Review : Taylor Series — Question 8

Question 8 – Solution

Strategy. Turn measured derivatives into algebraic constraints before predicting higher coefficients.

Step 1: Recover the parameters. Differentiation at $0$ gives $f'(0)=a$ and $f''(0)=a^2-b^2$. Hence $a=1$ and $b^2=4$, so $$\boxed{(a,b)=(1,2)\ \text{or}\ (1,-2).}$$ The sign cannot be recovered because cosine is even; both pairs produce the same function $e^x\cos 2x$.

Step 2: Predict the Taylor polynomial. Using real parts of $(1+2i)^n$, the derivative values for $n=0,\ldots,4$ are $1,1,-3,-11,-7$. Dividing by $n!$ yields $$\boxed{T_4(x)=1+x-\frac 32x^2-\frac{11}{6}x^3-\frac 7{24}x^4.}$$ For example, $(1+2i)^2=-3+4i$ and $(1+2i)^3=-11-2i$.

Step 3: Reject inconsistent exact data. Every pair compatible with the first two measurements forces $f'''(0)=-11$. The reported value $-10$ is incompatible. Changing the sign of $b$ conjugates the complex powers and leaves their real parts unchanged, so it cannot repair the discrepancy.

Step 4: Bound the fifth derivative uniformly. For real $t$, $$|f^{(5)}(t)|=\left|\operatorname{Re}\big((1+2i)^5e^{(1+2i)t}\big)\right| \le 5^{5/2}e^t.$$ Taylor’s theorem on each segment from $0$ to $x\in[-0.1,0.1]$ therefore gives $$|f(x)-T_4(x)|\le\frac{5^{5/2}e^{0.1}}{120}\,10^{-5}<5.2\times 10^{-6}.$$ For a purely rational check of the last inequality, use $\sqrt 5<9/4$ and $e^{0.1}<1.106$: the latter follows by bounding the tail after $1+0.1+0.1^2/2$ by $(0.1^3/6)/(1-0.025)$. These bounds give less than $5.185\times 10^{-6}$.