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Evaluating $\int \frac{1}{x^{7} - x} ~ d{x} $. How do I evaluate the following indefinite integral?
$$
\int \frac{1}{x^{7} - x} ~ d{x}.
$$
Could someone give me some advice as to what method I should use or the steps that I should take?
Note: The OP originally requested for help in evaluating $ \displaystyle \int \left( \frac{1}{x^{7}} - x \right) ~ d{x} $, which may not have been his/her actual intention.
| $\displaystyle \int\frac{1}{x^7-x}dx = \int\frac{1}{x^7.\left(1-\frac{1}{x^6}\right)}dx$
Put $\displaystyle \left(1-\frac{1}{x^6}\right) = t$ and $\displaystyle \frac{6}{x^7}dx = dt$
$\displaystyle = \frac{1}{6}\int\frac{1}{t}dt = \frac{1}{6}\ln \mid t \mid+C$
$\displaystyle = \frac{1}{6}\ln \left|\frac{x^6-1}{x^6}\right|+C$
| {
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probability of drawing marbles based on coin flip
An urn contains 4 red and 3 white marbles. A random marble $M_1$ is drawn and a fair coin is flipped. If the flip is heads then $M_1$ is put back into the urn. On the other hand, if the flip is tails, the marble $M_1$ is not put back into the urn. Now another random marble $M_2$ is drawn from the urn.
(i) What is $\mathrm{Pr}(M_2 = \text{red})$?
(ii) What is $\mathrm{Pr}(M_1 = \text{red}\mid M_2 = \text{red})$?
(iii) What is $\mathrm{Pr}(\text{flip is heads}\mid M_2 = \text{white})$?
Solutions:
(i) $\dfrac{4}{7}$; (ii) $\dfrac{15}{28}$; (iii) $\dfrac{1}{2}$
(iii)'s answer is a half because the the marble color is independent of the coin flip, correct?
Why is (i)'s answer $\dfrac{4}{7}$, why not $\dfrac{3}{6}$ or $\dfrac{4}{6}$ if the coin flip outcome is tails?
As for (ii), I'm completely lost.
Can someone explain please?
| (i) Every marble is just as likely to be picked for $M_2$ as any other marble. That is why the probability is $\frac{4}{7}$. If you want to see this in more detail, lets look at all the possible ways of getting $M_2$ = red. Flip heads: white then red, red then red, flip tails: White then red, red then red. Total probability:
$$\frac{1}{2} \frac{3}{7} \frac{4}{7} + \frac{1}{2} \frac{4}{7} \frac{4}{7} + \frac{1}{2} \frac{3}{7} \frac{4}{6} + \frac{1}{2} \frac{4}{7} \frac{3}{6} = \frac{4}{7}$$
(ii) Using $P(A|B) = P(A \cap B) / P(B)$
$$ P(M_1 = \text{red} \cap M_2 = \text{red}) = \frac{1}{2} \frac{4}{7} \frac{3}{6} + \frac{1}{2} \frac{4}{7} \frac{4}{7} = \frac{15}{49}$$
Giving $\frac{15}{49}/\frac{4}{7} = \frac{15}{28}$ as the answer.
(iii)
$$ P(\text{flip = heads} \cap M_2 = \text{white}) = \frac{1}{2} \left(\frac{4}{7} \frac{3}{7} + \frac{3}{7} \frac{3}{7}\right) = \frac{3}{14}$$
$P(M_2 = \text{white}) = \frac{3}{7}$ by similar argument to (i) and so using the conditional probability rule, the final answer is $\frac{3}{14} / \frac{3}{7} = \frac{1}{2}$
| {
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jenny farm and the dozen egg ??? Farmer Jenny decides to expand her business interests and starts to package and sell the eggs produced by her chooks to a local shop.
The cost of producing $x$ dozen eggs per day is given by, in dirhams:
$$C=( 5/12 x^2+ 4x-3)$$
and the selling price of one dozen eggs is given by, in dirhams$$(6- 1/4 x)$$
How many dozen eggs should be produced each day to maximize the total profit?
| Expanding on the answer by icurays1, the goal is indeed to maximize the quadratic, and the easiest way to do this uses calculus.
But in case you haven't learned calculus, you can maximize a quadratic just by completing the square - no calculator necessary!
The profit for selling $x$ dozen eggs is given by
\begin{align}
p(x) &= x\cdot \text{(price per dozen)} - \text{cost to make $x$ dozen}\\
&= x(6-\frac{1}{4}x) - (\frac{5}{12}x^2+4x-3)\\
&= -\frac{2}{3}x^2 + 2x + 3\\
&= -\frac{2}{3}(x^2 - 3x) + 3\\
&= -\frac{2}{3}(x-\frac{3}{2})^2 + \frac{3}{2} + 3\\
&= -\frac{2}{3}(x-\frac{3}{2})^2 + \frac{9}{2}
\end{align}
Now the term $-\frac{2}{3}(x-\frac{3}{2})^2$ is always negative or zero, so the quadratic achieves its maximum value when $-\frac{2}{3}(x-\frac{3}{2})^2 = 0$, i.e. when $x = \frac{3}{2}$.
Farmer Jenny should produce $\frac{3}{2}$ dozen eggs each day.
The trick here is to get rid of the linear term by "completing the square" so that you can easily see where the maximum is.
| {
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Ratio word problem with eggs? The eggs in a certain basket are either white or brown. If the ratio of the number of white eggs to the number of brown eggs is $2/3$ , each of the following could be the number of eggs in the basket except.....
A 10
B 12
C 15
D 30
E 60
| $$\dfrac 23 = \dfrac 46 = \dfrac 69 = \dfrac{8}{12} = \dfrac{10}{15} = \dfrac {12}{18} = \dfrac {14}{21} = \dfrac {16}{24} = \dfrac{18}{27} = \dfrac{20}{30} = \dfrac{22}{33} = \dfrac{24}{36}$$
For each ratio, the total number of eggs is given by the sum of the numerator with the denominator, ordered as follows $$\text{sum}\;\dfrac {2n}{3n}: 2n + 3n = 5n: \quad 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, ...$$
Note that $12$ does not appear among them.
| {
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Solving for $b$ in $25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$ What are the steps to get from:
$$25\left(\frac{\sqrt{10}-2\sqrt{5}}{50}\right) + 5b = \sqrt{5}$$
to:
$$b = \frac{\sqrt{5}}{5} + \frac{2\sqrt{5} - \sqrt{10}}{10}$$
Thanks.
| First do the easy simplification on the lefthand side:
$$25\left(\frac{\sqrt{10}-2\sqrt5}{50}\right)=\frac{25}{50}\left(\sqrt{10}-2\sqrt5\right)=\frac12\left(\sqrt{10}-2\sqrt5\right)\;,$$
so the equation can be rearranged to
$$5b=\sqrt5-\frac12\left(\sqrt{10}-2\sqrt5\right)\;.$$
Now divide through by $5$ to get
$$b=\frac{\sqrt5}5-\frac1{10}\left(\sqrt{10}-2\sqrt5\right)=\frac{\sqrt5}5-\frac{\sqrt{10}-2\sqrt5}{10}\;.$$
| {
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How to find last items of order to get its sum Find the sum of order:
$$\sum_{n=1}^{∞}\left(\frac{\frac{3}{2}}{2n+3}-\frac{\frac{3}{2}}{2n-1}\right)$$
There is how they count it in book:
$$s_{n} = \left(\frac{3}{10}-\frac{3}{2}\right)+\left(\frac{3}{14}-\frac{1}{2}\right)+\left(\frac{1}{6}-\frac{3}{10}\right)+\left(\frac{3}{22}-\frac{3}{14}\right)+...+\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$
$$s_{n} = \frac{-3}{2}+\frac{-1}{2}+\frac{3}{4n+2}+\frac{3}{4n+6}$$
$$s = \lim_{n->∞}s_{n} = \lim_{n->∞}\left [\frac{-3}2 - \frac12 + \frac3{4n+2} + \frac3{4n+6}\right ] = -2$$
I understand how to solve the lim, I just dont understand, how to get those last items from order, I mean this items:
$$\left(\frac{3}{4n-2}-\frac{3}{4n-10}\right)+\left(\frac{3}{4n+2}-\frac{3}{4n-6}\right)+\left(\frac{3}{4n+6}-\frac{3}{4n-2}\right)$$
UPDATE
Now I understand how to get those "last" items. But I'm confused now, why are they even in the sum inside of lim of $s_n$? If I would keep counting next items, they would get canceled. For example, there is $\frac3{4n+2}$ in $s_n$, if I count n+1 item, this would get canceled. So why do we count them in $s_n$ if only first two fractals {$\frac{-3}2, \frac12 $} couldn't be canceled (if not thinking of negative n).
Could anyone explain please?
| The 3 terms you have mentioned above saying you didnot understand are the (n-2),(n-1) and nth terms of that series.
So if you just substitute 'n' with 'n-2' you get : $$\frac{3}{4n-2} - \frac{3}{4n-10}$$
if you substitute 'n' with 'n-1' you get : $$\frac{3}{4n+2} - \frac{3}{4n-6}$$
| {
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Factorization of three variables Prove that : $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$
Since $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a) $
Therefore the equation becomes : $2(a^3+b^3+c^3)+3(a+b)(b+c)(c+a) - [(c+a)^3 +(a+b)^3 +(b+c)^3]$
Putting $A:=(c+a)$ ; $B:=(a+b)$ ; $C:=(b+c)$
$[(c+a)^3 +(a+b)^3 +(b+c)^3] $ becomes $(A^3+B^3+C^3)$ now again using the formulae:$ a^3+b^3+c^3 = (a+b+c)^3-3(a+b)(b+c)(c+a)$
Could you please guide if any other easier way of doing this...
| OK - if you need a really shorter way (which may work depending on the equation you have):
You have $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$
Looking at the LHS as a cubic polynomial in $a$, we immediately can notice $a=0$ is a root and hence $a$ is a factor. By symmetry, $b$ and $c$ are factors. As $abc$ must be a factor of the LHS (which is homogeneous and also of degree $3$), the only other factor possible is a scalar. Hence LHS is of form $k\cdot abc$ and the scalar $k$ can be obtained by checking for easy values, say $a = b = 1$ and $c = -1$.
| {
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Proving that $ 2 + 3\sqrt{-2} $ is reducible in $ \mathbb{Z}[\sqrt{-2}] $
Prove that $ 2 + 3\sqrt{-2} $ is irreducible in $
\mathbb{Z}[\sqrt{-2}] $
So far, I have let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $
I then took the norm and got $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2) $
I think I must then split 22 into $ (2)(11) $ but I don't know how to proceed from there.
Help is much appreciated!
Note: I originally posed the question as proving it was *ir*reducible. Apologies if I sent people down the wrong track in the answers below! Thank you again for the help.
|
Let $ 2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2}) $ then $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2)$
That is a good way to start!
Now we just need to show $a^2+2b^2 = 2$ and $c^2+2d^2 = 11$ is impossible, but the first part is possible so we need to show $c^2+2d^2 = 11$ is impossible:
This is easy, let's just write out all numbers of the form $x^2+2y^2$:
$$\begin{array}{|c|c|c|} \hline
0 & 2 & 8 & 18 \\ \hline
1 & 3 & 9 & 19 \\ \hline
4 & 6 & 12 & \\ \hline
9 & \color{red}{11} & & \\ \hline
\end{array}$$
So we have a factorization from the $x=1,y=3$ box which is $3^2 + 2\cdot 1^2 = 11$.
$$2(3+\sqrt{-2})(3-\sqrt{-2})$$
| {
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Completing a Cayley table with few given spaces \begin{array}{ccc}
* & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6} \\
\textbf{1} & 1 & & & & & \\
\textbf{2} & & & & & 1 & \\
\textbf{3} & & & 1 & & 2 \\
\textbf{4} & & & & & & & \\
\textbf{5} & & & & & & \\
\textbf{6} & & & & & & 1
\end{array}
need help figuring out the rest of the table
| Hint: 1 is the identity because the identity is the only idempotent element in a group.
Edit: (I had some nonsense here before.)
Don't forget that the table for a group can only have symbols appear once in each column and row. Also don't forget that the only orders possible are 1,2,3,6. (You can use this to eliminate the possibility that $5^3=1$, thus this group is cyclic.)
This question: About group multiplication table has helpful tips in the answers, along these lines.
OK, I'm starting to not believe in this problem (if we're in a group). Looking only at the powers of 5, we know $5^2\neq1$ (since $5*2=1$) and we know that $5^3\neq1$ (since $5^3=1\implies 2*5^3=5^2=2$, a contradiction with $3*5=2$) thus $5^6=1$.
But then, multiplying with 2's, $3*5=2 \implies3=2^2$. But then $5^6=1\implies 5^2=2^4\implies 5^2=3^2$, but $3^2=1$, a contradiciton with $5^2\neq 1$.
TaraB pointed out that $5^6=1$ regardless of the order of $5$, and so you can skip to the final contradiction directly!
| {
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Finding Laurent series
Find the Laurent series for $\frac{(z+1)}{z(z-4)^3}$ in $0<|z-4|<4$.
How can I do this series and what is the most efficent way of doing it? Do I find the partial fractions or since we are given the region, which is defined by $0<|z-4|<4$, then I know its form must correlate accordingly by $\frac{1}{1+(\frac{1}{a})}$, where $a$ is the series where we must manipulate so that we can apply the geometric series in the specified region?
| Let $\frac{z-4}{4}=t\implies z=4t+4$
then $|t|<1$ and $$\frac{(1+z)}{z(z-4)^3}=\frac{4t+5}{64(4t+4)(t^3)}=\frac{1}{64t^3}+\frac{1}{256t^3(1+t)}\big(=\frac{1}{256t^3}(1-t+t^2-t^3+\cdots)\big)$$ $$=\frac{1}{64t^3}+\frac{1}{256t^3}-\frac{1}{256t^2}+\frac{1}{256t}-\frac{1}{256}+\frac{t}{256}+\cdots$$ $$=\frac{5}{256t^3}-\frac{1}{256t^2}+\frac{1}{256t}-\frac{1}{256}+\frac{t}{256}+\cdots$$
Now, put back $t=\frac{(z-4)}{4}$
| {
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Trouble with factorising a polynomial I'm supposed to show that:
$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$
and the required answers are: $$ P=5, Q=4, R=-4 $$
I tried to solve this with partial fractions like so:
$$5(x-1)(x+2) = A(x+3) + B(x-2)$$
$\implies$ $A$=4, $B$=-4
$\implies$ $Q$=4, R=-4
But where does $P$=5 come from?
Or should I have first multiplied out the numerator and denominator and then used long division to solve?
| $$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$ gives
$$5(x-1)(x+2)=P(x-2)(x+3)+Q(x+3)+R(x-2)........(1)$$
Now put $x=2, x=-3,x=0$ respectively on both sides of $(1)$ to get, $Q=4,R=-4,P=5$ respectively. Hence you can reach the desired result.
| {
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Simplifying repeating square roots. How can I solve the equation:
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}} = \frac{1+\sqrt{53}}{2}$$
| Set $y = \sqrt{x + \sqrt{x + \dots}}$.
$y^2 = x+\sqrt{x+\dots} = x + y$
So $y^2 - y = x$.
But $\displaystyle y = \frac{1 + \sqrt{53}}{2}$
Now just substitute.
Addendum for people who want to see something a bit more rigorous.
We will take the infinite square root expression to be defined as the limit of the sequence defined recursively:
$$s(1) = \sqrt{x}$$
$$s(n+1) = \sqrt{x + s(n)}$$
We want to take $\lim_{n \to \infty} s(n)$.
For $x > 0, s(n) \le s(n+1)$ by induction on $n$:
$s(1) \le s(2)$ because $s(2) = \sqrt{x + s(1)} > \sqrt{x}$ since $s(1) > 0$.
If $s(n) \le s(n+1)$, then $s(n+2) = \sqrt{x + s(n+1)} \ge \sqrt{x+s(n)} = s(n+1)$.
So $s$ is monotonic.
Moreover, $s$ is bounded above by $\max(x,2)$. This will also be proven by induction.
$s(1) = \sqrt{x} \le \max(x,1) \le \max(x,2)$
If $s(n) \le 2$, then $s(n+1) = \sqrt{x+2} \le \sqrt{4} \le 2$.
Similarly, if $2 < s(n) \le x$, then $s(n+1) = \sqrt{x+s(n)} \le \sqrt{2x} \le \sqrt{x^2} = x$
Since $s$ is bounded and monotonic, it has a limit as $n \to \infty$.
Since $f(y) = \sqrt{ x + y }$ is continuous, $$\lim_{n \to \infty} s(n) = \lim_{n \to \infty} s(n+1) = \lim_{n \to \infty} \sqrt{ x + s(n) } = \sqrt{ x + \lim_{n \to \infty} s(n)}$$
Justifying the reasoning above.
| {
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Finding the sum $\sum\limits_{n=1,n\neq m^2}^{1000}\left[\frac{1}{\{\sqrt{n}\}}\right]$, Find the value
$\displaystyle\sum_{n=2,n\neq m^2}^{1000}\left[\dfrac{1}{\{\sqrt{n}\}}\right]$,
by $\{x\}=x-[x]$,
$[x]$was bracket function,for example:$[5.4]=5, [2.9]=2,[-1.1]=-2 $and so on.
| The answer is 3843 as computed by brute force.
The rest is not the answer but interesting pieces I find related to this sum.
For $n^2 < x < (n+1)^2$, let $x = n^2 + k$, we have:
$$\left\lfloor\frac{1}{\{\sqrt{x}\}}\right\rfloor
= \left\lfloor\frac{1}{\sqrt{n^2+k}-n}\right\rfloor
= \left\lfloor\frac{\sqrt{n^2+k} + n}{k}\right\rfloor
= \left\lfloor\frac{2n}{k}\right\rfloor
$$
This gives:
$$\sum_{x = n^2+1 }^{(n+1)^2-1} \left\lfloor\frac{1}{\{\sqrt{x}\}}\right\rfloor
= \sum_{k=1}^{2n}\left\lfloor\frac{2n}{k}\right\rfloor = D(2n)$$
where $D(n) = \sum_{k=1}^{n} \left\lfloor\frac{n}{k}\right\rfloor = \sum_{k=1}^{n} d(k)$ is the Divisor summatory function
which is basically a sum over $d(k)$, the number of divisors of $k$. The value of $D(n)$ is covered by the OEIS sequence A006218.
Notice for $n = 31$, the range $n^2+1, \ldots, (n+1)^2 - 1$ over cover the tail of our range of summation $1,\ldots, 1000$. For $1000 < x < 1024$, $k = x - n^2 > 39 \implies \left\lfloor\frac{2n}{k}\right\rfloor = 1$.
As a result:
$$\begin{align}\sum_{x=2,x\neq m^2}^{1000}\left\lfloor\dfrac{1}{\{\sqrt{x}\}}\right\rfloor
&= \sum_{x=2,x\neq m^2}^{1023}\left\lfloor\dfrac{1}{\{\sqrt{x}\}}\right\rfloor - 23
= \sum_{n=1}^{31} \sum_{x=n^2+1}^{(n+1)^2-1}\left\lfloor\frac{1}{\{\sqrt{x}\}}\right\rfloor -23\\
&= \sum_{n=1}^{31}D(2n) - 23 = 3843
\end{align}$$
About the asymptotic behavior of the following sum as a function of $N$
$$\sum_{x=2,x\neq m^2}^{N}\left\lfloor\dfrac{1}{\{\sqrt{x}\}}\right\rfloor$$
Not much is known about that. At the heart of the summation, the exact asymptotic behavior of $D(x)$ is the famous unsolved Dirichlet Divisor Problem. The only thing we know
is for large $x$,
$$D(x) = x\log x + x(2\gamma - 1) + O(x^\theta)$$
The best bound for $\theta$ known today is $131/464$ by Huxley (2003).
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Generating Function Example from class Example: Consider the sequence $(h_n)$ where $h_n$ is the number of nonnegative integer solutions to $$a_1+a_2+a_3+a_4+a_5=n.,$$ where $a_1$ is even, $a_2$ is odd, $a_3$ is a multiple of $5$, $a_4$ is 1,3, or 7, and $a_5$ is 0 or 1.
Without clear explaination, my professor writes $$(1+x^2+x^4\dots)(x+x^3+x^5+\dots)(1+x^5+x^{10}+\dots)(x+x^3+x^7)(1+x)=\sum_{a_1+a_2+a_3+a_4+a_5=0}^{\infty} x^{a_1+a_2+a_3+a_4+a_5},$$ where the $a_i$s respect the conditions stated above. We can simplify to $\frac{x(1+x)(x+x^3+x^7)}{(1-x^2)^2(1-x^5)}.$
My question: where did the first five generating functions come from? I see the relation between the exponents and the $a_i$s but how does one formally obtain these new generating functions and why are all of their coefficients 0 or 1?
| In this generating function, the things representing the solutions are the exponents. The condition that $a_1$ is even means that it can be $0, 2, 4, 6, \ldots$. Your professor therefore associates it to the even powers $(1 + x^2 + x^4 + \ldots)$. It's similar for the other cases.
Now, if you understand how multiplication works (distribution) and think about it, you can recognize that the coefficient of $x^n$ in the product is exactly the number of ways of adding $a_1 + \ldots + a_5$ to get $n$. Why? Let's look at a few of the small cases.
