the function f(x) = 7/x can also be written as f(x) = (7)1/x, which has derivatives f(x) = (7)/x^2, f(x) =…
the function f(x) = 7/x can also be written as f(x) = (7)1/x, which has derivatives f(x) = (7)/x^2, f(x) = (7)/x^3, f(x) = (7)/x^4, f^(4)(x) = (7)24/x^5. step 2 with a = -3, f(-3) = (7)/3, f(-3) = (7)/3^2, f(-3) = (7)/3^3, f(-3) = (7)/3^4, and f^(4)(-3) = (7)/3^5. submit skip (you cannot come back) need help? read it submit answer 3. -/3 points details my notes scalcet9 11.10.043.mi.sa. this question has several parts that must be completed sequentially. if you skip a part of the question, you will not receive any points for the skipped part. tutorial exercise use a maclaurin series in this table to obtain the maclaurin series for the given function. f(x) = 11x cos(1/15 x^2)
Answer
Explanation:
Step1: Recall the Maclaurin series for $\cos t$
The Maclaurin series for $\cos t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n)!}t^{2n}=1-\frac{t^{2}}{2!}+\frac{t^{4}}{4!}-\frac{t^{6}}{6!}+\cdots$
Step2: Substitute $t=\frac{1}{15}x^{2}$ into the $\cos t$ series
We get $\cos(\frac{1}{15}x^{2})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}(\frac{1}{15}x^{2})^{2n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}\frac{1}{15^{2n}}x^{4n}$
Step3: Multiply by $11x$
$f(x)=11x\cos(\frac{1}{15}x^{2})=11x\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}\frac{1}{15^{2n}}x^{4n}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}\times11}{(2n)!}\frac{1}{15^{2n}}x^{4n + 1}$
Answer:
$\sum_{n = 0}^{\infty}\frac{(-1)^{n}\times11}{(2n)!}\frac{1}{15^{2n}}x^{4n+1}$