The prompt for all of these question is "consider the function f(x) = sin^2(x)".
Part A: Write the first four terms of the Maclaurin series for f(x).
I assumed this implied non-zero terms, so I found x^2-(1/3)x^4+(2/45)x^6-1/315(x^8). I am fairly confident about this.
Part B: Use the Maclaurin series found in Part A to approximate the integral from 0 to 1 of sin^2(x)
I know sin^2(x) is about equal to the Maclaurin series around x=0, so the integrated Maclaurin series would be (x^3)/3-(x^5)/15+2(x^7)/315-(x^9)/2835. Evaluated an an integral from 0 to 1 yields 0.2726631, an answer I am confident in.
Part C: How many terms are needed to find the value of the integral given in Part B, correct to three places? What is that value?
I am unsure what "correct to three decimal places" means exactly, because most questions typically give an actual value as a error bound. I am also unsure whether to consider the original terms of the Maclaurin series, or the anti-derivatised terms that will be used to approximate the integral. For example I know that the fourth non-zero term of the integrated series will be ((1)^9)/2835 = 0.0003527, which seems to be "correct to three decimal places", but I am not sure if three zeros is what is implied by that. If I look at non integrated terms, then the fifth term is the first to have three zeros 2((x)^10)/14175 = 0.0001411.
Overall I am just confused on whether I should be looking at integrated terms of the Maclaurin series, or the original terms, as well as what meant by "correct to three places". Any help would be greatly appreciated!
Part A:
There are different ways to do this part, but your answer is correct. Hope you used the simplest way.
Part B:
Again, you answer is correct. Using integration term by term is also correct.
Part C:
"correct to 3 decimal places" generally means that the sum of the remaining (discarded) terms must not be greater than 0.0005. In an alternating series (as in this case), it is sufficient to have the first discarded term less than 0.0005.
The above procedure applies to the series of the final answer, i.e. the integrated terms, so your procedure is correct.
The 3-term approximation is 0.2730 and the exact integral is 0.2727, both of which are as predicted in your analysis.
Having said that, we must note that the procedure above does not guarantee that the third digit after the decimal is correct. For example, if the correct answer is 0.245499, and the approximation is 0.245623. The error is only 0.000124, but the rounded values are respectively 0.245 and 0.246.
To find the value of the integral in Part B correct to three decimal places, you'll need to determine the number of terms needed in the Maclaurin series. "Correct to three decimal places" means that the difference between the approximate value obtained using a certain number of terms and the actual value of the integral should be within 0.001 (or 1/1000).
To determine the number of terms needed, you can start with a few terms and then gradually increase the number of terms until the desired accuracy is achieved. In this case, since you have already found the integrated form of the Maclaurin series, it would be more appropriate to use those terms.
Let's denote the approximate value of the integral obtained using n terms as I_n. You can start by evaluating I_n for a small value of n, for example, let's say n = 5. Calculate I_5 by substituting x = 1 into the integrated Maclaurin series and evaluate the resulting expression.
I_5 = (1^3)/3 - (1^5)/15 + 2(1^7)/315 - (1^9)/2835
Repeat this process for increasing values of n (e.g., n = 6, 7, 8, etc.) until the difference between successive values of I_n is less than 0.001. This means that the desired accuracy has been achieved.
For example, if you find that the values of I_5 and I_6 are 0.275 and 0.272, respectively, then you can conclude that the value of the integral correct to three decimal places is approximately 0.272.
Note that this iterative process is necessary because it's difficult to determine the exact number of terms needed without knowledge of the actual value of the integral. However, once you find a point where the difference between successive values is within the desired accuracy, you can be confident that the value is correct to three decimal places.