Dannygem
Badges: 1
Rep:
?
#1
Report Thread starter 6 years ago
#1
Image

(d) I am confused as to how to go about simplifying this to get a answer for n in terms of E - still stuck in terms of simplification
0
reply
james22
Badges: 16
Rep:
?
#2
Report 6 years ago
#2
d): Try to write it as the product of 5 different terms with a single n from the numerator, and a single term from the denominator in each. 4 of these should tend to 1, and the other to 0.

2. x need not be an integer. Suppose x was not 0, what would that mean?

3. Use part 2.
0
reply
Dannygem
Badges: 1
Rep:
?
#3
Report Thread starter 6 years ago
#3
(Original post by james22)
d): Try to write it as the product of 5 different terms with a single n from the numerator, and a single term from the denominator in each. 4 of these should tend to 1, and the other to 0.

2. x need not be an integer. Suppose x was not 0, what would that mean?

3. Use part 2.
Thanks again James,

I split up the (n^4)/n! into n/n x n/(n-1) x n/(n-2) x n/(n-3) x 1/(n-4)!

Although i'm unsure where to go from there because although they tend to 1 (aparts from 1/(n-4)!, in order to simplify the expression I need to make sure it becomes larger. Since n > 0, 1 is not larger than n/(n-1). Is it allowed to simplify it to something larger (such as 2, instead of 1?), even though it tends to 1? I don't think I can let 1/(n-4)! become 0 either as I need to get a value for n in terms of epsilon
0
reply
james22
Badges: 16
Rep:
?
#4
Report 6 years ago
#4
(Original post by Dannygem)
Thanks again James,

I split up the (n^4)/n! into n/n x n/(n-1) x n/(n-2) x n/(n-3) x 1/(n-4)!

Although i'm unsure where to go from there because although they tend to 1 (aparts from 1/(n-4)!, in order to simplify the expression I need to make sure it becomes larger. Since n > 0, 1 is not larger than n/(n-1). Is it allowed to simplify it to something larger (such as 2, instead of 1?), even though it tends to 1? I don't think I can let 1/(n-4)! become 0 either as I need to get a value for n in terms of epsilon
Do you need to prove this using the epsilon definition of limits? Have to proven the result about the product of limits?
0
reply
Dannygem
Badges: 1
Rep:
?
#5
Report Thread starter 6 years ago
#5
(Original post by james22)
Do you need to prove this using the epsilon definition of limits? Have to proven the result about the product of limits?
Yeah I do, I haven't done much about the product of limits I don't think
0
reply
james22
Badges: 16
Rep:
?
#6
Report 6 years ago
#6
(Original post by Dannygem)
Yeah I do, I haven't done much about the product of limits I don't think
OK, you managed to split the fraction up how I would, now consider which terms are bounded. If you can find an upper bound for all but 1 of the terms, you can effectively ignore them (still need to include in your proof though). For example if |a_n| is bounded by 2, and b_n tends to 0, then a_n*b_n will tend to 0 (not too difficult to prove with epsilons).
1
reply
Dannygem
Badges: 1
Rep:
?
#7
Report Thread starter 6 years ago
#7
(Original post by james22)
OK, you managed to split the fraction up how I would, now consider which terms are bounded. If you can find an upper bound for all but 1 of the terms, you can effectively ignore them (still need to include in your proof though). For example if |a_n| is bounded by 2, and b_n tends to 0, then a_n*b_n will tend to 0 (not too difficult to prove with epsilons).
I'm not sure if I understand what you mean but if I do this:

