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July 29, 2025, 11:39:35 pm
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Author Topic: Dekoyl's Questions  (Read 14538 times)  Share 

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dekoyl

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Re: Dekoyl's Questions
« Reply #30 on: March 15, 2009, 10:08:13 pm »
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Quote from: Mao on March 15, 2009, 10:03:18 pm
mmm small question, why ?
Hmmm... :-\

Well it appears they're the same thing.

http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning

I just had in my notebook. :P

Actually I don't know.
I got everything mixed up.
« Last Edit: March 15, 2009, 10:12:41 pm by dekoyl »
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Mao

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Re: Dekoyl's Questions
« Reply #31 on: March 15, 2009, 10:22:42 pm »
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oooh, i see!

is taking the absolute value of the scalar, not the modulus of a vector. It's so  you don't arrive at a negative volume.
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dekoyl

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Re: Dekoyl's Questions
« Reply #32 on: March 15, 2009, 10:28:24 pm »
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Quote from: Mao on March 15, 2009, 10:22:42 pm
is taking the absolute value of the scalar, not the modulus of a vector. It's so  you don't arrive at a negative volume.
Yeah that's it :P
Are you baffled by the "parallelpiped with nonparallel sides", Mao?
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Mao

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Re: Dekoyl's Questions
« Reply #33 on: March 15, 2009, 10:31:55 pm »
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Quote from: dekoyl on March 15, 2009, 10:28:24 pm
Quote from: Mao on March 15, 2009, 10:22:42 pm
is taking the absolute value of the scalar, not the modulus of a vector. It's so  you don't arrive at a negative volume.
Yeah that's it :P
Are you baffled by the "parallelpiped with nonparallel sides", Mao?

I'm guessing it must mean is a linearly independant set [otherwise the volume will be zero].
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dekoyl

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Can I use P(0)?
« Reply #34 on: March 23, 2009, 07:54:42 pm »
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Sorry let me try again. I don't know what the heck I'm doing. Oh interesting. Yay I proved something. =D

Just to clarify: Is it okay to use in the basis step like Wikipedia shows when doing proofs by mathematical induction? http://en.wikipedia.org/wiki/Mathematical_induction#Example
My teacher said it was trivial but it's still accepted right?

Thanks
« Last Edit: March 23, 2009, 08:12:11 pm by dekoyl »
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kamil9876

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Re: Dekoyl's Questions
« Reply #35 on: March 23, 2009, 08:30:28 pm »
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lol damn! i just proved it now and u already solved it ;p

edit: logic fail :/

If you used P(0), it would be like saying that you're adding 2+7+12... until u get to -3. But you don't get to -3 so P(0) is undefined. How did u interpret P(0)? 2-3?
« Last Edit: March 23, 2009, 08:35:11 pm by kamil9876 »
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dekoyl

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Re: Dekoyl's Questions
« Reply #36 on: March 23, 2009, 08:33:05 pm »
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Quote from: kamil9876 on March 23, 2009, 08:30:28 pm
lol damn! i just proved it now and u already solved it ;p

Yes it is ok to use P(0) instead of say P(1). because the second part would be used to prove that its true for P(1) based on P(0) being true, which would take you to the exact same situation as if you were to start at P(1). So you shouldn't worry unless you have some philosophical objection to negative numbers.
Sorry about that kamil! I was a bit hasty on posting up that question.
Yep thanks for the confirmation - I'll continue to use P(0). :)

Also.. maybe a yes or no answer would suffice.  ::)

When finding such that , would this just be trial and error?

* Mao fixed LaTeX
« Last Edit: March 23, 2009, 09:38:45 pm by Mao »
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kamil9876

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Re: Dekoyl's Questions
« Reply #37 on: March 23, 2009, 08:40:58 pm »
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Lol ok. I'm feeling a bit confused now, don't trust me on what I said about P(0) being used. I am sure it would be allowed to be used if the sequence was defined by 2+7+12+...(2+5n). Since 2 would be the 0th(n=0) term. But unsure about whether P(-3) would actually be interpredted 2-3-8-13 and whether it is an obvious logical step if that the same formula would apply if we extend the definition of P(n) the way we did.
« Last Edit: March 23, 2009, 09:05:33 pm by kamil9876 »
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kamil9876

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Re: Dekoyl's Questions
« Reply #38 on: March 23, 2009, 09:08:31 pm »
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I cant see that equation in your post. The one with the factorials, but strangely i can see it when i quote what u said i got n!>2n^3.

SO the first part is trial and error:

it happens when n=6.

The next bit is more sophisticated.

f(n)=n!, g(n)=2n^3

In order to get to f(n+1) i need to multiply f(n) by n+1. But in order to get to g(n+1) i hav to multiply g(n) by which is the same as But this number is obviously less than n+1 for n>6 so f(n) keeps being multiplied by numbers that are bigger than what g(n) keeps being multiplied by and so f(n) increases more but f(n) is already greater than g(n) at n=6 so it remains bigger than g(n) once u go past 6.
« Last Edit: March 23, 2009, 09:19:03 pm by kamil9876 »
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Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

/0

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Re: Dekoyl's Questions
« Reply #39 on: March 23, 2009, 09:50:34 pm »
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Normally they would tell you where to start.
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dekoyl

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Re: Dekoyl's Questions
« Reply #40 on: March 23, 2009, 11:42:03 pm »
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^Hmm it doesn't but thanks anyways guys I got it now =P I did the first step wrong.

I think my brain is fried. I can't even find something so simple.


What results when adding to both sides?

Thank you.  :-[

Is it:
« Last Edit: March 23, 2009, 11:44:52 pm by dekoyl »
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/0

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Re: Dekoyl's Questions
« Reply #41 on: March 24, 2009, 12:09:33 am »
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Don't think you should add to both sides, just add on the next term

After you have checked P(0) and made the assumption that then



And eventually get to (you might have to use the factor theorem on the quartic)
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dekoyl

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Re: Dekoyl's Questions
« Reply #42 on: March 24, 2009, 12:12:39 am »
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Quote from: /0 on March 24, 2009, 12:09:33 am
just add on the next term
Yes I was confused about that. Haha.

Thank you oh saviour /0.

« Last Edit: March 24, 2009, 12:20:46 am by dekoyl »
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/0

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Re: Dekoyl's Questions
« Reply #43 on: March 24, 2009, 12:17:34 am »
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Lol np, but I probably need to answer the actual question rather than ramble on with the problem. I have the same issue in my tests :P
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dekoyl

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Re: Dekoyl's Questions
« Reply #44 on: March 24, 2009, 12:21:32 am »
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Hmm, Instead of:


should it be:


Not sure if it matters or not but I'm just wondering. (Then again I may be entirely wrong. Ha. =| )
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