How do you write the taylor series for #f(x)=sqrt(x)# at #a=16# and find the radius of convergence.?

Question

Charles Arocha · Accepted Answer

#f(x)=4+1/8(x-16)-1/512(x-16)^2+1/16384(x-16)^3+cdots#

and radius of convergence when

#L=abs(1-x/16)<1# A Taylor series is a series expansion of a function about a point. A one-dimensional Taylor series is an expansion of a real function #f(x)# about a point #x=a# is given by #f(x)=f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f^(3)(a))/(3!)(x-a)^3+cdots+(f^(n)(a))/(n!)(x-a)^n+cdots# We are given #f(x)=sqrt(x)# and #a=16# Here are the first four derivatives #f'(x)=1/(2sqrt(x))# #f''(x)=-(1)/(4x^(3//2))# #f^((3))(x)=(3)/(8x^(5//2))# #f^((4))(x)=-(15)/(16x^(7//2))# Plug these derivatives into the Taylor series expansion above #f(x)=sqrt(a)+1/(2sqrt(a))(x-a)-(1)/(2!4a^(3//2))(x-a)^2+(3)/(3!8a^(5//2))(x-a)^3-(15)/(4!16a^(7//2))(x-a)+cdots# Plug in #a=16# #f(x)=sqrt(16)+1/(2sqrt(16))(x-16)-(1)/(2!4(16)^(3//2))(x-16)^2+(3)/(3!8(16)^(5//2))(x-16)^3-(15)/(4!16(16)^(7//2))(x-16)+cdots# Simplify #sqrt(16)=4# #f(x)=4+1/(2(4))(x-16)-(1)/(2!4(4)^(3))(x-16)^2+(3)/(3!8(4)^(5))(x-16)^3-(15)/(4!16(4)^(7))(x-16)+cdots# Simplify the fractions #f(x)=4+1/8(x-16)-1/512(x-16)^2+1/16384(x-16)^3-1/2097152(x-16)^4+cdots# Finally, the radius of convergence is found when #L=abs(1-x/16)<1#

Elijah Abram · Answer

undefined To write the Taylor series for $ f(x) = \sqrt{x} $ at $ a = 16 $, we first need to find the derivatives of $ f(x) $ at $ x = 16 $ up to the desired order. Then, we can use the Taylor series formula to express $ f(x) $ as a sum of these derivatives multiplied by powers of $ x - a $.

First, let's find the derivatives of $ f(x) = \sqrt{x} $:

1. $ f(x) = \sqrt{x} $
2. $ f'(x) = \frac{1}{2\sqrt{x}} $
3. $ f''(x) = -\frac{1}{4x^{3/2}} $
4. $ f'''(x) = \frac{3}{8x^{5/2}} $
5. $ f^{(4)}(x) = -\frac{15}{16x^{7/2}} $

Now, evaluate these derivatives at $ x = 16 $:

1. $ f(16) = \sqrt{16} = 4 $
2. $ f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8} $
3. $ f''(16) = -\frac{1}{4\cdot16^{3/2}} = -\frac{1}{32} $
4. $ f'''(16) = \frac{3}{8\cdot16^{5/2}} = \frac{3}{512} $
5. $ f^{(4)}(16) = -\frac{15}{16\cdot16^{7/2}} = -\frac{15}{8192} $

The Taylor series centered at $ a = 16 $ is:

$$ f(x) = 4 + \frac{1}{8}(x-16) - \frac{1}{32}(x-16)^2 + \frac{3}{512}(x-16)^3 - \frac{15}{8192}(x-16)^4 + \cdots $$

The general term of the series is given by:

$$ \frac{f^{(n)}(a)}{n!}(x-a)^n $$

The radius of convergence $ R $ of a Taylor series can be determined using the ratio test:

$$ R = \lim_{n \to \infty} \frac{|a_{n}|}{|a_{n+1}|} $$

Where $ a_n $ is the $ n $-th term of the series. In this case, the series converges for all real numbers $ x $, so the radius of convergence is $ R = \infty $.