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?
f' = 2x cos(x^2)
f" = 2cos(x^2) - 4x^2 sin(x^2)
f"' = -12x sin(x^2) - 8x^3 cos(x^2)
so the series is
x^2 -x^6/3! + x^10/5! - x^14/7!
The rest should now present no difficulty
I actually do not know how to do the rest, I am completely lost on this. For part A, when writing the first terms, are they just: x^2 -x^6/3! + x^10/5! - x^14/7! or do I have to plug in numbers 1 through 4, and then solve to get the first four terms?
I think you should look this problem up online or ask a teacher.
To answer these questions, we need to understand the Maclaurin series expansion for the function f(x) = sin(x^2). The Maclaurin series is a special case of the Taylor series, where we expand a function around x = 0.
A) Writing the first four terms of the Maclaurin series for f(x):
The Maclaurin series expansion for sin(x) is as follows:
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
In our case, we have f(x) = sin(x^2). So, we substitute x with x^2 in the Maclaurin series for sin(x):
sin(x^2) = (x^2) - ((x^2)^3)/3! + ((x^2)^5)/5! - ((x^2)^7)/7! + ...
Expanding the terms, we get:
sin(x^2) = x^2 - (x^6)/3! + (x^10)/5! - (x^14)/7! + ...
So, the first four terms of the Maclaurin series for f(x) = sin(x^2) are:
f(x) = x^2 - (x^6)/6 + (x^10)/120 - (x^14)/5040
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, we need to integrate the first four terms of the Maclaurin series for f(x) over the interval from 0 to 1:
∫[0 to 1] f(x) dx = ∫[0 to 1] (x^2 - (x^6)/6 + (x^10)/120 - (x^14)/5040) dx
By integrating each term, we get:
∫[0 to 1] f(x) dx = (1/3)x^3 - (1/42)x^7 + (1/1320)x^11 - (1/90720)x^15
Evaluating the integral from 0 to 1, we substitute x = 1 and x = 0:
∫[0 to 1] f(x) dx = [(1/3)(1)^3 - (1/42)(1)^7 + (1/1320)(1)^11 - (1/90720)(1)^15] - [(1/3)(0)^3 - (1/42)(0)^7 + (1/1320)(0)^11 - (1/90720)(0)^15]
= 1/3 - 1/42 + 1/1320 - 1/90720
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?
To determine the number of terms needed for the desired accuracy, we need to keep adding terms until the difference between the sum of the series and the target value is within the desired tolerance.
Let's denote the value of the integral obtained in part B as I. We keep adding terms to the series until the difference between the sum of the series and I is less than or equal to the desired tolerance of three decimal places.
Starting with the value obtained in part B, we can add terms one by one to see if the difference falls within the desired tolerance:
I = 1/3 - 1/42 + 1/1320 - 1/90720
By adding the next term, we get:
I2 = I + (-1/409,080)
= 1/3 - 1/42 + 1/1320 - 1/90720 - 1/409,080
Checking the difference between I2 and I, we see if it falls within the desired tolerance. If not, we continue adding terms until the difference satisfies the tolerance.
Repeat this process until the difference between the last two sums is within the desired tolerance. The number of terms needed to achieve this represents the answer to part C, and the value of the integral correct to three decimal places is the final sum.