Search found 308 matches
- Tue Sep 19, 2017 7:48 pm
- Forum: Multivariate Calculus
- Topic: A tough limit
- Replies: 1
- Views: 3455
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...
- Thu Sep 14, 2017 11:19 am
- Forum: Multivariate Calculus
- Topic: A tough limit
- Replies: 1
- Views: 3455
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.
NOTE: I've got an (almost) solution.
- Wed Aug 09, 2017 3:16 pm
- Forum: Real Analysis
- Topic: A limit
- Replies: 4
- Views: 5516
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...
- Sun Jul 30, 2017 7:22 pm
- Forum: Real Analysis
- Topic: A limit
- Replies: 4
- Views: 5516
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...
- Tue Jun 27, 2017 8:29 am
- Forum: Complex Analysis
- Topic: Line integral 01
- Replies: 1
- Views: 4773
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...
- Sun Jun 25, 2017 5:18 am
- Forum: Multivariate Calculus
- Topic: Triple integral and ellipsoid
- Replies: 1
- Views: 3416
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*}...
- 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: 3474
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...
- Wed Jun 14, 2017 5:42 am
- Forum: Calculus
- Topic: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
- Replies: 2
- Views: 3625
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}{...
- Tue Jun 13, 2017 3:57 pm
- Forum: Calculus
- Topic: \(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
- Replies: 2
- Views: 3625
\(\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\)
And the cause of this:
Evaluate $$\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\,,\quad {n}\in\mathbb{N}\,,\, n\geqslant3\,.$$
Evaluate $$\int_{1}^{+\infty}{\frac{\log(x+1)}{x^{n}}\,dx}\,,\quad {n}\in\mathbb{N}\,,\, n\geqslant3\,.$$
- 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: 4128
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)...