Random math questions

[kamil9876]

I don't get the bolded bit. Is there any significance to saying (s-1) is in N? Are they not all in N?

If then and so doesn't make sense because of the hypothesis:

For each positive integer n, let P(n) be a statement.

[#1procrastinator]

But didn't we already say s>=2?

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IN evaluating , why don't the substitutions give the same answer?

I get when using the substitutions involving and when using the ones involving . May well have made a mistake somewhere but if I did I don't see it, I've gone over the calculations a poopload of times now.

[lzxnl]

Firstly, e^x > 0. I'll denote e^x with y and y>0.
Now cot y = 1/(tan y)
Therefore arccot (cot y) = y = arcccot(1/tan y) = arccot (cot y)
But tan y = 1/ cot y so y=arctan(1/cot y)
So arctan(1/cot y) = arccot (cot y)
arctan 1/y = arccot y = pi/2 - arctan y
Therefore arctan 1/y and -arctan y differ by a constant, so when integrating these two forms are equivalent.

[kamil9876]

But didn't we already say s>=2?

Yes that was the reason why .

Since s >=2, s-1 is a natural number...

[#1procrastinator]

Firstly, e^x > 0. I'll denote e^x with y and y>0.
Now cot y = 1/(tan y)
Therefore arccot (cot y) = y = arcccot(1/tan y) = arccot (cot y)
But tan y = 1/ cot y so y=arctan(1/cot y)
So arctan(1/cot y) = arccot (cot y)
arctan 1/y = arccot y = pi/2 - arctan y
Therefore arctan 1/y and -arctan y differ by a constant, so when integrating these two forms are equivalent.

Thanks - so the missing C has been the culprit all along? 

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Show that the sequence converges to zero by arguing directly from the definition of convergence.



Can we solve for explicitly?

The solution given does it another way, which was demand that from which follows. How would you show that the second inequality follow from n>2? It's fairly obvious but how would you prove it, and also, could we had ? Or 4n? Likewise, can we 'demand' n be greater than any arbitrary natural number rather than 1?

[kamil9876]

for

And sure, since we only care about how it behaves as is large we could have restricted ourselves to and obtained a similair inequality. But that isn't necessary, this was good enough.

[Deleted User]

How do I prove that the sequence defined by s_n+1 = (s_n^3 + 2)/7 is contractive?

I know that I need to show that |s_n+1 - s_n| ≤ k|s_n - s_n-1|.

So I've managed to get up to this point
s_n+1 - s_n = (s_n^3 - s_n-1^3)/7.
How do I proceed from here? And what would the value of k be?

Thanks in advance.

[kamil9876]

what's the initial value?

There's a factorization:



So we want to study and show that it is bounded below some number strictly less than . So it would be good to know the initial values.

Alternatively using a bit more theory (the mean value theorem). One knows that


for some . Thus if we can bound then we know that:



and of course we are thinking here of being the function

[Deleted User]

s₁ is ∈ (0,1).

And sorry I don't really understand the contraction theorem either so could you explain why has to be strictly less than 7? Thakns.

[#1procrastinator]

for

And sure, since we only care about how it behaves as is large we could have restricted ourselves to and obtained a similair inequality. But that isn't necessary, this was good enough.

Thanks...if we didn't use this 'trick', is it possible to solve for ? How messy would it be? :p

[#1procrastinator]

Does it make sense to talk of a union of a set with one of its subsets?

[BubbleWrapMan]

Does it make sense to talk of a union of a set with one of its subsets?

It makes sense, it's just redundant because .

[#1procrastinator]

thanks Timmeh. just realised i misread some definitions relating to the problem i was attempting and so defining redundant sets won't be necessary

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How do you evaluate ?

[#1procrastinator]

When proving by induction, when you assume P(n) is true, this only applies for P(n), right?
You can't assume P(n-1) as well...just want to make sure cause only way it's the only way I know how do one of the assignment questions   

[Lasercookie]

When proving by induction, when you assume P(n) is true, this only applies for P(n), right?
You can't assume P(n-1) as well...just want to make sure cause only way it's the only way I know how do one of the assignment questions   

There's the idea of strong induction where you show that P(k), P(k+1), ..., P(n) are true, then P(n+1) is true for some . Page 4 and 5 of the first set of linear algebra lecture notes for 1116 also has an explanation there too.

I like this analogy from here http://mathcircle.berkeley.edu/BMC4/Handouts/induct/node6.html

The difference is in what you need to assume in the induction step. For ordinary induction--in the ladder metaphor--you simply go from the rung you are on up to the next one. For strong induction, you need to know that all the rungs below the rung you are on are solid in order to step up. As a practical matter, both have the same logical strength when you apply them - since as you climb up the ladder from the bottom rung, you sweep through all the intermediate rungs anyway.