# Taylor Series

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

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

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

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

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 $c$, we can simply write the summation into above form?

Note by Vilakshan Gupta
1 year ago

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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 $a,x$ and $n$, there exists $c$ between $a$ and $x$ such that $f(x)=\sum_{j=0}^{n-1}\frac{f^{(j)}(a)}{j!}(x-a)^j +\frac{f^{(n)}(c)}{n!}(x-a)^n$

- 1 year ago

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.

- 1 year ago

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

- 1 year ago