb$ for some prime $p \ge 2$ and integers $a$ and $b$. Now $$1 = (x-2)-(x-3) = pa-pb = p(a-b).$$ This implies $p \mid 1$, contradicting the hypothesis that $p$ is a prime.
Now, by the Fundamental Theorem of Arithmetic, $$(x-2)(x-3)=k^2$$
implies $x-2=k_1^2$ and $x-3=k_2^2$ for coprime integers $k_1$ and $k_2$, such that $k=k_1k_2$. But the only two consecutive square integers are $0$ and $1$. Hence $x-2=1$ and $x-3=0$ is the only solution where $x-2$ is positive. The second solution comes through the power of zero to absorb everything: $x-2=0$ and $x-3=-1$ means $(x-2)(x-3)=0\cdot(-1)=0$, and since $0=0^2$ is a square, it counts.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| It seems you may not know l'Hôpital's rule. It's also possible without the need for differentiation. Multiply numerator and denominator with the so called conjugate expression of the numerator:
$$\frac{x^2 \sqrt{x+2} - 8}{4-x^2}
= \frac{\left( x^2 \sqrt{x+2} - 8 \right)\left( x^2 \sqrt{x+2} + 8 \right)}{\left(4-x^2\right) \left( x^2 \sqrt{x+2} - 8 \right)}$$
Then simplify via $(a-b)(a+b) =a^2-b^2$ in the numerator:
$$\frac{x^4(x+2)-64}{\left(4-x^2\right) \left( x^2 \sqrt{x+2} - 8 \right)}$$
Now you should factorise so you have a common factor $x-2$ in the numerator and denominator. In the denominator that's easy via $4-x^2 = (2-x)(2+x)=-(x-2)(x+2)$. For the nominator, you know that the polynomial divides $(x-2)$ since $x=2$ is a root of the polynomial. You can use long division or Horner's method for this:
$$x^4(x+2)-64 = x^5+2x^4-64=(x-2) (x^4+4 x^3+8 x^2+16 x+32)$$
So back to the limit:
$$\lim_{x \to 2} \frac{x^2 \sqrt{x+2} - 8}{4-x^2}
=\lim_{x \to 2} \frac{(x-2) (x^4+4 x^3+8 x^2+16 x+32)}{-(x-2)(x+2) \left( x^2 \sqrt{x+2} + 8 \right)}$$
Cancel the common factor $x-2$ and plug in $x=2$, simplify:
$$\lim_{x \to 2} \frac{x^4+4 x^3+8 x^2+16 x+32}{-(x+2) \left( x^2 \sqrt{x+2} + 8 \right)} = \ldots = -\frac{9}{4}$$
If you have seen l'Hôpital's rule, the other answers give you an alternative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) = \frac{b^{n + 1} - a^{n + 1}}{(b-a)}$
What can I do from here?
| By the mean value theorem, there exists $a<c<b$ such that
$$
\frac{b^{n+1}-a^{n+1}}{b-a}=(n+1)c^n.
$$
Hence,
$$
(n+1)a^n<\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| Let $u=x^3+1$ then you have your $du=3x^2dx$, then from your original equation $x^2dx$ is just equal to $\frac{du}{3}$. Now solving for the whole equation that just equal to $\int\frac{du}{3u}$.
$\int\frac{du}{3u}=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}lnu+ C= \frac{1}{3}ln(x^3+1)+C.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make sense:
Let $n = 1.$ Then $1 + a + a^2 + \ldots +1 < \frac {1 + a + a^2 + \ldots +1 + a}{2} = \frac{2 + 2a + a^2+ \ldots}{2} = 1 + a + \frac{a^2}{2} + \ldots$
Then I did this below, but it's unclear if the difference is positive:
$\frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1} - \frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} = \frac {n + na +na^2+ \ldots + na^{n - 1} + na^n - n - 1 - an - a- na^2 - a^2 - \ldots -na^{n - 1} - a^{n - 1}}{n(n + 1)} = \frac{n(1 + a + a^2+ \ldots + a^{n - 1} + a^n - 1 -a - a^2 -a^{n - 1}) - 1 - a - a^2 - \ldots - a^{n - 1}}{n(n+1)} $
$ = \frac{- 1 + a(n(a^{n - 1}) - 1 - a - \ldots - a^{n - 2})}{n(n+1)}$
What can I try now?
| $$\frac{1+a+\ldots+a^{n-1}}n<\frac{1+a+\ldots+a^n}{n+1}\iff$$
$$\frac1n\,\frac{1-a^n}{1-a}<\frac1{n+1}\,\frac{1-a^{n+1}}{1-a}\iff$$
$$(n+1)(1-a^n)\stackrel{\text{since}\;1-a<0}>n(1-a^{n+1})\iff-na^n+1-a^n>-na^{n+1}\iff$$
$$n(a^{n+1}-a^n)>a^n-1\iff na^n(a-1)>(a-1)(a^{n-1}+a^{n-2}+\ldots+a+1)\iff$$
$$na^n>1+a+\ldots+a^{n-1}$$
and the last inequality is trivial since $\;a^n>a^k\;,\;\;\forall\,k=0,1,2,...,n-1\;$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this integral $\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$ How to evaluate this integral
$$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$
Mathematica gave me the answer below
$$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left ( 2+\sqrt{3} \right )-\frac{\mathbf{G}}{3}$$
where $\mathbf{G}$ is Catalan's constant.
| We can attack this integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2}$$
by considering the complex contour integral
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{1+z^2}$$
where $C$ is the following contour
This is a keyhole contour about the positive real axis, but with additional keyholes about the branch points at $z=e^{i \pi/3}$, $z=-1$, and $z=e^{i 5 \pi/3}$. There are simple poles at $z=\pm i$.
I will outline the procedure for evaluation. The integral about the circular arcs, large and small, go to zero as the radii go to $\infty$ and $0$, respectively. Each of the branch points introduces a jump of $i 2 \pi$ due to the logarithm in the integrand. By the residue theorem, we have
$$-i 2 \pi \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} - i 2 \pi \int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} \\ - i 2 \pi \int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} - i 2 \pi \int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \\ i 2 \pi \sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} $$
Without going into too much detail, I will illustrate how the integrals are done by evaluating one of them. Consider
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -\int_1^{\infty} dy \frac{\log{y}+i \pi}{1+y^2}$$
Now,
$$\int_1^{\infty} \frac{dy}{1+y^2} = \int_{\pi/4}^{\pi/2} d\theta = \frac{\pi}{4}$$
$$\begin{align}\int_1^{\infty} dy\frac{\log{y}}{1+y^2} &= G\end{align}$$
so that
$$\int_{e^{i \pi}}^{\infty e^{i \pi}} dt \frac{\log{t}}{1+t^2} = -G - i \frac{\pi^2}{4} $$
Along similar lines,
$$\int_{e^{i \pi/3}}^{\infty e^{i \pi/3}} dt \frac{\log{t}}{1+t^2} = \frac{2}{3} G + \frac{\pi}{6} \log{(2+\sqrt{3})}$$
$$\int_{e^{i 5 \pi/3}}^{\infty e^{i 5 \pi/3}} dt \frac{\log{t}}{1+t^2} = \frac{2}{3} G - \frac{5 \pi}{6} \log{(2+\sqrt{3})} + i \frac{\pi^2}{2}$$
Combining the integrals, I get
$$\frac{G}{3} - \frac{2 \pi}{3} \log{(2+\sqrt{3})} + i \frac{\pi^2}{4}$$
The sum of the residues on the RHS is relatively simple to evaluate; I get
$$\sum_{\pm} \left[\frac{\log{(1+z^3)} \log{z}}{2 z} \right]_{z=\pm i} = \frac{(1/2 \log{2} -i \pi/4)(i \pi/2)}{2 i} + \frac{(1/2 \log{2} + i \pi/4)(i 3 \pi/2)}{-2 i}\\ = -\frac{\pi}{4} \log{2}-i \frac{\pi^2}{4}$$
The integral we seek is then the negative of the sum of the combined integrals and the sum of the residues, which gives us
$$\int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2} = -\frac{G}{3} + \frac{\pi}{4} \log{2} +\frac{2 \pi}{3} \log{(2+\sqrt{3})} $$
which agrees with Mathematica.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Solve the following questions: Following are two questions and their respective answers(as given in textbook):
Q1. If $$ k=\sum_{r=0}^n \frac{1}{n \choose r}$$, then write $$\sum_{r=0}^n \frac{r}{n \choose r}$$ in terms of k.
Ans. $$\frac {nk}{2}$$
Q2. If $$x+y=1$$ find $$\sum_{r=0}^n r^{2} y^{n-r} x^r$$.
Ans. $$n(nx+y)$$
The thing is, I have tried solving both the problems, but that hasn't been much fruitful.
I faced difficulties from the first one as the expressions are sum of reciprocals of binomial coefficients. Because I never came across such reciprocals earlier.
The problem with the second is because of the two terms x and y. Had either x or y not been there, I could have easily differentiated $$ (1+x)^n$$ or $$ (1+y)^n$$ and their respective expansions and derived the result.
So, I would be thankful to anyone who provides a proper solution for each of the above problems.
Note: I wanted to add 'homework' tag, but wasn't allowed to do so.
| For the second one, start with:
$\begin{align}
\sum_{0 \le r \le n} x^r y^{n - r}
&= y^n \sum_{0 \le r \le n} \left( \frac{x}{y} \right)^r \\
&= y^n \frac{1 - (x/y)^{n + 1}}{1 - x/y} \\
&= \frac{y^{n + 1} - x^{n + 1}}{y - x} \\
x \frac{\mathrm{d}}{\mathrm{d} x}
\left(
x \frac{\mathrm{d}}{\mathrm{d} x}
\frac{y^{n + 1} - x^{n + 1}}{y - x}
\right)
&= \sum_{0 \le r \le n} r^2 x^r y^{n - r}
\end{align}$
The result is a quite ugly expression:
$\begin{align}
&\frac{y^n (x^2 y + x y^2)
- (n + 1)^2 x^{n + 1} y^2
+ (2 n^2 + 2 n - 1) x^{n + 2} y
- n^2 x^{n + 3}}
{(y - x)^3} \\
&\qquad = \frac{y^{n + 1} x (x + y)
- x^{n + 1}
((n^2 + 2 n + 1) y^2
- (2 n^2 + 2 n - 1) x y
+ n^2 x^2)}
{(y - x)^3} \\
&\qquad = \frac{y^{n + 1} x (x + y)
- x^{n + 1} (y - x)
((n + 1)^2 y - n^2 x)
}
{(y - x)^3} \\
&\qquad = \frac{y^{n + 1} x
- x^{n + 1} (y - x)
((n + 1)^2 y - n^2 x)
}
{(y - x)^3}
\end{align}$
Yes, it is fine to consider $x$ and $y$ as separate variables, as mentioned. The relation $x + y = 1$ can be used to simplify the end result, however.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there another way to solve this equation? My problem is Solve the equation
$$2 \left(\sqrt{x^3-7 x^2+17x-14}+\sqrt{x^4-7 x^3+23x^2-37 x+28}\right)=4x^2-17 x+25.$$
And my solve.
We have
$$\sqrt{x^3-7 x^2+17x-14}= \sqrt{(x-2) \left(x^2-5 x+7\right)}\leqslant \dfrac{(x-2) + (x^2-5 x+7)}{2}= \dfrac{x^2 - 4x + 5}{2}.$$
Another way
\begin{align*}
\sqrt{x^4-7 x^3+23x^2-37 x+28} &= \sqrt{\left(x^2-4 x+7\right)\left(x^2-3 x+4\right)}\\
&\leqslant \dfrac{(x^2-4 x+7)+(x^2-3 x+4)}{2} \\
&= \dfrac{2x^2-7x+11}{2}.
\end{align*}
Therefore,
$$\text{LHS} \leqslant (x^2 - 4x + 5) + (2x^2-7x+11) = 3x^2 -11x + 16.$$
From the given equation, we have
$$4x^2-17 x+25 \leqslant 3x^2 -11x + 16 \Leftrightarrow x^2 -6x + 9 \leqslant 0 \leqslant (x-3)^2 \leqslant 0 \Leftrightarrow x = 3. $$
We see that, $x = 3$ satisfies the given equation.
Thus, the the given equation has only solution is $x = 3.$
Is there another way to solve this equation?
| There is a very laborious way of doing it using successive squarings (which will create extra roots.
Starting with $$2(\sqrt A+\sqrt B)=C$$ a first squaring leads to $$4(A+B)-C^2=8\sqrt{AB}$$ and a second squaring leads to $$\Big(4(A+B)-C^2\Big)^2=64AB$$ Using $$A=x^3-7 x^2+17 x-14\qquad B=x^4-7 x^3+23 x^2-37 x+28\qquad C=4 x^2-17 x+25$$ this leads to $$144 x^8-2752 x^7+23640 x^6-119376 x^5+387945 x^4-831620 x^3+1149958 x^2-939876
x+348849=0$$ By inspection, this equation shows a double root $x=3$. Then, what is left is $$144 x^6-1888 x^5+11016 x^4-36288 x^3+71073 x^2-78590 x+38761=0$$
As Laplacian Fourier commented, this "last equation has no real roots by Descartes rule of signs".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge\frac{100}{3}$ $a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that
$$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\geq\frac{100}{3}$$
| The function $f(x) =(x+x^{-1} )^2 $ is convex on $(0,\infty) $ hence $$ \frac{1}{3} (f(a) +f(b) +f(c))\geq f(\frac{a+b+c}{3}) =(3+3^{-1} )^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Hermite's identity for sum of floor function In Hermite's 1884 paper "Sur quelques conséquences arithmétiques des formules de la théorie des fonctions elliptiques", volume 5 of Acta Mathematica, pages 310-315, he proves what is often called "Hermite's identity" differently than the usual proof you'll find by Googling. Hermite's identity is, for real x and positive integer n,
$\sum_{k=0}^{n-1} E(x+k/n) = E(nx)$,
where $E(x)$ is the greatest integer $\leq x$. Hermite first establishes that for nonnegative integers $a$ and $b$,
$\frac{z^b}{(1-z)(1-z^a)} = \sum_{n=0}^\infty \left( 1 + \left[\frac{n}{a} \right]\right) z^{n+b}$.
Then using $\frac{z^a}{(1-z)(1-z^a)} = \frac{z^a(1+z^a+z^{2a}+\cdots+z^{(n-1)a})}{(1-z)(1-z^{na})}$ he says that the above identity follows. I'm not seeing how to get this and I'd be glad to hear if anyone sees how the deduction works.
| Assume that a function $f$ is given as a power series
$$ f(x) = A_0 + A_1x + A_2x^2 + \ldots; $$
then
$$ \frac{f(x)}{1-x} = A_0 + (A_0+A_1)x + (A_0 + A_1 + A_2)x^2 + \ldots $$
as can be verified easily by multiplying through by $1-x$.
Next
$$ f(x^a) = A_0 + A_1x^a + A_2x^{2a} + \ldots, $$
hence
$$ \frac{f(x^a)}{1-x} = \sum_{n \ge 0} (A_0 + A_1 + \ldots + A_\nu) x^n, $$
where $\nu = \lfloor \frac na \rfloor$. For $f(x) = \frac1{1-x}$ this implies
\begin{equation}\label{Her1}
\frac1{(1-x)(1-x^a)}
= \sum_{n \ge 0} \Big[ 1 + \Big\lfloor \frac na \Big\rfloor \Big] x^n .
\end{equation}
This implies
$$ \frac1{(1-x)(1-x^{ma})} =
\sum_{n \ge 0} \Big\lfloor \frac {n+ma}{ma} \Big\rfloor x^n $$
since $1 + \lfloor \frac n{ma} \rfloor = \lfloor \frac {n+ma}{ma} \rfloor$.
Multiplying this last equation through by $x^{ka}$ we obtain
$$ \frac{x^{ka}}{(1-x)(1-x^{ma})}
= \sum_{n \ge 0} \Big\lfloor \frac {n+ma}{ma} \Big\rfloor x^{n + ka}
= \sum_{n \ge ka} \Big\lfloor \frac {n+(m-k)a}{ma} \Big\rfloor x^n. $$
Now the identity
$$ \frac{1 - x^{ma}}{1-x^a} = 1 + x^a + x^{2a} + \ldots + x^{(m-1)a} $$
implies
\begin{align*}
\frac{x^a}{(1-x)(1-x^a)}
& = \frac{x^a(1 + x^a + a^{2a} + \ldots + x^{m-1})}{(1-x)(1-x^{ma})} \\
& = \sum_{k=1}^m \frac{x^{ka}}{(1-x)(1-x^{ma})} \\
& = \sum_{k=1}^m \sum_{n \ge ka}
\Big\lfloor \frac {n+(m-k)a}{ma} \Big\rfloor x^n.
\end{align*}
Replacing the summation index $k$ by $m-k$ we find
\begin{align*}
\frac{x^a}{(1-x)(1-x^a)}
& = \sum_{k=0}^{m-1}\sum_{n \ge ka}
\Big\lfloor \frac {n+ka}{ma} \Big\rfloor x^n.
\end{align*}
On the other hand we know
\begin{align*}
\frac{x^a}{(1-x)(1-x^a)}
& = \sum_{n \ge 0} \Big\lfloor \frac {n+a}a \Big\rfloor x^{n+a}
= \sum_{n \ge a} \Big\lfloor \frac {n}a \Big\rfloor x^n.
\end{align*}
Comparing the coefficient of $x^n$ and setting $z = \frac n{ma}$ we find
$$ \lfloor mz \rfloor = \Big \lfloor \frac na \Big\rfloor
= \sum_{k=0}^{m-1} \Big\lfloor \frac {n+ka}{ma} \Big\rfloor
= \sum_{k=0}^{m-1} \Big\lfloor z + \frac {k}{m} \Big\rfloor. $$
This proves Hermite's identity for rational values of $z$.
By piecewise continuity it holds for all real numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integration of $\frac{dx}{x\sqrt{\ln x^2 +6\ln x}}$ $\require{cancel}$
Hello everybody so this is my integration question. I have reached a point in my picture. I don't know how to continue. I appreciate the help.
Also , am I going in the right direction ??
Please note, square root is over $\ln x^2 +6\ln x$.
\begin{align*}
\int\frac{1}{x\sqrt{\ln x^2+6\ln x}}dx\underset{\substack{\big| \\ u=\ln t \\ du=\frac{dt}{t}}}{=}{}&\int\frac{1}{\sqrt{u^2+6u}}du\underset{\substack{\big| \\ \text{Complete} \\ \text{square.}}}{=}\int\frac{1}{\sqrt{(u+5)^2-9}}du={} \\
{}\underset{\substack{\big| \\ s=u+5 \\ ds=du}}{=}{}&\int\frac{1}{\sqrt{s^2-9}}ds\underset{\substack{\big| \\ s=3\sec z \\ ds=3\tan z\sec zdz}}{=}\int\frac{1}{\sqrt{9\sec^2z-9}}3\tan z\sec zdz={} \\
{}\underset{\substack{\big| \\ \sec^2z-1=\tan^2z}}{=}{}&\int\frac{1}{\sqrt{9\tan^2z}}dz=\int\frac{3\tan z\sec z}{3\tan z}dz=\int\sec zdz={} \\
{}={}&\int\frac{\sec z(\tan z+\sec z)}{\tan z+\sec z}dz=\int\frac{\cancel{\sec z\tan z}+\sec^2z}{\cancel{\sec z\tan z}}dz=\int\sec^2zdz.
