f(x) = ((e^x) - 1)/ x if x =/= 0

f(x) = 1 if x = 0

Find the Taylor Polynomial of degree n for f at 0 and an estimate for the remainder term.

Detailed answer would be greatly appreciated but all help is much appreciated.

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- Apr 1st 2013, 06:07 PMpatrickmanningTaylor Polynomial question
f(x) = ((e^x) - 1)/ x if x =/= 0

f(x) = 1 if x = 0

Find the Taylor Polynomial of degree n for f at 0 and an estimate for the remainder term.

Detailed answer would be greatly appreciated but all help is much appreciated. - Apr 1st 2013, 06:08 PMProve ItRe: Taylor Polynomial question
At 0 isn't the function just 1, so the Taylor Polynomial would be 1...

- Apr 1st 2013, 06:21 PMpatrickmanningRe: Taylor Polynomial question
I thought so too, but that seemed off for a question worth 20 marks on an Analysis exam. I still didn't know how to go about with the remainder part, though.

- Apr 1st 2013, 06:22 PMProve ItRe: Taylor Polynomial question
Assuming that the question is very badly worded, they probably want you to get the Taylor Polynomial for the function everywhere else. How are you going with that?

- Apr 1st 2013, 06:56 PMSworDRe: Taylor Polynomial question
That's not true at all =/ the derivatives of that function are NOT 0 at x=0 - just because you obtained the piece of information that the function is defined there, that doesn't mean the limit corresponding to the dervative is 0 there. The definition of derivatives deals with how the function changes as x changes - it doesn't care how the function was defined.

Edit - I suppose you could keep differentiating and apply L'Hopital's to find the derivatives. But they still won't be zero.

And the question isn't badly worded, either. They meant they want the series centered at x=0, so the form will be , instead of

@OP: In this specific case, you start with the expansion for e^x:

Subtract 1:

(notice how i simply adjusted the index to ommit the constant term)

then divide by x:

or equivalently

Can you work from there?

For the remainder, unfortunately I haven't worked too much with that so I can't confidently provide an answer, but this might help:

Taylor's theorem - Wikipedia, the free encyclopedia