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Finding value of $x$ in an A.P. If $1$ , $\log_{9}(3^{x+1} + 2)$, $\log_{3}(4⋅3^{x}-1)$ are in A.P. , then $x$ equals ?
| $$2\log_{9}(3^{x+1} + 2)=1+\log_{3}(4⋅3^{x}-1)$$
We have
$$\log_{3}(3^{x+1} + 2)=\log_{3}(4⋅3^{x+1}-3)$$
therefore
$$3^{x+1} + 2=4⋅3^{x+1}-3\implies 3^{x+2}=5$$
$$x=-2+\log_{3}(5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the generating function of a recurrence relation in dependence of a variable Given this inhomogeneous linear recurrence relation of 2nd order :
$F_n = F_{n-2} + a$ for $n \geq 2$
with $F_1 = 1$ and $F_0 = 0$
How do I find the generating function of this recurrence relation in dependence of the variable a? I tried solving it but the solution that I got ($\frac{x + a \cdot x^2}{1-x^2} + \frac{a}{1-x}$) doesn't seem to be right. I hope someone can help.
| You can split this 2nd order recurrence relation into two 1st order recurrence relations. Define $b_0=0$ and $b_n=b_{n-1}+a$ for $n\geq 1$. In the same way define $c_0=1$ and $c_{n}=c_{n-1}+a$ for $n\geq 1$. Now we get
$F_{2n}=b_n$ and $F_{2n+1}=c_n$, the first couple of terms of the generating function look like
$$
F_0 + F_{1} x + F_{2} x^2 + F_{3} x^3 + \cdots = b_0 + c_0 x + b_1 x^2 + c_1 x^3 + \dots.
$$
If you let $B(x), C(x)$ and $F(x)$ correspond to the generating functions of $b_n,c_n$ and $F_n$ respectively you see that
$$
F(x) = B(x^2) + x\cdot C(x^2).
$$
You can easily solve $b_n= a\cdot n$ from which you should be able to get
$$
B(x)= \frac{a x}{(x-1)^2}.
$$
The other recurrence relation can just as easily be solved as $c_n=1+a\cdot n.$ So we get
$$
C(x) = (1 + x^2 + x^3 + \cdots) + B(x) = \frac{1}{1-x} + \frac{a x}{(x-1)^2}.
$$
Now put it all together
$$
F(x) = \frac{a x^2}{(x^2-1)^2} + x\left(\frac{1}{1-x^2} + \frac{a x^2}{(x^2-1)^2}\right).
$$
This can be simplified to
$$
F(x)=\frac{x \left((a-1) x+1\right)}{(x-1)^2 (x+1)}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Use of substitutions in solving equations I am currently working on this problem, I am asked to solve the following
$x^2 - 4 - x\sqrt{x^3 + 3x} = 7$.
I am able to manipulate the above to obtain
$x^5 - x^4 +3x^3 + 22x^2 - 121 = 0$.
The problem asks us to use a substitution but I am unable to figure out what to substitute.
I tried letting $y=x^2 - 11$ and hence $y+14=x^2 + 3$, however I now have an $x^\frac{3}{2}$ which I am not sure how to work with that.
If anyone can offer a hint on where to go from there, especially a substitution that I might not be seeing, that would be extremely helpful.
| The original equation may be written $$x^2=11+x\sqrt{x^3+3x}\,.$$ For the square root to be defined means $x^3+3x\geqslant0$, which implies $x\geqslant 0$. Now, if $x\leqslant1$, then $11+x\sqrt{x^3+3x}\geqslant11>x^2$; on the other hand, if $x>1$, then $11+x\sqrt{x^3+3x}>x\sqrt{x^3}>x^2$. Either way, our equation can have no real solution.
So now we are looking for a complex solution. For convenience, let us rewrite the variable as $z$, so that the equation to which we seek a solution is $$z^2=11+z\sqrt{z^3+3z}.$$Again, in the complex case, for the square root to be meaningful, we must still have $$z^3+3z\in\Bbb R_{\geqslant0}.$$To see what this condition means, write $z=x+\mathrm iy$, where $x,y\in\Bbb R$. We have $$z^3+3z=x(x^2-3y^2+3)+\mathrm iy(3x^2-y^2+3)\in\Bbb R_{\geqslant0}.$$So, after eliminating the case when $z$ is real (i.e. $y=0$ and $z=x$), earlier shown impossible, it follows that $3x^2-y^2+3=0$. Thus $$y=\pm\sqrt{3x^2+3}.$$ Hence $z^3+3z=-x(8x^2+6)\in\Bbb R_{\geqslant0}$, which implies that $x\leqslant0$. Now, after substituting $z=x\pm\mathrm i\sqrt{3x^2+3}$ into the base equation for $z$, and considering the real coefficient, a little simplification yields $$-2x^2=14+x\sqrt{-x(8x^2+6)}\quad \text{with}\quad x\leqslant0.$$This is impossible, since the RHS is always positive while the LHS can never be positive, which shows that our equation has no solution, real or complex.
Note: The fifth-degree polynomial equation mentioned in the question, which is obtained by squaring both sides of the equation $x^2-11=x\sqrt{x^3+3x}\,,$ introduces spurious solutions: those of the equation $$x^2-11=-x\sqrt{x^3+3x}\,,$$which has one real and two (conjugate) complex solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857036",
"timestamp": "2023-03-29T00:00:00",
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Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.
| Condition A: The first two rolls musn't be both 6:
$$A=1-(\frac{1}{6}\times\frac{1}{6})=\frac{35}{36}$$
Condition B: The third roll mustn't be 6:
$$B=1-\frac{1}{6}=\frac{5}{6}$$
Condition C: The last two rolls must be 6:
$$C=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$
Probability of A, B and C being true at the same time:
$$A\bigcup B\bigcup C=A\times B\times C=\frac{175}{6^5}=\frac{175}{7776}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many different 4 letter words can be selected from the word ADVANCED? My attempt : $A-2 , D -2 , V - 1, N -1 , C -1 , E -1 $
$XXXX$ words $=0 $
$XXXY$ words $=0 $
$XXYY$ words $= \binom{2}{2}\times \frac{4!}{2!2!} = 6$
$XXYZ$ words $= \binom{2}{1}\times \binom{5}{2} \times \frac{4!}{2!} =240$
$XYZW$ words $= \binom{6}{4}\times 4!=360$
So the answer is $606$ ?
Is it correct ?
| $606$ is correct.
A g.f. way is to find the coefficient of $x^4$ in $4!(1+x)^4(1+x+\frac{x^2}{2!})^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859239",
"timestamp": "2023-03-29T00:00:00",
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If $\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)=l$, then what is the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$? If the function
$$\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)$$ exists and has a value equal to $l$ then what will be the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$
| You can write the limit as
$$
\lim_{x\to0}\frac{1+bx-(1+ax)\sqrt{1+x}}{x^3}\frac{1}{\sqrt{1+x}(1+bx)}
$$
Since the second factor has limit $1$, we can disregard it. Use Taylor expansion up to order $3$:
$$
\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)
$$
Then the numerator is
$$
1+bx-1-\frac{1}{2}x+\frac{1}{8}x^2-\frac{1}{16}x^3
-ax-\frac{a}{2}x^2+\frac{a}{8}x^3+o(x^3)\\=
\left(b-\frac{1}{2}-a\right)x+
\left(\frac{1}{8}-\frac{a}{2}\right)x^2+
\left(-\frac{1}{16}+\frac{a}{8}\right)x^3+o(x^3)
$$
In order the limit is finite, we need
\begin{cases}
b-\dfrac{1}{2}-a=0
\\[4px]
\dfrac{1}{8}-\dfrac{a}{2}=0
\end{cases}
and the limit is
$$
l=-\frac{1}{16}+\frac{a}{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860434",
"timestamp": "2023-03-29T00:00:00",
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Minimum value of algebraic expression.
If $0\leq x_{i}\leq 1\;\forall i\in \left\{1,2,3,4,5,6,7,8,9,10\right\},$ and $\displaystyle \sum^{10}_{i=1} x^2_{i}=9$
Then $\max$ and $\min$ value of $\displaystyle \sum^{10}_{i=1} x_{i}$
$\bf{My\; Try::}$
Using Cauchy-Schwarz Inequality
$$\left(x^2_{1}+x^2_{2}+.......+x^2_{10}\right)\cdot (1^2+1^2+....1^2)\geq \left(x_{1}+x_{2}+....+x_{10}\right)^2$$
So we get $$\left(x_{1}+x_{2}+....+x_{10}\right)\leq \sqrt{90}$$
Now How can I calculate for its minimum value, Help required, Thanks
| Clearly the minimum values is $9$ (since $a_i\geq a_i^2$), we can obtain it by taking $9$ values of $a_i$ to be $1$ and the rest to be $0$.
We flip the question around:
Given non negative reals $a_1,a_2\dots a_n$ with $a_1+a_2+\dots + a_n=A$
What is the minimum possible value of $a_1^2+a_2^2+\dots + a_n^2$?
Since the function $f(x)=x^2$ is convex the minimum possible value is reached when all values are equal, so the minimum is $\frac{A^2}{n}$
So if $a_1+a_2+\dots + a_n> 3\sqrt{n}$ we have $a_1^2+\dots + a_n^2>\frac{9n}{n}=9$
The maximum value is therefore $3\sqrt{n}$, reachable when each number is $\frac{3}{\sqrt{n
}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solutions of the equation $x^3-y^3=z!-18$ What are the solutions of the equation $x^3-y^3=z!-18$? Here $x,y,z$ are non-negative integers. I have tried brute force but is there a better method?
| We can write equation in this form: $(x-y)^3+3xy(x-y)=z!-18$.
For $z\leq 2$:
If $x$ and $y$ give different residues modulo 3, then $z\in \{0,1,2\}$ (otherwise LS would be 1 or 2 modulo 3 and RS would be divisible by 3).
It can be easily check that only solution is $z=2$, $x=-2$, $y=2$.
For $z\geq 3$:
From previous part, we have $3|(x-y)$. Therefore, $9|z!$, so it must be $z\geq 6$.
Also, since $9|x^3-y^3$, we have $3|x-y$ and $27|x^3-y^3$.
For $z\geq 9$, we have $18=y^3-x^3+z!$ which is not possible, since LS is not divisible by 27 and RS is divisible by 27. Therefore, only left solutions are from set $z\in\{6,7,8\}$.
For $z=6$ solutions are $x=-3$, $y=-9$ and $x=9$, $y=3$. Checking other solutions for $z\in\{6,7,8\}$ is left to reader.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861983",
"timestamp": "2023-03-29T00:00:00",
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Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to ${2^n}{-1}$. Please explain in quotations!
"Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to $2^n-1$." In a very simple case, for $n = 3, 1 + 2 + 4 = 7 = 8 - 1$.
| Since we don't know what $1+2+4+\cdots+2^{n-1}$ is yet, we'll call that number $S$. So we have $S=1+\cdots+2^{n-1}$. Multiplying by $2$ on both sides, we get
$$
2S=2+4+8+\cdots+2^{n-1}+2^n
$$
Here comes the trick: calculate $2S-S$:
$$
\begin{align}
2S-S=&2+4+8+\cdots+2^{n-1}+2^n\\-(1+&2+4+8+\cdots+2^{n-1})
\end{align}
$$
and we see that all but two of the terms disappear, and we're left with $2S-S=2^n-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding a tricky composition of two piecewise functions I have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:
$$f(x) =
\begin{cases}
2x+1, & \text{if $x \le 0$} \\
x^2, & \text{if $x > 0$}
\end{cases}$$
$$g(x) =
\begin{cases}
-x, & \text{if $x < 2$} \\
5, & \text{if $x \ge 2$}
\end{cases}$$
My main question lies in how to approach finding the formula for the composition $g(f(x))$. I have seen a couple of other examples of questions like this online, but the domains of each piecewise function were the same, so the compositions weren't difficult to determine.
In this case, I have assumed that, in finding $g(f(x))$, one must consider only the domain of $f(x)$. Thus, I think it would make sense to test for individual cases: for example, I would try to find $g(f(x))$ when $x <= 0$. $g(f(x))$ when $x <= 0$ would thus be $-2x-1$, right? However, I feel like I'm missing something critical, because I'm just assuming that the condition $x < 2$ for $g(x)$ can just be treated as $x <= 0$ in this case. Sorry for my rambling, and many thanks for anyone who can help lead me to the solution.
| $$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases} $$
Therefore
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases} $$
So now we need to know when $f(x) < 2$ and when $f(x) \ge 2$.
$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases} $$
Let's look at one piece of $f$ at a time. On the first piece, $2x + 1 \ge 2$ means $x \ge 1/2$. But this is impossible because $x \le 0$ on the first piece. Also on the first piece, $2x + 1 < 2$ means $x < 1/2$. Well, on the first piece we have $x \le 0 < 1/2$, therefore $f(x) < 2$ on the entire first piece, i.e., $f(x) < 2$ if $x \le 0$.
On the second piece, $x^2 \ge 2$ means $x \ge \sqrt{2}$ or $x \le -\sqrt{2}$. On the second piece we always have $x > 0$, therefore we have $x^2 \ge 2$ when $x \ge \sqrt{2}$. Also on the second piece, $x^2 < 2$ means $-\sqrt{2} < x < \sqrt{2}$. And since on the second piece we always have $x > 0$, then we must have $0 < x < \sqrt{2}$ in order to have $x^2 < 2$.
Putting it all together so far, we have the following:
$$ f(x) \ge 2 \text{ if and only if } x \ge \sqrt{2}$$
$$ f(x) < 2 \text{ if and only if } x \le 0 \text{ or } 0 < x < \sqrt{2} $$
Notice that this last one can be simplified but we need to keep them separate. Why is this? We'll see as we continue. Recall:
$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases} $$
Therefore:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0 \text{ or } 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases} $$
Separating the conditions gives us:
$$g(f(x)) = \begin{cases} -f(x), & x \le 0\\ -f(x), & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases} $$
And we need to do this because $f(x)$ itself is different for $x \le 0$ and $0 < x < \sqrt{2}$. Finally, we end up with:
$$ h(x) := g(f(x)) = \begin{cases} -(2x+1), & x \le 0 \\ -x^2, & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Find $\lim_{x\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}$
Find $$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}.$$
My attempt:
$$\lim_{n\to \infty}\left(\frac{n+2}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{3}{n-1}\right)^{2n+3}=\lim_{n\to \infty}\left(1+\frac{1}{\frac{n-1}{3}}\right)^{2n+3}$$
Now we should do something to change the power to $\frac{n-1}{3}$ because:
$\lim_{x\to \infty}(1+\frac{1}{x})^x=e$
But I cannot get the answer(the answer is $e^2$). Please give small hints not full answers.
Here is a picture from my answer:
Note that in persian $2=۲$ and $3=۳$.
edit:The answer is mistaked and take $n+1$ instad of $n+1$.
| $$\begin{align}\left(\frac{n+2}{n-1}\right)^{2n+3}&= \left(1+\frac{3}{n-1}\right)^{2n+3}\\ &= \left(1+\frac{3}{n-1}\right)^{2(n-1)+5}\\ &= \left(\left(1+\frac{3}{n-1}\right)^{n-1}\right)^2\left(1+\frac{3}{n-1}\right)^5\end{align}$$
If you like, you can replace $n-1$ with $n$ as $n\rightarrow \infty$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$
Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$
Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?
| We may even look out for integer solutions of $x^3+y^3+z^3=8$ alone. There is a huge literature on the sum of three cubes, and several data bases available. The equation has infinitely many solutions, for example
$$
\displaystyle (18t^3 + 2)^3 + (18t^4)^3 + (-18t^4 - 6t)^3 = 8
$$
for all integers $t$.
Among other solutions we find
$$
15^3+(-16)^3+9^3=8,
$$
which qualifies for $x+y+z=8$. Of course, we should solve the question as you did, but at least we already know a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factorize the polynomial $a^6+8a^3+27$?
I would like to factorize $a^6+8a^3+27$.
I got different answers but one of the answers is
$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$
Can someone tell me how to get this answer? Thanks.
| Here's a slightly sneaky way to arrive at the factorization.
Let $P(a)=a^6+8a^3+27$. Then
$$P(10)=1008027=3\cdot3\cdot31\cdot3613=93\cdot3\cdot3613=(10^2-10+3)\cdot3\cdot3613$$
so we might guess that $a^2-a+3$ is a factor of $a^6+8a^3+27$. If it is, then the other factor would have to take the form $a^4+a^3+\cdots+9$ (so that the product with $a^2-a+3$ has no $a^5$ term and ends with a $27$), and we can see that
$$3\cdot3613=10839=10^4+10^3-2\cdot10^2+3\cdot10+9$$
suggests $a^4+a^3-2a^2+3a+9$ as the other factor.
Note, one might also try grouping $P(10)=9\cdot31\cdot3613=(10-1)\cdot31\cdot3613$, but $a-1$ is clearly not a factor. (Note also that the hard work that's been swept under the rug here is the work that goes into factoring $1008027$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of $a_0+a_1+\cdots + a_n$ If $a_0=2 , a_1=5 $and for $n>1 $
$$a_n=5a_{n-1}-6a_{n-2}$$
then
$$a_0+a_1+\cdots+a_n=?$$
I know telescoping series but
$$a_n=5(a_{n-1}-a_{n-2})-a_{n-2}$$
Thanks for your help.
Mani
| Clearly
$$ a_n-2a_{n-1}=3(a_{n-1}-2a_{n-2}). $$
So
$$ a_n-2a_{n-1}=3^{n-1}(a_{1}-2a_{0})=3^{n-1}. \tag{1}$$
Similarly
$$ a_n-3a_{n-1}=2^{n-1}(a_{1}-2a_{0})=-2^{n-1}. \tag{2}$$
(1)(2) give
$$ a_n=3^n+2^n.$$
So
$$ a_0+a_1+a_2+\cdots+a_n=(1+3+3^2+\cdots+3^n)+(1+2+2^2+\cdots+3^n)=\frac12\cdot(3^{n+1}-1)+(2^{n+1}-1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\cos^n x + \sin^n x =1 $ the solutions of this equation as a function of the value of $n$??
\begin{align}
\cos^n x + \sin^n x =1
\end{align}
I already found the solution if n is odd,
| If $\cos x\sin x\not=0$, then $|\cos x|=\sqrt{1-\sin^2x}$ and $|\sin x|=\sqrt{1-\cos^2x}$ are both strictly less than $1$, which implies $\cos^nx+\sin^nx\lt\cos^2x+\sin^2x=1$ for $n\gt2$. For $n=1$, $\cos x+\sin x=1$ implies $\cos^2x+2\cos x\sin x+\sin^2x=1$, which implies $\cos x\sin x=0$. In sum, when $n$ is a positive integer other than $2$, $\cos^nx+\sin^nx=1$ implies $x$ is a multiple of $\pi/2$. If $n$ is even, any (integer) multiple of $\pi/2$ is a solution. If $n$ is odd, the multiple must be congruent to $0$ or $1$ mod $4$.
For negative integers $n$, we cannot have $\cos x\sin x=0$. If $n\lt0$ is even, then $\cos x\sin x\not=0$ implies $\cos^n+\sin^n\gt\cos^2+\sin^2=1$, so the equation has no solutions. If $n\lt0$ is odd, however, there are solutions. In particular, there is always a solution with $-\pi/2\lt x\lt 0$, since $\sec x\to\infty$ as $x\to-\pi/2^+$ while $\csc x\to-\infty$ as $x\to0^-$.
Finally, if $n=0$, the equation presumably has no solutions, since $\cos^0x+\sin^0x=1+1=2\not=1$ if $\cos x\sin x\not=0$ and isn't clearly defined if $\cos x\sin x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?
| It suffices to compute:
$$\int \frac{y}{y^3 - r^3}dy$$
(putting $r = - \sqrt[3]{\frac{b}{a} }$). Let's assume that $a, b > 0$.
