Скачать презентацию 05 Series Today s Objectives To understand
05 Method of Differences.pptx
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Скачать презентацию 05 Series Today s Objectives To understand
Скачать презентацию 05 Series Today s Objectives To understand
05 Series
Today’s Objectives • To understand the principles of the Method of Differences (also known as the method of telescoping sums). • To use the method of differences to evaluate the sum of a series. • To use the method of differences to prove given results.
The principles of the method of differences • As the name suggests we will consider the summation of a difference. • Consider this summation:
The principles of the method of differences • Starting with r = n and working down to r = 1: r =n r differs by 1 r =n– 2 Since r is decreasing by 1 on the LHS, it is neater to start with r = n on the RHS rather than with r = 1 (starting with r = 1 will give the same result). r =2 r =1
The principles of the method of differences • Cancelling the middle terms gives: • So the final simplified result is the upper limit substituted into the first function f(r + 1) and the lower limit substituted into the second function f(r).
The principles of the method of differences • If on the LHS, r increases by 1, then it would be neater to start substituting with r = 1 and we would obtain this result: • i. e. substitute the lower limit into the first function f(r) and the upper limit into the second function f(r+1).
The principles of the method of differences • In practice, rather than try and remember which limit to substitute where, it is often easier to just start expanding the LHS to see which terms cancel and what you are left with.
Example 1 • By using partial fractions and the method of differences, find an expression in terms of n for (an exam question may just ask you to find the expression):
Example 1 Solution • To use partial fractions we must factorise the denominator:
Example 1 Solution • So we can write our fraction as a difference. It is of the form f(r) – f(r + 1) but if we did not see that we can just begin expanding and notice which terms will cancel:
Example 2 • If f(r) = r(r + 1)! then simplify f(r) – f(r – 1) and hence sum the series:
Example 2 Solution • Simplifying f(r) – f(r – 1):
Example 2 Solution • How does this help to sum the given series? • If we write the given series in sigma notation we have: • And the term inside sigma on the RHS equals f(r) – f(r – 1) where f(r) = r(r + 1)!.
Example 2 Solution • Therefore, we have: • And using the method of differences:
Proving results • We can use the method of differences to prove given results, provided that we can write a function in the form f(r) – f(r + 1) or f(r) – f(r – 1). • We will now prove a couple of standard results:
Example 3 • Prove the following using the method of differences:
Example 3 Solution • We shall start by considering the following difference: • Now we will sum both sides:
Example 3 Solution • We shall start by considering the following difference: • Now we will sum both sides:
Example 3 Solution • We can use the method of differences to sum the LHS where f(r) = r 3 and f(r + 1) = (r + 1)3: • Now separate the sum on the RHS and rearrange:
Example 3 Solution • We can now take constants outside of the sigma notation and use standard results:
Example 3 Solution • Now simplifying: • Hence,
Proving results • The same method can be used to prove the standard result: • By considering (r + 1)4 – r 4. • In fact, it doesn’t have to be r + 1 and r as long as it is in the form f(r + 1) – f(r), e. g. (r + 2)4 – (r + 1)4 would also work but the arithmetic would be harder.
Example 4 • Find an expression in terms of n for the following sum: • Hence show that:
Example 4 Solution • Using the method of differences: • Now expanding the LHS:
Example 4 Solution • Putting these two results together: • Therefore:
Example 4 Solution • Using standard results and simplifying:
Example 4 Solution • Continuing to simplify gives: • Hence,
Example 5 • Use the identity: • To find:
Example 5 Solution • Using the identity we have: • The RHS is in the form:
Example 5 Solution • Using the method of differences:
Example 5 Solution • So, • Therefore,
Example 6 • Using the method of differences, prove that:
Example 6 Solution • Firstly we must rewrite the LHS as a difference. Since the denominator is a surd we will rationalise the denominator:
Example 6 Solution • This is now in the form f(r) – f(r – 1) where f(r) = √r. Now we can use the method of differences: • Hence,
In Summary • The “formula” for the method of differences is: • We can use the method to prove a given sum of a series by considering a difference or to find the sum of a series in terms of n.
Скачать презентацию 05 Series Today s Objectives To understand
05 Method of Differences.pptx