\end{align*}
Image of work by OP
Thank you.
| Set $t=\ln x$ and $dt=\frac{dx}{x}$
$$=\int\frac{dt}{\sqrt{t^2+6t}}\overset{\text{complete the square}}{=}\int \frac{dt}{\sqrt{(t+3)^2-9}}$$
Set $s=t+3$ and $ds=dt$
$$=\int\frac{ds}{\sqrt{s^2-9}}$$
Set $s=3\sec(p)$ and $ds=3\tan p \sec p dp$, then $\sqrt{s^2-9}=\sqrt{9\sec^2 p-9}=3\tan p$ and $p=\sec^{-1}\left(\frac s 3 \right)$
$$=\int \sec pdp=\ln(\tan p +\sec p)+\mathcal C=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrating a function of tan inverse How would I carry out the following integration?
$$\int_0^12\arctan x^2$$
I tried to substitute $x^2 = \tan\theta$
From there I wrote the integral as:
$$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$
Now, is my only option to use integration by parts? Or is there a better method to solve this integral?
| Notice,you can use integration by parts as follows
$$\int_0^1 2\tan^{-1}(x^2)\ dx=2\int_0^1 \tan^{-1}(x^2)\cdot 1\ dx$$
$$=2\left(\tan^{-1}(x^2)\int 1\ dx-\int \left(\frac{d}{dx}(\tan^{-1}(x^2))\int 1\ dx \right)\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{2x^2}{1+x^4}\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{2}{x^2+\frac{1}{x^2}}\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{\left(1+\frac{1}{x^2}\right)+\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}\ dx-\int \frac{\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{\left(1+\frac{1}{x^2}\right)dx}{\left(x-\frac{1}{x}\right)^2+2}-\int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)^2-2}\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+(\sqrt 2)^2}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^2-(\sqrt 2)^2}\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x-\frac 1x}{\sqrt 2}\right)-\frac{1}{2\sqrt 2}\ln\left|\frac{x+\frac 1x-\sqrt 2}{x+\frac 1x+\sqrt 2}\right|\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x^2-1}{x\sqrt 2}\right)-\frac{1}{2\sqrt 2}\ln\left|\frac{x^2+1-x\sqrt 2}{x^2+1+x\sqrt 2}\right|\right)_0^1$$
$$=2\left(1\cdot \frac{\pi}{4}-\frac{1}{\sqrt 2}\tan^{-1}\left(0\right)-\frac{1}{2\sqrt 2}\ln\left|\frac{2-\sqrt 2}{2+\sqrt 2}\right|-0-\frac{1}{\sqrt 2}\cdot \frac{\pi}{2}+0\right)$$
$$=\color{blue}{\frac{\pi}{2}-\frac{\pi}{\sqrt 2}-\frac{1}{\sqrt 2}\ln(3-2\sqrt 2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Isomorphism of a linear map of polynomials Let $V$ be the space of polynomials in one variable of degree $\leq3$.
Let $T$ be a map assigning to $p(x)$ the polynomial $ap(x)+(bx+c)p'(x)+(dx^2+ex+f)p''(x)$, for some real parameters $a, b, . . . , f.$
What are the set of parameters when $T$ is not an isomorphism?
I think I need to show that Ker$T$ contains only the zero vector, but I'm not sure whether this is correct or if it is, how to do this.
In trying to find the kernel I have found $T(a_0+a_1x+a_2x^2+a_3x^3)$ and simplified this giving me $$T(a_0+a_1x+a_2x^2+a_3x^3)=aa_0+ca_1+2fa_2+x(aa_1+a_1b+2ca_2+2ea_2+6fa_3)+x^2(aa_2+2a_2b+3ca_3+2da_2+6ea_3)+x^3(aa_3+3ba_3+6da_3)$$
I'm not really sure where to go now with it though.
| Let $T$ be $T(p(x))=a \cdot p(x)+(bx+c)\cdot p'(x)+(dx^{2}+ex+f)\cdot p''(x)$
Then, $T$ is a linear map as $T(k\cdot p(x))=k\cdot T(p(x))$ and $T(p(x)+q(x))=T(p(x))+T(q(x))$. So, $T$ can be expressed in the form of matrix(with respect to the basis $1, x, x^{2}, x^{3}$)
Thus, if we calculate $T$ with the fact that differentiation operator is expressed as
$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3\\ 0 & 0 & 0& 0 \end{pmatrix}$
We can say that
$T= a\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0& 1 \end{pmatrix}+b\begin{pmatrix} 0& 0&0&0&\\0&1&0&0\\0&0&2&0\\0&0&0&3\end{pmatrix}+c\begin{pmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{pmatrix}+d\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&2&0\\0&0&0&6\end{pmatrix}+e\begin{pmatrix}0&0&0&0\\0&0&2&0\\0&0&0&6\\0&0&0&0\end{pmatrix}+f\begin{pmatrix}0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\end{pmatrix}=\begin{pmatrix}a&c&2f&0\\0&a+b&2c+2e&6f\\0&0&a+2b+2d&3c+6e\\0&0&0&a+3b+6d\end{pmatrix}$
Thus, $\det(T)=a(a+b)(a+2b+2d)(a+3b+3d)$ should not be zero if and only if $T$ is an isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| If you only need to know what $d$ is, then consider $z=\cos\theta+i\sin\theta$, in which case $$\left(z-\frac 1z\right)^7=(2i\sin\theta)^7=...+35z^4\left(-\frac 1z\right)^3+35z^3\left(-\frac 1z\right)^4+...$$
Hence
$$2^7i^7\sin^7\theta=...-35\left(z- \frac 1z\right)...=...-35.2i\sin\theta+...$$
therefore the required $$d=\frac{35}{64}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Factorize $(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$ I am unable to factorize this over $\mathbb{Z}:$
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$$
Since, this from an exercise of a book (E. J. Barbeau, polynomials) it must have a neat factorization.
I tried to guess some factors by putting $x-y=0$, $x+y=0$, $x=0$.
I also notice that the expression is symmetric with respect to $x, y, z$.
I am trying to factor it without expanding the whole thing though the process of expansion can be eased by observing that $(x+y+z)(x+y-z)(y+z-x)(z+x-y)$ is a well known expression (Heron's formula), so one already knows it is
$$2(x^2y^2+y^2z^2+z^2x^2)-x^4-y^4-z^4$$
But I think there should be a way to do this without expanding the thing.
So, can someone tell me the factorization (preferably with the steps) or any hints as to how should I approach this problem.
| Note that
\begin{eqnarray}
&&(x^2+y^2+z^2)(x+y+z) (x+y-z) (y+z-x) (z+x-y)-8 x^2 y^2 z^2\\
&=&(x^2+y^2+z^2)(-x^4+2 x^2 y^2+2 x^2 z^2-y^4+2 y^2 z^2-z^4)-8x^2y^2z^2\\
&=&(x^2+y^2+z^2)[(-x^4+2 x^2 y^2-y^4)+(2 x^2 z^2+2 y^2 z^2-z^4)]-8 x^2 y^2 z^2\\
&=&(x^2+y^2+z^2)[z^4-(x^2-y^2)^2+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&[(x^2+y^2-z^2)+2z^2][(z^2+x^2-y^2)(z^2-x^2+y^2)+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&[(x^2+y^2-z^2)+2z^2][(x^2-y^2+z^2)(-x^2+y^2+z^2)+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&(x^2+y^2-z^2)(x^2-y^2+z^2)(-x^2+y^2+z^2)+2z^2[(x^2-y^2+z^2)(-x^2+y^2+z^2)\\
&&+(x^2+y^2-z^2)^2+2(x^2+y^2-z^2)z^2-4 x^2 y^2]\\
&=&(x^2+y^2-z^2)(x^2-y^2+z^2)(-x^2+y^2+z^2).
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$
I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how do I find the sum?
| $$\sum^{\infty}_{n=2} \frac{3}{10^n}=\frac{3}{100}+\frac{3}{1000}+...=0.03+0.003+...=0.0333...=\frac{0.333...}{10}=\frac{\frac{1}{3}}{10}=\frac{1}{30}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Value of an element of a symmetric matrix with given eigenvalue. Let \begin{bmatrix}
a & b & c \\
b & d & e \\
c & e & f
\end{bmatrix} be a real matrix with eigenvalues $1$,$0$ and $3$. If the eigenvectors corresponding to $1$ and $0$ are $(1,1,1)^{T}$ and$(1,-1,0)^{T}$ respectively, then what is the value of $3f?$ How to find value of $3f?$
From the given conditions on eigenvalues we have $$a+b+c=1\\b+d+e=1\\c+e+f=1\\a-b=0\\b-d=0\\c-e=0.$$ Is there any short way to find value of $f?$ Please suggest me. Thanks a lot.
| We know trace $a+d+f=4$
Take help from equations you have got from $(A-\lambda I)X=0$
$a=b=d$ and $c=e$
Now $a+d+f=2a+f=4-----(1)$
Find $a$ which is not very hard
$a+b+c=2a+c=1------(2)$
$c+e+f=2c+f=1------(3)$
Solving $(2),(3)$ we have $4a-f=1----(4)$
Solving $1$ and $4$ we have $f=\dfrac{7}{3}$
$3f=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $\sum_{n=-2}^{\infty}\cos^n x=8$, then find $x.$ Let $0<2x<\pi$. If $$\sum_{n=-2}^{\infty}\cos^n x=8,$$ then please find $x.$
I tried $\sum_{n=-2}^{\infty}\cos^n x=\frac{1}{1-\cos x}+\frac{1}{\cos x}+\frac{1}{\cos^2 x}=\frac{1+\cos x}{\cos^2 x \sin^2 x}=8.$ But I cant find $x.$
| Hint: If $-1 \lt \cos(x)\lt 1$ and $\cos(x)\ne 0$, we get that:
$$
\begin{align}
\sum_{n=-2}^{\infty} \cos(x)&=\frac{1}{\cos^2(x)}+\frac{1}{\cos(x)}+\frac{1}{1-\cos(x)}\\
&=\frac{1-\cos(x)+(1-\cos(x))\cos(x)+\cos^2(x)}{(1-\cos(x))\cos^2(x)}\\
&=\frac{1}{(1-\cos(x))\cos^2(x)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Maximizing $\sqrt6xy+4yz$ Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$.
Let $A$ be maximum value of $\sqrt6xy+4yz$. Find $2A^2-4$.
An initial approach was trying out the inequality $\text{RMS-AM-GM}$. Then I tried a parametric substitution: $x=\cos\alpha, y=\sin\alpha cos\beta, z=\sin\beta \sin\alpha$.
How would you go about this?
| As $y$ is repeated in $\sqrt6xy+4yz,$
WLOG choose $y=\cos\alpha, x=\sin\alpha\cos\beta, z=\sin\beta\sin\alpha$
$\sqrt6xy+4yz=\cos\alpha\sin\alpha(\sqrt6\cos\beta+4\sin\beta)$
$=\dfrac{\sin2\alpha}2\cdot\sqrt{22}\cos\left(\beta-\arccos\sqrt{\dfrac6{22}}\right)$
$\le\dfrac{\sqrt{22}}2$
The equality occurs if $\sin2\alpha=1$ and $\beta=\arccos\sqrt{\dfrac6{22}}$
$\iff\alpha=180^\circ n+45^\circ$ and $\cos\beta=\sqrt{\dfrac6{22}};\sin\beta=\dfrac4{\sqrt{22}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$f(z)=\frac{1}{z}=u(x,y)+iv(x,y)$ - Circles of equations $u(x,y)=a$ and $v(x,y)=b$ are orthogonal curves. Let $f(z)=\frac{1}{z}=u(x,y)+iv(x,y)$. Show that the circles of equations $u(x,y)=a$ and $v(x,y)=b$ ($a\not=0$,$b \not=0$, constants) are orthogonal curves.
I am stuck for a while on this problem. How could do this question?
| One has $$\frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2+y^2} +i \,\frac{-y}{x^2+y^2}.$$ Hence $u(x,y) = \frac{x}{x^2+y^2}$ and $v(x,y)=\frac{-y}{x^2+y^2}.$ The equation $u(x,y)=a$ gives $$ax^2+ay^2-x = 0.$$ Completing the square, you get $$\left(x-\frac{1}{2a}\right)^2+y^2 = \left(\frac{1}{2a}\right)^2.$$ Do the same with the other circle and you should get the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to prove $\int_{0} ^{1}{\sqrt{1-x^4}dx}=\frac{1}{12}\sqrt {\frac{2}{\pi}}(\Gamma\left(\frac{1}{4}\right))^2$? To prove ,
$$\int_{0} ^{1}{\sqrt{1-x^4}dx}=\frac{1}{12}\sqrt {\frac{2}{\pi}}(\Gamma\left(\frac{1}{4}\right))^2$$
When we substitute $x^4$ with t
we get the equation
$$\frac{1}{4}\int_{0}^1t^{\frac{-3}{4}}(1-t)^{\frac{1}{2}}dt$$
the expression is in the forms of the beta functions, But I am not able to solve ahead.
| $$I=\frac { 1 }{ 4 } \int _{ 0 }^{ 1 } t^{ \frac { -3 }{ 4 } }(1-t)^{ \frac { 1 }{ 2 } }dt$$
Now, on using Beta function, we get $$I=\frac { 1 }{ 4 } B\left( \frac { 1 }{ 4 } ,\frac { 3 }{ 2 } \right) $$
Now, we use the relation of Beta and gamma function. $$I=\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 2 } \right) }{ \Gamma \left( \frac { 7 }{ 4 } \right) } $$
Now, we use the formula: $\Gamma \left( n+1 \right) =n\Gamma \left( n \right) $
$$I=\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 1 }{ 2 } \right) \times \frac { 1 }{ 2 } }{ \frac { 3 }{ 4 } \Gamma \left( \frac { 3 }{ 4 } \right) } $$
We use this value: $\Gamma \left( \frac { 1 }{ 2 } \right) =\sqrt { \pi } $
$$I=\frac { \sqrt { \pi } }{ 6 } \frac { \Gamma \left( \frac { 1 }{ 4 } \right) }{ \Gamma \left( \frac { 3 }{ 4 } \right) } $$
$$I=\frac { \sqrt { \pi } }{ 6 } \frac { { \left( \Gamma \left( \frac { 1 }{ 4 } \right) \right) }^{ 2 } }{ \Gamma \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 3 }{ 4 } \right) } $$
Now, by Euler's Reflection Formula, we get $$I=\frac { \sqrt { \pi } }{ 6 } \frac { { \left( \Gamma \left( \frac { 1 }{ 4 } \right) \right) }^{ 2 } }{ \frac { \pi }{ \sin { \left( \frac { \pi }{ 4 } \right) } } } $$
$$\boxed{\therefore \int _{ 0 }^{ 1 }{ \sqrt { 1-x^{ 4 } } dx } =\frac { 1 }{ 12 } \sqrt { \frac { 2 }{ \pi } } { \left( \Gamma \left( \frac { 1 }{ 4 } \right) \right) }^{ 2 }}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Palindromes on Keypad and divisibility by $111$ The integers 1 through 9 are arranged as follows on a rectangular keypad:
$\begin{array}{c c c}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{array}$
Consider the 6-digit palindromes formed by going back and forth across a diagonal, row, or column of the grid. (ex: 123321, 159951, 357753, 456654, ...).
Why are all these numbers divisible by 111? In other words, what property of these numbers makes them divisible by 111? I've tried casework on whether the number is formed on a row, diagonal, or column, but haven't gotten anywhere.
In addition, can we generalize the result (keypads with different combinations of numbers that still have the above property)?
| Here's the keypad :
$$\begin{array}{c c c}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{array}$$
Take a look at rows, columns and diagonals. What do they have in common? If you take any 3 numbers in a row, columns or diagonal, say $a\, b\, c$, then $b-a=c-b$. Let's call this difference $\epsilon:=b-a=c-b$.
Now take any palindrome formed with a column, row or diagonal, its general form is
$$P=10^5a+10^4b+10^3c+10^2c+10b+a$$
Replace $b=a+\epsilon$ and $c=a+2\epsilon$
$$\begin{array}{rcl}
P&=&10^5a+10^4(a+\epsilon)+10^3(a+2\epsilon)+10^2(a+2\epsilon)+10(a+\epsilon)+a\\
&=&a(10^5+10^4+10^3+10^2+10+1)+\epsilon(10^4+2\cdot 10^3+2\cdot 10^2+10)\\
&=&111111a+12210\epsilon\\
&=&11\cdot 111\cdot 91\cdot a + 11\cdot 111\cdot 10\cdot \epsilon
\end{array}$$
Both the $a$ term and the $\epsilon$ term are divisible by $111$ and $11$.
Note: $\epsilon$ can be positive or negative or even null. It can even work for $a\,b\,c$ that aren't in the keypad.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$ If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval
$(A)(-1,0)\cup(0,1)$
$(B)(1,2)$
$(C)(-2,-1)$
$(D)(-\infty,-2)\cup(2,\infty)$
My try:
Let $x-\lfloor x \rfloor = \{x\}= t$, where $ 0 \leq \{x\}<1\Rightarrow 0\leq t<1$. Then
$$\Rightarrow -3t^2+2t+a^2 = 0\Rightarrow 3t^2-2t-a^2 = 0$$
$$\displaystyle \Rightarrow t = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$
I do not know how to solve further.
| First, let us see for which values of $a$ there will be solutions:
Since $\,0\leq t = \left\lfloor x \right\rfloor<1\,$ we have
\begin{align}
t &= \dfrac{1\pm \sqrt{1+3a^2}}{3} \in \left[0,1\right)
&\implies&& 0\le1\pm\sqrt{1+3a^2} < 3
\\
&\big(\text{subtract } 1 \text{ from each inequality}\big)
&\implies&& -1\le\pm\sqrt{1+3a^2} < 2
\end{align}
Now consider two cases separately:
*
*$\,-1\le + \sqrt{1+3a^2} < 2\,$
\begin{align}
&-1\le \sqrt{1+3a^2} < 2&\implies&& 0\le\sqrt{1+3a^2} < 2
\\
&\big(\text{square both sides of each inequality}\big)
&\implies&& 0\leq 1+3a^2 <4
\\
&\big(\text{subtract } 1\big)
&\implies&& 0\leq 3a^2<3
\\
&\big(\text{divide by } 3\big)
&\implies&& 0\leq a^2 < 1
\\
&&\implies&& \bbox[1ex, border:solid 1.5pt #e10000]{a\in\left(-1,1\right)}
\end{align}
*$\,-1\le - \sqrt{1+3a^2} < 2\,$
\begin{align}
&-1\le- \sqrt{1+3a^2} < 2&\implies&& 0 < \sqrt{1+3a^2} \leq 1
\\
&\big(\text{square both sides of each inequality}\big)
&\implies&& 0\lt 1+3a^2 \leq 1
\\
&\big(\text{subtract } 1 \text{ again}\big)
&\implies&& -1\lt 3a^2 \leq 0
\\
&&\implies&& \bbox[1ex, border:solid 1.5pt #e10000]{a = 0}
\end{align}
Second, find which values of $a$ correspond to the integral solution of the equation. The answer will be complement to the set of these values in $\mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $tan^2 \theta = \frac{x}{y}$ how can we construct the angle $\theta$? If we are given the values of $x$ and $y$ and we know that $\tan^2 \theta = \dfrac{x}{y}$ is it possible for us to construct the angle $\theta$?
| Continuing from @egreg ‘s work
$\sin ^2 \theta = \dfrac {x}{x + y}$ and $\cos ^2 \theta = \dfrac {y}{x + y}$
Then, $\dfrac {y}{x + y} - \dfrac {x}{x + y} = \cos ^2 \theta - \sin ^2 \theta $
$∴\dfrac {y - x}{y + x} = \cos 2\theta$
Construct the above angle and then bisect it.