After decomposing into partial fractions, i.e.:
$$\frac{y}{y^3 - r^3} = \frac1{3r} \left( \frac1{y - r} - \frac{y - r}{y^2 + ry + r^2} \right)$$
we find:
$$\int\frac{y}{y^3 -r^3} dy = \frac1{3r}\left[ \log\left( \frac{y-r}{\sqrt{y^2 + ry + r^2}} \right) + \sqrt 3 \arctan \left( \frac{2y + r}{r \sqrt 3} \right) \right]$$
Hence,
$$\int_0^{\infty} \frac{y}{y^3- r^3} dy = \frac{1}{r\sqrt 3} \left( \arctan (-\infty) - \arctan\left( \frac1{\sqrt 3} \right) \right)= -\frac{2\pi}{3r\sqrt 3}$$
and so,
$$\int_{-\infty}^{\infty} \frac{e^{3x}}{ae^{2x} + b} dx = \frac{2\pi}{3\sqrt 3 \sqrt[3]{a^2 b}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Finding the value of an expression using substitution How to find the value of $x^{4000} + \frac{1}{x^{4000}}$ if $x+\frac{1}{x}=1$.
I think binomial theorem will be useful in this.
Here's my proceedings:
$$x+\frac{1}{x}=1$$
Raising both sides to the power of 4000
$$(x+\frac{1}{x})^{4000} = 1^{4000}$$
$${4000 \choose 0} x^{4000} + {4000 \choose 1} x^{3999} \frac{1}{x} + {4000 \choose 2} x^{3998} \frac{1}{x^2} + ... +{4000 \choose 4000} \frac{1}{x^{4000}} = 1$$
$$x^{4000} + \frac{1}{x^{4000}} = 1 - ({4000 \choose 1} x^{3999} \frac{1}{x} + {4000 \choose 2} x^{3998} \frac{1}{x^2} + ... + {4000 \choose 3999} x \frac{1}{x^{3999}})$$
$$x^{4000} + \frac{1}{x^{4000}} = 1- \sum\limits_{r=1}^{3999} {4000 \choose r} x^{4000-r} \frac{1}{x^r}$$
Now how to proceed further to get to my answer?
| Hint: Our number $x$ is a primitive sixth root of unity. Taking high powers of it is exceptionally easy.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| Let $$I = \int\frac{x^2}{(x\sin x+\cos x)(x\cos x-\sin x)}dx$$
Now Put $x=2y\;,$ Then $dx = 2dy$
So $$I = \int \frac{2y^2}{(2y\sin 2y+\cos 2y)(2y\cos 2y-\sin 2y)}dy$$
So $$I = \int\frac{2y^2}{(y^2-1)\sin 2y+2y \cos 2y}dy$$
$$\bullet\; (y^2-1)\sin 2y+2y\cos 2y = (y^2+1)\left[\cos 2y\cdot \frac{2y}{1+y^2}+\sin 2y\cdot \frac{y^2-1}{1+y^2}\right]$$
$$\displaystyle = (y^2+1)\cos \left(2y-\alpha\right) = (y^2+1)\cos \left(2y-\tan^{-1}\left(\frac{y^2-1}{2y}\right)\right)$$
So $$I = \int \sec \left(2y-\tan^{-1}\left(\frac{y^2-1}{2y}\right)\right)\cdot \frac{2y^2}{1+y^2}dy$$
Now Put $\displaystyle \left(2y-\tan^{-1}\left(\frac{y^2-1}{2y}\right)\right)=z\;,$ Then $\displaystyle \frac{2y^2}{y^2+1}dy = dz$
So $$I = \int \sec z dz = \ln \left|\sec z+\tan z\right|+\mathcal{C} = \ln \left|\tan \left(\frac{\pi}{4}+\frac{z}{2}\right)\right|+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
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"source": "stackexchange",
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How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums?
How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums?
My Try: I have Solved It using Limit as a Sum (Reinman Sum of Integral.)
But I did not understand How can I solve it Using Sequeeze Theorem or any other way.
| For $0 \le k \le n$, let $t_k = \frac{k}{n}$.
Divide the interval $[0,1]$ into $n$ sub-intervals $[t_{k-1},t_k]$ for $1 \le k \le n$ and apply MVT to $\tan^{-1} x$ on the intervals. We find for each $k$, there is a $x_k \in ( t_{k-1}, t_k )$ such that
$$\tan^{-1}t_k - \tan^{-1} t_{k-1} = \frac{t_k-t_{k-1}}{1+x_k^2}$$
Since $\frac{1}{1+x^2}$ is monotonic decreasing for $x \ge 0$, we have the bound
$$\frac{n}{n^2 + (k-1)^2} = \frac{t_k-t_{k-1}}{1+t_{k-1}^2} \ge \tan^{-1}t_k - \tan^{-1}t_{k-1} \ge \frac{t_k-t_{k-1}}{1+t_k^2} = \frac{n}{n^2+k^2}$$
Summing this over $k$ from $1$ to $n$, we get
$$\frac{1}{2n} + \sum_{k=1}^n \frac{n}{n^2+k^2} = \sum_{k=1}^n \frac{n}{n^2+(k-1)^2}
\ge \tan^{-1}t_n - \tan^{-1}t_0 = \tan^{-1} 1 = \frac{\pi}{4} \ge \sum_{k=1}^n\frac{n}{n^2+k^2}$$
This leads to
$\displaystyle\;\left|\sum_{k=1}^n \frac{n}{n^2+k^2} - \frac{\pi}{4}\right| \le \frac{1}{2n}$
and hence $\displaystyle\;\lim_{n\to\infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \frac{\pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show $1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$, where $z=\cos x+i \sin x$ Let $z=\cos x+i \sin x$. Show that
$$1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$$
| That identity has a rather simple geometric interpretation. Consider the following diagram:
where $x$ is the red angle and
\begin{align}
z &= \cos x + i\cdot\sin x, \\
m &= \cos \frac{x}{2} + i\cdot\sin \frac{x}{2}, \\
n &= m \cdot \cos\frac{x}{2}.
\end{align}
In other words, $z$ is any point with $|z| = 1$, and $x$ is the angle it forms with line $01$. Point $m$ indicates the bisector of $x$ and $n$ is the orthogonal projection of $1$ onto line $0m$ (starting at $0$ we would need to go in the direction of $m$ for $\cos\frac{x}{2}$). That happens to be the orthogonal projection of $z$ onto $0m$ and thus also an intersection of segments $0m$ and $1z$.
We want to prove that $2\cdot n$ and $z+1$ represent the same point.
Observe, that because $|z|=1$, $\{0, 1, z, z+1\}$ constitute four corners of a rhombus in which bisectors are the diagonals, what implies that $0m$ actually passes through $z+1$.
Now we only need to prove that $|2\cdot n| = |z+1|$. But that is also true, because $n$ happens to be at the intersection of diagonals, which is also their midpoint.
I hope this helps $\ddot\smile$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$
| By writing the integrand function as its Taylor series centered at $x=0$ and performing termwise integration we get:
$$ I=\int_{0}^{1}\frac{dx}{1+x^6}=\sum_{k\geq 0}\frac{(-1)^k}{6k+1}=\sum_{n\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+7}\right).\tag{1} $$
The last series can now be computed through the discrete Fourier transform (DFT).
Let $\left\{\omega=\exp\left(\frac{\pi i}{6}\right),\omega^5,\omega^7,\omega^{11}\right\}$ be the set of the primitive twelth roots of unity and $\chi(n)$ be the arithmetic function that equals $+1$ if $n\equiv\!1\!\!\pmod{12}$, $-1$ if $n\equiv\!7\!\!\pmod{12}$ and zero otherwise.
We have:
$$\chi(n) = \frac{1}{12}\left(-2\omega^5 \omega^n -2i \omega^{3n}-2\omega \omega^{5n}+2\omega^5\omega^{7n}+2i\omega^{9n}+2\omega \omega^{11n}\right)\tag{2}$$
and for every $k\in\{1,2,\ldots,11\}$ the equality:
$$ \sum_{n\geq 1}\frac{\omega^{kn}}{n}=-\log(1-\omega^k)\tag{3} $$
holds. By putting everything together,
$$ I = \sum_{n\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+7}\right) =\sum_{n\geq 1}\frac{\chi(n)}{n}= \color{red}{\frac{\pi+\sqrt{3}\log(2+\sqrt{3})}{6}}\tag{4} $$
easily follows from $2+\sqrt{3}=\cot\frac{\pi}{12}$.
Another approach is to use the substitution $\frac{1}{1+x^6}=y$ to get:
$$ I = \frac{1}{6}\int_{\frac{1}{2}}^{1}y^{-1/6}(1-y)^{-5/6}\,dy \tag{5}$$
then, by setting
$$ J=\frac{1}{6}\int_{0}^{\frac{1}{2}}y^{-1/6}(1-y)^{-5/6}\,dy = \frac{1}{6}\int_{\frac{1}{2}}^{1}(1-y)^{-1/6}y^{-5/6}\,dy \tag{6} $$
noticing that:
$$ I+J=\frac{1}{6}\int_{0}^{1}y^{-1/6}(1-y)^{-5/6}\,dy=\frac{\pi}{3}\tag{7}$$
by Euler's beta function and the reflection formulas for the $\Gamma$ function, while $I-J$ provides the logarithmic contribute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
If $f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
If $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
I tried to replace $x$ by $x+1, x-1$ in the equality,to get something like $f(x + k)=f(x)$ but without success.
Any help is appreciated.
| From the given equation, $$f(x+4)=\sqrt{3}f(x+3)-f(x+2),\\\sqrt{3}f(x+3)=3f(x+2)-\sqrt{3}f(x+1),\\f(x+2)=\sqrt{3}f(x+1)-f(x).$$ Adding those, we obtain that $$f(x+4)=f(x+2)-f(x),$$ and also $$f(x+6)=f(x+4)-f(x+2),$$ therefore $f(x+6)=-f(x)$. Hence, $$f(x+12)=-f(x+6)=f(x),$$ therefore $f$ is periodic with period $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
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Check convergence and find the sum $\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}$
$$\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2}$$
I have starting an overview about series, the book starts with geometric series and emphasizing that for each series there is a corresponding infinite sequence.
For convergence I can look at the partial sums, but how can I find the sum of the series?
| Note that
$$ \frac{1}{9n^2 + 3n - 2} = \frac{1}{(3n + 2)(3n - 1)} = \frac{1}{3} \left( \frac{1}{3n - 1} - \frac{1}{3n + 2} \right) $$
for all natural numbers $n$.
The partial sums of the sum that you are interested in are then given by
$$ \sum_{n=1}^N \frac{1}{9n^2 + 3n - 2} = \frac{1}{3} \sum_{n=1}^N \left( \frac{1}{3n - 1} - \frac{1}{3n + 2} \right) = \frac{1}{3} \sum_{n=1}^N \left( \frac{1}{3n - 1} - \frac{1}{3(n + 1) - 1} \right) $$
One sees that the sum telescopes; that is to say all of the terms cancel except for the first and last, leaving us with
$$ \frac{1}{3} \left( \frac{1}{3 \cdot 1 - 1} - \frac{1}{3(N + 1) - 1} \right) = \frac{1}{6} - \frac{1}{9N + 6} $$
As $N$ tends to $\infty$, this tends to $\frac{1}{6}$, and so we see that the sum converges and that its value is $\frac{1}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of $n$ terms and infinite terms of series
The sum of $n$ terms of the series $$\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....$$
And also calculate sum of $\infty$ terms.
$\bf{My\; Try::}$ We can write above series as $$ 1\underbrace{-\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....}_{S_{n}}$$
So here $$\bf{r^{th}}\; terms \; of \; above\; series (T_{r}) = \frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}$$
So $$T_{r} = \frac{1}{3}\left[\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-5)}{(r-1)!}\cdot \frac{1}{2^{r-1}}-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}\right]$$
So $$S_{n} = \sum^{n}_{r=1}T_{r} = \frac{1}{3}\sum^{n}_{r=1}\left[\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-5)}{(r-1)!}\cdot \frac{1}{2^{r-1}}-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}\right]$$
So $$S_{n} = \frac{1}{3}\left[-3-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}\right] = -1+\frac{1}{3}\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}$$
So our Sum is $$=\frac{1}{3}\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}$$
Is my process is right or not, If not the how can i calculate it, Thanks
|
This series looks like
\begin{align*}
\frac{1}{2}&+\sum_{r=2}^\infty\frac{1}{r!}\left(\frac{1}{2}\right)^r(2r-3)!!\tag{1}\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}(2r-1)!!\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{(2r)!!}\tag{2}\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{2^rr!}\tag{3}\\
&=\frac{1}{2}+\frac{1}{2}\sum_{r=1}^\infty\frac{1}{r+1}\binom{2r}{r}\left(\frac{1}{4}\right)^r\tag{4}\\
&=\frac{1}{2}\sum_{r=0}^\infty\frac{1}{r+1}\binom{2r}{r}\left(\frac{1}{4}\right)^r\\
&=\frac{1}{2}\left.\left(\frac{1-\sqrt{1-4x}}{2x}\right)\right|_{x=\frac{1}{4}}\tag{5}\\
&=\frac{1}{2}\cdot 2\\
&=1
\end{align*}
Comment:
*
*In (1) we use double factorial notation $(2r-3)!!=(2r-3)\cdot(2r-5)\cdots 5\cdot 3\cdot 1$
*In (2) we use $r!=r!!\cdot(r-1)!!$
*In (3) we use $(2r)!!=2^r r!$
*In (4) we obtain the Catalan numbers $C_r=\frac{1}{r+1}\binom{2r}{r}$
*In (5) we use the generating function of the Catalan numbers
| {
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"url": "https://math.stackexchange.com/questions/1887095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of the integral: $\int_{0}^{\frac{\pi}{4}} \frac{1}{3\cos^2\theta + \sin^2\theta}d\theta$
Using the substitution $t = \tan\theta$ , find the value of the integral:
$$\int_{0}^{\pi/4} \cfrac{1}{3\cos^2\theta + \sin^2\theta}d\theta$$
I isolated $\theta$ in $t = \tan\theta$, and then substituted the expression into the integral. I then replaced $d\theta$ by $dt$, which resulted in a very ugly, large expression.
The following is the answer given:
I do not understand how the first integral was obtained.
| You could also use that
$\displaystyle\int_{0}^{\pi/4} \cfrac{1}{3\cos^2\theta + \sin^2\theta}d\theta=\int_0^{\pi/4}\frac{\frac{1}{\cos^2\theta}}{\frac{3\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\theta}{\cos^2\theta}}d\theta=\int_0^{\pi/4}\frac{\sec^2\theta}{3+\tan^2\theta} d\theta=\int_0^1\frac{1}{3+t^2}dt$
| {
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"url": "https://math.stackexchange.com/questions/1888761",
"timestamp": "2023-03-29T00:00:00",
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Stirling formula, O-Notation In lecture we proof that
: $ \sqrt{2 \pi n} (\frac{n}{e})^n< n! \le \sqrt{2 \pi n} (\frac{n}{e})^n e^{\frac{1}{12n}}$
But how we came form this to the formula $n!= \sqrt{2 \pi n} (\frac{n}{e})^n (1+ O(1/n)) $ with $0< O(1/n) \le \frac{1}{12n}$
I can rewirte the second formula as that $\lim\sup_{n\rightarrow \infty}| \frac{\frac{n!}{\sqrt{2 \pi n}*(\frac{n}{e})^n} -1}{1/n}|$ exists, what means that $\frac{\frac{n!}{\sqrt{2 \pi n}*(\frac{n}{e})^n} -1}{1/n}$ is bounded.
When i use the first formula i get $0=\frac{1-1}{n} \le \frac{\frac{n!}{\sqrt{2 \pi n}*(\frac{n}{e})^n} -1}{1/n} \le \frac{1}{12}*\frac{(e^{\frac{1}{12n}}-1)}{1/12 n}$
I know $\lim_{n\rightarrow \infty} \frac{1}{12}*\frac{(e^{\frac{1}{12n}}-1)}{1/12 n}= \frac{1}{12}*1=\frac{1}{12}$ so the right side is bounded and so the limessuperior exists.
I need to show that $ \frac{n!}{\sqrt{2 \pi n} (\frac{n}{e})^n} -1 \le \frac{1}{12n}$, that means $ \frac{\frac{n!}{\sqrt{2 \pi n} (\frac{n}{e})^n} -1}{\frac{1}{n}} \le \frac{1}{12}$. But this formula i only get for $n$ big enough and not for all n?
| You have proved that, as $n\to \infty$,
$$
\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n< n! \le \sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^ne^{\large \frac{1}{12n}}\tag1
$$ from wich one deduces that, as $n \to \infty$,
$$
1< \frac{n!}{\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n} \le e^{\large \frac{1}{12n}} \tag2
$$ by the Taylor series expansion, as $n \to \infty$, one has
$$
e^{\large \frac{1}{12n}}=1+O\left(\frac1n\right)\tag3
$$ giving
$$
1< \frac{n!}{\sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n} \le 1+O\left(\frac1n\right) \tag4
$$ or, as $n \to \infty$,
$$
n!= \sqrt{2 \pi n} \:\left(\frac{n}{e}\right)^n\left(1+ O\left(\frac1n\right)\right) \tag5
$$
as announced.
| {
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"url": "https://math.stackexchange.com/questions/1888868",
"timestamp": "2023-03-29T00:00:00",
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} |
Combinatorial identity $\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}$ How to prove the combinatorial identity?
$$\prod_{j=1}^n {n\choose j} =\prod_{k=1}^n {k^k\over k!}$$
I took $\ln$ of the left hand side
$$\sum_{j=0}^n \ln(j^{n+1}) - \ln((j!)^2)$$
but not going anywhere from here. any help is welcome
| We have, by product operator $\prod$ properties
$$
\prod_{j=1}^{n}{ n \choose j}
=
\prod_{j=1}^{n}{ n! \over (n-j)!j! }
=
{\prod_{j=1}^n n! \over \prod_{j=1}^n (n-j)!\cdot j!}
=
{(n!)^n \over \prod_{j=1}^n (n-j)!\prod_{j=1}^nj!}
$$
The result can be obtained by induction on n. The proposition $P(n):\prod_{j=1}^{n}\binom{n}{j}=\prod_{j=1}^{n}\frac{j^j}{j!}$ can be verified for $n = 2,3$ by simple inspection. Let us prove that for all $n\geq 3$ the proposition $P(n):\prod_{j=1}^{n}\binom{n}{j}=\prod_{j=1}^{n}\frac{j^j}{j!}$ is true whenever the proposition $P(n-1):\prod_{j=1}^{n-1}\binom{n-1}{j}=\prod_{j=1}^{n-1}\frac{j^j}{j!}$ is true. For this, we prove three equalities.
\begin{align}
(\ast)\quad (n!)^n
=&
n^n\cdot(n-1)^n\cdot (n-2)^n\cdot \ldots\cdot 3^n\cdot 2^n\cdot 1^n
\\
=&
n^n\cdot\color{red}{(n-1)}\cdot\color{blue}{(n-1)^{n-1}}\cdot \color{red}{(n-2)}\cdot\color{blue}{(n-2)^{n-1}}\cdot \ldots\cdot \color{red}{3}\cdot \color{blue}{3^{n-1}}\cdot \color{red}{2}\cdot \color{blue}{2^{n-1}}\cdot \color{red}{1}\cdot \color{blue}{1^{n-1}}
\\
=& n^n\cdot \color{red}{(n-1)!}\cdot
\color{blue}{(n-1)^{n-1}\cdot (n-2)^{n-1}\cdot \ldots\cdot 3^{n-1}\cdot 2^{n-1}\cdot 1^{n-1}}
\\
=&
n^n\color{red}{(n-1)!}\color{blue}{\big((n-1)!\big)^{n-1}}
\end{align}
\begin{align}
(\ast\ast)\quad\prod_{j=1}^{n}(n-j)!
=&
\color{red}{(n-1)!}\cdot\color{blue}{(n-2)!\cdot(n-3)!\cdot \ldots \cdot (n-(n-3))!\cdot(n-(n-2))!(n-(n-1))!(n-(n))!}
\\
=&
\color{red}{(n-1)!}\cdot\color{blue}{((n-1)-1)!\cdot((n-1)-2)!\cdot \ldots \cdot 3!\cdot 2!\cdot 1!}
\\
=&
\color{red}{(n-1)!}\color{blue}{\prod_{j=1}^{(n-1)}((n-1)-j)!}
\end{align}
and
\begin{align}
(\ast\ast\ast)\quad\prod_{j=1}^{n}j!
