MATH105 April 2012
• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q8 (e) • Q8 (f) • Q8 (g) • Q8 (h) • Q8 (i) •
Question 01 (c)
Express the Taylor series of the function
about x = 0 in summation notation.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
First, use the fact from part (b) that
provided |r| < 1, to get the power series of per integration. Then make a proper substitution for r.
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
- If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
- If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.
Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies.
For |r|<1 we use that
and integrate both sides to obtain
Since |r| < 1 we have |1-r| = 1-r, therefore
To solve for the integration constant C we plug in r = 0:
so C = 0.
So we obtain the intermediate result that
To arrive at the summation notation, we noted that every term has the same sign (so there are no terms like floating around), every integer power of is included starting from , and each is divided by k.
As a last step, the Taylor series for can be found by replacing by in the series above. We find:
Click here for similar questions
MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Taylor series, Pages using DynamicPageList parser function, Pages using DynamicPageList parser tag