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 Post subject: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Fri, 6 Jul 2012 21:49:48 UTC 
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Use the principle of mathematical induction to show that:

a_n = n/(n^2 + 1) is decreasing.

BASE CASE:
Show that n = 1 is true.
I did this

INDUCTION HYPOTHESIS
Assume n = k is true

=> a_k < a_k+1
=> (k)/(k^2 + 1) < (k+1)/((k+1)^2 + 1)

a_k = (k)/(k^2 + 1)

SHOW:
Show that n = k + 1 is true:
=> a_k+1 < a_k+2

=> (k+1)/((k+1)^2 + 1) < (k+2)/((k+2)^2 + 1)

I am stuck on this induction step. How do I "revive" the a_k expression on my LHS or RHS of the above expresion?


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Fri, 6 Jul 2012 22:20:44 UTC 
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As written you have the induction steps backwards, that is increasing instead of decreasing.


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Fri, 6 Jul 2012 22:32:23 UTC 
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A-R-Q wrote:
Use the principle of mathematical induction to show that:

a_n = n/(n^2 + 1) is decreasing.

BASE CASE:
Show that n = 1 is true.
I did this

INDUCTION HYPOTHESIS
Assume n = k is true

=> a_k > a_k+1
=> (k)/(k^2 + 1) < (k+1)/((k+1)^2 + 1)

a_k = (k)/(k^2 + 1)

SHOW:
Show that n = k + 1 is true:
=> a_k+1 > a_k+2

=> (k+1)/((k+1)^2 + 1) > (k+2)/((k+2)^2 + 1)

I am stuck on this induction step. How do I "revive" the a_k expression on my LHS or RHS of the above expresion?


mathematic wrote:
As written you have the induction steps backwards, that is increasing instead of decreasing.
OK, fixed it I believe.


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Sat, 7 Jul 2012 10:01:51 UTC 
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I don't think you need induction.

Just show a_n - a_{n+1} > 0 for all n \geq 1.

Spoiler:
a_n - a_{n+1} = \frac{n}{n^2+1} - \frac{n+1}{(n+1)^2+1}

= \frac{n(n^2+2n+2) - (n+1)(n^2+1)}{(n^2+1)(n^2+2n+2)} 
= \frac{n^2+n-1}{(n^2+1)(n^2+2n+2)}
=\frac{(n+\frac{1}{2})^2-\frac{5}{4}}{(n^2+1)(n^2+2n+2)}

Both numerator and denominator of this last fraction are positive for n \geq 1, so a_n - a_{n+1} > 0 and the sequence is decreasing.


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Sat, 7 Jul 2012 23:37:53 UTC 
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Easy answer: Take the reciprocal and show that it is an increasing sequence.

(n + 1/n) < (n+1 + 1/(n+1))


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Sun, 8 Jul 2012 17:48:11 UTC 
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Oh? What if there are negative terms? Or worse, a zero term?

a-b>0 is definition for a>b and is usually the most generally direct way to do things. In this particular case the alternative can work with the right caveats, but without them things are incomplete.

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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Sun, 8 Jul 2012 21:57:42 UTC 
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Shadow wrote:
Oh? What if there are negative terms? Or worse, a zero term?

a-b>0 is definition for a>b and is usually the most generally direct way to do things. In this particular case the alternative can work with the right caveats, but without them things are incomplete.

My comment was a reference to the specific problem, not as a general approach.


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Sun, 8 Jul 2012 22:09:11 UTC 
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mathematic wrote:
Shadow wrote:
Oh? What if there are negative terms? Or worse, a zero term?

a-b>0 is definition for a>b and is usually the most generally direct way to do things. In this particular case the alternative can work with the right caveats, but without them things are incomplete.

My comment was a reference to the specific problem, not as a general approach.


Naturally, but that's no reason not to be careful. Remember, the original poster cannot read your mind, and just because you have checked the requirements to use the method you describe in your head, doesn't mean the op has, and if not advice like that can do more harm than good.

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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Mon, 9 Jul 2012 22:52:12 UTC 
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Shadow wrote:
mathematic wrote:
Shadow wrote:
Oh? What if there are negative terms? Or worse, a zero term?

a-b>0 is definition for a>b and is usually the most generally direct way to do things. In this particular case the alternative can work with the right caveats, but without them things are incomplete.

My comment was a reference to the specific problem, not as a general approach.


Naturally, but that's no reason not to be careful. Remember, the original poster cannot read your mind, and just because you have checked the requirements to use the method you describe in your head, doesn't mean the op has, and if not advice like that can do more harm than good.

I don't understand why you are making such a big deal. The original question was for a specific problem and my response was to that specific problem. I see no reason why the proposer would think I was trying to give an answer for all series.


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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Tue, 10 Jul 2012 01:20:54 UTC 
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mathematic wrote:
Shadow wrote:
mathematic wrote:
Shadow wrote:
Oh? What if there are negative terms? Or worse, a zero term?

a-b>0 is definition for a>b and is usually the most generally direct way to do things. In this particular case the alternative can work with the right caveats, but without them things are incomplete.

My comment was a reference to the specific problem, not as a general approach.


Naturally, but that's no reason not to be careful. Remember, the original poster cannot read your mind, and just because you have checked the requirements to use the method you describe in your head, doesn't mean the op has, and if not advice like that can do more harm than good.

I don't understand why you are making such a big deal. The original question was for a specific problem and my response was to that specific problem. I see no reason why the proposer would think I was trying to give an answer for all series.


But your original "Easy Answer" was technically incorrect. If someone took your solution verbatim and wrote:

mathematic wrote:
Take the reciprocal and show that it is an increasing sequence.

(n + 1/n) < (n+1 + 1/(n+1))


then there would be a good chance that the OP would lose a significant number of points on that problem. The reason is because this is not a mathematically sound proof, as pointed out by Shadow in the first line of this quote:

Shadow wrote:
Oh? What if there are negative terms? Or worse, a zero term?

a-b>0 is definition for a>b and is usually the most generally direct way to do things. In this particular case the alternative can work with the right caveats, but without them things are incomplete.


However, I would like to point out that although you have to check that everything is positive (which is very trivial), mathematic's trick is quicker and it simplifies a lot of the messy algebra. So, in my opinion, going the alternative route in this case is preferable.~

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 Post subject: Re: [!] MATHEMATICAL INDUCTION AND SEQUENECES
PostPosted: Tue, 10 Jul 2012 06:53:49 UTC 
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Of course it is, if the proper technicalities are observed, and they must be. Math is not a half-assed subject. You can be in context in your head, but unless you include that in your answer your logic is incomplete. The method is fine, but I can make any ludicrous claim I want and it might be right, but without a complete argument to back it up, that claim is useless as far as math is concerned, no matter how correct it is.

I don't claim to be perfect at this either, but it doesn't mean I'm wrong to mention how these things need to be said.

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