Note that this method can effectively avoid the square root construction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find distance from point to line I am asked to find the distance between a point ( 5,1,1 ) and a line
$\displaystyle \left\{\begin{matrix} x + y + z = 0\\ x - 2y + z = 0 \end{matrix}\right.$
What ive done so far is to simplify by gauss elimination and I get to:
$\displaystyle \begin{bmatrix} 1 &1 &1 &0 \\ 0 &1 &0 &0 \end{bmatrix}$
In turn I put this back in the form of an equation
$\left\{\begin{matrix} x + y + z = 0\\ y = 0 \end{matrix}\right.$
What I need to find the distance between the point and line is any point on the line and a directional vector, right?
Am i right in assuming that we have
$\begin{pmatrix} x,y,z \end{pmatrix} = \begin{pmatrix} t,0,-t \end{pmatrix}$
So a point on this line could be (1,0,-1) or (12,0,-12)? If this is correct, how would I go on about finding the directional vector so I can put this all on the form of
$\begin{Vmatrix} \overrightarrow{PQ} \times \overrightarrow{v} \end{Vmatrix} \div \begin{Vmatrix} \overrightarrow{v} \end{Vmatrix}$
So to sum up, Q is given, the line is given in a system of equations, how do I extract a point P on the line and the vector v?
| A variational method :
Let $(x,y,z)$ the point on the line the closest from $(5,1,1)$. The distance is $D$.
$$D^2=(x-5)^2+(y-1)^2+(z-1)^2$$
For the smallest $D$, the differentiation leads to :
$$(x-5)dx+(y-1)dy+(z-1)dz=0$$
And on the line :
$$dx+dy+dz=0$$
$$dx-2dy+dz=0$$
This must be true any $dx,dy,dz$ so :
$\begin{Vmatrix}
1 & 1 & 1 \\
1 & -2 & 1 \\
(x-5) & (y-1) & (x-1)
\end{Vmatrix} =0$ which leads to :
$$x-z=4$$
Note : instead of considering the determinant, one can eliminate $dx,dy,dz$ from the three above equations and obtain $x-z=4$ as well.
Then, solving :
$\begin{cases}
x+y+z=0 \\
x-2y+z=0 \\
x-z=4 \\
\end{cases}$ gives :
$\begin{cases}
x=2 \\
y=0 \\
z=-2 \\
\end{cases}$
$$D=\sqrt{ (2-5)^2+(0-1)^2+(-2-1)^2 } = \sqrt{19}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stuck. The general idea I know is to get each of the 3 terms above into the form $D\cdot \frac{1}{1-z}$, where $D$ is a constant. But for the first term, it would be $\frac{1}{4}\cdot\frac{1}{(0-z)}$. I can't get rid of the zero.
|
The function
\begin{align*}
f(z)&=\frac{1}{z(z-2)^2} = \frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
\end{align*}
has a simple pole at $z=0$ and a pole of order two at $z=2$.
Since we want to find a Laurent expansion with center $2$, we look at the other pole $0$ and see we have to distinguish two regions.
\begin{align*}
0<|z-2|<2\qquad\quad\text{and}\qquad\qquad
2<|z-2|
\end{align*}
*
*The first region $ |z-2|<2$ is a disc with center $2$, radius $2$ and the pole $0$ at the boundary of the disc. In the interior of this disc the fraction with pole $0$ admits a representation as power series at $z=2$, while the fractions with pole $2$ admit a representation as principal part of a Laurent series at $z=2$.
*The second region $|z-2|>2$ containing all points outside the disc with center $2$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=2$.
A power series expansion of $\frac{1}{z+a}$ at $z=2$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{(a+2)+(z-2)}=\frac{1}{a+2}\cdot\frac{1}{1+\frac{z-2}{a+2}}\\
&=\frac{1}{a+2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z-2}{a+2}\right)^n\tag{1}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(a+2)^{n+1}}(z-2)^n\tag{2}
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=2$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{(z-2)+(a+2)}=\frac{1}{z-2}\cdot\frac{1}{1+\frac{a+2}{z-2}}\\
&=\frac{1}{z-2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{a+2}{z-2}\right)^n\tag{1}\\
&=\sum_{n=0}^{\infty}(-1)^n(a+2)^n\frac{1}{(z-2)^{n+1}}\\
&=\sum_{n=1}^{\infty}(-1)^{n-1}(a+2)^{n-1}\frac{1}{(z-2)^{n}}\tag{3}
\end{align*}
We can now obtain the Laurent expansion of $f(z)$ at $z=2$ for both regions
*
*Region: $|z-2|<2$
\begin{align*}
f(z)&= \frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\cdot\frac{1}{2+(z-2)}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-2)^n-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\tag{4}\\
&=\frac{1}{8}\sum_{n=-2}^{\infty}\frac{(-1)^n}{2^n}(z-2)^n
\end{align*}
*
*Region: $2<|z-2|$
\begin{align*}
f(z)&=\frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\frac{1}{(z-2)+2}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\sum_{n=1}^{\infty}(-1)^{n-1}2^{n-1}\frac{1}{(z-2)^n}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\tag{5}\\
&=\frac{1}{8}\sum_{n=3}^{\infty}(-1)^{n-1}2^{n}\frac{1}{(z-2)^n}
\end{align*}
Comment:
*
*In (1) we use the geometric series
\begin{align*}
\frac{1}{1+z}=\frac{1}{1-(-z)}=\sum_{n=0}^\infty (-z)^n=\sum_{n=0}^\infty (-1)^nz^n\qquad\qquad |z|<1
\end{align*}
*In (4) we apply the series expansion of (2)
*In (5) we apply the principal part representation of (3)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Limit of $x(2\pi - 4\arctan(x))$ I am trying to decide the value of the limit stated below. But I am running out of ideas, since I can't figure out how to rewrite it in a beneficial way.
$$\lim_{x \to \infty} x(2\pi - 4\arctan(x))$$
My attempt
$\lim\limits_{x \to \infty} x(2\pi - 4\arctan(x))$
rewrite $2\pi$ as $4\arctan(0)$
Substitute $x = \frac{1}{t}$
$\lim\limits_{t \to 0} \frac{1}{t}(4\arctan(t) - 4\arctan(\frac{1}{t}))$
$\lim\limits_{t \to 0} \left(\frac{4\arctan(t)}{t} - \frac{4\arctan(\frac{1}{t})}{t}\right)$
The limit of the first term is $4$, because of the standard limit $\lim\limits_{x \to 0} \left(\frac{\arctan(t)}{t}\right) = 1$. But after that I have no idea.
I would like some hints how to solve this problem, and problems similair to this. I currently have problems with limits approaching $0$.
| L'Hopitals rule is more than sufficient here so this is overkill but you could also have arrived at the result by knowing the first term in the Laurent series for $\arctan(x)$ which says that
$\arctan(x) \approx \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} + ...$ for large $x$. This is sort of like a Maclaurin series only with inverse powers.
$\arctan(x)$ is deligtful because it is rather simple to derive the full formula. You start with
$$\pi/2 - \arctan(x) = \arctan(\infty) - \arctan(x) = \int_x^\infty \frac{ds}{1 + s^2} ds $$
and then make a rearrangement and expansion of the integrand with the geometric series in such a way that we get inverse powers as those are integrable to infinity
$$\frac{1}{s^2 +1} = \frac{1}{s^2}\frac{1}{1 + (-s^{-2})} = \frac{1}{s^2}\sum_{k = 0}^\infty (-1)^ks^{-2k} = \sum_{k = 0}^\infty (-1)^k \frac{1}{s^{2(k + 1)}}$$
You integrate
$$\int_x^\infty \frac{ds}{1 + s^2} ds = \sum_{k = 0}^\infty \int_{x}^\infty (-1)^k\frac{1}{s^{2(k + 1)}}ds = \sum_{k = 0}^\infty \frac{(-1)^{k}}{2k +1}\frac{1}{s^{2k + 1}}$$
and got
$$\arctan(x) = \frac{\pi}{2} + \sum_{k = 0}^\infty \frac{(-1)^{k + 1}}{2k +1}\frac{1}{s^{2k + 1}} = \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + ...$$
This neat formula is just a novelty though but taking the first two terms we get
$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) = \lim_{x \to \infty}x\left (2\pi - 4\left ( \frac{\pi}{2} - \frac{1}{x} + O\left (\frac{1}{x^3}\right )\right ) \right )}$$
$$\boxed{\lim_{x \to \infty}x (2\pi - 4\arctan(x)) = \lim_{x \to \infty} 4 + O\left ( \frac{1}{x^2}\right ) = -4 }$$
| {
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Insights needed for the following Lagrange Multipler problem Find the point of the paraboloid $z = \frac{x^2}{4} + \frac{y^2}{25}$ that is closest to the point $(3, 0, 0)$.
So this seems like a pretty standard question of Lagrange multiplier, except I ran into some problems but cant figure out where I went wrong.
Attempt:
I maximuse $D^2$ instead of D to remove the square-root
1.) Constraint equation : $\frac{x^2}{4} + \frac{y^2}{25} - z$
2.) $D^2 = (x-3)^2 + y^2 + z^2$
3.) Do the usual Lagrange procedure
4.) $2z = -\lambda \\ 25y = \lambda y \\4(x-3) = \lambda x$
5.) Assume $x \neq 0, y\neq0$, we get $z = -\frac{25}{2}$
6.) This is where I got stuck, how can z be negative where is the sum of 2 positive number?
Which part of my attempt did I commit a mistake, and how do I rectify it?
Any insights and help is deeply appreciated.
| if $y=0$ then $z = \frac{x^2}{4}$ so $\lambda = -\frac{x^2}{2}$
so $$ 4(x-3) = -\frac{x^3}{2} \\ \implies x^3+8x-24 = 0 \\ \implies(x-2)(x^2+2x+12)=0$$
so your solution is the point $(2,0,1)$
| {
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Prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$ If the points $\big(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}),(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}) ,\big(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\big)$ are collinear for three distinct values of $a,b,c$ and $a,b,c\neq1$, then prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$.
What should be the smart approach to this question. If I try equating area of triangle formed by three points equal to $0$, then it gets too complicated. Please provide some insight.
| Let:
$$F=abc-(ab+bc+ca)+3(a+b+c) \ \ (0)$$
Let us consider the determinant
$$D=\begin{vmatrix}\frac{a^3}{a-1}&\frac{a^2-3}{a-1}&1\\\frac{b^3}{b-1}&\frac{b^2-3}{b-1}&1\\ \frac{c^3}{c-1}& \frac{c^2-3}{c-1}&1\end{vmatrix}=k\begin{vmatrix}a^3&a^2-3&a-1\\b^3&b^2-3&b-1\\ c^3&c^2-3&c-1\end{vmatrix} \ \ \ (1)$$
with $k=\dfrac{1}{(a-1)(b-1)(c-1)}$.
(a method you have mentionned), equal to twice the area of the corresponding triangle. It is zero iff the corresponding points are collinear (mathworld.wolfram.com/Collinear.html).
One can factorize $D$ (through a Computer Algebra system or by hand computation, by factoring expressions $(a-c)$, $(b-c)$ and $(a-b)$, or by using the simpler method described at the bottom of this text), under the form:
$$D=F \times \frac{(a-b)(a-c)(b-c)}{(a-1)(b-1)(c-1)} \ \ (2)$$
with $F$ as defined by (0). So: $$D=0 \ \ \ \Leftrightarrow \ \ \ F=0$$
Edit : An important remark for problems of a similar kind
I was puzzled by the fact to find back in (2) the classical factorization of the Vandermonde determinant:
$$V=V(a,b,c)=\begin{vmatrix}a^2&a&1\\b^2&b&1\\ c^2&c&1\end{vmatrix}= (a-b)(a-c)(b-c) \ \ (3).$$
After some googling, I realized that the last determinant in formula (1) is a combination of "Generalized Vandermonde Determinants" (caution: the term can be employed with a different meaning), all having $V$ in factor.
It is a consequence of Theorem 1 in E.R. Heineman (1929): "Generalized Vandermonde Determinants" https://projecteuclid.org/download/pdf_1/euclid.bams/1183496754 generalized Vandermonde determinant heineman .
It is maybe interesting to postpone the reading of this theorem now ; it is better to understand the key idea: the elementary symmetric polynomials in $n$ variables can be expressed, as generalized Vandermonde determinants divided by $V$. Here, for $n=3$:
$$\begin{vmatrix}a^3&a&1\\b^3&b&1\\ c^3&c&1\end{vmatrix} = V\times(a+b+c), \ \
\begin{vmatrix}a^3&a^2&1\\b^3&b^2&1\\ c^3&c^2&1\end{vmatrix} = V\times(ab+bc+ca), \ \
\begin{vmatrix}a^3&a^2&a\\b^3&b^2&b\\ c^3&c^2&c\end{vmatrix} = V\times(abc)
$$
Where $V=V(a,b,c)$ has been defined in (3).
Using these formulas, the development of the last determinant in (1) becomes very straightforward.
Remark: the rule behind the choices of these determinants is as follows: in the (left) augmented matrix:
$$\begin{bmatrix}a^2&a&1\\b^2&b&1\\ c^2&c&1\end{bmatrix} \ \ \rightarrow \ \ \begin{bmatrix}a^3&a^2&a&1\\b^3&b^2&b&1\\ c^3&c^2&c&1\end{bmatrix}
$$
delete, one at a time, one of the three last columns and take the determinant. This process is analogous in any dimension.
Bibliography: See also the more recent paper of Thorsten Werther: "Generalized Vandermonde Determinants over the Chebyshev Basis", 1993, International Computer Science Institute: ftp://ftp.icsi.berkeley.edu/pub/techreports/1993/tr-93-024.pdf
| {
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Evaluate $\int_a^{2a}\sqrt{2ax-x^2}\:dx$ where $a$ is a constant. Evaluate $$\int_a^{2a}\sqrt{2ax-x^2}\:dx$$ where $a$ is a constant.
I used substitution, $x=a(1-\sin k)$. This is my working
But I think I made some mistake since answer isn't correct. Please help.
|
Assume $a\ge0$:
$$\int_{a}^{2a}\sqrt{2ax-x^2}\space\text{d}x=\int_{a}^{2a}\sqrt{a^2-\left(x-a\right)^2}\space\text{d}x=$$
Substitute $u=x-a$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=a-a=0$ and upper bound $u=2a-a=a$:
$$\int_{0}^{a}\sqrt{a^2-u^2}\space\text{d}u=$$
Substitute $u=a\sin(s)$ and $\text{d}u=a\cos(s)\space\text{d}s$.
Then $\sqrt{a^2-u^2}=\sqrt{a^2-a^2\sin^2(s)}=a\cos(s)$ and $s=\arcsin\left(\frac{u}{a}\right)$.
This gives a new lower bound $s=\arcsin\left(\frac{0}{a}\right)=0$ and upper bound $s=\arcsin\left(\frac{a}{a}\right)=\frac{\pi}{2}$:
$$a^2\int_{0}^{\frac{\pi}{2}}\cos^2(s)\space\text{d}s=$$
Use $\cos^2(y)=\frac{\cos(2y)+1}{2}$:
$$a^2\int_{0}^{\frac{\pi}{2}}\left[\frac{1}{2}+\frac{\cos(2s)}{2}\right]\space\text{d}s=a^2\left[\frac{1}{2}\int_{0}^{\frac{\pi}{2}}1\space\text{d}s+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos(2s)\space\text{d}s\right]=$$
$$a^2\left[\frac{1}{2}\left[s\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos(2s)\space\text{d}s\right]=a^2\left[\frac{\pi}{4}+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos(2s)\space\text{d}s\right]=$$
Substitute $p=2s$ and $\text{d}p=2\space\text{d}s$.
This gives a new lower bound $p=2\cdot0=0$ and upper bound $p=2\cdot\frac{\pi}{2}=\pi$:
$$a^2\left[\frac{\pi}{4}+\int_{0}^{\pi}\cos(p)\space\text{d}p\right]=a^2\left[\frac{\pi}{4}+\left[\sin(p)\right]_{0}^{\pi}\right]=$$
$$a^2\left[\frac{\pi}{4}+\sin(\pi)-\sin(0)\right]=a^2\left[\frac{\pi}{4}+0-0\right]=a^2\left[\frac{\pi}{4}\right]=\frac{a^2\pi}{4}$$
| {
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Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is $P(|Y| > 1)$?
Suppose that $X_1,X_2,\ldots,X_{25}$ are independent random variables from $\mathcal{N}(1, 4)$. Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is the probability that $|Y| > 1$?
We know that $Y$ is a standard normal random variable.
I keep trying to transform it into normal distribution, but I keep getting the wrong answer. The answer is $0.6826$. How do I solve this?
| $Y$ is not a standard normal random variable!
If $X_i \sim N(1,4)$, then $X=\frac{1}{25}\sum_{i=1}^{25} X_i \sim N(1,\frac{4}{25}).$ So $Y$ is normal with mean
$$EY=\frac{5}{2} -\frac{2}{5}=\frac{21}{10},$$
and variance
$$\textrm{var}(Y)=\left(\frac{5}{2}\right)^2 \cdot \frac{4}{25} =1.$$
Can you finish it?
| {
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The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$
$x^2+y^2+6x-24y+72=0$ is a circle and $x^2-y^2+6x+16y-46=0$ is a hyperbola.These cut at four points.
$x^2+y^2+6x-24y+72=0$ becomes $(x+3)^2+(y-12)=9^2$.
$x^2-y^2+6x+16y-46=0$ becomes $(x+3)^2-(y-8)^2=-9$.
When i solved these two equations to find the points of intesection,$y=10\pm \sqrt{41}$ and $x$ is $-3\pm\sqrt{36\pm 4\sqrt{41}}$
Now it is very difficult to find the sum of distances of these points from $(-3,2)$.Answer is $40$ given in the book.
| We don't have to find the coordinates of the four points.
