05 Series

Содержание

1. 05 Series
2. Today’s ObjectivesTo understand the principles of the
3. The principles of the method of differencesAs
4. Starting with r = n and working
5. Cancelling the middle terms gives:So the final
6. If on the LHS, r increases by
7. In practice, rather than try and remember
8. Example 1By using partial fractions and the
9. Example 1 SolutionTo use partial fractions we must factorise the denominator:
10. Example 1 SolutionSo we can write our
11. Example 2If f(r) = r(r + 1)!
12. Example 2 SolutionSimplifying f(r) – f(r – 1):
13. Example 2 SolutionHow does this help to
14. Example 2 SolutionTherefore, we have:And using the method of differences:
15. Proving resultsWe can use the method of
16. Example 3Prove the following using the method of differences:
17. Example 3 SolutionWe shall start by considering the following difference:Now we will sum both sides:
18. Example 3 SolutionWe shall start by considering the following difference:Now we will sum both sides:
19. Example 3 SolutionWe can use the method
20. Example 3 SolutionWe can now take constants outside of the sigma notation and use standard results:
21. Example 3 SolutionNow simplifying:Hence,
22. Proving resultsThe same method can be used
23. Example 4Find an expression in terms of n for the following sum:Hence show that:
24. Example 4 SolutionUsing the method of differences:Now expanding the LHS:
25. Example 4 SolutionPutting these two results together:Therefore:
26. Example 4 SolutionUsing standard results and simplifying:
27. Example 4 SolutionContinuing to simplify gives:Hence,
28. Example 5Use the identity:To find:
29. Example 5 SolutionUsing the identity we have:The RHS is in the form:
30. Example 5 SolutionUsing the method of differences:
31. Example 5 SolutionSo,Therefore,
32. Example 6Using the method of differences, prove that:
33. Example 6 SolutionFirstly we must rewrite the
34. Example 6 SolutionThis is now in the
35. In SummaryThe “formula” for the method of
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Today’s ObjectivesTo 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

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Слайд 105 Series

Слайд 2Today’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.

Today’s ObjectivesTo understand the principles of the Method of Differences (also known as the method of telescoping

Слайд 3The principles of the method of differences
As the name suggests

we will consider the summation of a difference.
Consider this

summation:

Слайд 4Starting with r = n and working down to r

= 1:
The principles of the method of differences
r differs by

r = n – 2

r = 2

r = 1

r = n – 1

r = n

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).

Starting with r = n and working down to r = 1:The principles of the method of

Слайд 5Cancelling 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

Cancelling the middle terms gives:So the final simplified result is the upper limit substituted into the first

Слайд 6If 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

If on the LHS, r increases by 1, then it would be neater to start substituting with

Слайд 7In 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.

The principles of the method of differences

In practice, rather than try and remember which limit to substitute where, it is often easier to

Слайд 8Example 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):

$Example 1By using partial fractions and the method of differences, find an expression in terms of n$

Слайд 9Example 1 Solution
To use partial fractions we must factorise the

denominator:

$Example 1 SolutionTo use partial fractions we must factorise the denominator:$

Слайд 10Example 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:

$Example 1 SolutionSo we can write our fraction as a difference. It is of the form f(r)$

Слайд 11Example 2
If f(r) = r(r + 1)! then simplify f(r)

– f(r – 1) and hence sum the series:

Example 2If f(r) = r(r + 1)! then simplify f(r) – f(r – 1) and hence sum

Слайд 12Example 2 Solution
Simplifying f(r) – f(r – 1):

Слайд 13Example 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 SolutionHow does this help to sum the given series?If we write the given series in

Слайд 14Example 2 Solution
Therefore, we have:

And using the method of differences:

Слайд 15Proving 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:

Proving resultsWe can use the method of differences to prove given results, provided that we can write

Слайд 16Example 3
Prove the following using the method of differences:

Слайд 17Example 3 Solution
We shall start by considering the following difference:

Now

we will sum both sides:

Слайд 18Example 3 Solution
We shall start by considering the following difference:

Now

we will sum both sides:

Слайд 19Example 3 Solution
We can use the method of differences to

sum the LHS where f(r) = r3 and f(r +

1) = (r + 1)3:

Now separate the sum on the RHS and rearrange:

Example 3 SolutionWe can use the method of differences to sum the LHS where f(r) = r3

Слайд 20Example 3 Solution
We can now take constants outside of the

sigma notation and use standard results:

Слайд 21Example 3 Solution
Now simplifying:

Hence,

Слайд 22Proving results
The same method can be used to prove the

standard result:

By considering (r + 1)4 – r4.
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.