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#13 I surrender....
Numbers are the essence of the Universe
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#15 This is an easy one.
Prove that any even power of any odd integer leaves a remainder of 1 when divided by 4.
Last edited by JaneFairfax (2007-06-01 02:15:24)
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This one is easier
Last edited by Stanley_Marsh (2007-06-01 07:39:28)
Numbers are the essence of the Universe
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Yah , it comes from mutiplication law of mod
Numbers are the essence of the Universe
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#16
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#17 Here is another exercise I made up myself.
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#18
Last edited by JaneFairfax (2007-08-07 10:48:00)
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This result is important because I used it in proving Tonys question: http://www.mathsisfun.com/forum/viewtopic.php?id=7807
Last edited by JaneFairfax (2007-08-11 03:10:04)
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#19
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#20
I just made this problem up myself; hopefully Ive got the maths correct.
Last edited by JaneFairfax (2007-09-13 11:26:02)
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#21
Consider the sequence
2, 5, 9, 12, 16, 19
The first term is 2 and successive terms are formed by alternately adding 3 and 4 to previous terms:
Prove that
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a more powerful result of 15:
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Hi Jane;
Answer for #21
Don't suppose anyone will ever see this but here goes:
The recurrence formula for your set of numbers is.
this has a characteristic equation of
x^3-x^2-x+1=0 which has roots of {-1,1,1}
This then is the general form of the solution:
we need to determine c1,c2, and c3 from the initial conditions
So the general solution is:
This agrees with your general solution at the bottom of problem 21
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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bobbym. looks like youve got the answer. Well done!
My solution:
Let
Then
and are APs with common difference 7 and first terms 2 and 5 respectively. ThusNote that
can be written as:The result follows.
Last edited by JaneFairfax (2009-04-14 01:22:51)
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Hi Jane;
Thanks for showing me yours, I am not that happy with mine, what justification do I have for the first step? I haven't proven that a(n+3)=a(n+2)+a(n+1)-a(n) is the recursion for that sequence. Can you provide some rigor to my arguments?
bobbym
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Well, Im not that familiar with techniques on solving difference equations, but your solution looked right to me so I gave you full credit for it.
Im pretty sure your first step was correct.
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Hi Jane;
You say you are not very familiar with methods of solving difference equations but yet you came up with the idea of splitting the sequence into 2 coupled difference equations, each handling every other term. I salute you.
bobbym
Last edited by bobbym (2009-04-14 11:48:04)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Aw, thanks. Well, I thought up that problem myself, so for this particular problem I happen to have have my own solution handy.
At least, I discovered the following handy lemma myself:
If a sequence A has properties that alternate between its odd and even terms, let B be the subsequence of its odd-numbered terms and C the subsequence of its even-numbered terms. Find formulas for B and C separately. Then combine the formulas by rewriting A in the manner described above.
Id love to call this Janes lemma if no-one else has discovered it before.
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Hi Jane;
I discovered that about 10 years ago but didn't notice your method of combining them and promptly forgot about the whole idea. So I guess its yours and I will always remember it as Jane's lemma.
bobbym
Last edited by bobbym (2009-04-19 18:24:25)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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