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 find the Taylor polynomial with degree n for the given function f(x) = e^(2x^2) at the number a = 0, we can use the formula for the nth degree Taylor polynomial:
Tn(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ... + f^n(a)(x-a)^n/n!
1. Calculating the derivatives:
f(x) = e^(2x^2)
f'(x) = (d/dx) [e^(2x^2)] = 4x e^(2x^2)
f''(x) = (d^2/dx^2) [e^(2x^2)] = (4 + 16x^2) e^(2x^2)
f'''(x) = (d^3/dx^3) [e^(2x^2)] = (32x + 96x^3) e^(2x^2)
2. Evaluating the derivatives at a = 0:
f(0) = e^(2*0^2) = e^0 = 1
f'(0) = 4*0*e^(2*0^2) = 0
f''(0) = (4 + 16*0^2)*e^(2*0^2) = 4
f'''(0) = (32*0 + 96*0^3)*e^(2*0^2) = 0
3. Plugging the values into the formula:
T(3)(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + f'''(0)(x-0)^3/3!
= 1 + 0(x) + 4(x^2)/2 + 0(x^3)/6
= 1 + 2x^2
Therefore, the Taylor polynomial of degree 3 at a = 0 for f(x) = e^(2x^2) is T(3)(x) = 1 + 2x^2.
To estimate the accuracy of the approximation f ≈ Tn(x) using Taylor's Inequality, we need to find an expression for the remainder term Rn(x) and then derive an upper bound for its absolute value.
The remainder term Rn(x) is given by:
Rn(x) = f(x) - Tn(x)
1. Expressing f(x) - Tn(x):
Rn(x) = e^(2x^2) - (1 + 2x^2)
2. Using Taylor's Inequality:
|Rn(x)| ≤ (M/((n+1)!)) * |x-a|^(n+1)
where M is an upper bound on the absolute value of the (n+1)st derivative of f(x) in the interval [a, x].
3. Finding an upper bound for |Rn(x)|:
To bound |Rn(x)|, we need to find an upper bound for the (n+1)st derivative of f(x), which is f^(n+1)(x).
f^(n+1)(x) = (d^(n+1)/dx^(n+1)) [e^(2x^2)]
In this case, n = 3, and we need to find an upper bound for the fourth derivative, f^(4)(x), since it is the highest derivative in our expression.
4. Calculating the fourth derivative:
f^(4)(x) = (d^4/dx^4) [e^(2x^2)] = (32 + 192x^2 + 384x^4) e^(2x^2)
5. Finding an upper bound for |f^(4)(x)| in the interval [0, 0.1]:
To find an upper bound, we evaluate the derivative at the endpoints of the interval and choose the higher value:
|f^(4)(x)| = max{|f^(4)(0)|, |f^(4)(0.1)|}
|f^(4)(0)| = |32|
|f^(4)(0.1)| = |32 + 192(0.1)^2 + 384(0.1)^4|
Calculating the values:
|f^(4)(0)| = 32
|f^(4)(0.1)| ≈ |32 + 192(0.01) + 384(0.0001)| = 32.0192
6. Evaluating the bound for |Rn(x)|:
|Rn(x)| ≤ (M/((n+1)!)) * |x-a|^(n+1)
Using the maximum value for |f^(4)(x)| and plugging in the values for n, a, and x:
|R(3)(x)| ≤ (32.0192 / (4!)) * (0.1 - 0)^(3+1)
Simplifying the expression:
|R(3)(x)| ≤ (32.0192 / 24) * 0.1^4
= 32.0192 / 24 * 0.0001
= 0.01334133 (rounded to 8 decimal places)
Therefore, when x lies in the interval [0, 0.1], we have |R(3)(x)| ≤ 0.01334 (rounded to five decimal places).