Here is a cold one. Taylor Series

Question

For f(x)=x^3/4. How large should n be so that for x∈[1,2], |f(x)-T_n(x)| < 0.001?

i got n=15

anyone else have similar answer?

thanks! 

Answer 1

f(x) = x^3/4

f'(x) = 3/4 x^-1/4

f''(x) = -1/4*3/4 x^-5/4

f'''(x) = -5/4 * -1/4 * 3/4 x^-9/4 ...

The power of x on fⁿ(x) is x^(-n+3/4)

Labeling the derivative coefficients c_n, c_n= - (4n -7)/4 c_n-1, with c₀ = 1

While this is true for n = 1 as well, as 4n - 7 < 0 for n = 1, this expression does not begin to alternate until n = 2

Of course, we are going to do the Taylor series about 1, so the x^(-n+3/4) will equal 1 for all terms, and we only need to concern ourselves with the coefficient.

As the Taylor series is ∑fⁿ(x)/n!, our estimate of x^(3/4) at x = 2 is, as 2 - 1 = 1, ∑c_n/n!

Thus, the ratio of successive terms is - (4n - 7)/4n

The sum will converge, as it is a decreasing alternate series and we can show by induction that all terms have magnitude less than or equal to 1/n so they converge to 0 (for n = 1 1 <= 1; for n = 2, 1/4 <= 1/n; and for k = n+1, 1/k * (4k - 3)/(4(k+1) < 1/k * 4k/(4(k+1)) = 1/k+1 )

The convergence is very slow, however. There is no nice computation of the first n such that the magnitude of our coefficient is less than .001, so the easiest way to keep track of these terms is through use of a spreadsheet.

Note that a much better estimate would be to average two successive estimates.

For n = 21, our term is 0.00103667904842778 and for n = 22, our term is -0.000954215942302846;

Thus, n = 21 is sufficient