Taylor and Maclaurin Series
The Essentials
The Taylor Series and the Maclaurin Series are series approximations for any function. The Taylor series finds an nth polynomial to approximate values for any equations f(x) centered at a value a. This is denoted as...
\[
\sum_{i=0}^n\frac{f^n(a)}{n!}(x-a)^n
\]
If a = 0, the series is called the Maclaurin series of the function. As n approaches infinity, the Taylor series approaches more accurately the exact value of the function f(x).
\[
\lim_{n \to \infty}\sum_{i=0}^n\frac{f^n(a)}{n!}(x-a)^n = f(x)
\]
Example
Find the 5th order Taylor Series for the equation $f(x)=3x+5x^2+4x^{\frac{1}{2}}$ centered at $a=4$. Evaluate at $x=5$.
-
First we will use the equation for the Taylor Series to find...
\[ \sum_{i=0}^5\frac{f^n(4)}{n!}(x-4)^n \]
-
Plug in the a=4 for the series
\[ (\frac{3(4)+5(4)^2+4(4)^{\frac{1}{2}}}{0!})(x-4)^0+(\frac{3+10(4)+2(4)^{\frac{-1}{2}}}{1!})(x-4)^1+(\frac{10-1(4)^{\frac{-3}{2}}}{2!})(x-4)^2+(\frac{\frac{3}{2}(4)^{\frac{-5}{2}}}{3!})(x-4)^3+(\frac{\frac{-15}{4}(4)^{\frac{-7}{2}}}{4!})(x-4)^4+(\frac{\frac{105}{8}(4)^{\frac{-9}{2}}}{5!})(x-4)^5 \]
- Simplify
\[ \frac{100}{1}+\frac{44}{1}(x-4)+\frac{9.875}{2}(x-4)^2+\frac{0.046875}{6}(x-4)^3+\frac{-0.02929}{24}(x-4)^4+\frac{0.02563}{120}(x-4)^5 \]
- Evaluate at x=5
\[ \frac{100}{1}+\frac{44}{1}(5-4)+\frac{9.875}{2}(5-4)^2+\frac{0.046875}{6}(5-4)^3+\frac{-0.02929}{24}(5-4)^4+\frac{0.02563}{120}(5-4)^5 \]
- Solution: 148.94
Practice
Evaluate these expressions:
- f(x) = sin(x) to the 4th order Taylor Series centered at a = 0
- g(y) = $\sqrt{y^7}$ to the 3rd order Taylor series centered at a=3. Evaluate at y=5
Solutions:
- \( f(x)\approx x+\frac{-1}{6}x^3 \)
- \( g(5)\approx 277.13 \)