In how many ways can we get $a_1 + \ldots + a_5 = 2$? Looking at them, we know there is exactly one (when $a_2 = a_4 = 1$). Similarly, in $$(1+x^2+x^4\dots)(x+x^3+x^5+\dots)(1+x^5+x^{10}+\dots)(x+x^3+x^7)(1+x)$$
we see that the only way to get the exponent $2$ is to choose $1$ from the first, $x$ from the second, $1$ from the third, $x$ from the fourth, and $1$ from the last. Thus the coefficient of $x^2$ in the answer will be $1$. If we were to do this for another, say $n=4$ (or in our problem, the coefficient of $x^4$), you can directly show that for every solution in the $a_i$ there is a corresponding choice of terms whose exponents yield that solution in the product.
Thus the number of solutions for $\sum a_i = n$ will be the coefficient of $x^n$ in the expansion of $(1+x^2+x^4\dots)(x+x^3+x^5+\dots)(1+x^5+x^{10}+\dots)(x+x^3+x^7)(1+x)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why does the $ p $-series diverge when $ p = 1 $? We’re currently working with $ p $-series in my Calculus class, and I’ve fallen for the apparently common misconception that the infinite sum $ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where $ p = 1 $, should converge, whereas in reality, it diverges. Can anyone demonstrate or link me to a proof or explanation of why this is so?
| To answer your question, the integral test says that if $a_n$ is a positive decreasing sequence, and there exists a decreasing positive $f$ such that $f(n)=a_n$ for each $n$, then $$\sum_{n=1}^\infty a_n$$ converges if and only if $$\int_1^\infty f(x) dx$$ does. You have correctly found that with $f(x)=\frac 1 x $ and $a_n=\frac 1 n$ we can prove divergence. Using the integral test, you can also show that any $p$ series converges for $p>1$ and diverges for $p\leq 1$.
Another elegant proof is the following. The $1$-series (a.k.a. Harmonic series), goes as follows. Let
$$H_{n}=\sum_{k=1}^n \frac 1 k $$
Then $$\eqalign{
& {H_2} = \sum\limits_{k = 1}^2 {{1 \over k}} = 1 + 1 \cdot {1 \over 2} \cr
& {H_{{2^2}}} = \sum\limits_{k = 1}^4 {{1 \over k}} = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} > 1 + {1 \over 2} + {2 \over 4} = 1 + 2 \cdot {1 \over 2} \cr
& {H_{{2^3}}} = \sum\limits_{k = 1}^8 {{1 \over k}} = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7} + {1 \over 8} > 1 + {1 \over 2} + {2 \over 4} + {4 \over 8} = 1 + 3 \cdot {1 \over 2} \cr
& {H_{{2^4}}} = \sum\limits_{k = 1}^{16} {{1 \over k}} = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7} + {1 \over 8} + \cdots + {1 \over {16}} > 1 + {1 \over 2} + {2 \over 4} + {4 \over 8} + {8 \over {16}} = 1 + 4 \cdot {1 \over 2} \cr} $$
In general, we then show $${H_{{2^n}}} > 1 + {n \over 2}$$ which proves the Harmonic or $1$-series must diverge.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove that $-\frac{\sqrt{x}}{1+x}\log{x} \leq \log{2}$ for $0 < x < 1$ Graphically and numerically it is obvious but I'm looking for an analytical reasoning.
Just maximizing the left hand side does not yield an analytical expression for the maximum.
I also tried some known bounds for $\log$, but all of them had "overshoot".
| Put $y=\frac{1}{\sqrt{x}}$ to get $(y+\frac{1}{y})\log 2-2\log y \geq 0, y>1$.
Differentiating gives $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$, so we have critical points when $(y-\frac{1}{y})\log 2=2$. Solving the quadratic in $y$ and taking the positive root gives $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. It suffices to check the inequality for this value and the end points. ($y \to 1$ and $y \to +\infty$).
Put $y=1$ to get $2\log 2-2 \log(1)$ which is clearly non-negative. If the inequality holds for $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$, then since the gradient $\frac{1}{y}((y-\frac{1}{y})\log 2 -2)$ is clearly non-negative when $y>\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$, the inequality holds as $y \to +\infty$.
Finally we prove the inequality when $y=\frac{1+\sqrt{(\log 2)^2+1}}{\log 2}$. Rewrite the inequality as $2y\log 2 \geq (y-\frac{1}{y})\log 2+2\log y$ and substitute in the value to get $2(1+\sqrt{(\log 2)^2+1}) \geq 2+2\log(\frac{1+\sqrt{(\log 2)^2+1}}{\log 2})$. Put $z=\sqrt{(\log 2)^2+1}$, so we get $z+\log(\log 2) \geq \log (1+z)$, or equivalently, $e^z\log 2 \geq 1+z$. The function $e^x\log 2-(1+x)$ is an increasing function for $x \geq 1$ (The gradient is $e^x\log 2-1 \geq e\log 2-1>0$), and $z>\sqrt{1+0.69^2}>1.2$, so it suffices to show that $e^{1.2}\log 2 \geq 2.2$. Note that $\log 2>0.69$ and $e^{1.2}>1+1.2+\frac{1.2^2}{2}+\frac{1.2^3}{6}=3.208>3.2$, and $3.2(0.69)=2.208>2.2$, so we are done.
| {
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Series involving log $\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$ Does anybody know how to prove this series?
$$\sum_{n=1}^{\infty} \left( n\log \left(\frac{2n+1}{2n-1}\right)-1\right) = \frac{1-\log 2}{2}$$
I arrived at this through Mathematica.
I tried writing $\log \left(\frac{2n+1}{2n-1}\right)$ as $\int_0^1 \frac{1}{\frac{2n-1}{2}+x}dx$ and $-\sum_{k=1}^\infty \frac{(-2)^k}{k(2n-1)^k}$ but none of them worked.
| $$n \log \left(\dfrac{2n+1}{2n-1}\right) = n \left(\log(1+1/2n) - \log(1-1/2n)\right)$$
\begin{align}
\log(1+x) - \log(1-x) & = \left(x - \dfrac{x^2}2 + \dfrac{x^3}3 \mp \cdots\right) - \left(-x - \dfrac{x^2}2 - \dfrac{x^3}3 - \cdots\right)\\
& = 2\left(x + \dfrac{x^3}3 + \dfrac{x^5}5 + \cdots \right)
\end{align}
Hence,
$$n \left(\log(1+1/2n) - \log(1-1/2n)\right) = 2n \left(\sum_{k=0}^{\infty} \dfrac1{(2k+1)(2n)^{2k+1}}\right) = \sum_{k=0}^{\infty} \dfrac1{(2k+1)(2n)^{2k}}$$
Hence,
$$n \log \left(\dfrac{2n+1}{2n-1}\right) - 1 = \sum_{k=1}^{\infty} \dfrac1{(2k+1)(2n)^{2k}}$$
$$\sum_{n=1}^{\infty} \left(n \log \left(\dfrac{2n+1}{2n-1}\right) - 1\right) = \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \dfrac1{(2k+1)(2n)^{2k}} = \sum_{k=1}^{\infty} \dfrac{\zeta(2k)}{(2k+1) \cdot 2^{2k}}$$
\begin{align}f(m) & = \sum_{n=1}^m \left(n \log\left(\dfrac{2n+1}{2n-1}\right)-1\right)\\
& = \left( 1\log(3) - 1 + 2 \log(5) - 2 \log(3) + 1 + \cdots \right)\\
& = m - \log(3) - \log(5) - \cdots - \log(2m-1) + m \log(2m+1)
\end{align}
Now
$$\log(1) + \log(3) + \log(5) + \cdots + \log(2m-1) = \log \left(\dfrac{(2m)!}{2^m \cdot m!}\right)$$
Hence,
$$f(m) = m + m \log(2m+1) - \log \left(\dfrac{(2m)!}{2^m \cdot m!}\right)$$
Now use Stirling to get your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Partial Fraction Decomposition Problem... half answered... $$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$
I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:
$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{Cx+D}{x^2+3}$$
Multiplying by the LCD : $(x+3)^2(x^2+3)$
I am left with :
$$5x^3+19x^2+27x-3=A(x+3)(x^2+3)+B(x^2+3)+(Cx+D)(x+3)^2$$
By setting $x=-3:B=-4$
Now is where I am running into trouble. Now that I can substitute B into the original decomposition equation, There is no value of x that will leave only one variable to solve for. Please lend me a hand you guys(and girls). Thanks!
| I'll denote $f(x)$ this rational function. Here is an alternative to the expanding/equating strategy.
Limit trick: multiply both sides by $x$, compare the degrees, and let $x$ tend to $+\infty$:
$$
\lim_{x\rightarrow+\infty}xf(x)=5=A+C.
$$
Substitution trick: pick a small value in the domain of $f$. Here $0$ is perfect:
$$
f(0)=\frac{-3}{27}=\frac{A}{3}+\frac{B}{3}+\frac{D}{3}
$$
We need one more equation. Let's try $-2$, it's not too bad.
$$
f(-2)=\frac{-21}{7}=A+B+\frac{D-2C}{7}.
$$
Now, no miracle, there is a $4\times 4$ system to solve using also $B=-4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Partial Fractions - combinatorics - trouble Having a very hard time with this question:
Q: Use partial fractions to find the power series expansion of $$\frac{1+5x}{1-2x-3x^2}$$
| I just want to show how to apply Partial Fraction Decomposition
$$\frac{1+5x}{1-2x-3x^2}=\frac{1+5x}{(1-3x)(1+x)}=\frac A{1-3x}+\frac B{1+x}$$
$$\implies 1+5x=A(1+x)+B(1-3x)=x(A-3B)+A+B$$
Comparing the coefficients of the different powers of $x,A+B=1,A-3B=5$
$\implies B=-1,A=2$
Alternatively,
putting $x+1=0\implies x=-1,\quad 1+5\cdot(-1)=A\cdot0+B\cdot\{1-3(-1)\}\implies B=-1 $
putting $1-3x=0\implies x=\frac{1}{3},\quad 1+5\cdot\left(\frac13\right)=A\cdot\left(1+\frac13\right)+B\cdot0\implies A=2 $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Legendre Polynomials: proofs $\int_{-1}^1P_n^2(x)dx=\frac{2}{(2n+1)}$ Does any one know, how to compute any of those two things?
*
*The relationship between Legendre polynomials and Shifted Legendre Polynomials.
*$\displaystyle\int_{-1}^1P_n^2(x)dx=\dfrac{2}{(2n+1)}$ for $n\geq0$.
I tried to use Bonnet's equation:
$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$
but I couldn't move. Thanks :)
Edit: The second problem refers to regular Legendre Polynomials.
| I will answer your question about determining the value of $\int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
$$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$
$\newcommand{\partial}[1]{\left[#1\right]}$
$\newcommand{\bracket}[1]{\left(#1\right)}$
\begin{equation}
\begin{split}
I&=\int_{-1}^1 P_n(x)^2dx
\\ &=\frac{1}{2^{2n}n!^2}\int_{-1}^1 [(x^2-1)^n]^{(n)} \cdot [(x^2-1)^n]^{(n)}dx
\\ &=\frac{1}{2^{2n}n!^2} \bracket{\partial{[(x^2-1)^n]^{(n)}\cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-\int_{-1}^1 [(x^2-1)^n]^{(n+1)}\cdot [(x^2-1)^n]^{(n-1)}dx}
\\ &=\frac{1}{2^{2n}n!^2} \bracket{0-\int_{-1}^1 [(x^2-1)^n]^{(n+1)}\cdot [(x^2-1)^n]^{(n-1)}dx}
\\ &=...
\\ &=\frac{(-1)^n}{2^{2n}n!^2} \int_{-1}^1 [(x^2-1)^n]^{(2n)}\cdot [(x^2-1)^n]^{(0)}dx
\\ &=\frac{(-1)^n}{2^{2n}n!^2} \cdot (2n)! \cdot \int_{-1}^1 (x^2-1)^ndx
\end{split}
\end{equation}
Now
\begin{equation}
\begin{split}
\int_{-1}^1 (x^2-1)^ndx&=\int_{-1}^1 (x+1)^n(x-1)^ndx
\\ &=\bracket{\partial{\frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - \int_{-1}^1 \frac{(x+1)^{n+1}}{n+1}\cdot n(x-1)^{n-1}dx}
\\ &=\bracket{0 - \int_{-1}^1 \frac{(x+1)^{n+1}}{n+1}\cdot n(x-1)^{n-1}dx}
\\ &=...
\\ &=(-1)^n\int_{-1}^1 \frac{n!\cdot (x+1)^{2n}}{(2n)!}\cdot n!(x-1)^0dx
\\ &=(-1)^n\frac{n!^2}{(2n)!}\int_{-1}^1 (x+1)^{2n}dx
\\ &=(-1)^n\frac{n!^2}{(2n)!}\partial{\frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
\\ &=(-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1}
\end{split}
\end{equation}
So finally we get our desired value:
\begin{equation}
\begin{split}
I&=\int_{-1}^1 P_n(x)^2dx
\\ &=\frac{(-1)^n}{2^{2n}n!^2} \cdot (2n)! \cdot (-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1}
\\ &=\frac{2}{2n+1}
\end{split}
\end{equation}
Q.E.D.
| {
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$\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$ Find the value of $\frac{\sqrt{1+\sqrt{x}}+\sqrt{1+\sqrt{1-x}}}{\sqrt{1-\sqrt{x}}+\sqrt{1-\sqrt{1-x}}}$, if $x\in \left(0,\frac{1}{2}\right)$. I know it is equal to $\sqrt{2}+1$, but I don't know how to prove it?
| a$\sqrt{1-\sqrt{x}}=a$
$\sqrt{1+\sqrt{x}}=b$
$ab=\sqrt{1-x}$
Now your expression will look something like :
$\frac{b+\sqrt{1+ab}}{a
+\sqrt{1-ab}}$
| {
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"source": "stackexchange",
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Finding real roots of $ P(x)=x^8 - x^7 +x^2 -x +15$ Let $ P(x)=x^8 - x^7 +x^2 -x +15 $, Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as $$ P(x)= x(x-1)(x^6+1) +15 $$ we find that $ P(x) $ does not have any positive roots.
Where did I err ?
| Just for "fun" :
$\displaystyle x^8-x^7+x^2-x+15 = \frac{(128x^4-64x^3-16x^2-8x-9)^2+4(16x^2-11x+121)^2+60(x-49)^2+43055}{2^{14}}>0\;\forall x\in \mathbb{R}$
Another Solution::
If $x<0,$ note that $x^8+(-x^7)+x^2+(-x)>0,$ so the polynomial cannot have any negative roots.
If $x\geq 0,$ then note that from AM-GM inequality we have:
$\left\{\begin{aligned}& \frac 78 x^8+\frac 18\geq x^7\\& x^2+\frac 14\geq x\end{aligned}\right\} ;$
Thus $\displaystyle\frac 78x^8-x^7+x^2-x+\frac 38>0;$
| {
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Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$
We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.
Any idea? Or any other method to establish the convergence?
| To elaborate on Ross Millikan's answer and André Nicolas's comment, denote the $n$-th block sum (ignoring the sign) by $B_n$. Then
\begin{align}
B_n &= \sum_{i=1}^n \frac1{k_n+n},\\
k_n &= \frac{n(n-1)}{2},\\
k_{n+1} &= k_n+n = \frac{n(n+1)}{2} \le n^2.\tag{1}
\end{align}
Therefore
\begin{align}
B_n - B_{n+1}
&= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^{n+1} \frac1{k_{n+1}+i}\\
&= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^n \frac1{k_{n+1}+i} - \frac1{k_n+2n+1}\\
&= \sum_{i=1}^n \frac{n}{(k_n+i)(k_{n+1}+i)} - \frac1{k_n+2n+1}\ \text{ by } (1)\\
&\ge \sum_{i=1}^n \frac{n}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\
&= \frac{n^2}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\
&\ge \frac{1}{k_n+2n} - \frac1{k_n+2n+1}\ \text{ by } (1)\\
&> 0.
\end{align}
Hence the block sum is indeed monotonic decreasing and the alternating series test applies.
| {
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Proving $\pi(\frac1A+\frac1B+\frac1C)\ge(\sin\frac A2+\sin\frac B2+\sin\frac C2)(\frac 1{\sin\frac A2}+\frac 1{\sin\frac B2}+\frac 1{\sin\frac C2})$ Let $\Delta ABC$, prove that
$$\pi\left(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\right)\ge \left(\sin{\dfrac{A}{2}}+\sin{\dfrac{B}{2}}+\sin{\dfrac{C}{2}} \right) \left(\dfrac{1}{\sin{\dfrac{A}{2}}}+\dfrac{1}{\sin{\dfrac{B}{2}}}+\dfrac{1}{\sin{\dfrac{C}{2}}}\right)$$
| Maybe this will help you,
$$\sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \le \dfrac {1}{8}$$(With equality iff $A=B=C$)
Well here's a proof of it:
We have
$$\sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2}$$
$$=\dfrac{1}{2}[\cos \dfrac{A-B}{2}-\cos \dfrac {A+B}{2}]\sin \dfrac{C}{2}$$
$$\le \dfrac{1}{2}[1-\sin \dfrac{C}{2}]\sin \dfrac{C}{2}$$
$$=\dfrac{1}{2}[\sin \dfrac{C}{2}-\sin^2 \dfrac{C}{2}]$$
$$=\dfrac{1}{2}[ \dfrac{1}{4} -(\dfrac{1}{2}- \sin \dfrac{C}{2})^2] \le \dfrac{1}{8}$$
$$A+B+C= \pi$$
Maximum value$(ABC)=\dfrac{\pi^3}{27}$(From AM-GM inequality)
Maximum value of $(AB+BC+CA) =\dfrac{\pi^2}{9}$
$$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\dfrac{AB+BC+CA}{ABC}$$
We have
$$\dfrac{1}{ABC} \ge \dfrac{27}{\pi^3}$$
$$(AB+BC+CA) \ge \dfrac{\pi^2}{3}$$
(You can multiply inequalities, when they are both positive)
We get, $$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C} \ge \dfrac{\pi^2}{3}.\dfrac{27}{\pi^3}$$
$$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge 9 \dfrac{1}{\pi}$$
(Multiplying $\pi$)
$$ \pi. (\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}) \ge 9$$
You can carry on from here. (Manipulation of inequalities, nothing much.)
| {
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Having trouble showing that these series are the same. $$\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n+1)}{2}+1}\frac{(x-\pi/4)^n}{n!} $$
$$= \frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n-1)}{2}}\frac{(x-\pi/4)^{n+1}}{(n+1)!} + 1$$
*(added the +1) sorry didn't see this before, my answer guide is a poor quality photocopy
The second one is in the answer guide. The first one is my answer. The problem from the book is
Write the taylor series for $\sin x$ centered at $\frac{\pi}{4}$
my work:
$f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$f^{\prime}\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$f^{\prime\prime}\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$f^{\prime\prime\prime}\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$f^{\left(4\right)}\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$\sin x = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(\pi/4)}{n!}(x -\pi/4)^n$
how did the book get its answer (the second one listed at start of this question).
| Edit: None of these is the Taylor series of $\sin $ at $\pi/4$...There is a problem with the sign of the coeffcients in both. Here is a safe way to get the sign: use the floor function.
$$
\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\lfloor n/2\rfloor}\frac{(x-\pi/4)^{n}}{n!}.
$$
Previous answer: They are not the same. The first one starts with $\frac{\sqrt{2}}{2}(-1+(x-\pi/4)+\ldots)$. The second one with $\frac{\sqrt{2}}{2}((x-\pi/4)+\ldots)$.
The second one should start at $-1$, if you want them to be equal. Adding this term and shifting your index by $1$ (i.e. replacing $n+1$ by $n$) in the second one, it becomes
$$
\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{(n-1)(n-2)}{2}}\frac{(x-\pi/4)^{n}}{n!}.
$$
Now
$$
\frac{(n-1)(n-2)}{2}=\frac{n^2-3n+2}{2}=\frac{n^2+n}{2}+1-2n\equiv \frac{n(n+1)}{2}+1 \;\mbox{mod}\;2.
$$
So the coefficients of the two series are indeed the same for every term of degree $n$ with $n\geq 0$.
Now the sign of the coefficients, starting from $n=0$ in the formula of the first one (the $+1$ term is artificial in the second one, and does not correspond to the formula for $n=0$), is: $-1,1,1,-1, -1, \ldots$ when it should be $1,1,-1,-1,1\ldots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| the sequence $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+.....$$
is an $arithmetico geometric series$
the sum of an$AG$ series of the form
$$S_{\infty}=a+(a+d)r+(a+2d)r^2+(a+3d)r^3+.....\infty=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$converting the above series inti this form
$$\frac{1}{2}( \frac{1}{1} +\frac{2}{2}+ \frac{3}{4}+...... )$$
here$a=1$,$d=1$,$r=\frac{1}{2}$.
you get the answer $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 11
} |
Find minimum of $\sin^4 x + \cos^4 x + \sec^4 x$ I tried manipulating the terms but I couldn't get anywhere.
