Maths Help : Othor Games

» Tue Apr 05, 2011 7:25 pm

My tutorial sheets asks me to calculate log₃ 24 in terms of log₂ 3

While I can get it in terms log₃ 2 (see below), I can't seem to get it in terms of log₂ 3

log₃ 24 = log₃ (3 * 8) = log₃ 3 + log₃ 8 = 1 + 3 log₃ 2

» Tue Apr 05, 2011 8:28 pm

My tutorial sheets asks me to calculate log₃ 24 in terms of log₂ 3

While I can get it in terms log₃ 2 (see below), I can't seem to get it in terms of log₂ 3

log₃ 24 = log₃ (3 * 8) = log₃ 3 + log₃ 8 = 1 + 3 log₃ 2

Base switch, perhaps?

log₃ 24 = (log₂ 24)/(log₂ 3) = (log₂ [3*8])/(log₂ 3) = (log₂ 3 + log₂ 8)/(log₂ 3) = (log₂ 3)/(log₂ 3) + (log₂ 8)/(log₂ 3) =

1 + 3/(log₂ 3)

» Tue Apr 05, 2011 11:12 am

Now I feel stupid.

Thanks

» Tue Apr 05, 2011 6:11 pm

I have another problem. Where would you suggest I start with this? (I know I will feel stupid after having read the answer, but I may as well get it over with)

Show by induction on n

Dⁿ((1-x)^-2) = (n+1)!(1-x)^-n-2

for 0 < x < 1 and n ∈ ? where D = d/dx

» Tue Apr 05, 2011 9:38 am

I have another problem. Where would you suggest I start with this? (I know I will feel stupid after having read the answer, but I may as well get it over with)

Show by induction on n

Dⁿ((1-x)^-2) = (n+1)!(1-x)^-n-2

for 0 < x < 1 and n ∈ ? where D = d/dx

Well, I dunno. Just start by taking the first derivative, then second derivative and so and see where it takes you after you have collection of results that look a little like what they want you to get to.. You can probably make the required conclusion from the result.

» Tue Apr 05, 2011 12:07 pm

Prove it for n=1, see what it is for n=k and n=k+1, then add the results for n=k and n=1. If it's the same as for n=k+1, you're right.

» Tue Apr 05, 2011 10:32 am

I have another problem. Where would you suggest I start with this? (I know I will feel stupid after having read the answer, but I may as well get it over with)

Show by induction on n

Dⁿ((1-x)^-2) = (n+1)!(1-x)^-n-2

for 0 < x < 1 and n ∈ ? where D = d/dx

Basically what Simmura said.

Spoiler

Dⁿ((1-x)^-2) = (n+1)!(1-x)^-n-2

n=1:

D((1-x)^-2) = (-2)(-1)(1-x)^-3 = 2(1-x)^-3 = (1+1)!(1-x)^-1-2; True for n=1.

Assume true for some n = k.

Test n=k+1

D^k+1[(1-x)^-2] = D[D^k[(1-x)^-2]] = D[(k+1)!(1-x)^-k-2] = (k+1)!*D[(1-x)^-k-2]

= (k+1)!*(-k-2)*(-1)*(1-x)^-k-3 = (k+1)!(k+2)(1-x)^-k-3 = (k+2)!(1-x)^-(k+1)-2 = ([k+1]+1)!(1-x)^-(k+1)-2 Q.E.D.

» Tue Apr 05, 2011 2:47 pm

Yay. Another question.

How do I show that:

(log_a n)^c / n -> 0 as n -> infinity. a,c are constants.

» Tue Apr 05, 2011 9:44 am

You could start by http://www.wolframalpha.com/input/?i=lim((log_123+n)^10+/+n),+n-%3Einfinity and see their steps (click Show steps). These show a step for each power (10 in this case) - you can try to generalize for any power.

» Tue Apr 05, 2011 10:00 pm

You could start by http://www.wolframalpha.com/input/?i=lim((log_123+n)^10+/+n),+n-%3Einfinity and see their steps (click Show steps). These show a step for each power (10 in this case) - you can try to generalize for any power.

Ah. I didn't manage to get WolframAlpha to display that sort of detail. That looks the way to go. Many thanks.

» Tue Apr 05, 2011 7:47 pm

Yay. Another question.

How do I show that:

(log_a n)^c / n -> 0 as n -> infinity. a,c are constants.

The rule of thumb "Logarithms grow more slowly than any other power or root of x" comes to mind.

Spoiler

Given that http://en.wikipedia.org/wiki/List_of_logarithmic_identities#Limits,
lim_n→∞ (log_a n)^c/n =
lim_n→∞ [(log_a n)/n^(1/c)]^c =
[lim_n→∞(log_a n)/n^(1/c)]^c =
0^c = 0

Q.E.D.

» Tue Apr 05, 2011 12:57 pm

The rule of thumb "Logarithms grow more slowly than any other power or root of x" comes to mind.

That was what I was trying to prove: (log_a n)^500 = o(n^2). Now you mention it, it probably would be easier to prove it to two stages though, as I think the proof for log_a n = o(n) and n = o(n^2) is fairly simple (5 lines at max).

» Tue Apr 05, 2011 7:46 pm

(log_a n)^500 = o(n^2)
log_a n = o(n) and n = o(n^2)

Err, neither of these are true. I think you must have mixed something up.

» Tue Apr 05, 2011 4:48 pm

Err, neither of these are true. I think you must have mixed something up.

They should be. For example, n^2 grows much faster than n. This can also be shown by the fact that

http://texify.com/?$lim_{n->\infty} \frac{n}{n^2} \rightarrow 0$

» Tue Apr 05, 2011 10:16 pm

Ah, my mistake. It's just that in practice when a programmer says something is o(x), one tends to assume they really mean Theta(x), which does not make sense in your context.