The prompt for this question is f(x) =sin(x^2)
A)Write the first four terms of the Maclaurin series for f(x)
B)Use the Maclaurin series found in Part A to approximate the integral from 0 to 1 of sin(x^2) dx
C)How many terms are needed to find the value of the integral given in Part B, correct to three decimal places? What is that value?
I have no idea what to do for this whole problem, please write out the steps
recall that
sin(x) = x - x^3/3! + x^5/5! - x^7/7! for the first 4 terms
so replacing x with x^2
sin(x^2 = x^2 - x^6/6 + x^10/120 + x^14/5040
ʃsin(x^2) = ʃ(x^2 - x^6/6 + x^10/120 + x^14/5040) dx from 0 to 1
= [(1/3)x^3 - (1/42)x^7 + (1/1320)x^11 - (1/75600)x^15] from 0 to 1
= (1/3 - 1/42 + 1/1320 - 1/75600) - 0
= .310268
confirmation
https://www.wolframalpha.com/input/?i=%CA%83+sin(x%5E2)+dx+from+0+to+1
So my answer using 4 terms is correct to 6 decimal places
for c) take one less term and see if the answer changes correct to 3 decimals.
f(x) = f(0) + x f'(0) + (x^2/2)f''(0) +(x^3/6)f'''(0) ......
f(0) = sin(x^2) = 0
f'(0) = 2xcosx^2 = 2*0 = 0
f''(0) = -2x(2x)sinx^2 +2cosx^2 = 2
f'''(0) = -8x^3cosx^2 -8xsinx^2-4xcosx^2
= -8x^3
so
f(x) = 2(x^2/2) -8 x^3(x^3/6)
= x^2 -(4/3)x^6
now integrate
x^3/3 - (4/3)x^7/7
= x^3/3 - 4x^7/21
CHECK MY ARITHMETIC
Okay I changed C) and used one less term and got 0.31028 as the answer. What does that mean?
To solve this problem, we'll break it down into three parts: A, B, and C. Let's go step by step:
A) Writing the first four terms of the Maclaurin series for f(x):
The Maclaurin series is a Taylor series expansion centered at x = 0. The general formula for the Maclaurin series expansion of a function f(x) is given by:
f(x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + f'''(0)(x-0)^3/3! + ...
To find the first four terms of the Maclaurin series for f(x) = sin(x^2), we need to evaluate the function and its derivatives at x = 0:
f(x) = sin(x^2)
First term (n = 0): f(0) = sin(0^2) = sin(0) = 0
Second term (n = 1): f'(x) = d(sin(x^2))/dx = 2x*cos(x^2) --> f'(0) = 2(0)*cos(0^2) = 0
Third term (n = 2): f''(x) = d(2x*cos(x^2))/dx = 2*cos(x^2) - 4x^2*sin(x^2) --> f''(0) = 2*cos(0^2) - 4(0)^2*sin(0^2) = 2
Fourth term (n = 3): f'''(x) = d(2*cos(x^2) - 4x^2*sin(x^2))/dx = -8xcos(x^2) - 8x^3*cos(x^2) - 4x^2*sin(x^2) --> f'''(0) = -8(0)cos(0^2) - 8(0)^3*cos(0^2) - 4(0)^2*sin(0^2) = 0
Therefore, the Maclaurin series for f(x) = sin(x^2) is:
f(x) ≈ 0 + 0(x-0) + 2(x-0)^2/2! + 0(x-0)^3/3!
≈ 0 + x^2 + 0
So the first four terms of the Maclaurin series for f(x) are: 0, x^2, 0.
B) Using the Maclaurin series found in Part A to approximate the integral from 0 to 1 of sin(x^2) dx:
To approximate the integral using the Maclaurin series, we can substitute the Maclaurin series of f(x) into the integral:
∫[0 to 1] sin(x^2) dx ≈ ∫[0 to 1] (0 + x^2 + 0) dx
≈ ∫[0 to 1] x^2 dx
Integrating x^2 from 0 to 1:
∫[0 to 1] x^2 dx = [x^3/3] evaluated from 0 to 1
= (1^3/3) - (0^3/3)
= 1/3
Therefore, the approximation of the integral from 0 to 1 of sin(x^2) dx using the Maclaurin series is 1/3.
C) Determining the number of terms needed for the value of the integral to be correct to three decimal places:
To determine the number of terms needed, you can keep adding more terms to the Maclaurin series until the difference between consecutive approximations becomes less than the desired accuracy.
Let's continue adding terms to our Maclaurin series:
f(x) ≈ 0 + x^2 + 0 + (R4(x)), where R4(x) represents the remainder term.
Adding the next term (n = 4):
f''''(x) = d(-8xcos(x^2) - 8x^3*cos(x^2) - 4x^2*sin(x^2))/dx
= 8cos(x^2) - 24x^2*cos(x^2) + 24x^4*sin(x^2) - 8x*sin(x^2)
To estimate the integral to the desired accuracy, we need to determine the value of x such that the remainder term, R4(x), becomes less than the desired error (epsilon).
|R4(x)| ≤ epsilon
Let's use a numerical approximation by substituting x = 1 into R4(x):
|R4(1)| = |8cos(1^2) - 24(1)^2*cos(1^2) + 24(1)^4*sin(1^2) - 8(1)*sin(1^2)|
Using a calculator or computer to evaluate this expression, we get:
|R4(1)| ≈ 3.711
To find the number of terms needed, we need to keep adding terms until the remainder term is less than epsilon. Let's set epsilon = 0.001 (three decimal places accuracy).
|R4(1)| ≤ 0.001
From the calculations above, it is evident that four terms in the Maclaurin series are not sufficient to achieve the desired accuracy. We would need to add more terms to the series.
Unfortunately, without further calculations, it is difficult to determine exactly how many terms are needed to achieve the desired accuracy. It would require further calculation of the remainder term, Rn(x), for larger values of n until we find the value of n for which |Rn(1)| ≤ 0.001.
However, we can conclude that at least four terms are not enough to achieve the desired accuracy.