=&
n!\cdot \color{blue}{(n-1)!\cdot(n-2)!\cdot \ldots \cdot 3!\cdot 2!\cdot 1!}
\\
=&
n!\color{blue}{(n-1)!\cdot((n-1)-1)!\cdot((n-1)-2)!\cdot \ldots \cdot 3!\cdot 2!\cdot 1!}
\\
=&
n!\color{blue}{\prod_{j=1}^{(n-1)}j!}
\end{align}
Then
$$
\prod_{j=1}^{n}{ n \choose j}
=
{(n!)^n \over \prod_{j=1}^n (n-j)!\prod_{j=1}^nj!}
=
{n^n\color{red}{(n-1)!}\color{blue}{\big((n-1)!\big)^{n-1}} \over \color{red}{(n-1)!}\color{blue}{\prod_{j=1}^{(n-1)}((n-1)-j)!} n!\color{blue}{\prod_{j=1}^{(n-1)}j!}}
= {n^n \over n!} \prod_{j=1}^{n-1}{ n-1 \choose j}
$$
To obtain the desired result simply replace $P(n-1):\prod_{j=1}^{n-1}\binom{n-1}{j}=\prod_{j=1}^{n-1}\frac{j^j}{j!}$ in the last equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the reciprocal of the following infinite product I hae to evaluate the reciprocal of the following product to infinity
$$\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19}{2 \cdot 2 \cdot 6 \cdot 6 \cdot 10 \cdot 10 \cdot 14 \cdot 14 \cdot 18 \cdot 18}\cdot\ldots $$
I am guessing you express each number as a sum. Example, the first $\frac{1}{2}$ can be expressed as $\frac{2-1}{2}$ and the next $\frac{3}{2}$ as $\frac{2+1}{2}$, hence their product equals $1-\frac{1}{2^2}$.
| Since $\frac{1\cdot 3}{2\cdot 2}=1-\frac{1}{2^2}$, $\frac{5\cdot 7}{6\cdot 6}=1-\frac{1}{6^2}$ and so on, we want to compute:
$$ P = \prod_{n\geq 1}\left(1-\frac{1}{(4n-2)^2}\right)^{-1} $$
while the Weierstrass product for the sine function $\frac{\sin(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2}\right)$ gives
$$ \prod_{n\geq 1}\left(1-\frac{1}{(4n)^2}\right)^{-1}=\frac{\pi}{2\sqrt{2}},\qquad \prod_{n\geq 1}\left(1-\frac{1}{(2n)^2}\right)^2=\frac{\pi}{2}$$
hence
$$\boxed{\, P = \frac{\frac{\pi}{2}}{\frac{\pi}{2\sqrt{2}}} = \color{red}{\sqrt{2}}\;}$$
since every number of the form $(4n-2)$ is an even number that is not a multiple of $4$.
An alternative approach, following Did's comment, is to notice that
$$ \prod_{n=1}^{N}\left(1-\frac{1}{(4n-2)^2}\right)^{-1} =\frac{4^N\,((2N)!)^3}{(N!)^2\,(4N)!}, $$
hence $P=\sqrt{2}$ also follows from Stirling's approximation. That is no wonder since we already know that Wallis product and Stirling's approximation are closely related.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$ Solve differential equation
$$2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$$
I know that we have to use some smart substitution here, so that the equation becomes linear.
The only thing I came up with is a smart guessed particular solution: $y = x^2$. If we plug this function in, we get:
$$2\cdot2\cdot x^2 - 3(2x)^2 +8x^2 = 4x^2 - 12x^2 + 8x^2= 0$$
I made a mistake. The coefficients where different in the exam:
$$
\begin{cases}
3\ddot{y}y + 3(\dot{y})^2 - 2x^2 = 0, \\
y(0) = 1, \\
\dot{y}(0) = 0.
\end{cases}
$$
Does it make the solution easier?
| $$
\begin{cases}
3y''y + 3(y')^2 - 2x^2 = 0, \\
y(0) = 1, \\
y'(0) = 0.
\end{cases}
$$
$$y''y+y'^2=(y'y)'\quad\to\quad 3(y'y)'=2x^2$$
$$3y'y=\frac{2}{3}x^3+c_1$$
$y'(0)=0\quad\to\quad c_1=0$
$$y'y=\frac{2}{9}x^3$$
$$2y'y=(y^2)'=\frac{4}{9}x^3$$
$$y^2=\frac{1}{9}x^4+c_2$$
$y(0)=1\quad\to\quad c_2=1$
$$y=\sqrt{\frac{1}{9}x^4+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Derivation of Variance of Discrete Uniform Distribution over custom interval I'm trying to prove that the variance of a discrete uniform distribution is equal to $\cfrac{(b-a+1)^2-1}{12}$. I've looked at other proofs, and it makes sense to me that in the case where the distribution starts at 1 and goes to n, the variance is equal to $\cfrac{(n)^2-1}{12}$. I want to find the variance of $unif(a,b)$, that is a uniform distribution that goes from $a$ to $b$, where $a<b$ and $a$ does not necessarily equal 1.
I also realize that you can add / subtract to the distribution, and the variance will not change; hence, you can simply plug in the value $n=b-a+1$. However, I was wondering if there was any way to directly derive the first form of the variance using the property that $V(X) = E(X^2)-\mu^2$. This is the work I've done thus far, but I'm at an impasse. Additionally, is there a easier way to extend the result for the variance of $unif(1,n)$ to $unif(a,b)$? I suppose an easier way would be to show that the $\mathbb{V}(X) = \mathbb{V}(X+a)$, where $a$ is a constant, and then plug in the new range $(b-a+1)$
$
\begin{split}
\mathbb{V}(X) &= \mathbb{E}(X-\mu)^2
\\
&= \mathbb{E}(X^2) - \mu^2
\\
&= \frac{1}{b-a+1}\sum_{x=a}^{b}x^2 - \frac{(a+b)^2}{4}
\\
&= \frac{1}{b-a+1} \left(\frac{(b)(b+1)(2b+1) - (a-1)((a-1)+1)(2(a-1)+1)}{6}\right) -
\frac{(a+b)^2}{4}
\\
&= \frac{1}{b-a+1} \left(\frac{(b)(b+1)(2b+1)-(a-1)(a)(2a-1)}{6}\right)-
\frac{(a+b)^2}{4}
\\
&= \frac{1}{b-a+1} \left(\frac{(2b^3+3b^2+b)-(2a^3-3a^2+a)}{6}\right) -
\frac{(a+b)^2}{4}
\\
&= \frac{1}{b-a+1} \frac{(4b^3+6b^2+2b)-(4a^3-6a^2+2a)-(3a^2+6ab+3b^2)}{12}
\\
&= \frac{1}{b-a+1} \frac{4b^3+3b^2+2b-4a^3+3a^2-2a-6ab}{12}
\\
&= \dots
\\
&= \frac{(b-a+1)^2-1}{12}
\end{split}
$
| So much time has passed, but at least now we know it can be done: https://en.wikibooks.org/wiki/Statistics/Distributions/Discrete_Uniform
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help solving $1 < \frac{x + 3}{x - 2} < 2$ I worked a lot of inequalities here in MSE and that greatly helped me.
I've seen a similar inequality [here] [1], but the one I have today is significantly different, in that I'll end up with a division by 0, which is not possible.
[1] [Simple inequality
$$1 < \frac{x + 3}{x - 2} < 2$$
$$\implies 1 < \frac{x - 2 + 2 + 3}{x - 2}$$
$$\implies 1 < 1 + \frac{5}{x - 2} < 2$$
$$\implies 0 < \frac{5}{x - 2} < 1$$
$$\implies 0 < \frac{1}{x - 2} < \frac{1}{5}$$
I can't continue without falling into the division by 0. So my many thanks for your help.
| $$\frac { 1 }{ x-2 } -\frac { 1 }{ 5 } <0\\ \frac { 5-x+2 }{ 5\left( x-2 \right) } <0\\ \frac { 7-x }{ \left( x-2 \right) } <0\\ \frac { \left( 7-x \right) \left( x-2 \right) }{ { \left( x-2 \right) }^{ 2 } } <0\\ \left( 7-x \right) \left( x-2 \right) <0\\ x\in \left( -\infty ;2 \right) \cup \left( 7;+\infty \right) $$
on the other hand $0<\frac { 5 }{ x-2 } \Rightarrow x>2$ so the answer will be $$\left( -\infty ;2 \right) \cup \left( 7;+\infty \right) \cap \left( 2;+\infty \right) \Rightarrow \left( 7;+\infty \right) \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find The Last Two Digits Of $9^{8^7}$ Find the last two digits of $9^{8^7}$.
I tried finding a secure pattern for the last two digits of powers of $9$ but that didn't work. Any answers?
| Alternatively
By FLT
$3^{\phi(100) = 40}= 9^{20} \equiv 1 \mod 100$
$9^{8^7} = 9^{2^{21}} \equiv 9^{2^{21} \mod 20} \mod 100$
$2^{\phi(5)=4} \equiv 1 \mod 5$ so $2^{21=4*5 + 1} \equiv 2 \mod 5$
$2^{21} \equiv 0 \mod 4$ os $2^{21} \equiv 2 + 5k \mod 20$ where $4|2 + 5k$.
i.e. $2^{21} \equiv 12 \mod 20$.
so $9^{2^{21}} \equiv 9^{12} \mod 100$
$9^{12} = (10 - 1)^{12} = 10^{12} - .... -12*10 + 1 \equiv -120 + 1 \equiv 81 \mod 100$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_0^1 x(1+x^2)^n dx = \dfrac{1}{2n+2}$ $\int_0^1 x(1+x^2)^n dx = \dfrac{1}{2n+2}$
How is this calculated?
This is an interesting case because as a consequence of this, $\int_0^1 n^2x(1+x^2)^n dx$ converges to infinity with n to infinity, but the integrant converges point wise to 0.
(see Rudin "Real and complex Analysis" example 7.6.)
| We use substitution rule with $1 - x^2 = t$. We have
\begin{align}
&dt = -2x dx \nonumber \\
&x dx = - \frac{1}{2} dt \nonumber
\end{align}
\begin{align}
&x = 0, \quad t = 1 - 0 = 1 \nonumber \\
&x = 1, \quad t = 0 \nonumber
\end{align}
Pulling it all into one single expression.
$$
\int_{0=x}^{1=x} x(1-x^2)^n dx = \int_{0=t}^{1=t} \cdot \left( - \frac{1}{2} \right) t^n dt = \frac{1}{2} \int_{0}^{1} t^n dt = \frac{1}{2} \cdot \frac{t^{n+1}}{n+1} \Bigg\rvert^1_0 = \frac{1}{2} \frac{1}{n+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Squaring Infinite Series Expansion Of e^x $Fact$:$$\lim\limits_{n \to \infty}\frac{x^0}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots+\frac{x^n}{n!}=e^x$$
so
$$\lim\limits_{n \to \infty}\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}=e$$
also
$$\lim\limits_{n \to \infty}e^2=\frac{2^0}{0!}+\frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\dots+\frac{2^n}{n!}$$
can I safely say...
$$L.H.S=(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots)^2=\frac{2^0}{0!}+\frac{2}{2!}+\frac{4}{2!}+\frac{8}{3!}+\dots=R.H.S$$
if yes, how to prove L.H.S=R.H.S (without using the above fact)?
It is not a problem from textbook I was just thinking about it.
| Let $$f(z) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\tag{1}$$ then it is easy to see that the series on right is absolutely convergent for all values of $z \in \mathbb{C}$ and hence the function $f(z)$ is well defined for all complex values of $z$. Using Cauchy's product rule for multiplication of infinite series we see that $$f(z)f(w) = \sum_{n = 0}^{\infty}\frac{z^{n}}{n!}\cdot\sum_{n = 0}^{\infty}\frac{w^{n}}{n!} = \sum_{n = 0}^{\infty}c_{n}\tag{2}$$ where $$c_{n} = \sum_{k = 0}^{n}\frac{z^{n - k}}{(n - k)!}\cdot\frac{w^{k}}{k!} = \frac{1}{n!}\sum_{k = 0}^{n}\binom{n}{k}z^{n - k}w^{k} = \frac{(z + w)^{n}}{n!}\tag{3}$$ and hence from $(2), (3)$ we get $$f(z)f(w) = \sum_{n = 0}^{\infty}\frac{(z + w)^{n}}{n!} = f(z + w)\tag{4}$$ for all complex $z, w$. Putting $z = w = 1$ we get $$f(1)\cdot f(1) = f(2)$$ which is the result asked in question.
Using $(4)$ and little bit of algebra it is possible to prove that $\{f(z)\}^{n} = f(nz)$ for all complex $z$ and rational $n$. Putting $z = 1$ and replace $n$ by $x$ we see that $f(x) = \{f(1)\}^{x}$ where $x$ is rational or in grand fashion $$\left(1 + \frac{1}{1!} + \frac{1}{2!} + \cdots\right)^{x} = 1 + x + \frac{x^{2}}{2!} + \cdots$$ where $x$ is rational (this is the fact you mention in the beginning of your question).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find large power of a non-diagonalisable matrix
If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$.
The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresponding to eigenvalue $-1$. So, it is not diagonalizable.
Is there any other way to compute the power?
| Generalized eigenvector $\;u_1=\begin{pmatrix}a\\b\\c\end{pmatrix}\;$for $\;\lambda=1\;$ :
$$Au_1=1\cdot u_1+v_1\iff \begin{pmatrix}a\\a+c\\b\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}+\begin{pmatrix}0\\1\\1\end{pmatrix}\implies \begin{cases}b=c+1\\{}\\a+c=b+1\end{cases}\implies \begin{cases}a=2\\{}\\b=1\\{}\\c=0\end{cases}$$
$$\implies u_1=\begin{pmatrix}2\\1\\0\end{pmatrix}\;,\;\;\text{so define}\;\;P:=\begin{pmatrix}0&2&0\\1&1&1\\1&0&\!\!-1\end{pmatrix}\implies P^{-1}=\frac14\begin{pmatrix}\!\!-1&2&2\\2&0&0\\\!\!-1&2&\!\!-2\end{pmatrix}$$
and now:
$$P^{-1}AP=\frac14\begin{pmatrix}\!\!-1&2&2\\2&0&0\\\!\!-1&2&\!\!-2\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix}0&2&0\\1&1&1\\1&0&\!\!-1\end{pmatrix}=\begin{pmatrix}1&1&0\\0&1&0\\0&0&\!\!-1\end{pmatrix}\implies$$
$$A^{30}=P\begin{pmatrix}1&1&0\\0&1&0\\0&0&\!\!-1\end{pmatrix}P^{-1}$$
You can now check that taking powers of the above quasi-diagonal matrix ( i.e., the Jordan Canonical Form for $\;A\;$) is very easy, though not as easy as with diagonal matrices, of course.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares:
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$
Let's consider $a,b>0$ for now.
$$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$
I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then:
$$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$
$$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$
$$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$
$$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$
$$|p-1| \leq 1$$
$$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$
Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears:
$$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$
Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$
Finally, we rewrite:
$$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$
Or, using the Pochhammer symbol:
$$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$
My questions are:
Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form?
Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it?
If the method I used is correct, it can be used for any integral of the form ($m \geq 2$):
$$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$
| It was already shown that
$$
I_1(p)=\int_0^\infty \frac{dx}{\sqrt{(x^3+1)(x^3+p)}}=\frac{2 \pi}{3 \sqrt{3}} {_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right).
$$
By transformation 2.11(5) from Erdelyi, Higher transcendental functions (put $z=\frac{1-\sqrt{p}}{1+\sqrt{p}}$)
$$
{_2F_1} \left(\frac{1}{2},\frac{2}{3};1;1-p \right)=\left(\frac{2}{1+\sqrt{p}}\right)^{4/3}{_2F_1} \left(\frac{2}{3},\frac{2}{3};1;\left(\frac{1-\sqrt{p}}{1+\sqrt{p}}\right)^{2} \right).
$$
By Pfaff's transformation
$$
{_2F_1} \left(\frac{2}{3},\frac{2}{3};1;\left(\frac{1-\sqrt{p}}{1+\sqrt{p}}\right)^{2} \right)=\left(\frac{(1+\sqrt{p})^2}{4\sqrt{p}}\right)^{2/3}{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;\frac{(1-\sqrt{p})^2}{-4\sqrt{p}} \right).
$$
As a result
$$
I_1(p)=\frac{2 \pi}{3 \sqrt{3}p^{1/3}}{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;\frac{(1-\sqrt{p})^2}{-4\sqrt{p}} \right).
$$
Now we will use a generalization of AGM found by Borwein and Borwein, A Cubic Counterpart of Jacobi's Identity and the AGM, Transactions of the American Mathematical Society, Vol. 323, No. 2, (1991), pp.691-701 (after correcting for some typos):
$$
a_{n+1}=\frac{a_n+2b_n}{3} ,\quad b_{n+1}=\sqrt[3]{b_n\frac{a_n^2+a_nb_n+b_n^2}{3}},\quad a_0=1,\quad b_0=s,
$$
$$
\quad AG_3(1,s)=\lim_{n\to\infty} a_n=\frac{1}{{_2F_1} \left(\frac{1}{3},\frac{2}{3};1;1-s^3 \right)}.
$$
Using this we get
\begin{align}
I_1(p)=\frac{2 \pi}{3 \sqrt{3}~p^{1/3}\cdot AG_3\left(1,\left(\frac{1+\sqrt{p}}{2~\sqrt[4]{p}}\right)^{2/3}\right)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
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Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| You can solve this without much thinking.
By expanding the product, we find $m(m+1)(m+2)(m+3) = m^4 + 6m^3 + 11m^2 + 6m$. That's a bit bigger than $(m^2)^2 = m^4$. $(m^2+c)^2 = (m^4 + 2m^2c + c^2)$ is still too small because there is no term $m^3$. $(m^2+cm)^2 = m^4 + 2cm^3+c^2m^2$ looks better, especially if we let c = 3.
$(m^2+3m)^2 = m^4 + 6m^3 + 9m^2$ is still just a little bit too small, by about $2m^2$. So we add another c: $(m^2+3m+c)^2 = m^4 + 6m^3 + 2cm^2 + 9m^2 + 6cm + c^2$. We let c = 1 to match the term $11m^2$ and get $(m^2+3m+1)^2 = m^4 + 6m^3 + 11m^2 + 6m + 1$. Exactly 1 more than the product, just what we wanted. So
$m(m+1)(m+2)(m+3) = (m^2+3m+1)^2 - 1$.
That's the answer, without much thought at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
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} |
Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $\tan B$ but that doesn't help in getting the value of $\tan A + \tan B$. The value comes in terms of $\tan A$ or $\tan B$ but the expected answer is independent of any variable .
Also
$$\frac{\sin(A)\cdot\cos(B) + \sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)} = \tan(A) + \tan(B)$$
We could get a value only if instead of $\cos A$ there was $\sin B$ in the relation(which we get on adding the ratios)
| Hint:
Let $\dfrac{\sin A}{\sqrt3}=\dfrac{\sin B}2=p$ and $\dfrac{\cos A}{\sqrt5}=\dfrac{\cos B}3=q$
$\implies(\sqrt3p)^2+(\sqrt5q)^2=1$
and $(2p)^2+(3q)^2=1$
Solve for $p^2,q^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$ Let $a_1, a_2,....,a_n, b_1, b_2,...,b_n$, let $\frac{b_1}{a_1} = max \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ ,
$\frac{b_n}{a_n} = min \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ show that:
$$\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$$
| Note that
\begin{align*}
&\ \sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2 - \left(\sum_{i=1}^na_ib_i \right)^2-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2\left(\sum_{i=1}^na_i^2 \right)^2\\
=&\ \sum_{i,j=1}^n\left[a_i^2b_j^2- a_ib_ia_jb_j-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\
=&\ \frac{1}{2}\sum_{i,j=1}^n\left[a_i^2b_j^2 + a_j^2b_i^2- 2a_ib_ia_jb_j-\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\
\le&\ \frac{1}{2}\sum_{i,j=1}^n\left[a_i^2b_j^2 + a_j^2b_i^2- 2a_ib_ia_jb_j-\left(\frac{b_i}{a_i}-\frac{b_j}{a_j}\right)^2 a_i^2a_j^2\right]\\
=&\ 0.
