91Ó°ÊÓ

Suppose that the function \(f\) is infinitely differentiable on an oper: interval that contains \(0,\) and suppose that \(f^{\prime}(x)=\) \(-2 f(x)\) and \(f(0)=1 .\) Express \(f(x)\) as a power series in \(x\) What is the sum of this series?

Let \(f\) be a continuous, positive, decreasing function on [1, \infty) for which \(\int_{1}^{\infty} f(x) d x\) converges. Then we know that the series \(\sum_{k=1}^{\infty} f(k)\) also converges. Show that $$0

Suppose that the function \(f\) is infinitely differentiable on an often interval that contains \(0,\) and suppose that \(f^{\prime \prime}(x)=\) \(-2 f(x)\) for all \(x\) and \(f(0)=0 . f^{\prime}(0)=1 .\) Express \(f(x)\) as a power series in \(x .\) What is the sum of this series?

Show that $$\sinh x=\sum_{k=0}^{\infty} \frac{1}{(2 k+1)} x^{2 k+1} \text { for all real } x$$

Expand \(f(x) . f^{\prime}(x),\) and \(\int f(x) d x\) in power series (a) \(f(x)=x 2^{-1}\) (b) \(f(x)=x \arctan x\)

Show that $$\cosh x=\sum_{k=0}^{\infty} \frac{1}{(2 k) !} x^{2 k} \quad \text { for all real } x$$

Estimate within 0.001 by series expansion and check your result by carrying out the integration directly. $$\int_{0}^{1 / 2} x \ln (1-x) d x$$

Derive a series expansion in \(x\) for the function and specify the numbers \(x\) for which the expansion is valid. Take \(a>0\). $$f(x)=e^{2 x}$$

Estimate within 0.001 by series expansion and check your result by carrying out the integration directly. $$\int_{0}^{1} x \sin x d x$$

Derive a series expansion in \(x\) for the function and specify the numbers \(x\) for which the expansion is valid. Take \(a>0\). $$f(x) \equiv \sin a x$$