Search found 308 matches

by Grigorios Kostakos
Tue Sep 19, 2017 7:48 pm
Forum: Multivariate Calculus
Topic: A tough limit
Replies: 1
Views: 2973

Re: A tough limit

The function $f(x,y)=x^y\,,\; (x,y)\in{\mathbb{R}}^2\,,$ is at least twice continuously differentiable in a open disk with centre $(1,1)$ and its 2nd degree Taylor polynomial is \begin{align*} P_{2,f,(1,1)}(x,y)&=f(1,1)+({\rm{grad}}\,{f})(1,1)\cdot(x-1,y-1)+\frac{1}{2}\,(x-1,y-1)\,H_f(1,1)\,(x-1...
by Grigorios Kostakos
Thu Sep 14, 2017 11:19 am
Forum: Multivariate Calculus
Topic: A tough limit
Replies: 1
Views: 2973

A tough limit

Examine if the limit $$\displaystyle\mathop{\lim}\limits_{(x,y)\to(1,1)}{\frac{x^y-y\,(x-1)-1}{x^2+y^2-2\,(x+y-1)}}$$ exists and if does exists, calculate it.


NOTE: I've got an (almost) solution.
by Grigorios Kostakos
Wed Aug 09, 2017 3:16 pm
Forum: Real Analysis
Topic: A limit
Replies: 4
Views: 4644

Re: A limit

...I am unable to check for any particular typos that may have occured during typesetting... The above note comes after an interchange of private messages. Let's make it more clear: There is some problem with limit $L$ : To be equal to $\xi$ (as been given), $\mathop{\lim}\limits_{n\to+\infty} \pro...
by Grigorios Kostakos
Sun Jul 30, 2017 7:22 pm
Forum: Real Analysis
Topic: A limit
Replies: 4
Views: 4644

Re: A limit

I think you need to give an initial value though. It's not necessary! Obviously assuming that $x_1\stackrel{(*)}{>}-1$, in any case the sequence $\{x_n\}_{n=1}^{\infty}$ is monotonic and bounded. $(*)$ If $x_1=-1$, then the sequence is the zero sequence and the fraction $\frac{\sqrt{1-\xi^2}}{\prod...
by Grigorios Kostakos
Tue Jun 27, 2017 8:29 am
Forum: Complex Analysis
Topic: Line integral 01
Replies: 1
Views: 3172

Re: Line integral 01

We give and a solution to that -not so difficult- exercise: The complex function ${\rm{Log}}\,\big(z-\tfrac{1}{2}\big)$ is holomorphic on ${\mathbb{C}}\setminus\big\{z\in{\mathbb{C}}\;|\;\Re(z)\leqslant\tfrac{1}{2}\,,\; \Im({z})=0\big\}$. comlineint.png [/centre] Because the curve $\gamma$ with par...
by Grigorios Kostakos
Sun Jun 25, 2017 5:18 am
Forum: Multivariate Calculus
Topic: Triple integral and ellipsoid
Replies: 1
Views: 2963

Re: Triple integral and ellipsoid

Considering the change of cooordinates \begin{align*} \left\{\begin{array}{l} x=a\,r\sin\vartheta\cos\varphi\\ y=b\,r\sin\vartheta\sin\varphi\\ z=c\,r\cos\vartheta \end{array}\right\}\,,\quad r\in[0,+\infty)\,,\; \vartheta\in[0,\pi], \; \varphi\in[0,2\pi]\,, \end{align*} with Jacobian \begin{align*}...
by Grigorios Kostakos
Wed Jun 14, 2017 1:05 pm
Forum: Calculus
Topic: \(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)
Replies: 1
Views: 2966

Re: \(\int_{0}^{+\infty}\frac{\log^2{x}}{(x+\alpha)(x+\beta)}\,dx\)

For the sake of discussion I give the next step: Considering the complex function $$f(z)=\dfrac{{\rm{Log}}^3{z}+\pi^2{\rm{Log}}{z}}{(z-\alpha)(z-\beta)}\,,\quad z\in\mathbb{C}\setminus\{x+0i\;|\;x\leqslant0\},$$ by the same procedure as in \(\int_{0}^{+\infty}\frac{\log{x}}{(x+\alpha)(x+\beta)}\,dx...
by Grigorios Kostakos
Wed Jun 14, 2017 5:42 am
Forum: Calculus
Topic: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
Replies: 2
Views: 3205

Re: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)

Nice solution mathofusva! Here is my attempt for $n>2$. \begin{align*} \int_{1}^{+\infty}{\frac{\log(x+1)}{x^n}\,dx} &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\frac{1}{x}}\\ {dx\,=\,-\frac{1}{t^2}dt}\\ \end{subarray}}\,-\int_{1}^{0}{t^n\log\big(\tfrac{1}{t}+1\big)\frac{1}{...
by Grigorios Kostakos
Tue Jun 13, 2017 3:57 pm
Forum: Calculus
Topic: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
Replies: 2
Views: 3205

\(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)

And the cause of this:

Evaluate $$\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\,,\quad {n}\in\mathbb{N}\,,\, n\geqslant3\,.$$
by Grigorios Kostakos
Tue Jun 13, 2017 1:02 pm
Forum: Calculus
Topic: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)
Replies: 3
Views: 3783

Re: \(\psi\big(\frac{n+1}{2}\big)-\psi\big(\frac{n}{2}\big)\)

I suppose you don't want Gauss digamma theorem to be used, e? , but rather a more elegant one. Toiis, I seek something about the difference $\psi\big(\tfrac{n+1}{2}\big)-\psi\big(\tfrac{n}{2}\big)$ and the more elegant I find is $$\displaystyle\psi\big(\tfrac{n+1}{2}\big)-\psi\big(\tfrac{n}{2}\big)...