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Sigma Function watch

1. The function, f(n), means the sum of the factors of n

I started by saying, x is a prime number who has factors 1 and x.
x^n would have factors 1 and x^n and the number of factors would be equal to n + 1. (I don't know how to show this)

Majorly messed up my Latex, was gona show what I've done so far, just adding it now...

f(x^n) = 1 + x^n / x^n-1 + x^n / x^n-2 + x^n / x^n-3 + ... + x^n

f (x^n) = 1 + x + x^2 + x^3 + ... + x^n

then basically x = 2

But I know I haven't SHOWN anything but I can't think of anything else
2. Then it says find an expression for a prime number p for f(p^n)

And show that is equals
3. So yeah thanks a lot for anyone who can help me out here plus rep obvs.
4. I'm not sure how you've done that to be honest but could it not be proved fairly simply by induction?
5. (Original post by corney91)
I'm not sure how you've done that to be honest but could it not be proved fairly simply by induction?
I assumed there would be an easier way than induction because it's from a GCSE level text book.
6. (Original post by SahilB91)
x^n would have factors 1 and x^n
Are you still assuming is prime here? Because then is a root of ... I'm not sure what you're trying to say...

(Original post by SahilB91)
I assumed there would be an easier way than induction because it's from a GCSE level text book.
Yeah, induction is definitely not in the GCSEs but this looks way harder than anything I ever did at GCSE
7. The question you need to answer: What are the factors of 2^n?

It's easy to "see" what they are, and I'm not sure you really need to justify it - particularly at GCSE. But if you do want to prove it: Suppose k divides 2^n. Let p be a prime dividing k. Can p be anything other than 2? So, we can write k in the form 2^m for some m. And then m must be <= n.

For the 2nd problem, at the end:

Spoiler:
Show
You will want to show that . Easiest way to do this is multiply both sides by (p-1) and notice that an awful lot of stuff on the LHS cancels.
8. (Original post by corney91)
Are you still assuming is prime here? Because then is a root of ... I'm not sure what you're trying to say...

Yeah, induction is definitely not in the GCSEs but this looks way harder than anything I ever did at GCSE
Yeah it is a bit abstract. Well I don't think x is always a root of n whereas I think x^n always is. Because with this function say n = 1 then the same factor would be summed twice, but I may be wrong. So I think as long as n > 2 then x would be.
9. (Original post by DFranklin)
The question you need to answer: What are the factors of 2^n?

It's easy to "see" what they are, and I'm not sure you really need to justify it - particularly at GCSE. But if you do want to prove it: Suppose k divides 2^n. Let p be a prime dividing k. Can p be anything other than 2? So, we can write k in the form 2^m for some m. And then m must be <= n.

For the 2nd problem, at the end:

Spoiler:
Show
You will want to show that . Easiest way to do this is multiply both sides by (p-1) and notice that an awful lot of stuff on the LHS cancels.
I really like the first bit you wrote. I am not to sure about the spoiler contents because:

multiplying both sides by (p-1) seems to always leave something like...

p = p^n

1 + p^n = (p^n+1 - 1) / p -1

p + p^n+1 - 1 - p^n = p^n+1 - 1

p - p^n = 0

p = p^n
10. (Original post by SahilB91)
I really like the first bit you wrote. I am not to sure about the spoiler contents because:

multiplying both sides by (p-1) seems to always leave something like...

p = p^n

1 + p^n = (p^n+1 - 1) / p -1

p + p^n+1 - 1 - p^n = p^n+1 - 1

p - p^n = 0

p = p^n
I don't see where any of these lines have come from. I think you are making some algebra errors. Write it down for some simple case - n = 2, say.
11. (Original post by DFranklin)
I don't see where any of these lines have come from. I think you are making some algebra errors. Write it down for some simple case - n = 2, say.
Sorry, if you look at your earlier post, although multiplying both sides by (p-1) should work it doesn't in practice.

As you always end up with p^n = p^a

where a is dependant on the highest power of p in the equation.

i.e. if there a p^3 then a = 3, if there is just p then a = 1.

I am really stuck here because it works when there is a n = any number, but I can't show with n.
12. It does work. I suggest you post your working.
13. (Original post by DFranklin)
It does work. I suggest you post your working.
Right ok (thanks btw).

This is where I am going wrong...

When it says show that

I am not sure about the LHS and what to leave out when I multiply by (p-1).

I mean do I multiply (p-1) against the whole LHS how it appears now, or do I leave out the p^2 term and only use 1 + p + p^n?
14. The whole LHS. (I don't understand how you could think you could leave anything out without ending up with the wrong answer).
15. (Original post by SahilB91)
Sorry, if you look at your earlier post, although multiplying both sides by (p-1) should work it doesn't in practice.

As you always end up with p^n = p^a

where a is dependant on the highest power of p in the equation.

i.e. if there a p^3 then a = 3, if there is just p then a = 1.

I am really stuck here because it works when there is a n = any number, but I can't show with n.

It does work. You can easily prove it by induction using the formula that DFranklin gave you (S= [(p^n+1)-1]/(p-1) for S=1+p+p²+...+p^n )

or you can just multiply by (p-1).

Another proof, just so that you are convinced, is to do it this way:

For n a fixed positive integer, you have:

Write down your expression S(n) = 1 + p +...+p^n
multiply by p, and you get pS(n) = p + p² +...+p^n + p^(n+1)

Substract S(n)-pS(n)
If you write S(n) and pS(n) one under the other, you will easily see that in the substraction, every term cancels out, EXCEPT 1 and p^n+1

Therefore, S(n)-pS(n) = 1 - p^(n+1)
Therfore, S(n) = (1-p^n+1)/(1-p)

Multiply again by -1, and you get the formula that DFranklin gave you
16. Sigh. A perfect example of someone jumping in when I was waiting for the OP to try to do it himself.
17. (Original post by DFranklin)
The whole LHS. (I don't understand how you could think you could leave anything out without ending up with the wrong answer).
This is why:

p^3 - p^n = 0

p^3 = p^n

Is my algebra wrong?
18. In "1 + p + p^2 + ... + p^n", the "..." means 'and so on' - that is, that you continue the pattern until you get to the end term.

In other words: "1 + p + p^2 + ... + p^5" = "1 + p + p^2 + p^3 + p^4 + p^5".
19. (Original post by paronomase)
It does work. You can easily prove it by induction using the formula that DFranklin gave you (S= [(p^n+1)-1]/(p-1) for S=1+p+p²+...+p^n )

or you can just multiply by (p-1).

Another proof, just so that you are convinced, is to do it this way:

For n a fixed positive integer, you have:

Write down your expression S(n) = 1 + p +...+p^n
multiply by p, and you get pS(n) = p + p² +...+p^n + p^(n+1)

Substract S(n)-pS(n)
If you write S(n) and pS(n) one under the other, you will easily see that in the substraction, every term cancels out, EXCEPT 1 and p^n+1

Therefore, S(n)-pS(n) = 1 - p^(n+1)
Therfore, S(n) = (1-p^n+1)/(1-p)

Multiply again by -1, and you get the formula that DFranklin gave you
I feel so stupid because I've done further maths and I didn't even consider induction because this is from a GCSE textbook.
20. (Original post by DFranklin)
In "1 + p + p^2 + ... + p^n", the "..." means 'and so on' - that is, that you continue the pattern until you get to the end term.

In other words: "1 + p + p^2 + ... + p^5" = "1 + p + p^2 + p^3 + p^4 + p^5".
Yeah I shouldn't have wasted so much time and just used a multiplicative factor

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