Since
$$(x^2+y^2+6x-24y+72)-(x^2-y^2+6x+16y-46)=0$$
$$\Rightarrow 2y^2-40y+118=0$$
we can set the four points as $(-3\pm\alpha,\beta),(-3\pm\gamma,\omega)$ where $$\beta+\omega=-(-40)/2=20,\qquad\beta\omega=118/2=59.$$
Now, noting that
$$(-3+\alpha)^2+\beta^2+6(-3+\alpha)-24\beta+72=0\Rightarrow \alpha^2+(\beta-2)^2=20\beta-59$$
the sum of the distances can be represented as
$$2\sqrt{\alpha^2+(\beta-2)^2}+2\sqrt{\gamma^2+(\omega-2)^2}=2\left(\sqrt{20\beta-59}+\sqrt{20\omega-59}\right)\tag1$$
Then,
$$\begin{align}&\left(\sqrt{20\beta-59}+\sqrt{20\omega-59}\right)^2\\&=20(\beta+\omega)-2\cdot 59+2\sqrt{20^2\beta\omega-20\cdot 59(\beta+\omega)+59^2}\\&=20\cdot 20-2\cdot 59+2\sqrt{20^2\cdot 59-20\cdot 59\cdot 20+59^2}\\&=20^2\end{align}$$
The result follows from $(1)$.
| {
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Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in sum notation:
$$\sum_{i=1}^n\left(i^3-(i-1)^3\right) = n^3$$
If I expand: $(i-1)^3 = i^3 - 3i^2 + 3i - 1$
Thus
$$\left(i^3 - (i-1)^3\right) = 3i^2 - 3i +1$$
My sum is now:
$$3\sum_{i=1}^n i^2 -3 \sum_{i=1}^n i + n = n^3$$
$$\sum_{i=1}^n i^2 = \frac{1}{3} \left(n^3 + 3 \frac{n(n-1)}{2} -n\right)$$
And the expression on the RHS above is not $\frac{1}{6} n (n+1) (2n+1)$
I don't want to solve the above using forward difference, I want to keep it backward.
| Your error is when you evaluate
$$\sum_{i=1}^n i = \frac{n(n-1)}{2}$$
The correct formula is
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
plugging this, everything works.
| {
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Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step. Thank you.
| Since $a_n=n^4-n^2=n^2(n-1)(n+1)$, we can also write
$$
a_{n+1}=(n+1)^4-(n+1)^2=(n+1)^2n(n+2)
$$
Suppose $a_n=n^4-n^2$ is divisible by $3$; then $3$ divides one among $n$, $n-1$ and $n+1$. Since $n$ and $n+1$ appear in the decomposition of $a_{n+1}$. The only remaining case is when $3$ divides $n-1$; but in this case we can use $n+2=(n-1)+3$ and we are done.
Suppose $4$ divides $a_n$. Then either $2\mid n$ or $2\mid(n+1)$ (and also $n-1$). In the case $2\mid(n+1)$, we have $4\mid(n+1)^2$. In the case $2\mid n$, we have $2\mid(n+2)$, so $4\mid n(n+2)$.
Using Fermat's little theorem is of course easier; since $n^3\equiv n\pmod{3}$,
$$
n^4-n^2\equiv n\cdot n^3-n^2\equiv n^2-n^2\equiv0\pmod{3}
$$
Since $n^2\equiv n\pmod{2}$, we have
\begin{align}
n^2-n&\equiv n-n\equiv0\pmod{2}\\[4px]
n^2+n&\equiv n+n\equiv2n\equiv0\pmod{2}
\end{align}
Is this a proof by induction? Well, yes! One uses induction for proving Fermat's little theorem. ;-)
| {
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Finding the limit as $n \to \infty $ of $n\ln\left(1+\frac{\ x}{n^2}\right)$ Find
$$\lim_{n\to \infty} n\ln\left(1+\frac{\ x}{n^2}\right)$$
My attempt: $\lim_{n\to \infty} n \left[\ln\left(\frac{\ n^2 +x}{n^2}\right)\right]$
= $\lim_{n\to \infty} n [\ln (n^2 +x) - \ln(n^2)]$
But I'm not sure how to get this out of indeterminate form.
| $n\ln\left(1+\frac{\ x}{n^2}\right) = \frac{1}{n}n^2\ln\left(1+\frac{\ x}{n^2}\right) = \frac{1}{n} \ln \left( \left( 1+\frac{x}{n^2} \right)^{n^2} \right)$
If $n\to\infty$, then $ \frac{1}{n} \ln \left( \left( 1+\frac{x}{n^2} \right)^{n^2} \right) \equiv \frac{1}{n} \ln \left( e^x \right) \equiv \frac{x}{n} \equiv 0$
| {
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The graph of the equation $x+y=x^3+y^3$ is the union of The graph of the equation $x+y=x^3+y^3$ is the union of
$(A)$line and an ellipse$(B)$line and a parabola$(C)$line and hyperbola$(D)$line and a point
I tried to factorize the given equation.
$x^3-x+y^3-y=0$
$x(x^2-1)+y(y^2-1)=0$
The answer given is a line and an ellipse but i do not understand how this equation is split into a line and an ellipse equation.
| $$(x+y)=x^3+y^3=(x+y)(x^2-xy+y^2)$$
It can be $\;y=-x\;$ , a straight line, or else, after cancelling:
$$x^2-xy+y^2=1\iff x^2-xy+y^2-1=0\iff$$
$$\left(x-\frac y2\right)^2+\frac34y^2=1\;\;(**)$$
Put now:
$$\begin{cases}u:=x-\frac y2\\{}\\v:=y\end{cases}\iff x=u+\frac v2\;,\;\;y=v$$
and you get the ellipse $\;u^2+\frac34v^2=1\;$
| {
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Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reasoning?
Question
Calculate the limit of:
$$x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
when $n$ goes to infinity.
Answer
This can also be written as:
$$\lim_{n \to \infty} \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
The denominator can be written as:
$$\ln(1 + \sqrt[3]{n} + \sqrt[4]{n}) = \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) + \ln(\sqrt[3]{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) \rightarrow 0$$
The numerator can be written as:
$$\ln(1 + \sqrt{n} + \sqrt[3]{n}) = \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) + \ln(\sqrt{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) \rightarrow 0$$
This means that we have the following limit:
$$\lim_{n \to \infty} \frac{\ln(\sqrt{n})}{\ln(\sqrt[3]{n})} = \lim_{n \to \infty} \frac{\ln(n^{\frac{1}{2}})}{\ln(n^{\frac{1}{3}})} = \lim_{n \to \infty} \frac{\frac{1}{2}\ln(n)}{\frac{1}{3}\ln(n)} = \lim_{n \to \infty} \frac{3 \ln(n)}{2\ln(n)} \rightarrow \frac{3}{2}$$
The limit converges towards $\frac{3}{2}.$
| It is much shorter using equivalents:
$1+\sqrt n+\sqrt[3]n\sim_\infty\sqrt n$, hence $\;\ln(1+\sqrt n+\sqrt[3]n)\sim_\infty\frac12\ln(n)$.
Similarly, $\ln(1+\sqrt[3] n+\sqrt[4]n)\sim_\infty\frac13\ln(n)$, whence
$$\frac{\ln(1+\sqrt n+\sqrt[3]n)}{\ln(1+\sqrt[3] n+\sqrt[4]n)}\sim_\infty\frac{\frac12\ln(n)}{\frac13\ln(n)}=\frac32.$$
| {
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Applying Euler's formula I try to solve the following task:
Show with Euler's formula (from complex analysis) that for $z\neq 2k\pi$ where $k\in\mathbb{Z}$ the following is true: $$\sum\limits_{v=0}^n \cos(vz) = \frac{1}{2}+\frac{1}{2}\frac{\sin(nz+\frac{z}{2})}{\sin \frac{z}{2}}$$
My attempt:
$$2\sum\limits_{v=0}^n \cos(vz) = 1+\frac{\sin(nz+\frac{z}{2})}{\sin \frac{z}{2}}$$
$$\sum\limits_{v=0}^n 2\cos(vz)-\sum\limits_{v=0}^n\frac{1}{n} = \frac{\sin(nz+\frac{z}{2})}{\sin \frac{z}{2}}$$
$$\sum\limits_{v=0}^n 2\cos(vz)\sin \frac{z}{2}-\frac{\sin \frac{z}{2}}{n} = \sin(nz+\frac{z}{2})$$
Now I tried using the sinus identity $$\sin ( x \pm y ) = \sin x \cos y \pm \cos x \sin y $$ on the term $\cos(vz)\sin \frac{z}{2}$ but It doesn't seem to make it easier nor help.
My second thought to write cosinus as a series doesn't look like it will help either:
$$2\sum\limits_{v=0}^n \sum_{m=0}^\infty (-1)^m\frac{(vz)^{2m}}{(2m)!} = 1+\frac{\sin(nz+\frac{z}{2})}{\sin \frac{z}{2}}$$
I also don't see where I can apply the Euler formula.
Thanks for help
| $2\sum\limits_{v=0}^n \cos(vz) =\sum\limits_{v=0}^n (e^{ivz}+e^{-ivz}) = \frac{1-e^{(n+1)iz}}{1-e^{iz}}+\frac{1-e^{-(n+1)iz}}{1-e^{-iz}} = \frac{e^{\frac{n+1}{2}iz}}{e^{\frac{1}{2}iz}}(\frac{{e^{\frac{n+1}{2}iz}}-e^{\frac{-(n+1)}{2}iz}}{{e^{\frac{1}{2}iz}}-e^{\frac{-1}{2}iz}})+
\frac{e^{\frac{-(n+1)}{2}iz}}{e^{\frac{-1}{2}iz}}(\frac{{e^{\frac{n+1}{2}iz}}-e^{\frac{-(n+1)}{2}iz}}{{e^{\frac{1}{2}iz}}-e^{\frac{-1}{2}iz}}) =
2\cos{(\frac{n}{2}z)}\frac{\sin{\frac{n+1}{2}z}}{\sin{\frac{1}{2}z}}=\frac{\sin(nz+\frac{1}{2}z)+\sin(\frac{z}{2})}{sin{\frac{z}{2}}} = 1+\frac{\sin(nz+\frac{z}{2})}{\sin \frac{z}{2}}$
| {
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find the limit of $\lim_{x \to 0} \frac{1}{8}\sin(x)^{1/x^2}$
$$\lim_{x \to 0} \frac{\sin(x)}{8}^{\frac{1}{x^2}}$$
$y=\lim_{x \to 0}\frac{1}{x^2}\cdot \lim_{x \to 0} \ln(\frac{\sin(x)}{8})=-\infty$
So we have $e^{-\infty}=0$?
| The limit is $0$.
We have for $0 < x < 1$ the inequality
$$-\log x = \int_x^1 \frac{dt}{t} \geqslant \frac{1-x}{1} \\ \implies \log x \leqslant x - 1$$
If $x > 0$, then as $x \to 0+$
$$\frac{\log \sin x}{x^2} \leqslant \frac{\sin x - 1}{x^2} \to -\infty.$$
If $x < 0$, then, extending into the complex plane, as $x \to 0-$
$$\frac{\log \sin x}{x^2} = \frac{\log\left( -\sin |x|\right)}{|x|^2} = \frac{\log\left( \sin |x|\right) + \log(e^{i\pi})}{|x|^2} \to -\infty.$$
Hence
$$\lim_{x \to 0} (\sin x)^{1/x^2} = \exp\left(\lim_{x \to 0}\frac{\log \sin x}{x^2} \right) = \exp(-\infty) = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit of the sequence $\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right]$ How can we evaluate the following limit
$$
\lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right]
$$
| Using the first principle,
$$\begin{align} \\ & \lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right]
\\ & =\lim_{n\to\infty}\frac{1}{n}\left[\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{2}{n}\right)+\dots+\log\left(1+\frac{n}{n}\right)\right]
\\ & =\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}\log\left(1+\frac{r}{n}\right)
\\ & =\lim_{h\to 0} h \cdot \sum_{r=1}^{n}\log\left(1+rh\right)
\\ & =\int_0^1 \log\left(1+x\right) dx
\\ & =\frac{1}{\ln 10}\int_0^1 \ln\left(1+x\right) dx
\\ & =\frac{1}{\ln 10}\left[x\ln\left(1+x\right)-x+\ln\left(1+x\right)\right]_0^1
\\ & =\frac{1}{\ln 10}\left[(x+1)\ln\left(1+x\right)-x\right]_0^1
\\ & =\frac{2\ln\left(2\right)-1}{\ln 10} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How one can show $(r!)^s$ divides $(rs)!$? I would like if anybody have suggestions to prove:
if $n = rs$
with: $r > 0$ and $s > 0$
then $(r!)^s \mid n!$
suggestions?
| Let illustrate it with an example. Say to prove that
$$
3!^4 | 12!
$$
We may analyze the products as follows:
$$
\eqalign
{
1 \cdot 2 \cdot 3 \cdot \ \ \ &\ \ \ \ \ 1 \cdot 2 \cdot 3 \cdot \cr
1 \cdot 2 \cdot 3 \cdot \ \ \ &\ \ \ \ \ 4 \cdot 5 \cdot 6 \cdot \cr
1 \cdot 2 \cdot 3 \cdot \ \ \ &| \ \ \ \ 7 \cdot 8 \cdot 9 \cdot \cr
1 \cdot 2 \cdot 3 \cdot \ \ \ &\ \ \ \ \ 10 \cdot 11 \cdot 12 \cr
}
$$
Now you should observe that the following hold :
$$
\eqalign{
(1 \cdot 2 \cdot 3 ) \ &| \ \ (1 \cdot 2 \cdot 3) \ obvious \cr
(1 \cdot 2 \cdot 3 ) \ &| \ \ (4 \cdot 5 \cdot 6) \ \ because\ \ \binom{6}{3}\ is \ integer\cr
(1 \cdot 2 \cdot 3 ) \ &| \ \ (7 \cdot 8 \cdot 9) \ \ because\ \ \binom{9}{3}\ is \ integer\cr
(1 \cdot 2 \cdot 3 ) \ &| \ \ (10 \cdot 11 \cdot 12) \ \ because\ \ \binom{12}{3}\ is \ integer \cr
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Let $BD$ be the internal angle bisector of $\Delta ABC$ with $D$ on $AC$. The incentre of $\Delta ABC$ is $(0,4)$ and $D$ is at $(1,3)$
Let $BD$ be the internal angle bisector of $\Delta ABC$ with $D$ on $AC$. The incentre of $\Delta ABC$ is $(0,4)$ and $D$ is at $(1,3)$. If $a,b,c$ are in arithmetic progression, find the point $B$. ($a,b,c$ represents the sides $BC,AC,AB$).
Incentre of a triangle lies on the angle bisector. I can get the parametric point of $B$.
The line joining incentre and $D$ is $$y-4=\frac{4-3}{0-1}(x-0)$$
$$y-4=-x$$
$$y=-x+4$$
Hence $B=(h,4-h)$
I could not solve the problem further. How should I use the information about $a,b,c$?
| From the Bisector Theorem we have
\begin{align*}
\frac{|AD|}{|DC|}&=\frac{|AB|}{|BC|}=\frac{c}{a}\qquad\implies\qquad |AD|=\frac{c}a|DC|=\frac{c}a\left(b-|AD|\right)
\end{align*}
By solving for $|AD|$ the last equation we get
$$|AD|=\frac{bc}{a+c}$$
Since $a,b$ and $c$ are in arithmetic progression it follows that $a+c=2b$, then
$$|AD|=\frac{bc}{2b}=\frac{c}{2}\tag{1}$$
Let $I$ be the incenter of the triangle $ABC$, from the Bisector Theorem again we get
$$\frac{|BI|}{|ID|}=\frac{|AB|}{|AD|}=\frac{c}{c/2}=2\qquad\implies\qquad |BI|=2|ID|=2\sqrt2$$
Which means
\begin{align*}
|BI|&=2\sqrt{2}\\
\sqrt{h^2+(4-(4-h))^2}&=2\sqrt{2}\\
\sqrt{h^2+h^2}&=2\sqrt{2}\\
|h|\sqrt{2}&=2\sqrt{2}\\
\end{align*}
It follows that $h=-2$ in order to make sure that $I$ is inside the triangle $ABC$, then the coordinates of $B$ are $(-2,6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relation between eigenvalues of symmetric $2\times2$ non-singular matrices The eigenvalues of a symmetric $2\times2$ non-singular matrix $$A=\left(\begin{matrix} a_{11} & a_{12}\\ a_{12} & a_{22}\end{matrix}\right)$$ are $\alpha_1$ and $\alpha_2$. How are the eigenvalues of
$$B=\left(\begin{matrix} b\cdot a_{11} & c\cdot a_{12}\\ c\cdot a_{12} & b\cdot a_{22}\end{matrix}\right)$$ and $$C=\left(\begin{matrix} c\cdot a_{11} & c\cdot a_{12}\\ c\cdot a_{12} & b\cdot a_{22}\end{matrix}\right)$$
in which $b$ and $c$ are positive constants, related with $\alpha_1$ and $\alpha_2$.
| Let $\alpha_{1,2}(b,c)$ denote the eigenvalues of $B$. You have
$$
\alpha_1(b,c) + \alpha_2(b,c) = \operatorname{tr}(B) = b\operatorname{tr}(A) = b(\alpha_1+\alpha_2)
$$
and (which is the more messy part)
\begin{align*}
\alpha_1(b,c)\alpha_2(b,c)
&= \det(B) = b^2a_{11}a_{22} - c^2a_{12}^2 = b^2\left(\det(A)+a_{12}^2\right) - c^2a_{12}^2\\
&= b^2\alpha_1\alpha_2 + (b^2-c^2)a_{12}^2.
\end{align*}
Now, we solve the first equation for $\alpha_2(b,c)$ and insert it into the second one:
$$
\alpha_1(b,c)\left(b(\alpha_1+\alpha_2)-\alpha_1(b,c)\right) = b^2\alpha_1\alpha_2 + (b^2-c^2)a_{12}^2.
$$
By symmetry, the same equation holds with $\alpha_1(b,c)$ replaced by $\alpha_2(b,c)$. Now, we solve for it and obtain
$$
\alpha_{1,2}(b,c) = \frac b 2\left[\alpha_1+\alpha_2\pm\sqrt{(\alpha_1-\alpha_2)^2 + 4(t^2-1)a_{12}^2}\,\right],
$$
where $t = c/b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Product of its Prime Factors Given that $4095 = 8^4 - 1$ write $4095$ as a product of its prime factors.
I know how I could separate $4095$ into prime factors however I'm not sure how I could use $8^4 - 1$ to help me.
I could perhaps move the $1$ over to get $4096$ and then work from there...
| To be more systematic:
\begin{align*}
x^{12}-1 &= (x^6-1)(x^6+1)\\
&=(x^3-1)(x^3+1)(x^2+1)(x^4-x^2+1)\\
&=(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1)
\end{align*}
(As a polynomial it could be factored further).
