Taylor Series

0. Can we approximate a function using a power series?

With the calculus’ help, we can!

1. Consider the function
(1)\displaystyle  f(x) = \sum^{\infty}_{n=0}b_{n}(x-a)^{n} = b_{0} + b_{1}(x-a) + \dots.
We suppose it is smooth (i.e., has infinitely many derivatives). We see
(2)\displaystyle  \begin{aligned} f'(x) &= \sum^{\infty}_{n=1}b_{n}n(x-a)^{n-1}\\ f'(a) &= b_{1} \end{aligned}
and
(3)\displaystyle  \begin{aligned} f''(x) &= \sum^{\infty}_{n=2}b_{n}n(n-1)\cdot(x-a)^{n-2}\\ f''(a) &= 2b_{2}. \end{aligned}
We also have
(4)\displaystyle  \begin{aligned} f'''(x) &= \sum^{\infty}_{n=3}b_{n}n(n-1)(n-2)\cdot(x-a)^{n-3}\\ f'''(a) &= 3\cdot2\cdot1 b_{3} = 6b_{3}. \end{aligned}
The general case appears to be
(5)\displaystyle  \left.\frac{\mathrm{d}^{n}}{\mathrm{d} x^{n}}f(x)\right|_{x=a}=n! b_{n}
Thus the coefficients are
(6)\displaystyle  b_{n} = \left.\frac{1}{n!}\frac{\mathrm{d}^{n}}{\mathrm{d} x^{n}}f(x)\right|_{x=a}
and we can reconstruct the function {f(x)}.

2. Definition. Let {f(x)} be any function. The “Taylor Series about {x=a} is a series
(7)\displaystyle  \sum^{\infty}_{n=0}f^{(n)}(a)\cdot(x-a)^{n}
When {a=0}, it’s called a MacLaurin Series.

Remark 1. If we have the MacLaurin series for a function, and if the series converges for any value of {x}, then we can use the MacLaurin series as a synonym for the original function. That’s the usefulness of MacLaurin series.

Remark 2. The Taylor series helps us compute {f(x+h)} when {h\ll x} and when {f(x)} is known. For example, {\sqrt{1.001}} can be computed using the Taylor series of {\sqrt{1+x}} about {x=1}.

Example 1. Consider the function {\exp(x)}. We see that
(8)\displaystyle  \frac{\mathrm{d}\exp(x)}{\mathrm{d} x}=\exp(x).
Thus the MacLaurin series for the exponential function is
(9)\displaystyle  \exp(x)=\sum^{\infty}_{n=0}\frac{x^{n}}{n!}
Where does this series converge?

Using the absolute ratio test, we find
(10)\displaystyle  \lim_{n\rightarrow\infty}\frac{x^{n+1}/(n+1)!}{x^{n}/n!} = \lim_{n\rightarrow\infty}\frac{x}{n+1}=0.
So the series converges for any value of {x}.

Example 2. Consider the sine function {\sin(x)}. What is its MacLaurin series? Writing
(11)\displaystyle  \sum^{\infty}_{n=0}c_{n}x^{n}=\sin(x)
we have
(12)\displaystyle  \begin{aligned} c_{0}&=\sin(0)=0\\ c_{1}&=\cos(0)=1\\ c_{2}&=\frac{-\sin(0)}{2!}=0\\ c_{3}&=\frac{-\cos(0)}{3!}=-1/3! \end{aligned}
The coefficients are sort of “periodic” in the sense that only the odd ones remain, and their sign alternates. We have
(13)\displaystyle  c_{n} = c_{2k-1} = \frac{(-1)^{n}}{n!} = \frac{(-1)^{k+1}}{(2k-1)!}
thus the MacLaurin series is
(14)\displaystyle  \sin(x)=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}x^{n}}{(2n-1)!}.
Where does this converge?

