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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

Hello,

First sorry for my english, i'm not good in it.

i've got a little problem when learning Taylor Series,

Somewhere i read that the n derivation of a function is same as n derivation of functions taylor series but that here are few excuses.

For example this function f(x)=0 x=0, f(x)=e^−(1/x^2) x!=0 is supposed to doesnt equal its taylor series? Is there any proof or explanation for it?

Thanks a lot

Calculus beginner

*Last edited by Ricsie (2012-11-19 08:13:06)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi;

Is the the function you want to Taylorize?

That is x factorial on the end? This is highly unlikely that you would be asked for the Taylor series of such a function.

It is more likely this is what you want:

Yes?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

Its this function:

i wanted to say if x is 0 then the f(x) is 0 and if x is not 0 then the f(x) is the one above

Our professor of math told me that the taylor series arent same as this function, but he didnt explain me why and how. He told me a hint to proof that this function and its taylor series arent same should be to make sme basic derivation npot for numbers but globally for n. which is pretty hard and im not sure this way i cna make the proof that the taylor series arent same as the function

By taylor series not being same as function i mean that usually if you derivate your taylor series its looking more and more like the function from derivation to derivation and in n infinite derivation it should be same as the function, and in this function it doesnt work, its taylor series doesnt going to look like the function

Hope wrote it easy to understand :-) and hope i havent made any theoretical mistakes

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi;

If you are trying to expand that around 0 it is called a Mclaurin series instead of a Taylor series.

Also you can say a series converges or does not converge to the function.

In this case that function has a singularity at zero, so you can not expand as a Taylor series but you can expand it as a Laurent series.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

I thought the taylor series for this function are(on image)

So theres no taylor series for this function? And just laurent series?

Because the question was, give me a proof that taylor series of this function doesnt converges into the function so the answer is it has no taylor series so they couldnt converge into the function if they not exist?

I need to look deeper into laurent series, have read about them just a little bit,

Thanks for all your help and corrections

Rick

Slovakia

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi Ricsie;

Those look like Laurent series because they have negative powers?

Check this page it has a better explanation for what you want.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

Okay now i hopefully understand it after reading the link yo usent me and also something about Laurent series, So if im right the function has some singularity point then it needs Laurent series,

The function

has singularity in 0 but when we define the function in x=0 that it is 0. Now we have a function which is defined everywhere, in this case it is still for a Laurent series?Thanks

*Last edited by Ricsie (2012-11-19 21:04:18)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

When it has been defined as 0? Then it is 0 but it's Taylor series does not represent e^(- 1 / x^2 ). That is what they are saying about a third of the way down the page. I would say 0 does not have a Laurent series. That is the best I can understand it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

Yes you are right, This function has laurent series but becuase it is defined in 0 too as 0 then the Laurent series converges into the function except in the f(0)=0 there it never converges,

Found the same example as our professor gave us, on this page:

`en.wikipedia.org/wiki/Laurent_series`

Tanks for helping me with a solution and also helping me in my self-learning, yesterday i was totally confused about series(we learn just taylor at university) but now i hopefully know something about them :-)

Thanks a lot!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi;

Unfortunately they do not teach very much about series to students. They go on and on about convergence and the 2 million tests that determine it. They never go into the all important finding what the series converges to.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

Yes thats true, our professor just a little bit explained us Taylor series and told us that its just a bonus we doesnt need to know it, but who wants to know more about series should do this exercise, and gave us this function, and i just found that series are a very interesting theory,

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

its just a bonus we doesnt need to know it

Yikes! They think that if it isn't on some kaboobly doo exam it is not important!

The Taylor series is your best friend in math. It allows us to compute things by turning functions into polynomials. We know very little about functions in general for computation. We know lots of things about polynomials!

When you get a job you will be using Taylor series for mostly everything and why is that unimportant?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Ricsie****Member**- Registered: 2012-11-19
- Posts: 7

Yes we doesnt even have exams from Taylor series, we are examined from derivatives, integrations, and limits. Nothing about series just really some basics of infinite series, but nothing more, but hopefully theres lots of information on internet, and lots of helpful people like you who can help people to solve and understand things :-)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi;

Thanks to Cauchy you will always see more about continuous math than discrete. Lots of people can integrate but few can sum!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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