\end{align*}
The inequality follows immediately.
| {
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"url": "https://math.stackexchange.com/questions/1904463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} ( \text{agm} (\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} ))^{-1}$ The following definite integral turns out to be expressible as the Arithmetic-Geometric Mean: $$I_4(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^4+a^4)(x^4+b^4)}}=\frac{\pi}{2 \sqrt2 a b} \left( \text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)\right)^{-1}$$
$$I_4(1,1)=\frac{\pi}{2 \sqrt2}$$
I would like to remind you that:
$$I_2(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2} \left( \text{agm} \left(a,b \right)\right)^{-1}$$
$$I_2(1,1)=\frac{\pi}{2}$$
Which is why I have two questions:
How do we prove the identity for $I_4(a,b)$?
Is it possible to also express other integrals of this type using agm? Such as $I_8(a,b)$?
Because of the relation of the agm to elliptic integrals, we can also write:
$$I_4(a,b)=\frac{1}{a b \sqrt{a^2+b^2}} K \left( \frac{a-b}{\sqrt{2(a^2+b^2)}} \right)$$
Here the parameter convention is $$K(k)=\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2 t^2)}}$$
This seems to be the best way to prove the identity, but I don't know which substitution to use.
Another way to express this integral would be through the hypergeometric function:
$$I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right)$$
And for every integral of this type we have:
$$I_m(a,b)=\frac{I_m(1,1)}{a^{m-1}} {_2F_1} \left(\frac{1}{2},\frac{m-1}{m};1;1-\frac{b^m}{a^m} \right)$$
The outline for the proof can be found in this question for $m=3$ and is easily adapted to the general case.
Here we assume $a \geq b$.
As for the arithmetic geometric mean, I tried to get it into a simpler form, but the only transformation I was able to achieve is this:
$$\text{agm} \left(\frac{a+b}{2},\sqrt{\frac{a^2+b^2}{2}} \right)=\text{agm} \left(\frac{a+b}{2}+i \frac{a-b}{2},\frac{a+b}{2}-i \frac{a-b}{2} \right)$$
Since we have complex conjugates, it's quite obvious, that they will give real numbers at first iteration, an it would give the left hand side.
| Since $K(k)=\frac{\pi}{2} ~{}_2F_1(1/2,1/2;1;k^2)$ and it was shown in the question that
$$
I_4(a,b)=\frac{\pi}{2 \sqrt2 a^3} {_2F_1} \left(\frac{1}{2},\frac{3}{4};1;1-\frac{b^4}{a^4} \right),
$$
the first question is equivalent to proving the equality
$$
\frac{\sqrt{2} }{b \sqrt{b^2+1}}{_2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{(1-b)^2}{2 \left(b^2+1\right)}\right)}={_2F_1\left(\frac{1}{2},\frac{3}{4};1;1-b^4\right)}\tag{1}.
$$
$\it{Proof}$. The proof can be accomplished by combining 3 transformations for hypergeometric function.
By Pfaff's transformation
$$
{_2F_1\left(\frac{1}{2},\frac{3}{4};1;1-b^4\right)}=\frac{1}{b^2}{~_2F_1\left(\frac{1}{2},\frac{1}{4};1;\frac{b^4-1}{b^4}\right)}.
$$
By quadratic transformation 2.11(5) from Erdelyi, Higher transcendental functions (put $z=\frac{b^2-1}{b^2+1}$)
$$
{~_2F_1\left(\frac{1}{2},\frac{1}{4};1;\frac{b^4-1}{b^4}\right)}=\sqrt{\frac{2b^2}{1+b^2}}{~_2F_1\left(\frac{1}{4},\frac{1}{4};1;\left(\frac{b^2-1}{b^2+1}\right)^2\right)}.
$$
By transformation 2.11(2) (put $z=\frac{(1-b)^2}{2 \left(b^2+1\right)}$)
$$
{~_2F_1\left(\frac{1}{4},\frac{1}{4};1;\left(\frac{b^2-1}{b^2+1}\right)^2\right)}={~_2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{(1-b)^2}{2 \left(b^2+1\right)}\right)}.
$$
| {
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"url": "https://math.stackexchange.com/questions/1905349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational? If $a+b+c=0$ then the roots of $ax^2+bx+c=0$ are rational ?
Is it a "If and only if " statement or "only if " statement ?
For $a,b,c \in \mathbb Q$ , I think it is a "if and only if" statement . Am I correct ?
I can prove that if $a+b+c=0$ and $a,b,c \in \mathbb Q$ , then roots are rational.
But I can not prove that if roots are rational and $a,b,c \in \mathbb Q$ , then $a+b+c=0$.
Any help ? or any conditions that $ax^2 +bx+c =0 $
should satisfy in order to have rational roots ?
My clarifications :
1) What is a rational number ? It is a number which can be written in the form $\frac{p}{q} $ , where $q \neq 0$ and $p , q \in \mathbb Z$
2)
In this case : Let's take the quadratic equation $ax^2+bx+c=0$ where $a \neq 0$.
We all know that the roots are given by , $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
Case 1 : Suppose that $a,b,c \in \mathbb Q$.
Then $x$ is rational if and only if $b^2-4ac$ is a perfect square or zero $(0,1,4,9,16,...)$.
Now we need to make $b^2-4ac$ a perfect square !
Now observe that if $a+b+c=0$ , then $b^2-4ac = (-a-c)^2-4ac=(a-c)^2$
That is if $a,b,c \in \mathbb Q$ and $a+b+c=0$ then the solutions are rational.
Case 2 : Suppose that $a,b,c \in \mathbb Q$.
If $c=0$ , then all the roots are rational. (This is easy if all $a,b,c$ are rationals)
Case 3 : Suppose that $b,c \in \mathbb R- \mathbb Q$.(If $a$ is irrational we can divide by $a$ )
$a+b+c=0$ condition does not satisfy.
Ex : $(1-\sqrt{2})x^2-2x+(1+\sqrt{2})=0$
| If $a+b+c= 0 \implies x = 1$ is a root and is a rational number, and the other root is $x = \dfrac{c}{a}$ also a rational number since $a, c \in \mathbb{Q}$. To see a counter example for the other part, take $a = c = 3, b = 10$, then $a+b+c = 16 \neq 0$, yet the equation $3x^2+10x+3 = 0$ has rational roots $x = -3, -\dfrac{1}{3}$. Thus it is an "IF" statement and not "IF AND ONLY IF".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rationalizing denominator with cube roots Rationalize the denominator of $$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}$$ and simplify.
I already have an answer. I just want to compare answers with others. Maybe someone has different solutions? Also, I really disagree with the answer found at the back of the questionnaire.
| Use $x+x^2+x^3=x\cdot(x^3-1)/(x-1)$ with $x=\sqrt[3]{4}$.
Hence
$$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}=\frac{6(\sqrt[3]{4}-1)}{\sqrt[3]{4}(4-1)}=2-2^{1-2/3}=2-\sqrt[3]{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-4^n=9m,\text{where $m$ is an integer.}$$
What I am thinking in the $n+1$ step is,
\begin{align}
& 5^{n+2}-4^{n+2} \\
= {} & 5^2(5^n-4^n)+5^24^n-4^{n+2} \\
= {} & 5^2(5^n-4^n)+4^n(5^2-4^2) \\
= {} & 5^29m+4^n9 \\
= {} & 9(5^2m+4^n)
\end{align}
Does this approach make sense?
*Show that $a+b$ is a factor of $a^n+b^n$ where $n$ is a positive odd number.
I am thinking this in the $n+1$ step.
$$a^{2n+1}+b^{2n+1}$$
But then I cannot get it further.
| For #3 use the technique of #2: For $n\in \mathbb N \cup \{0\}$ we have $$a^{2n+3}+b^{2n+3}=a^2(a^{2n+1}+b^{2n+1})+(-a^2+b^2)b^{2n+1}=$$ $$=a^2(a^{2n+1}+b^{2n+1})+(a+b)(b-a)b^{2n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A proof of the identity $ \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)} $. I have to prove that
$$
\forall n \in \mathbb{N}_{0}, ~ \forall x \in \mathbb{R} \setminus \mathbb{N}_{0}:
\qquad
\sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k}
= \frac{n!}{(x + 0) (x + 1) \cdots (x + n)}.
$$
However, I was unable to find a proof. I have tried to use the binomial expansion of $ (1 + x)^{n} $ to get the l.h.s., by performing a suitable multiplication followed by integration, but I was unable to obtain the required form. Please help me out with the proof. Thanks in advance.
| An induction on $n$ will work. Let
$$f(x,n)=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\;,$$
and for the induction step suppose that
$$f(x,n)=\frac{n!}{x(x+1)\ldots(x+n)}\;.$$
Then
$$\begin{align*}
\frac{(n+1)!}{x(x+1)\ldots(x+n+1)}&=\frac{n+1}{x+n+1}f(x,n)\\
&=\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\\
&=(n+1)\sum_{k=0}^n(-1)^k\binom{n}k\frac1{(x+k)(x+n+1)}\\
&=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1-k}\binom{n}k\left(\frac1{x+k}-\frac1{x+n+1}\right)\\
&=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\frac1{x+k}\\
&\qquad-\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\\
&=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n+1}k-\frac1{x+n+1}\color{red}{\sum_{k=0}^n(-1)^k\binom{n+1}k}\\
&=f(x,n+1)-\frac{(-1)^{n+1}}{x+n+1}-\frac1{x+n+1}\color{red}{\left(0-(-1)^{n+1}\right)}\\
&=f(x,n+1)\;,
\end{align*}$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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The integral $\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy$ What is $$\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy?$$ I split it as $\frac{y^{2}}{(y^2+1)^{0.5}} + \sqrt{y^2+1}.$ Now I substituted $y^{2}=u $ thus $2y\,dy=du$ so we get $0.5 \sqrt{\frac{u}{u + 1}} + 0.5 \sqrt{\frac{1 + u}{u}}$ but now what to do? Another idea was doing $+1-1$ in original question but that too doesn't lead anywhere. Now $y=\tan{x} $ as suggested below is an easy way but I am seeking for a purely algebraic way. Thanks.
| Let $\displaystyle y=\frac{1}{2}\left(t-\frac{1}{t}\right),\;dy=\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$,
so that $\displaystyle\sqrt{y^2+1}=\frac{1}{2}\left(t+\frac{1}{t}\right),\;\;t=y+\sqrt{y^2+1},\;\;\frac{1}{t}=\sqrt{y^2+1}-y.$
Then $\displaystyle\int \frac{2 (2y^2+1)}{(y^2+1)^{0.5}} dy=2\int\frac{\frac{1}{2}\left(t^2-2+\frac{1}{t^2}\right)+1}{\frac{1}{2}\left(t+\frac{1}{t}\right)}\cdot\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$
$\displaystyle=2\int\frac{t^2+\frac{1}{t^2}}{t+\frac{1}{t}}\cdot\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt=\int\frac{t^4+1}{t(t^2+1)}\cdot\frac{t^2+1}{t^2}dt$
$\displaystyle=\int\frac{t^4+1}{t^3}dt=\int\left(t+t^{-3}\right)dt=\frac{1}{2}\left(t^2-\frac{1}{t^2}\right)+C=\frac{1}{2}\left(\big(t-\frac{1}{t}\big)\big(t+\frac{1}{t}\big)\right)+C$
$\displaystyle=\frac{1}{2}\left(\big(2y\big)\big(2\sqrt{y^2+1}\big)\right)+C=\color{blue}{2y\sqrt{y^2+1}+C}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to factorise this expression $ x^2-y^2-x+y$ This part can be factorised as $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?
:)
| This is a far less elegant method than the other answers but I'll keep it up for posterity: set it up as a quadratic in $x$.
$$x^2-x+(y-y^2)=0$$
$$x=\frac{1\pm\sqrt{1+4(y^2-y)}}{2}$$
$$\left(x-\frac{1+\sqrt{1+4(y^2-y)}}{2}\right)\left(x-\frac{1-\sqrt{1+4(y^2-y)}}{2}\right)$$
$$\frac{1}{4}\left(2x-1+\sqrt{(2y-1)^2}\right)\left(2x-1-\sqrt{(2y-1)^2}\right)$$
$$\left(x+y-1\right)\left(x-y\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution of an Inequality Show that $(ab + bc + ca)(ab^{-1}+ bc^{-1} + ca^{-1}) \geq (a + b + c)^{2}.$
We, at first, change the given expression $(ab + bc + ca)(ab^{-1} + bc^{-1} +ca^{-1})$ into $(ab^2 + bc^2 + ca^2 )(1/a + 1/b + 1/c)$ . Now we will show that this term $\geq (a + b+ c)^{2}.$ Again by Chebyshev's inequality we showed $(1/a + 1/b + 1/c) \geq \frac{9}{a + b + c}.$ But after that we face some problem. Please help or give some hint so that we can solve this problem.
| It follows from the Cauchy-Schwarz inequality:
$$(ab + bc + ca)(ab^{-1} + bc^{-1} + ca^{-1}) \ge \left(\sqrt{ab}\cdot \sqrt{ab^{-1}} + \sqrt{bc}\cdot \sqrt{bc^{-1}} + \sqrt{ca}\cdot \sqrt{ca^{-1}}\right)^2 = \cdots$$
Now finish the rest.
| {
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
| For the LHS, use $\cos(120^\circ +A)=\cos 120^\circ \cos A - \sin 120^\circ \sin A$
and $\cos(240^\circ+A)=\cos 240^\circ \cos A - \sin 240^\circ \sin A$
Then $\cos^3(120^\circ+A)=\left(\cos 120^\circ \cos A - \sin 120^\circ \sin A \right)^3$
and $\cos^3(240^\circ+A)=\left( \cos 240^\circ \cos A - \sin 240^\circ \sin A \right)^3$
Pop in values of $\cos 120^\circ$, $\sin 120^\circ$, etc before expanding brackets.
$\cos^3(120^\circ+A)=\left(-1/2 \cos A -\sqrt 3/2 \sin A \right)^3=-\frac 18 \left(\cos A +\sqrt 3 \sin A \right)^3$
and $\cos^3(240^\circ+A)=-\frac 18\left( \cos A - \sqrt3\sin A \right)^3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of $75\leq n\leq 300$ not divisible by any of $9,12,15$ - check work I need to find the number of $75\leq n\leq 300$ not divisible by any of $9,12,15$.
I thought about to solve this for $1\leq n\leq 300$ and $1\leq n\leq 74$ and then subtract. For each of them I thought to use inclusion-exclusion. I just want to check whether I'm right.
For the first problem, the answer is:
$300-\left(\left\lfloor\frac{300}{9}\right\rfloor+\left\lfloor\frac{300}{12}\right\rfloor+\left\lfloor\frac{300}{15}\right\rfloor-\left\lfloor\frac{300}{\operatorname{lcm}(9,12)}\right\rfloor-\left\lfloor\frac{300}{\operatorname{lcm}(9,15)}\right\rfloor-\left\lfloor\frac{300}{\operatorname{lcm}(12,15)}\right\rfloor+\left\lfloor\frac{300}{\operatorname{lcm}(9,12,15)}\right\rfloor\right)$
and for the second the same expression with $74$ replacing each $300$. Finally, subtract them. Is this correct?
| Let $f(n)$ denote the amount of integers in the range $[1,n]$ not divisible by $9$ nor $12$ nor $15$:
$f(n)=n-\left\lfloor\frac{n}{9}\right\rfloor-\left\lfloor\frac{n}{12}\right\rfloor-\left\lfloor\frac{n}{15}\right\rfloor+\left\lfloor\frac{n}{lcm(9,12)}\right\rfloor+\left\lfloor\frac{n}{lcm(9,15)}\right\rfloor+\left\lfloor\frac{n}{lcm(12,15)}\right\rfloor-\left\lfloor\frac{n}{lcm(9,12,15)}\right\rfloor$
We can simplify it as $f(n)=n-\left\lfloor\frac{n}{9}\right\rfloor-\left\lfloor\frac{n}{12}\right\rfloor-\left\lfloor\frac{n}{15}\right\rfloor+\left\lfloor\frac{n}{36}\right\rfloor+\left\lfloor\frac{n}{45}\right\rfloor+\left\lfloor\frac{n}{60}\right\rfloor-\left\lfloor\frac{n}{180}\right\rfloor$
And the answer to your question is $f(300)-f(74)$:
*
*$f(300)=240$
*$f(74)=60$
Hence the amount of such integers is $240-60=180$.
Python verification: print sum([0 not in [n%9,n%12,n%15] for n in range(75,301)]).
| {
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"question_score": "1",
"answer_count": 1,
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Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words.
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}n_{22} + n_{11} + n_{22} + 1 - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{11} + n_{22} + 1 = n_{11} + n_{22} + 1$
Which shows that it's true for a 2*2 matrix.
Looking at other 2*2 matrices
$\begin{vmatrix}
n_{11} + 1 & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = n_{11}n_{22} + n_{22} - n_{12}n_{21} = \begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} + n_{22} = n_{22}$
Similarly:
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22} + 1
\end{vmatrix} = n_{11}$
$\begin{vmatrix}
n_{11} & n_{12} + 1 \\
n_{21} & n_{22}
\end{vmatrix} = -n_{21}$
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} + 1 & n_{22}
\end{vmatrix} = -n_{12}$
Positive if on the main diagonal, negative if on the side diagonal.
Using that, we can show that
$\begin{vmatrix}
n_{11} + 1 & n_{12} & n_{13} \\
n_{21} & n_{22} + 1 & n_{23} \\
n_{31} & n_{32} & n_{33} + 1
\end{vmatrix} = (n_{11} + 1)(n_{22} + n_{33} + 1) - n_{12}n_{21} - n_{13}n_{31} = n_{11}n_{22} - n_{12}n_{21} + n_{11}n_{33} - n_{13}n_{31} + n_{11} + n_{22} + n_{33} + 1 = n_{11} + n_{22} + n_{33} + 1$
And in a similar fashion we can apply this to higher order matrices.
But I haven't the slightest clue how to word or show this "similar fashion".
| Since $\texttt{rank}(A) = 1$ $A = u v^T$ for some $n \times 1$ matrices $u$ and $v$ and consequently $\texttt{trace}(A) = v^Tu.$
We have $$\begin{pmatrix} I & 0 \\ v^T & 1 \end{pmatrix} \begin{pmatrix}I & u \\ -v^T & 1 \end{pmatrix} = \begin{pmatrix} I & u \\ 0 & 1 + v^Tu \end{pmatrix}. \tag{1}$$
And we also have $$\begin{pmatrix} I & -u \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I & u \\ -v^T & 1 \end{pmatrix} = \begin{pmatrix} I + uv^T & 0 \\ -v^T & 1 \end{pmatrix}. \tag{2}$$
Taking determinant of $(1)$ and $(2)$ the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Convergence/divergence of $\sum_{k=1}^\infty\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}$ A problem asks me to determine if the series
$$\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$$
converges or diverges.
(from the textbook Calculus by Laura Taalman and Peter Kohn (2014 edition); section 7.6, p. 639, problem 33)
I am allowed to use the ratio test first and then any other convergence/divergence test if the former test does not work. In my original work, I attempted the ratio test and it was rendered inconclusive.
$$ a_k = \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)} $$
$$ a_{k + 1} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2(k+1))}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2(k+1)-1)} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} $$
$$ \frac{a_{k + 1}}{a_k} = \frac{\frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)}}{\frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}} = \frac{2 \times 4 \times 6 \times \cdots \times (2k) \times (2k+2)}{1 \times 3 \times 5 \times \cdots \times (2k-1) \times (2k+1)} \times \frac{1 \times 3 \times 5 \times \cdots \times (2k-1)}{2 \times 4 \times 6 \times \cdots \times (2k)} = \frac{2k+2}{2k+1}$$
Evaluating $\rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k}$ will determine if $\sum_{k=1}^\infty \frac{2 \times 4 \times 6 \times \cdots \times (2k)}{1 \times 3 \times 5 \times \cdots \times (2k-1)}$ converges or diverges. The conclusions for the ratio test are as follows:
$\circ$ If $\rho < 1$, then $\sum_{k=1}^\infty a_k$ converges.
$\circ$ If $\rho > 1$, then $\sum_{k=1}^\infty a_k$ diverges.
$\circ$ If $\rho = 1$, then the test is inconclusive.
$$ \rho = \lim_{x \to \infty} \frac{a_{k + 1}}{a_k} = \lim_{x \to \infty} \frac{2k+2}{2k+1} = 1$$
Since $\rho = 1$, the ratio test is rendered inconclusive, as I stated earlier. I will have to use other convergence/divergence tests to solve the problem.
