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This question already has an answer here:

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.

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marked as duplicate by marty cohen, hardmath, R_D, Rohan, tomi Dec 30 '16 at 7:20

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  • $\begingroup$ Maybe this answers your question (see the example for $\exp$)? You can consider the residue and prove that it converges to zero when $k\to\infty$ (if $f\in\mathcal{C}^\infty$). $\endgroup$ – anderstood Nov 2 '16 at 23:04
  • $\begingroup$ Multiply the nth term of the series by $x^n$ and then write down a general differential equation of low order with arbitrary low degree polynomial coefficients and then see if you can make the unknown series satisfy that differential equation by choosing the undetermined coefficients that define the polynomials in the right way. $\endgroup$ – Count Iblis Nov 2 '16 at 23:12
  • $\begingroup$ @anderstood you mean if $f$ is analytic (that is its Taylor converges and $f$ is equal to its Taylor series for $|x|$ small enough) $\endgroup$ – reuns Nov 2 '16 at 23:13
  • $\begingroup$ @user1952009 Yes, I was considering the provided examples. $\endgroup$ – anderstood Nov 2 '16 at 23:51
  • $\begingroup$ Is there a reason you chose the largest Heegner Number as your example? $\endgroup$ – Brevan Ellefsen Nov 3 '16 at 0:00
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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:

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

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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.


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.

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  • $\begingroup$ This inevitably begs the question "What happens when you try this with something that can't be written in this form?". Great answer, though this doesn't really address the second half of the OP's question, namely "how to show convergence". +1 nevertheless $\endgroup$ – Brevan Ellefsen Nov 3 '16 at 0:25
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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.

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  • $\begingroup$ I believe you meant to refer to algebraic function and transcendental function. Furthermore, according to the Abel-Ruffini theorem, there are algebraic functions not expressible in a finite amount of radicals and basic arithmetic operations. $\endgroup$ – Simply Beautiful Art Nov 3 '16 at 21:47
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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 $e^x$.$$e^x=1+x+\dfrac {x^2}{2!}+\dfrac {x^3}{3!}+\dfrac {x^4}{4!}+\dfrac {x^5}{5!}+\ldots\&c\tag2$$

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