Saturday, April 13, 2013
Discovering Maclaurin Series
By the fundamental theorem of calculus,
f(n)=f(0)+∫n0f′(n−x)dx |
f(n)=f(0)+n⋅f′(0)+∫n0x⋅f″ |
Give it a try and it's not too hard to show that:
\begin{aligned} \displaystyle f(n) = f(0) + n \cdot f'(0) + \frac{n^2f''(0)}{2!} + \frac{n^3f'''(0)}{3!} + \cdots + \frac{n^kf^{(k)}(0)}{k!} + \cdots \end{aligned} |
\begin{aligned} \displaystyle f(n) = \sum_{i=0}^\infty \frac{n^i f^{(i)}(0)}{i!} \end{aligned} |
For example, as Professor Jim Fowler pointed out, "just the behavior of cosine near x=0 is enough to recover the value of cosine at any other point --- it's nuts".
Special thanks to Jim for his generous insight and humorous teaching of Calculus One at Coursera.
Note:
\begin{aligned} \dfrac{d}{dx} f(n-x) &= - \, f'(n-x) \\ f(n-x) &= - \int f'(n-x) \, dx \\ \biggl[ \, {f(n-x)}\biggr]_n^0 &= - \int_n^0 f'(n-x) \, dx \\ f(n) - f(0) &= \int_0^n f'(n-x) \, dx \\ \therefore \; f(n)&= f(0) + \int_0^n f'(n-x) \, dx \end{aligned} |