The only other thing I can observe is that the minimum must be greater than $1$ since all the terms are non-negative and the range of $\sec^4 x$ is $[1, \infty)$. Also it can't be $1$ since then $\sin^4 x + \cos^4$ must be $0$ which means $\sin x = \cos x = 0$ which is untrue for all $x$.
| $\sin^{4}{x}+\cos^{4}{x}+\sec^{4}{x}=\sin^{4}{x}+\left(\cos^{2}{x}-\sec^{2}{x}\right)^2+2$
Minimum value of $\sin^4{x}$ is zero and is attained at $x=n\pi$
The minimum value of $\cos^{2}{x}-\sec^{2}{x}$ is zero and attained when $\cos^{2}{x}=\sec^{2}{x}$ or $x=n\pi$. Both minimum's coincide at $x=n\pi$. And the minimum of the expression is $2$
Note: If the region where minimum is attained had not coincided , the calculus methods would be the best try.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
The generating function for the Fibonacci numbers Prove that $$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...=\frac{1}{1-(z+z^2)}$$
The coefficients are Fibonacci numbers, i.e., the sequence $\left\{1,1,2,3,5,8,13,21,...\right\}$.
| The proof is quite simple. Let's write our sum in a compact format:
$$
1+z+2z^2+3z^3+5z^4+8z^5+... = \sum_{n=0}^\infty F_nz^n
$$
Where $F_n$ is the $n$th Fibonacci number, starting with $F_0=F_1=1$, and $F_{n+2}=F_n+F_{n+1}$. It is from here that we will prove what needs to be proven.
$$\begin{align}
(1-z-z^2)\sum_{n=0}^\infty F_nz^n &= \sum_{n=0}^\infty F_nz^n - \sum_{n=0}^\infty F_nz^{n+1} - \sum_{n=0}^\infty F_nz^{n+2}\\
&= \sum_{n=0}^\infty F_nz^n - \sum_{n=1}^\infty F_{n-1}z^n-\sum_{n=2}^\infty F_{n-2}z^n\\
&= F_0 + (F_1-F_0)z + \sum_{n=2}^\infty (F_n-F_{n-1}-F_{n-2})z^n
\end{align}$$
Now, $F_1=F_0$ and $F_n=F_{n-1}+F_{n-2}$. Therefore,
$$
(1-z-z^2)\sum_{n=0}^\infty F_nz^n = F_0 = 1
$$
And thus
$$
\sum_{n=0}^\infty F_nz^n = \frac{1}{1-(z+z^2)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
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"answer_id": 0
} |
Prove $\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\ge\sum_{i=1}^{n}\frac{1-a_{i+1}}{1-a_{i}}$ if $a_{i}>0$ and $a_{1}+a_{2}+\cdots+a_{n}=1$ Let $a_{i}>0,i=1,2,\cdots,n$, and $a_{1}+a_{2}+\cdots+a_{n}=1$.
How can we prove that
$$\displaystyle\sum_{i=1}^{n}\dfrac{a_{i}}{a_{i+1}}\ge\displaystyle\sum_{i=1}^{n}\dfrac{1-a_{i+1}}{1-a_{i}}$$
where $a_{n+1}=a_{1}$?
I think this can be done using the AM-GM inequality.
| Below is a proof for $n=3$ (which, unfortunately, does not seem to
generalize).
When $n=3$, the inequality to be shown can be rewritten as
$$
\frac{a_1}{a_2}+\frac{a_2}{a_3}+\frac{a_3}{a_1} \geq
\frac{a_1+a_3}{a_2+a_3}+\frac{a_1+a_2}{a_1+a_3}+\frac{a_2+a_3}{a_1+a_2}
\tag{1}
$$
or equivalently,
$$
\bigg(\frac{a_1}{a_2}-\frac{a_1+a_3}{a_2+a_3}\bigg)+
\bigg(\frac{a_2}{a_3}-\frac{a_1+a_2}{a_1+a_3}\bigg)+
\bigg(\frac{a_3}{a_1}-\frac{a_2+a_3}{a_1+a_2}\bigg) \geq 0
\tag{2}
$$
In other words,
$$
\bigg(\frac{a_1}{a_2}-\frac{a_1+a_3}{a_2+a_3}\bigg)+
\bigg(\frac{a_2}{a_3}-\frac{a_1+a_2}{a_1+a_3}\bigg)+
\bigg(\frac{a_3}{a_1}-\frac{a_2+a_3}{a_1+a_2}\bigg) \geq 0
\tag{3}
$$
Or
$$
\bigg(\frac{\frac{a_1}{a_2}-1}{\frac{a_2}{a_3}+1}\bigg)+
\bigg(\frac{\frac{a_2}{a_3}-1}{\frac{a_3}{a_1}+1}\bigg)+
\bigg(\frac{\frac{a_3}{a_1}-1}{\frac{a_1}{a_2}+1}\bigg) \geq 0
\tag{4}
$$
So if we put $x_k=\frac{a_k}{a_{k+1}}+1$, this is equivalent to
$$
\bigg(\frac{x_1-2}{x_2}\bigg)+
\bigg(\frac{x_2-2}{x_3}\bigg)+
\bigg(\frac{x_3-2}{x_1}\bigg) \geq 0
\tag{5}
$$
or
$$
\frac{x_1}{x_2}+
\frac{x_2}{x_3}+
\frac{x_3}{x_1} \geq
\frac{2}{x_1}+
\frac{2}{x_2}+
\frac{2}{x_3}
\tag{6}
$$
Now, by AM-GM we have
$$
\frac{2x_k}{x_{k+1}}+\frac{x_{k+2}}{x_k} \geq 3\bigg(\frac{x_kx_{k+2}}{x_{k+1}^2}\bigg)^{\frac{1}{3}} \tag{7}
$$
Also, Holder’s inequality implies that for any positive $w_1,w_2,w_3$,
$$
(1^3+w_1^3)(1^3+w_2^3)(1^3+w_3^3) \geq (1\times 1 \times 1+w_1w_2w_3)^3
$$
Taking $w_k=(x_k-1)^{\frac{1}{3}}=(\frac{a_k}{a_{k+1}})^{\frac{1}{3}}$, we see
that $x_1x_2x_3 \geq 8$, and hence (7) implies that
$$
\frac{2x_k}{x_{k+1}}+\frac{x_{k+2}}{x_k} \geq \frac{6}{x_{k+1}} \tag{8}
$$
Summing on $k=1,2,3$, we deduce (6) from (8), qed.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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} |
compute line integral where line is boundary of $(x-5)^2 + (y-2)^2 \le 1$ Line integral of $$\frac{-y^3}{(x^2+y^2)^2} dx + \frac{xy^2}{(x^2+y^2)^2} dy$$
Using greene's theorem, this integral is equal to zero. Am i right? $P(x,y)$ and $Q(x,y)$ are not continuous at $(0,0)$. Since the circle does not contain $(0,0)$, green's theorem applies?
| Using Green's Theorem we get:
$$\int_C\frac{-y^3}{(x^2+y^2)^2}dx+\frac{xy^2}{(x^2+y^2)^2}dy=\int\int_D\left(\frac{8xy^3}{(x^2+y^2)^3}-\frac{2x^2y}{(x^2+y^2)^2}\right)dA$$
Then switch to polor coordinates, since $D$ is a unit circle, (just shifted) we get:
$$=\int_0^{2\pi}\int_0^1\left(8\cos\theta \sin^2\theta\cdot r-2\cdot r\cos^2\theta\sin\theta\right)drd\theta=0$$
These are fairly simply integrals to integrate. For the first simply let $u=\sin\theta$ and the second let $u=\cos\theta$, after we integrate with respect to $r$, of course. So yes, you're right.
| {
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"url": "https://math.stackexchange.com/questions/341306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
How to prove ${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$ ?thanks. How to prove
$${{a}^{a}}{{b}^{b}}\ge {{\left(\frac{a+b}{2}\right)}^{a+b}}$$
$a>0$,$b>0$,
thanks.
| Let $P=(1+x)^{1+x} (1-x)^{1-x}$ , where $0 ≤ x <1$ ; then
$\log P=(1+x)\log(1+x)+(1-x)\log(1-x)$
$\space$$\space$ $\space$$=x${$\log(1+x)-\log(1-x)$ } +$\log(1+x)+\log(1-x)$ $\space$$=2x\bigg(x+\frac{x^3}{3} + \frac{x^5}{5}+...\bigg) - 2\bigg(\frac{x^2}{2}+\frac{x^4}{4} + \frac{x^6}{6}+...\bigg)$ $=2\bigg(\frac{x^2}{1×2}+\frac{x^4}{3×4} + \frac{x^6}{5×6}+...\bigg)$$≥0$
Hence $\log P≥0$ i.e. $P ≥1$ that is $(1+x)^{1+x} (1-x)^{1-x}≥1$ ; put $x=z/u$ , where $u>z$ ; then
$\bigg(1+z/u\bigg)^{1+z/u} \bigg(1-z/u\bigg)^{1-z/u}\space≥1$ $\space$ i.e. $\bigg(\frac{u+z}u\bigg)^{u+z} \bigg(\frac{u-z}u\bigg)^{u-z}≥1^u=1$
i.e. $\big(u+z\big)^{u+z} \big(u-z\big)^{u-z}≥u^{2u}$ ; now put $u+z=a , u-z=b$ , so that
$u=\frac{a+b}2 , a>0,b>0$ ;
then $a^ab^b ≥ \bigg(\frac{a+b}2\bigg)^{a+b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/341439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
} |
factorisation of algebra homework question that asks to perform the multiplication and division and simplification.
$$\frac{x^2+7x+10}{x^2+5x+4} \times \frac{x^2+3x+2}{x^2+4x+4} =$$
$$\frac{(x+5)(x+2)}{(x+4)(x+1)} \times \frac{(x+1)(x+2)}{(x+2)(x+2)} =$$
$$ = \frac{(x+5)}{(x+4)}$$
is my working out and answer correct?
I am typing this extra sentence in order to meet your quality standards!!!
| Yes, it is correct because you are able to cancel out two of the binomials leaving $(x+5)$ on the top and $(x+4)$ on the bottom.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A condition for a triangle to be isosceles in $\Delta ABC$,and
$$\dfrac{\sin{(\dfrac{B}{2}+C)}}{\sin^2{B}}=\dfrac{\sin{(\dfrac{C}{2}+B)}}{\sin^2{C}}$$
prove that $B=C$
I think $\sin{(\dfrac{B}{2}+C)}\sin^2{C}=\sin{(\dfrac{C}{2}+B)}\sin^2{B}$
then
$$\sin{(\dfrac{B}{2})}\cos{C}\sin^2{C}+\cos{\dfrac{B}{2}}\sin^3C=\sin{\dfrac{C}{2}}\cos{B}\sin^2B+\cos{\dfrac{C}{2}}\sin^3B$$
so
$$(\sin{\dfrac{B}{2}}-\sin{\dfrac{C}{2}})f(B,C)=0$$
my question:How can prove $f(B,C)\neq 0$ ？
|
Its just kind of a hint, I hope its of some help.
Using sine-rule on $\triangle ABC$, you get :
$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}= 2R$$
$$\dfrac{\sin B}{\sin C}= \dfrac{b}{c} \implies \dfrac{\sin^2 B}{\sin^2 C}= \dfrac{b^2}{c^2}$$
Use sine-rule for $\triangle ADB$ and $\triangle AEC$
$$\dfrac{c}{\sin (\frac{B}{2}+C)}=\dfrac{AD}{\sin \frac{B}{2}}=\dfrac{BD}{\sin A}= 2R'$$
$$\dfrac{b}{\sin (\frac{C}{2}+B)}=\dfrac{AE}{\sin \frac{C}{2}}=\dfrac{EC}{\sin A}= 2R''$$
$\sin A=( \dfrac{c}{\sin (\frac{B}{2}+C) \times BD} )^{-1}$
$\sin A= (\dfrac{b}{\sin (\frac{C}{2}+B ) \times EC})^{-1}$
Equate them:
$( \dfrac{c}{\sin (\frac{B}{2}+C) \times BD} )=(\dfrac{b}{\sin (\frac{C}{2}+B ) \times EC})$
$\dfrac{c \times EC}{b \times BD} = \dfrac{\sin (\dfrac{B}{2}+C)}{\sin (\dfrac{C}{2}+B)}$
$\dfrac{c \times EC}{b \times BD}=\dfrac{b^2}{c^2} \implies \dfrac{EC}{BD}=\dfrac{b^3}{c^3}$
In $\triangle BDC$ and $\triangle BEC$
$\dfrac{a}{\sin (\dfrac{B}{2}+C)}=\dfrac{BD}{\sin C}$ .....$1$
$\dfrac{a}{\sin (\dfrac{C}{2}+B)}=\dfrac{EC}{\sin B}$......$2$
Dividing ($1$) and ($2$), you get:
$\dfrac{\sin (\dfrac{C}{2}+B)}{\sin (\dfrac{B}{2}+C)}=\dfrac{BD \cdot \sin B}{EC \cdot \sin C}$
I couldn't get the conclusion right. Maybe this kinda approach is useful.:)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To show that a partial dertivative (of a piecewise function) is continuous at $0$
$$f(z)=\cases{\frac{x^4-6x^2y^2+y^4}{x^2+y^2}
+i\frac{4xy(x^2-y^2)}{x^2+y^2},& $z\ne0$\cr 0, &$z=0$}$$
Let $u=\Re(f)$.
I have shown from first principles that $\frac{\partial u}{\partial x} = 0$ at $0$. To go on to show that $\frac{\partial u}{\partial x}$ is continuous at $0$, I have to show that $\frac{\partial u}{\partial x}\to 0$ near $0$, right? But while the numerator of this partial clearly approaches $0$ near $0$, so does the denominator...
It would be most helpful if someone can demonstrate how to prove that
$\frac{\partial u}{\partial x}$ is continuous at $0$.
P.S. Will someone also help to link this post to my earlier one: Show that this piecewise function is differentiable at $0$
| $$\frac{\partial u}{\partial x}=\frac{(x^2+y^2)(4x^3-12xy^2)-2x(x^4-6x^2y^2+y^4)}{(x^2+y^2)^2}=$$
$$=\frac{2x^5-14xy^4+4x^3y^2}{(x^2+y^2)^2}=2x\frac{x^4+2x^2y^2-7y^4}{(x^2+y^2)^2}\implies$$
$$\left|\;\frac{\partial u}{\partial x}\;\right|\le|2x|\frac{x^4+2x^2y^2+7y^4}{(x^2+y^2)^2}=|2x|\left(1+6\frac{y^4}{(x^2+y^2)^2}\right)\xrightarrow[(x,y)\to(0,0)]{}0$$
Hint for the last part: what's inside the parentheses is bounded. Use polar coordinates, for example.
| {
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Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.
I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this.
Thank you!
Sorry, my approach:
$$\begin{align}
\frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\
\sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\
\theta + 45^\circ &= 45^\circ,\ 135^\circ \\
\theta &= 0^\circ, \ 90^\circ
\end{align}$$
| We have $$\sin(\theta) + \cos(\theta) = 1 \text{ and } \sin^2(\theta) + \cos^2(\theta) = 1$$
If we have $$a+b = 1 \text{ and } a^2 + b^2 =1 \tag{$\star$}$$then either $a=0,b=1$ or $a=1, b=0$. This gives us $$\sin(\theta) = 0, \cos(\theta) = 1 \text{ (or) }\sin(\theta) = 1, \cos(\theta) = 0$$Hence, $$\theta = 0^{\circ},90^{\circ}$$
EDIT(Expanding out the implications of $(\star)$).
We shall show here that $(\star) \implies$ either $a=0,b=1$ or $a=1, b=0$.
We have $a=1-b$. Plugging this in $a^2 + b^2 = 1$, we get that $$(1-b)^2 + b^2 = 1 \implies 1-2b+b^2 + b^2 = 1 \implies 2b^2-2b = 0 \implies 2b(b-1) = 0$$
Hence, $b=0$ or $b=1$. Plugging in $b=0$ in $a=1-b$ gives us $a=1$. Similarly, plugging in $b=1$ in $a=1-b$ gives us $a=0$. Hence, the possible solutions are
$$(a,b) = (0,1) \text{ or }(1,0)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
} |
Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
| We can use discrete calculus! Let $x^{\overline{k}}$ denote the $k$th rising factorial power of $x$. That is: $$x^{\overline k} = \underbrace{x(x+1)(x+2)\cdots(x+k-1)}_{k\text{ factors}}$$
Then, $S_n = \sum_{k=0}^{n} k(k+1) = \sum_0^{n+1}x^{\overline 2}\,\delta x$. (Notice that I started the summation at $k=0$; this makes it easier to plug in the lower limit, but does not affect the value of the summation since the first term is $0$.)
Using the power rule for summation, we have:
\begin{align}
S_n &= \sum_0^{n+1}x^{\overline 2}\,\delta x\\
&= \frac{(x-1)^{\overline 3}}{3}\Bigg|_0^{n+1}\\
&= \frac{n^{\overline 3}}{3}\\
&= \frac{n(n+1)(n+2)}{3}
\end{align}
| {
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Prove that if $6 \mid (2a + 4)$ and $9 \mid (12 + 3b)$ then $3 \mid (a + b)$ Could you help me with the problem below?
Prove that if $6 \mid (2a + 4)$ and $9 \mid (12 + 3b)$ then $3 \mid (a + b)$.
Thank you!
| Since $6 \mid (2a+4)$, let $$(2a+4) = 6p.$$
$$a = (6p-4)/2.$$
$$a = (3p-2) \tag{1}$$
Again since $9 \mid (12+3b)$, let $$(12+3b) = 9q.$$
$$b = (9q-12)/3.$$
$$b = (3q-4)\tag{2}$$
(1) + (2) gives you: $$(a+b) = 3(p+q)-6 \quad \Rightarrow \quad (a+b) = 3(p+q-2).$$
Hence $3$ divides $(a+b)$ or $3 \mid(a+b)$.
Hope the answer is clear!
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| Whenever you have an "$x$ divides $y$" problem, start by stating it in modular form.
$$
a^2+ab+b^2 \equiv 0 \mod 9
$$
Now, we can also write this as
$$
a^2-2ab+b^2 \equiv -3ab \mod 9
$$
Therefore, we must have that $(a-b)^2\equiv 0 \pmod3$, and this can only be satisfied if $a\equiv b\pmod 3$.
So let $b=a+3k$ and substitute back into the original equation:
$$
a^2+a^2+3ak+a^2+6ak+9k^2\equiv 0 \mod 9
$$
Which can be simplified to
$$
3a^2 + 9ak + 9k^2 \equiv 0 \mod 9\\
3a^2 \equiv 0 \mod 9\\
a^2 \equiv 0 \mod 3
$$
And this is only satisfied if $a\equiv 0\pmod3$. And because $a\equiv b\pmod3$, we also have that $b\equiv 0\pmod3$. So 3 divides $a$ and 3 divides $b$.
| {
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Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$ Define $\sin x $ and $\cos x$ via their infinite series:
$$
\sin x = \sum_n (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}, \qquad
\cos x = \sum_n (-1)^n \frac{x^{2n}}{(2n)!}.
$$
Is there a short, clever proof that $\cos(x+y) = \cos x \cos y - \sin x \sin y$ for all real $x,y$? I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = \cos y$, $f'(0) = - \sin y$. Does anyone know other (preferably slick!) proofs?
| $$\begin{array} {rcl}
\cos(x + y) + i \sin(x + y)& = & e^{i(x + y)} \\
&=& e^{ix}e^{iy} \\
&=& (\cos(x) + i\sin(x))(\cos(y) + i\sin(y)) \\
&=& (\cos(x)\cos(y) - \sin(x)\sin(y)) + i(\sin(x)\cos(y) + \sin(y)\cos(x)) \\
\end{array}$$
Equating real and imaginary parts you get
$$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$
$$\sin(x + y) = \sin(x)\cos(y) + \sin(y)\cos(x)$$
| {
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"url": "https://math.stackexchange.com/questions/349435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 7
} |
How do I prove that $2\cos (2\theta + {\pi \over 3}) \equiv - 2\sin(2\theta - {\pi \over 6})$ Using the identity $\cos (\theta + {\pi \over 2}) \equiv - \sin\theta $
| Using the identity $\quad\cos\left(\theta + \frac{\pi}{2}\right)\equiv−\sin\theta$,
Let $\;\color{green}{\bf x = 2\theta - \dfrac{\pi}{6}} \implies\left(\implies \color{blue}{\bf x + \dfrac{\pi}{2}}\; = \;\left(2\theta - \dfrac{\pi}{6}\right) + \dfrac{\pi}{2} \;= \;\color{blue}{\bf 2\theta + \dfrac{\pi}{3}}\right)$
$$2\cos \left(\color{blue}{\bf 2\theta + {\pi \over 3}}\right)\; = \;2\cos\left(\color{blue}{\bf x + \dfrac{\pi}{2}}\right) \;\equiv\; - 2\sin(\color{green}{\bf x})\; = \;-2\sin\left(\color{green}{\bf 2\theta - \dfrac{\pi}{6}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the remainder of $(2012^{2013}+2013^{2012}):(2012\times 2013)$? What is the remainder if the sum $$(2012^{2013}+2013^{2012})$$
is divided by $$2012\times 2013$$
| We don't need Euler Totient Theorem
As $a+1\equiv1 \pmod a,$
$$N=a^{a+1} + (a+1)^a\equiv (a+1)^a\pmod a\equiv1^a \equiv a$$
$\implies N=1+a\cdot A$ for some integer $A$
Similarly, as $a\equiv-1 \pmod{a+1},$
$$N=a^{a+1} +(a+1)^a\equiv a^{a+1}\pmod{a+1}\equiv(-1)^{a+1}\equiv
\begin{cases}
1 & \text{ if } a \text{ is odd} \\
-1 & \text{ if } a \text{ is even}
\end{cases}$$
If $a$ is odd, $N\equiv1\pmod {a+1}\implies N=1+(a+1)B$ for some integer $B$
Then $1+a\cdot A=N=1+(a+1)B\implies \frac{(a+1)B}a=A$ which is an integer
$\implies a$ divides $(a+1)B$
$\implies a$ divides $B$ as $(a+1,a)=1$
So, $B=a\cdot C$ where $C$ is some integer
Then $N=1+(a+1)B=1+(a+1)\cdot a\cdot C\equiv 1\pmod{a(a+1)}$
If $a$ is even, $1+a\cdot A=N=-1+(a+1)B\implies (a+1)B-a\cdot A=2=2(a+1-a)$
$\implies \frac{a(A-2)}{a+1}=B-2$ which is an integer
$\implies (a+1)$ divides $a(A-2)$
$\implies (a+1)$ divides $(A-2)$ as $(a,a+1)=1$
So, $A=2+(a+1)D$ for some integer $D$
So, $N=1+a\cdot A=1+a\{2+(a+1)D\}=2a+1\pmod{a(a+1)}$
Here $a=2012$ which is even, so the remainder will be $2a+1=2\cdot2012+1=4025$
| {
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"url": "https://math.stackexchange.com/questions/352873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Factoring into Linear Factors How would one factor the following expression:
$(b - a)(c^2 - a^2) - (c-a)(b^2 - a^2)$
into the set of linear factors:
$(b - a)(c - a)(c - b)$
(This is not for homework but rather exam review. I ran into this issue when required to calculate a matrix's determinant in linear factor form).
