Alternating Series Error Bound in Taylor Representation of $f(x) = \sin(x^2) + \cos(x)$

Question

Hello, I was solving the problem in the above image:

Text in the image:

Let $P_4(x)$ be the fourth-degree Taylor Polynomial for $f(x) = \sin(x^2) + \cos(x)$ about $x = 0$. Using information from the graph of $y = |f^{(5)}(x)|$ shown above, show that $$\bigg|P_{4}\bigg(\frac{1}{4}\bigg) – f\bigg(\frac{1}{4}\bigg)\bigg| \leq \frac{1}{3000}$$

The sample solution uses Lagrange Error Bound, but I was wondering if a solution like this was correct:

$$f(x) = \sin(x^2) + \cos(x)$$
$$=1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{121x^6}{720} \text{(first four terms)}$$

From the Alternating Series Error Bound, the fourth degree Taylor polynomial (which is just the first three terms), $$\text{difference} \leq \bigg|\frac{121}{720} \cdot \bigg(\frac{1}{4}\bigg)^6\bigg| \leq \frac{1}{3000}$$

I'd appreciate if someone could tell me whether my solution is correct or not?

Thanks

Answer 1

A couple problems:

• We actually have $f(x) = 1 {\,\color{red}+\,} \frac{x^2}{2!} + \frac{x^4}{4!}-\frac{121x^6}{720}+\cdots.$
• The alternating series error bound test you reference requires you to show that the absolute values of the terms are decreasing to zero.

Therefore, in order to correctly apply your argument, you'd need to show that, when evaluated at $x=1/4$, we have three positive terms, followed by one negative term, followed by three positive terms, followed by one negative term, etc. You'd also need to show that after summing consecutive positive terms, the corresponding alternating series has terms for which the absolute values are decreasing to zero.

Can it be made into a valid solution in this manner? Yes, it seems so.

Is your solution currently valid? No.