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An expert is a man who has made all the mistakes, which can be made, in a very narrow field. (Niels Bohr)


Jan
15
comment convergence of $\prod_{n=1}^\infty (1-\frac{z}{n!})$
@corciacandy What do you mean by 'order'? If you are talking about convergence order towards the convergence of the product is uniform or punctually you can use the M test Weiersstrass to achieve uniform convergence.
Jan
4
comment if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =?
@user21820 yes. I did not express myself properly. I wanted to say that for each root a new root is added. Thanks for pointing.
Jan
3
comment if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =?
What do you think strange is not strange. The truth is that every time you raise squared each algebraic expressions you enter one (1) new root in the equation. Example: $x=1\implies x^2=1\implies x=1 \mbox{or } x=-1$
Jan
3
comment C*-algebras: Literature?
@Freeze_S No problem. Added a more complete list.
Jan
2
comment $ \|u(x)+u(x)-x +o(\|x\|^p)\|<r $?
@Behaviour I edited my attempt. I think it is now clearer.
Dec
31
comment A question regarding a double series.
see jstor.org/stable/1967602?seq=1#page_scan_tab_contents
Dec
28
comment $ C^p_0(B,\mathbb{X})$ is a Banach space with the norm of $C^p\!\!$-topology?
@aes It Would Be $u$ is continuous with respect to norm $\|\;\cdot\;\|_{\mathbb{X}}$ and the $i$-th derivative $D^{(i)}u(x)$ is continuous with respect to norm $\|\;\cdot\;\|_{\mathcal{L}(\mathbb{X}^i,\mathbb{X})}$?
Dec
28
comment $ C^p_0(B,\mathbb{X})$ is a Banach space with the norm of $C^p\!\!$-topology?
@aes You mean you need to $\| u \|_{C^{\,0}}<\infty? $
Dec
4
comment find the upper bound on a vector
@user189013 yes, if $f:\mathbb{R}^d\to \mathbb{R}^d$ is $C^r$ for $r\geq 2$.
Dec
1
comment Prove that the sequence converges to zero
@daren Calculate and check. $\sqrt{n+3}-\sqrt n=\frac{\sqrt{n+3}+\sqrt n}{\sqrt{n+3}+\sqrt n}\cdot \sqrt{n+3}-\sqrt n= \ldots = \frac3{\sqrt{n+3}+\sqrt n}$
Nov
26
comment find the upper bound on a vector
Their function $f$ has all the partial derivatives of second order $\frac{\partial}{\partial x_i\partial x_j}f(x) $continuous to $x=(x_1,\ldots,x_d)$?
Nov
26
comment Help differentiate long equation
Do you want to derive twice with respect to alpha, correct? You got the first derivative with respect to alpha? It would be interesting if you write here the way you are trying to follow. Is not there some symmetry in their function or special property that make it easier the calculations enoug?
Nov
26
comment find the upper bound on a vector
If $x$ and $y$ are vectors in $\Re^d$ then what it means$(y-x)^2$?
Nov
21
comment How can we calculate $(x^x)'$
@Pauly B I made some comments. His response provides a better and more elegant expression for the question. +1
Nov
21
comment How can we calculate $(x^x)'$
Third, I did not ignore the warnings of the user AndersKaseorg. Simply say that a definition is wrong without pointing precisely where the misses does not help much. I gave the proper return. As I said recursive definition differs from that given only answer. But in essence is the same.
Nov
21
comment How can we calculate $(x^x)'$
Second, my answer must be looked at in its entirety. Using recursion is the main idea of my answer. Recursion is the key idea of the answer. Both the idea of my response was used in another answer.
Nov
21
comment How can we calculate $(x^x)'$
First, my definition of power towers differs from the definition given by the author in your question. But she's not wrong. Only different is the definition suggested in the question. Yes I admit that was a mistake on my part. A lack of attention.
Nov
21
comment How can we calculate $(x^x)'$
Hello guys. I would like to clarify three things. For lack of space in the comments I will divide the explanations into three comments. Already took a +1 in this response.
Nov
16
comment Extending the domain of the “greatest common divisor” function.
@SteveKass Thanks for the pointers. I corrected my little mistake. I meant for continuous $x>0$ and $y> 0$.
Nov
16
comment Extending the domain of the “greatest common divisor” function.
@JMoravitz Thanks for the pointers. I corrected my little mistake.