Now set $x=2$. You get
$$4095=1\cdot 7\cdot3 \cdot 3 \cdot 5 \cdot13=3^2\cdot5\cdot 7\cdot 13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{x \to 5} \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}=2$ using $\delta-\epsilon$ method. I need to prove that Prove that $\lim_{x \to 5} \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}=2$ using $\delta-\epsilon$ method. So far, what I have is $\forall \epsilon >0$, there exists a $\delta>0$ such that if $0<|x-5|<\delta$, then $\left| \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}-2 \right| = \left| \frac{-2x+10+\sqrt{x-3}-\sqrt{7-x}}{x-4} \right|$. At this point, I don't know how to deal with the two radicals in the numerator.
| HINT:
$$\sqrt{x-3}-\sqrt{7-x}=\frac{2(x-5)}{\sqrt{x-3}+\sqrt{7-x}}$$
so that we have
$$\left|\frac{(-2x+10)+\sqrt{x-3}-\sqrt{7-x}}{x-4}\right|=2\frac{|x-5|}{|x-4|}\left|1-\frac{1}{\sqrt{x-3}+\sqrt{7-x}}\right|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric substitution $\tan{\frac{x}{2}}=t$. What is $\cos{x}$ then? For example, the integral is:
$$\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}dx$$
And we use the substitution: $\tan{\frac{x}{2}}=t$
Now, to get $\cos{x}$ in terms of $\tan\frac{x}{2}$, I first expressed $\cos^2\frac{x}{2}$ and $\sin^2\frac{x}{2}$ in temrs of $\tan\frac{x}{2}$:
$$\cos^2\frac{x}{2}=\frac{1}{\frac{1}{\cos^2\frac{x}{2}}}=\frac{1}{\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}}{\cos^2\frac{x}{2}}}=\frac{1}{1+\tan^2\frac{x}{2}}=\frac{1}{1+t^2}$$
$$\sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}=1-\frac{1}{1+t^2}=\frac{t^2}{1+t^2}$$
Now, using trig idendity: $\cos{2x}=\cos^2{x}-\sin^2{x}$ we have:
$$\cos{x}=\bigg(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\bigg)=\frac{1}{1+t^2}-\frac{t^2}{1+t^2}=\frac{1-t^2}{1+t^2}$$
which is good, but, I don't know what is wrong with this next procedure (first, getting $\sin^2{x}$, then using trig identity $\sin^2{x}+\cos^2{x}=1$ getting $\cos{x}$).
So, first, expressing $\sin{x}$ in terms of $\tan\frac{x}{2}$ using trig identity $\sin{2x}=2\sin{x}\cos{x}$:
$$\sin^2{x}=4\sin^2\frac{x}{2}\cos^2\frac{x}{2}=4\cdot\frac{t^2}{1+t^2}\cdot\frac{1}{1+t^2}=\frac{4t^2}{(1+t^2)^2}$$
So:
$$\cos^2{x}=1-\sin^2{x}=1-\frac{4t^2}{(1+t^2)^2}=\frac{t^4-2t^2+1-4t^2}{(1+t^2)^2}=\frac{(t^2-1)^2}{(t^2+1)^2}$$
And, $\cos{x}$ is then:
$$\cos{x}=\pm\frac{t^2-1}{t^2+1}$$
Why do I get $\pm$ here? Did I make a mistake somewhere? Thank you for your time.
| I hope it's better to write $$\sin x=A(3\sin x+4\cos x)+B\cdot\dfrac{d(3\sin x+4\cos x)}{dx}$$
and find $A,B$ by comparing the coefficients $\sin x,\cos x$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$ I was wondering if anyone could give a hint on how to prove this expression, I have been stuck on it for hours. Thanks in advance!
Proving that $$\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$$ where $S(n,n-2)$ are the Stirling numbers of the second kind.
| Given the known recurrence for Stirling No. of the $2$nd kind
$$
\left\{ \matrix{
n \cr
m \cr} \right\} = m\left\{ \matrix{
n - 1 \cr
m \cr} \right\} + \left\{ \matrix{
n - 1 \cr
m - 1 \cr} \right\}
$$
we obtain
$$
\eqalign{
& \left\{ \matrix{
n \cr
n - 1 \cr} \right\} = \left( {n - 1} \right)\left\{ \matrix{
n - 1 \cr
n - 1 \cr} \right\} + \left\{ \matrix{
n - 1 \cr
n - 2 \cr} \right\} = \left( {n - 1} \right) + \left\{ \matrix{
n - 1 \cr
n - 2 \cr} \right\}\quad \Rightarrow \quad \cr
& \Rightarrow \quad \left\{ \matrix{
n \cr
n - 1 \cr} \right\} = \left( \matrix{
n \cr
n - 2 \cr} \right) = \left[ {0 \le n} \right]\left( \matrix{
n \cr
2 \cr} \right) \cr}
$$
where $[P]$ is the Iverson bracket: $
\left\{ \matrix{
[TRUE] = 1 \hfill \cr
[FALSE] = 0 \hfill \cr} \right.
$
and then
$$
\eqalign{
& \left\{ \matrix{
n \cr
n - 2 \cr} \right\} = \left( {n - 2} \right)\left\{ \matrix{
n - 1 \cr
n - 2 \cr} \right\} + \left\{ \matrix{
n - 1 \cr
n - 3 \cr} \right\} = \cr
& = \left( {n - 2} \right)\left( \matrix{
n - 1 \cr
n - 3 \cr} \right) + \left\{ \matrix{
n - 1 \cr
n - 3 \cr} \right\} = \left[ {1 \le n} \right]\left( {n - 2} \right)\left( \matrix{
n - 1 \cr
2 \cr} \right) + \left\{ \matrix{
n - 1 \cr
n - 3 \cr} \right\} \cr}
$$
Therefrom we can write (we can take the summation index to start from $0$)
$$
\eqalign{
& F(x) = \sum\limits_{0\, \le \,n} {\left\{ \matrix{
n \cr
n - 2 \cr} \right\}x^{\,n} } = \cr
& = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
n - 1 \cr
n - 3 \cr} \right)x^{\,n} } + x\sum\limits_{0\, \le \,n} {\left\{ \matrix{
n - 1 \cr
n - 3 \cr} \right\}x^{\,n - 1} } = \cr
& = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
n - 1 \cr
n - 3 \cr} \right)x^{\,n} } + x\,F(x) \cr}
$$
which finally gives
$$
\eqalign{
& \left( {1 - x} \right)F(x) = \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
n - 1 \cr
n - 3 \cr} \right)x^{\,n} } = x^{\,3} \sum\limits_{0\, \le \,n} {\left( {n - 2} \right)\left( \matrix{
n - 1 \cr
n - 3 \cr} \right)x^{\,n - 3} } = \cr
& = x^{\,3} {d \over {d\,x}}\sum\limits_{0\, \le \,n} {\left( \matrix{
n - 1 \cr
n - 3 \cr} \right)x^{\,n - 2} } = x^{\,3} {d \over {d\,x}}{1 \over x}\sum\limits_{1\, \le \,n} {\left( \matrix{
n - 1 \cr
2 \cr} \right)x^{\,n - 1} } = x^{\,3} {d \over {d\,x}}{1 \over x}{{x^{\,2} } \over {\left( {1 - x} \right)^{\,3} }} = \cr
& = x^{\,3} {d \over {d\,x}}{x \over {\left( {1 - x} \right)^{\,3} }} = x^{\,3} {{1 + 2x} \over {\left( {1 - x} \right)^{\,4} }} \cr}
$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Average number of throws of 2 dice in order for the sum to be a prime number If I throw 2 dice, what's the average number of throws that I need to do in order for the sum of the 2 dice to be a prime number?
My attempt:
The probability that the sum will be a prime number is $\frac{4}{9}$. And, by applying the formula of mathematical expectation($\frac{1}{p}$) I get $\frac{1}{\frac{4}{9}}$ then $\frac{9}{4} > 1$. Which is not correct.
| Prime numbers between $2$ and $12$ are $2,3,5,7,11$
which can be obtained in $15$ ways as $\boxed{1-1}\boxed{ 1-2/2-1}\boxed{1-4/4-1,\; 2-3/3-2}\boxed{1-6/6-1,2-5/5-2,3-4/4-3}\boxed{ 5-6/6-5} $
Thus P(prime number) = $\dfrac{15}{36} = \dfrac5{12}$
and $E[X] = 1/p = \dfrac{12}5 = 2.4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
IVP solved with Laplace transformation - mistake? I want to solve following IVP with Laplace transformation.
\begin{align}
x''(t) + 2x'(t) + x(t) &=
\begin{cases}
0, & t < 0\\
1, & t \in (0;2)\\
3, & t > 2
\end{cases}\\
x(0) &= 0\\
x'(0)&=0\\
\end{align}
I've transformed left side of the equation like this:
\begin{align}
\mathcal{L}(x''(t) + 2x'(t) + x(t)) = s^2X(s) - sx(0) - x'(0) + 2(sX(s) - x(0)) + X(s)
\end{align}
With conditions applied:
$$
s^2X(s)+2sX(s)+X(s)
$$
Then I transformed right side of the equation like this:
\begin{align}
\mathcal{L}(f(t)) &= F(s)\\
&= \int^0_{-\infty}0\cdot e^{-st} dt + \int^2_{0}1\cdot e^{-st} dt + \int^\infty_{2}3\cdot e^{-st} dt\\
&= 0 + \left[ - \frac{1}{s}e^{-st} \right]_{t=0}^2 + 3 \left[ - \frac{1}{s}e^{-st} \right]_{t=2}^\infty\\
&= \frac{1-e^{-2s}}{s} + 3 \frac{e^{-2s}}{s}\\
&= \frac{1+2e^{-2s}}{s}
\end{align}
So, now the transformed equation with applied conditions looks like this:
\begin{align}
s^2X(s)+2sX(s)+X(s) &= \frac{1+2e^{-2s}}{s}\\
X(s)&= \frac{1+2e^{-2s}}{s(s^2 +2s +1)}
\end{align}
Now I attempted to decompose the right side to partial fractions:
\begin{align}
X(s)&= \frac{1+2e^{-2s}}{s(s^2 +2s +1)}\\
&= (1+2e^{-2s})\frac{1}{s(s+1)^2}\\
&= (1+2e^{-2s})\left( \frac{A}{s} + \frac{B}{s+1} + \frac{C}{(s+1)^2} \right)\\
&= (1+2e^{-2s})\left( \frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} \right)\\
&= \frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} + 2\frac{1}{s}e^{-2s} - 2\frac{1}{s+1}e^{-2s} - 2\frac{1}{(s+1)^2}e^{-2s}
\end{align}
Now I used Inversed Laplace Transformation (u is Heaviside unit-step function):
$$
x(t) = 1- e^{-t} - te^{-t} + 2u(t-2) -2 e^{2-t}u(t-2) - 2e^{2-t}(t-2)u(t-2)
$$
The problem is, that this solution doesn't work for $t < 0$:
$x''(-1) = 2e$
$x'(-1) = -e$
$x(-1) = 1$
$x''(-1) + 2x'(-1) + x(-1) = 2e -2e + 1 = 1$, but $-1 < 0$, so the result should be 0.
Can you see, what I'm doing wrong? I've tried to check my steps with WolframAlpha, but still I don't see the error.
The other situations ($t \in (0;2)$ and $t > 2$) seem to work ok.
| $$
\frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} + 2\frac{1}{s}e^{-2s} - 2\frac{1}{s+1}e^{-2s} - 2\frac{1}{(s+1)^2}e^{-2s}
$$
The inverse Laplace is :
$$
x(t) = u(t)- e^{-t}u(t) - te^{-t}u(t) + 2u(t-2) -2 e^{2-t}u(t-2) - 2e^{2-t}(t-2)u(t-2)
$$
Then
$$x(t) = 0 \forall t < 0 $$
$$
\int_{-\infty (or 0)}^{\infty} u(t) e^{-st} dt = \int_{0}^{\infty} u(t) e^{-st}dt =\int_{0}^{\infty} e^{-st} dt = \frac{e^{-st}}{-s}|_{0}^{\infty} = \frac{1}{s}
$$
$$
\rightarrow \mathcal{L^{-1}}\left\{\frac{1}{s}\right\} = u(t)
$$
you can do the same with the others.
Also you can check a Laplace transform table online to verify this.
| {
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"url": "https://math.stackexchange.com/questions/1725322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find maximum $2\sin 5x-3\cos x$. Is it possible to find the maximum of $2\sin 5x-3\cos x $ without using calculus nor numerical methods?
I suspect there is a way to play around with trig identities until the expression is only in terms of either $\cos$ or $\sin$. I have done some trials but have only managed to make the expression messier.
| We have
$$2\sin(5x)-3\cos(x)=\\2\sin(3x+2x)-3\cos(x)=\\2\sin(3x)\cos(2x)+2\cos(3x)\sin(2x)-3\cos(x)$$
We make use of the identities: $$\sin(3x)=3\sin(x)-4\sin^3(x)\\ \cos(3x)=4\cos^3(x)-3\cos(x)$$
and $$\sin(2x)=2\cos(x)\sin(x)\\\cos(2x)=1-2\sin^2(x)$$
Thus we obtain
$$2[(3\sin(x)-4\sin^3(x))(1-\sin^2(x))]+2[(4\cos^3(x)-3\cos(x))2\sin(x)\cos(x)]-3\cos(x)=\\16\sin^5(x)-20\sin^3(x)+6\sin(x)+16\cos^4(x)\sin(x)-12\sin(x)\cos^2(x)-3\cos(x)$$
We can now make use of the rather obscure identities $$\sin(x)=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$$ and $$\cos(x)=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}$$
to get an expression solely on terms of $u=\tan{\frac{x}{2}}$
Still,I am not sure if this will help though (in terms of not using calculus or numerical methods per request).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine a matrix given two other matrices
Define the following matrices $C,P\in M_2(\mathbb{Z}_{26})$:
$$
C=\begin{pmatrix}22&13\\10&2\end{pmatrix},\quad P=\begin{pmatrix}6&21\\8&4\end{pmatrix}.
$$
Find an invertible matrix $A\in M_2(\mathbb{Z}_{26})$ such that $C=AP$.
My first attempt is to obtain $A$ as $CP^{-1}$. Unfortunately, $\det{P}=12$ and $\gcd(\det{P},26) = 2$. Hence $P$ is not invertible and this approach doesn't work.
Yet, if $C=AP$ is solvable with an invertible $A$ over $\mathbb Z_{26}$, it must also be solvable with an invertible $A$ over $\mathbb Z_{13}$. So, my second attempt is to solve the equation
$$
\tilde{C}=\tilde{A}\tilde{P}\tag{1}
$$
over $\mathbb Z_{13}$, where
\begin{align}
\tilde{C} &= \begin{pmatrix}22&13\\10&2\end{pmatrix}
= \begin{pmatrix}9&0\\10&2\end{pmatrix},\\
\tilde{P} &= \begin{pmatrix}6&21\\8&4\end{pmatrix}
=\begin{pmatrix}6&8\\8&4\end{pmatrix},
\end{align}
and then set
$$A = \tilde{A} + 13A_1\tag{2}$$
for some $A_1\in M_2(\mathbb Z_{26})$ whose entries are either zero or one. (We abuse the notation $\tilde{A}$ here to mean also a matrix with entries in $\mathbb Z_{26}$.) Since $\det\tilde{P} = 12 = -1$ in $\mathbb Z_{13}$, we have
$$
\tilde{P}^{-1} = -\begin{pmatrix}4&-8\\-8&6\end{pmatrix}
=\begin{pmatrix}9&8\\8&7\end{pmatrix}
$$
and hence
$$
\tilde{A} =\tilde{C}\tilde{P}^{-1}
=\begin{pmatrix}9&0\\10&2\end{pmatrix}\begin{pmatrix}9&8\\8&7\end{pmatrix}
=\begin{pmatrix}3&7\\2&3\end{pmatrix}.
$$
Now I'm struggling on finding the correct choice for $A_1$. There are $2^4=16$ different choices and I feel like checking the determinant for all these matrices is incredibly time consuming. Is there a quicker way to do this?
We want the new $A$ to satisfy $\gcd(\det{A},26)=1$ and therefore $\gcd(\det{A},2)=1$ so that we can find $A^{-1}$.
| I would rewrite this as $P^TA^T=C^T$ so it is more consistent with the way we normally view Gaussian elimination. Then
$$\begin{align}\begin{bmatrix}6&8&22&10\\21&4&13&2\end{bmatrix} & \xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}21&4&13&2\\6&8&22&10\end{bmatrix}\\
& \xrightarrow{R_1\times5}\begin{bmatrix}1&20&13&10\\6&8&22&10\end{bmatrix}\\
& \xrightarrow{R_2+20R_1}\begin{bmatrix}1&20&13&10\\0&18&22&2\end{bmatrix}\\
& \xrightarrow{R_2\times3}\begin{bmatrix}1&20&13&10\\0&2&14&6\end{bmatrix}\\
& \xrightarrow{R_1+16R_2}\begin{bmatrix}1&0&3&2\\0&2&14&6\end{bmatrix}\end{align}$$
So now we can read the solution as $A_{11}\equiv3\pmod{26}$, $A_{21}\equiv2\pmod{26}$, $2A_{12}\equiv14\pmod{26}$, and $2A_{22}\equiv6\pmod{26}$. So one of the $4$ possible solutions is
$$A=\begin{bmatrix}3&7\\2&3\end{bmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluate $\int_0^1 \sqrt{x^2+4x+1}\,dx$
Evaluate $\displaystyle\int_0^1 \sqrt{x^2+4x+1}\,dx$
I tried simplify by doing this:
$$\int_0^1 \sqrt{x^2+4x+1}\:dx=\displaystyle\int_0^1 \sqrt{(x+2)^2-3}\:dx$$
Then, by letting $t=x+2$, $dt=dx$
$$\int_0^1 \sqrt{t^2-3}\:dt=\displaystyle\int_0^1 \sqrt{(t-\sqrt 3)(t+\sqrt 3)}\:dt$$
I tried solving $\displaystyle\int_0^1 \sqrt{t^2-3}\:dt$ by substituting: $$k=t^2-3, dt=\frac{dk}{2\sqrt{|k+3|}}$$
But that leads me nowhere:
$$\frac{1}{2}\displaystyle\int_0^1 \frac{\sqrt{k}}{\sqrt{|k+3|}}\:dk$$
Any hints?
| HINT:
Set $x+2=\sqrt3\sec u\implies\sqrt3\tan u=\sqrt{x^2+4x+1}$
$\int\sqrt{(x+2)^2-3}dx=\int\sqrt3\tan u(\sqrt3\sec u\tan u)du=3\int(\sec^3u-\sec u)\ du$
See How to integrate $\sec^3 x \, dx$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731891",
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"source": "stackexchange",
"question_score": "3",
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} |
Trigonometric Indentities $\dfrac{1}{\cos x}+ \dfrac{2\cos x}{\cos 2x}$
If $x=\dfrac{\pi}{7}$ Prove that above expression is equal to $4$.
I worked a few steps and reached here:
$\dfrac{(4\cos^2 x -1)}{\cos x (2\cos^2 x-1)}$
Not able to proceed after this!!!
| For $x=\frac{\pi}{7}$ it follows
$$\cos (4x)=-\cos(3x)\tag{1}$$
Let $y=\cos x$, using two times the cosine double-angle formula we get
$$\cos (4x)=2\cos^2 (2x)-1=2(2y^2-1)^2-1=8y^4-8y^2+1\tag{2}$$
And, from the identity
$$\cos (3x)=4\cos^3x-3\cos x\qquad\text{i.e.}\qquad \cos(3x)=4y^3-3y\tag{3}$$
we get
\begin{align*}
&&8y^4-8y^2+1&=-4y^3+3y\\
\iff&&8y^4+4y^3-8y^2+3y+1&=0\\
\iff&&(y+1)(8y^3-4y^2-4y+1)&=0
\end{align*}
Since $y=\cos \frac{\pi}7\neq -1$ it follows from the last equation
$$8y^3-4y^2-4y+1=0\qquad\iff\qquad4y^2-1=8y^3-4y=4y(2y^2-1)\tag{4}$$
On the other hand, the given expression can be written as
\begin{align*}
\frac1{\cos x}+\frac{2\cos x}{\cos (2x)}&=\frac{4\cos^2 x-1}{(\cos x)(2\cos^2x -1)}\\
&=\frac{4y^2-1}{y(2y^2-1)}\\
&=\frac{4y(2y^2-1)}{y(2y^2-1)}\qquad\text{from }(4)\\
&=4
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737060",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a,b,c>0\;,$ Then value of $\displaystyle \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\rfloor $,
If $a,b,c>0\;,$ Then value of $\displaystyle \bigg \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfloor $, Where $\lfloor x \rfloor $ Rep. floor function of $x$.
