How do you use Taylor series to estimate the accuracy of approximation for #f(x)=sqrt(x)# with #a=1# and #n=3# with #0.9

Question

How do you use Taylor series to estimate the accuracy of approximation for #f(x)=sqrt(x)# with #a=1# and #n=3# with #0.9<=x<=1.1#?

Elijah Abram · Accepted Answer

Taylor's Theorem guarantees such an estimate will be accurate to within about 0.00000565 over the whole interval #[0.9,1.1]#.

Let #f(x)=sqrt{x}=x^{1/2}# so that #f'(x)= 1/2 x^{-1/2}#, #f''(x)=-1/4 x^{-3/2}#, and #f'''(x)=3/8 x^{-5/2}#. Since #f(1)=1#, #f'(1)=1/2#, #f''(1)=-1/4#, and #f'''(1)=3/8#, the Taylor polynomial of degree #n=3# for #f# centered at #a=1# is #P_{3}(x)=1+1/2 (x-1) - (1/4)/(2!) (x-1)^{2}+ (3/8)/(3!) (x-1)^{3}# #=1+1/2 (x-1) - 1/8 (x-1)^{2} + 1/16 (x-1)^{3}#. If we let #E_{3}(x)=f(x)-P_{3}(x)# be the error in using #P_{3}(x)# to approximate #f(x)# for #x approx a = 1#, then Taylor's Theorem guarantees #|E_{3}(x)|=|E_{n}(x)|\leq M/((n+1)!) * |x-a|^{n+1}=M/(4!)|x-1|^{4}#, where #|f''''(t)|\leq M# for all #t# between #x# and #a=1#. Since #f''''(x)=-15/16 x^{-7/2}#, we have #|f''''(t)| = 15/16 t^{-7/2}#, which is maximized for #t\in [0.9,1.1]# at #t=0.9#. Therefore, we can take #M=|f''''(0.9)|=15/16 * 0.9^{-7/2}approx 1.35557#. Hence, rounding up gives #|E_{3}(x)|\leq 1.3556/24 * 0.1^{4}approx 0.00000565# for #x in [0.9,1.1]#. You can double-check this by noting that #sqrt{.9}-P_{3}(.9) approx -0.00000420# and #sqrt{1.1}-P_{3}(1.1) approx -0.00000365#.

Christopher Allers · Answer

undefined To estimate the accuracy of the approximation for $ f(x) = \sqrt{x} $ using Taylor series with $ a = 1 $ and $ n = 3 $ within the range $ 0.9 \leq x \leq 1.1 $, you can follow these steps:

1. Write down the Taylor series expansion for $ f(x) = \sqrt{x} $ centered at $ a = 1 $, up to the third term ($ n = 3 $).
   
   $$ f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 $$

2. Compute the derivatives of $ f(x) = \sqrt{x} $ up to the third order, evaluated at $ a = 1 $.

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

3. Evaluate these derivatives at $ a = 1 $ to get $ f'(1) $, $ f''(1) $, and $ f'''(1) $.

$$ f'(1) = \frac{1}{2} $$
   $$ f''(1) = -\frac{1}{4} $$
   $$ f'''(1) = \frac{3}{8} $$

4. Plug these values into the Taylor series expansion formula.

$$ f(x) \approx f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 $$

$$ \approx 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 + \frac{1}{16}(x-1)^3 $$

5. Use the interval $ 0.9 \leq x \leq 1.1 $ to estimate the accuracy of the approximation. Calculate the maximum error by finding the maximum value of the absolute value of the fourth derivative within this interval.

$$ |f^{(4)}(x)| = |-15x^{-7/2}| = \frac{15}{x^{7/2}} $$

The maximum value occurs at the endpoints of the interval:

$$ \frac{15}{(0.9)^{7/2}} $$ and $$ \frac{15}{(1.1)^{7/2}} $$

6. Evaluate these values to find the maximum error.

How do you use Taylor series to estimate the accuracy of approximation for #f(x)=sqrt(x)# with #a=1# and #n=3# with #0.9&lt;=x&lt;=1.1#?

How do you use Taylor series to estimate the accuracy of approximation for #f(x)=sqrt(x)# with #a=1# and #n=3# with #0.9<=x<=1.1#?