Search found 597 matches
- Tue Nov 10, 2015 4:05 pm
- Forum: Calculus
- Topic: Logarithmic integral
- Replies: 2
- Views: 3113
Re: Logarithmic integral
Hey Grigoris, thank you for your solution. Here is another one which actually is able to crack with more ease the more general case: $$\int_0^1 \frac{\ln (1-x) \ln^k x}{1-x}\, {\rm d}x, \;\;\; k \geq 1$$ We have successively: $$\begin{align*} \int_{0}^{1}\frac{\ln (1-x)\ln x}{1-x}\, {\rm d}x &=-...
- Tue Nov 10, 2015 4:02 pm
- Forum: Number theory
- Topic: Irrational number
- Replies: 1
- Views: 7386
Irrational number
Let $N \in \mathbb{N} \mid N>1$. Prove that the number:
$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$
is irrational.
$$\mathcal{N}= \sqrt{1\cdot 2 \cdot 3 \cdot 4 \cdots (N-1)\cdot N}$$
is irrational.
Hidden message
- Tue Nov 10, 2015 3:59 pm
- Forum: Number theory
- Topic: Five last digits of number
- Replies: 4
- Views: 4529
Five last digits of number
Find the five last digits of the number:
$$\mathscr{N}=\underbrace{9^{9^{9^{\cdot^{\cdot^{\cdot^{\cdot^{9}}}}} }}}_{1001 \;\; \rm {nines}}$$
$$\mathscr{N}=\underbrace{9^{9^{9^{\cdot^{\cdot^{\cdot^{\cdot^{9}}}}} }}}_{1001 \;\; \rm {nines}}$$
- Tue Nov 10, 2015 2:09 pm
- Forum: General Mathematics
- Topic: There do not exist points
- Replies: 2
- Views: 2938
There do not exist points
Prove that there do not exist four points in $\mathbb{R}^2$ whose pairwise distances are all odd integers.
- Tue Nov 10, 2015 2:03 pm
- Forum: Calculus
- Topic: Logarithmic integral
- Replies: 2
- Views: 3113
Logarithmic integral
Evaluate the integral:
$$\int_0^1 \frac{\ln (1-x) \ln x}{1-x}\, {\rm d}x$$
$$\int_0^1 \frac{\ln (1-x) \ln x}{1-x}\, {\rm d}x$$
- Mon Nov 09, 2015 5:18 pm
- Forum: General Mathematics
- Topic: A sum!
- Replies: 4
- Views: 6600
A sum!
Evaluate the following sum:
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}}$$
Source
- Mon Nov 09, 2015 5:16 pm
- Forum: Real Analysis
- Topic: Convergence of a series
- Replies: 1
- Views: 2884
Convergence of a series
Let \( a_n =\underbrace{\sin \left ( \sin \left ( \sin \cdots (\sin x) \cdots \right ) \right )}_{n \; \rm {times}} \) and \( x \in (0, \pi/2) \). Examine if the series: $$ \mathcal{S}=\sum_{n=1}^{\infty} a_n $$ converges. Do the same question for the series: \( \displaystyle \mathcal{S}=\sum_{n=1}^...
- Mon Nov 09, 2015 5:08 pm
- Forum: Calculus
- Topic: Double integral
- Replies: 1
- Views: 2590
Double integral
Let $\displaystyle D= \left\{ \dfrac{1}{2} \leq x^2+y^2\leq 1, \; x^2+y^2-2x \leq 0 \;\;\; y \geq 0 \right\}$. Evaluate the double integral:
$$\iint \limits_{D} \frac{\log (x^2+y^2)}{\sqrt{x^2+y^2}}\; {\rm d} x\; {\rm d} y$$
$$\iint \limits_{D} \frac{\log (x^2+y^2)}{\sqrt{x^2+y^2}}\; {\rm d} x\; {\rm d} y$$
Hidden Message
- Mon Nov 09, 2015 5:04 pm
- Forum: General Mathematics
- Topic: Finite sums
- Replies: 0
- Views: 1913
Finite sums
These are two well known sums, but let them exist here as well.
a) $\displaystyle \sum_{k=1}^{m} \tan^2\left(\frac{k\pi}{2m+1}\right) = m(2m+1)$
b) $\displaystyle \sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$
a) $\displaystyle \sum_{k=1}^{m} \tan^2\left(\frac{k\pi}{2m+1}\right) = m(2m+1)$
b) $\displaystyle \sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$
- Mon Nov 09, 2015 2:57 pm
- Forum: General Topology
- Topic: Rendezvous value
- Replies: 3
- Views: 3557
Re: Rendezvous value
Hi Demetres, how about if we replace complete with compact? I think that the exercise now would be correct. I was thinking about this in the last $3$ days and I came up to the same conclusion that this cannot hold in $\mathbb{R}$ as you said. But I think I have a huntch that replacing complete with ...