$\bf{My\; Try::}$ Using $b+c>a$ and $c+a>b$ and $a+b>c$
So $\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<3$
And Using $\bf{A.M\geq H.M}\;,$ we get
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>\frac{9}{\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}}$$
I did not Understand How can I calculate its Lower bond, Help Required
Thanks
| Hint a small trick is that if you take $3$ numbers greater than $0$ you will always have ywo such pairs whose sum is greater than $3$rd number so division of two pairs will give answer of division as $0<x<1$ thus the sum would always be less than $2$ implying iys sums floor function would be $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.
When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely.
But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.
| If $a=b$ then you can modify the technique for $a=b=1$ slightly to obtain:
$\int \frac{dx}{b+b\cos x}=\int\frac{b-b\cos x}{(b+b\cos x)(b-b\cos x)}dx$
$=\int\frac{b-b\cos x}{b^2-b^2\cos^2 x}dx=\int\frac{b-b\cos x}{b^2(1-\cos^2 x)}dx=\frac{1}{b}\int\frac{1-\cos x}{\sin^2 x}dx$
Splitting the numerator, and further simplifying:
$\frac{1}{b}\int\frac{1}{\sin^2 x}dx-\frac{1}{b}\int\frac{\cos x}{\sin^2 x}dx=\frac{1}{b}\int\csc^2 x\:dx-\frac{1}{b}\int\frac{\cos x}{\sin^2 x}dx$
The integral on the left is $-\cot x$ and the one on the right is an easy $u$-sub with $u=\sin x$.
So what happens with $a\neq b$?
$\int \frac{dx}{a+b\cos x}=\int\frac{a-b\cos x}{(a+b\cos x)(a-b\cos x)}dx=\int\frac{a-b\cos x}{a^2-b^2\cos^2 x}dx$
Now, add and subtract $b^2$ to the denominator and group the $+b^2$ with $-b^2\cos^2x$.
$=\int\frac{a-b\cos x}{a^2-b^2+b^2-b^2\cos^2 x}dx=\int\frac{a-b\cos x}{(a^2-b^2)+b^2(1-\cos^2 x)}dx$
Split the numerator again, and use pythagorean identity.
$\int\frac{a-b\cos x}{(a^2-b^2)+b^2(\sin^2 x)}dx$
and here I am stuck.
| {
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"url": "https://math.stackexchange.com/questions/1740458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| A lot of these answers involve taking the limit as the term number approaches $\infty$, but this is pretty tedious sometimes, so I thought an alternative might help with the evaluation of the sum.
$$\sum_{n=1}^\infty \frac{1}{n(n+3)}=\sum_{n=1}^\infty \bigg(\frac{A}{n}+\frac{B}{n+3}\bigg)=\sum_{n=1}^\infty \bigg(\frac{1/3}{n}-\frac{1/3}{n+3}\bigg)=\frac{1}{3}\sum_{n=1}^\infty \bigg(\frac{1}{n}-\frac{1}{n+3}\bigg)=\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n+3}\bigg)=\frac13\bigg(\sum_{n=1}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=4}^\infty \frac{1}{n}\bigg)$$$$=\frac13\bigg(\sum_{n=1}^3\frac{1}{n}+\sum_{n=4}^\infty \frac{1}{n}\bigg)-\frac13\bigg(\sum_{n=4}^\infty \frac{1}{n}\bigg)=\frac13\sum_{n=1}^3\frac{1}{n}=\frac13\bigg(\frac11+\frac12+\frac13\bigg)=$$$$\frac13\bigg(\frac66+\frac36+\frac26\bigg)=\frac13\bigg(\frac{11}6\bigg)=\frac{11}{18} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what series to compare. I've tried to compare with $\sum \frac{1}{4n+1} - \frac{1}{4n + 3}$ because this serie converges to $\frac{\pi}{4}$, but we have exactly the opposite inaquality $\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} > \frac{1}{4n + 1} - \frac{1}{4n + 3}$.
Can you help me?
| Asmptotics.
As $n \to +\infty$:
$$\begin{align}
(4n+1)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{1}{4n}\right)^{-1/2}
=\frac{1}{2\sqrt{n}}\left(1-\frac{1}{8n}+O(n^{-2})\right)
\\
(4n+3)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{3}{4n}\right)^{-1/2}
=\frac{1}{2\sqrt{n}}\left(1-\frac{3}{8n}+O(n^{-2})\right)
\\
(4n+1)^{-1/2} - (4n+3)^{-1/2}&=
\frac{1}{2\sqrt{n}}\left(0+\frac{2}{8n}+O(n^{-2})\right)
=\frac{1}{8n^{3/2}}+O(n^{-5/2})
\end{align}$$
The series converges by comparison with $$\sum\frac{1}{8n^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form for the area of a convex cyclic n-gon, given the set of edge lengths Let's say we are given a set of positive reals, and we're told that these are the edges of a convex cyclic $n$-gon, and we must compute it's area.
For $n = 3$ there is the famous Heron's formula:
$$A = {1\over 4}\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2}$$
For $n = 4$ there is Brahmagupta's formula:
$$A = {1\over 4}\sqrt{(a^2 + b^2 + c^2 + d^2)^2 + 8abcd - 2(a^4 + b^4 + c^4 + d^4)}$$
Is there a generalization for arbitrary $n$? Does it even exist?
| In case the links go down (shout-out to Martin Buettner), at least for the pentagon we can triangulate it trigonometrically. But beware: it gets a bit complicated. What we see for the pentagon heralds the complexity of the problem for larger numbers of sides.
Let $ABCDE$ be the pentagon, with sides $AB=a, BC=b, CD=c, DE=d, EA=e$. Form the diagonals $AC=x, AD=y$. Now, apply the Law of Cosines to triangles $ABC$ and $ACD$. We get:
$\cos(ABC)=\frac{a^2+b^2-x^2}{2ab}$
$\cos(CDA)=\frac{c^2+y^2-x^2}{2cy}$
The angles on the left side of the equations must be supplementary if quarilateral $ABCD$ is to be cyclic. So the left sides sum to zero, and therefore the right sides do the same. Form this sum and isolate the first power of $y$ leaving $y^2$ in the "solved" expression. This gives:
$y=(\frac{ab}{c})(\frac{c^2+y^2-x^2}{x^2-a^2-b^2}) ... Eq. 1$
We now do the same thing using triangles $CDE$ and $ACD$, focusing on a different angle of the latter triangle.
$\cos(CDE)=\frac{d^2+e^2-y^2}{2de}$
$\cos(ACD)=\frac{c^2+x^2-y^2}{2cx}$
Again the angles are supplementary, their cosines add up to zero and so do the right sides of the equations above. This time solve the sum for $y^2$; do not take the square root:
$y^2=\frac{cx(d^2+e^2)+de(c^2+x^2)}{cx+de} ... Eq. 2$
Now comes the good part. Take the right side of Eq. 2 and substitute that for $y^2$ in the right side of Eq. 1. Square the resulting equation and match the resulting expression for $y^2$ with the right side of Eq. 2. You now have a rational equation for $x$, which can be turned to a (complicated) polynomial equation. That equation turns out to be degree 7:
$x^7(cde)+$
$x^6(c^2d^2+c^2e^2+d^2e^2-a^2b^2)+$
$x^5(cde)[c^2+d^2+e^2-2(a^2+b^2)]+$
$x^4[c^2d^2e^2-2(a^2+b^2)(c^2d^2+c^2e^2+d^2e^2)+2a^2b^2(c^2+d^2+e^2)]+$
$x^3(cde)[-2(a^2+b^2)(c^2+d^2+e^2)+(a^4+6a^2b^2+b^4)^2]+$
$x^2[-2(a^2+b^2)c^2d^2e^2+(a^2+b^2)^2(c^2d^2+c^2e^2+d^2e^2)-a^2b^2(c^2+d^2+e^2)^2]+$
$x(cde)(a^2-b^2)^2(c^2+d^2+e^2)+$
$(a^2-b^2)^2c^2d^2e^2=0,$
and (unless the pentagon is "rigged") we expect it to be irreducible. So given a set of side lengths we cannot generally construct a cyclic pentagon -- even if we are allowed a marked ruler!
For each root obtained (in general, numerically) for $x$, we find $y^2$ using Eq. 2 and then $y$ with Eq. 1. Of course $y$ from Eq. 1 must be a square root of $y^2$ from Eq. 2, but Eq. 1 also determines a specific sign. At least for side lengths where a pentagon can be drawn in real space, there is one root where both $x$ and $y$ are positive. That is your convex cyclic pentagon. The other roots correspond to various "crossed" pentagon configurations -- the pentagram, the "crystal ball" pentagon where a quadrilateral sits on top of a triangle, and the "vampire" pentagon where three triangles look like fangs. The reversed sign of $x$ incusive-or $y$ in these roots come from sides crossing each other so they're effectively reversed relative to the $x$ or $y$ diagonal.
Now that you have $x$ and $y$, you can use Heron's Formula on each triangle and add them up to get the area of the convex pentagon. For the crossed pentagon roots you get a net or signed area by properly giving each triangle a $+$ sign (if the triangle has zero or two "negative" diagonals) or $-$ sign (one "negative" diagonal) and then doing the addition.
Yes, it's complicated. But, actually, it's simpler than a direct area calculation, and you learn more about the geometry of the pentagon.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How to compute the following series using taylor expansion manipulation? How to compute $\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}$ and $\sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}$ using taylor expansion manipulation?
$1.\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}=\frac{1}{x^2}\sum^{\infty}_{n=0} \frac{x^{n+2}}{(n+2)n!}=\frac{1}{x^2}\int\sum^{\infty}_{n=0} \frac{x^{n+1}}{n!}=\frac{x}{x^2}\int\sum^{\infty}_{n=0} \frac{x^{n}}{n!}=\frac{x}{x^2}e^x=\frac{e^x}{x}$
$2. \sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}=\frac12\sum^{\infty}_{n=0}(-1)^n \frac{(2n+2)x^{2n+1}}{(2n+1)!}=\frac12 \frac{d}{dx}\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+2}}{(2n+1)!}=\frac{x}{2} \frac{d}{dx}\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}=\frac{x\cos{x}}{2}$
Is my process and result correct?
| There were some flaws in the development in the OP. Following that development, we write instead
$$\begin{align}
\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}&=\frac{1}{x^2}\sum^{\infty}_{n=0} \frac{x^{n+2}}{(n+2)n!}\\\\
&=\frac{1}{x^2} \sum^{\infty}_{n=0} \int_0^x\frac{t^{n+1}}{n!}\,dt\\\\
&= \frac{1}{x^2}\int_0^x \sum^{\infty}_{n=0} \frac{t^{n+1}}{n!}\,dt\\\\
&=\frac{1}{x^2}\int_0^x t\sum^{\infty}_{n=0} \frac{t^{n}}{n!}\,dt\\\\
&=\frac{1}{x^2}\int_0^x te^t\,dt\\\\
&=\frac{1}{x^2}\left(e^x(x-1)+1\right)
\end{align}$$
For the second one, we have
$$\begin{align}
\sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}&=\frac12\sum^{\infty}_{n=0}(-1)^n \frac{(2n+2)x^{2n+1}}{(2n+1)!}\\\\
&=\frac12 \frac{d}{dx}\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+2}}{(2n+1)!}\\\\
&=\frac12 \frac{d}{dx}\left(x \sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\right)\\\\
&=\frac12 \frac{d}{dx}\left(x \sin(x)\right)\\\\
&=\frac12 (x\cos(x)+\sin(x))
\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/1746636",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Lucas numbers proof I'm running through some example problems and encountered this one:
Define a sequence of integers $L_n$ by $L_1=1, L_2=3, L_{n+1}=L_n+L_{n-1}.$ Show that $L_n = a\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n + b\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n$ and find the values of $a$ and $b$.
I wasn't able to prove the formula using induction first since the LHS has $a$ and $b$ in it so I cannot show the base case. How can I show the equation holds, and then find $a$ and $b$?
| We, as usual, assume that there is a solution $L_n=\alpha^n$ so we have $\alpha^{n+1}=\alpha^n+\alpha^{n-1}$ hence that $$\alpha^2=\alpha+1$$ It follows
$$\begin{cases} \alpha_1=\frac{1+\sqrt 5}{2}\\ \alpha_2=\frac{1-\sqrt 5}{2}\end{cases}$$
Hence $L_n=(\frac{1+\sqrt 5}{2})^n$ and $L_n=(\frac{1-\sqrt 5}{2})^n$ are solutions.
A general solutions is $$L_n=a(\frac{1+\sqrt 5}{2})^n+b(\frac{1-\sqrt 5}{2})^n$$ what is straightforward to verify. Two initial conditions determine the values $a$ and $b$ in each case. For the given values we have to solve
$$\begin{cases}a(\frac{1+\sqrt 5}{2})+b(\frac{1-\sqrt 5}{2})=1\\a(\frac{1+\sqrt 5}{2})^2+b(\frac{1-\sqrt 5}{2})^2=3\end{cases}$$ which is an easy linear system in $a$ and $b$
| {
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"answer_id": 1
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Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition).
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and made this:
$$\sum_{k=0}^2 \frac {n}{n^2+k} = \frac {n}{n^2+1} + \frac{n}{n^2+2} = \frac {n(n^2+2)+n(n^2+1)}{(n^2+1)(n^2+2)} = \frac {2n^3+...}{n^4+...}$$
And it's limit is $0$, for $k$ up to $3$ is $0$ too, so I assumed that the original limit must be $0$. Wrong, my book says that is $1$, I graphed the function and indeed that is correct. How should I solve it?
| Note: $\frac{n}{n^2+n}\leq\frac{n}{n^2+k}\leq\frac{n}{n^2+1}\leq\frac{1}{n}$ for all $k$ between $1$ and $n$. Hence, $\frac{1}{n+1}\leq\frac{n}{n^2+k}\leq\frac{1}{n}$. Note that $\sum_{k=0}^{n}\frac{1}{n}=\frac{n+1}{n}$ and $\sum_{k=0}^{n}\frac{1}{n+1}=\frac{n+1}{n+1}=1$. Therefore: $$1\leq\lim_{n\to\infty}\sum_{k=0}^{n}\frac{n}{n^2+k}\leq\lim_{n\to\infty}\frac{n+1}{n}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$ I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\dfrac{\pi}{14}+\sin\dfrac{5\pi}{14}\right)=\\
=\sin\dfrac{3\pi}{14}-\left(2\sin\dfrac{3\pi}{14}\cdot\cos\dfrac{\pi}{7}\right)=\\
=\sin\dfrac{3\pi}{14}\left[1-2\cdot\cos\dfrac{\pi}{7}\right]=...
\end{gather}
Tried everything... Here deadlock. I really do not know what to do next. Help somebody, please.
| Let $14x=\pi$
$$S=\sin3x-\sin x-\sin5x=\sin3x+\sin(-x)+\sin(-5x)$$
Using How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? ,
$$\sin(-2x)\cdot S=\cos5x-\cos7x$$
$$-\sin2x\cdot S=\cos5x=\sin2x$$
As $\cos7x=0,\cos5x=\sin2x$ as $5x+2x=\dfrac\pi2$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
There does not exist any integer $m$ such that $3n^2+3n+7=m^3$ I have this really hard problem that I am working on and I just don't seem to get it. The question is: let $n$ be a positive integer; prove that there does not exist any integer $m$ such that $3n^2+3n+7=m^3$.
Please explain in detail otherwise I might not understand. I am also open for any methods that might help. Thank you. Please help.
| I am making this community wiki because I'm really just unpacking the comments. With the right method, this problem may seem trivially easy, but it still requires a few tedious calculations and the answer is not instantaneously obvious, despite what some people may want you to believe.
We know that $m^3 \equiv 0, 1, 8 \pmod 9$ (if you doubt this, take the cube of any positive integer, add up its digits and repeat adding up digits until you only have a single digit left; a 9 is equivalent to 0).
Then $3n^2 \equiv 0, 3 \pmod 9$. And $3n \equiv 0, 3, 6 \pmod 9$, which suggests that $3n^2 + 3n \equiv 0, 3, 6 \pmod 9$ as well. You add 7 to these and you get $3n^2 + 3n \equiv 7, 1, 4 \pmod 9$.
Therefore we need $3n^2 + 3n \equiv 3 \pmod 9$ so that $3n^2 + 3n + 7 \equiv 1 \pmod 9$.
Enumerating the cases one by one:
*
*If $n \equiv 0 \pmod 9$, then $3n^2 + 3n + 7 \equiv 0 + 0 + 7 = 7 \pmod 9$.
*If $n \equiv 1 \pmod 9$, then $3n^2 + 3n + 7 \equiv 3 + 3 + 7 = 4 \pmod 9$.
*If $n \equiv 2 \pmod 9$, then $3n^2 + 3n + 7 \equiv 3 + 6 + 7 = 7 \pmod 9$.
*If $n \equiv 3 \pmod 9$, then $3n^2 + 3n + 7 \equiv 0 + 0 + 7 = 7 \pmod 9$.
*If $n \equiv 4 \pmod 9$, then $3n^2 + 3n + 7 \equiv 3 + 3 + 7 = 4 \pmod 9$.
*If $n \equiv 5 \pmod 9$, then $3n^2 + 3n + 7 \equiv 3 + 6 + 7 = 7 \pmod 9$.
*If $n \equiv 6 \pmod 9$, then $3n^2 + 3n + 7 \equiv 0 + 0 + 7 = 7 \pmod 9$.
*If $n \equiv 7 \pmod 9$, then $3n^2 + 3n + 7 \equiv 3 + 3 + 7 = 4 \pmod 9$.
*If $n \equiv 8 \pmod 9$, then $3n^2 + 3n + 7 \equiv 3 + 6 + 7 = 7 \pmod 9$.
As it turns out, $3n^2 + 3n \equiv 3 \pmod 9$ is impossible. Thus we have shown that $3n^2 + 3n + 7$ lacks one of the characteristics of cubes and therefore can't be a cube.
If we had found an instance of the desired congruence modulo 9, that would have been insufficient to prove the equation does have solutions, even though the absence of the desired congruence does prove there are no solutions.
Also note that it's unnecessary to restrict $n$ to positive integers, since if $n$ is negative, then $n^2$ is positive, while $n = 0$ gives us 7, which is quite clearly not a cube.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Solving $\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$ The integral is:
$$\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$$
This is a Putnam question from my calculus book. It was in the section about trig substitution. However, I see no way of how mixing a trig function with a natural logarithm would make it any easier. I've tried the substitution $x=\tan \theta$ and got $$\dfrac 12\int \ln(1+\tan \theta)^2 d \theta$$
However, I wasn't able to simplify it to any integrable form.
| A substitution other than a trigonometric substitution can be used to evaluate this integral.