Thank you!
| Recall that
$$x^2-y^2=(x-y)(x+y)$$
then
\begin{array}((b - a)(c^2 - a^2) - (c-a)(b^2 - a^2)&=(b-a)(c-a)(c+a)-(c-a)(b-a)(b+a)\\&=(b-a)(c-a)[(c+a)-(b+a)]\\&=(b-a)(c-a)(c-b)\end{array}
| {
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"url": "https://math.stackexchange.com/questions/353446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \frac{x^2 + x+3}{x^2+2x+5}\ dx$ How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$
To be honest, I'm embarrassed. I decomposed it and know what the answer should be but
I can't get the right answer.
| $$I=\int \frac{x^2 + x+3}{x^2+2x+5} dx=\int \frac{x(x+1)+3}{(x+1)^2+2^2} dx$$
Putting $x+1=2\tan\theta,dx=2\sec^2\theta d\theta,$
$$I= \frac{2\tan\theta(2\tan\theta-1)+3}{4\sec^2\theta d\theta}2\sec^2\theta d\theta$$
$$2\tan\theta(2\tan\theta-1)+3=4\tan^2\theta-2\tan\theta+3=4\sec^2\theta-2\tan\theta-1$$
$$\text{So,} I=\frac12\int(4\sec^2\theta-2\tan\theta-1)d\theta=2\tan\theta-\log|\sec\theta+\tan\theta|-\theta+C $$ where $C$ is an arbitrary constant for indefinite integration.
As $\tan\theta=\frac{x+1}2,$
$\sec^2\theta =1+\left(\frac{x+1}2\right)^2=\frac{x^2+2x+5}4$
$\theta=\arctan \left(\frac{x+1}2\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$ Why doesn't this work?
Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$
going to cylindrical this should be easy. $z=(4-r^{2})^{\frac {1}{2}}$ and $z=1$
Clearly $z=1$ yields $x^{2}+y^{2}=3 \to r=(3)^{\frac {1}{2}}$
$$\int^{2\pi}_{0}\int^{0}_{(3)^{\frac {1}{2}}} [(4-r^{2})^{\frac {1}{2}} -1] \, r\,dr\,d\theta$$
$$\int^{2\pi}_{0}\left[-\frac {1}{3} (4-r^{2})^{\frac {3}{2}}-\frac{r^{2}}{2}\right]\Bigg|^{0}_{(3)^{\frac {1}{2}}} d\theta$$
$$-\frac{8}{3}-\left[-\frac{1}{3}-\frac{3}{2}\right] \to -\frac{16}{6} +\frac{11}{6} \to -\frac{5}{6} \to -\frac{5\pi}{3}$$
I don't get why the bound isn't supposed to be $\int^{0}_{(3)^{\frac {1}{2}}}$ - why is it $\int^{(3)^{\frac {1}{2}}}_{0}?$ Clearly when I start integrating, my radius is root 3 and it reduces to 0?
|
Here is a cute picture of the question.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Largest element of the set $\{ \sin{1}, \sin{2}, \sin{3}\}$ I have to find the largest element of the following set $\{ \sin{1}, \sin{2}, \sin{3}\}$.
I converted every element to the first quadrant so I can use the monotony of cosine, the set becomes:
$$\Big\{ \cos{\frac{\pi-2}{2}}, \cos{\frac{\pi-4}{2}}, \cos{\frac{\pi-6}{2}}\Big\}$$.
$$\Big|\frac{\pi-6}{2}\Big| > \Big|\frac{\pi-4}{2}\Big|$$
$$\frac{6-\pi}{2} - \frac{\pi-2}{2} > 0 \Rightarrow \frac{6-\pi}{2} > \frac{\pi-2}{2}$$
$$ \frac{4-\pi}{2} - \frac{\pi-2}{2} < 0 \Rightarrow \frac{4-\pi}{2} < \frac{\pi-2}{2}$$
$$ \frac{4-\pi}{2} < \frac{\pi-2}{2} < \frac{6-\pi}{2} \Rightarrow \cos{\frac{\pi-4}{2}} < \cos{\frac{\pi-2}{2}}< \cos{\frac{\pi-6}{2}}$$ so since cos is monotonically decreesing in the first quadrant, $\cos{\frac{\pi-4}{2}} = \sin 2$ is the biggest element in the set.
Is this correct and is there any easier solution?
| (Let's see how rusty my TeX is...)
We could also use trig identities:
$$\sin 2 = \sin(2·1) = 2 \sin 1 \cos 1 > 2 \sin 1 \cos \frac{\pi}{3} = \sin 1 ,$$
$$\sin 3 = \sin (2+1) = \sin 2 \cos 1 + \cos 2 \sin 1 < \sin 2 \cos \frac{\pi}{4} + \cos \frac{\pi}{2} \sin 1 = \frac{\sqrt{2}}{2} \cdot \sin 2 < \sin 2
$$
Thus, sin 2 is the largest of the three values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the improper integral?
| As this one has been solved using a keyhole contour I thought I would show that it can also be done using a slice contour. Let $$f(z) = \frac{1}{1+z^3}.$$
The slice consists of three parts parameterized by $R$, which is real and goes to infinity. The first part $\Gamma_1$ is a line going from zero to $R$ along the real axis. The next one is a counterclockwise arc $\Gamma_2$ going from $R$ to $Re^{2i\pi/3}.$ The last one namely $\Gamma_3$ is a line back to the origin from $Re^{2i\pi/3}.$ along the ray at angle $2i\pi/3.$
Now we have the following bound on the integral along the arc:
$$\left|\int_{\Gamma_2} f(z) dz\right| \le 2\pi R/3 \frac{1}{R^3-1}
\in O(1/R^2) \rightarrow 0 \quad \text{as} \quad R \rightarrow\infty.$$
Moreover along $\Gamma_3$ setting $z = te^{2i\pi/3}$ we have
$$\int_{\Gamma_3} f(z) dz = \int_R^0 \frac{1}{1+t^3 e^{2i\pi}} e^{2i\pi/3} dt =
- e^{2i\pi/3} \int_0^R \frac{1}{1+t^3} dt.$$
Now with $Q$ being the integral we are looking for, we thus have in the limit
$$\int_{\Gamma_1} f(z) dz = Q \quad \text{and}\quad
\int_{\Gamma_3} f(z) dz = - e^{2i\pi/3} Q.$$
Applying the Cauchy residue theorem to the slice contour, we obtain
$$ Q (1 - e^{2i\pi/3} ) = 2\pi i
\operatorname{Res}(f(z); z = e^{i\pi/3}) $$
where $$\operatorname{Res}(f(z); z = e^{i\pi/3}) =
\lim_{z\to e^{i\pi/3}} \frac{z-e^{i\pi/3}}{1+z^3} =
\lim_{z\to e^{i\pi/3}} \frac{1}{3z^2} =\frac{1}{3} e^{- 2i\pi/3}.$$
It follows that
$$ Q= \frac{1}{3} 2\pi i \frac{e^{- 2i\pi/3}}{1 - e^{2i\pi/3}} =
\frac{1}{3}2\pi i \frac{e^{- 3i\pi/3}}{e^{-i\pi/3} - e^{i\pi/3}} =
\frac{1}{3}\pi \frac{1}{\sin(\pi/3)} =
\frac{1}{3}\pi \frac{2}{\sqrt 3} = \frac{2\pi}{3\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358262",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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$x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$?
Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?
I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit. Are there some good shortcuts?
This is not homework; I'm preparing for an examination.
| Another way:
Since $x= 3^\frac{1}{3} + 3^\frac{-1}{3}$,
$$ \Rightarrow x - 3^\frac{-1}{3} = 3^\frac{1}{3}$$
Take cubes on both sides
$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}(x-3^\frac{-1}{3}) = 3$$
$$ x^3 - \frac{1}{3} -3x.3^\frac{-1}{3}3^\frac{1}{3} = 3$$
$$x^3 - \frac{1}{3} -3x = 3$$
Multiply by 3 to get your answer :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the limit of $\frac{1-\cos(2x)}{1-\cos(3x)}$ for $x \to 0$ As $x$ goes to $0$, what is the limit of
$$\frac{1-\cos(2x)}{1-\cos(3x)}$$
Thanks.
| One way of doing it is expanding in Maclaurin series:
$$
\lim_{x \to 0} \frac{1-\cos 2x}{1- \cos 3x} = \lim_{x \to 0} \frac{1-1 +\frac{4x^2}{2} - O(x^4)}{1- 1 + \frac{9x^2}{2} - O(x^4)} = \frac{\frac{4}{2}}{\frac{9}{2}}=\frac{4}{9}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Minimum value of inverse trigo function (cubic) The minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$ is equal to ( following options)
a) $\displaystyle \frac{\pi^3}{32}$
b) $\displaystyle\frac{5\pi^3}{32}$
c) $\displaystyle\frac{9\pi^3}{32}$
d) $\displaystyle\frac{11\pi^3}{32}$
Can we go like this :
$ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$ Therefore minimum value of $(\sin^{-1}x)^3 = (\frac{-\pi}{2})^3$= -$\frac{\pi}{8}$
Please guide..
| Hint: Use the fact that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$
| {
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Very curious sequence of integrals $I_n=\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2}\mathrm dx$ I was studying the behaviour of very curious sequence of integrals
$$I_n=\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx$$ which gives a very beautiful result for $n=1$; I tried to calculate for different values of $n$ but every time what I get is a $4^{n-1}$ times $\pi$ along with a fraction that in the denominator has almost a product a consecutive primes, Can we generalize this pattern? Any help would be appreciated!
Here are few calculations:
$$
I_1=22/7-\pi
$$
$$
I_2=-\frac {2^2 \cdot 43\cdot 1097} {3\cdot 5\cdot 7\cdot 11 \cdot 13} +4\pi
$$
$$
I_3=\frac {13\cdot 31\cdot 13912991} {3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23}-16\pi
$$
| Well, I guess I can't compete with this research cited by Norbert, but maybe this will be of any help.
One can try to expand $(1-x)^n = \sum_{k=0}^n {n \choose k}(-x)^k$, so
$$\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx= \sum_{k=0}^n {n \choose k}(-1)^k\int_0^1 \frac {x^{4n+k}} {1+x^2} \,\mathrm dx$$
Then one can change variable so $$\int_0^1 \frac {x^{4n+k}} {1+x^2} \,\mathrm dx=\frac{1}{2}\int_0^1 \frac {t^{2n+\frac{k-1}{2}}} {1+t} \,\mathrm dt$$
The last integral (in the indefinite form) is:
$$\int_0^1 \frac {t^{2n+\frac{k-1}{2}}} {1+t} \,\mathrm dt=\frac{2 t^{\frac{1}{2} (k+4 n+1)} \, _2F_1\left(1,\frac{1}{2} (k+4 n+1);\frac{1}{2} (k+4 n+3);-t\right)}{k+4 n+1}$$
Plugging the limits will give:
$$\int_0^1 \frac {(x(1-x))^{4n}} {1+x^2} \,\mathrm dx=\frac{1}{4}\sum_{k=0}^n {n \choose k}(-1)^k\left(\psi ^{(0)}\left(\frac{k}{4}+n+\frac{3}{4}\right)-\psi ^{(0)}\left(\frac{k}{4}+n+\frac{1}{4}\right)\right)$$
where $\psi ^{(0)}\left(\frac{k}{4}+n+\frac{1}{4}\right)$ is the $0$-derivative of the digamma function $\psi(z)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$ , and find the point on the curve closest to $P$.
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$, and find the point on the curve closest to $P$.
What I did so far is :
plot the y var from $x^2-y^2=1 \implies y=\sqrt{1-x^2}$
Create a distance equation : $d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$ d = \sqrt{x^2+(1-y)^2}$
$ d = \sqrt{2-2\sqrt{1-x^2}}$
$\displaystyle \frac d{dx} (d) =\frac x{\sqrt{1-x^2}\sqrt{2-2\sqrt{1-x^2}}}$
I need to find the max or min?
Any suggestions? thanks!
| Any point$(Q)$ on the hyperbola $x^2-y^2=1$ can be represented as $(\sec t,\tan t)$
The distance of $Q(\sec t,\tan t)$ from $P(0,1)$ is $$\sqrt{(0-\sec t)^2+(1-\tan t)^2}=\sqrt{2-2\tan t+2\tan^2t}=\sqrt{\frac{(2\tan t-1)^2+3}2}\ge \sqrt{\frac32} $$
We can also apply Second derivative test on $\sqrt{2-2\tan t+2\tan^2t}$ to find the minimum distance
$\sqrt{2-2\tan t+2\tan^2t}$ will be minimum iff $2-2\tan t+2\tan^2t$ is minimum positive
Let $f(t)=2-2\tan t+2\tan^2t$
$f'(t)=-2\sec^2t+4\tan t\sec^2t=2\sec^2t(2\tan t-1)$
For the extreme values of $f(t),f'(t)=0$
$\implies \sec^2t(2\tan t-1)=0$
$\implies 2\tan t-1=0$ as $\sec^2t\ge1$
Now, $f''(t)=2\sec t(\sec t\tan t)(2\tan t-1)+\sec^2t(2\sec^2t)$
$$\text{ At }\tan t=\frac12, f''(t)=2\sec^4t>0$$
$$\text{ At }\tan t=\frac12,f(t)=\frac32$$
| {
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"url": "https://math.stackexchange.com/questions/364241",
"timestamp": "2023-03-29T00:00:00",
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infinite sum limit how to find the following Hi what is the limit of the following sum:
$$\lim \limits_{n\rightarrow\infty}\frac{2}{n^2}\sum\limits_{j=0}^{n-1}\sum\limits_{k=j+1}^{n-1}\frac{k}{n}$$
Thanks a lot!
| Assuming the problem is correctly written, we have
$$\begin{align}
\lim_{n\rightarrow\infty}
\frac{2}{n^2}\sum_{j=0}^{n-1}\sum_{k=j+1}^{n-1}\frac{j}{n}
&=
\lim_{n\rightarrow\infty}
\frac{2}{n^3}\sum_{j=0}^{n-1}j\sum_{k=j+1}^{n-1}1 \\
&=
\lim_{n\rightarrow\infty}
\frac{2}{n^3}\sum_{j=0}^{n-1}j(n-j-1) \\
&=
\lim_{n\rightarrow\infty}
\frac{1}{n^3}\left\{
(n-1)\sum_{j=0}^{n-1}j-\sum_{j=0}^{n-1}j^2
\right\} \\
&=
\lim_{n\rightarrow\infty}\frac{1}{n^3}\left\{
(n-1)\frac{n(n-1)}{2}-\frac{n(n-1)(2n-1)}{6}
\right\} \\
&= \frac{1}{2}-\frac{1}{3} \\
&= \frac{1}{6}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Constrained optimization max $ f(x,y) = x+y$ subject to $x^2+y^2 \leq 4, x \geq0, y \geq0$ max $ f(x,y) = x+y$ subject to $x^2+y^2 \leq 4, x \geq0, y \geq0$
I need to solve this by the Kuhn Tucker conditions without using concavity of the Lagrangian.
| The constraint $x^2 + y^2 \leq 4$ may be replaced by $x^2 + y^2 = 4$ since if $f$ were maximized when $x^2 + y^2 < 4$ we could simply increase $x$ or $y$ so that $x^2 + y^2 = 4$, contradicting the fact that $x+y$ was maximized.
As such we may write $y = \sqrt{4-x^2}$. Now we need to maximize $f = x + (4-x^2)^{1/2}$. The maximum occurs when $f'(x)=0$, that is when:
$$f'(x) = 1 + -x(4-x^2)^{-1/2} = 0 $$
This is equivalent to the fact that:
$$x = \sqrt{4-x^2}$$
Which after simplifying gives $2x^2 = 4$ and so $ x = \sqrt{2}$. As the equations are symmetric $y = \sqrt{2}$ also.
| {
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"timestamp": "2023-03-29T00:00:00",
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Confusion over a limit. Different ways of solving give different answers? Qn: If it is given that
$$
\lim_{x\to\infty} \frac{x^2 - x - 2}{x + 1} - ax - b = 1
$$
then a and b must be?
Now, I tried doing this by 2 methods.
Method 1:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = (x - 2) - ax - b $$
Since the limit is finite, $a$ must be $= 1$ and so, $b = -3$
Method 2:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = \frac{x^2(1 - \frac 1 x - \frac2 {x^2})}{x(1 + \frac1x)} - ax - b $$
$$ = \frac{x - 1 - \frac2x}{(1 + \frac1x)} - ax - b $$
as $x \to \infty$, we have the above expression
$$ = x - ax - b$$
So, $a = 1$ and $b = -1$
Which of the above is correct?
| In method 2 you are losing too much information. In particular, you lose the linear and constant term in the numerator of the fraction.
To be more particular,
you need to write
$\frac{x^2 - x - 2}{x + 1} - ax - b
=\frac{x^2 - x - 2 - (x+1)(ax+b)}{x + 1}
=\frac{x^2 - x - 2 - (ax^2+x(a+b)+b)}{x + 1}
=\frac{x^2(1-a) - x(1+a+b) - 2 -b}{x + 1}
$.
If you want the limit of this to exist,
you must have $a = 1$, or else the $x^2$ term
will cause the fraction to be unbounded.
The fraction then becomes
$\frac{- x(2+b) - 2 -b}{x + 1}
$.
Any value of $b$ allows the limit to exist,
since it is of the form
$\frac{linear}{linear}$.
To have the limit be $1$,
we will ignore the fortuitous factorization
that can be done with the numerator
and just look at the coefficients of $x$.
For the limit to be $1$,
the coefficients of $x$ is the numerator and denominator must be equal,
so that $1 = -(2+b)$,
or $b = -3$.
Note that the constant terms in this fraction do not matter.
In other words,
if $c$ and $d$ are any real numbers,
to make $\lim_{x \to \infty} \frac{- x(2+b)+c}{x + d} = 1$,
you must choose $b = -3$.
The limit then becomes
$\lim_{x \to \infty} \frac{x+c}{x + d}$,
and this limit is $1$ for any $c$ and $d$.
To see this,
note that
$\frac{x+c}{x + d}-1
= \frac{x+c-(x+d)}{x + d}
= \frac{c-d}{x + d}
$
and this goes to $0$
as $x \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/367080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?
| If we take $$x=n(n^2-3)$$ $$y=3n^2-1$$ for $n\in\mathbb{N},$ $$x^2+y^2=(n^6-6n^4+9n^2)+(9n^4-6n^2+1)=n^6+3n^4+3n^2+1=(n^2+1)^3$$ then the above equation has infinitely many integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 12,
"answer_id": 6
} |
What's the Maclaurin series for $\arcsin(x)$? I solved the problem by using a known series: $\frac{1}{\sqrt{1-x^2}}$, but the solution I got is wrong. Also, I'm not sure what to do with the constant of integration $C$. Where is my mistake?
$$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} +... $$
$$ \int\frac{1}{\sqrt{1-x^2}}dx = \int1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16} +... dx$$
$$ \arcsin(x) + C = x + \frac{2x^3}{3} + \frac{3x^5}{24} + \frac{5x^7}{112}+... \tag{what happens to $C$?}$$
The right solution is:
$$ x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} +... $$
| You're integrating the right hand side incorrectly. The integral of $\dfrac{x^2}{2}$ isn't $\dfrac{2x^3}{3}$, it's $\dfrac{x^3}{2\cdot 3} = \dfrac{x^3}{6}.$ :)))
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Binary Division IF I convert the dividend and divisor into decimal, perform the division and convert the remainder and quotient back in to binary will I get correct answer?
I'm doing this:
*
*$630 ÷ 13$
*Quotient=$48= (110000)_2$
*Remainder=$6= (110)_2$
But if I use this online calculator here, I get
*
*Quotient=$(1111110)_2$
*Remainder=$0$
So my question is what is the correct way to perform binary division, the exclusive-or operator one or by simply subtracting the numbers?
| The answer is that "it depends".
If your binary numbers are integers, then the first result is correct:
$$
630=48\cdot13+6,
$$
so the integer part of the quotient is, indeed, 48 and the remainder is six.
The link that you gave specifically states that it is doing polynomial division.
