Class 25
\(\displaystyle \frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1 + x + x^2 + x^3 + \cdots\) for \(-1 < x < 1\).
\(\displaystyle \frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots\) for \(-1 < x < 1\).
\(\displaystyle \frac{1}{(1-x)^2} = \sum_{n=0}^\infty (n+1)x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots\) for \(-1 < x < 1\).
\(\displaystyle \ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\) for \(-1 < x \le 1\).
\(\displaystyle \arctan(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\) for \(-1 \le x \le 1\).
What about other functions?
Let
\[f(x) = \sum_{n=0}^\infty c_n(x-a)^n\]
on some open interval containing \(a\).
Then
\[c_n = \frac{f^{(n)}(a)}{n!}\]
Suppose
\[\sum_{n=0}^\infty c_n(x-a)^n = \sum_{n=0}^\infty d_n(x-a)^n\]
for every \(x\) from some open interval \(I\) containing \(a\). Then \(c_n = d_n\) for every \(n = 0, 1, 2, \dots\).
Let \(f\) be infinitely differentiable at \(a\). Then the power series
\[\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n\]
is called the Taylor series of \(f\) centered at \(a\).
Taylor series centered at \(0\) is also sometimes called Maclaurin series.
Big question: Does the Taylor series of \(f\) converge, and if it does, which function does it converge to?
If \(f\) has at least \(n\) derivatives at \(a\), then the polynomial
\[p_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{6} + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n\]
is called the \(n\)-th Taylor polynomial of \(f\) at \(a\).
If \(f\) is infinitely differentiable at \(a\) then \(p_n(x)\) is the \(n\)-th partial sum of its Taylor series at \(a\).
\(f(x) = \exp x\) at \(a = 0\)
\(f(x) = \sin x\) at \(a = 0\)
\(f(x) = \sin x\) at \(a = \frac{\pi}{4}\)
\(f(x) = \ln x\) at \(a = 1\)
Let \(f\) has \(n+1\) derivatives on an open interval \(I\) containing \(a\). Define
\[R_n(x) = f(x) - p_n(x)\]
where \(p_n\) is the \(n\)-th Taylor polynomial of \(f\) at \(a\).
Then for every \(x \in I\) there exists a number \(c\) between \(a\) and \(x\) such that
\[R_n(x) = \frac{f^{(n+1)}({\color{red} c})}{(n+1)!}(x-a)^{n+1}\]