Taylor Series - How to evaluate and prove it?

Question

Question: How would you prove the Taylor series for a function, and how would you find the exact value of it?

I know that$$e^x=\sum_{k=0}^{\infty}\frac {x^k}{k!}\\\frac 1{1-x}=\sum_{k=0}^\infty x^k\\\sin x=\sum_{k=0}^\infty \frac {(-1)^k}{(2k+1)!}x^{2k+1}\\\vdots$$

My question is: How would you prove the taylor expansion for any polynomial $f(x)$, and how would you evaluate the convergence?

For Example: With $e^x$, letting $x$ be a simple integer like $2$, we have\begin{align*}e^2=1+2+\frac 4{2!}+\frac 8{3!}+\ldots\tag1\end{align*} How would you solve for the exact value of $e^2$ when you have the infinite sequence? (Note that it does converge) And $x$ doesn't have to be an integer. It can also be a number such as $\pi\sqrt{19}$.\begin{align*}e^{\pi\sqrt{19}}=1+\pi\sqrt{19}+\frac {19\pi^2}{2!}+\frac {19\pi^3\sqrt{19}}{3!}+\frac {19^2\pi^4}{4!}+\ldots\tag2\end{align*} Which is intimating.

Answer 1

Assume the form: $$f(x) = \sum_{n=0}^{ \infty} c_n (x-a)^n$$ Differentiate both sides and set $x=a$ and equate the coefficients.

All but one of the $c_n$ coefficients will be multiplied by zero from $(x-a)^n$ terms.

$$f(a) = c_0$$

$$f^{(k)}(x) = \sum_{n=0}^{ \infty} c_n \frac{n!}{(n-k)!}(x-a)^{n-k}$$

$$f^{(k)}(a) = c_k \, k!$$

only the $(x-a)^0$ term survives so $n=k$.

$$c_k = \frac{f^{(k)}(a)}{k!}$$

$$f(x) = \sum_{n=0}^{ \infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$$

Exact value: it might not be possible to find an exact value. e.g. infinite series.

Convergence: that's a large topic, one method is to find the $x$ range such that the limit of consecutive terms tends to zero:

$$\lim_{n\to \infty} \frac{f^{(n+1)}(a) (x-a)}{f^{(n)}(a)(n+1)} \to 0$$

Answer 2

With regard to solving for exact values of $e^x$: For any $\alpha\in\mathbb{R}$, the number $e^\alpha$ cannot, in general, be written in a simpler way than the evaluation of the Taylor series of $e^x$ at $x=\alpha$. By this, I mean that we cannot write an expression for $e^\alpha$ that does not necessarily have infintely-many terms. If we can write $e^\alpha$ as a summation of finitely-many terms, then we say that $e^\alpha$ is algebraic. Otherwise, it is transcendental.

Answer 3

Assume $e^x$ is a straight line of the form $c_0+c_1x$ at $x=0$.

$$e^x=c_0+c_1x$$

$$e^0=c_0+c_1(0)\implies c_0=1$$

$$e^x=1+c_1x$$

$$e^x=c_1\tag{differentiate both sides}$$

$$e^0=c_1\implies c_1=1\tag{This is about $x=0$}$$

so

$$e^x=1+x$$

That was nice, but it'd be better if we assumed $e^x=c_0+c_1x+c_2x^2$. Proceeding in the same manner, we get

$$e^x=1+x+\frac12x^2$$

Assuming, say, that $e^x$ were a polynomial of $3^{\text{rd}}$ degree, we seem to get a pattern, and by (easy) induction, we find that

$$e^x=1+x+\frac12x^2+\dots+\frac1{n!}x^n+\dots$$

This method can be made to work for the others too, though it assumes that $e^x$ can be written in this form. It goes to show not all functions can be expanded this way. A most famous example of this is the expansion of $e^{-1/x^2}$ at $x=0$, which comes out to give the false expansion:

$$e^{-1/x^2}\ne0+0x+0x^2+\dots=0$$

Indeed, these functions are non-analytic, as opposed to analytic functions, which can be written in a power series.

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For convergence, one needs to show that

$$\left|e^x-\sum_{k=0}^n\frac1{k!}x^k\right|<\epsilon\ \forall \ x$$

This is known as uniform convergence. We need to show that for all $\epsilon>0$, there exists some $N$ such that for all $n>N$, the above holds. If for $N$ changes depending on the value of $x$, this is known as pointwise convergence.

There are plenty of ways for showing this with different functions. For the exponenial funcion, we find it converges everywhere via the Weierstrass M-test.

Answer 4

Just a quickie, but we do have this formula:$$a^x=1+cx+\dfrac {c^2x^2}{2!}+\dfrac {c^3x^3}{3!}+\dfrac {c^4x^4}{4!}+\dfrac {c^5x^5}{5!}+\ldots\&c\tag1$$ Where $c=\ln a$. Setting $a=e$ gives $c=1$ which gives the Taylor Expansion Series for $e^x$.$$e^x=1+x+\dfrac {x^2}{2!}+\dfrac {x^3}{3!}+\dfrac {x^4}{4!}+\dfrac {x^5}{5!}+\ldots\&c\tag2$$