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HSC 2013 MX2 Marathon (archive) (3 Viewers)

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seanieg89

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Re: HSC 2013 4U Marathon

If you do manage to properly prove the Taylor Series formula (for all R and C), then Euler's formula should follow immediately with no issues because the radius of convergence for sin(x) and cos(x) are infinity.

However, the issue is HOW you are going to properly prove the formula using methods within Extension 2?
Same issue though, you would be trying to prove that the infinite series converge to e,sin,cos respectively. But to talk about this happening on C we need to define e,sin and cos on C in the first place!
 

Carrotsticks

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Re: HSC 2013 4U Marathon

The only treatment of Taylor Series I have ever seen in the HSC has only been in R because it has been very well-defined in the course. So the infinite series for e = 1+1/1!+1/2! +... is fine, since it's in R.

Sy123, might want to stick within R when having problems to do with functions, though polynomial functions have been defined in C.
 

seanieg89

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Re: HSC 2013 4U Marathon

The only treatment of Taylor Series I have ever seen in the HSC has only been in R because it has been very well-defined in the course. So the infinite series for e = 1+1/1!+1/2! +... is fine, since it's in R.

Sy123, might want to stick within R when having problems to do with functions, though polynomial functions have been defined in C.
Agreed.

You would need something like the mean value theorem to actually prove convergence of the taylor series though.
 

Shadowdude

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Re: HSC 2013 4U Marathon

Plus no doubt you'll learn about them when you continue your mathematical study at tertiary level...

hint hint
 

Sy123

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Re: HSC 2013 4U Marathon

Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved:



It isn't that difficult I guess but I reckon it was neat.
 

bleakarcher

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Re: HSC 2013 4U Marathon

Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved:



It isn't that difficult I guess but I reckon it was neat.
Very nice question Sy.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Heh don't worry I won't do anything that extravagant anymore, here is a neat question I just solved:



It isn't that difficult I guess but I reckon it was neat.
Let which denotes the j'th triangular number. Then the sum in the n'th bracket is just:



Now when we write this out we get:



Usin the sum of a series formula we get:









As required.
 

Sy123

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Re: HSC 2013 4U Marathon

Let which denotes the j'th triangular number. Then the sum in the n'th bracket is just:



Now when we write this out we get:



Usin the sum of a series formula we get:









As required.
Nice work guys, I will post another question soon. (unless someone else could contribute heh, instead of me just giving questions out)
 
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GoldyOrNugget

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Re: HSC 2013 4U Marathon







I can justify this, since I know the limit:

Hence the product (10000/9999)^ 9999 is always less than e < 9999
Sneaky... I haven't seen that approach before. What if I changed it to ? The method I have in mind can compare any pair of large exponents regardless of the values of the bases and exponents.
 

Sy123

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Re: HSC 2013 4U Marathon

Sneaky... I haven't seen that approach before. What if I changed it to ? The method I have in mind can compare any pair of large exponents regardless of the values of the bases and exponents.
Well we can define the actual definition of e^x to do something similar:



Now looking back at the new thing you gave us:









So I too can do it for any large exponents :D
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

Hmph. Fine, I'll accept it :p the solution I have in mind uses logs.
 

Sy123

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Re: HSC 2013 4U Marathon

Hmph. Fine, I'll accept it :p the solution I have in mind uses logs.
I originally thought of using a^x > log_a x somehow but this came to me first.

Do you mind posting your method? I'm curious
 
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cutemouse

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Re: HSC 2013 4U Marathon

1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

New question: Expand and simplify (x-a)(x-b)(x-c)...(x-z).
 
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Sy123

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Re: HSC 2013 4U Marathon

1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

New question: Expand and simplify (x-a)(x-b)(x-c)...(x-z).


 

Carrotsticks

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Re: HSC 2013 4U Marathon

1/n! = 1/[n(n-1)(n-2)...(3)(2)1] < 1/[2.2.2...2.2.1] (as n, n-1, ... , 3 > 2 for n >= 3) = 1/2^(n-1)

New question: Expand and simplify (x-a)(x-b)(x-c)...(x-z).
0

You are thinking too hard.

EDIT: Just saw your x =/= x, but I think that was the intended result.
 
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