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# FP2 help watch

1. I'm stuck on question number 7. http://www.ocr.org.uk/Images/136137-...hematics-2.pdf

I've done part 1 a=2 b=n c=1 d=n-1

I'm not really too sure how to do part 2. I've integrated y=1/x using the limits from 1 to n, to get ln(n), but then I'm not too sure what to do.
2. (Original post by Music99)
I'm stuck on question number 7. http://www.ocr.org.uk/Images/136137-...hematics-2.pdf

I've done part 1 a=2 b=n c=1 d=n-1

I'm not really too sure how to do part 2. I've integrated y=1/x using the limits from 1 to n, to get ln(n), but then I'm not too sure what to do.
Substituting those values in we have

I'll start by finding the upper bound:

We have a lower bound for so replacing with that lower bound will make bigger:

Can you use this to find the upper bound? Then follow a similar procedure to find the lower bound.
3. (Original post by notnek)
Substituting those values in we have

I'll start by finding the upper bound:

We have a lower bound for so replacing with that lower bound will make bigger:

Can you use this to find the upper bound? Then follow a similar procedure to find the lower bound.
So everything apart from the 1 on the RHS cancels so we get f(n)<1 so the upper bound is 1? Also if that is correct can you explain why this is the upperbound I would have thought it is the lower bound... (Sorry if I'm being stupid).
4. (Original post by Music99)
So everything apart from the 1 on the RHS cancels so we get f(n)<1 so the upper bound is 1? Also if that is correct can you explain why this is the upperbound I would have thought it is the lower bound... (Sorry if I'm being stupid).
you can subtract ln n from both sides of the first statement notnek gave you and then rearrange slightly to give you something that looks like

f(n) - 1 < 0 < f(n) - something else

This form gives you the upper and lower bounds for f(n)
5. (Original post by Music99)
So everything apart from the 1 on the RHS cancels so we get f(n)<1 so the upper bound is 1? Also if that is correct can you explain why this is the upperbound I would have thought it is the lower bound... (Sorry if I'm being stupid).
1 is correct.

If you're looking to make something as large as possible and subtracting something, you want the number you're subtracting to be as small as possible.

e.g.

2<3<4

5-4 < 5-3 < 5-2
6. (Original post by notnek)
1 is correct.

If you're looking to make something as large as possible and subtracting something, you want the number you're subtracting to be as small as possible.

e.g.

2<3<4

5-4 < 5-3 < 5-2
Ah brilliant thank you! I need to practice these! :P
7. (Original post by notnek)
1 is correct.

If you're looking to make something as large as possible and subtracting something, you want the number you're subtracting to be as small as possible.

e.g.

2<3<4

5-4 < 5-3 < 5-2
So i reattempted the question and I've confused myself.

the integral comes out as ln(n) so we can rewrite the inequalities as

. So then we can rearrnage to get

. Which is the same as

. From here I get stuck.
8. (Original post by Music99)
So i reattempted the question and I've confused myself.

the integral comes out as ln(n) so we can rewrite the inequalities as

. So then we can rearrnage to get

. Which is the same as

. From here I get stuck.
If you're going to subtract from one side of the inequality, you need to subtract it from all other sides as well:

Try continuing from here.
9. (Original post by notnek)
If you're going to subtract from one side of the inequality, you need to subtract it from all other sides as well:

Try continuing from here.
So we get ? I'm not too sure on the RHS.
10. (Original post by Music99)
So we get ? I'm not too sure on the RHS.
The LHS is not f(n), it's missing a 1. And the RHS is missing a 1/n term from f(n). So

I assume because you are using this method, you weren't comfortable with the method I gave?
11. (Original post by notnek)
The LHS is not f(n), it's missing a 1. And the RHS is missing a 1/n term from f(n). So

I assume because you are using this method, you weren't comfortable with the method I gave?
I initially thought I was, but when i tried it I got really confused, I'll look over it again though.
12. (Original post by Music99)
I initially thought I was, but when i tried it I got really confused, I'll look over it again though.
I got the same as notnek when I did it, which gives the bounds 1/n < f(n) < 1, although I actually thought I was doing the same as him but in 1 stage rather than 2, as opposed to giving you a completely different method
13. (Original post by davros)
I got the same as notnek when I did it, which gives the bounds 1/n < f(n) < 1, although I actually thought I was doing the same as him but in 1 stage rather than 2, as opposed to giving you a completely different method
Yeah I looked at the wrong bit of my working on paper, so didn't see the 1 was meant to be there for f(n) :P. I'll just practice a few and it will be fine .

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