What that means is that it interprets the individual bits as coefficients of powers of $x$ in such a way that the least signficant bit are the lower degree terms. This polynomial is an element of the ring $\mathbb{F}_2[x]$. In this case it interprets
$$
630=1001110110_2=x^9+x^6+x^5+x^4+x^2+x
$$
and
$$
13=1101_2=x^3+x^2+1.
$$
This time it happens that the polynomial division has zero remainder, and the output is that the quotient is $111110_2=x^6+x^5+x^4+x^3+x^2+x$. Let's check:
$$
\begin{align}
(x^3+x^2+1)(x^6+x^5+x^4+x^3+x^2+x)&=x^3(x^6+x^5+x^4+x^3+x^2+x)+\\
&+x^2(x^6+x^5+x^4+x^3+x^2+x)+\\
&+1(x^6+x^5+x^4+x^3+x^2+x)\\
&=x^9+x^8+x^7+x^6+x^5+x^4+\\
&+x^8+x^7+x^6+x^5+x^4+x^3+\\
&+x^6+x^5+x^4+x^3+x^2+x\\
&=x^9+2x^8+2x^7+3x^6+3x^5+3x^4+2x^3+x^2+x\\
&=x^9+x^6+x^5+x^4+x^2+x,
\end{align}
$$
because the arithmetic of the coefficients of the polynomials is done modulo two.
The latter type of division is very important to mast for example when CRC-checks are tagged to data blocks.
So the answer depends on how your sequence of bits should be interpreted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim_{x \to +\infty} \left(1+\frac{\cos x}{2\sqrt{x}}\right)$ Let $f(x)=x+\sin(\sqrt{x})$. I want to find $\lim_{x \to +\infty} f'(x)$.
Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \rightarrow \infty$, hence $\frac{1}{2\sqrt{x}} \rightarrow 0$. Then $1 + \frac{1}{2\sqrt{x}} \rightarrow 1$. Therefore by the Sandwich Theorem $f'(x) \rightarrow 1$.
Lemma
$\lim_{x \to \infty} g(x)=l$ if and only if $\lim_{n \to \infty} g(x_{n})=l$ for all sequences $(x_{n}) \subset E$ with $\lim_{n \to \infty} x_{n} = \infty$, where $E$ is the domain of $g$.
Attempt 2
$f'(x)=1+\frac{\cos x}{2\sqrt{x}}$. Now let $g(x)=f'(x)$ and $x_{n}=n^2$. Then $\lim_{n \to \infty} (x_{n})=\infty$. We have $g(x_{n})=1+\frac{\cos(n)}{2n}$. Then as $n \rightarrow \infty$, $g(x_{n}) \rightarrow 1.$ Therefore by the Lemma above, $\lim_{x \to +\infty} g(x)=1$.
Question
Are the attempts above correct?
Thank you for your time.
Edited Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ For all $x \geq 0$, $1+\frac{\cos x}{2\sqrt{x}} \geq 0$. So $$0 \leq 1+\frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \rightarrow \infty$, hence $\frac{1}{2\sqrt{x}} \rightarrow 0$. Then $1 + \frac{1}{2\sqrt{x}} \rightarrow 1$. Therefore by the Sandwich Theorem $f'(x) \rightarrow 1$.
Question
Is this correct? Also, I would like to know if it is necessary to show that $$1-\frac{1}{2\sqrt{x}} \leq 1 + \frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}} \tag{1}$$
instead of $$0 \leq 1+\frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}}. \tag{2}$$
In my attempt to show inequality $(1)$, I got as far as $$1-\left|\frac{\cos x}{2\sqrt{x}}\right| \le \left|1 -\left(-\frac{\cos x}{2\sqrt{x}}\right)\right| \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ Could you please help me show that $$1-\frac{1}{2\sqrt{x}} \leq 1 + \frac{\cos x}{2\sqrt{x}}.$$
Thank you.
| The first is almost correct, I guess you intend to do the right thing.
You should rather show
$$|f'(x)-1|\le \frac1{2\sqrt x}.$$
(What you write does not exclude that $f'(x)$ might become awfully negative).
In your second attempt you use only one special sequence instead of all sequences (i.e. you should use an arbitrary sequence). Therefore, unless you know that the limit exists (but you merely don't know the value), this attempt is not complete.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is every answer of $5^k - 2^k$ divisible by 3? We have the formula $$5^k - 2^k$$
I have noticed that every answer you get from this formula is divisible by 3. At least, I think so. Why is this? Does it have to do with $5-2=3$?
| Let's use induction to prove that:
\begin{equation}
5^k - 2^k
\end{equation}
is divisible by 3
Let's check the proposition for k = 0 is divisible by 3
\begin{equation}
5^0 - 2^0 = 1 - 1 = 0
\end{equation}
which is divisible by 3
Let's assume that for k = n the proposition
\begin{equation}
5^n - 2^n
\end{equation}
is divisible by 3 and let's prove that
\begin{equation}
5^{(n+1)} - 2^{(n+1)}
\end{equation}
is also true, so
\begin{equation}
5^{(n+1)} - 2^{(n+1)} = 5{(5^n)} - 2{(2^n)}
\end{equation}
\begin{equation}
= (3 + 2)(5^n) - 2(2^n)
\end{equation}
\begin{equation}
= 3(5^n) + 2(5^n) - 2(2^n)
\end{equation}
\begin{equation}
= 3(5^n) + 2(5^n - 2^n)
\end{equation}
So in the above expression
\begin{equation}
3(5^n)
\end{equation}
is divisible by 3
\begin{equation}
2(5^n - 2^n)
\end{equation}
is divisible by 3 as per our assumption, so the finale expression
\begin{equation}
3(5^n) + 2(5^n - 2^n)
\end{equation}
is also divisible by 3
So we can conclude that
\begin{equation}
5^{(n+1)} - 2^{(n+1)}
\end{equation}
is divisible by 3
Meaning
\begin{equation}
5^n - 2^n
\end{equation}
is divisible by 3 for
\begin{equation}
n \in \mathbb{N}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 7,
"answer_id": 4
} |
Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$ how to find the Laurent expansion of $\frac{z}{z^2+1}$ valid for $|z-3|>2$.
$$\frac{z}{z^2+1} = \frac{1}{2}\left( \frac{1}{z-3 + (3 + i)} + \frac{1}{z-3 + (3 - i)}\right)$$
Given $|z-3| < \sqrt{10}$ or $|z-3| > \sqrt{10}$, I think we have two different expansions, how do I proceed? Do I need to consider both cases and thus have two answers?
| A related problem. First convert to the partial fraction form
$$ \frac{z}{z^2+1}=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}. $$
Then, we consider the first term, since the second is the same
$$ \frac{1}{2}\frac{1}{z-i}=\frac{1}{2}\frac{1}{(z-3)+(3-i)}= \frac{1}{2}\frac{1}{z-3}\frac{1}{1+\frac{3-i}{z-3}}$$
$$ = \frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k\,(3-i)^k}{(z-3)^{k+1}}\, $$
where
$$ \Big|\frac{3-i}{z-3}\Big| <1 \implies |z-3| > |3-i| \implies |z-3| > {\sqrt{10}}. $$
The same can be done with the other term which has the form
$$ = \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k(3+i)^k}{(z-3)^{k+1}}\quad |z-3|>\sqrt{10}. $$
So, we have
$$ \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k(3-i)^k}{(z-3)^{k+1}}+\frac{1}{2}\frac{1}{z-3}\sum_{k=0}^{\infty} \frac{(-1)^k(3+i)^k}{(z-3)^{k+1}} $$
$$ =\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k((3-i)^k+(3+i)^k)}{(z-3)^{k+1}}\,,\quad |z-3|> \sqrt{10}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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question in number theory Let $p$ is an odd prime and $n$ is an even natural number. It is clear that $2$ divides $p^n+1$. I would like to know Is the following claim true?
$4$ does not divides $p^n+1$.
| If $p=3, 3^n+1=(4-1)^n+1\equiv1+(-1)^n\pmod 4\implies 4|(3^n+1)$ if $n$ is odd
Other primes can be written as $(6a\pm1)$ where $a$ is any integer
Now, $(6a+1)^n+1\equiv 1+6nk+1\pmod 4\equiv2(nk+1)\pmod 4$
$\implies 4|\{(6a+1)^n+1\}$ if $n\cdot k$ is odd
Again, $(6a-1)^n+1\equiv(-1)^n-6nk+1\pmod 4$
If $n$ is odd $=2m+1$(say),$(6a-1)^n+1\equiv-1+6(2m+1)k+1\equiv2k\pmod 4$ will be divisible by $4$ if $k$ is even
If $n$ is even $=2m$(say),$(6a-1)^n+1\equiv1-6(2m)k+1\equiv2\pmod 4$ will not be divisible by $4$
Alternatively, $(2a+1)^2=4a^2+4a+1\equiv1\pmod 4$
$\implies (2a+1)^{2n}\equiv1\pmod4$
$\implies (2a+1)^{2n}+1\equiv2\pmod4$ will not be divisible by $4$
and $(2a+1)^{2n+1}\equiv(2a+1)\pmod4$
$\implies (2a+1)^{2n+1}+1 \equiv2(a+1)\pmod4$ will be divisible by $4$ if $a$ is odd
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
A problem on range of a trigonometric function: what is the range of $\frac{\sqrt{3}\sin x}{2+\cos x}$? What is the range of the function
$$\frac{\sqrt{3}\sin x}{2+\cos x}$$
| Note that $y=f(x)=\dfrac{\sqrt 3 \sin x}{2+\cos x}$.
Note that $y^2=\dfrac{3 \sin ^2 x}{(2+\cos ^2 x)^2}=\dfrac{3 (1- \cos ^2 x)}{(2+\cos ^2 x)^2}$
Put $\cos x =u$, we have
\begin{align}
y^2(2+u)^2 =3(1-u^2) \\ \implies (3+y^2)u^2+4y^2u+4y^2-3=0
\end{align}
For this to have a solution, we need discriminant of the quadratic to be non-negative. Hence, we get
$$(4y)^2-4(3+y^2)(4y^2-3) \ge 0 \\ \implies y^2 \le 1\\ \implies-1 \le y \le 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382136",
"timestamp": "2023-03-29T00:00:00",
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} |
Integral of $\int\frac{(x^4+1)\,dx}{x^3+4x}$ I followed the steps to solve this integral and want to know if I did it right and if $C=0? $
$$\int\frac{(x^4+1)\,dx}{x^3+4x} = \int\frac{(x^4+1)\,dx}{x(x^2+4)} = \frac{A}{x}+\frac{Bx+C}{x^2+4}$$
$$(x^2+4)A+x(Bx+C)=x^4+1$$
$$x=0 => 4A=1 => A=\frac{1}{4}$$
$$Ax^2+4A+Bx^2+Cx=x^4+1 = > (A+B)x^2+4A+Cx=x^4+1$$
$$A+B=0 => B=-\frac{1}{4}, C=0$$
Thanks!
| There are no constants $A,B,C$ such that
$$\frac{x^4+1}{x^3+4x} = \frac{A}{x}+\frac{Bx+C}{x^2+4},$$
because the integrand is a rational function in $x$ and the degree of the polynomial in the numerator is greater than the degree of the polynomial in the denominator. So prior to expanding it into partial fractions, the standard technique is to rewrite it as
$$\frac{x^4+1}{x^3+4x}=x+\frac{1-4x^2}{x^3+4x}$$
by using polynomial long division or Ruffini's rule.
Now you can proceed by expanding $\frac{1-4x^2}{x^3+4x}$ into partial fractions.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
| $$ LHS =\frac {1 - \cos A} {\sin A} + \frac {\sin A} {1 - \cos A} $$
$$ = \frac {2 \sin^2 \frac A2} {2\sin \frac A2 \cos \frac A2} + \frac {2\sin \frac A2 \cos \frac A2}{2 \sin^2 \frac A2}$$
$$ = \frac {\sin \frac A2} {\cos \frac A2} + \frac {\cos \frac A2} {\sin \frac A2} $$
Now just cross multiply and you get the answer.
| {
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"url": "https://math.stackexchange.com/questions/385537",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
Determine the Maclaurin series expansion of $\frac{\mathrm{exp}(z)}{(z+1)}$ Determine the Maclaurin series expansion of $\frac{\mathrm{exp}(z)}{(z+1)}$.
This is the composition of the series expansion of the exponential function centered about $z = -1$. We can rectify the expansion about $\mathrm{exp}(z)$ by writing:
$$\begin{aligned}
\frac{\mathrm{exp}(z + 1)}{e (z+1)} &= \frac{1}{e(1+z)} \sum_{n=0}^\infty \frac{(z+1)^n}{n!} \\
&= \frac{1}{e} \sum_{n=0}^\infty \frac{(z+1)^{(n-1)}}{n!}
\end{aligned}$$
Is this expansion correct? There is a slight uneasiness here that I have with calling this a Maclaurin expansion due to the centering and the negative power when $n = 0$.
Any thoughts?
| The function $e^z$ is entire and has Maclaurin series:
$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!} + \cdots$$
The function $\frac{1}{1+z}$ has a pole at $z=-1$ and has a Maclaurin series which converges for $|z|<1$:
$$\frac{1}{1+z}=1-z+z^2-z^3+ \cdots$$
Their product has a series that converges for $|z|<1$:
$$\frac{e^z}{1+z} = 1+\frac{z^2}{2} -\frac{z^3}{3}+ \frac{3z^4}{8} -\frac{11z^5}{30}+ \frac{53z^6}{144} - \cdots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An easier way to find the integral of: $\int {x\sqrt {2 + x} {\rm{ }}dx} $, where ${u^2} = 2 + x$ My attempt at the question:
$\eqalign{
& \int {x\sqrt {2 + x} {\rm{ }}dx} \cr
& {u^2} = 2 + x \cr
& 2u{{du} \over {dx}} = 1 \cr
& {{du} \over {dx}} = {1 \over {2u}} \cr
& u = \sqrt {2 + x} \cr
& x = {u^2} - 2 \cr
& so: \cr
& \int {x\sqrt {2 + x} {\rm{ }}dx} = \int {x\sqrt {2 + x} } {\rm{ }}{{dx} \over {du}}du \cr
& = \int {x\sqrt {2 + x} } {\rm{ }} \times 2\sqrt {2 + x} du \cr
& = \int {2x} (2 + x)du \cr
& = \int {4x} + 2{x^2}du \cr
& = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du \cr
& = \int {4{u^2} - 8} + 2({u^4} - 4{u^2} + 4)du \cr
& = \int {2{u^4} - 4{u^2}} du \cr
& = {2 \over 5}{u^5} - {4 \over 3}{u^3} + C \cr
& = {2 \over 5}{(\sqrt {2 + x} )^5} - {4 \over 3}{(\sqrt {2 + x} )^3} + C \cr
& = {2 \over 5}{(2 + x)^{{5 \over 2}}} - {4 \over 3}{(2 + x)^{{3 \over 2}}} + C \cr} $
A few questions I have:
Given ${u^2} = 2 + x$, $u = \pm \sqrt {2 + x} $, so why is it that we only take the principal square root and not the negative one for substitution?
The second question I have is a general one; is there an easier way of finding the integral? Have I done things in a manner that isn't overly longwinded? If so please suggest ways that would allow me to reach an answer quicker.
I'm on shakey grounds with integration at the moment so I was wondering if I could integrate this part of my working out without expanding out:
$ = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du$
Thank you for all your help!
| Let's use your substitution, without the unnecessary manipulations.
Let $u^2=x+2$. Then $2u\,du=dx$ and $x=u^2-2$. Substitute, getting rid of all $x$ all at once. We get
$$\int (u^2-2)(u)(2u)\,du=\int (2u^4-4u^2)\,du=\frac{2u^5}{5}-\frac{4u^3}{3}+C.$$
| {
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A sum with binomial coefficients Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$
| Using these three identities for $n^{\text{th}}$ differences:
$$
\begin{align}
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^n&=n!\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+1}&=n!\binom{n+1}{2}\\
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^{n+2}&=n!\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)
\end{align}
$$
and the fact that the $n^{\text{th}}$ difference of a polynomial of degree less than $n$ is $0$, we get
$$
\begin{align}
&\sum_{k=0}^n(-1)^k\binom{n}{k}(n-2k)^{n+2}\\
&=2^{n+2}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(k-n/2)^{n+2}\\
&=2^{n+2}\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\left(k^{n+2}-\frac{(n+2)n}{2}k^{n+1}+\frac{(n+2)(n+1)n^2}{8}k^n+\dots\right)\\
&=2^{n+2}n!\left(\left(3\binom{n+2}{4}+\binom{n+2}{3}\right)-\frac{(n+2)n}{2}\binom{n+1}{2}+\frac{(n+2)(n+1)n^2}{8}\right)\\
&=2^nn!\binom{n+2}{3}\\[6pt]
&=\frac{2^nn(n+2)!}{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Inequality concerning the pairwise correlation coefficients of three random variables I was asked to prove:
The correlation coefficients, $\rho_{12}$, $\rho_{23}$, $\rho_{13}$ between three random variables $X_1$, $X_2$, $X_3$ obey
$$(1+\rho_{12})(1+\rho_{13})(1+\rho_{23})\ge\frac{1}{2}(1+\rho_{12}+\rho_{23}+\rho_{13})^2$$
I'm able to prove (using triangle inequality) a weaker result:
$$(1+\rho_{12})(1+\rho_{13})(1+\rho_{23})\ge\frac{1}{8}(1+\rho_{12}+\rho_{23}+\rho_{13})^3$$
but have no clue in attacking the original one. Anyone have an idea? Thanks in advance!
| For ease of notation, I'll let $a = \rho_{12}$, $b = \rho_{13}$ and $c = \rho_{23}$. The matrix
$$
\begin{pmatrix}
1 & a & b \\
a & 1 & c \\
b & c & 1
\end{pmatrix}
$$
is positive semi-definite, so its determinant is non-negative:
$$
1 + 2abc \ge a^2 + b^2 + c^2.
$$
Now consider $(1 + a + b + c)^2$:
\begin{align*}
(1 + a + b + c)^2 & = 1 + 2(a + b + c) + (a^2 + b^2 + c^2) + 2(ab + ac + bc) \\
& \le 1 + 2(a + b + c) + 1 + 2abc + 2(ab + ac + bc) \\
& = 2(1 + a + b + c + ab + ac + bc + abc) \\
& = 2(1 + a)(1 + b)(1 + c).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$
It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
| $\displaystyle\sum_{k=0}^{n} k^{3} + (n+1)^{3} = 0^{3} + \sum_{k=1}^{n+1}k^{3} = \sum_{k=0}^{n} (k+1)^{3} = \sum_{k=0}^{n} (k^{3} +3k^{2}+3k+1)$
$\displaystyle = \sum_{k=0}^{n} k^{3} + \sum_{k=0}^{n} k^{2} + 3 \frac{n(n+1)}{2} + (n+1)$
$ \displaystyle \implies 3 \sum_{k=0}^{n} k^{2} = 3 \sum_{k=1}^{n} k^{2} = (n+1)^{3} -3 \frac{n(n+1)}{2} -(n+1) = \frac{n(n+1)(2n+1)}{2}$
Similarly, the sum $\displaystyle \sum_{k=1}^{n} k^{3} $ can be evaluated by starting with the equation $\displaystyle\sum_{k=0}^{n} k^{4} + (n+1)^{4}= 0^{4} + \sum_{k=1}^{n+1} k^{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
A binomial identity from Mathematical Reflections Here is the problem:
Let $m,n$ be positive integers with $n>m$. Prove that
$\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$
This problem is O243 of Mathematical Reflections. A solution had been published using complex integration (https://www.awesomemath.org/assets/PDFs/MR5sol(1).pdf). However, I would like to see a solution using the difference operator, if any, since the form of the summand brings this to mind.
| First of all, it is worth stating explicitly that the problem assumes that $\binom{n}{k}$ is zero when $n < 0$ or $n>k$, even for $k\geqslant 0$. Indeed, otherwise
In[61]:= Table[
Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1], {k, 0,
n}], {n, 1, 5}, {m, 0, n - 1}]
Out[61]= {{0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0, 0}}
With those restrictions in place the claimed result is indeed reproduced:
In[65]:= Table[
Sum[(-1)^k Binomial[n, k] Binomial[n + m - 2 k, n - 1] Boole[
0 <= n - 1 <= n + m - 2 k], {k, 0, n}], {n, 1, 5}, {m, 0, n - 1}]
Out[65]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}
In[66]:= Table[Binomial[n, m + 1], {n, 1, 5}, {m, 0, n - 1}]
Out[66]= {{1}, {2, 1}, {3, 3, 1}, {4, 6, 4, 1}, {5, 10, 10, 5, 1}}
With this said, the upper bound of summation over $k$ is $m_\ast = \left\lfloor \frac{m+1}{2} \right\rfloor$:
$$
\mathcal{S}(n,m)= \sum_{k=0}^{m_\ast} (-1)^k \binom{n}{k} \binom{n+m-2k}{n-1}
$$
The summand is the hypergeometric term, meaning that
$$
r(k) = \frac{c_{k+1}}{c_k} = \frac{(-1)^{k+1} \binom{n}{k+1} \binom{n+m-2(k+1)}{n-1} }{(-1)^k \binom{n}{k} \binom{n+m-2k}{n-1} } = \frac{-n+k}{k+1} \frac{-\frac{m+1}{2} + k}{-\frac{m+n-1}{2} + k} \frac{-\frac{m}{2} + k}{-\frac{m+n}{2}+k}
$$
and therefore
$$
c_k = c_0 \prod_{q=1}^{k} r(q) = \binom{m+n}{n-1} \frac{(-n)_k}{k!} \frac{\left(-\frac{m}{2}\right)_k \cdot \left(-\frac{m+1}{2}\right)_k}{\left(-\frac{m+n}{2}\right)_k \cdot \left(-\frac{m+n-1}{2}\right)_k}
$$
And thus, since $r(m_\ast) = 0$ we have
$$
\mathcal{S}(n,m) = \binom{n+m}{n-1} \cdot {}_3F_2\left(\left.\begin{array}{cll} -n & -m/2 & -(m+1)/2 \\ & -(m+n)/2 & -(m+n-1)/2 \end{array} \right| 1 \right)
$$
Now, per this identity:
$$
{}_3F_2\left(\left.\begin{array}{cll} -n & a & b \\ & d & 1+a+b-d-n \end{array} \right| 1 \right) = \frac{(d-a)_n \cdot (d-b)_n}{(d)_n \cdot (d-a-b)_n}
$$
Using this identity for $a = -m/2$, $b = -(m+1)/2$ and $d=\epsilon-(m+n)/2$ with the intent to consider the limit of $\epsilon \to 0$ we have
$$
\mathcal{S}(n,m) = \binom{n+m}{n-1} \lim_{\epsilon \to 0} \frac{ \left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n}{ \left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n }
$$
Using
$$
\left(-\frac{n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n}{2} + \epsilon \right)_n = \frac{1}{2^{2n}} \frac{\Gamma(n+2 \epsilon)}{\Gamma(-n+2\epsilon)} = (-1)^n \frac{\Gamma(1+n-2\epsilon) \Gamma(n+2\epsilon)}{2^{2n} \pi} \sin(2 \pi \epsilon)
$$
$$
\left(-\frac{m+n}{2} + \epsilon \right)_n \cdot \left(\frac{1}{2}-\frac{n-m}{2} + \epsilon \right)_n = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{\Gamma\left(\frac{-n-m}{2} + \epsilon\right)\Gamma\left(\frac{1+m-n}{2} + \epsilon\right)} = \frac{\Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) }{-\frac{2\pi^2}{\sin(\pi m) + \sin(\pi n - 2 \pi \epsilon)}} \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right) = (-1)^n \frac{\sin(2\pi \epsilon)}{2 \pi^2} \Gamma\left(\frac{n-m}{2} + \epsilon\right) \Gamma\left(\frac{n+m+1}{2} + \epsilon\right) \Gamma\left(1 + \frac{m+n}{2} - \epsilon\right) \Gamma\left( \frac{1+n-m}{2} - \epsilon\right)
$$
Combining, and using the duplication formula we get
$$
\mathcal{S}(n,m) = \binom{n+m}{n-1} \frac{n! (n-1)!}{(n+m)! (n-m-1)!} = \binom{n}{m+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Simple algebra simplification question? Hello everyone I have the following question.
