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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

Hello.

I am trying to teach myself infinite series. I recently posted on this forum questions involving the relationships between rational expressions and hyperbolas and because I am still in that mindset, I am using that example.

I have managed to work out that the series expansion or infinite series or infinite sum or whatever it is called for a function

will be of the form

such that

if computed will yield the desired function. I have confirmed this in Maple.

How do I go about doing this by hand so as to show all my work? Computing it in Maple is easy but I want to know how to do it for myself. I attempted the problem myself, treating it as an integration problem but that didn't work and it makes sense that it didn't work because I am counting to infinity instead of finding the area under a curve.

Any help is appreciated.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

Hi;

Series expansions only converge for specific ranges. Do you want to expand around x = 0 or something else?

**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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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

Whatever works; whatever gets the desired result. If x=0 works then yes.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

Hi;

A Taylor polynomial is a truncated Taylor series. In numerical mathematics we use approximations for computation. If you intend to plug in small values into your function then you would produce a Taylor series around 0.

It is easy to demonstrate the work to produce such a formula.

**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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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

I see what you are saying and I already know that. When it is around zero it is a Maclaurin series. My original question is, how do I go from

to

without the aid of a computer program like Maple? Is there a way to compute it by hand like an integral? That is the information I am lacking.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

It is easy to go from the function to the series but to go back might not be as easy since I have never seen it done. If you were to undo the differentiations that still would not produce the original function just another polynomial.

When you take the Mclaurin series of sin(x) you get :

Now it is obvious that no matter what we do to that we will never get sin(x) back.

**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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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

So the Taylor (Maclaurin) series is just a way to represent a function and isn't necessarily meant to be solved except by computer. Got it.

Thanks for your help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

Hi;

Yes, they are the same thing for certain values only. Sometimes when you see the series on the right you are able to recognize which function it belongs to.

In most cases you will not. You could always curve fit but you must again already know the form on the left.

So the Taylor (Maclaurin) series is just a way to represent a function and isn't necessarily meant to be solved except by computer.

The Taylor series is very useful in numerical work. It is used to approximate functions and to develop new methods

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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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

Thanks for all the info.

Concerning the notation, is it just as proper to write

in the form

or is the second form superfluous? Are either one considered *more* correct?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

Hi;

I would say both are correct but I prefer the first one.

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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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

Yeah, I think I do too. The second one feels unnecessarily elongated only to say the same exact thing.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

To remember that might be useful if you could get the sum in analytical form for an upper bound of n. Then you could use the laws of limits to get the sum to infinity. This is only theoretical as far as I am concerned. It is almost never easier to get the sum to n rather than infinity.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,812

bobbym wrote:

It is easy to go from the function to the series but to go back might not be as easy since I have never seen it done. If you were to undo the differentiations that still would not produce the original function just another polynomial.

When you take the Mclaurin series of sin(x) you get :

Now it is obvious that no matter what we do to that we will never get sin(x) back.

We can form a second order differential equation and get it from there.

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**Reuel****Member**- Registered: 2010-11-28
- Posts: 176

anonimnystefy wrote:

We can form a second order differential equation and get it from there.

How?

bobbym: Which notation of the initial statement would you say is the most correct?

or

or

or something else? They can all be interpreted to mean the same thing but I do not know if any of them are better than the rest.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,435

Hi;

I would use the first one.

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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