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#1 2015-02-18 11:27:16

Au101
Member
Registered: 2010-12-01
Posts: 353

Calculus Proof

Another tricky little question :S

I wasn't really sure where I was going with this so I decided to just dive right in and differentiate:

As for the second part of the first proof, I'm pretty stumped. We know that k > 0, so for k < 1, k - 1 will be negative and an element of the set of real numbers between 0 and 1 (exclusive). In other words:

So you will have a fraction and the exponent of x will not be an integer. Whereas, for k > 1 you will have a positive exponent, which may be an integer and may, in fact, be any positive real number. That's the closest thing to a brainwave I've had really - do you have any ideas?

Last edited by Au101 (2015-02-18 11:29:51)

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#2 2015-02-18 20:34:11

Olinguito
Member
Registered: 2014-08-12
Posts: 649

Re: Calculus Proof

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This is < 0 when k < 1 and > 0 when k > 1.


Bassaricyon neblina

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#3 2015-02-19 03:00:53

Au101
Member
Registered: 2010-12-01
Posts: 353

Re: Calculus Proof

Perfect! thank you!

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#4 2015-02-19 06:19:25

Au101
Member
Registered: 2010-12-01
Posts: 353

Re: Calculus Proof

I was wondering if anybody could suggest a starting point for the second proof, just to get me going smile

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#5 2015-02-19 14:38:34

Au101
Member
Registered: 2010-12-01
Posts: 353

Re: Calculus Proof

Okay, so I've been experimenting with this a little bit and I think I have the answer, but I'm not quite sure I'm laying this out properly (or, to be honest, that I'm right). What do you think?

Last edited by Au101 (2015-02-19 14:39:50)

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#6 2015-02-19 14:40:05

Au101
Member
Registered: 2010-12-01
Posts: 353

Re: Calculus Proof

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