I have the following fraction
$f(x)=-\frac{4}{x^2}+\frac{1}{(x-1)^2}$
But how would I reduce it I know I have to use the multiply the opposite numerator by denominator and I got.
$(x-1)^2(4)=4x^2-8x+4$
$1(x^2)=x^2$
So I got got $f(x)=-\frac{4x^2-8x+4+x^2}{x^2(x-1)^2}$
But this is incorrect.What am I doing wrong.
| You are close.
Your mistake was applying the "-"
to both terms instead of just one.
You wrote
$f(x)=-\frac{4x^2-8x+4+x^2}{x(x-1)^2}$.
What you should have is
$f(x)=\frac{-(4x^2-8x+4)+x^2}{x(x-1)^2}
=\frac{-4x^2+8x-4+x^2}{x(x-1)^2}
=\frac{-3x^2+8x-4}{x(x-1)^2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$ Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$
I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.
I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.
| Here's one way to go.
First, note that
$$\begin{eqnarray*}
\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx
&=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx
+\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx.
\end{eqnarray*}$$
For now I'll simply claim that
\begin{equation*}
\int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx = \pi\log 64.\tag{1}
\end{equation*}
(I would be surprised if this integral has not been handled somewhere on this site.)
But
$$\begin{eqnarray*}
\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx
&=& \int_0^\pi\frac{3x}{\sin x} \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k}} \sin^{2k}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\int_0^\pi x \sin^{2k-1}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\frac{\pi^{3/2}\Gamma(k)}{2\Gamma(k+1/2)} \\
&=& -\pi \sum_{k=1}^\infty \frac{1}{3^{2k-1}2k(2k-1)} \\
&=& -\pi \log \frac{32}{27}.
\end{eqnarray*}$$
(The last sum can be found by standard methods.
Schematically, $\sum \frac{a^{2k-1}}{2k(2k-1)} = \sum \int {\mathrm da} \frac{a^{2k-2}}{2k}$.)
Thus, the integral is $\pi \log 54$ as claimed.
Proof of (1):
We have
$$\begin{eqnarray*}
\int_0^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx
&=& \int_{0^+}^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx \\
&=& 3\int_{0^+}^\pi x \cot\frac{x}{2} \ \mathrm dx
\hspace{5ex}\textrm{(double angle formulas)} \\
&=& 12 \int_{0^+}^{\pi/2} t\cot t \ \mathrm dt
\hspace{5ex} (t = x/2) \\
&=& -12\int_{0^+}^{\pi/2} \log\sin t \ \mathrm dt
\hspace{5ex}\textrm{(integrate by parts)} \\
&=& -6\int_{0^+}^{\pi/2} \log\sin^2 t \ \mathrm dt \\
&=& -6\int_{0^+}^{\pi/2} \log(1-\cos^2 t) \ \mathrm dt \\
&=& 6 \int_{0^+}^{\pi/2} \sum_{k=1}^\infty \frac{1}{k}\cos^{2k}t \ \mathrm dt
\hspace{5ex}\textrm{(series for log)} \\
&=& 6\sum_{k=1}^\infty \frac{1}{k} \int_{0^+}^{\pi/2} \cos^{2k}t \ \mathrm dt
\hspace{5ex} \textrm{(Tonelli's theorem)}\\
&=& 6\sum_{k=1}^\infty \frac{1}{k}
\frac{\sqrt{\pi}\Gamma(k+1/2)}{2\Gamma(k+1)} \\
&=& 3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}\frac{2k-1}{k} \\
&=& \pi \log 64.
\end{eqnarray*}$$
Note that
$$\begin{eqnarray*}
6\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}
&=& -6\pi \left[\sum_{k=0}^\infty {1/2 \choose k}(-1)^{k} - 1\right] \\
&=& -6\pi[(1-1)^{1/2} - 1] \\
&=& 6\pi
\end{eqnarray*}$$
and
$$\begin{eqnarray*}
-3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k}
&=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k \int_0^1 x^{k-1} \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x} \left[
\sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1
\right] \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x} \left(
\sqrt{1-x} -1
\right) \ \mathrm dx \\
&=& 3\pi(-2+\log 4) \\
&=& -6\pi + \pi\log 64.
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 1
} |
Test of convergence of $\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$ I am having some trouble with this problem and don't know if I am doing it right:
$$\int_{-\infty}^{\infty} \dfrac{x^6+6}{x^8+8}dx$$
so the steps I have taken so far are, I split it into
$$\int_0^\infty \dfrac{x^6+6}{x^8+8} + \int_{-\infty}^0 \dfrac{x^6+6}{x^8+8}$$
for the second integral I made a change of variable $x = -u$
so it would look the same as the first integral and then use $$ \lim_{x\to \infty} \dfrac{f(x)}{g(x)}$$ where $g(x) = \dfrac1{x^2}$.
Just wondering if my approach is correct.
| Your approach is fine, as long as you compare with $1/x^2$ when $|x|\gt1$ and then use the continuity of $\frac{x^6+6}{x^8+8}$ for $|x|\le1$. We can illustrate this with a slightly different comparison function.
First, note that
$$
\lim_{|x|\to\infty}\frac{(x^6+6)(x^2+1)}{x^8+8}=1\tag{1}
$$
Thus, $(1)$ says that there is an $m$ so that if $|x|\ge m$, then $\frac{(x^6+6)(x^2+1)}{x^8+8}\le2$. Since $\frac{(x^6+6)(x^2+1)}{x^8+8}$ is continuous on the compact set $[-m,m]$, it is bounded there. Therefore, there is an $M$ so that
$$
\frac{(x^6+6)(x^2+1)}{x^8+8}\le M\tag{2}
$$
Therefore,
$$
\frac{x^6+6}{x^8+8}\le\frac{M}{1+x^2}\tag{3}
$$
Now, simply use
$$
\int_{-\infty}^\infty\frac{\mathrm{d}x}{1+x^2}=\pi\tag{4}
$$
to show that
$$
\int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x\tag{5}
$$
converges by comparison.
Although this answer uses contour integration, it does give the value for
$$
\int_{-\infty}^\infty\frac{x^6+6}{x^8+8}\,\mathrm{d}x=\frac{\pi}{8}\csc\left(\frac{\pi}{8}\right)\left(2^{5/8}+3\cdot2^{-5/8}\right)\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Recurrence Relation How do I solve:
$k(k+1)a_{k}=2(\lambda k-1)a_{k-1}+(a-\lambda^2)a_{k-2}$
where $\lambda$ and $a$ are constants, and similar other recurrence relations?
| Alright, let us substitute for $k-2$ . then the recurrence relation reads:
\begin{equation}
(k+2)(k+3) a_{k+2} - 2 \left( \lambda (k+2)-1 \right) a_{k+1} - \left( a+\lambda^2 \right) a_k = 0
\end{equation}
for $k=0,1,2\cdots$. Here for the sake of simplicity we will make some additional assumption that $a_1 = (\lambda-1) a_0$. We will present the solution in this case and then we will show how to wave this assumption. Let us get it over and done with now.
We define the Z-transform as $A[z]:= \sum\limits_{n=0}^\infty a_n z^{-n}$. Then simple algebra shows that the quantity in question satisfies the following ODE:
\begin{equation}
\frac{d^2 A[z]}{d z^2} + \frac{2\lambda}{z^2} \frac{d A[z]}{d z} - \frac{a+\lambda^2-2 z+2 \lambda z}{z^4} \cdot A[z] = 0
\end{equation}
Now we use standard techniques to eliminate the coefficient at the first derivative. We have $A[z] = \exp(-1/2 \int (2 \lambda/z^2) \quad dz) \cdot v(z) = \exp(\lambda/z) \cdot v(z)$ and the function $v(z)$ satisfies the following ODE:
\begin{equation}
\frac{d^2 v(z)}{d z^2} - \frac{a+\lambda^2-2 z}{z^4} v(z) = 0
\end{equation}
It is not hard to see that the ODE above can be obtained from the confluent hypergeometric ODE by changing both the abscissa and the ordinate accordingly. The final result is given below:
\begin{eqnarray}
a_n &:=& \left. \frac{1}{n!} \frac{d^n }{d x^n}
F_{1,1} \left[1- \frac{1}{\sqrt{a+2 \lambda^2}}, 2, 2 \sqrt{a+2 \lambda^2} x \right] \cdot \exp\left( (\lambda-\sqrt{a+2 \lambda^2}) x\right)
\right|_{x=0}\\
&=&
\frac{2^n \left(a+2\lambda^2\right)^{n/2} \Gamma \left(n-\frac{1}{\sqrt{2\lambda^2+a}}+1\right)}{\Gamma (n+1) \Gamma (n+2) \Gamma
\left(1-\frac{1}{\sqrt{2\lambda^2+a}}\right)}
\, _2F_1\left(-n-1,-n;\frac{1}{\sqrt{2\lambda^2+a}}-n;\frac{1}{2}-\frac{\lambda}{2 \sqrt{2\lambda^2+a}}\right)
\end{eqnarray}
for $n=0,1,2,\cdots$.
The Mathematica code below demonstrates that the results are correct:
In[862]:= {a, l, z} = RandomReal[{0, 1}, 3, WorkingPrecision -> 50];
aa = Table[((2^n) ((a + 2 l^2)^(n/2))
Gamma[1 - 1/Sqrt[a + 2 l^2] + n] )/(
Gamma[1 - 1/Sqrt[a + 2 l^2]] Gamma[1 + n] Gamma[2 + n])
Hypergeometric2F1[-1 - n, -n, 1/Sqrt[a + 2 l^2] - n,
1/2 - l/(2 Sqrt[a + 2 l^2])], {n, 0, 10}] // Simplify;
(aa[[2]] - (l - 1) aa[[1]]) // Simplify
Table[(k + 2) (k + 3) aa[[k + 3]] -
2 (l (k + 2) - 1) aa[[k + 2]] - (a + l^2) aa[[k + 1]], {k, 0,
Length[aa] - 3}] // Simplify
Out[864]= 0.*10^-50
Out[865]= {0.*10^-49, 0.*10^-50, 0.*10^-50, 0.*10^-50, 0.*10^-50,
0.*10^-51, 0.*10^-51, 0.*10^-52, 0.*10^-52}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Cubing a simple thing I am trying to expand $\quad (x + 2)^3 $
I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.
I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this
$$(x^2 + 4x + 4)(x + 2)$$
This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with
$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$
Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.
| Your idea to do this:
$$(x^2 + 4x + 4)(x + 2)$$
was correct, because $(x + 2)^2 = (x^2 + 4x + 4)$. So it follows that $(x + 2)^3 = (x^2 + 4x + 4)(x + 2)$.
Let's try just multiplying it out:
To multiply $(x^2 + 4x + 4)$ by $(x+2)$, we'll multiply every term in $(x^2 + 4x + 4)$ by every term in $(x+2)$.
That's easy to do, and easy to visualise, because $(x^2 + 4x + 4)\cdot (x + 2)$ is the same as $(x \cdot(x^2 + 4x + 4)) + (2 \cdot (x^2 + 4x + 4)).$
$$(x \cdot(x^2 + 4x + 4)) = x^3 + 4x^2 + 4x$$
$$(2 \cdot (x^2 + 4x + 4)) = 2x^2 + 8x + 8$$
Add the two:
$$x^3 + \color{red}{4x^2} + \color{blue}{4x} + \color{red}{2x^2} + \color{blue}{8x} + 8 = x^3 + 6x^2 + 12x + 8$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
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} |
Limit of a recursive sequence with $u_n$ It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.
The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.
| Writing $u_n$ as $\frac{p_n}{q_n}$ for two sequences $(p_n)$, $(q_n)$ to be determined. We have:
$$u_{n+1} = 1 + \frac{1}{u_n} \iff \frac{p_{n+1}}{q_{n+1}} = 1 + \frac{q_n}{p_n} = \frac{p_n+q_n}{p_n}$$
Normalize $(p_n)$ and $(q_n)$ appropriately, we can turn the non-linear recurrence equation into a linear one:
$$\begin{pmatrix}p_{n+1}\\q_{n+1}\end{pmatrix}
= \begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}p_n\\q_n\end{pmatrix}$$
The charateristic polynomial of the matrix $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ is
equal to:
$$\lambda (\lambda - 1 ) - 1^2 = \lambda^2 - \lambda - 1 = (\lambda - \varphi)(\lambda + \varphi^{-1})$$
where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. The matrix
has eigenvalues $\varphi$ and $-\varphi^{-1}$ with corresponding eigenvectors:
$$\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}\varphi\\1\end{pmatrix} = \varphi \begin{pmatrix}\varphi\\1\end{pmatrix}
\quad\text{ and }\quad
\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix} = -\varphi^{-1}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}
$$
Since $u_1 = 1$, we will choose $p_1 = q_1 = 1$. Expressing $\begin{pmatrix}1\\1\end{pmatrix}$ in terms of the eigenvectors:
$$\begin{pmatrix}1\\1\end{pmatrix} = \alpha \begin{pmatrix}\varphi\\1\end{pmatrix} + \beta
\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}
\implies \alpha = \frac{\varphi+1}{\varphi+2} \;\text{ and }\; \beta = \frac{1}{\varphi+2}
$$
We get
$$ \begin{pmatrix}p_n\\q_n\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}p_1\\q_1\end{pmatrix} = \alpha \varphi^{n-1} \begin{pmatrix}\varphi\\1\end{pmatrix} + \beta (-\varphi^{-1})^{n-1}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}$$
This allow us to derive a closed form expression for $u_n$:
$$u_n = \frac{p_n}{q_n} = \frac{\alpha \varphi^n + \beta (-\varphi^{-1})^n}{\alpha \varphi^{n-1} + \beta (-\varphi^{-1})^{n-1}} = \varphi \frac{\alpha + \beta (-\varphi^{-2})^n}{\alpha + \beta (-\varphi^{-2})^{n-1}}$$
Since $| -\varphi^{-2} | < 1$, this implies $\lim_{n\to\infty} u_n = \varphi \frac{\alpha + \beta*0}{\alpha + \beta*0} = \varphi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Composition of systems of equations Suppose $$2x + 3y = u$$ $$x - 4y = v$$
and further that
$$3u - 5v = c$$ $$2u + 3v = d$$
Express c and d in terms of $x$ and $y$ by matrix multiplication.
It's quite easy by direct substitution but I can;t work out how to use matrix multiplication. Any ideas? Thanks in advance!
| $$\begin{align*}
2x + 3y& = u\\
x - 4y &= v
\end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}u\\ v\end{bmatrix}$$
$$\begin{align*}
3u - 5v &= c\\
2u + 3v &= d
\end{align*}\quad \underset{\substack{\text{convert to}\\ \text{matrix language}}}{\leadsto}\quad \begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\begin{bmatrix} u\\ v\end{bmatrix}=\begin{bmatrix}c\\ d\end{bmatrix}$$
$$\begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\begin{bmatrix} u\\ v\end{bmatrix}=\begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\Bigg(\begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}\Bigg)=\begin{bmatrix}c\\ d\end{bmatrix}\implies\begin{bmatrix} p & q\\ r &s\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}c\\ d\end{bmatrix}\\[0.4in]
\quad\text{ where }\quad \begin{bmatrix} p & q\\ r &s\end{bmatrix}=\begin{bmatrix} 3 & -5\\ 2 &\hphantom{-}3\end{bmatrix}\begin{bmatrix} 2 & \hphantom{-}3\\ 1 &-4\end{bmatrix}.\\[0.4in]
\begin{bmatrix} p & q\\ r &s\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}=\begin{bmatrix}c\\ d\end{bmatrix}\quad\underset{\substack{\text{convert to}\\ \text{equation language}}}{\leadsto}\quad \begin{align*}
px + qy& = c\\
rx + sy &= d
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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1st derivative of $\frac{2x}{\sqrt{x^2 + 1}}$ Another simple question that I can't work out today, yet I would work it out two weeks ago!
I need to find the 1st derivative of $$\frac{2x}{\sqrt{x^2 + 1}}$$.
So I use the Quotient rule and I get: $$\frac{(x^2 + 1)^.5 (2) - (2x)(0.5x^2 + 0.5)^-5}{x^2+1}$$
Am I heading in the correct direction and do I just need to multiply and try to get rid of the exponents somehow?
Thanks
| Without messing up with nominator and denominators:
$x=\tan { u } $ then draw right triangle, see
$\\ \frac { 1 }{ \sqrt { x^{ 2 }+1 } } =\cos { u } \\$
$\\ \frac { x }{ \sqrt { x^{ 2 }+1 } } =\sin { u } \\$
$ \frac { du }{ dx } =\frac { 1 }{ x^{ 2 }+1 } \\$
$ f=\frac { 2x }{ \sqrt { x^{ 2 }+1 } } =2\sin { u } \\$
$$ f'\quad =2\cos { u } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 } } \frac { du }{ dx } =\frac { 2 }{ \sqrt { x^{ 2 }+1 } } \frac { 1 }{ x^{ 2 }+1 } $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying a Product of Summations I have, for a fixed and positive even integer $n$, the following product of summations:
$\left ( \sum_{i = n-1}^{n-1}i \right )\cdot \left ( \sum_{i = n-3}^{n-1} i \right )\cdot \left ( \sum_{i = n-5}^{n-1}i \right )\cdot ... \cdot \left (\sum_{i=5}^{n-1}i \right )\cdot \left (\sum_{i=3}^{n-1}i \right )\cdot \left (\sum_{i=1}^{n-1}i \right )$
Where there are $\frac{n}{2}$ groups of summations multiplied together.
For example, consider the case where $n=4$ :
$\left ( \sum_{i = 3}^{3}i \right )\cdot \left ( \sum_{i = 1}^{3} i \right ) = \left ( 3 \right )\left ( 1+2+3 \right ) = 18$
I have tried in vain to simplify the product. Perhaps there are identities I could make use of.
Edit : I can expand the product to clarify:
$$\left ( n-1 \right )\cdot \left [ (n-3)+(n-2)+(n-1) \right ]\cdot \left [ (n-5)+...+(n-1) \right ]\cdot ... \cdot\left [3+4+...+(n-1)\right ]\cdot \left [1+2+...+(n-1) \right ]$$
From where I can see a $(n-1)^{\frac{n}{2}}$ term, but the others are quite jumbled.
| Let $n=2m$. Your expression is then
$$\begin{align*}
&(2m-1)\cdot\frac32(2n-4)\cdot\frac52(2n-6)\cdot\ldots\cdot\frac{n-1}2n\\\\
&\qquad=1\cdot(2m-1)\cdot3(2m-2)\cdot5(2m-3)\cdot\ldots\cdot(2m-1)m\\\\
&\qquad=\prod_{k=1}^m(2k-1)(2m-k)\\\\
&\qquad=(2m-1)!!\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^mm!}\cdot\frac{(2m-1)!}{(m-1)!}\\\\
&\qquad=\frac{(2m)!}{2^m}\binom{2m-1}m\\\\
&\qquad=\frac{n!}{2^{n/2}}\binom{n-1}{n/2}\;.
