Consider the following function.
f(x) = e^(2 x^2), a = 0, n = 3, 0 <= x <= 0.1
(a) Approximate f by a Taylor polynomial with degree n at the number a.
T(3)(x) = ?
(b) Use Taylor's Inequality to estimate the accuracy of the approximation f ≈ Tn(x) when x lies in the given interval. (Round the answer to five decimal places.)
|R(3)(x)| </ ?
Note: Please make sure to answer both a & b by filling in "?" show me answer and work because I need help solving them out
To approximate the function f(x) = e^(2x^2) using a Taylor polynomial with degree n at the number a, we need to find the nth-degree Taylor polynomial Tn(x).
(a) Let's start by calculating the derivatives of f(x) up to the third degree and evaluating them at x = a = 0.
f(x) = e^(2x^2)
f'(x) = 4x * e^(2x^2)
f''(x) = (4 + 16x^2) * e^(2x^2)
f'''(x) = (64x + 48x^3) * e^(2x^2)
Evaluate these derivatives at x = 0:
f(0) = e^(2(0^2)) = 1
f'(0) = 4(0) * e^(2(0^2)) = 0
f''(0) = (4 + 16(0^2)) * e^(2(0^2)) = 4
f'''(0) = (64(0) + 48(0^3)) * e^(2(0^2)) = 0
Now, we can construct the Taylor polynomial Tn(x) using the results of these evaluations:
T(3)(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3
= 1 + 0(x-0) + (4/2)(x-0)^2 + 0(x-0)^3
= 1 + 2x^2
Therefore, the third-degree Taylor polynomial for f(x) at x = a = 0 is T(3)(x) = 1 + 2x^2.
(b) To estimate the accuracy of the approximation f ≈ Tn(x) using Taylor's Inequality, we need to find an upper bound for the remainder term Rn(x).
The remainder term Rn(x) is given by the formula:
Rn(x) = (f^(n+1)(t))(x-a)^(n+1) / (n+1)!
where t lies between a and x.
Since we have already calculated the derivatives up to the third degree, we need to find the maximum value of the fourth derivative f^(4)(x) in the interval [0, 0.1]. To do this, we can evaluate the fourth derivative at the endpoints of the interval and choose the larger value.
f''''(x) = (192 + 48x^2) * e^(2x^2)
Evaluate f''''(x) at x = 0 and x = 0.1:
f''''(0) = (192 + 48(0^2)) * e^(2(0^2)) = 192
f''''(0.1) = (192 + 48(0.1^2)) * e^(2(0.1^2)) ≈ 192.485
The maximum value of f''''(x) in the interval [0, 0.1] is approximately 192.485.
Now, let's calculate the upper bound for the remainder term Rn(x) using Taylor's Inequality:
|R(3)(x)| ≤ (192.485)(0.1 - 0)^(3+1) / (3+1)!
= (192.485)(0.1^4) / 4!
≈ 0.000128
Therefore, the estimated accuracy of the approximation f ≈ Tn(x) when x lies in the interval [0, 0.1] is |R(3)(x)| ≤ 0.000128.