My issue is that I'm not sure which other convergence/divergence test to use. Any suggestions?
Many thanks for the help.
| By Stirling approximation, we have that
$$a_k=\frac{2\times 4\times 6\times\cdots\times(2k)}{1\times 3\times 5\times\cdots\times(2k-1)}=\frac{4^k\cdot (k!)^2}{(2k)!}=\frac{4^k}{\binom{2k}{k}}\sim \sqrt{\pi k}.$$
Hence the series $\sum_{k\geq 1} a_k$ diverges.
More generally the series $\sum_{k\geq 1} \frac{a_k}{k^r}$ converges iff $r-1/2>1$, that is $r>3/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}\le3\sqrt{\frac{2(a^2+b^2+c^2)}3}$$
| The triangle condition is equivalent to $0<a<1/2$, $0<b<1/2$ and $0<c<1/2$. Assume $c=\min(a, b, c)$, then
$$\sqrt{a^2+b^2} + \sqrt{b^2+c^2} + \sqrt{a^2+c^2} \le \sqrt{a^2+b^2} + \sqrt{b^2+bc} + \sqrt{a^2+ac} < \sqrt{a^2 + b^2} + (b+c/2) + (a+c/2) = \sqrt{a^2+b^2} + 1 < \sqrt{(1/2)^2+(1/2)^2} + 1 = 1+\frac{\sqrt{2}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$? Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$?
I tried graphing but it didn't help. Could anyone answer? Thanks!
| \begin{align}
f(a)
&= \dfrac{a+13}{3a+7} \\
&= \dfrac 13 \dfrac{3a+39}{3a+7} \\
&= \dfrac 13 \left(1 + \dfrac{32}{3a+7} \right) \\
\end{align}
This implies that $f(a)$ is strictly decreasing for positive integers.
Hence the maximum value must be $f(1) = \dfrac 75$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Recurrence relation of type $a_{n+1} = a^2_{n}-2a_{n}+2$
A sequence $\{a_{n}\}$ is defined by $a_{n+1} = a^2_{n}-2a_{n}+2\forall n\geq 0$ and $a_{0} =4$
And another sequence $\{b_{n}\}$ defined by the formula $\displaystyle b_{n} = \frac{2b_{0}b_{1}b_{2}..........b_{n-1}}{a_{n}}\forall n\geq 1$ and $\displaystyle b_{0}=\frac{1}{2}$,Then
$(a)$ value of $a_{10}$
$(b)\;\; $ The value of $n$ for which $\displaystyle b_{n} = \frac{3280}{3281}$
$(c)$ The Sequence $\{b_{n}\}$ satisfy the recurrence formula
$\bf{Options::}$
$(1)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1-b^2_{n}}\;\;\;\;\;\; (b)\; \displaystyle b_{n+1} = \frac{2b_{n}}{1b^2_{n}}\;\;\;\;\;\; (c)\; \displaystyle b_{n+1} = \frac{b_{n}}{1+2b^2_{n}}\; (d)\; \displaystyle b_{n+1} = \frac{b_{n}}{1-2b^2_{n}}$
$\bf{My\; Try::}$ Given $a_{n+1} = a^2_{n}-2a_{n}+2 = \left(a_{n}-1\right)^2+1$
So $a_{1} = (a_{0}-1)^2+1=(4-1)^2+1=10$
Similarly $a_{2} = (a_{1}-1)^2+1 = 9^2+1 = 82$
Similarly $a_{3} = (81)^2+1 = $
But Calculation like this is very complex, Plz help me how can i solve above problems, Thanks
| For the second recurrence
$$
b_{\,n + 1} = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{b_{\,n} }}
$$
we have
$$
\left\{ \begin{gathered}
b_{\,0} = 1/2 \hfill \\
b_{\,1} = \frac{2}
{{b_{\,0} }} = 4 \hfill \\
b_{\,n + 1} = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{b_{\,n} }} = \frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {b_{\,k} } }}
{{\frac{{2\prod\limits_{0\, \leqslant \,k\, \leqslant \,n - 2} {b_{\,k} } }}
{{b_{\,n - 1} }}}} = b_{\,n - 1} ^2 \hfill \\
\end{gathered} \right.
$$
assuming that, in deducing $b_1$, we can adopt empty product = $1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do this?
My method:
$$
{f'(x)}=\left [\ln \left (\frac{x}{(x^2+1)^\frac{1}{2}} \right) \right ]'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right)\left (\frac{x}{(x^2+1)^\frac{1}{2}} \right)'=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[(x^2+1)^\frac{1}{2}]'}{[(x^2+1)^\frac{1}{2}]^2} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(x^2+1)']}{\left | x^2+1 \right |} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x[\frac{1}{2}(x^2+1)^{-\frac{1}{2}}(2x)]}{x^2+1} \right)=\left (\frac{(x^2+1)^\frac{1}{2}}{x} \right) \left (\frac{(x^2+1)^\frac{1}{2}-x^2(x^+1)^{-\frac{1}{2}}}{x^2+1} \right)=\frac{(x^2+1)^{(\frac{1}{2}+\frac{1}{2})}-x^2(x^2+1)^{\frac{1}{2}+-\frac{1}{2}{}}}{x(x^2+1)}=-\frac{x^2}{x}=-x
$$
and
$$
f''(x)=(-x)'=-1\
$$
This took me much more than 1.5 hours just to type into LaTex :'(
| Let $f(x)=\log\left(\frac{x}{\sqrt{x^2+1}}\right)$ and
$$ g(t)=f(\sinh t) = \log\tanh t = \log\sinh t-\log\cosh t. $$
We have
$$ \frac{d}{dt}g(t) = \frac{\cosh t}{\sinh t}-\frac{\sinh t}{\cosh t} = \frac{1}{\sinh(t)\cosh(t)}$$
but the LHS also equals $\cosh(t)\,f'(\sinh t)$, hence
$$ f'(\sinh t) = \frac{1}{\sinh(t)\cosh^2(t)} $$
and
$$ f'(x) = \color{red}{\frac{1}{x(1+x^2)}}.$$
We may freely assume $x>0$ since otherwise $f(x)$ makes no sense (and $f'(x)$ as well).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Solving system of $9$ linear equations in $9$ variables I have a system of $9$ linear equations in $9$ variables:
\begin{array}{rl}
-c_{1}x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} - c_{2}x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} - c_{3}x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} - c_{4}x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} - c_{5}x_{5} + x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} - c_{6}x_{6} + x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} - c_{7}x_{7} + x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} - c_{8}x_{8} + x_{9} &= 0 \\
x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} - c_{9}x_{9} &= 0
\end{array}
I want to find a general non-trivial solution for it. What would be the easiest and least time consuming way to find it by hand? I don't have a lot of background in maths, so I would very much appreciate if you actually found the solution and explained briefly.
Thanks in advance!
EDIT: Very important to mention is that always any $c_{i} > 1$ and any $x_{i} \geq 20$. Also it would be nice if someone posted how would a general non-trivial solution look in the form of $$S = \left \{( x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}\right )\}$$
| obviously $x_i = 0$ is a solution. If you want a non-trivial solution
Subtract any two lines and you get $(c_i+1)x_i = (c_j+1)x_j = 0$
If any $c_j = -1$ we have all $(c_i + 1)x_i = 0$ so if $c_i \ne -1$ then $x_i = 0$ and if $c_i = -1$ then $x_i$ can be anything at all.
If none of the $c_i = -1$ then let $(c_i + 1)x_i = M \ne 0$.
Then $x_i = M/(c_i+1)$. But is that possible?
Each line is $(\sum_{i= 0}^9 M/(c_i+1) ) - M = 0$
So $\sum_{i=0}^9 1/(c_i + 1) = 1$.
There are no non-zero solutions unless that very unlikely criteria is met.
So
(recap)
1) $x_i= 0$ is a solution.
2) If any $c_i = -1$ then $x_i = \begin{cases} anything; c_i = -1 \\ 0; c_i \ne -1 \end{cases}$
3) If $c_i \ne -1 \forall i$ then $x_i = M/(c_i +1)$ will be a solution IF any of the lines add to 0, which will happen if and only if $\sum \frac 1{c_i + 1} = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
As $\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)}$, is this is a function of only $x$ or both $x$ and $n$? $$\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)}$$
Now, the LHS seems to be a function of only $x$, whereas the RHS seems to be a function of both $x$ as well as $n$. Please remove this ambiguity for me as I don't understand why is it that by simply choosing a value of $x$ on LHS is immaterial of what value $n$ holds.
| This is equality ,not a function
$$\frac1{(1-x)^2} = {1+2x+3x^2 +...nx^{n-1}} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$\frac1{(1-x)^2} = \frac{d}{dx}\{x+x^2+x^3 +\cdots x^{n}\} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$x+x^2+x^3 +\cdots x^{n}=\frac{x(1-x^n)}{1-x}$$
$$\frac1{(1-x)^2} = \frac{d}{dx}\left\{\frac{x(1-x^n)}{1-x}\right\} + \frac{x^n}{(1-x)^2} + \frac{nx^n}{(1-x)^2}$$
$$\frac1{(1-x)^2}=\frac1{(1-x)^2}$$
This hold true for $$\forall \space x \in \mathbb{R} -\{1\} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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$ \int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$ $$I=\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx$$
I can't think of a substitution to solve this problem, by parts won't work here. Can anyone tell how should I solve this problem?
| Another way you might be interested .
Notice that $$\frac{d}{dx}\left(\frac{1}{x+\sqrt x+1}\right)=\frac{-(1+\frac{1}{2\sqrt x})}{(x+\sqrt x+1)^2}$$
Put then $$\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx=\frac{A+B\sqrt x+Cx}{x+\sqrt x+1}$$ and take the derivative in both sides. You get at once $C=0$ and the system $$2\sqrt x(2+\sqrt x)=-A(2\sqrt x+1)+B(1-x)$$ which gives $A=B=-2$.
Thus $$\int\frac{2+\sqrt{x}}{\left(x+\sqrt{x}+1\right)^2}dx=\frac{-2(\sqrt x+1)}{x+\sqrt x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the generating function for $c_r = \sum^r_{i=1}i^2$ Find the generating function of
where $c_0 = 0, c_r = \sum^r_{i=1}i^2$.
Hence show that
$\sum^r_{i=1}i^2 = C^{r+1}_3 + C^{r+2}_3$
Attempt:
$c_r = \sum^r_{i=1}i^2$
= $x + 4x^2 + 9x^3 + ... + r^2x^r$
= $x(1 + 4x + 9x^2 + ... + r^2x^{r-1})$
= $x(\frac{1}{1-2x})$
How do i proceed from here? Is this even right?
| We are looking for a generating function
\begin{align*}
\sum_{r=0}^\infty c_r x^r=\sum_{r=0}^\infty \left(\sum_{i=1}^r i^2\right) x^r
\end{align*}
Note that Cauchy multiplication of a series with $\frac{1}{1-x}$ transforms the coefficient $a_r$ of a series to $\sum_{i=0}^r a_r$.
\begin{align*}
\frac{1}{1-x}\sum_{r=0}^\infty a_r x^r=\sum_{r=0}^\infty \left(\sum_{i=0}^r a_r\right)x^r
\end{align*}
We obtain
\begin{align*}
\sum_{r=0}^\infty c_r x^r&=\sum_{r=0}^\infty \left(\sum_{i=1}^r i^2\right) x^r\\
&=\frac{1}{1-x}\sum_{r=0}^\infty r^2x^r\tag{1}\\
&=\frac{x^2}{1-x}\sum_{r=0}^\infty r(r-1)x^{r-2}+\frac{x}{1-x}\sum_{r=0}^\infty r x^{r-1}\tag{2}\\
&=\frac{1}{1-x}D_x^2\left(\frac{1}{1-x}\right)+\frac{1}{1-x}D_x\left(\frac{1}{1-x}\right)\tag{3}\\
&=\frac{2x^2}{(1-x)^4}+\frac{x}{(1-x)^3}\tag{4}\\
&=2x^2\sum_{r=0}^\infty \binom{-4}{r} (-x)^r+x\sum_{r=0}^\infty \binom{-3}{r} (-x)^r\tag{5}\\
&=2\sum_{r=0}^\infty \binom{r+3}{3}x^{r+2}+\sum_{r=0}^\infty \binom{r+2}{2}x^{r+1}\tag{6}\\
&=2\sum_{r=2}^\infty \binom{r+1}{3}x^{r}+\sum_{r=1}^\infty \binom{r+1}{2}x^{r}\tag{7}\\
&=\sum_{r=1}^\infty \left(\binom{r+1}{3}+\binom{r+2}{3}\right)x^{r}\tag{8}\\
\end{align*}
and the claim follows.
Comment:
*
*In (1) we factor out $\frac{1}{1-x}$.
*In (2) we rearrange the series to apply differentiation in the following.
*In (3) we write the expression using the differential operator $D_x:=\frac{d}{dx}$.
*In (4) we apply the differentiation.
*In (5) we use the binomial series expansion.
*In (6) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (7) we shift the index of the series accordingly to obtain $x^r$ in both series.
*In (8) we collect the series and use the binomial identity $\binom{r+2}{3}=\binom{r+1}{3}+\binom{r+1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rational solutions to a quadratic-type equation. For any fixed rational $t$, are there any nonzero rational solutions to the equation
$$y^{2} = (1-t^2)x^2 + \frac{1}{4} \text{ ?} $$
My attempt: consider the line $L$ whose gradient is $m$ and passes through the point $(0, \frac{1}{2})$. Then equation of $L$ is $y=mx + 1/2$, then consider its intersection with the quadratic in question and simplify. However, i'm yet to find any idea beyond this stage.
| To simplify, rewrite the original equation as $(2y)^2+4(t^2-1)x^2=1$ and substitute $2y$ with $2mx+1$:
\begin{align}&(2mx+1)^2+4(t^2-1)=4(m^2+t^2-1)x^2+4mx+1=1\\
\iff & 4x\bigl((m^2+t^2-1)x+m\bigr)=0,
\end{align}
so the non-zero solution is
$$x=\frac{m}{1-m^2-t^2},\qquad y=\frac{m^2}{1-m^2-t^2}+\frac12=\frac{1+m^2-t^2}{2(1-m^2-t^2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving equation in three variables please help me understand how the following equation with 3 variables and power of 2 is solved and what solution approach is the quickest.
$$3y^2 - 3 = 0$$
$$4x - 3z^2 = 0$$
$$-6xz+ 6z = 0 $$
| $$
\begin{cases}
3y^2-3=0\\
4x-3z^2=0\\
6z-6xz=0
\end{cases}\Longleftrightarrow
\begin{cases}
3y^2-3=0\\
x=\frac{3z^2}{4}\\
6z(1-x)=0
\end{cases}\Longleftrightarrow
\begin{cases}
3y^2-3=0\\
x=\frac{3z^2}{4}\\
6z\left(1-\frac{3z^2}{4}\right)=0
\end{cases}
$$
So, when you solve:
$$6z\left(1-\frac{3z^2}{4}\right)=0$$
We get three solutions for $z$, $z=0$ or $z=\pm\frac{2}{\sqrt{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what value c is $f(x) = \frac{x^2+5x-14}{x^3-x^2-x-2}$ at 2 continuous? Im pretty new to continuinity and it seems for me the function can as easily be repaired with $c = 2$ so that $g(x) = \begin{cases}
\frac{x^2+5x-14}{x^3-x^2-x-2}, & x > 2\\
0, & x = 2 \\ \frac{x^2+5x-14}{x^3-x^2-x-2}, & x < 2
\end{cases} $
, because $f(2) = \frac{0}{0}$, which is a gap in the function and the function. But plotting the function proves it otherwise, because the value is ~1,5. So I'd have to take the limes from the left and right to this point, $left-\lim\limits_{n \to 2} \frac{x^2+5x-14}{x^3-x^2-x-2} = \frac{2^2+5\cdot2-14}{2^3-2^2-2-2} = \frac{0}{0}$... Help is appreciated!
| Instead of plugging $2$ directly into the expression, first factor the numerator and denominator as
$$f(x) = \frac{x^2 + 5x - 14}{x^3 - x^2 - x - 2} = \frac{(x + 7)(x - 2)}{(x - 2)(x^2 + x + 1)} = \frac{x + 7}{x^2 + x + 1}.$$
Then substituting $x = 2$ tells you that
$$\lim_{x \to 2}f(x) = \frac{2 + 7}{2^2 + 2 + 1} = \frac{9}{7}.$$
So the function is continuous if $g(c) = 9/7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$
I don't know how to begin to solve this problem
| Also we can use the following reasoning.
Since $\prod\limits_{cyc}(a+b)-(a+b+c+d)(abc+abd+acd+bcd)=(ac-bd)^2\geq0$,
it remains to prove that
$$(a+b+c+d)(abc+abd+acd+bcd)\geq\prod\limits_{cyc}(a+b+c-d)$$
which is also true by BW.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer
Given a sequence $(a_n)$, with $a_1 = 4$ and $a_{n+1} = a_n^2-2$ for all $n \in \mathbb{N}$, prove that there is a triangle with side lengths $a_n-1,a_n,a_n+1$ and that its area is equal to an integer.
We must have that \begin{align*}(a_n-1)+a_n &> a_n+1\\a_n+(a_n+1) &> a_n-1\\(a_n-1)+(a_n+1) &> a_n,\end{align*} which gives us equivalently that $a_n > 2$. Now using Heron's Formula we get \begin{align*}S = \sqrt{\dfrac{3a_n}{2}\left(\dfrac{3a_n}{2}-(a_n-1)\right)\left(\dfrac{3a_n}{2}-a_n\right)\left(\dfrac{3a_n}{2}-(a_n+1)\right)} &= \sqrt{\dfrac{3a_n}{2} \cdot \dfrac{a_n+2}{2} \cdot \dfrac{a_n}{2} \cdot \dfrac{a_n-2}{2}}\\&=\dfrac{1}{4}\sqrt{3a_n^2(a_n-2)(a_n+2)}.\end{align*} How do we continue from here?
| We see that $3a_n^2(a_n^2-4)$ must be a perfect square divisible by $4$. We notice that $a_n \equiv 2 \pmod{4}$ and so $a_n^2$ is a multiple of $4$, similarly with $a_n-2$ and $a_n+2$. Thus we just need $3a_n^2(a_n^2-4)$ to be a perfect square. To prove this, we prove the following: $$a_n^2(a_n^2-4) = 12(a_1 \cdots a_n)^2 \quad \text{for } n \geq 1.$$ We prove this by induction on $n$. For $n = 1$ the result holds trivially since $a_1^2(a_1^2-4) = 4^2(4^2-4) = 12(a_1)^2 = 12(4)^2$. Now suppose the result holds for some $n$. Then noting that $$a_n^2(a_n^2-4) = a_{n+1}^2-4,$$ we see that $a_{n+1}^2(a_{n+1}^2-4) = 12(a_1 \cdots a_{n+1})^2$, completing the induction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solutions of $\sin2x-\sin x>0$ with $x\in[0,2\pi]$ What are the solutions of this equation with $x\in[0,2\pi]$?
$$\sin2x-\sin x>0$$
I took this to
$$(\sin x)(2\cos x-1)>0$$
Now I need both terms to be the same sign. Can you please help me solve this?
| Using Prosthaphaeresis Formula,
$$\sin2x-\sin x=2\sin\dfrac x2\cos\dfrac{3x}2$$
Check for $x=0,2\pi$
Else for $0<x<2\pi,\sin\dfrac x2>0$
So we need $$\cos\dfrac{3x}2>0$$ which is possible
if $2m\pi\le\dfrac{3x}2<2m\pi+\dfrac\pi2\iff\dfrac{4m\pi}3\le x<\dfrac{(4m+1)\pi}3$
or if $2n\pi+\dfrac{3\pi}2<\dfrac{3x}2\le2n\pi+2\pi\iff\dfrac{(4n+3)\pi}3<x\le\dfrac{4(n+1)\pi}3$
where $m,n$ are arbitrary integers.
Again we need $$\dfrac{4m\pi}3\ge0\iff m\ge0, \dfrac{(4m+1)\pi}3\le2\pi\iff m\le\dfrac54\implies m=0,1$$
$$\dfrac{(4n+3)\pi}3\ge0\iff n\ge-\dfrac34;\dfrac{4(n+1)\pi}3\le2\pi\iff n\le\dfrac12\implies n=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to set up equation to find the unknown values if a limit exist.
