Find the 12th-degree Taylor polynomial for $x \sin \left(x^{2}\right)$ centered at x = 0.Suppose you use the first two non-zero terms of the polynomial to approximate $x \sin \left(x^{2}\right)$ for 0

Question

Find the 12th-degree Taylor polynomial for $x \sin \left(x^{2}\right)$ centered at x = 0.Suppose you use the first two non-zero terms of the polynomial to approximate $x \sin \left(x^{2}\right)$ for 0 < x < 1.Is your approximation too big or too small?
A)Too small
B)Too big

Quizplus · Accepted Answer

The Taylor series expansion of $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$. When we substitute $x^2$ for $x$, we get $\sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \cdots$. Multiplying this by $x$ gives $x\sin(x^2) = x^3 - \frac{x^7}{3!} + \frac{x^{11}}{5!} - \cdots$. The first two non-zero terms are $x^3$ and $-\frac{x^7}{3!}$. Since the second term is negative, using only the first two terms for $0 < x < 1$ results in an approximation that is too small, because the actual function has alternating positive and negative terms, and the approximation misses the positive contributions from higher-order terms.

Find the 12th-Degree Taylor Polynomial For xsin⁡(x2)x \sin \left(x^{2}\right)xsin(x2) Centered at X = 0

Find the 12th-Degree Taylor Polynomial For $x \sin \left(x^{2}\right)$ Centered at X = 0