Using a so-called self-similar substitution of $u = \dfrac{1 - x}{1 + x}$, one has $x = \dfrac{1 - u}{1 + u}$. Thus $dx = -\dfrac{2}{(1 + u)^2} \, du$ and for the limits of integration when $x = 0, u = 1$ and when $x = 1, u = 0$.
Noting that $1 + x = \dfrac{2}{1 + u}$ and $1 + x^2 = \dfrac{2(1 + u^2)}{(1 + u)^2}$ the integral becomes
\begin{align*}
I &= \int^1_0 \frac{\ln (1 + x)}{1+ x^2} \, dx\\
&= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \cdot \frac{(1 + u)^2}{2 (1 + u^2)} \cdot \frac{2}{(1 + u)^2} \, du\\
&= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \frac{du}{1 + u^2}\\
&= \int^1_0 \left [\ln 2 - \ln (1 + u) \right ] \frac{du}{1 + u^2}\\
&= \ln 2 \int^1_0 \frac{du}{1 + u^2} - \int^1_0 \frac{\ln (1 + u)}{1 + u^2} \, du\\
\Rightarrow 2I &= \ln 2 \cdot \Big{[}\tan^{-1} u \Big{]}^1_0
\end{align*}
giving
$$\int^1_0 \frac{\ln (1 + x)}{1 + x^2} \, dx = \frac{\pi}{8} \ln 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculating integrals using method of Cauchy residues Calculate by the method of residues
$$\int_{0}^{\infty} \frac {x^2 dx}{x^4+6x^2+13}.$$
I am stuck in finding the residues and the contour, any hint?
| Let $I$ be given by
$$I=\int_0^\infty \frac{x^2}{x^4+6x^2+13}\,dx$$
Enforce the substitution $x\to 1/x$ to reveal
$$I=\int_0^\infty \frac{1}{13x^4+6x^2+1}\,dx$$
Then, noting that the integrand is an even function, we can write
$$I=\frac12 \int_{-\infty}^\infty \frac{1}{13x^4+6x^2+1}\,dx \tag 1$$
Now, we move to the complex plane and analyze the integral
$$\begin{align}
J&=\frac12 \oint_{C}\frac{1}{13z^4+6z^2+1}\,dz\\\\
&=\frac12 \int_{-R}^R \frac{1}{13x^4+6x^2+1}\,dx+\int_0^\pi \frac{1}{13R^4e^{i4\phi}+6R^2e^{i2\phi}+1}\,iRe^{i\phi}\,d\phi \tag 2\\\\
&=\pi i \left(\text{Res}\left(\frac{1}{13z^4+6z^2+1}, z=z_1\right)+\text{Res}\left(\frac{1}{13z^4+6z^2+1}, z=z_2\right)\right) \tag 3\\\\
&=\pi i \left(\frac{1}{52z_1^3+12z_1}+\frac{1}{52z_2^3+12z_2}\right) \tag 4
\end{align}$$
where $z_1=\sqrt{\frac{-3+i2}{13}}$ and $z_2=-\sqrt{\frac{-3-i2}{13}}$ are the roots of $13z^4+6z^2+1$ in the upper-half plane.
NOTES:
In going from $(2)$ to $(3)$, we used the Residue Theorem.
In going from $(3)$ to $(4)$, we applied L'Hospital's Rule to find the residues as
$$\lim_{z\to z_{1,2}}\frac{(z-z_{1,2})}{13z^4+6z^2+1}=\frac{1}{52z_{1,2}^3+12z_{1,2}}$$
As $R\to \infty$, the first integral on the right-hand side of $(2)$ approaches $I$ while the second approaches zero. Therefore, we find that $I$ as given in $(1)$ is given by
$$\begin{align}
I&=\frac{\pi i}{4} \left(\frac{1}{z_1(13z_1^2+3)}+\frac{1}{z_2(13z_2^2+3)}\right)\\\\
&=\frac{\pi}{8}\left(\frac{1}{z_1}+\frac{1}{z_2}\right)\\\\
&=\frac{\pi}{4}\text{Re}\left(\sqrt{-3+i2}\right)\\\\
&=\frac{\pi}{4}\sqrt{\frac{\sqrt{13}-3}{2}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show convexity of a function via inequalities I am stuck with deriving the convexity of the function
$$
f(x) = \sqrt{1 + x^2}
$$
from first principles, that is I would like to show that for any $x,y \in \mathbb R$ and $\lambda \in (0,1)$ we have
$$
f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda) f(y)
$$
The fact that it is a convex function is easy to see from the second derivative test so I am ok with the statement, its just that I cannot derive it using basic inequalities!
Here is what I tried so far:
Attempt 1
Use the convexity of the squaring function to write
\begin{align*}
\sqrt{1 + (\lambda x + (1 - \lambda)y)^2} &\le \sqrt{1 + \lambda x^2 + (1 - \lambda)y^2} \\
&= \sqrt{\lambda (1 + x^2) + (1 - \lambda)(1 + y^2)}
\end{align*}
Now use the fact that
$$
\sqrt{A + B} \le \sqrt{A} + \sqrt{B}
$$
to bound the the last expression by
\begin{align*}
\sqrt{\lambda (1 + x^2) + (1 - \lambda)(1 + y^2)} &\le \sqrt{\lambda (1 + x^2)} + \sqrt{(1 - \lambda)(1 + y^2)}
\end{align*}
but now I'm stuck because I cannot take out the factors and bound it above as both $\lambda$ and $1 - \lambda$ are less than $1$ .. :(
Attempt 2:
Don't use the convexity of the squaring function, instead expand. Then
\begin{align*}
\sqrt{1 + (\lambda x + (1 - \lambda)y)^2} &= \sqrt{1 + \lambda^2x^2 + 2\lambda(1 - \lambda)xy + (1 - \lambda)^2y^2} \\
&\le \sqrt{1 + 2\lambda^2x^2 + 2(1 - \lambda)^2y^2}
\end{align*}
so now I need to find a way to show that
$$
1 + \lambda^2x^2 + (1 - \lambda)^2y^2 \le \lambda^2 + (1 - \lambda)^2
$$
but somehow I am too blind / tired to find the right argument .. :(
Thanks for hints!
| We need to check
$$\sqrt{1 + (\lambda x + (1-\lambda)y)^2} \leq \lambda \sqrt{1+x^2} + (1-\lambda) \sqrt{1+y^2}.$$
This is equivalent to
$$1 + (\lambda x + (1-\lambda)y)^2 \leq \lambda ^2 (1+x^2) + (1-\lambda)^2(1+y^2) + 2\lambda (1-\lambda) \sqrt{1+x^2}\sqrt{1+y^2},$$
or equivalently
$$1 + xy \leq \sqrt{1+x^2}\sqrt{1+y^2}.$$
The last inequality is easy to check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How do I simplify $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ using long division? The infinite series $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ is supposed to simplify to $1-\frac{x^2}{2}+\frac{7x^4}{48}+\frac{19x^6}{576}\cdots$ but I don't know how this was calculated.
Did I make an error or did Zill make an error on the $x^6$ term (I got $\frac{11}{576x^6}$.)
| Assume that:
$$\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}+\ldots} = 1+a_2 x^2+a_4 x^4 + a_6 x^6 + \ldots $$
and multiply both sides by $1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}+\ldots$. If, after that, we consider the coefficient of $x^2$ in both sides, we get $\frac{1}{2}+a_2=0$, so $a_2=-\frac{1}{2}$. If we consider the coefficient of $x^4$, we get $\frac{5}{48}+\frac{1}{2}a_2+a_4=0$, hence $a_4=\frac{7}{48}$. If we consider the coefficient of $x^6$, we get:
$$ \frac{7}{576}+\frac{5}{48}a_2 + \frac{1}{2}a_4 + a_6 = 0, $$
hence $\color{red}{a_6=-\frac{19}{576}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find and solve simultaneous recurrence relations for determining n-digit ternary sequences whose sum of digits is a multiple of 3 I'm studying recurrence relations, and I ran into the following problem:
Find and solve simultaneous recurrence relations for determining $n$-digit ternary sequences whose sum of digits is a multiple of 3.
I wish I could say that I have a general idea as far as how to solve this, but I don't. I've never dealt with simultaneous recurrence relations. It's in the section of the book that covers using generating functions, so I'm assuming I will need to be using one of those. Any help would be appreciated.
| Here is a different method that can help you verify your answer once
you have found it. The generating function of these sequences is
$$f(z) = (1+z+z^2)^n.$$
With $\rho = \exp(2\pi i/3)$ we can extract the multiples of three
using
$$\left.\frac{1}{3}(f(z)+f(\rho z)+f(\rho^2 z))\right|_{z=1}.$$
We get
$$\frac{1}{3} (3^n + (1+\rho+\rho^2)^n + (1+\rho^2+\rho)^n)
= 3^{n-1}.$$
Adddendum. As pointed out by @Brian. M. Scott generating
functions are not the best approach here. If you insist it can be done
as follows.
Let $a_n$ be the number of ternary sequence with digit sum
congruent to zero modulo three, $b_n$ congruent to one and $c_n$
congruent to two. We thus have as our initial condition $a_0 = 1$ and
$b_0 = c_0 = 0.$ Introduce
$$A(z) = \sum_{n\ge 0} a_n z^n, \quad
B(z) = \sum_{n\ge 0} b_n z^n, \quad
\text{and}\quad
C(z) = \sum_{n\ge 0} c_n z^n.$$
Now we have the recurrences
$$a_n = a_{n-1} + b_{n-1} + c_{n-1}
\\ b_n = a_{n-1} + b_{n-1} + c_{n-1}
\\ c_n = a_{n-1} + b_{n-1} + c_{n-1}.$$
E.g. we obtain a ternary sequence on $n$ symbols with digit sum
congruent to one modulo three by appending a one to a sequence with
digit sum congruent to zero, a zero to one congruent to one and a two
to one congruent to two.
Next multiply these recurrences by $z^{n-1}$ and sum over $n$
ranging from one to infinity to get
$$\sum_{n\ge 1} z^{n-1} a_n
= \sum_{n\ge 1} z^{n-1} a_{n-1}
+ \sum_{n\ge 1} z^{n-1} b_{n-1}
+ \sum_{n\ge 1} z^{n-1} c_{n-1}$$
which is
$$\frac{1}{z} \sum_{n\ge 1} z^{n} a_n
= A(z) + B(z) + C(z)$$
or $$\frac{1}{z} (A(z) - 1) = A(z) + B(z) + C(z).$$
Similarly we get
$$\frac{1}{z} B(z) = A(z) + B(z) + C(z)$$
and
$$\frac{1}{z} C(z) = A(z) + B(z) + C(z)$$
These three equations may be re-written as
$$-1/z = A(z) (1-1/z) + B(z) + C(z)
\\ 0 = A(z) + B(z) (1-1/z) + C(z)
\\ 0 = A(z) + B(z) + C(z) (1-1/z).$$
This yields
$$-1/z = A(z)(-1/z) + B(z) 1/z
\\ 0 = A(z)(-1/z) + B(z)((1-1/z)^2 - 1).$$
We have $$-1/z = B(z) (1/z+2/z-1/z^2)$$
or $$B(z) = \frac{-1/z}{3/z-1/z^2}
= \frac{-z}{3z-1} = \frac{z}{1-3z}.$$
We also have $-1 = -A(z) + B(z)$ so
$$A(z) = 1 + B(z) = \frac{1-2z}{1-3z}$$
and finally $0 = \frac{1-z}{1-3z} + C(z) (z-1)/z$
so that $$C(z) = \frac{z}{1-3z}.$$
Therefore the answer is
$$A(z) = \frac{1-2z}{1-3z},\quad
B(z) = \frac{z}{1-3z},\quad\text{and}\quad
C(z) = \frac{z}{1-3z}.$$
Extracting coefficients we get
$$[z^n] A(z) = 3^n - 2\times 3^{n-1} = 3^{n-1},\quad
[z^n] B(z) = 3^{n-1},\quad\text{and}\quad
[z^n] C(z) = 3^{n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
proving the inequality $ (\frac{n}{e})^n \leq n! \leq en ( \frac{n}{e})^n$ by induction I want to prove $ (\frac{n}{e})^n \leq n! \leq en ( \frac{n}{e})^n$ by induction. For this prove I want to use the inequality $(\frac{n+1}{n})^n < e <(\frac{n+1}{n})^{n+1}$.
for $n=1$ the inequality is obvious. I Assume it is true for a fixed $n\in \mathbb{N}$ and want to show that
$
(\frac{n+1}{e})^{n+1} \leq (n+1)! \leq e(n+1) ( \frac{n+1}{e})^{n+1}
$
However I have problems with using both the induction hypothesis and the inequalities $(\frac{n+1}{n})^n < e <(\frac{n+1}{n})^{n+1}$ to obtain my answer.
| I had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
$$\begin{align}
a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\
b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n}
\end{align}$$
and
$$\underset{n \rightarrow \infty}{\lim} a_n ~~~=~~~ \underset{n \rightarrow \infty}{\lim} b_n ~~~=~~~ e \\ ~ \\$$
While both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import rcParams
rcParams.update({'figure.autolayout': True})
pts = np.arange(0, 20, 1)
a_n = lambda n: (1+1/n)**n
b_n = lambda n: (1-1/n)**(-n)
plt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = "bx", markersize = "5", markeredgewidth = "2", label = "$a_n$")
plt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = "rx", markersize = "5", markeredgewidth = "2", label = "$b_n$")
plt.plot(pts, [np.exp(1)]*len(pts), color = "black", linewidth = 2, label = "$e$")
plt.xlim(1.5, 14.5)
plt.ylim(2.0, 3.5)
plt.legend(loc = "best")
plt.setp(plt.gca().get_legend().get_texts(), fontsize = "22")
plt.show()
So we're going to use the following inequality:
$$\forall n \in \mathbb{N} ~ : ~~~~~ \left ( 1 + \frac{1}{n} \right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \tag*{$\circledast$} \\ ~ \\$$
Thesis
$$\forall n \in \mathbb{N}, ~ n \geq 2 ~ : ~~~~~ e \cdot \left ( \frac{n}{e} \right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \cdot e \cdot \left ( \frac{n}{e} \right )^n \\ ~ \\$$
Proof By Induction
Base Case
We begin with $n = 2$ and get
$$\begin{align}
& ~ && e \cdot \left ( \frac{2}{e} \right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \cdot e \cdot \left ( \frac{2}{e} \right )^2 \\ ~ \\
& \Leftrightarrow && e \cdot \frac{4}{e^2} ~~~~&&<~~~~ 1 \cdot 2 ~~~~&&<~~~~ 2 \cdot e \cdot \frac{4}{e^2} \\ ~ \\
& \Leftrightarrow && \frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \frac{8}{e} \\ ~ \\
&\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\
\end{align}
$$
Which is a true statement.
Inductive Hypothesis
Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step
$$\begin{align}
& ~ && e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
& = && (n+1) \cdot \frac{1}{e} \cdot e \cdot \left ( \frac{n+1}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( \frac{n+1}{n} \right )^n\\ ~ \\
& = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( 1 + \frac{1}{n} \right )^n\\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot e\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n!\\ ~ \\
& = && (n+1)!\\ ~ \\
& = && (n+1) \cdot n!\\ ~ \\
& \overset{\text{I.H.}}{<} && (n+1) \cdot n \cdot e \cdot \left ( \frac{n}{e} \right )^n\\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n}{e} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( \frac{n}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot e \\ ~ \\
& \overset{\circledast}{<} && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{-(n+1)} \\ ~ \\
& = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\
\end{align}
$$
Conclusion
Therefore the statement holds $\forall n \in \mathbb{N}, ~ n \geq 2$.
$$\tag*{$\square$}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ is equal to :
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ is equal
to :
*
*$\frac{9}{e^2}$
*$3 \log3−2$
*$\frac{18}{e^4}$
*$\frac{27}{e^2}$
My attempt :
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(n+2n)}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2n/n)\}}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2)\}}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(3)\}}{n^{2n}})^{\frac{1}{n}}$
$=\lim_{n \to \infty} (\{(1+1/n)(1+2/n)\dots(3)\})^{\frac{1}{n}}$
I'm stuck here.
Can you please explain?
| Similarly as in this question we can use the fact that:
If $a_n$ is a sequence of positive real numbers and the limit $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists, then $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ also exists and
$$\lim\limits_{n\to\infty} \sqrt[n]{a_n} = \lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}.$$
For the proof see, for example Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$ or How to show that $\lim_{n \to \infty} a_n^{1/n} = l$?
Now if we apply the above to the sequence
$$a_n=\frac{(n+1)(n+2)\dots(3n)}{n^{2n}}$$
we get
\begin{align*}
\frac{a_{n+1}}{a_n}
&= \frac{(3n+1)(3n+2)(3n+3)}{n+1} \cdot \frac{n^{2n}}{(n+1)^{2(n+1)}}\\
&= \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot \frac{n^{2n}}{(n+1)^{2n}}\\
&= \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot \frac{1}{\left(1+\frac1n\right)^{2n}}.
\end{align*}
We now see that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{3^3}{e^2}.$$
| {
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"url": "https://math.stackexchange.com/questions/1760925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Help with finding limit of function with 2 variables Find $$\lim_{(x,y)\to(0,0)} \frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}$$
I have tried taking limits when $y=x$ and when $y=-x$ but failing to get close to an answer.
| *
*Method 1. One may write, as $x^2+y^2 \to 0$, by the use of Taylor series expansions, $$ \begin{align}
\sin(x^2+y^2)&=x^2+y^2+O((x^2+y^2)^3)
\\\\e^{\sin(x^2+y^2)}&=1+x^2+y^2+O((x^2+y^2)^2) \end{align} $$
giving, as $x^2+y^2 \to 0$, $$
\frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}=1+O(x^2+y^2). $$
*Method 2. One may write, as $x^2+y^2 \to 0$, $$ \frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}=\frac{\color{red}{e^{\sin(x^2+y^2)}-1}}{\color{red}{\sin(x^2+y^2)}}\times
\frac{\color{blue}{\sin(x^2+y^2)}}{\color{blue}{x^2+y^2}}. $$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$. Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$.
Since we know that the minimum and maximum values of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$
I applied this formula to get the minimum and maximum values of $(2\sqrt3+4)\sin x+4\cos x$ are $-\sqrt{(2\sqrt3+4)^2+(4)^2}$ and $\sqrt{(2\sqrt3+4)^2+(4)^2}$
$\sqrt{(2\sqrt3+4)^2+(4)^2}=\sqrt{12+16+16+16\sqrt3}=\sqrt{44+16\sqrt3}=2\sqrt{11+4\sqrt3}\neq 2(2+\sqrt5)$
I do not know where i am wrong or is there some other method possible?Please help.
| You want to show $\sqrt{11+4\sqrt3} < 2 + \sqrt5 \iff 11+4\sqrt3 < 9+4\sqrt5 \iff 1 < 2(\sqrt5-\sqrt3) \\ \iff 1 < 4 (8-2\sqrt{15}) \iff 8\sqrt{15} < 31 \iff 960 < 961$
| {
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"source": "stackexchange",
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Why can we use the following integration method? In order to integrate the following function, our professor taught us the following method:
$\dfrac{x^4+1}{x^3+x^2} = \dfrac{x^4+1}{x^2(x+1)}=\dfrac{A}{x^2}+\dfrac{B}{x}+\dfrac{C}{x+1}$
I don't understand why we can do that, in reality, $\dfrac{x^4+1}{x^3+x^2} = \dfrac{x^4+1}{xx(x+1)}=\dfrac{A}{x}+\dfrac{B}{x}+\dfrac{C}{x+1}$
Although I understand why we cannot find $A, B$and$C$ that way (Indeed to find A we evaluate $\dfrac{x^4+1}{x^2(x+1)}$ at x=0)
| You can see that your partial fraction decomposition cannot work simply noting that the numerator of the fraction: $A(x+1)+Bx(x+1)+Cx^2$ cannot be a polynomial of degree $4$. In other words, if the denominators are the factors of a polynomial of degree $n$, and the numerators of the partial fractions are numbers, that the numerator of their sum cannot be of degree $\ge n$.