For what values of the constants $a$ and $b$ does the following limit exist?
$$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$
for this question, $$f(x) = \frac{-(x+3)(\sqrt{ax+b}-2)}x,x<-3$$
$$f(x) = \frac{(x+3)(\sqrt{ax+b}-2)}x,x>=-3$$
Firstly, I found difficulties in setting up two equation solving for a and b.
Secondly, since $|x+3|=x+3$ only when $x\to0$, I can only use the follwoing to set up the equation. $$\lim_{x\to0}\frac{(x+3)(\sqrt{ax+b}-2)}x$$
However, I cannot simplified it to cancel out the denominator $x$ by some method like rationalization. It gets $$\lim_{x\to0}\frac{(x+3)(ax+b-4)}{(\sqrt{ax+b}-2)x}$$.
Therefore, how can we set up the equation to find $a$ and $b$ or it needs other methods to do this type of questions?
| Note that we can write
$$\begin{align}
\frac{|x+3|(\sqrt{ax+b}-2)}{x}&=\left(\frac{|x+3|(\sqrt{ax+b}-2)}{x}\right)\left(\frac{(\sqrt{ax+b}+2)}{(\sqrt{ax+b}+2)}\right)\\\\
&=\frac{|x+3|(ax+b-4)}{x(\sqrt{ax+b}+2)}\\\\
&=\left(\frac{|x+3|}{(\sqrt{ax+b}+2)}\right)\left(a+\frac{b-4}{x}\right)\tag 1
\end{align}$$
The first parenthetical term on the right-hand side of $(1)$ approaches $\frac{3}{2+\sqrt b}$ as $x\to 0$. The limit of the second term does not exist unless $b=4$. If $b=4$, then that limit is $a$, and we find
$$\lim_{x\to 0}\frac{|x+3|(\sqrt{ax+4}-2)}{x}=\frac{3a}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964631",
"timestamp": "2023-03-29T00:00:00",
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Combinations - Calculating the outcomes for a 15-coin collection I searched a lot to find some sort of guidance on how to approach solving this problem, but I couldn't find an answer anywhere. Read the book many times, still no luck The problem as follows:
Cecil has a 15-coin collection. Four coins are quarters, seven coins are dimes, three are nickels and one is a penny. For each scenario, calculate the total possible outcomes if Cecil randomly selects five coins.
Cecil selects no quarters
(11,5) 11! / 6! = 55440
Cecil selects at least 1 dime
(9,5) = 9! / 4! = 15120
Cecil selects an equal number of nickels and dimes
(6,5) = 6! / 1! = 720
The cent amount of Cecil’s selection is an even number
(7,5) = 7! /5! 2! = 2520
Are my answers to each scenario correct? If not, what am I missing? Thanks in advance for any answers.
| You computed the number of ordered selections in which Cecil selects no quarters. However, we are interested in the number of subsets, so the selections are not ordered.
Cecil selects no quarters.
The number of ways Cecil can select five coins from the eleven coins that are not quarters is
$$\binom{11}{5} = \frac{11!}{5!6!}$$
Cecil selects at least one dime.
Since there are seven dimes and eight other coins, one way to do this would be to add the number of ways Cecil could select one dime and four of the other eight coins, two dimes and three of the other eight coins, three dimes and two of the other eight coins, four dimes and one of the other eight coins, and five dimes.
$$\binom{7}{1}\binom{8}{4} + \binom{7}{2}\binom{8}{3} + \binom{7}{3}\binom{8}{2} + \binom{7}{4}\binom{8}{1} + \binom{7}{5}\binom{8}{0}$$
However, it is simpler to subtract the number of selections in which Cecil selects five coins that are not dimes from the total number of selections of five coins. Since seven of the fifteen available coins are dimes, this can be done in
$$\binom{15}{5} - \binom{8}{5}$$
ways.
Cecil selects an equal number of nickels and dimes.
There are seven dimes and three nickels. Since Cecil is selecting five of the fifteen coins, it is possible that he selects no nickels or dimes, one nickel and one dime, and two nickels and two dimes. He cannot select three of each since that would require the selection of at least six coins. If Cecil selects no nickels or dimes, he must select all five of the other coins. If he selects one nickel and one dime, he must select three of the other five coins. If he selects two nickels and two dimes, he must select one of the other five coins. Therefore, the number of ways in which Cecil can select an equal number of nickels and dimes is
$$\binom{7}{0}\binom{3}{0}\binom{5}{5} + \binom{7}{1}\binom{3}{1}\binom{5}{2} + \binom{7}{2}\binom{3}{2}\binom{5}{1}$$
The last question depends on knowing the values of the coins. They are $25$ cents for a quarter, $10$ cents for a dime, $5$ cents for a nickel, and $1$ cent for a penny.
Cecil selects an even number of cents.
For this to occur, the number of coins with odd cent values that are selected must be even, which means Cecil must select an odd number of dimes. Since there are seven dimes and he selects five coins, this means that he selects one dime and four of the other eight coins, three dimes and two of the other eight coins, or five dimes. Therefore, the number of selections of five coins which result in an even number of cents is
$$\binom{7}{1}\binom{8}{4} + \binom{7}{3}\binom{8}{2} + \binom{7}{5}\binom{8}{0}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding the other factors of the polynomial and $K$ If $(2x-1)$ is a factor of the polynomial $$p(x)=2x^3-5x^2-kx+3,$$ find $k$ and the other two factors of $p(x)$.
| by using the long division
$$\frac{2x^3-5x^2-kx+3}{2x-1}=x^2-2x-\frac{k+2}{2}+\frac{3-\frac{k+2}{2}}{2x-1}$$
the remainder or last term must be zero
or
$$3-\frac{k+2}{2}=0$$
hence $$k=4$$
the other factors
$$x^2-2x-3=(x-3)(x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965864",
"timestamp": "2023-03-29T00:00:00",
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Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$ Let $a,b,c$ and $n$ be natural numbers and $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=1$.
Does it possible to find all tuples $(a,b,c,n)$ such that:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$
| A linear diophantine equation
$$ax+by=c$$
where $c$ is divisible by $d=gcd(a,b)$, has solution
$$a=a_{0}+\frac{b}{d}\cdot k$$
$$b=b_{0}-\frac{a}{d}\cdot k$$
However, your equation is not linear, since
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=n$$
$$\frac{\left(a+b\right)ab+\left(b+c\right)bc+\left(a+c\right)ac}{abc}=n$$
Note here you are interested in combinations of $\left(a,b,c\right)$ such that this quotient is an integer $n$. Also,
$$a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}=abcn$$
proves it's not possible to represent $\left(a,b,c\right)$ in the same terms of $k$ as a linear diophantine equation, so they don't share a straight-forward relationship, but this last equation is very interesting, since it is contained in the cubic expansion $\left(a+b+c\right)^{3}$.
Then,
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3\left(a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}\right)+6abc$$
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3\left(abcn\right)+6abc$$
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3abc\left(n+2\right)$$
$$\left(a+b+c\right)^{3}-\left(a^{3}+b^{3}+c^{3}\right)=3abc\left(n+2\right)$$
As you can see, this difference is divisible by 3, so using congruence is an excellent idea:
$$\left(a+b+c\right)^{3}\equiv a^{3}+b^{3}+c^{3}\;\left(mod\;\;3\right)$$
So, given that $gcd\left(a,b,c\right)=1$, every combination of $\left(a,b,c\right)$ where the cube of the sum is congruent to the sum of the cubes $\left(mod\;\;3\right)$ is a possible answer.
For instance, $\left(1,1066,3977\right)$ and $\left(1,1598,4182\right)$ are valid answers, but it's difficult to trace an easy linear relation between $1$ and the other values.
| {
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"timestamp": "2023-03-29T00:00:00",
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A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 1 & 1 \\
2^2 & \left(\dfrac{3}{2}\right)^2 & \left(\dfrac{4}{3}\right)^2 \\
3^2 & \left(\dfrac{4}{2}\right)^2 & \left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}&\left[\begin{array}{11}C_1\rightarrow\dfrac{C_1}{1} \\
C_2\rightarrow\dfrac{C_2}{2^2}\\
C_3\rightarrow\dfrac{C_3}{3^2}\end{array}\right]\\
&=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \left(\dfrac{3}{2}\right)^2-2^2 & \left(\dfrac{4}{3}\right)^2-2^2 \\
3^2 & 2^2-3^2 & \left(\dfrac{5}{3}\right)^2-3^2 \\
\end{vmatrix} &\left[\begin{array}{11}C_2\rightarrow C_2-C_1 \\
C_3\rightarrow C_3-C_1\end{array}\right]\\
&=
(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & 2^2-\left(\dfrac{3}{2}\right)^2 & 2^2-\left(\dfrac{4}{3}\right)^2 \\
3^2 & 3^2-2^2 & 3^2-\left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}\\
&=(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & \dfrac{20}{9} \\
3^2 & 5 & \dfrac{56}{9} \\
\end{vmatrix}\\
&=(1^2\times2^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & 20 \\
3^2 & 5 & 56 \\
\end{vmatrix}\\
&=4\times(-2)\\
&=-8
\end{align}
$$
As you can see, my solution is a not a very promising one. If I encounter such questions again, so would you please suggest a better method which doesn't include this ridiculous amount of calculations.
| Create two zeroes in the first row ($C_2 \to C_2-\color{red}{4}C_1$ and $C_3 \to C_3-\color{blue}{9}C_1$) and expand:
$$\begin{vmatrix}
1 & \color{red}{4} & \color{blue}{9} \\
4 & 9 & 16 \\
9 & 16 & 25 \\
\end{vmatrix}=
\begin{vmatrix}
1 & 0 & 0 \\
4 & 9-16 & 16-36 \\
9 & 16-36 & 25-81 \\
\end{vmatrix} = \begin{vmatrix}
-7 & -20 \\
-20 & -56 \\
\end{vmatrix} = 7 \cdot 56 - 20^2 = -8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
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If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$, prove that:...... 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 Attempt;
Here,
$$e^{iA}=\cos A+i\sin A$$
$$e^{iB}=\cos B+i\sin B$$
$$e^{iC}=\cos C+i\sin C$$
Then,
$$e^{iA}+e^{iB}+e^{iC}=0$$
Now, what should I do further. Please help.
| let $a = e^{iA}, b = e^{iB}, c = e^{iC}$. Also note that $a+b+c = 0$
$$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc -ca) = 0$$
$$a^3 + b^3 + c^3 = 3abc$$
$$\mathrm{cis}\ 3A + \mathrm{cis}\ 3B + \mathrm{cis}\ 3C = 3\ \mathrm{cis}\ (a+b+c)$$
equating the real parts will give you the required expression:
$$\mathrm{cos}\ 3A + \mathrm{cos}\ 3B + \mathrm{cos}\ 3C = 3\ \mathrm{cos}\ (a+b+c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proof by induction that $4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}$ I am stuck on this problem for my discrete math class.
Prove the equation by induction for all integers greater than or equal to $3$:
$$4^3+4^4+4^5+⋅⋅⋅+4^n = \frac{4(4^n-16)}{3}.$$
I know that base case $n=3$:
$4^3=64$ as well as $4(4^3-16)/3 = 64$
My confusion is on the induction step where:
$4^3+4^4+4^5+⋅⋅⋅+4^n+4^{(n+1)} = 4(4^{(n+1)}-16)/3$.
I don't know what to do next.
| When doing a proof by induction, I suggest you say so up front.
Label the base case and the inductive hypothesis. Here is your template.
Prove:
$4^3 + 4^4 + 4^5 \cdots 4^n = 4(4^n - 16)/3$
Proof by induction.
Base case:
$4^3 = 4(4^3-16)/6$
Inductive hypothesis:
Suppose, $4^3 + 4^4 + 4^5 \cdots 4^n = 4(4^n - 16)/3$
We must show that,
$4^3 + 4^4 + 4^5 \cdots 4^{n+1} = 4(4^{n+1} - 16)/3$
based on the inductive hypothesis.
$4^3 + 4^4 + 4^5 \cdots 4^{n+1}\\
4^3 + 4^4 + 4^5 \cdots 4^n + 4^{n+1}$
$\frac {4(4^n-16)}{3} + 4^{n+1}$ from the inductive hypothesis.
$\frac {4(4^n-16) + 3(4^{n+1})}{3}\\
\frac {4(4^n-16 + 3\cdot4^{n})}{3}\\
\frac {4(4^{n+1}-16)}{3}$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Any other cases similar to $\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac 12 (3+\sqrt[4]{5}+\sqrt{5}+\sqrt[4]{125})$ Ramanujan gave many curious identities, one of which was $$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}=\frac 12(3+\sqrt[4]5+\sqrt5+\sqrt[4]{125})\tag1$$
And another one:$$\sqrt[3]{\sqrt[5]{\frac {32}5}-\sqrt[5]{\frac {27}5}}=\sqrt[5]{\frac 1{25}}+\sqrt[5]{\frac {3}{25}}-\sqrt[5]{\frac {9}{25}}\tag2$$
Question: Is there some algebraic method to prove $(1)$ and $(2)$, and can that method be generalized? And more specifically, can both $(1)$ and $(2)$ be represented as a sum of radicals of the same degree? Such as $\sqrt[4]{A+B\sqrt[4]{C}}=\sqrt[4]{X}+\sqrt[4]{Y}+\sqrt[4]{Z}$.
Also, I do realize that for $(1)$, you can raise everything to the fourth power and simplify to get an identity (such as $1=1$), but it doesn't provide insight on why or how the equation holds.
| The first identity of (1) can not be generalized, namely $5$ is the only integer satisfying (1). To see this, let
$$ \sqrt[4]{\frac{a+b\sqrt[4]{p}}{a-b\sqrt[4]{p}}}=\frac{\sqrt[4]{p}+q}{\sqrt[4]{p}-q} \tag{3}$$
where $p,q$ are integers such that $\sqrt[4]{p},\sqrt{p}$ are not integers and $a,b,$ are rationals. Then (3) is equivalent to
$$ (a+b\sqrt[4]{p})(\sqrt[4]{p}-q)^4-(a-b\sqrt[4]{p})(\sqrt[4]{p}+q)^4=0 $$
or
$$ \sqrt[4]{p}\bigg\{2q(-2a+3qb)\sqrt{p}+\big[-4q^3a+(p+q^4)b\big]\bigg\}=0. $$
Let
$$ \bigg\{\begin{array}{ll}
-2a+3qb&=&0,\\
-4q^3a+(p+q^4)b&=&0,
\end{array}$$
which is a system of linear homogeneous equations. It has nonzero solutions iff
$$ \det\bigg(\begin{matrix}
-2&3q\\
-4q^3&p+q^4
\end{matrix}\bigg)=0 $$
from which one obtains $p=5q^4$ and $a=\frac32qb$. Thus (3) becomes
$$ \sqrt[4]{\frac{3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}$ and $ S_i =\sum _{n=1}^{\infty} \frac{i} {(36n^2-1)^i}$ . Find $S_1 + S_2 $ I know to find sum of series using method of difference. I tried sum of write the term as (6n-1)(6n+1). i don't know how to proceed further.
| Assuming that you know about the polygamma functions, first write
$$(36n^2-1)^2=(6n+1)^2(6n-1)^2$$ and perform a first partial fraction decomposition
$$\frac 2{(36n^2-1)^2}=\frac{1}{2 (6 n+1)}-\frac{1}{2 (6 n-1)}+\frac{1}{2 (6 n+1)^2}+\frac{1}{2 (6 n-1)^2}$$
Computing the partial sum
$$\Sigma_p=\sum_{n=1}^p\frac 2{(36n^2-1)^2}=$$ $$\Bigg[\frac{1}{12} \left(\psi ^{(0)}\left(p+\frac{7}{6}\right)-\psi
^{(0)}\left(\frac{7}{6}\right)\right) \Bigg]-\Bigg[\frac{1}{12} \left(\psi ^{(0)}\left(p+\frac{5}{6}\right)-\psi
^{(0)}\left(\frac{5}{6}\right)\right) \Bigg]+\Bigg[\frac{1}{72} \left(\psi ^{(1)}\left(\frac{7}{6}\right)-\psi
^{(1)}\left(p+\frac{7}{6}\right)\right) \Bigg]+\Bigg[\frac{1}{72} \left(\psi ^{(1)}\left(\frac{5}{6}\right)-\psi
^{(1)}\left(p+\frac{5}{6}\right)\right) \Bigg]$$
Now, using the asymptotics
$$\Sigma_p=\frac{1}{72} \left(6 \psi ^{(0)}\left(\frac{5}{6}\right)-6 \psi
^{(0)}\left(\frac{7}{6}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)+\psi
^{(1)}\left(\frac{7}{6}\right)\right)-\frac{1}{1944 p^3}+O\left(\frac{1}{p^4}\right)$$ Thus
$$S_2=\frac{1}{72} \left(6 \psi ^{(0)}\left(\frac{5}{6}\right)-6 \psi
^{(0)}\left(\frac{7}{6}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)+\psi
^{(1)}\left(\frac{7}{6}\right)\right)=-1+\frac{\pi }{4 \sqrt{3}}+\frac{\pi ^2}{18}$$
$$S_1+S_2=\frac{2 \pi ^2}{9}+\frac{\pi }{4 \sqrt{3}}-1\sim 1.64670 $$
Obviously, this is not the same result as in other answers.
I must say that I do not understand how, adding positive terms to $S_1$, we could have a vlaue smaller then $\frac{ \pi ^2}{6} \sim 1.64493$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
trouble with double/iterated integration of $\int^1_0[\int^1_0v(u+v^2)^4du]dv$ I have:
For $\int^1_0v(u+v^2)^4du$:
u substitution (using x instead since there's a u in there already) with $x=(u+v^2), dx/du=1$
$v\int^1_0x^4=v\frac{1}{5}x^5=v\frac{(u+v^2)^5}{5}-\frac{v(0+v^2)^5}{5}|^1_0=\frac{(v+v^3)^5}{5}-\frac{v^{15}}{5}$
then
$ \frac{1}{5}[\int^1_0(v+v^3)^5dv-\int^1_0v^{15}dv]$
used substitution again with $x= (v+v^3),dx/du=(1+3v^2)$ for the first one
$\int^1_0\frac{x^5dx}{(1+3v^2)}->\frac{1}{1+3v^2}\int^1_0x^5dx=\frac{1}{1+3v^2}\frac{1}{6}x^6|^1_0->\frac{1}{1+3v^2}\frac{1}{6}(v+v^3)^6|^1_0$
$= \frac{1}{24}*1-0$
for the second:
$\int^1_0 v^{15}=\frac{1}{16}v^{16}|^1_0=\frac{1}{16}$
$\frac{1}{5}[\frac{1}{24}-\frac{1}{16}] = -\frac{1}{240}$
The answer is supposed to be 31/30.
| Note that $v(1+v^2)^5\ne (v+v^3)^5$. Rather, we have upon integrating with respect to $u$,
$$ \int_0^1 v(u+v^2)^4\,du=\frac v5 \left( (1+v^2)^5-v^{10} \right)$$
Now, note that the integration over $v$ is facilitated by the substitution $w=1+v^2$. This is left as an exercise for the reader.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the system of inequalities Solve the system:
$$|3x+2|\geq4|x-1|$$
$$\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$$
So for $|3x+2|\geq4|x-1|$ I got the solution $x=6$ and $x=2/7$ and Wolframalpha agrees with me. I'm having troubles writing the final solution for $\frac{x^{2}+x-2}{2+3x-2x^{2}}\leq 0$.
$1)$ $x^{2}+x-2\leq0$ and $-2x^{2}+3x+2>0$
$x^{2}+x-2\leq0$ $\Rightarrow x\in [-2,1]$
$-2x^{2}+3x+2>0$ $\Rightarrow x\in (-1/2,2)$
Now I need the intersection of those two sets, which is $(-1/2,1]$. From the first inequality I got $x=6$ and $x=2/7$ so the final solution is $x=2/7$.
