All Questions
Tagged with sequences-and-series logarithms
454
questions
3
votes
3
answers
388
views
$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$ as a limit of a sum
Working on the same lines as
This/This and
This
I got the following expression for the Dilogarithm $\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$:
$$\operatorname{Li}_{2} \left(\frac{1}{e^{\...
8
votes
1
answer
276
views
Approximating $\log x$ by a sum of power functions $a x^b$
Let's approximate $\log x$ on the interval $(0,1)$ by a power function $a x^b$ to minimize the integral of the squared difference
$$\delta_0(a,b)=\int_0^1\left(\log x-a x^b\right)^2dx.\tag1$$
It's ...
2
votes
2
answers
186
views
Closed form of the integral $\int_{0}^{1} \log^n \left (\frac {1-x}{1+x}\right )dx$
I found this nice integral
$$i=\int_{0}^{1} \log^3\left (\frac {1-x}{1+x}\right)\;dx\tag{1}$$
on youtube but I don't remember where.
Let us generalize a bit to a power $n=0, 1, 2, ...$ and ask for the ...
-1
votes
2
answers
43
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Show that if $x>1$, $\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^4}-\dfrac{1}{6x^6}-\cdots$
Show that if $x>1$, $\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-\dfrac{1}{4x^4}-\dfrac{1}{6x^6}-\cdots$
$\log_e\sqrt{x^2-1}=\dfrac{1}{2}\log_e(x^2-1)=\dfrac{1}{2}[\log_e(x+1)+\log_e(x-1)]$
I know:
...
3
votes
1
answer
82
views
Showing the quintessential logarithm property using the Maclaurin series of $\log$
For $-1\le x<1$, we have
$$\log(1-x) = -\sum_{k=1}^{\infty} \frac{x^{k}}{k}\\$$
Taking $a,b$ with $|a|,|b|<1$ and $(1-a)(1-b)\le2$, on the function side clearly we have
$$\log(1-a)+\log(1-b) = \...
0
votes
1
answer
56
views
Equality after exponential function swap
The equation $ 2^x = x^2 $ has two real obvious solutions $(x=2,4)$ and another root not so obvious $x\approx- 0.766665$;
Assuming existence and uniqueness, extending equality $f(x)= g(x)$ between ...
3
votes
0
answers
102
views
How to tell if an infinite series sum will be rational or irrational?
I plugged in the following series into a calculator: $$\sum_{n=1}^\infty \ln(1+\frac{1}{n^2})$$ and got a result of approximately $1.29686$. That's nice and all, but I want to know: is this result ...
0
votes
1
answer
93
views
Alternating sum of logarithms [duplicate]
I'm trying to solve the product $\prod_{k=1}^{\infty} \frac{2k}{2k+1}$:
I've started by letting it be equal to P:
P = $\prod_{k=1}^{\infty} \frac{2k}{2k+1}$ $\implies$ ln(P) = $\sum_{k=1}^{\infty} \...
2
votes
2
answers
146
views
Is there a closed form solution to $\sum_{n=1}^{\infty}\frac{\ln\left(\frac{n+1}{n}\right)}{n}$ [duplicate]
While coming up with an idea for another way to milk the integral in my previous question, I got stuck at this summation: $$\sum_{n=1}^{\infty}\frac{\ln\left(\frac{n+1}{n}\right)}{n}$$
I do not know ...
1
vote
2
answers
70
views
Asympototic estimation of log log function
Let $N>0$ be an integer and consider the sum $\sum_{n=3}^N(\log \log N-\log \log n)$. It is not hard to see that this sum has the complexity $O(N^2)$ since $\log \log x <\log x<x$, so that ...
2
votes
1
answer
158
views
Iterating $\log(x\log(x\log(...)))$
For a real positive $x$, let $F(x)$ denote the sequence
$$
\left(\log x,\log(x\log(x)),\log(x\log(x\log(x))),\log(x\log(x\log(x\log(x)))),...\right),
$$
stopping at the first nonpositive value or ...
1
vote
1
answer
125
views
What is this "periodic" sequence called?
I came across this weird "periodic" sequence (containing only natural numbers) where the first 15 elements are
$$1, 1, 2, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8.$$
The sequence is not finite, ...
0
votes
1
answer
88
views
Series of functions $\log(1+x^{2n})$
Does the series of functions $\sum_{n=0}^{+\infty} \log(1+x^{2n})$ converges for $|x|<1$?
Obviously I've observed that for $x=0$ the series is convergent and for $x\ge1$ is divergent.
How should I ...
6
votes
0
answers
338
views
When does this iterated logarithm series converge / diverge?
Question: Let $b > 1$ be a positive real, let $\ell_b(n) = \max(1, \lfloor \log_b n \rfloor)$, and let $f_b(n) = n \ell_b(n) \ell_b^2(n) \dots $ (where we iterate $\ell_b$ until we hit $1$). For ...
1
vote
1
answer
48
views
solving equation with limit: $1+x =\lim_{n\to\infty}(1+\frac{y}{n})^n => y = \lim_{n\to\infty}n((1+x)^\frac{1}{n} - 1)$
I'm trying to understand a part of logarithmic series expansion proof below:
The logarithmic function is the inverse of the exponential function, so that
$y=ln(1+x)$ implies $1+x=e^y = \lim_{n\to\...