Timeline for Multivariate Taylor Series equals iterated univariate Taylor Series
Current License: CC BY-SA 4.0
14 events
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| May 10 at 18:46 | comment | added | Christian Remling | @WaveL: That doesn't matter here since (for differentiable functions, which your example is not) $\partial f(x,y)/\partial x$ at $(x,y)=(0,0)$ is the same whether you first take the derivative and then set $x=y=0$ or first fix $y=0$, differentiate the resulting function of $x$ and finally take $x=0$. | |
| May 10 at 14:45 | comment | added | WaveL | @ChristianRemling most likely I do not understand the topic in enough detail, but why is it „obvious“? My issue is, that in the iterated version substitutions occur in sequence whereas in in the multivariate setting they happen simultaneously. So for $\frac{x}{y}$ the simultaneous substitution with zero is not defined, the one where I do y first also not but the one where I first set x to zero works. Couldn’t this lead to an issue? | |
| May 10 at 14:34 | comment | added | WaveL | @Iosif: I am sorry that I am somehow not able to answer your questions precisely. I think my issue really is about points where the actual function f is not defined. So for my example comment, U would be all points where the denominator does not become zero. However I „still want to develop a Taylor series at this singularity“ and I thought this might lead into problems, that the expressions above are not well-defined anymore. | |
| May 10 at 13:38 | comment | added | Christian Remling | We would surely want to assume at the very least that $f\in C^{\infty}$ near zero, and then the two methods obviously produce the same formal object. Convergence is a separate issue of course. | |
| May 10 at 13:36 | history | edited | YCor |
edited tags
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| May 10 at 13:06 | comment | added | Iosif Pinelis | Your "For example" comment does not answer my questions. | |
| May 10 at 12:58 | comment | added | WaveL | @Iosif For example for $f(x,y) = \frac{xy^3}{x^2+y^2}$ the first iteration yields $T_f(x,y) = 0 + yx - \frac{6}{y} x^3 + \frac{120}{y^3} x^5 ...$. Then the second iteration $T_{T_f}(x,y)$ does not exist as the first derivative of $\frac{-6}{y} x^3$ with respect to $y$ at $y=0$ does not exist hence we cannot construct the second iteration as described in the above scheme. | |
| May 10 at 12:52 | history | edited | WaveL | CC BY-SA 4.0 |
added missing factorials.
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| May 10 at 12:42 | comment | added | Iosif Pinelis | What is $U$ here? What do you mean by "series [...] does not exist"? Exist in what sense? | |
| May 10 at 12:23 | comment | added | Iosif Pinelis | Factorials are missing here. | |
| May 10 at 11:53 | history | edited | WaveL | CC BY-SA 4.0 |
fixed some invalid sentences; fixing typos
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| May 10 at 11:47 | history | edited | WaveL | CC BY-SA 4.0 |
added 42 characters in body
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| S May 10 at 11:40 | review | First questions | |||
| May 10 at 11:59 | |||||
| S May 10 at 11:40 | history | asked | WaveL | CC BY-SA 4.0 |