As noted in the comments, the first step is to use long division to find:
$$
\frac{x^4+1}{x^3+x^2}=x-1+\frac{x^2+1}{x^3+x^2}
$$
than we can use partial fractions for:
$$
\frac{x^2+1}{x^2(x+1)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C}{x+1}
$$
and find $A=1,B=-1,C=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Explicit functions evaluated (a) Defined $f$ by $f(y):=\int_0^\infty\frac{xy}{(x^4+y^4)^{3/4}}dx$. Prove $f(y)$ is defined (i.e integral exists) for every $y\in\mathbb{R}$.
(b)Prove that actually $f(y)=c\operatorname{sign} y$ for some positive number $c$. *So in particular $f$ is not continuous at $0$, therefore a substitution may help and no need to evaluate $c$ explicitly.
(c)Prove that $g(y):=\int_0^\infty\frac{xy}{(x^4+y^4+x^2)^{3/4}}dx$ DOES define a function that is continuous for all $y\in \mathbb{R}$.
These are a few examples I ran across in the midst of a self study. Please, any help is appreciated.
| Part (a)
Clearly $f(0) = 0$.
For $x \geqslant 1,$
$$\left|\frac{xy}{(x^4+y^4)^{3/4}} \right| = \frac{x|y|}{(x^4+y^4)^{3/4} } \leqslant \frac{x|y|}{x^3 } = \frac{|y|}{x^2},$$
and, since,
$$\int_1^\infty \frac{dx}{x^2} = 1 < \infty,$$
it follows by the comparison test that
$$\int_1^\infty\frac{xy}{(x^4+y^4)^{3/4}} \, dx < \infty$$
For $0 < x \leqslant 1$ and $y \neq 0,$
$$\left|\frac{xy}{(x^4+y^4)^{3/4}} \right| = \frac{x|y|}{(x^4+y^4)^{3/4} } \leqslant \frac{x|y|}{|y|^3 } = \frac{x}{|y|^2},$$
and, since, $x$ is integrable over $[0,1],$ it follows, again, by the comparison test that
$$\int_0^1\frac{xy}{(x^4+y^4)^{3/4}} \, dx < \infty$$
Therefore, $f(y)$ is defined for all $y \in \mathbb{R}.$
Part (b)
We have
$$f(y) = \int_0^\infty \frac{xy}{(x^4+y^4)^{3/4}} \, dx = \int_0^\infty \frac{xy}{|y|^3\left(1+ x^4/|y|^4\right)^{3/4}} \, dx. $$
Changing variables with $x = |y|u$ we have
$$f(y) = \frac{y}{|y|^3} \int_0^\infty \frac{|y|u}{\left(1+ u^4\right)^{3/4}}|y| \, du \\ = \frac{y}{|y|} \int_0^\infty \frac{u}{\left(1+ u^4\right)^{3/4}} \, du \\ = \frac{y}{|y|} \int_0^\infty \frac{u}{\left(1+ u^4\right)^{3/4}} \, du \\ = c\text{sign}(y),$$
where
$$c = \int_0^\infty \frac{u}{\left(1+ u^4\right)^{3/4}} \, du.$$
Therefore, $f$ is not continuous at $y =0$ since $f(0) = 0$ and
$$ \lim_{ y \to 0+}f(y) = c, \\ \lim_{ y \to 0-}f(y) = -c.$$
Part (c)
Again we have $g(0) = 0.$ For $y$ in any compact interval $[-b,b],$ we find
$$\left|\frac{xy}{(x^4+y^4+x^2)^{3/4}}\right| \leqslant h(x) = \begin{cases} \frac{|b|}{x^2}, \,\,\,x \geqslant 1 \\ \frac{|b|}{\sqrt{x}}, \,\, 0 < x \leqslant 1\end{cases}.$$
Since $h$ is integrable over $[0, \infty)$ it follows by the Lebesgue dominated convergence theorem that $g$ is continuous for all $y \in \mathbb{R},$ since for any $y_0 \in [-b,b]$ where $b$ can be as large as we like,
$$\lim_{y \to y_0} g(y) = \int_0^\infty \lim_{y \to y_0} \frac{xy}{(x^4+y^4+x^2)^{3/4}}\, dx = g(y_0).$$
| {
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"url": "https://math.stackexchange.com/questions/1770278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from Mathematica
| We employ of the rearrangement inequality.
First, since $x\mapsto x^2$ preserves the order for $x>0,$ we have $$a^2b+b^2c+c^2a\le a^3+b^3+c^3.$$
Next write the condition as $$a\ln a+b\ln b+c\ln c=0.$$ Again using the rearrangement inequality, as $x\mapsto\ln x$ preserves the order for $x>0,$ we have the inequalities: $$\begin{cases}a\ln b+b\ln c+c\ln a\le0\\
a\ln c+b\ln a+c\ln b\le0\\
a\ln a+b\ln b+c\ln c=0\end{cases}$$
Adding these together, we have $(a+b+c)\ln(abc)\le0,$ hence $abc\le1.$
Now let $\ln a+\ln b+\ln c=k\le0.$ Apply the Lagrange multiplier method with condition $g(a,b,c):=\ln a+\ln b+\ln c=k$ for a fixed $k\le0,$ to maximize $f(a,b,c):=a^3+b^3+c^3.$
Thus the Lagrange multiplier gives that the extreme of $f$ occurs when $$\begin{cases}3a^2-\lambda\frac{1}{a}=0\\
3b^2-\lambda\frac{1}{b}=0\\
3c^2-\lambda\frac{1}{c}=0\end{cases},$$
i.e. when $a^3=b^3=c^3=\lambda;$ then $a=b=c$ and hence $f(a,b,c)=3abc=3e^k\le3.$
$\square$
Hope this helps.
Edit:
As pointed out in the comment, the final part about Lagrange multipliers is incorrect; in fact, given the constraint $abc<1,$ it does not follow that $a^2b+b^2c+c^2a\le3.$
Also pointed out in the comment is to use Jensen's inequality to obtain $a+b+c\le3,$ but we are not getting answers yet.
Everything we tried so far to fix this fails. We shall update if we find a way to work around.
| {
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"url": "https://math.stackexchange.com/questions/1774794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.
Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.
First, I took
$$2n+1 \equiv x^2 \equiv 0, 1 \pmod 4$$
which showed that $n$ was even.
Now,
$$3n + 1 \equiv y^2 \equiv 0, 1, 4 \pmod 8$$ But since $n$ is even, we get that $8|n$.
So, now any square $\equiv 0, 1, 4, 5, 6, 9 \pmod{10}$. So, I tested $2n+1$ and $3n+1$ for all numbers from 0 to 9. For only two, 0 and 5, were both of them ending with the residues above mentioned. So, I finally proved that $5|n$.
So, $40|n$.
Is my proof correct?
Also, my proof is too roundabout and lengthy. I had to write many programs to take different modulos. Can anyone suggest a more elegant proof especially for the second part when I have to show that $5|n$?
Thanks.
| Your proof is good.
I suppose you could be more direct by pointing out:
For any natural $m$, $m^2 \equiv 0,1,4 \mod 8$ and $m^2 \equiv 0,1,-1 \mod 5$.
So $2n+1$ and $3n+1$ being perfect squares implies:
a) $2n \equiv -1, 0 ,3 \mod 8$.
As $2n$ is even $2n \equiv -1, 3 \mod 8$ are impossible so $2n \equiv 0 \mod 8$ so $n \equiv 0, 4 \mod 8$.
b) $3n \equiv -1, 0, 3 \mod 8$.
As $n \equiv 0, 4 \mod 8$, $3n \equiv 0, 4 \mod 8$ (respectively). As $3n \equiv 4 \mod 8$ isn't possible $n \equiv 4 \mod 8$ isn't possible so $n \equiv 0 \mod 8.$
c) $2n \equiv -1, 0, -2 \mod 5 \equiv 4, 0, -1 \mod 5$.
So $n \equiv 2, 0, -1 \mod 5$
d) $3n \equiv -1, 0, -2 \mod 5 \equiv -6, 0, 3 \mod 5$
So $n \equiv -2, 0, 1 \mod 5$
The only compatibility between c) and d) is $n \equiv 0 \mod 5$
So we have $8|n$ and $5|n$ so $40|n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A rational function integration Evaluate $$\int \frac{3x^2+1}{(x^2-1)^3}dx$$ I tried breaking the numerator in terms of the denominator but it didn't help much. I also tried a few substitutions buy most of them were useless. Please give some hints. Thanks.
| Partial fractions will do the trick.
Try write $$\frac{(3x^2+1)}{(x^2-1)^3}=\frac{ax^2+bx+c}{(x-1)^3}+\frac{ex^2+fx+g}{(x+1)^3}.$$
Further note that $(x+1)^3-(x-1)^3$ leaves only the quadratic and constant terms as non-zero.
So $$\frac{1}{(x-1)^3}-\frac{1}{(x+1)^3}= \frac{-6x^2-2}{(x-1)^3(x+1)^3} =\frac {-1}{2} \cdot \frac{(3x^2+1)}{(x-1)^3(x+1)^3}.$$
The rest is trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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When are we able to find a quadratic with roots that are a function of another quadratic? Motivation:
Given the roots of the quadratic $2x^2+6x+7=0$ find a quadratic with roots $\alpha^2-1$ and $\beta^2-1$
I was able to solve this problem in two ways:
Method 1:
Sum of the roots $\alpha+\beta=-\frac{b}{a}$
Product of roots $\alpha\beta=\frac{c}{a}$
Hence $\alpha+\beta=-3$ and $\alpha\beta=\frac{7}{2}$
We want an equation with roots $\alpha^2-1$ and $\beta^2-1$
The sum of the roots of the new quadratic will be $\alpha^2-1+\beta^2-1=\alpha^2+\beta^2-2$
The product of the roots of the new quadratic will be $(\alpha^2-1)(\beta^2-1)=\alpha^2\beta^2-(\alpha+\beta)+1$
We are able to compute $\alpha^2+\beta^2$ as it is $(\alpha+\beta)^2-2\alpha\beta$ and so the problem is solved.
Plugging the numbers gives $4u^2+45=0$
Method 2
Let $u=\alpha^2-1\implies\alpha=\sqrt{u+1}$
but we know that $\alpha$ solves the original equation so:
$$\begin{align}2\alpha^2+6\alpha+7&=0\\2(u+1)+6\sqrt{u+1}+7&=0\\\sqrt{u+1}&=\frac{-2u-9}{6}\\u+1&=\frac{1}{36}(-2u-9)^2\\36u+36&=4u^2+36u+81\\0&=4u^2+45\end{align}$$
Question:
The first method clearly uses the values of $\alpha$ and $\beta$ but the second seemingly only requires $\alpha$. How is this possible? Sure, one man's $\alpha$ is another's $\beta$ and so you could relabel as the choice of $\alpha$ and $\beta $ is arbitrary. This is believable because of the symmetry involved in the new roots $\alpha^2-1$ looks much like a $\beta^2-1$ but I feel there must be more to this. Supposing one root of the new quadratic was $\alpha^2-1$ but the other was $\beta^3-2\beta$ or something worse? How would the second method know? This leads me to a more fundamental question.
Are there only certain functions of the roots of an old quadratic that can we can find a new quadratic in this way? I suppose we could consider $(x-f(\alpha)(x-g(\beta)=0$ where $f$ and $g$ are the functions of the old roots but then could we alway compute these numerically?
Thanks for taking the time to read this and for any contributions.
| This is my solution:
Note that $2x^2+6x+7=0$ is equivalent to $x^2= -3x-\dfrac 72$
So
*
*$\alpha + \beta = -3$
*$\alpha \, \beta = \dfrac 72$.
*$\alpha^2 - 1 = -3\alpha-\dfrac 92$
*$\beta^2 - 1 = -3\beta-\dfrac 92$
We compute
\begin{align}
\hline
(\alpha^2 - 1) + (\beta^2 - 1)
&= (-3\alpha-\dfrac 92)+(-3\beta-\dfrac 92)\\
&= -3(\alpha + \beta) - 9 \\
&= 0 \\
\hline
(\alpha^2 - 1)(\beta^2 - 1)
&=(-3\alpha-\dfrac 92)(-3\beta-\dfrac 92) \\
&= 9\alpha\beta+\dfrac{27}{2}(\alpha + \beta) + \dfrac{81}{4}\\
&= \dfrac{63}{2} -\dfrac{81}{2}+\dfrac{81}{4} \\
&= \dfrac{45}{4}\\
\hline
\end{align}
So the polynomial with roots $\alpha^2-1$ and $\beta^2-1$ is $4x^2 + 45$.
another way to look at your second method
Let $\alpha$ and $\beta$ be the roots of $Ax^2 + Bx + C$.
Let $u = \alpha^2 - 1$
Then $\alpha = \pm \sqrt{u+1}$.
\begin{align}
Ax^2 + Bx + C &= 0 \\
A\alpha^2 + B\alpha + C &= 0 \\
A(u+1) \pm B\sqrt{u+1} + C &= 0 \\
\pm B\sqrt{u+1} &= -A(u+1) - C \\
\pm B\sqrt{u+1} &= -Au -(A+C) \\
B^2u + B &= A^2u^2 +2A(A+C)u + (A+C)^2 \\
A^2u^2 +(2A^2-B^2+2AC)u + (A+C)^2 - B &= 0
\end{align}
Letting $\beta = v^2 - 1$ would have resulted in
$A^2v^2 +(2A^2-B^2+2AC)v + (A+C)^2 - B = 0$
So it seems that just using $\alpha$ will work.
| {
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"url": "https://math.stackexchange.com/questions/1776534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| Classical approach:
$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ whereas $u$ and $v$ are functions in $x$.
So when we have
$$\frac{2}{x^2+3^3}$$
We have
$$u(x) = 2$$
$$v(x) = x^2+ 3^3$$
And from that it follows
$$u'(x) = 0$$
$$v'(x) = 2x$$
Using the above formula
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{2}{x^2+3^3}\right) = \frac{0 -2\cdot2x}{(x^2+3^3)^2} = \frac{-4x}{x^4+2\cdot 3^3 x^2 + 3^6} = \frac{-4x}{x^4+54x^2+729}$$
| {
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"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
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For real numbers $a,b,c$ calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ if we have... For real numbers $a,b,c$ we have: $a+b+c=11$ and $\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=\frac{13}{17}$, calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$
I think we should use a trick to solve this,because doing algebraic operations on this problem are too tedius!
| $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=$$
$$=(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)-3=$$
$$=11\cdot\frac{13}{17}-3=\frac{92}{17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the Lagrange multipliers with one constraint: $f(x,y,z) = xyz$ and $g(x,y,z) = x^2+2y^2+3z^2 = 6$ Where
$f(x,y,z) = xyz$
and the constraint is $g(x,y,z) = x^2+2y^2+3z^2 = 6$
I have tried this problem like three or four times and not gotten the solution, I even asked this question once and got the wrong answer from my one and only answerer.
The correct answer is {$\pm\sqrt{2}$,$\pm1$,$\pm\frac{\sqrt{2}}{\sqrt{3}}$} This problem just makes no sense from an algebraic standpoint this is my 4th problem off the odd number exercises and I could do the previous 3 just fine.
My attempt:
Let the function $f$ be defined as
$f(x,y,z) = xyz$
find the maximum and minimum values subject to the constraint:
$g(x,y,z) = x^2+2y^2+3z^2$
$$F=x y z +\lambda \left(x^2+2 y^2+3 z^2-6\right)$$ Computing derivatives $$F'_x=y z+2 \lambda x=0\tag 1$$ $$F'_y=x z+4 \lambda y=0\tag 2$$ $$F'_z=x y+6 \lambda z=0\tag 3$$ $$F'_\lambda=x^2+2 y^2+3 z^2-6=0\tag 4$$ Now, I should consider equations $(1,2,3)$ and solve them for $x,y,z$ in terms of $\lambda$.
Multiplying equations 1,2,3 by $x,y,z$ we obtain that $2x^2=4y^2=6z^2$. From here I found the corresponding multiples that $x^2$ and $y^2$ are in terms of $z$ and plugged into equation 4 to solve for $z$. I found that $x$ was a multiple of $z$ by 3 and $y$ was a multiple of $z$ by $\frac{3}{2}$ I found this by setting $4y^2=6z^2$ and got $\frac{6}{4}$ = $\frac{3}{2}$ Now plugging these into equation 4 I obtained $3+\frac32+3z^2-6=0$ My algebra lead me to $z= \pm\frac{1}{\sqrt{2}}$ but if done right $z$ should equal $\pm\frac{\sqrt{2}}{\sqrt{3}}$
I got this by adding 6 over to the right then subtracted 3 leaving me: $\frac{3}{2}$+$3z^2$=$3$ then subtracting $\frac{3}{2}$ lead me to : $3z^2$= $\frac{3}{2}$ and dividing by 3 gives $\frac{3}{2} \div \frac{3}{1}$ which is equivalent to $\frac{3}{2} \times \frac{1}{3}$ = $\frac{3}{6}$ and you can see that really leaves me with $z^2$= $\frac{1}{3}$ which is equivalent to $z=$ $\pm$ $\frac{1}{\sqrt{3}}$
What did I do incorrectly and how do I proceed from here once I have found $z$. Also how is it that you can write the two functions as two functions added together giving you $F$.
| You asked where exactly you have made a mistake:
Multiplying equations 1,2,3 by $x,y,z$ we obtain that $2x^2=4y^2=6z^2$.
This is correct. (And, since you know what the solution should be, you can check for yourself, that the solution fulfills this equation.)
From here I found the corresponding multiples that $x^2$ and $y^2$ are in terms of $z$ and plugged into equation 4 to solve for $z$. I found that $x$ was a multiple of $z$ by 3 and $y$ was a multiple of $z$ by $\frac{3}{2}$ I found this by setting $4y^2=6z^2$ and got $\frac{6}{4}$ = $\frac{3}{2}$
This is not true.
From $2x^2=6z^2$ you get $x^2=3z^2$.
From $4y^2=6z^2$ you get $2y^2=3z^2$ or $y^2=\frac32z^2$.
Now by plugging this into the equation $x^2+2y^2+3z^2=6$ you get $3z^2+3z^2+3z^2=6$, i.e.
$$9z^2=6$$
and $z^2=\frac23$, $z=\pm\sqrt{\frac23}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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