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FP2: Method of Differences - Where have I gone wrong?

tumblr_m8fhc6QMya1rww3jzo1_500.png

I've done part a and for part b I did the whole let n=1, n=2, n=3 blah blah and nothing cancelled out :/

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Original post by creativebuzz
tumblr_m8fhc6QMya1rww3jzo1_500.png

I've done part a and for part b I did the whole let n=1, n=2, n=3 blah blah and nothing cancelled out :/


What did you get for n=1, n=2 and n=3?

Think back to FP1 and this may simplify the problem.
Original post by SeanFM
What did you get for n=1, n=2 and n=3?

Think back to FP1 and this may simplify the problem.


n=1:
1/2

n=2:
7/6

n=3:
25/12

Ah, I can't remember much of FP1 :/
Original post by creativebuzz
n=1:
1/2

n=2:
7/6

n=3:
25/12

Ah, I can't remember much of FP1 :/


Okay, don't worry :smile:

Think about the first two things: r and -1. Thinking about what I said about FP1, is there any way to remove those from your calculations (for now) so that you're left with (1/r) - (1/(r+1), something you're more familiar with from this chapter?
Original post by SeanFM
Okay, don't worry :smile:

Think about the first two things: r and -1. Thinking about what I said about FP1, is there any way to remove those from your calculations (for now) so that you're left with (1/r) - (1/(r+1), something you're more familiar with from this chapter?



Erm can you seperate them so you can have (sigma)r which is 1/2n(n+1) and (sigma)-1 which is just -1 (I think)
Original post by SeanFM
Okay, don't worry :smile:

Think about the first two things: r and -1. Thinking about what I said about FP1, is there any way to remove those from your calculations (for now) so that you're left with (1/r) - (1/(r+1), something you're more familiar with from this chapter?


Why do you always torture people instead of giving the answer?
Original post by creativebuzz
Erm can you seperate them so you can have (sigma)r which is 1/2n(n+1) and (sigma)-1 which is just -1 (I think)


Almost :smile:

Your first sum (n(n+1))/2 is right. Now think about the -1. If each 'r' contributes -1 to the overall sum.. how many -1's do you have? (In terms of n).
Original post by mrno1324
Why do you always torture people instead of giving the answer?


It'll help them to learn more, and giving the answer is against the spirit of this forum. :smile:
Original post by SeanFM
Almost :smile:

Your first sum (n(n+1))/2 is right. Now think about the -1. If each 'r' contributes -1 to the overall sum.. how many -1's do you have? (In terms of n).


Hmm would you have r many -1s
Original post by creativebuzz
Hmm would you have r many -1s


Technically you'll have 'n' -1s but I know that that is what you meant. (As if you're summing the first 3 terms (n=3), when r=1 you get -1 from the -1, when r=2 you get another -1 all the way up to n.

Now you can progress with the other half of the question :smile:
Original post by SeanFM
Technically you'll have 'n' -1s but I know that that is what you meant. (As if you're summing the first 3 terms (n=3), when r=1 you get -1 from the -1, when r=2 you get another -1 all the way up to n.

Now you can progress with the other half of the question :smile:

But doesn't it just give you the same results

WIN_20150916_203440.JPG
Original post by creativebuzz
But doesn't it just give you the same results

WIN_20150916_203440.JPG


Yes, if you just calculate without splitting up the terms.

But think about why I've asked about the sum of r and -1 and what I mean by 'you're familiar with how to deal with (1/r)-(1/(r+1) when it's a sum'.
Original post by SeanFM
Yes, if you just calculate without splitting up the terms.

But think about why I've asked about the sum of r and -1 and what I mean by 'you're familiar with how to deal with (1/r)-(1/(r+1) when it's a sum'.


Hmm I'm not entirely sure.. :/
Original post by creativebuzz
Hmm I'm not entirely sure.. :/


From part a you can see that the sum of that one fraction is the same as summing it up when it's broken down into four terms.

The sum of all four terms can then be split up into the sum of the sum of each term.

The sum of (1+2+3) is the same as the sum of (1) + the sum of (2) + the sum of (3).
Original post by SeanFM
From part a you can see that the sum of that one fraction is the same as summing it up when it's broken down into four terms.

The sum of all four terms can then be split up into the sum of the sum of each term.

The sum of (1+2+3) is the same as the sum of (1) + the sum of (2) + the sum of (3).


Yeah I understand that but I don't see how it's allowing my values to cancel out like they should
Original post by creativebuzz
Yeah I understand that but I don't see how it's allowing my values to cancel out like they should


Because you already know the sum of r and -1 for whatever n you choose (from FP1), then you're just concerned about (1/r) - (1/r+1), the sum of which you find by the method of differences.
Original post by SeanFM
Because you already know the sum of r and -1 for whatever n you choose (from FP1), then you're just concerned about (1/r) - (1/r+1), the sum of which you find by the method of differences.


Wait so n and r aren't the same thing?
Reply 17
Avatar for Zacken
Zacken
22
Original post by creativebuzz
Wait so n and r aren't the same thing?


What SeanFM is saying might be slightly clearer to you by writing it out using eqations instead.

He's saying that:

∑r=1n(r−1+1r−1r−1)=∑r=1nr−∑r=1n1+∑r=1n(1r−1r−1)\displaystyle \sum_{r=1}^n \left(r - 1 + \frac{1}{r} - \frac{1}{r-1}\right) = \sum_{r=1}^n r - \sum_{r=1}^n 1 + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

This reduces to

n(n+1)2−n+∑r=1n(1r−1r−1)\displaystyle \frac{n(n+1)}{2} - n + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

Now apply the method of differences to the last sum only, that is, sum up/evaluate the below sum individually.

∑r=1n(1r−1r−1)\displaystyle \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right) and then add that result to the first two terms in the above equation.
(edited 8 years ago)
Original post by Zacken
What SeanFM is saying might be slightly clearer to you by writing it out using eqations instead.

He's saying that:

∑r=1n(r−1+1r−1r−1)=∑r=1nr−∑r=1n1+∑r=1n(1r−1r−1)\displaystyle \sum_{r=1}^n \left(r - 1 + \frac{1}{r} - \frac{1}{r-1}\right) = \sum_{r=1}^n r - \sum_{r=1}^n 1 + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

This reduces to

n(n+1)2−n+∑r=1n(1r−1r−1)\displaystyle \frac{n(n+1)}{2} - n + \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right)

Now apply the method of differences to the last sum only, that is, sum up/evaluate the below sum individually.

∑r=1n(1r−1r−1)\displaystyle \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r-1}\right) and then add that result to the first two terms in the above equation.


Ah I see! But just to make clear, is n and r the same thing? If not, what's the difference?
Original post by creativebuzz
Ah I see! But just to make clear, is n and r the same thing? If not, what's the difference?


r represents what we are putting in r=1,2,3,4....,n-1,n
For each term! Here it is represnting the general term and r takes all the values listed in the sum and n is one of them


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