Question

1) Find the Taylor polynomial of order n for f(x) = ln 2x about x = 1?

2) Find the Taylor polynomial of order 3 for f(x) = 2√x about a = 4?

3)The area bounded by the curves x = 4y^2 , the line y = 1 and the y-axis in the first quadrant is rotated around the line x = 4: Determine the volume of resulting solid of revolution.

Answer 1

1). We remind that the Taylor series expansion for the ln y around 1 (http://www.math.com/tables/expansion/log.htm):

$ln(y)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}(y-1)}{n}$

After substitution $y=2x$ we receive:

$ln(2x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}(2x-1)}{n}$

2). We remind the Taylor formula:

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+...+\frac{f^{(k)}(a)}{k!}(x-a)^{(k)}+h_k(x)(x-a)^{(k)},$

where $h_k(x)\rightarrow0,x\rightarrow a$ .

We compute the derivatives of $f(x)=2\sqrt{x}$ and get:

$f'(x)=\frac{1}{\sqrt{x}},\qquad f''(x)=-\frac{1}{2}\frac{1}{x\sqrt{x}},\qquad f'''(x)=\frac{3}{4}\frac{1}{x^2\sqrt{x}}$

Setting $k=3$ and $a=4$ we get:

$2\sqrt{x}=4+\frac{1}{2}(x-4)-\frac{1}{32}(x-4)^2+\frac{1}{256}(x-4)^3+h_3(x)(x-4)^3$ ,

$h_3(x)\rightarrow0, x\rightarrow4$

3). The volume of the body, which is obtained after rotation of the graph is:

$\int_{0}^1\pi (4-4y^2)^2dy=\int_{0}^116\pi (1-2y^2+y^4)dy=16\pi(y-\frac{2}{3}y^3+\frac{y^5}{5})|_0^1=8\pi$

Answer: 1). $ln(2x)=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}(2x-1)}{n}$

2). $2\sqrt{x}=4+\frac{1}{2}(x-4)-\frac{1}{32}(x-4)^2+\frac{1}{256}(x-4)^3+h_3(x)(x-4)^3$ ,

$h_3(x)\rightarrow0, x\rightarrow4$

3). $8\pi$