Using the ratio test, we have
(15)\displaystyle  \lim_{n\rightarrow\infty}\left|\frac{\left(\displaystyle\frac{x^{2n+3}}{(2n+3)!}\right)}{\left(\displaystyle\frac{x^{2n+1}}{(2n+1)!}\right)}\right| =\lim_{n\rightarrow\infty}\frac{x^{2}}{(2n+1)(2n+2)} = 0.
Thus it converges for any value of {x}.

Example 3. Lets find the MacLaurin series for {\cos(x)}. We see that
(16)\displaystyle  \cos(x)=\sum^{\infty}_{n=0}b_{n}x^{n},
we need to find the {b_{n}}. We see that
(17)\displaystyle  \begin{aligned} b_{0} &= \cos(0) = 1\\ b_{1} &= -\sin(0) = 0\\ b_{2} &= \frac{-1}{2!}\cos(0) = -1/2\\ b_{3} &= \frac{1}{3!}\sin(0) = 0\\ b_{2k} &= \frac{(-1)^{k}}{(2k)!} \end{aligned}
So we have the MacLaurin series be
(18)\displaystyle  \cos(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}x^{2n}.
Where will it converge? Using the absolute ratio test, we have
(19)\displaystyle  \begin{aligned} \lim_{n\rightarrow\infty}\left|\frac{\left(\displaystyle\frac{(-1)^{n+1}x^{2n+2}}{(2n+2)!}\right)}{\left(\displaystyle\frac{(-1)^{n}x^{2n}}{(2n)!}\right)}\right| &=\lim_{n\rightarrow\infty}\frac{x^{2}}{(2n+1)(2n+2)}\\ &=0. \end{aligned}
Thus the MacLaurin series for {\cos(x)} converges for any value of {x}.

3. Euler’s Formula. Recall Euler’s formula states
(20)\displaystyle  \exp(\mathrm{i}\theta)=\cos(\theta)+\mathrm{i}\sin(\theta)
and we have the MacLaurin series for {\exp(x)}. Setting {x=\mathrm{i}\theta}, we find
(21)\displaystyle  \begin{aligned} \exp(\mathrm{i}\theta) &= \sum^{\infty}_{n=0}\frac{(\mathrm{i}\theta)^{n}}{n!}\\ &=1+\mathrm{i}\theta-\frac{\theta^{2}}{2!}-\frac{\mathrm{i}\theta^{3}}{3!}+\frac{\theta^{4}}{4!}+\dots \end{aligned}
Gathering the real and imaginary parts together we get
(22)\displaystyle  \exp(\mathrm{i}\theta) = \left(\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n)!}\right) +\mathrm{i}\left(\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\right)
But look: the imaginary part is precisely the MacLaurin series for {\sin(x)}! And the real part is the MacLaurin series for {\cos(x)}, too! So what did we do? We just derived Euler’s formula.