$2)$ $x^{2}+x-2\geq 0$ and $-2x^{2}+3x+2<0$
$x^{2}+x-2\geq 0$ $\Rightarrow x\in(-\infty, -2]\cup[1,\infty]$
$-2x^{2}+3x+2<0$ $\Rightarrow x\in(-\infty, -1/2)\cup(2,\infty)$
Now I'm having troubles finding the intersection of those sets.
| Consider $$f(x)=\frac{x^{2}+x-2}{2+3x-2x^{2}}=\frac{(x-1)(x+2)}{-(2x+1)(x-2)}$$ It follows $$f(x)=\frac{-(x-1)(x+2)}{(2x+1)(x-2)}\le0\iff\frac{(x-1)(x+2)}{(2x+1)(x-2)}\ge0$$
One has at once the solution set for $f(x)\le 0$ is $$S_1=(-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)$$
Now consider $$g(x)=\left|\frac{3x+2}{x-1}\right|=\left|\frac{3+\frac 2x}{1-\frac 1x}\right|$$ it follows $$\begin{cases}\lim_{x\to\pm\infty}=3\\\lim_{x\to-1}g(x)=\infty\\g\text { decreasing on } (-\infty,-\frac 23)\cup(-1,\infty)\\g\text { increasing on } (-\frac 23,1)\end{cases}$$ Furthermore $$\left|\frac{3x+2}{x-1}\right|=4\iff\begin{cases}3x+2=4(x-1)\iff x=6\\3x+2=-4(x-1)\iff x=\frac 27\end{cases}$$ Hence the solution set for $g(x)\ge 4$ is $$S_2=[\frac 27,1)\cup(1,6]$$
Thus our solution is $$S_1\cap S_2=\left((-\infty,-2]\cup(-\frac 12,1]\cup(2,\infty)\right)\cap\left([\frac 27,1)\cup(1,6]\right)=\color{red}{(\frac 27,1)\cup (2,6]}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$
Positive real numbers $a,b,c$ satisfy $abc=1$. Prove
$$\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1$$
I tried AM–GM, Cauchy–Schwarz and Jensen's but they all failed.
| Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.
Hence, by C-S and Rearrangement we obtain $$\sum\limits_{cyc}\frac{1}{a^2+a+1}=\sum\limits_{cyc}\frac{y^2}{x^2+xy+y^2}=\sum\limits_{cyc}\frac{y^2(y+z)^2}{(x^2+xy+y^2)(y+z)^2}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(x^2+xy)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2)(y^2+2yz+z^2)}=\frac{\sum\limits_{cyc}(x^4+2x^3y+2x^3z+3x^2y^2+4x^2yz)}{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)}=$$
$$=\frac{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)+xyz\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}-x-y-z\right)}{\sum\limits_{cyc}(x^4+2x^3y+x^3z+3x^2y^2+5x^2yz)}\geq1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
value of $a$ for which $25^x+(a+2)5^x-(a+3)<0$for at least one real $x$
Find the values of $a$ for which the inequality is satisfied for $25^x+(a+2)5^x-(a+3)<0$
for at least one real value of $x$
$\bf{My\; Try::}$ We can write it as $a(5^x-1)<-\left[25^x+2\cdot 5^x-3\right]$
So $$a < -\left(\frac{25^x-5^x+3\cdot 5^x-3}{5^x-1}\right) = -\frac{(5^x+3)(5^x-1)}{5^x-1} = -(5^x+3)$$
Now how can i solve after that help required, Thanks
| $$(5^x)^2+(a+2)5^x-(a+3)=(5^x+a+3)(5^x-1)<0$$
If $5^x-1>0,1<5^x<-(a+3)$ which is possible if $1<-(a+3)\iff a<-4$
If $5^x-1<0,-(a+3)<5^x<1$ which is possible if $1>-(a+3)\iff a>-4$
So, we need $a\ne4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Taylor Series of trigonometric function I searched quite a bit online but could only find the MACLAURIN SERIES of $\sin x$ and $\cos x$:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}+\cdots$$
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+\cdots$$
Can anyone explain how we can express the TAYLOR SERIES of $\sin x$ and $\cos x$, and also show me the derivation?
Thanks in advance!
| Observe
\begin{align}
\sin x =&\ \sin(x-a+a) = \sin(x-a) \cos a +\cos(x-a) \sin a\\
=&\ \cos a \sum^\infty_{n=0} (-1)^n\frac{(x-a)^{2n+1}}{(2n+1)!} + \sin a\sum^\infty_{n=0}(-1)^n\frac{(x-a)^{2n}}{(2n)!}.
\end{align}
I will leave it as an exercise for the reader to figure out the Taylor expansion for $\cos x$ at $x=a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| $P(x)=1+(ax^2+bx+c)(x-1)^3=1-(c+bx+ax^2)(1-3x+3x^2-x^3)$
$=1-c+x(3c-b)+x^2(-a+3b-3c)+x^3(\cdots)$
We need $1-c=3c-b=-a+3b-3c=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $
Prove
$$ \frac{1}{\sin 2x} + \frac{1}{\sin 4x } + \cdots + \frac{1 }{\sin 2^n x} = \cot x - \cot 2^n x $$
where $n \in \mathbb{N}$ and $x$ not a multiple of $\frac{ \pi }{2^k} $ for any $k \in \mathbb{N}$.
My try. If $n=2$, we have
$$\begin{align}
\frac{1}{\sin 2x} + \frac{ 1}{\sin 4 x} &= \frac{1}{\sin 2x} + \frac{1}{2 \sin 2x \cos 2x } \\[6pt]
&= \frac{ 2 \cos 2x + 1 }{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{2 \cos^2 x - 2 \sin^2 x + \cos^2 x + \sin^2 x}{2 \sin 2x \cos 2x} \\[6pt]
&= \frac{3 \cos^2 x - \sin^2 x}{2 \sin 2x \cos 2x}
\end{align}$$
but here I got stuck. I am on the right track? My goal is to ultimately use induction.
| Let $S_n$ be the sum $\sum_{k=1}^n\frac{1}{\sin(2^kx)}$.
we want prove by induction that
$\forall n\geq1 \;\; S_n=cot(x)-cot(2^nx)$.
the formula is true for $n=1$.
let $n\geq1 : S_n true$.
we must prove that
$S_{n+1}-S_n=cot(2^nx)-cot(2^{n+1})$.
$cot(2^nx)-cot(2^{n+1}x)=$
$\frac{\cos(2^nx)}{\sin(2^nx)}-\frac{\cos(2^{n+1}x)}{2\sin(2^nx)\cos(2^nx)}=$
$\frac{1}{\sin(2^{n+1})}=S_{n+1}-S_n$.
using $\sin(2X)=2\sin(X)\cos(X)$
and
$\cos(2X)=2\cos^2(X)-1$.
thus
$\color{green}{S_{n+1}=S_n+cot(2^nx)-cot(2^{n+1})
=cot(x)-cot(2^{n+1})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$ where $0\leq x\leq 360$ Solve
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot \sin 2x \cdot \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x \cdot \cos x ). (4 \sin x \cdot \cos x \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I proceed further?
| Using $\sin3B=3\sin B-4\sin^3B$
$$32\sin^3x\cdot\cos^2x\cdot\cos2x=\sin x(3-4\sin^2x)$$
If $\sin x=0,x=180^\circ n$ where $n$ is any integer
Else using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$32\cdot\dfrac{1-\cos2x}2\cdot\dfrac{1+\cos2x}2\cdot\cos2x=3-2(1-\cos2x)$$
$$\iff8(c-c^3)=1+2c\iff4c^3-3c=-\dfrac12$$
$$\implies\cos6x=-\dfrac12,6x=360^\circ m\pm120^\circ\iff x=60^\circ m\pm20^\circ$$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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real values of $a$ for which the range of function $ f(x) = \frac{x-1}{1-x^2-a}$ does not contain value from $\left[-1,1\right]$
All real values of $a$ for which the range of function $\displaystyle f(x) = \frac{x-1}{1-x^2-a}$ does not
contain any value from $\left[-1,1\right]$
$\bf{My\; Try::}$ Let $\displaystyle y = \frac{x-1}{1-x^2-a}\Rightarrow y-x^2y-ay=x-1$
So $$x^2y+x+y(1-a)-1=0$$
Now for real values of $y<-1\cup y>1$ equation has real roots
So $$1-4y^2(1-a)\geq 0\Rightarrow 1\geq 4y^2(1-a)\Rightarrow y^2\leq \frac{1}{4(1-a)}\;, a\neq 1$$
Now how can i solve it after that, Help required, Thanks
| We want to find $a$ such that $|f(x)|=\bigg|\frac{x-1}{1-x^2-a}\bigg|\gt 1$ for every $x$ in the domain.
*
*If $a\not=0$, then $f(1)=\frac{0}{-a}=0$.
*If $a=0$, then $f(-2)=\frac{-2-1}{1-4-0}=1$.
Therefore, there is no such $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991546",
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How many #8 digits can you build?
How many different 8-digit numbers can be formed using two 1s, two 2s, two 3s, and two 4s such that no two
adjacent digits are the same?
So we have $0 -> 9 = 10$ options for the digits but with the constraints we have something different.
Cases:
$1$ _ _
$2$ _ _
$3$ _ _
$4$ _ _
We can do recursion in the form $A(N), B(N),.. D(N)$ as to how many sequences can be built from the $1, 2, .. 4$ for $N$ letters.
We have $T(n) = A(N-1) + B(N-1) + C(N-1) + D(N-1)$
We are after $T(8)$. We have that for instance $A(n-1) = B(n-2) + C(n-2) + D(n-2) = [A(n-3) + C(N-3) + D(n-3)] + [A(n-3) + B(n-3) + D(n-3)] + [A(n-3) + B(n-3) + C(n-3)] = 3A(n-3) + 2B(n-3) + 2C(n-3) + 2D(n-3)$.
But this gets long fast.
| Use inclusion/exclusion principle:
*
*Include the total number of sequences, which is $\frac{8!}{2!2!2!2!}$
*Exclude the number of sequences containing $11$, which is $\frac{7!}{1!2!2!2!}$
*Exclude the number of sequences containing $22$, which is $\frac{7!}{1!2!2!2!}$
*Exclude the number of sequences containing $33$, which is $\frac{7!}{1!2!2!2!}$
*Exclude the number of sequences containing $44$, which is $\frac{7!}{1!2!2!2!}$
*Include the number of sequences containing $11$ and $22$, which is $\frac{6!}{1!1!2!2!}$
*Include the number of sequences containing $11$ and $33$, which is $\frac{6!}{1!1!2!2!}$
*Include the number of sequences containing $11$ and $44$, which is $\frac{6!}{1!1!2!2!}$
*Include the number of sequences containing $22$ and $33$, which is $\frac{6!}{1!1!2!2!}$
*Include the number of sequences containing $22$ and $44$, which is $\frac{6!}{1!1!2!2!}$
*Include the number of sequences containing $33$ and $44$, which is $\frac{6!}{1!1!2!2!}$
*Exclude the number of sequences containing $11$ and $22$ and $33$, which is $\frac{5!}{1!1!1!2!}$
*Exclude the number of sequences containing $11$ and $22$ and $44$, which is $\frac{5!}{1!1!1!2!}$
*Exclude the number of sequences containing $11$ and $33$ and $44$, which is $\frac{5!}{1!1!1!2!}$
*Exclude the number of sequences containing $22$ and $33$ and $44$, which is $\frac{5!}{1!1!1!2!}$
*Include the number of sequences containing $11$ and $22$ and $33$ and $44$, which is $\frac{4!}{1!1!1!1!}$
Hence the number of such sequences is $\sum\limits_{n=0}^{4}(-1)^n\cdot\binom4n\cdot\frac{(8-n)!}{(1!)^n(2!)^{4-n}}=864$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$. I'm trying to figure out how to get $\frac{2}{3}(3x-5)^{2}+\frac{19}{3}$ from this expression $6x^{2}-20x+23$.
Hints?
| HINT:
Complete the square, which can be done by noting that:
$$6x^2 - 20x + 23 = 6(x^2 - \frac{10}{3}x + \frac{25}{9}) + \frac{19}{3}$$
Then multiply $(x^2 - \frac{10}{3}x + \frac{25}{9})$ by $9$ and divide $6$ by $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find this diophantine equation intgers $x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$ Find the solutions of the diophantine equation
$$x^4+4x^3y-6x^2y^2-4xy^3+y^4=1$$
I have found $(x,y)=(\pm 1,0),(0,\pm 1)$: how to find all solutions?
Thanks
| $$\overbrace{(x+y)^4}^A-4\overbrace{xy^2(3x+2y)}^B=1$$
As $x,y$ are integers, a possible solution is $A=1, B=0$ i.e.
$$A=1\Rightarrow x+y=\pm 1\quad\cdots (1)\\
B=0\Rightarrow xy^2=0 \Rightarrow x=0 \text{ or } y=0\quad\cdots (2a)\\
\text{or}\\
3x+2y=0\Rightarrow 3x=-2y\quad\cdots (2b)$$
From $(1),(2a)$,
$$(x,y)=(0,\pm1), (\pm1,0)$$
From $(1),(2b)$,
$$(x,y)=(2,-3), (-2,3)..$$
Also, another solution is
$$(x,y)=(\pm3,\pm2)$$
Hence, integer solutions are
$$(x,y)\;=\;(0,\pm 1),\; (\pm1,0),\; (\pm2,\mp3),\; (\pm3,\pm2)\quad\blacksquare $$
NB:
$$\begin{align}
4xy^2(3x+2y)&=(x+y)^4-1\\
&=(x+y-1)(x+y+1)((x+y)^2+1)\\
&=(2n)(2n+2)(4n^2+4n+2)&&(x+y=2n+1)\\
xy^2(3x+2y)
&=2n(n+1)(2n^2+2n+1)\\
&=2n(n+1)(2n(n+1)+1)\\
&=u(u+1)&&(u=2n(n+1))
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
| There's no need for induction, the answer is immediate with modular arithmetic!
$$ 5\cdot10^n+10^{n-1}+3\equiv 5\cdot1^n+1^{n-1}+3=5+4=9\equiv0\pmod{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
} |
Calculate $ \lim\limits_{n \to \infty} \sum\limits_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $
Let $a \in (1, \infty)$. Calculate $$ \lim\limits_{n \to \infty} \sum_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $$
Here's what I tried:
Let $ x_n = \sum_{k=0}^n \frac{2^{k-1}}{a^{2k-1}+1} $. Let $a_n, b_n$ so that $a_n < x_n < b_n$. If $a_n \to l, b_n \to l$ with $ l \in \mathbb{R}$ then $x_n \to l$. I tried saying that $$ {{1+2+2^2+\dots+2^{n-1}}\over{a^{2n-1}+1}} < x_n < {{1+2+2^2+\dots+2^{n-1}}\over{a+1}} $$ but the limit of the lefthand-side fraction is $0$ and the one for the righthand-side is $\infty$. How can I solve this?
| Let $a>\sqrt2$, we have that
\begin{align}
\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}&=\frac a{2a+2}+\sum_{k=0}^{\infty} \frac{2^k}{a^{2k+1}}(1+a^{-1-2k})^{-1}\\
&=\frac a{2a+2}+\sum_{k=0}^{\infty} \frac{2^k}{a^{2k+1}}\sum_{n=0}^{\infty}(-a^{-1-2k})^n
\end{align}
Applying Fubini's theorem,
\begin{align}
\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}
&=\frac a{2a+2}+\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}\sum_{k=0}^{\infty}(\frac2{a^{2n+2}})^k\\
&=\frac a{2a+2}+\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}(1-\frac2{a^{2n+2}})^{-1}\\
&=\frac a{2a+2}+\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}-\frac2{a^{n+1}}}\\
&=\frac a{2a+2}-\sum_{n=1}^{\infty}(-1)^n\frac{a^n}{a^{2n}-2}\tag{*}\label{*}\\
&=\frac a{2a+2}-\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}(1+\frac2{a^{2n}-2})\\
&=\frac a{2a+2}+\frac1{a+1}-\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}\frac2{a^{2n}-2}\\
&=\frac 1 2+\frac1 {2a+2}-\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}\frac2{a^{2n}-2}
\end{align}
This formula can give an approximation while $a\to\infty$.
We can see that the sum $\to \infty$ when $a\to \sqrt2_+$ :
$$\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}=\frac 1 2+\frac1 {2a+2}+\bbox[5px,border:2px solid red]{\frac2{a(a^2-2)}}-\sum_{n=2}^{\infty}\frac{(-1)^n}{a^n}\frac2{a^{2n}-2}$$
which gives :
$$\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}\operatorname*{\sim}_{a\to\sqrt2_+}\frac1{2(a-\sqrt2)}$$
More exactly, from $\eqref{*}$,
\begin{align}
\sum_{k=0}^{\infty} \frac{2^{k-1}}{a^{2k-1}+1}
&=\frac a{2a+2}-\frac12\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n+\sqrt2}-\frac12\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n-\sqrt2}\\
&=\frac a{2a+2}+\frac12\sum_{n=1}^{\infty}\frac{1}{a^n+\sqrt2}+\frac12\sum_{n=1}^{\infty}\frac{1}{a^n-\sqrt2}-\sum_{n=1}^{\infty}\frac{1}{a^{2n}+\sqrt2}-\sum_{n=1}^{\infty}\frac{1}{a^{2n}-\sqrt2}
\end{align}
One can then express this ugly sum using some uglier q-psi functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $\int_{|z|=4}\frac{\sin z}{z(z-2i)}\ dz$
Evaluate $$\int_{|z|=4}\frac{\sin z}{z(z-2i)}\ dz$$
The two singularities are included in inside the disk $|z|<4$, so I broke the integral in a sum of two integrals.
The first one:
$$\int \frac{\frac{\sin z}{z}}{z-2i}\ dz = 2i\pi f_1(2i)$$
where $f_1 = \frac{\sin z}{z} \implies f_1(2i) = \frac{\sin 2i}{2i}$.
The second one:
$$\int \frac{\frac{\sin z}{z-2i}}{z}\ dz = 2i\pi f_2(0)$$
where $f_2 = \frac{\sin z}{z-2i}\implies f_2(0) = 0$.
The result should be the sum of the two integrals.
However, a friend of mine did the following:
$$\int \frac{\sin z}{(z-0)(z-2i)}\ dz$$
Then
$$\frac{1}{(z-0)(z-2i)} = \frac{A}{z-0}+\frac{B}{z-2i}\implies
A = \frac{i}{2}, B = \frac{-i}{2}$$
and
$$\int_{|z|=4} \frac{\sin z}{z(z-2i)}\ dz = \int_{|z|=4} \frac{A}{z}\ dz+\int_{|z|=4}\frac{B}{z-2i}\ dz = 0 + 2i\pi$$
Which one is right? If both are wrong, at least which method is the right one? I've seen mine on the internet but I can't find anything wrong with my friend's answer.
| Both methods work, but there is a small error in the second.
It is true that
$$
\frac{1}{z(z-2i)} = \frac{A}{z} + \frac{B}{z-2i}
$$
for $A = i/2$ and $B = -i/2$, but the error comes in the next line where your friend essentially says
$$
\frac{\sin z}{z(z-2i)} = \frac{A}{z} + \frac{B}{z-2i}
\newcommand{\contour}{{|z|=4}}
$$
The left-hand side got multiplied by $\sin z$, but not the right. We can fix it by including $\sin z$ on the right:
\begin{align}
\int_\contour \frac{\sin z}{z(z-2i)} \, dz &= \int_\contour A \cdot \frac{\sin z}{z} \, dz + \int_\contour B \cdot \frac{\sin z}{z-2i} \, dz
\end{align}
The first integral evaluates to zero since the singularity at $z=0$ is removable (by assigning the value $1$). The second integral has a first-order pole at $z=2i$ and can be computed using Cauchy's Integral Formula:
\begin{align*}
\int_\contour B \cdot \frac{\sin z}{z-2i} \, dz &= 2\pi i \cdot B \sin(2i) \\
&= 2\pi i \cdot (-i/2) \cdot \sin(2i) \\
&= \pi \sin(2i) \\
&= 2\pi i \cdot f_1(2i)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fibonacci Identity with Binomial Coefficients A friend showed me this cool trick: Take any row of Pascal's triangle (say, $n = 7$):
$$1, 7, 21, 35, 35, 21, 7, 1$$
Leave out every other number, starting with the first one:
$$7, 35, 21, 1$$
Then these are backwards base-5 "digits", so calculate:
$$7 + 35 \cdot 5 + 21 \cdot 5^2 + 1 \cdot 5^3 = 7 + 175 + 525 + 125 = 832$$
and divide by $2^{7-1} = 2^6 = 64$:
$$\frac{832}{64} = 13$$
and $F_7 = 13$ (the seventh Fibonacci number)!
