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Author Topic: Mathematics Help Thread  (Read 227299 times)

Heron TSG

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Re: Mathematics Help Thread
« Reply #90 on: October 22, 2010, 08:04:43 am »

Ah, forgot to factor out of both sides when I factored one of the numbers. Gotta stop doing that. Thanks for your help, I went back and fixed that, ended with this.
(x/(x+10))1/2
I'm off to take a test in The Calculus! Factoring, you will not me this day! Even if I do every problem twice to make sure.
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Est Sularus Oth Mithas
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Vector

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Re: Mathematics Help Thread
« Reply #91 on: October 22, 2010, 08:10:05 am »

But...
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Heron TSG

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Re: Mathematics Help Thread
« Reply #92 on: October 22, 2010, 08:16:43 am »

But...?

What I mean is, I accidentally did something like this.

[(x2+2)+(x+3)] = x[(X+2)+(X+3)]

instead of this.

[(x2+2)+(x+3)] = x[(X+2)+(3)]
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Est Sularus Oth Mithas
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Virex

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Re: Mathematics Help Thread
« Reply #93 on: October 22, 2010, 09:54:31 am »

Not to burst your bubble, but
[(x2+2)+(x+3)] = x2 + x + 5 =/=  x[(X+2)+(3)] = x2 + 5x
« Last Edit: October 22, 2010, 10:00:37 am by Virex »
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Jim Groovester

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Re: Mathematics Help Thread
« Reply #94 on: October 22, 2010, 10:48:21 am »

But...?

What I mean is, I accidentally did something like this.

[(x2+2)+(x+3)] = x[(X+2)+(X+3)]

instead of this.

[(x2+2)+(x+3)] = x[(X+2)+(3)]

That is not how you factor.

If you're using the distributive property to pull out an x from a quantity, you have to do it to every term in the bracket. You did not do this.

Using your example,

[(x2+2)+(x+3)]

The proper way to pull an x out of this is the following:

[(x2 + 2) + (x + 3)] = x[((x + (2/x)) + (1 + (3/x)))]

Do you see the difference? Every term. (Also, pulling out an x from an x leaves you with 1, not 0 like what you have.)
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Virex

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Re: Mathematics Help Thread
« Reply #95 on: October 22, 2010, 03:16:49 pm »

Made another update to my solution to the converging sequence problem. Hopefully now the proof is sound. I'd really like for someone to run it trough and see if it's not a load of bogus, but I guess I'll have to wait till Vector's done with her answer.
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KaminaSquirtle

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Re: Mathematics Help Thread
« Reply #96 on: October 22, 2010, 03:33:35 pm »

I'd love to, however I'm nowhere near solving it on my own, so I can't.  Right now I'm running through some simpler stuff, like proving Newton's Method, etc.  I'd like to ask two questions though, just to see if I have enough prerequisite knowledge right now.

1) Did you need to use anything involving Cauchy Sequences?  I think we get to those next week.

2) Is there more to it than proving it is monotonic for all n>=n0, for some n0 in the positive integers, and showing that it is bounded?  Because we haven't gone over any other ways to solve recursive sequences.  Like I said, they were only introduced this week.

Of course, this is assuming it converges, but I'm fairly sure it does.  Please don't tell me either way though, I really want to get this one on my own.
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Virex

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Re: Mathematics Help Thread
« Reply #97 on: October 22, 2010, 03:52:53 pm »

I'd love to, however I'm nowhere near solving it on my own, so I can't.  Right now I'm running through some simpler stuff, like proving Newton's Method, etc.  I'd like to ask two questions though, just to see if I have enough prerequisite knowledge right now.

1) Did you need to use anything involving Cauchy Sequences?  I think we get to those next week.
I did not explicitly use anything related to Cauchy sequences. I also can't tell you if it's a Cauchy function...