4. Taylor Series Makes Approximations Consider the function
(23)\displaystyle  f(x)=\sqrt{x}.
Its Taylor series about {x=1} is the same as the MacLaurin series for the function
(24)\displaystyle  g(h)=\sqrt{1+h}.
Just write {x=1+h}. For small {0<h\ll1}, we can use the first couple of terms in the Taylor series as an approximate value. So what’s the first 6 nonzero terms in the Taylor series? We see
(25)\displaystyle  f(x)=x^{1/2}\quad\implies\quad f(1)=1
describes the constant term. The linear term has coefficient
(26)\displaystyle  f'(x)=\frac{x^{-1/2}}{2}\quad\implies\quad f'(1)=\frac{1}{2}.
The quadratic term
(27)\displaystyle  f''(x)=\frac{-x^{-3/2}}{2^{2}}\quad\implies\quad f''(1)=-1/2^{2}.
Observe how the exponent behaves when differentiating: {(1/2)-n}. The numerator is always odd, the denominator doesn’t change. So the {n^{th}} derivative would be
(28)\displaystyle  f^{(n)}(x)=\frac{1\cdot3\cdot5\cdot(\dots)\cdot(2n-1)}{2^{n}}(-1)^{n-1}x^{(1-2n)/2}
This gives us our coefficients! We then have
(29)\displaystyle  c_{n} = \frac{(-1)^{n-1}(2n)!}{(n!2^{n})^{2}}
for {n>0}. Observe the numerator appears odd, but we can justify it thus:
(30)\displaystyle  \begin{aligned} \frac{(2n)!}{n!2^{n}} &= \frac{1\cdot2\cdot3\cdot(\dots)\cdot(2n)}{(1\cdot2\cdot(\dots)\cdot n)2^{n}}\\ &= \frac{\bigl(1\cdot3\cdot(\dots)\cdot(2n-1)\bigr)\bigl(2\cdot4\cdot(\dots)\cdot 2n\bigr)}{(2\cdot4\cdot(\dots)\cdot2n)}\\ &= 1\cdot3\cdot(\dots)\cdot(2n-1)\frac{\bigl(2\cdot4\cdot(\dots)\cdot 2n\bigr)}{(2\cdot4\cdot(\dots)\cdot2n)}\\ &= 1\cdot3\cdot(\dots)\cdot(2n-1)\left(\vphantom{\frac{a}{a}}1\right)\\ &= 1\cdot3\cdot(\dots)\cdot(2n-1) \end{aligned}
At any rate, this gives us a polynomial expression for {f(1+h)} as
(31)\displaystyle  f(1+h) = \left(1+\sum^{6}_{n=1}\frac{(-1)^{n-1}(2n)!}{(n!2^{n})^{2}}h^{n}\right) + R_{6}(h)
where {R_{6}(h)} is the “error” term. What does the error term tell us? Precisely how good or bad our polynomial approximates {f(1+h)}. That is to say, how {(1+c_{1}h+\dots+c_{6}h^{6})} approximates {f(1+h)}.

Question: how do we find the error term?

Exercise 1 [Math.SE]. Using the MacLaurin series for sine and cosine, check that
\displaystyle\frac{\mathrm{d}\cos(x)}{\mathrm{d}x}=-\sin(x) and \displaystyle\frac{\mathrm{d}\sin(x)}{\mathrm{d}x}=\cos(x)

Exercise 2. What is the Taylor series for \ln(x) about x=1? I.e., what is the MacLaurin series for \ln(1+h)?

Exercise 3 [Math.SE]. Using Exercise 2, demonstrate using the MacLaurin series for \exp(x) that \exp\bigl(\ln(1+x)\bigr)=1+x. You can work up to x^{4} if you want.

Exercise 4 [Math.SE]. What is the MacLaurin series for the function 1/(1-x)?

Exercise 5 [Math.SE]. Find the MacLaurin series for \arctan(x) by first taking its derivative, then considering the MacLaurin series for this derivative, and then integrating the MacLaurin series.

Exercise 6 [Math.SE]. Using the MacLaurin series for sine and cosine, plug them into the formula \sin^{2}(x)+\cos^{2}(x), and what do we get?

Exercise 7 [Math.SE] What is the Taylor series for \ln(\tan(x))-\ln(x) about x=0?

Exercise 8. Use the MacLaurin series for sine to prove
\displaystyle\lim_{x\to0}\frac{\sin(x)}{x}=1.

Exercise 9 [Math.SE]. What is the MacLaurin series for \displaystyle [\cos(x)]^{\sin(x)}?

Exercise 10. What is the Taylor series for f(x)=\sqrt[3]{x} about x=1? Or if you prefer, find the MacLaurin series for f(1+h)=\sqrt[3]{1+h}.

Exercise 11. What is the MacLaurin series for {\displaystyle\frac{1}{\sqrt{1-x}}}?

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Analysis, Approximation, Calculus, Differential, Infinite Series, Power Series, Taylor Series and tagged , . Bookmark the permalink.

3 Responses to Taylor Series

  1. Pingback: Taylor Polynomials | My Math Blog

  2. Pingback: Taylor Series Uses #1: Complicated Integrals | My Math Blog

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