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Taylor series: remainder term

  • Thread starter in vissia
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I'm currently studying the Taylor series and I cannot figure out how the remainder term came to be. If anyone could clarify this for me, I would be really grateful ...!

I understand that the Taylor series isn't always equal to f(x) for each x, so we put Rn at the end as the remainder term (note that a + h = x).

f(a+h) = f(a) + [tex]\frac{h}{1!}[/tex]*f'(a) + [tex]\frac{h^2}{2!}[/tex]*f''(a)+⋯+[tex]\frac{h^n}{n!}[/tex] f[tex]^{(n)}[/tex] (a) +Rn

So Rn is f(x) minus it's Taylor series.

But then we try to approximate how big Rn actually is. For fixed values of a and x, we look at a new function that looks like this (t is any number, p [tex]\in[/tex] (1, 2, 3, ..., n)):

F(t) = f(x) - f(t) - [tex]\frac{x - t}{1!}[/tex]*f'(t) - [tex]\frac{(x - t)^{2}}{2!}[/tex]*f''(t) - ... - [tex]\frac{(x - t)^{^n}}{n!}[/tex]*f[tex]^{(n)}[/tex](t) - Rn(x)*([tex]\frac{(x - t)^{p}}{(x - a)^{p}}[/tex]^{p}

From here on we say that F(a) = F(x) = 0 and see that, following Rolle, there should be a value w between a and x for which F'(w) = 0 and that's how we end up with a formula for Rn.

What bothers me here: What is F(t)? And if I compare it to the first formula, where did that fraction of ((x-t)/(x-a))^p next to Rn come from?

Thank you for your time!
"I understand that the Taylor series isn't always equal to f(x) for each x, so we put Rn at the end as the remainder term"

I thought we put Rn as the remainder term becuase we are only looking at the first few (finitely many) terms in the Taylor series, which are in general pretty close to f(x), but still need some correction. ?

What book are you using?
I gathered this from the notes from our Math class and it also says so in one of the books we're supposed to use, Higher Mathematics 1 (Višja matematika) (it's Slovene, you probably don't know it) for our Analisys class.

By looking at your comment, I just figured something out ... Thanks!
The thing is, in our book it's explaned by having a function that can be derived n+1 times and constructing a Taylor series by using n derivations and what's left out (the n+1th derivation) is the remainder term. Which is what you explained in much plainer words :)

But I still don't get why the remainder in that F(t) function is multiplied by that fraction?

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