He said it works for any $n$. I have worked out that this would be to prove that:
$$\frac{1}{2^{n-1}}\sum_{k = 0}^{\lfloor{\frac{n}{2}}\rfloor}{\left(5^k {n \choose 2k + 1} \right)} = F_n $$
I'm not sure how to proceed from here. Is there a neat combinatoric or easy algebraic proof I am missing? Thanks!
| Starting with Binet's Formula:
$$F_n = \frac{1}{\sqrt{5}} \left( \left(\frac{1 + \sqrt{5}}{2} \right)^n - \left(\frac{1 - \sqrt{5}}{2} \right)^n \right) $$
Expand using Binomial Formula:
$$F_n = \frac{1}{2^n\sqrt{5}} \left( \sum_{k=0}^n \left({n \choose k} \left( \sqrt{5} \right)^k\right) - \sum_{k=0}^n \left({n \choose k} \left( \sqrt{5} \right)^k \left(-1\right)^k\right) \right) $$
Combine:
$$F_n = \frac{1}{2^n\sqrt{5}} \left( \sum_{k=0}^n \left({n \choose k} \left( \sqrt{5} \right)^k\left(1-\left(-1\right)^k\right)\right)\right) $$
But $1 - \left(-1\right)^k$ is $0$ for even $k$ and $2$ for odd $k$. So we can consider only the case where $k$ is odd, so $k = 2j + 1$:
$$F_n = \frac{1}{2^n\sqrt{5}} \left( \sum_{2j+1=1}^{2j+1=n} \left({n \choose 2j+1} \left( \sqrt{5} \right)^{2j+1}\cdot2\right)\right) $$
Simplify:
$$F_n = \frac{1}{2^{n-1}\sqrt{5}} \sum_{j=0}^{\lfloor{\frac{n}{2}}\rfloor} \left({n \choose 2j+1} \left( 5^j \sqrt{5} \right)\right) $$
$$F_n = \frac{1}{2^{n-1}} \sum_{j=0}^{\lfloor{\frac{n}{2}}\rfloor} \left(5^j {n \choose 2j+1} \right) $$
And we are done!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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What is $(-1)^{\frac{1}{7}} + (-1)^{\frac{3}{7}} + (-1)^{\frac{5}{7}} + (-1)^{\frac{9}{7}} + (-1)^{\frac{11}{7}}+ (-1)^{\frac{13}{7}}$? The question is as given in the title. According to WolframAlpha, the answer is 1, but I am curious as to how one gets that. I tried simplifying the above into $$6(-1)^{\frac{1}{7}}$$ however that does not give 1.
| Note that $(-1)^{\frac{1}{7}} = -1$
Also note that $(-1)^{\frac{9}{7}} = -(-1)^{\frac{2}{7}},\ (-1)^{\frac{11}{7}} = -(-1)^{\frac{4}{7}},\ (-1)^{\frac{13}{7}} = -(-1)^{\frac{6}{7}}$
You can then rewrite your addition as
$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}{7}} - (-1)^{\frac{6}{7}}$$
Now remembering that $a^{\frac{p}{q}} = (a^{\frac{1}{q}})^p = (\sqrt[q]{a})^p$ and substituting $q = 7$ and the appropriate $p$ you get:
$$(-1)^{\frac{1}{7}} - (-1)^{\frac{2}{7}} + (-1)^{\frac{3}{7}} - (-1)^{\frac{4}{7}} + (-1)^{\frac{5}{7}} - (-1)^{\frac{6}{7}} \\
= (-1) - (1) + (-1) - (1) + (-1) - (1) \\
= -6$$
Now what if instead of $a^{\frac{p}{q}} = (a^{\frac{1}{q}})^p$ you make $a^{\frac{p}{q}} = \sqrt[q]{a^p}$? Then for negative $a$, odd $p$ and even $q$ that would only be calculatable if you entered the realm of the complex numbers and made use of the imaginary unit. But when you enter the realm of the complex numbers, you get that $\sqrt[q]{x}$ has $q$ different values. That is, there are $q$ different numbers $x_i$ for which $(x_i)^q = x$.
WolframAlpha goes straight ahead and uses the complex plane to calculate the roots making use of the complex numbers. Now if you make use of the fact that the sum of all $q$ roots of $x$ is 0, you can do the following:
Call $x_i$ the ith root of (-1). Then $x_i = (-1)^{\frac{i}{7}}$ and if you notice that in your sum all roots are represented except for $x_0$ which is -1, then
$$\sum_{i=0}^{6} x_i = -1 + \sum_{i=1}^{6} x_i$$ but $$\sum_{i=0}^{6} x_i = 0$$ so $$0 = -1 + \sum_{i=1}^{6} x_i \iff \sum_{i=1}^{6} x_i = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
| Ignoring big theorems,
$$ f(n) = 1^4+2^4+\ldots+n^4 $$
is quite blatantly a polynomial in $n$ with degree $5$. It follows that it is enough to compute $f(0),f(1),\ldots,f(5)$ to recover the coefficients of such a polynomial by Lagrange interpolation:
$$ f(x) = \sum_{k=0}^{5}\,f(k)\!\!\!\!\prod_{\substack{j\in\{0,\ldots,5\}\\ j\neq k}}\frac{x-j}{k-j}.$$
Alternative approach: since
$$ \sum_{n\geq 0} n^4 e^{-nx} = \frac{d^4}{dx^4}\sum_{n\geq 0}e^{-nx} = \frac{d^4}{dx^4}\left(\frac{1}{1-e^{-x}}\right) =\frac{e^x(e^x+1)(e^{2x}+10 e^{x}+1)}{(e^x-1)^5}$$
it follows that for any $x\in(0,1)$ we have
$$ \sum_{n\geq 0}n^4 x^n = \frac{x \left(1+11 x+11 x^2+x^3\right)}{(1-x)^5} $$
and $f(N)$ is the coefficient of $x^N$ in $\frac{x \left(1+11 x+11 x^2+x^3\right)}{(1-x)^6}$. By stars and bars,
$$\boxed{\,f(N) = 1^4+2^4+\ldots+N^4= \color{red}{\frac{N(N+1)(2N+1)(3N^2+3N-1)}{30}} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find the dimensions of the rectangle that will give the minimum perimeter. a farmer wants to make a rectangular paddock with an area of $ 4000 m^2$ However, fencing costs are high and she wants the paddock to have a minimum perimeter.
I have found the perimeter:
$$x\cdot y = 4000\\
y = \frac{4000}{x}$$
$$\begin{align}\text{Perimeter} &= 2x + 2y\\
&= 2x + 2(4000/x)\\
&= 2x + (8000/x)\end{align}$$
How do I find the dimensions that will give the minimum perimeter?
| To minimise $2(x+y)$ subject to $xy=A$ note that: $$(x+y)^2=4xy+(x-y)^2=4A+(x-y)^2$$
Now since $x$ and $y$ are both positive, the minimum value of $2(x+y)$ occurs when $4(x+y)^2$ is a minimum and hence when $(x+y)^2$ is a minimum.
Since $(x-y)^2$ is non-negative, the minimum occurs when $x-y=0$ ie when you have a square and $x+y=2\sqrt A$ and $2(x+y)=4\sqrt A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding sum of infinite series $1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $ So the question is 'express the power series $$1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\ldots $$
in closed form'.
Now we are allowed to assume the power series of $e^x$ and also we derived the power series for $\cosh x$ using exponentials.
Now my question is the best way to approach the problem above. There is a hint that says 'use the fact that $\zeta^2+\zeta +1=0 $ where $\zeta $ is a cube root of unity'.
The way I solved this was to recognise that the third derivative of the series was equal to the series itself. So I just solved a linear ODE of order 3 and then found the constants (and so the hint made sense - in a way). But I don't think the question was designed for me to do this and so I feel as though I'm missing something obvious that makes this problem very easy.
Can anyone see any alternatives that make use of the hint in a more natural way?
| We evaluate the power series $e^{\zeta x}$, noting that $\zeta^2 = -(1+\zeta)$ and $\zeta^3 = 1$, and get
$$
e^{\zeta x} = 1 + \zeta x + \frac{\zeta^2x^2}{2!} + \frac{\zeta^3x^3}{3!} + \frac{\zeta^4 x^4}{4!} + \cdots\\
= 1 + \zeta x - \frac{(1+\zeta)x^2}{2!} + \frac{x^3}{3!} + \frac{\zeta x^4}{4!} -\cdots
$$
while evaluating the power series $e^{\zeta^2 x}$ gives
$$
e^{\zeta^2 x} = 1 + \zeta^2 x + \frac{\zeta^4x^2}{2!} + \frac{\zeta^6x^3}{3!} + \frac{\zeta^8 x^4}{4!} + \cdots\\
= 1 - (1+\zeta)x + \frac{\zeta x^2}{2!} + \frac{x^3}{3!} - \frac{(1+\zeta)x^4}{4!} + \cdots
$$
Adding these together gives us
$$
e^{\zeta x} + e^{\zeta^2 x} = 2 - x - \frac{x^2}{2!} + \frac{2x^3}{3!} - \frac{x^4}{4!} - \cdots
$$
Finally, we see that if we add the power series of $e^x$ to this one, we get $3$ times the series we were after. That means that the final answer is
$$
1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \cdots = \frac{e^x + e^{\zeta x} + e^{\zeta^2 x}}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009189",
"timestamp": "2023-03-29T00:00:00",
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Explain how and why the cancellation of $ 6 $’s in $ \dfrac{16}{64} $ to get $ \dfrac{1}{4} $ is a fallacious statement. Based on what we know from elementary and middle school teachers, most of us know that 16/64 correctly equals 1/4 because 16/64 is simplified with a common divisibility of 16. However, there is another way to prove that 16/64 equals 1/4 without dividing the numerator and denominator by 16. Who can explain how and why that method leads to a fallacious statement?
| The way someone might have justified that:
"Just remember that
$$\require {cancel}\frac {10}{20} = \frac{1\cancel {0}}{2\cancel {0}} = \frac{1}{2} $$
Therefore
$$\require {cancel}\frac{16}{64} = \frac{1\cancel {6}}{\cancel {6}4} = \frac{1}{4} $$
$\blacksquare$"
There is actually a problem on project euler regarding this type of fractions. Those that can be fallaciously simplified to something that holds as true.
Why it does not work:
There is a widely-used simplification that is
$$\require {cancel}\frac{a\cdot b}{a\cdot d} = \frac{\cancel {a}b}{\cancel {a}d} $$
That works because we have a product. The above fraction is just syntatic sugar for
$$a\cdot b \cdot \frac{1}{a} \cdot \frac{1}{d}$$
But the product is commutative and therefore we have
$$\require{cancel} a\cdot b \cdot \frac{1}{a} \cdot \frac{1}{d} = \cancel {a}\cdot b \cdot \cancel {\frac{1}{a}} \cdot \frac{1}{d} = \frac{b}{d}$$
The problem with the digits is that $16$ is not $1\cdot6$ just as $64 \not= 6\cdot4$. That means $\frac{1}{64} $ is not syntatic sugar for $\frac{1}{6}\cdot\frac{1}{4} $ and the 6s won't cancel. It only works when the numbers end in 0 because if $k $ and $j $ end in 0, then $k $ is the product of $k'$ with $10$ and $j $ is the product of $j'$ with $10$. Then we have:
$$\require {cancel}\frac{k}{j} = \frac{k'\cdot10}{j'\cdot10} = \frac{k'\cdot\cancel {10}}{j'\cdot\cancel {10}} = \frac{k'}{j'}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of Maclaurin series Find the sum of the infinite series
\begin{equation}
\sum_{n=2}^\infty\frac{7n(n-1)}{3^{n-2}}
\end{equation}
I think it probably has something to do with a known Maclaurin series, but cannot for the life of me see which one.. Any hints would be appreciated!
Edit: Using your hints, I was able to solve the problem. Solving it here in case someone is wondering about the same thing:
Manipulate \begin{equation}\sum_{n=2}^\infty x^n=\frac{1}{1-x}-x-1 \end{equation}
by differentiating both sides:
\begin{equation}\sum_{n=2}^\infty nx^{n-1}=\frac{1}{(1-x)^2}-1 \end{equation}
differentiate again:
\begin{equation}\sum_{n=2}^\infty (n-1)nx^{n-2}=\frac{2}{(1-x)^3} \end{equation}
plugging in $\frac{1}{3}$ for x:
\begin{equation}\sum_{n=2}^\infty n(n-1)(\frac{1}{3})^{n-2}=\frac{2}{(1-\frac{1}{3})^3} \end{equation}
which is equivalent to
\begin{equation}\sum_{n=2}^\infty \frac{n(n-1)}{3^{n-2}} \end{equation}
finally, multiplying by 7 gives us
\begin{equation}\begin{split}\sum_{n=2}^\infty \frac{7n(n-1)}{3^{n-2}}&=7\frac{2}{(1-\frac{1}{3})^3}\\
&=\underline{\underline{\frac{189}{4}}} \end{split} \end{equation}
| Hint: Notice
$$ \sum x^n = \frac{ 1}{1- x} $$
$$ \sum n x^{n-1} = \frac{1}{(1-x)^2} $$
$$ \sum n (n-1) x^{n-2} = \frac{2 }{(1-x)^3} $$
they all converge for $|x| < 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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} |
An differential equation$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$ How to solve this ODE $$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$$
I tried to find its integral factor, but failed.
Many thanks for your help.
| Hint:
Let $t=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=2x\dfrac{dy}{dt}$
$\therefore(x^2+y^2+3)2x\dfrac{dy}{dt}=2x\left(2y-\dfrac{x^2}{y}\right)$
$(x^2+y^2+3)\dfrac{dy}{dt}=2y-\dfrac{x^2}{y}$
$(t+y^2+3)\dfrac{dy}{dt}=2y-\dfrac{t}{y}$
Let $u=y^2$ ,
Then $\dfrac{du}{dt}=2y\dfrac{dy}{dt}$
$\therefore\dfrac{t+y^2+3}{2y}\dfrac{du}{dt}=2y-\dfrac{t}{y}$
$(t+y^2+3)\dfrac{du}{dt}=4y^2-2t$
$(t+u+3)\dfrac{du}{dt}=4u-2t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \frac{a}{b}\cdot \frac{\sqrt{3}}{2}$ and $\displaystyle \frac{c}{\sin C} = \frac{b}{\sin 60^0}\Rightarrow \sin C = \frac{c}{b}\cdot \frac{\sqrt{3}}{2}$
So $$\sin A\sin C = \frac{3}{4}\cdot \frac{ac}{b^2}$$
Now using Cosine formula:
$$\cos B = \cos 60^0 = \frac{1}{2}= \frac{a^2+c^2-b^2}{2ac}\Rightarrow b^2=a^2+c^2-ac$$
So $$\sin A\sin C = \frac{3}{4}\bigg[\frac{ac}{a^2+c^2-ac}\bigg] = \frac{3}{4}\bigg[\frac{1}{\frac{a}{c}+\frac{c}{a}-1}\bigg]\leq \frac{3}{4}$$
Using $\bf{A.M\geq G.M},$ We get $$\frac{a}{c}+\frac{c}{a}\geq 2\Rightarrow \frac{a}{c}+\frac{c}{a}-1\geq 1$$
$\bf{ADDED::}$ Using Jensen Inequality:: For $f(x) = \ln(\sin x)\;,$
$$\ln(\sin A)+\ln(\sin C)\leq 2\cdot \ln \sin \left(\frac{A+C}{2}\right) = \ln \cos 30^0 = 2\cdot \ln \frac{\sqrt{3}}{2} = \ln \frac{3}{4}$$
But I do not understand how to calculate the lower bound for $\sin A\sin C$.
Thanks in advance!
| Please check the following hint. Firstly we have
$$
R := \sin A \sin C = \sin A \sin \left( \pi - (B+A) \right) = \sin A \sin \left( \frac{2\pi}{3} - A \right)
$$
thus,
\begin{align}
R
&= \sin A \left( \frac{\sqrt{3}}{2} \cos A + \frac{1}{2} \sin A \right)
= \frac{\sqrt{3}}{4} \sin 2A + \frac{1}{4} \left( 1 - \cos 2A \right)
\\
&=
\frac{\sqrt{3}}{4} \sin 2A - \frac{1}{4} \cos 2A + \frac{1}{4}
=
\frac{1}{2}\left( \sin 2A \cos \frac{\pi}{6} - \cos 2A \sin \frac{\pi}{6} \right) + \frac{1}{4}
\\
&
=
\frac{1}{2}\sin \left( 2A - \frac{\pi}{6} \right) + \frac{1}{4}
\end{align}
Besides,
$$
C = \pi - (A+B) = \frac{2\pi}{3} - A > 0
$$
hence, $0 <A < \frac{2\pi}{3}$, and
$$
0 < 2A - \frac{\pi }{6} < \frac{4\pi}{3} - \frac{\pi }{6} = \pi + \frac{\pi }{6}
$$
Then
$$
- \frac{1}{2} < \sin \left( 2A - \frac{\pi}{6} \right) \le 1
$$
Lastly we obtain
$$
0 < R = \frac{1}{2} \sin \left( 2A - \frac{\pi}{6} \right) + \frac{1}{4}\le \frac{3}{4}
$$
In addition, $R = 3/4$ when $2 A - \frac{\pi}{6} = \frac{\pi}{2}$ or $A = \frac{\pi}{3}$ (i.e. $A=B=C=\pi/3$).
EDIT: There was a mistake right here: $0 <A < \frac{2\pi}{3}$, and
$$
- \frac{\pi }{6} < 2A - \frac{\pi }{6} < \frac{4\pi}{3} - \frac{\pi }{6} = \pi + \frac{\pi }{6}
$$
However, in this case, we also have
$$
- \frac{1}{2} < \sin \left( 2A - \frac{\pi}{6} \right) \le 1
$$
Then the remainder of solution is unchanged.
| {
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"url": "https://math.stackexchange.com/questions/2015166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $f(x)=4x^3+3x^2-x-1$, The Range of values $\frac{f(x_1)-f(x_2)}{x_1-x_2}$ can take is- My Attempt :-
$f'(x)=12x^2+6x-1$ where $f'(x) \ge \frac{-7}{4}$. So (I think) from LMVT we can directly say that
$$\frac{f(b)-f(a)}{b-a} \ge \frac{-7}{4}$$ But the answer Given is $$\frac{f(b)-f(a)}{b-a} > \frac{-7}{4}$$
So my Question is Is This Application of LMVT correct ,if yes then why $\frac{-7}{4}$ is excluded from the given range ?
| The mean value theorem does not tell you that every $c$ has a corresponding $a$ and $b$ with $\frac{f(a)-f(b)}{a-b}=f'(c)$. It tells you that given $a$ and $b$ you can find a $c$, but it doesn't say given $c$ you can find $a$ and $b$. In your case,
$\dfrac{f(a)-f(b)}{a-b}=4(a^2+ab+b^2)+3(a+b)-1=\dfrac{4(2a+2b+1)^2+(4a+1)^2+(4b+1)^2}{8}-\dfrac{7}{4}$.
The minimum of this thus occurs only for $a=b=-\frac{1}{4}$, but this is not allowed because then $a-b$ is $0$ in the denominator. So there is no such $a$ and $b$ that gives you $-\frac{7}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as the same solution.
I can only seem to find two; namely $(1, 0, 0)$ and $( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})$. Is there a method to finding a third or is it still just trial and error?
| There are infinitely many. The complete rational solution to
$$a^2+b^2+c^2=1$$
is given by
$$\left(\frac{p^2-q^2-r^2}s\right)^k+\left(\frac{2pq}s\right)^k+\left(\frac{2pr}s\right)^k=1\tag1$$
where $s=p^2+q^2+r^2$ and $k=2$. But eq $(1)$ is also satisfied for $k=1$ if
$$p=\frac{q^2+r^2}{q+r}$$
For example, if $q=1,\,r=5$, then,
$$\big({-}\tfrac{5}{31}\big)^k+\big(\tfrac{6}{31}\big)^k+\big(\tfrac{30}{31}\big)^k=1$$
for $k=1,2$, and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Subsets and Splits