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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)


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.
Nov
15
comment As fields $F$ not isomorphic with $\mathbb R$ , but as sets $F \sim \mathbb R$ , example ?
$\mathbb{R}$ and $\mathbb{C}$.
Nov
3
comment How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$?
@Winther Guiding me through the comments got a double set equal to the integral.
Nov
2
comment How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$?
@Winther Thank you for detailing your observations so readily. I think I can work with this idea and maybe get some interesting partial results.
Nov
2
comment How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$?
@Winther Around which point $x_o$ between $0$ and $1$ the Taylor expansion of $e^{x^n\log x}$ is centered?
Nov
2
comment Percentage of primes among the natural numbers
+1 The technique used in the addendum is simply wonderful. I saw the first time a variant of it here
Nov
2
comment How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$?
@Lucian +1 for your very interesting link.
Nov
2
comment How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$?
@Winther, thanks. But how you got this result?
Nov
2
comment How can we solve the integral $\int_\epsilon^1 e^{x^n\log x }\mathrm{d}x$?
@Winther I really do not know. Maybe for an appropriate value of $\epsilon>0$ we could discover a serie that converges to the same value of the integral.