Random math questions

[#1procrastinator]

In evaluating , is it valid to rewrite it as and say that it equals 0 cause the bottom goes to infinity?

[Leip]

I guess it is valid, but what argument are you using to show that the denominator goes to infinty?

If you are using the argument that is exponential in t and is polynomial in t (monomial even), couldn't you do that without rewriting it?

In any case, I would start by taking the x in the denominator outside the limit. From there, if you want to show things more explicitly, maybe you could consider applying l'Hopitals rule and induct (i.e. show that upon repetition of l'Hopitals rule the numerator becomes constant in t).

[#1procrastinator]

I guess it is valid, but what argument are you using to show that the denominator goes to infinty?

Bah, that gets me no where then.

If you are using the argument that is exponential in t and is polynomial in t (monomial even), couldn't you do that without rewriting it?

In any case, I would start by taking the x in the denominator outside the limit. From there, if you want to show things more explicitly, maybe you could consider applying l'Hopitals rule and induct (i.e. show that upon repetition of l'Hopitals rule the numerator becomes constant in t).

I was a bit hesitant in using l'Hopital's rule cause I thought that came after the chapter I'm on but turns out it was a couple of chapters back :p

So we would apply the rule x number of times until the numerator goes to 1 (taking out the constants of course) and then have 1/e^t, which would go to 0

[kamil9876]

What's your definition of ? If it is a power series then you can just notice it from the power series. For instance you can notice that where . Then we have a bound:

as since

[Leip]

I was a bit hesitant in using l'Hopital's rule cause I thought that came after the chapter I'm on but turns out it was a couple of chapters back :p

So we would apply the rule x number of times until the numerator goes to 1 (taking out the constants of course) and then have 1/e^t, which would go to 0

L'Hopital's is so often useful, you just have to be careful to remember what it does so that you don't overuse it. I like to think of it as comparing similar order terms in the Taylor expansion of both the numerator and denominator. Sometimes it is better to use the Taylor expansions directly.

But it's fairly straight forward in this case, and you can even leave the constants in since they work out nicely to x!.















Btw, I sure hope x is a positive integer!

[#1procrastinator]

excellent, thanks

@kamil: nah, it's from integrating the gamma function . am not up to power series yet

[#1procrastinator]

Could I please get some pointers on how to show that where Jn is an nxn matrix with all entries comtaining a 1 and I is the identity matrix. Thanks

Edit fixed missing power

[kamil9876]

Doesn't seem right, perhaps something is missing on the left of the of the parenthesis? (Otherwise why would they be there?)

If it was true it would imply that by cancelling out the

[#1procrastinator]

Sorry the lhs should be the inverse of the difference of the two matrices

[kamil9876]

I see, so in other words we want to see that the following is the identity matrix:






Now use the relation which follows because the entries of are

[#1procrastinator]

I see, so in other words we want to see that the following is the identity matrix:






Now use the relation which follows because the entries of are

Thanks - I thought we had to somehow derive that equation/identity(?) from scratch.

[kamil9876]

In other words, you are asking how would one solve the following problem without being given the answer: "What is the inverse of ?"

One possible way is to find the eigenspaces of . One can see that the rank of is so the nullspace is dimensional and so there is at most one other eigenspace and it is of dimension . By inspection is an eigenvalue with eigenvector . Thus there is a matrix such that



It follows that:



So

Now you just need to find what actually was from diagonalization. At least this gives a natural explanation of how appears!

[#1procrastinator]

I don't follow this proof of the principle of mathematical induction:

Theorem:
For each positive integer n, let P(n) be a statement. If
1) P(1) is true and
2) the implication if P(k), then P(k+1) is rue for every positive integer k,
then P(n) is true for every positive integer n.

Proof:
Assume that the theorem is false. Then conditions 1 and 2 are satisfied but there exists some positie integers n for which P(n) is a false statement. Let S={n∈N: P(n) is false}. Since S is a nonempty subset of N, it follows by the well-ordering principle that S contains a least element s. Since P(1) is true, 1∉S. Thus s≧2 and (s-1) is in N. Therefore (s-1)∉S and so P(s-1) is a true statement. By condition (2), P(s) is also true and s∉S. However, this contradicts our assumption that s∈S.

I don't get the bolded bit. Is there any significance to saying (s-1) is in N? Are they not all in N? I don't see how (s-1)∉S follows - I get it if s is 2 because 1 isn't in S but how do you know it holds for s>2?

[mark_alec]

s is the smallest natural number for which the statement P(s) is false. Since P(1) is true (statement #1), s must be greater than or equal to 2. Since s >=2, s-1 is a natural number and we know that P(s-1) is also true; but by statement #2 if P(s-1) is true, then P(s) must also be true. Hence we have a contradiction and S must be an empty set.

[#1procrastinator]

s is the smallest natural number for which the statement P(s) is false. Since P(1) is true (statement #1), s must be greater than or equal to 2. Since s >=2, s-1 is a natural number and we know that P(s-1) is also true; but by statement #2 if P(s-1) is true, then P(s) must also be true. Hence we have a contradiction and S must be an empty set.

How do we know P(s-1) is also true for all s>2? And why is it necessary to state s-1 is a natural number, aren't they all natural numbers?