Quote
2) Is there more to it than proving it is monotonic for all n>=n0, for some n0 in the positive integers, and showing that it is bounded?  Because we haven't gone over any other ways to solve recursive sequences.  Like I said, they were only introduced this week.
That could work. I used a method similar to that, but not quite the same, because you can reduce the function to a much simpler form with the appropriate method
« Last Edit: October 22, 2010, 03:55:54 pm by Virex »
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KaminaSquirtle

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Re: Mathematics Help Thread
« Reply #98 on: October 22, 2010, 04:10:45 pm »

I did not explicitly use anything related to Cauchy sequences. I also can't tell you if it's a Cauchy function, due to the method I used.
Excellent, I just wanted to make sure that I didn't need to know any of that.  Vector mentioning them worried me.

2) Is there more to it than proving it is monotonic for all n>=n0, for some n0 in the positive integers, and showing that it is bounded?  Because we haven't gone over any other ways to solve recursive sequences.  Like I said, they were only introduced this week.
That could work. I used a method similar to that, but not quite the same, because you can reduce the function to a much simpler form with the appropriate method
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Virex

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Re: Mathematics Help Thread
« Reply #99 on: October 22, 2010, 04:52:24 pm »

I did not explicitly use anything related to Cauchy sequences. I also can't tell you if it's a Cauchy function, due to the method I used.
Excellent, I just wanted to make sure that I didn't need to know any of that.  Vector mentioning them worried me.

2) Is there more to it than proving it is monotonic for all n>=n0, for some n0 in the positive integers, and showing that it is bounded?  Because we haven't gone over any other ways to solve recursive sequences.  Like I said, they were only introduced this week.
That could work. I used a method similar to that, but not quite the same, because you can reduce the function to a much simpler form with the appropriate method
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KaminaSquirtle

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Re: Mathematics Help Thread
« Reply #100 on: October 22, 2010, 05:50:05 pm »

I did not explicitly use anything related to Cauchy sequences. I also can't tell you if it's a Cauchy function, due to the method I used.
Excellent, I just wanted to make sure that I didn't need to know any of that.  Vector mentioning them worried me.

2) Is there more to it than proving it is monotonic for all n>=n0, for some n0 in the positive integers, and showing that it is bounded?  Because we haven't gone over any other ways to solve recursive sequences.  Like I said, they were only introduced this week.
That could work. I used a method similar to that, but not quite the same, because you can reduce the function to a much simpler form with the appropriate method
Spoiler (click to show/hide)
Spoiler (click to show/hide)
Thank you, that should be enough for me to crack this thing.  Hopefully.  Anyway, that's enough tips for now, I feel I've allowed myself enough outside help.
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Vector

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Re: Mathematics Help Thread
« Reply #101 on: October 22, 2010, 06:15:34 pm »

Virex,

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pronouns: prefer neutral ones, others are fine. height: 5'3".

Virex

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Re: Mathematics Help Thread
« Reply #102 on: October 22, 2010, 06:35:12 pm »

Virex,

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Vector

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Re: Mathematics Help Thread
« Reply #103 on: October 22, 2010, 06:44:52 pm »

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"The question of the usefulness of poetry arises only in periods of its decline, while in periods of its flowering, no one doubts its total uselessness." - Boris Pasternak

nonbinary/genderfluid/genderqueer renegade mathematician and mafia subforum limpet. please avoid quoting me.

pronouns: prefer neutral ones, others are fine. height: 5'3".

Heron TSG

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Re: Mathematics Help Thread
« Reply #104 on: October 22, 2010, 08:28:55 pm »

That is not how you factor.

If you're using the distributive property to pull out an x from a quantity, you have to do it to every term in the bracket. You did not do this.
Excuse me, I meant to say this.

[ (x2(2)) +  (x(3)) ]

becomes

x [ (x(2)) +  (3) ]

Durn multiplication/addition confusion. The simplest things are often the most easily missed.
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Est Sularus Oth Mithas
The Artist Formerly Known as Barbarossa TSG
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