Write the function $f(x)=e^x$ as a sum of infinite series using Taylor's Theorem with Lagrange's Form of Remainder.

My solution goes like this:

Taylor's Theorem with Lagrange's form of remainder states that, if $f$ is a function defined on an interval $[a,a+h]$ where $h>0$ and if $f$ satisfies the following properties i.e

(i) $f^{(n-1)}$ is continuous over $[a,a+h]$,

(ii) $f^{(n-1)}$ is derivable over $[a,a+h]$,

then, we can write, $f(a+h)=f(a)+hf(a)+\frac {h^2}{2!}f''(a)+\cdots +\frac {h^{n-1}}{(n-1)!}f^{(n-1)}(a)+\frac{h^n}{n!}f^n(a+\theta h),$ where $\theta \in (0,1)$.

If $h=x$ and $a=0$ we get, the Maclaurin's formula i.e, $f(x)=f(0)+xf(0)+\frac {x^2}{2!}f''(0)+\cdots +\frac {x^{n-1}}{(n-1)!}f^{(n-1)}(0)+\frac{x^n}{n!}f^n(\theta x),$ where $\theta \in (0,1)$ in the interval $[0,x].$

Here, $f(x)=e^x.$ Since, $f$ everywhere continous and differentiable so, we apply Maclaurin's formula. Also, $f^n(x)=e^x.$

We have, $\lim_{n\to\infty}f(x)=e^x.$

We note that, $\lim_{n\to\infty}\frac{x^n}{n!}f^n(\theta x)=\lim_{n\to\infty}\frac{x^n}{n!}e^{\theta x}=0.$ This is because, $e^{\theta x}$ is bounded (for, $1\leq e^{\theta x}\leq e^x$) and also, $\lim_{n\to\infty}\frac{x^n}{n!}=0.$

We write, $y_n=f(0)+xf(0)+\frac {x^2}{2!}f''(0)+\cdots +\frac {x^{n-1}}{(n-1)!}f^{(n-1)}(0).$

So, according to Maclaurin's formula in the interval, $[0,x]$ we have, $f(x)=y_{n-1}+R_n,$ where $R_n=\frac{x^n}{n!}f^n(\theta x).$ Since, $\{f(x)\}$ is (a constant sequence and hence,) a convergent sequence, $\{R_n\}$ is an convergent sequence as well, so, the sequence (of partial sums) $y_{n-1}$ is convergent as well and it's limit exists. This means, $\lim_{n\to\infty} f(x)=e^x=\lim_{n\to\infty} y_{n-1}+\lim_{n\to\infty}R_n=\lim_{n\to\infty}y_n.$

So, $e^x=1+x+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\cdots.$

As $x\in\Bbb R$ is an arbitrary so, the above is the general infinite series representation of $e^x.$

However, in a book, this problem is solved in a much complicated way i.e they first consider the sequence, $\{y_n\}$ just like I did, and then they used, D'Alambert's Ratio Test, to prove that the series $1+x+\frac{x^2}{2!}+\cdots +\frac{x^n}{n!}+\cdots$ is convergent and the rest part was similar to what followed my work i.e taking limits on both sides and all that. But what I want to know, whether my work is still valid or not? This is because I don't think there is any need to invoke any Ratio Test to show that the series is indeed a convergent one as it folows trivially from the convergence of sequence of partial sums $\{y_n\}.$ Any help regarding this issue will be highly appreciated.

Here's a picture of the thing given in the book. The book is a regional one.

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1 Answer 1


Convergence of a series can be proved directly if we can get some guess of the sum of series or have an estimate of the partial sums of the series.

In particular the root test and ratio tests for convergence of series are based on the convergence of $$1+r+r^2+\dots+r^n+\dots$$ And we can get the partial sums directly via the formula $$1+r+r^2+\dots+r^{n-1}=\frac{1-r^n}{1-r},r\neq 1$$ If $|r|<1$ we have $r^n\to 0$ and the above series converges to $1/(1-r)$.

For a typical use of Taylor (or Maclaurin) series one writes $$f(x) =\sum_{k=1}^{n} \frac{x^{k-1}}{(k-1)!}f^{(k-1)}(0)+R_n(x)$$ and there are ways to express remainder $R_n(x) $ in suitable forms which allow us to study the behavior of $R_n(x) $ as $n\to\infty$.

If for some range of values of $x$ we can prove that $R_n(x) \to 0$ then by definition (of convergence) it proves that the series $\sum_{k=0}^{\infty}\frac{x^k}{k!}f^{(k)}(0)$ is convergent and its sum is $f(x) $ for the values of $x$ in that range.

Your proof for the exponential series is thus correct and one doesn't need to add anything further.


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