find the bound on the magnitude of the error if we approximate sqrt 2 using the taylor approximation of degree three for sqrt 1+x about x=0
To find the bound on the magnitude of the error when approximating √2 using the Taylor approximation of degree three for √(1+x) about x=0, we can use Taylor's theorem.
Taylor's theorem states that for a function f(x) that is k+1 times differentiable in an interval containing a, the Taylor polynomial of degree k for f(x) about x=a is given by:
P_k(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ... + ((a)/k!)(x-a)^k
And the error term, E_k(x), is given by:
E_k(x) = (f^(k+1)(c))/(k+1)! * (x-a)^(k+1)
where c is some value between a and x.
In this case, let's consider f(x) = √(1+x) and a = 0.
The Taylor polynomial of degree three (k=3) for f(x) about x=0 is:
P_3(x) = √(1+0) + (1/2√1)(x-0) - (1/4√1)(x-0)^2/2 + (3/8√1)(x-0)^3/6
Simplifying, we have:
P_3(x) = 1 + (1/2)x - (1/8)x^2 + (1/16)x^3
To find the error bound, we need to find the maximum value of the absolute value of the error term, E_3(x), in the interval from 0 to 2 (since we want to approximate √2):
E_3(x) = (√(1+c))/(4 * 3!) * x^4
For the maximum value of E_3(x), we need to find the maximum value of (√(1+c))/24 in the interval from 0 to 2.
To find this maximum value, we can use optimization techniques. However, since this is a simple case, we can observe that (√(1+c)) is maximized when c=1. Therefore, we can substitute c=1 into E_3(x):
E_3(x) = (√2)/(24) * x^4
To find the bound on the magnitude of the error, we need to find the maximum value of |E_3(x)| in the interval from 0 to 2.
Substituting x=2 into |E_3(x)|, we have:
|E_3(2)| = (√2)/(24) * 2^4 = (√2)/3
Therefore, the bound on the magnitude of the error when approximating √2 using the Taylor approximation of degree three for √(1+x) about x=0 is (√2)/3.