# Validity of Maclaurin expansions

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#1
So, I was thinking about this and wasn't quite sure why the Maclaurin expansion is only valid for values between |x| < 1. I understand that the series diverges for values outside this range and converges for values inside it (easily checked by various convergence tests such as d'Alembert's). My question is, how do we know that it converges to the value we want it to? For all we know, it could converge to a completely different function, right?

Thanks for helping guys!
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6 years ago
#2
(Original post by majmuh24)
So, I was thinking about this and wasn't quite sure why the Maclaurin expansion is only valid for values between |x| < 1. I understand that the series diverges for values outside this range and converges for values inside it (easily checked by various convergence tests such as d'Alembert's). My question is, how do we know that it converges to the value we want it to? For all we know, it could converge to a completely different function, right?

Thanks for helping guys!
Firstly, not all Maclaurin series are only for |x| < 1. The Maclaurin series for e^x, sin x and cos x converge for all values of x!

Your general question is a very important one. Often we can prove a series converges using a comparison test or ratio test or something else, without knowing what it converges to!

The Maclaurin series is a special case of the Taylor series - I would suggest looking up one of the versions of the Taylor series with remainder because this gives some details about the sort of thing you're asking
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#3
(Original post by davros)
Firstly, not all Maclaurin series are only for |x| < 1. The Maclaurin series for e^x, sin x and cos x converge for all values of x!

Your general question is a very important one. Often we can prove a series converges using a comparison test or ratio test or something else, without knowing what it converges to!

The Maclaurin series is a special case of the Taylor series - I w)ould suggest looking up one of the versions of the Taylor series with remainder because this gives some details about the sort of thing you're asking
My original statement was a bit badly worded, I meant that they are definitely convergent for functions when |x|<1 (for most functions I've seen anyway, can't see an obvious counter example anywhere but I assume there is some )

Yep, that's what was confusing me!

Isn't that working on a similar principle? The Taylor series essentially 'shifts' the value that the series converges around so it still converges for |x|<1, just around that value if you know what I mean (sorry if this sounds a bit weird, I couldn't think of a better way to explain it )

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6 years ago
#4
(Original post by majmuh24)
My original statement was a bit badly worded, I meant that they are definitely convergent for functions when |x|<1 (for most functions I've seen anyway, can't see an obvious counter example anywhere but I assume there is some )

Yep, that's what was confusing me!

Isn't that working on a similar principle? The Taylor series essentially 'shifts' the value that the series converges around so it still converges for |x|<1, just around that value if you know what I mean (sorry if this sounds a bit weird, I couldn't think of a better way to explain it )

Posted from TSR Mobile
Grrr - I just typed a big update to this,then my browser crashed and lost the lot

For different convergence region just think of binomial series you see at A level e.g, (1+2x)^-1 requires |x| < 1/2 for convergence

You are correct about the Taylor series just shifting the origin of the series (rather than its radius of convergence). However, the Taylor series is more general so most of the proofs I've seen about a series converging to a function start from the Taylor expansion because that gives us more generality. Of course, these proofs can usually be simplified once written if you are dealing with a specific series e,g, one about x=0)
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