(n^4)/n! < E

n/n x n/(n-1) x n/(n-2) x n/(n-3) x 1/(n-4)! < E

I need some way of keeping the 'n' as I need to write 'n' in terms of epsilon, so I can't just say that 1/(n-4)! tends to 0 otherwise the whole expression becomes 0
0
reply
atsruser
Badges: 11
Rep:
?
#8
Report 6 years ago
#8
(Original post by Dannygem)
[IMG]
(2) I have no idea how to go about this, apart from the obvious that since 'x' is a integer then the only way that lxl can be less than E is for it to be 0 as E can be all values above 0.
Proof by contradiction will do the job here. Claim that |x|=\delta &gt; 0 and derive a contradiction from the fact that |x| &lt; \epsilon for all \epsilon &gt; 0
0
reply
Dannygem
Badges: 1
Rep:
?
#9
Report Thread starter 6 years ago
#9
(Original post by atsruser)
Proof by contradiction will do the job here. Claim that |x|=\delta &gt; 0 and derive a contradiction from the fact that |x| &lt; \epsilon for all \epsilon &gt; 0
Thanks, i've managed to do part 2 and 3 now!

3 uses a very similar contradiction unsure as to why it wasn't just a subpart of question (2) as question (1) is a lot longer than them both combined
0
reply
atsruser
Badges: 11
Rep:
?
#10
Report 6 years ago
#10
(Original post by Dannygem)
Thanks, i've managed to do part 2 and 3 now!

3 uses a very similar contradiction unsure as to why it wasn't just a subpart of question (2) as question (1) is a lot longer than them both combined
I think that you're meant to use (2) to prove (3). It's one of those "hence, show that .." questions without being explicit about it.
1
reply
Dannygem
Badges: 1
Rep:
?
#11
Report Thread starter 6 years ago
#11
(Original post by atsruser)
I think that you're meant to use (2) to prove (3). It's one of those "hence, show that .." questions without being explicit about it.
Yeah I was going to just say that since x = 0, An - L = 0 and so An = L
but I thought i'd just do the proof again with An - L because it was so short
0
reply
DFranklin
Badges: 18
Rep:
?
#12
Report 6 years ago
#12
(Original post by Dannygem)
I'm not sure if I understand what you mean but if I do this:

(n^4)/n! < E

n/n x n/(n-1) x n/(n-2) x n/(n-3) x 1/(n-4)! < E

I need some way of keeping the 'n' as I need to write 'n' in terms of epsilon, so I can't just say that 1/(n-4)! tends to 0 otherwise the whole expression becomes 0
If n > 8, then n-4, n-3, n-2 and n-1 are all bigger than n/2.
0
reply
pineapplechemist
Badges: 2
Rep:
?
#13
Report 6 years ago
#13
(Original post by Dannygem)
Yeah I was going to just say that since x = 0, An - L = 0 and so An = L
but I thought i'd just do the proof again with An - L because it was so short
Hi, I'm doing the same sheet (and hence are on the same course as you ha ha). For 2 is it literally a case of saying something like 'suppose that x does not equal zero. Then x is not less than or equal to all e>0. Hence it must be zero?
0
reply
Dannygem
Badges: 1
Rep:
?
#14
Report Thread starter 6 years ago
#14
(Original post by pineapplechemist)
Hi, I'm doing the same sheet (and hence are on the same course as you ha ha). For 2 is it literally a case of saying something like 'suppose that x does not equal zero. Then x is not less than or equal to all e>0. Hence it must be zero?
Hi there

For 2 basically what I did was:

Assume l x l < E for all values of E > 0
Suppose x was not equal to 0, this means that l x l > 0.
However, l x l < E and since E > 0, l x l must be =< 0.
This contradicts our statement that l x l > 0, and so 'x' must equal zero.


For 3 you can then think about the result in this question and apply it

(Original post by DFranklin)
If n > 8, then n-4, n-3, n-2 and n-1 are all bigger than n/2.
Thanks for the hint, i've now managed to get 6((n/2)!), I still have no idea how to get rid of the factorial

Not sure if I am actually just supposed to do what James22 said above and try to prove it that way, it'd be different to the way the first 3 subparts of the question were answered.
0
reply
james22
Badges: 16
Rep:
?
#15
Report 6 years ago
#15
(Original post by Dannygem)
I'm not sure if I understand what you mean but if I do this:

(n^4)/n! < E

n/n x n/(n-1) x n/(n-2) x n/(n-3) x 1/(n-4)! < E

I need some way of keeping the 'n' as I need to write 'n' in terms of epsilon, so I can't just say that 1/(n-4)! tends to 0 otherwise the whole expression becomes 0
Can you bound teh first 4 terms? If you can, call the product of those bounds x, and then you have |n^4/n!|<x/(n-4)! for all n. Now you just need to show that x/(n-4)! tends to 0.
0
reply
DFranklin
Badges: 18
Rep:
?
#16
Report 6 years ago
#16
(Original post by Dannygem)
Hi there

For 2 basically what I did was:

Assume l x l < E for all values of E > 0
Suppose x was not equal to 0, this means that l x l > 0.
However, l x l < E and since E > 0, l x l must be =< 0.
This contradicts our statement that l x l > 0, and so 'x' must equal zero.


For 3 you can then think about the result in this question and apply it



Thanks for the hint, i've now managed to get 6((n/2)!), I still have no idea how to get rid of the factorial
Pretty sure you've messed up somewhere there. (What you've written is going to be a valid bound for large enough n, but I very much doubt your reasoing to get there is correct).

But as for getting rid of factorials: In general, note that N! is bigger (typically a LOT bigger!) than N, so once you get to dividing a constant by a factorial it is very easy to show it tends to 0.
0
reply
Dannygem
Badges: 1
Rep:
?
#17
Report Thread starter 6 years ago
#17
(Original post by DFranklin)
Pretty sure you've messed up somewhere there. (What you've written is going to be a valid bound for large enough n, but I very much doubt your reasoing to get there is correct).

But as for getting rid of factorials: In general, note that N! is bigger (typically a LOT bigger!) than N, so once you get to dividing a constant by a factorial it is very easy to show it tends to 0.
Yeah I think i've done that now. I was just wondering about how to write n in terms of epsilon, as I had to do for the previous 3 subparts. I don't think I can :/
0
reply
atsruser
Badges: 11
Rep:
?
#18
Report 6 years ago
#18
(Original post by Dannygem)
Hi there

For 2 basically what I did was:

Assume l x l < E for all values of E > 0
Suppose x was not equal to 0, this means that l x l > 0.
However, l x l < E and since E > 0, l x l must be =< 0.
This contradicts our statement that l x l > 0, and so 'x' must equal zero.
I think that this is essentially correct, but it isn't expressed in a very standard form. I think that you need to be a little more explicit:

1. Assume that |x| = \delta &gt; 0

2. Write down a specific \epsilon &gt; 0 that is smaller than \delta (Hint: what can you say about the average of 0 and \delta?)

3. Write down a contradiction, again explicitly, in terms of \delta and your constructed \epsilon.
0
reply
DFranklin
Badges: 18
Rep:
?
#19
Report 6 years ago
#19
(Original post by Dannygem)
Yeah I think i've done that now. I was just wondering about how to write n in terms of epsilon, as I had to do for the previous 3 subparts. I don't think I can :/
It's hard to help when you're not providing working.

If you cannot give a value for "n" in terms of epsilon, you haven't shown anything. On the other hand, your value doesn't have to be optimal (i.e. it doesn't have to be the smallest possible choice).


E.g. to find N such that n &gt; N \implies 1/n! &lt; \epsilon, I can simply go:

n! is never smaller than n, so 1/n! \leq 1/n So if N &gt; 1/\epsilon then 1/n!  \leq 1/n &lt; 1/N &lt; \epsilon.

This is completely valid, despite the fact that if, say, epsilon is 1/1000 I end up with n &gt; 1000 \implies 1/n! &lt; \epsilon when in fact the result holds for any n > 6.
0
reply
Smaug123
  • Study Helper
Badges: 15
Rep:
?
#20
Report 6 years ago
#20
(Original post by DFranklin)
On the other hand, your value doesn't have to be optimal (i.e. it doesn't have to be the smallest possible choice).
I used n=10002! in a proof once, when it turned out n=187 worked. (I think those were the numbers.)
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (158)
14.62%
I'm not sure (46)
4.26%
No, I'm going to stick it out for now (320)
29.6%
I have already dropped out (30)
2.78%
I'm not a current university student (527)
48.75%

Watched Threads

View All