- Find the Maclaurin series expansion of \(f(x) = xe^{2x}\) (that is, the Taylor series expansion of \(f(x)\) about \(a = 0\)).
- If \(T_5(x)\) is the polynomial consisting of the terms of the Maclaurin series of \(f(x)\) through degree 5, and \(T_5(x)\) is used to approximate \(f(x)\) on the interval \([0,1/2]\), find a bound for the maximum error \(|f(x) - T_5(x)|\) on \([0,1/2]\).
Solution
Since \[ e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}, \] we have \[ xe^{2x}=\sum_{n=0}^\infty \frac{2^n}{n!}x^{n+1}. \] Thus \[ T_5(x)=x+2x^2+2x^3+\frac43x^4+\frac23x^5. \]
Taylor’s theorem gives, for some \(\xi\) between \(0\) and \(x\), \[ f(x)-T_5(x)=\frac{f^{(6)}(\xi)}{6!}x^6. \] From the derivative formula, \[ f^{(6)}(t)=64te^{2t}+192e^{2t}=(64t+192)e^{2t}. \] On \(0\le t\le1/2\), this is increasing, so \[ |f^{(6)}(t)|\le224e. \] Also \(x^6\le(1/2)^6=1/64\). Therefore \[ |f(x)-T_5(x)|\le \frac{224e}{6!\,64}=\frac{7e}{1440} \] for every \(x\in[0,1/2]\).