\end{align*}$$
| {
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"url": "https://math.stackexchange.com/questions/400480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Evaluating a summation of inverse squares over odd indices $$ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
I want to evaluate this sum when $n$ takes only odd values.
| This can be shown in a similar way to Euler's proof of $\zeta(2) = \frac{\pi^2}{6}$, which starts with the function $\frac{\sin(x)}{x}$
(i.e. the sinc function). Here we start with the cosine function which can be expressed as the infinite product
\begin{align}
\cos(x) &= \prod_{n=1}^\infty \left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right) \\
&= \left(1- \frac{4x^2}{\pi^2}\right)\left(1- \frac{4x^2}{9\pi^2}\right)\left(1- \frac{4x^2}{25\pi^2}\right) ... \\
&=1-x^2\left(\frac{4}{\pi^2}+\frac{4}{9\pi^2}+\frac{4}{25\pi^2}+...\right)+...
\end{align}
$\cos(x)$ can also be expressed by the following Maclaurin series expansion:
$$\cos(x) = \sum_{n=1}^\infty \frac{(-1)^n}{(2n)!}x^{2n} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +...$$
Comparing the $x^2$ coefficients gives:
$$-\frac{1}{2!} = -\frac{4}{\pi^2}\left(1+\frac{1}{9} + \frac{1}{25} + ...\right)$$
Thus,
$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \frac{\pi^2}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Indeterminate Limits I have been studying independently through various online courses and I still have trouble understanding what to do with certain limits. I am hoping for some guidance on the following two problems to help me solve them (I do not need to answer as much as help understanding where I am going).
$$ \lim_{x\to 0} \frac{\sqrt{16 + 4x} - \sqrt{16 - 4x}}{x} $$
$$ \lim_{x\to 0} \frac{\frac{1}{(x + 6)^2} - \frac{1}{36}}{x} $$
I can tell that these are both $ \frac{0}{0} $ indeterminate equations and I can simplify them in a number of ways but I cannot seem to get the 0 out of the bottom of the equation. Any help pointing me in the right direction would be greatly appreciated!
| I would assume that you typed in the expression for the limit correctly.
For the problem you provided, there is a short way to determine the limit. Given that the limit is:
$$\lim_{x\to 0} \frac{\sqrt{16 + 4x} - \sqrt{16 + 4x}}{x}$$
The numerator expression becomes $0$. Then, we have:
$$\lim_{x\to 0} \frac{0}{x} = \lim_{x \to 0} 0 = 0$$
~~
For the second problem, we can rewrite the expression as:
$$\lim_{x \to 0} \dfrac{\frac{1}{(x + 6)^2} - \frac{1}{36}}{x} \cdot \dfrac{36(x + 6)^2}{36(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{36 - (x + 6)^2}{36x(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{36 - x^2 - 12x - 36}{36x(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{-x(x + 12)}{36x(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{-(x + 12)}{36(x + 6)^2}$$
Thus,
$$\lim_{x \to 0} \dfrac{-(x + 12)}{36(x + 6)^2} = \dfrac{-12}{36(6)^2} = -\dfrac{1}{108}$$
Verified by Wolfram for:
First problem
Second problem
For the problem you edited
The given limit is:
$$\lim_{x \to 0} \dfrac{\sqrt{16 + 4x} - \sqrt{16 - 4x}}{x}$$
Multiply the top and bottom by the conjugate of the numerator expression, which is $\sqrt{16 + 4x} + \sqrt{16 - 4x}$. This gives:
$$\lim_{x \to 0} \dfrac{16 + 4x - (16 - 4x)}{x(\sqrt{16 + 4x} + \sqrt{16 - 4x})}$$
$$= \lim_{x \to 0} \dfrac{8x}{x(\sqrt{16 + 4x} + \sqrt{16 - 4x})}$$
$$= \lim_{x \to 0} \dfrac{8}{\sqrt{16 + 4x} + \sqrt{16 - 4x}}$$
Thus,
$$\lim_{x \to 0} \dfrac{8}{\sqrt{16 + 4x} + \sqrt{16 - 4x}}$$
$$= \dfrac{8}{\sqrt{16} + \sqrt{16}}$$
$$= \dfrac{8}{4 + 4}$$
$$= 1$$
Verified by WolframAlpha.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$ Prove that for $n\ge 3$,
$$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\dfrac{n(n-1)}{4}+1$$
where $\varphi$ is the Euler's totient function
I think we must use this
$$\sum_{k=1}^{n}\varphi(k)=\dfrac{1}{2}\left(1+\sum_{k=1}^{n}\mu(k)\left[\dfrac{n}{k}\right]^2\right)$$
| This is only a partial answer, just a possibly profitable approach.
On the rhs we have an expression which contains just the half of the sum of consecutive numbers (with a small deviation), so I would try the problem comparing the double of the sum of totients with that of the natural numbers.
The conjecture is
$$ \varphi(2) + \varphi(3) + ... + \varphi(n) \ge {n(n-1)\over 4} + 1 = {n(n+1)\over 2} \frac12 - \frac n2+1 $$
The $\varphi(\cdot)$ and the expression for the sum-of-consecutive-numbers expanded gives
$$
2\cdot(1-\frac12) + 3\cdot(1-\frac13)+4\cdot(1-\frac12)+...+ \varphi(n) \ge (1+2+3+4+...+n)(1-\frac12) - \frac n2 +1 $$
and we see, that most of the parentheses on the lhs evaluate to more than $(1-\frac12)=\frac12$
To simplify multiply both sides by 2 to get
$$
(2 + 3 \cdot\frac43+4 + 5 \cdot\frac85 + 6 \cdot\frac 23 + 7 \cdot\frac {12}7 +...+2 \varphi(n)) \ge (1+2+3+4+...+n) - n +2 $$
Then we get first
$$(2 + (3 +1)+4 + (5 +3)+ (6 - 6/3) + (7 + 5)+...+2 \varphi(n)) \ge (1+2+3+4+...+n) - n +2 $$ and separating the lhs in two parentheses
$$ \begin{eqnarray} && (2 + 3 +4 + 5 + 6 + 7+...+ n) \\
&+ & (0+1+0+3-2+5+0+3... + 2 \varphi(n)-n ) \\ &\ge & (1+2+3+4+...+n) & - n +2 \end{eqnarray} $$
and finally
$$ 0+1+0+3-2+5+0+3 + \ldots + 2\varphi(n) \ge 3 $$
The problem reduces then to show, that the partial sums of the lhs-expression are always positive and $\ge 3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$ can anyone tell me why
$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rules if I write the equation in the following way
$$\frac{6x^2+19x+15}{(x-1)(x-2)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-2}$$
Thanks
| An easy way to see why your proposal doesn't work is the following: Multiply both sides of your equation with $(x-1)(x-2)^2$ and extend the formula continuously to $x=1$ and $x=2$. We now have
$$6x^2+19x+15 = A(x-2)^2+B(x-1)(x-2)+C(x-1)(x-2).$$
If we now enter $x=2$, we get $24+38+15 = A\cdot 0^2+B\cdot 1\cdot 0 + C\cdot 1\cdot 0 = 0$, which is obviously false.
On the other hand
$$6x^2+19x+15 = A(x-2)^2 + B(x-1)(x-2)+C(x-1)$$
would have reduced to $24+38+15 = C$, which is perfectly fine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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sign charts and expressed in interval form Solve by using sign charts and express the solutions in interval form.
$$\text{(a)} \frac{(x+2)(x-3)^{2}}{x^{2}+x-2} \geq 0$$
$$\text{(b)} \frac{1}{x-1} < \frac{2}{x}$$
| Here is a full solution for $(b)$
$$ \frac{1}{x-1} < \frac{2}{x} \implies \frac{2}{x} - \frac{1}{x-1}>0 \implies \frac{x-2}{x(x-1)}>0 $$
$$ \implies \left\{x-2>0\quad \cap \quad x(x-1)>0 \right\} \cup \left\{x-2<0\quad \cap \quad x(x-1)<0 \right\} $$
$$ \implies \left\{x-2>0\, \cap\,(x-1)>0 \right\}\cup \left\{x-2<0\,\cap\left\{( x<0 \cap x-1>0 ) \cup ( x>0 \cap x-1<0 )\right\}\right\} $$
$$ \implies \left\{x-2 >0 \right\} \cup \left\{ x-2<0 \,\cap \, 0<x<1 \right\} $$
$$ \implies \left\{x-2 >0 \right\} \cup \left\{ 0<x<1 \right\} $$
$$ \implies (2,\infty) \cup ( 0,1 ) $$
Note: $\cup$ stands for union while $\cap$ stands for intersection.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $ has precisely two real roots Prove that given $ a < b < c $ this equation: $$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $$
has precisely 2 real roots.
I understand there are 3 point of discontinuities, but I have no idea how to prove this. Can you give me a hint?
Thanks in advance.
| HINT:
Rearrange the equation to get $3x^2-2(a+b+c)x+ab+bc+ca=0$
$$\text{The discriminant }\{2(a+b+c)\}^2-4\cdot3\cdot(ab+bc+ca)$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
$$=4\cdot\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2>0 \text{ for distinct real }a,b,c$$
What can we conclude from this?
| {
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Finding the constant term of a degree $3$ polynomial Let $p(x)=x^3+ax^2+bx+c$, where $a,b,c$ are real constants. If $p(-3)=p(2)=0$ and $p'(-3)<0$, which of the following is a possible value of $c$?
A) $-27$
B) $-18$
C) $-6$
D) $-3$
E) $-\dfrac{1}{2}$
$\textbf{My attempt at this problem}$:
I drew a rough sketch of the curve on the $xy$-plane. A portion of the degree $3$ curve has a local max to the left of $-3$ and a local min between $-3$ and $2$. When $x> 2$, $p$ is a strictly increasing function, no longer intersecting the $x$-axis.
Since $p(-3)=p(2)=0$, we plug in these integers to get
$$
p(-3)=(-3)^3 + a(-3)^2+b(-3)+c=0, \\
p(2) =(2)^3 + a(2)^2+b(2)+c=0.
$$
There is another root when $x<-3$.
Is there a trick to solve this problem without so much calculation?
| $p$ has three roots, $-3$, $2$ and $x_0$, say. We have
$$ p(x) = x^3 + ax^2 + bx + c = (x+3)(x-2)(x-x_0). $$
So $c = 6x_0$. As, as you said, $x_0 < -3$, A) is the only possibility.
| {
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Limit of double sum: $\lim\limits_{n\to\infty}n^{-2}\sum\limits_{k=1}^n\sum\limits_{m=k+1}^n\left(\frac{n-2k}{n+2k}\right)^2\frac{n-2m}{n+2m}$ Who is so kind to enlighten me about the steps I need to follow?
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\sum_{m=k+1}^n\left(\frac{n-2k}{n+2k}\right)^2\frac{n-2m}{n+2m}$$
| We have a Riemann sum:
$$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \left ( \frac{1-\frac{2k}{n}}{1+\frac{2k}{n}}\right)^2 \sum_{m=k+1}^n \frac{1-\frac{2m}{n}}{1+\frac{2m}{n}} = \int_0^1 dx \left (\frac{1-2 x}{1+2 x}\right )^2 \, \int_x^1 dy \frac{1-2 y}{1+2 y}$$
The evaluation of the above integral is straightforward but messy. The inner integral has an antiderivative
$$\begin{align}\int dy \frac{1-2 y}{1+2 y} &= \int \frac{dy}{1+2 y} - \int dy \frac{2 y}{1+2 y}\\ &= \frac12 \log{(1+2 y)} - \left [y - \frac12 \log{(1+2 y)} \right ]\\ &= \log{(1+2 y)}-y\end{align}$$
The integral is now a single integral when the inner integral is evaluated over its integration limits:
$$\int_0^1 dx \left (\frac{1-2 x}{1+2 x}\right )^2 \left [\log{3} - 1 + x - \log{(1+2 x)} \right ]$$
This integral may be evaluated by substituting $u=1+2 x$, $x=(u-1)/2$, to get
$$\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right ) \left [\log{3}-1+\frac{u-1}{2} - \log{u}\right]$$
Now,
$$\begin{align}\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right )(\log{3}-1) &= \left(\frac{7}{3}-2 \log{3}\right) ( \log{3}-1) \\ &= -\frac{7}{3} + \frac{13}{3} \log{3} - \log^2{3}\end{align}$$
$$\begin{align}\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right ) \log{u} &= 2 \int_1^3 du \frac{\log{u}}{u^2} - 2 \int_1^3 du \frac{\log{u}}{u} + \frac12 \int_1^3 du \, \log{u} \\ &= 2 \left [ - \frac{\log{u}}{u}\right]_1^3 + 2 \int_1^3 \frac{du}{u^2} - [\log^2{u}]_1^3 + \frac12 [u \log{u}-u]_1^3 \\ &= \frac13 + \frac{5}{6} \log{3} - \log^2{3}\end{align}$$
$$\begin{align}\frac12 \int_1^3 du \left (\frac{4}{u^2} - \frac{4}{u}+1\right ) \frac{u-1}{2} &= \frac14 \int_1^3 du \left (-\frac{4}{u^2} + \frac{8}{u} - 5+u \right )\\ &= 2 \log{3}-\frac{13}{6}\end{align}$$
Adding the above three results together, I get for the desired limit:
$$\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \left (\frac{n-2 k}{n+2 k} \right )^2 \, \sum_{m=k+1}^n \frac{n-2 m}{n+2 m} = -\log^2{3} + \frac{33}{6} \log{3} - \frac{29}{6}$$
which is about $0.002085$.
| {
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Fair and Unfair coin Probability I am stuck on this question.
A coin with $P(H) = \frac{1}{2}$ is flipped $4$ times and then a coin with $P(H) = \frac{2}{3}$ is tossed twice. What is the probability that a total of $5$ heads occurs?
I keep getting $\frac{1}{6}$ but the answer is $\frac{5}{36}$.
Attempt: $P($all heads on the four coins$)P($either one of the tosses is heads on the two coins$)+P(3$ heads on the four coins$)P($both coins are heads$)$
$P($all heads on the four coins$) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
$P($either one of the tosses is heads on the two coins$) = 1-P($no heads on both tosses$) = 1-\frac{1}{3}\cdot\frac{1}{3} = \frac{8}{9}$.
$P($exactly $3$ heads on the four tosses$) = \frac{1}{4}$.
$P($both coins are heads$) = \frac{2}{3}\cdot\frac{2}{3} = \frac{4}{9}$.
Final Equation: $\frac{1}{16}\cdot\frac{8}{9}+\frac{1}{4}\cdot\frac{4}{9} = \frac{1}{6}$.
Why am I off by $\frac{1}{36}$?
| The required probability will be
$P($exactly $4 $ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $1 $ head from the $2$ flips of $2$nd coin $)+$
$P($exactly $3 $ heads from the $4$ flips of $1$st coin$)\cdot P($exactly $2 $ heads from the $2$ flips of $2$nd coin $)$
Using Binomial Distribution, the required probability
$$\binom44\left(\frac12\right)^4\left(1-\frac12\right)^{4-4} \cdot\binom21\left(\frac23\right)^1\left(1-\frac23\right)^{2-1}$$
$$+\binom43\left(\frac12\right)^3\left(1-\frac12\right)^{4-3} \cdot\binom22\left(\frac23\right)^2\left(1-\frac23\right)^{2-2}$$
$$=\frac1{16}\cdot\frac49+\frac14\cdot\frac49=\frac5{36}$$
| {
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How to solve this integral $\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx$? I will be grateful if you would write me a solution procedure for this integral
$$\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx. $$
I am sure that an antiderivative is
$$\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I put $+\infty $ instead of $x$ I get
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right)-\frac{1}{4} \left( \frac{2 \log 1}{1}-\log 1 \right) \\
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right).
\end{align*}
As you can see, it is useless. Can you help me please? Thanks
Can I use this solution below?
Let $$I=\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I calculate the limit of I i get:
$$\lim_{x\to\infty}I=0$$
So the final result is
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}=0
\end{align*}
| Putting $x=\tan\theta$ we get,
$$I=\int_0^{\frac\pi2}\frac{\tan\theta\ln \tan \theta}{\sec^2\theta}d\theta$$
$$=\frac12\int_0^{\frac\pi2}\sin2\theta\ln \tan \theta d\theta$$
Now as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
$$I=\frac12\int_0^{\frac\pi2}\sin2\left(\frac\pi2+0-\theta\right)\theta\ln \tan \left(\frac\pi2+0-\theta\right) d\theta$$
$$=-\int_0^{\frac\pi2}\sin2\theta \ln\tan \theta d\theta=-I$$
as $\sin(\pi-2\theta)=\sin2\theta$
and $\tan \left(\frac\pi2+0-\theta\right)=\cot\theta=(\tan\theta)^{-1}\implies \ln \tan\left(\frac\pi2+0-\theta\right)=-\ln\tan \theta$
$$\implies I=0$$
| {
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Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed less than $20$?
| Start from $\sqrt{n}-\sqrt{n-1}$. Multiply top and "bottom" by $\sqrt{n}+\sqrt{n-1}$.
We get $\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$, which is $\gt \dfrac{1}{2\sqrt{n}}$. It follows that $\dfrac{1}{\sqrt{n}}\lt 2(\sqrt{n}-\sqrt{n-1})$.
Add up, $n=1$ to $n=100$. We get
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{100}}\lt 2(\sqrt{1}-\sqrt{0})+2(\sqrt{2}-\sqrt{1})+\cdots +2(\sqrt{100}-\sqrt{99}).$$
Observe the mass cancellation on the right: it collapses to $20$.
| {
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Rationalizing expressions In my precalc book, I have the following problem:
Calculate $a+b+c$ if $a,b,c\in\mathbb{Q}$ and
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$
I think that the RHS can stay untouched, while operating the LHS, but I can't find a way to factor $\sqrt[3]{2}-1$ as the third power of something. Any help is greatly appreciated.
With the help of Olegg, i got the solution
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$
$$\sqrt[3]{\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$
$$\sqrt[3]{\frac{1}{(\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}})^3}}$$
$$\frac{1}{\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}}}$$
$${\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}}$$
$$a+b+c=\frac{1}{3}$$
| Hint.
$(a,b,c) = \Bigl(\dfrac{1}{9},-\dfrac{2}{9},\dfrac{4}{9}\Bigr)$ $-$ one of rational solutions (ignoring permutations).
So, $a+b+c=\dfrac{1}{3}$.
| {
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$3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. Is my induction solution correct? Show using mathematical induction that $3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. I'm not sure whether what I did at the last is valid?
Basis step:
for all non-negative integers
$$P(n) = 3^{3n+1} < 2^{5n+6} $$
$$P(0) = 3^{3(0) + 1} = 3 < 64 = 2^{5(0) + 6}$$
$$P(0) = T$$
Inductive Step:
Assume: $3^{3k+1} < 2^{5k+6}$
Show: $3^{3(k+1)+1} < 2^{5(k+1)+6}$
$$ 3^{3(k+1)+1} = 3^{3k+4} = 3^3 \cdot 3^{3k+1}$$
By inductive hypothesis~
$$3^3 \cdot 3^{3k+1} < 3^3 \cdot 2^{5n+6} $$
This is the part where I'm not sure if you can do this in induction but it seems logically correct.
$$3^3 \cdot 2^{5n+6} = 27 \cdot 2^{5n+6}$$
$$2^{5(k+1)+6} = 2^{5k+5+6}= 2^5 \cdot 2^{5n+6} = 32 \cdot 2^{5n+6}$$
I'm not sure whether it should be $\le$ or $<$ but I used '$<$' for $3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $
Therefore:
$$3^{3(k+1)+1} < 3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $$
| Excellent work:
You can conclude, since you have shown
$$3^{3(k+1)+1}\;\; <\;\; 3^3 \cdot 2^{5n+6} \;\;{\color{blue}{\bf <}}\;\; 2^{5(k+1)+6}$$
or simply, $$3^{3(k+1)+1}\; {\color{blue}{\bf <}}\; 2^{5(k+1)+6}$$ as desired.
The "blue" strict inequality is all you need. You have shown, prior to your conclusion, that $$3^{3(k+1)+1}\;\; <\;\; 3^3 \cdot 2^{5n+6}$$ and $$3^3\cdot 2^{5n+6} \;\;<\;\; 2^{5(k+1)+6}$$
| {
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Understand integral from Gradshteyn and Ryzhik book "Table of integrals, series, products" I was checking useful integrals in this book. I have found one (6.298) that is what I need, but I don't understand how every step towards the final result works.
$$\int_0^{+\infty}\,\left[2\cosh(ab)-e^{-ab}\Phi\left(\frac{b-2ax^2}{2x}\right)-e^{ab}\Phi\left(\frac{b+2ax^2}{2x}\right)\right]\,x\,e^{-(\mu-a^2)x^2}\,\,dx=\frac{1}{\mu-a^2}e^{-b\sqrt{\mu}}$$
where $\Phi(x)=erf(x)$, $a,b>0$ and $Re\,\mu>0$.
Can anybody help me with the intermediate steps to get the final result?
