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
thats wrong a) is 1+2x^2 I need help with the answering b if you don't mind
To find the Taylor polynomial approximation T3(x) and the remainder R3(x), we will follow these steps:
(a) Finding the Taylor polynomial approximation:
Step 1: Determine the values of f(a), f'(a), f''(a), and f'''(a). These will serve as the coefficients of the Taylor polynomial.
In this case, a = 0, so we need to find the derivatives of f(x) up to the third order and evaluate them at x = 0.
f(x) = e^(2x^2)
f(0) = e^(2(0)^2) = e^0 = 1
f'(x) = 4x * e^(2x^2)
f'(0) = 4(0) * e^(2(0)^2) = 0
f''(x) = (4 + 8x^2) * e^(2x^2)
f''(0) = (4 + 8(0)^2) * e^(2(0)^2) = 4
f'''(x) = (16x + 16x^3) * e^(2x^2)
f'''(0) = (16(0) + 16(0)^3) * e^(2(0)^2) = 0
Step 2: Write the Taylor polynomial.
The general form of a Taylor polynomial of degree n centered at a is:
Tn(x) = f(a) + f'(a)*(x - a) + (f''(a) / 2!)*(x - a)^2 + (f'''(a) / 3!)*(x - a)^3 + ... + (fn(a) / n!)*(x - a)^n
Substituting the values we found:
T3(x) = 1 + 0*(x - 0) + (4 / 2!)*(x - 0)^2 + 0*(x - 0)^3
T3(x) = 1 + 2x^2
(a) Therefore, the Taylor polynomial approximation T3(x) for f(x) = e^(2x^2) is T3(x) = 1 + 2x^2.
(b) Using Taylor's Inequality:
To estimate the accuracy of the approximation f ≈ Tn(x) when x lies in the given interval [0, 0.1], we need to find the maximum value of the absolute value of the (n+1)th derivative of f(x) in that interval. Taylor's Inequality states that the remainder Rn(x) is bounded by the maximum value of |f(n+1)(c)| * [(x - a)^(n+1)] / (n+1)!, where c is some value between a and x.
In this case, n = 3, so we need to find the maximum value of |f(4)(c)| * [(x - a)^4] / 4!.
f(x) = e^(2x^2)
f^(4)(x) = (48x^4 + 96x^2 + 16) * e^(2x^2)
Since we're interested in the interval [0, 0.1], we can find the maximum value of |f^(4)(c)| in that interval. We'll perform numerical calculations to approximate the maximum.
For x = 0.1:
|f^(4)(c)| = (48c^4 + 96c^2 + 16) * e^(2c^2)
By evaluating |f^(4)(c)| at different values of c within [0, 0.1], such as c = 0 and c = 0.1, we can find the maximum.
|f^(4)(0)| = (48(0)^4 + 96(0)^2 + 16) * e^(2(0)^2) = 16
|f^(4)(0.1)| = (48(0.1)^4 + 96(0.1)^2 + 16) * e^(2(0.1)^2) ≈ 16.0842
Therefore, the maximum value of |f^(4)(c)| in the interval [0, 0.1] is approximately 16.0842.
Now, we can use Taylor's Inequality:
|R3(x)| ≤ M * |(x - a)^4| / 4!
where M is the maximum value of |f^(4)(c)| within the interval.
|R3(x)| ≤ 16.0842 * (0.1 - 0)^4 / 4!
|R3(x)| ≤ 16.0842 * 0.0001 / 24
|R3(x)| ≤ 0.00673608
(b) Therefore, based on Taylor's Inequality, |R3(x)| </ 0.00674 (rounded to five decimal places).
With a=0, this is just a MacLaurin series, so
f(0) = 1
f'(0) = 0
f"(0) = 4
f(3)(0) = 48
Now you can write the series and apply the inequality.