Taylor Series

Consider the taylor series expansion of f(x)f(x) about the point aa,

which is limxaf(x)=f(a)+f(a)1!(xa)+f(a)2!(xa)2+\displaystyle \lim_{x \to a} f(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\cdots

Now , I want to ask can we write this as limxaf(x)=f(a)+f(a)1!(xa)+f(c)2!(xa)2\displaystyle \lim_{x \to a}f(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(c)}{2!}(x-a)^2?

That is, we are compressing all the infinite terms into a single term, where the derivative is evaluated about any other point cc in the domain of ff.

In this case, for example I modified the third term, but it could be done for any finite term in the series.

Can we prove this that for any particular cc, we can simply write the summation into above form?

Note by Vilakshan Gupta
9 months, 2 weeks ago

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1 vote

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There are many forms of the exact remainder for a Taylor’s series. The one closest to what you want is this....

For a sufficiently differentiable function, and for any a,xa,x and nn, there exists cc between aa and xx such that f(x)=j=0n1f(j)(a)j!(xa)j+f(n)(c)n!(xa)n f(x)=\sum_{j=0}^{n-1}\frac{f^{(j)}(a)}{j!}(x-a)^j +\frac{f^{(n)}(c)}{n!}(x-a)^n

Mark Hennings - 9 months, 2 weeks ago

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Yes sir, this is what I wanted...but I wanted to know the proof of the claim or if you could write a proof of it...or you could provide a reference from where I can read more about it.

Vilakshan Gupta - 9 months, 2 weeks ago

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Put F(t)=j=0n1f(j)(t)j!(xt)jG(t)=(xt)n F(t)=\sum_{j=0}^{n-1}\frac{f^{(j)}(t)}{j!}(x-t)^j \hspace{2cm} G(t)=(x-t)^n Then F(a)=j=0n1f(j)(a)j!(xa)jF(x)=f(x)F(t)=f(n)(t)(n1)!(xt)n1 F(a)=\sum_{j=0}^{n-1}\frac{f^{(j)}(a)}{j!}(x-a)^j \hspace{1cm} F(x)=f(x)\hspace{1cm}F’(t)=\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1} While G(a)=(xa)nG(x)=0G(t)=n(xt)n1 G(a)=(x-a)^n\hspace{1cm}G(x)=0\hspace{1cm}G’(t)=-n(x-t)^{n-1} Now use the Generalised Mean Value Theorem, to find cc between aa and xx such that (F(x)F(a))G(c)=(G(x)G(a))F(c)(F(x)-F(a))G’(c)=(G(x)-G(a))F’(c) Most textbooks on real analysis should give proofs of Taylor’s Theorem with remainder...

Mark Hennings - 9 months, 1 week ago

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@Mark Hennings, @Chew-Seong Cheong, @David Vreken , @Chris Lewis

Vilakshan Gupta - 9 months, 2 weeks ago

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