Are there other conditions missing? Like $\mu-a^2>0$?
| This example is easier than I thought. Define:
\begin{equation}
f^{(a,b)}(x):= 2 \cosh(a b)- \exp(-a b) Erf[\frac{b- 2 a x^2}{2 x}] - \exp(a b) Erf[\frac{b+ 2 a x^2}{2 x}]
\end{equation}
Then
\begin{eqnarray}
&&\int\limits_0^\infty f^{(a,b)}(x) \cdot x \exp\left(-(\mu-a^2) x^2\right) dx=\\
&&
\int\limits_0^\infty \frac{1}{2} \frac{2 b e^{-\frac{4 a^2 x^4+b^2}{4 x^2}}}{\sqrt{\pi } x^2} \cdot \frac{\exp\left(-(\mu-a^2) x^2\right)}{\mu-a^2} dx=\\
&&\frac{b}{\sqrt{\pi }\left(\mu -a^2\right)} \int\limits_0^\infty\frac{e^{-\frac{b^2+4 \mu x^4}{4 x^2}}}{ x^2 } dx \quad (i)
\end{eqnarray}
where in the second line we integrated by parts and in the third line we simplified the result.
Now we have:
\begin{eqnarray}
(-b) \frac{e^{-\frac{b^2+4 \mu x^4}{4 x^2}}}{ x^2 } =
\left(-\frac{b}{2 x^2}+\sqrt{\mu}\right) e^{-\left(\frac{b}{2 x} + \sqrt{\mu} x\right)^2+b\sqrt{\mu}} +
\left(-\frac{b}{2 x^2}-\sqrt{\mu}\right) e^{-\left(\frac{b}{2 x} - \sqrt{\mu} x\right)^2-b\sqrt{\mu}}
\end{eqnarray}
Therefore by integrating both sides we get:
\begin{eqnarray}
(-b) \int \frac{e^{-\frac{b^2+4 \mu x^4}{4 x^2}}}{ x^2 } dx =
e^{b \sqrt{\mu}} \frac{\sqrt{\pi}}{2} Erf[\frac{b}{2 x}+\sqrt{\mu} x] + e^{-b\sqrt{\mu}} \frac{\sqrt{\pi}}{2} Erf[\frac{b}{2 x}-\sqrt{\mu} x]
\end{eqnarray}
Therefore the definite integral reads:
\begin{eqnarray}
(-b) \int\limits_0^\infty \frac{e^{-\frac{b^2+4 \mu x^4}{4 x^2}}}{ x^2 } dx =
e^{-b\sqrt{\mu}} \frac{\sqrt{\pi}}{2} (-2) \quad (ii)
\end{eqnarray}
because there is only a contribution from the second error function on the right hand side. Now $(i)$ along with $(ii)$ yields the required result.
| {
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Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$? Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$?
solution:
\begin{align*}
fgg(n) &= f(g(g(n)))\\
&= f(g(3n+4)))\\
&= f(3(3n + 4) + 4)\\
&= f(9n + 16)\\
&= 2(9n + 16)^2 - 3\\
&= 2(81n^2 + 288n + 256) - 3 \\
&= 162n^2 + 576n + 509
\end{align*}
Check: Try $n = 1$, $f(g(g(1)) = f(g(7)) = f(25) = 2(25)^2 - 3 = 1247$,
$162(1^2) + 576(1) + 509 = 1247$
The practice test multiple choice answers are...
A. $6n^3+8n^2-9n-12$
B. $n^4+4n^3+4n^2+16n$
C. $-6n^3+8n^2+9n-12$
D. $n^4+4n^3+4n^2+16n$
Which one is it? What am I doing wrong?
| Let's check what @amWhy noted by using Maple. It is a fun.
> f:=n-> 2n^2-3:
> g:=n-> 3n+4:
> s:=n->(g@g)(n):
> simplify((f@s)(n));
162n^2+576n+509
| {
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Narcissistic numbers in other bases It is well known that $153$ is a narcissistic number; that is, it is equal to the sum of the cubes of its digits since $153=1^3+5^3+3^3$.
Other bases have similar numbers. For example, in base $3$, seventeen is $122$; and in base $4$, thirty-five is $203$.
Let $B_3$ be the set of bases with no such [edit] three-digit numbers. The first two members of $B_3$ are $2$ and $72$.
Why is every member of $B_3$ except $2$ a multiple of $9$?
| Proof by magic:
$$
\begin{eqnarray}
(k+1)^3 & + & 0^3 & + & (2k+1)^3 & = & (3k+1)^2 (k+1) & + & (3k+1)0 & + & (2k+1) \\
k^3 & + & 0^3 & + & (2k+1)^3 & = & (3k+2)^2k & + & (3k+2)0 & + & (2k+1) \\
(5k+1)^3 & + & (4k+2)^3 & + & (6k+2)^3 & = & (9k+3)^2(5k+1) & + & (9k+3)(4k+2) & + & (6k+2) \\
(7k+5)^3 & + & (2k+1)^3 & + & (6k+4)^3 & = & (9k+6)^2(7k+5) & + & (9k+6)(2k+1) & + & (6k+4)\\
\end{eqnarray}
$$
The first two identities show that any base not divisible by $3$ admits at least one three-digit narcissistic number; with the sole exception of base $2$ (the number resulting from the second identity would be $001$; not a proper three-digit one). The other two lines cover bases which are multiples of three but not multiples of $9$; again showing that each of them admits at least one narcissistic number. Thus, only bases which are multiples of $9$ can possibly not admit any narcissistic number (well, other than base $2$, of course).
| {
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Operator Norm of a Matrix composed of Standard Basis and Fourier Basis Let $\mathbf{A}_n$ be an $n\times 2n$ matrix (where $n=2^k$) composed of Fourier basis and standard basis; that is,
$$\mathbf{A}_n = \begin{bmatrix}\mathbf{I}_n & \mathbf{F}_n\end{bmatrix}$$
where $\mathbf{I}_n$ is the identity matrix and $\mathbf{F}_n$ is the DFT matrix.
Generally, a DFT matrix $\mathbf{F}_n$ is defined as
$$
\mathbf{F}_n = \frac{1}{\sqrt{n}}
\begin{bmatrix}
\omega^0 & \omega^0 & \omega^0 & \cdots & \omega^0 \\
\omega^0 & \omega^1 & \omega^2 & \cdots & \omega^{n-1} \\
\omega^0 & \omega^2 & \omega^4 & \cdots & \omega^{2(n-1)} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\omega^0& \omega^{n-1}& \omega^{2(n-1)}& \cdots & \omega^{(n-1)(n-1)}
\end{bmatrix}
$$
where $\omega = \exp(-2\pi i / n)$ is the primitive $n^\mathrm{th}$ root of unity.
For example, if $n=2^2$, $\mathbf{A}_4$ is the concatenation of $\mathbf{F}_4$ and $\mathbf{I}_4$:
$$
\mathbf{A}_4=
\begin{bmatrix}
1 & 0 & 0 & 0 & 1/2 & 1/2 & 1/2 & 1/2 \\
0 & 1 & 0 & 0 & 1/2 & -i/2 & -1/2 & i/2 \\
0 & 0 & 1 & 0 & 1/2 & -1/2 & 1/2 & -1/2 \\
0 & 0 & 0 & 1 & 1/2 & i/2 & -1/2 & -i/2
\end{bmatrix}
$$
One can easily obtain $\mathbf{F}_4$ by plugging in $\omega = \exp(-2\pi i / 4) = -i$.
I'm interested in computing the operator norm of $\mathbf{A}$ defined as:
$$||\mathbf{A}||_{op} = \max_{||\mathbf{x}||_2 = 1} ||\mathbf{A}\mathbf{x}||_2$$
where $||\mathbf{x}||_2$ is the Euclidean norm of the vector $\mathbf{x}$.
Do we have a closed form of $||\mathbf{A}_n||_{op}$ for any $n$?
| The singular values of $(I,F)$ are the square roots of the eigenvalues of $(I,F)(I,F)^\ast=I+FF^\ast$. Since the DFT matrix $F$ is unitary, $I+FF^\ast=2I$. Consequently, all singular values and in turn the operator $2$-norm of $(I,F)$ are equal to $\sqrt{2}$, regardless of the size $n$.
| {
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Roots of $8x^3-4x^2-4x+1$ It is known that the roots of polynomial $8x^3-4x^2-4x+1$ are $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$ and $\cos\frac{5\pi}{7}$.
However this is what Wolfram Alpha/Wolfram Mathematica gives:
$$x = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
\sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i
\sqrt{3}\right)}$$
$$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i
\sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i
\sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$
$$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1-i
\sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1+i
\sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$
These must be the same roots, but is there a way to show it? Or at least a way to show that these roots given by Wolfram are real?
Can you show that
$$\cos\frac{\pi}{7} = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
\sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i
\sqrt{3}\right)}$$
| One can show for example, that by putting $a= 2 \cos(\pi/7)$, and $b=a^2-1$, that if $ab=a+b$ and $b^2=1+a+b=1+ab$, then you are indeed dealing with the chords of the heptagon, and that these values are correct.
It is relatively easy to show by using coins on a table, that these two relations correspond to the solutions that $a$ and $1/b$ solve $x^3 - x^2 - 2x + 1$, in which case $a/2$ solves the equation in the OP.
When i saw the expansion of the root into complex numbers, what immediately sprang to mind was the construction of the heptagon using a trisector, given in page 199 of Conway + Guy 'The book of Numbers'. The basic construction involves a hexagonal lattice (Eisenstein integers), with a circle of radius 2 centred on some point, this becomes the circum-circle.
One constructs the x-axis through the centre, and a 'y-axis', passing through the first point going through the point U at cis(120 degrees)(ie $(-1+\sqrt{-3})/2$.
The point V is an other lattice point, at $-3 \sqrt{-3})$. The line through U and V intersects the 'y-axis'. This minimal angle is trisected, the cut nearer the 'y-axis' is extended to the x-axis, as are the two more lines, that make 60-degrees with the first.
These strike the x-axis, and perpendiculars from these three points strike the vertex-curve at the vertices of the heptagon.
| {
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Prove two similiar integral equalities I am looking to show that
$$ \begin{align*}
I_1 & = \int_{ma}^{na} \frac{ \log(x - a) }{x^2 + a^2 }
= \frac{\log (2a^2)}{2a} \bigl[ \arctan n + \arctan m\bigr] \\
I_2 & = \int_{a/m}^{a/n} \frac{\log(x + a) }{ x^2 + a^2 }
= \frac{\log (2a^2)}{2a}\bigl[ \arctan \frac{1}{m} + \arctan \frac{1}{n} \bigr]
\end{align*} $$
Given that $nm = n + m + 1$. Now here the substitutions suggested to solve the integrals are given as
$$ I_1 \, \text{let} \, x = \frac{at + a^2}{t - a}
\quad \text{and} \quad
I_2 \, \text{let} \, x = \frac{-at + a^2}{t + a}
$$
but even after using these substitutions I am not closer to a solution.
Can anyone give some hints on evaluating the integrals? (I already have proved the equality using the method in the first post, but I was intrigued by the integrals in the second post.
Even more so by the fact that the poster seems to trivialize these integrals, and that they stumpled all of my CAS tools. Are there any clever, and smart ways to attack these integrals?
| Hint:
Integrate by parts:
$$\dfrac{1}{a}\int{\ln{(x-a)}\arctan{\left(\dfrac{x}{a} \right)}\ dx}$$
and
$$\dfrac{1}{a}\int{\ln{(x+a)}\arctan{\left(\dfrac{x}{a} \right)}\ dx}.$$
Addition
For example, in the second integral we put
$u=\arctan{\left(\dfrac{x}{a} \right)}, \;\;dv=\ln{(x+a)}\ dx.$ Then
$$\dfrac{1}{a}\int{\ln{(x+a)}\arctan{\left(\dfrac{x}{a} \right)}\ dx}= \\
=
\dfrac{1}{a}\arctan{\left(\dfrac{x}{a} \right)} \int{\ln{(x+a)}\ dx} - \int\ln{(x+a)} \cdot \dfrac{1}{a^2} \cdot \dfrac{1}{1+\frac{x^2}{a^2}}\ dx= \\
=\dfrac{1}{a}\arctan{\left(\dfrac{x}{a} \right)} \int{\ln{(x+a)}\ dx} - \int\ln{(x+a)} \dfrac{1}{x^2+{a^2}}\ dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $\int^1_0 \frac2{\sqrt{2-x^2}} dx$ $$\int^1_0 \frac2{\sqrt{2-x^2}} dx$$
using substitution $x=\sqrt 2 \sin \theta$
$$\int^{\pi/4}_0 \frac{2\cos \theta d\theta}{\sqrt{2-2\sin^2 \theta}} = \int^{\pi/4}_0 \frac{2\cos\theta d\theta}{\sqrt2 \cos\theta} = \int^{\pi/4}_0 \frac{2d\theta}{\sqrt2} = \int^{\pi/4}_0 \frac{2\cdot\sqrt2 d\theta}{\sqrt2\cdot\sqrt2} = \int^{\pi/4}_0 \sqrt2d\theta = \sqrt2 \theta = \sqrt2 (\frac\pi4 - 0) = \sqrt2 \frac\pi4 $$
The problem is that the answer is $\frac\pi2$. Where did I make a mistake?
UPDATE:
using substitution $x=\sqrt 2 \sin\theta \rightarrow dx=\sqrt2\cos\theta$
$$\int^{\pi/4}_0 \frac{2\sqrt2\cos\theta}{\sqrt{2-2\sin^2\theta}} = \int^{\pi/4}_0 \frac{2\sqrt2\cos\theta}{\sqrt2 cos\theta} = \int^{\pi/4}_0 \frac{2\sqrt2}{\sqrt2} = \int^{\pi/4}_0 2 = 2 \theta = 2 (\frac\pi4 - 0) = \frac\pi2 $$
| you should substitute $dx$ by $\sqrt{2}\cos t dt$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Continued fraction for $\tan(nx)$ I found this beautiful continued fraction expansion of $\tan(nx)$, $n$ being a positive integer, online but I don't remember the source now:
$\displaystyle \tan(nx) = \cfrac{n\tan x}{1 -\cfrac{(n^{2} - 1^{2})\tan^{2}x}{3 -\cfrac{(n^{2} - 2^{2})\tan^{2}x}{5 -\cfrac{(n^{2} - 3^{2})\tan^{2}x}{7 -\cdots}}}}$
the last term in the continued fraction being $\dfrac{(n^{2} - (n - 1)^{2})\tan^{2}x}{(2n - 1)} = \tan^{2}x$. For example for $n = 3$ we have
$\displaystyle \tan 3x = \cfrac{3\tan x}{1 -\cfrac{8\tan^{2}x}{3-\cfrac{5\tan^{2}x}{5}}}$
which is correct and can be verified easily. I would like to know a simple proof via elementary trigonometry and algebra.
I had asked this question long back on NRICH(https://nrich.maths.org/discus/messages/153904/145455.html) but did not get any helpful answer. The reason I wish to know the proof is that it will lead me to an elementary proof of continued fraction expansion of $\tan z$. Putting $nx = z$ where $z$ is kept constant and $n$ is made to tend to $\infty$ (so that $x \to 0$) we get
$\displaystyle \tan z = \cfrac{z}{1-\cfrac{z^{2}}{3 -\cfrac{z^{2}}{5-\cfrac{z^{2}}{7-\cdots}}}}$
instead of the complicated proof by Lambert given in my blog http://paramanands.blogspot.com/2011/04/continued-fraction-expansion-of-tanx.html
I have got another formula from an exercise in "A Course of Pure Mathematics" by G. H. Hardy:
$$\cos nx = 1 - \frac{n^{2}}{2!}\sin^{2}x - \frac{n^{2}(2^{2} - n^{2})}{4!}\sin^{4}x - \frac{n^{2}(2^{2} - n^{2})(4^{2} - n^{2})}{6!}\sin^{6}x - \cdots$$
I will try to find out if this formula can be used to prove the continued fraction expansion of $\tan(nx)$.
Update: This question was originally posed by Euler in 1813. See Chrystal's Algebra Vol 2 (page 526, problem 31).
Further Update: The following approach looks promising but I don't know how to complete it.
We have $\tan nx=A/B$ where $$A=\binom{n} {1}\tan x-\binom{n}{3}\tan ^3x+\dots$$ and $$B=\binom{n} {0}-\binom{n}{2}\tan ^2x+\dots$$ and thus $$\frac{\tan nx} {n\tan x} =\frac{C} {B} =\dfrac{1}{1+\dfrac{B-C}{C}}$$ where $$C=1-\frac{(n-1)(n-2)}{3!}\tan^2x+\dots$$ Next $$B-C=-\frac{n^2-1^2}{1!3}\tan^2x+\frac{(n^2-1^2)(n-2)(n-3)}{3!5}\tan^4x-\dots$$ and this can be rewritten as $$-\frac{n^2-1^2}{3}\tan^2x\left(1-\frac{(n-2)(n-3)}{2!5}\tan^2x+\dots\right)$$ If the expression in large parentheses above is $D$ then we have $$\frac{\tan nx} {n\tan x} =\dfrac{1}{1-\dfrac{(n^2-1^2) \tan^2x}{3+\dfrac{3C-3D}{D}}}$$ The problem now is to put $B, C, D$ into a common pattern so that we can easily get the next expressions like $E, F, \dots$.
Thanks to user Paul Enta (and his wonderful answer) the above approach is completed via the use of hypergeometric series. See my answer for details.
| This continued fraction can be obtained from the Gauss' continued fraction for the ratio of two hypergeometric fractions after some preparation. One can express
\begin{equation}
\tan nx = \frac{\sin nx}{\sin x}\frac{\cos x}{\cos nx}\tan x
\end{equation}
Using the Chebyshev polynomials of the first and of the second kind, with $z=\cos x$,
\begin{align}
T_n(z)&=\cos nx\\
U_{n-1}(z)&=\frac{\sin nx}{\sin x}
\end{align}
one may express
\begin{equation}
\tan nx = n\tan x\frac{zU_{n-1}(z)}{nT_n(z)}
\end{equation}
Denoting
\begin{equation}
A=\frac{zU_{n-1}(z)}{nT_n(z)}
\end{equation}
we introduce the representation of the polynomials in terms of hypergeometric functions here and here:
\begin{equation}
A=z\frac{{}_2F_1\left(-n+1,n+1;\frac{3}{2};\frac{1-z}{2} \right)}{{}_2F_1\left(-n,n;\frac{1}{2};\frac{1-z}{2} \right)}
\end{equation}
Now, to change the variable in order to obtain $-\tan^2x$, we use the quadratic transformation of the functions DLMF:
\begin{equation}
{}_2F_1\left(a,b;\frac{1}{2}(a+b+1);u\right)=(1-2u)^{-a}{}_2F_1\left(\frac{1}{2}a,%
\frac{1}{2}a+\frac{1}{2};\frac{1}{2}(a+b+1);\frac{4u(u-1)}{(1-2u)^{2}}%
\right)
\end{equation}
and thus
\begin{equation}
A=\frac{{}_2F_1\left( \frac{-n+1}{2},-\frac{n}{2}+1;\frac{3}{2};-\frac{1-z^2}{z^2} \right)}{{}_2F_1\left( \frac{-n}{2},\frac{-n+1}{2};\frac{1}{2};-\frac{1-z^2}{z^2} \right)}
\end{equation}
which can be written as
\begin{equation}
A=\frac{{}_2F_1\left( a,b+1;c+1;v \right)}{{}_2F_1\left(a,b;c;v\right)}
\end{equation}
where $a=(1-n)/2,b=-n/2,c=1/2,v=-\tan^2x$.
This ratio can be expressed as a Gauss's continued fraction:
\begin{equation}
\frac{{}_2F_1\left(a,b;c;z\right)}{{}_2F_1\left(a,b+1;c+1;z\right)}=t_{0%
}-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2}z}{t_{2}-\cfrac{u_{3}z}{t_{3}-\cdots}}}
\end{equation}
where
\begin{align}
t_k&=c+k\\
u_{2k+1}&=(a+k)(c-b+k)\\
u_{2k}&=(b+k)(c-a+k)
\end{align}
Here
\begin{align}
t_k&=\frac{2k+1}{2}\\
u_{2k+1}&=\frac{1}{4}\left( (2k+1)^2 -n^2\right)\\
u_{2k}&=\frac{1}{4}\left( 4k^2 -n^2\right)
\end{align}
then $u_p=\frac{1}{4}(p^2-n^2)$. We can express
\begin{equation}
A^{-1}=\frac{1}{2}-\cfrac{\frac{n^2-1^2}{4}\tan^2x}{\frac{3}{2}-\cfrac{(n^2-2^2)/4\tan^2x}{\frac{5}{2}-\cfrac{(n^2-3^2)/4\tan^2x}{\frac{7}{2}-\cdots}}}
\end{equation}
or
\begin{equation}
A^{-1}=1-\cfrac{(n^2-1^2)\tan^2x}{3-\cfrac{(n^2-2^2)\tan^2x}{5-\cfrac{(n^2-3^2)\tan^2x}{7-\cdots}}}
\end{equation}
Finally,
\begin{equation}
\tan nx=\cfrac{n\tan x}{1-\cfrac{(n^2-1^2)\tan^2x}{3-\cfrac{(n^2-2^2)\tan^2x}{5-\cfrac{(n^2-3^2)\tan^2x}{7-\cdots}}}}
\end{equation}
EDIT: The quadratic transform is valid for $(1-z)/2<1/2$ or $\cos x>0$. However the decomposition $\tan nx=An\tan x$ shows that $A$ must be invariant through the change $x\to \pi+x$, so we can assume $z>0$ in the transform.
| {
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"url": "https://math.stackexchange.com/questions/432771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Fractions in Questions and Answers
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