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October 11, 2025, 03:55:25 am

Author Topic: TT's Maths Thread  (Read 148180 times)  Share 

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TrueTears

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Re: TT's Maths Thread
« Reply #75 on: November 15, 2009, 05:52:53 pm »
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hahaha that splitting was the crux of the question, now I got it!

Let



We require to find

Telescoping the first sum yields:



Telescoping the second sum yields:



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TrueTears

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Re: TT's Maths Thread
« Reply #76 on: November 15, 2009, 06:55:13 pm »
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Here's a fun one... any idea? I'm fiddling around for now haha

For each permutation of the integers , form the sum to find the average value of all these sums.

Something I just came up with:

Consider a permutation like this: (Just notice the first 'term' ie, for each row)



















Now ignore the other terms, and just consider the first term of each row, now instead of adding "horizontally" why not try adding vertically?

Thus we get:

Now if we make another set of rows starting with as we yield the "vertical" sum of

Now if we can compute this sum:

Then we found the total sum! Now my problem is how to compute it...



Actually that's not right... that doesn't give the sum of all the permutations... back to drawing board.



Gonna come back to this one, can't get it.
« Last Edit: November 15, 2009, 08:51:54 pm by TrueTears »
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TrueTears

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Re: TT's Maths Thread
« Reply #77 on: November 15, 2009, 09:15:25 pm »
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Evaluate the infinite product:

So I factored it then I don't know how to continue.
« Last Edit: November 15, 2009, 09:19:15 pm by TrueTears »
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supdawg

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Re: TT's Maths Thread
« Reply #78 on: November 15, 2009, 09:23:13 pm »
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|a-b|
18 ways to equal 1
16 ways to equal 2
14 ways to equal 3
...
2 ways to equal 9

E(|a-b|) = 11/3
E(Sum) = 55/3

TrueTears

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Re: TT's Maths Thread
« Reply #79 on: November 15, 2009, 09:59:03 pm »
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Okay I'm not sure if this method is rigorous, seems rather primitive without some formal definitions, but anyways:

Writing the first few terms of the product out you get:



A pattern can be seen with the canceling:

Say we have the fractions in the form of

Then the cancelling looks like this: (Ignore the first two fractions of the product series, consider from onwards...)

To cancel out the top right number, it will correspond with the bottom right number of the following fraction:



'' will always cancel.

To cancel the top left number, it will always correspond with the bottom left number of 2 fractions BEFORE it.

where represents 1 other fraction in between.

'' will always cancel.

Therefore since the top 2 numbers of any fraction will cancel out the bottom 2 numbers of another fraction, then ALL fractions in the product series to infinity will cancel.

Now analyzing the first two fractions we see some terms which can not be 'canceled'



The 1 can be omitted as it does nothing.

Using the "pattern" we formed earlier, both the top right numbers of both fractions will cancel. So we are left with:



Now the top left number, ie the 2 in the second fraction, can not cancel as there is only 1 fraction preceding it, since we realised we need TWO fractions preceding it to cancel it.

Now both the 3 and 4 in the respective fractions can be cancelled since they are the bottom left of the each fraction which means there must be another 2 fractions FOLLOWING it which can cancel them out. Thus 3 and 4 are gone.

Thus we are only left with





Is there any other methods with this question? The canceling is not bad but is there any more rigorous methods of doing it?

Thanks!
« Last Edit: December 20, 2009, 02:11:01 pm by TrueTears »
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kamil9876

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Re: TT's Maths Thread
« Reply #80 on: November 15, 2009, 10:06:00 pm »
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Evaluate the infinite product:

So I factored it then I don't know how to continue.

Split the product into two. Now notice the simpler product:






Now notice how 3 cancels, 4 cancels, 5 cancels etc. and in general, the denominator of the ith term cancels with the numerator of the (i+2)th term, this should cancel a lot of in between factors and leave only some factors on the ends which will give a nice closed expression of which we can take the limit of.

A similair thing can be done with the second factor since it can be rewritten as:




which also has this recursive nature due to n-1,n,n+1, being consecutive, so same strategy can be applied.

LOL TT beat me to it :P
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."

TrueTears

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Re: TT's Maths Thread
« Reply #81 on: November 15, 2009, 10:09:20 pm »
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lol yeah thanks kamil, you way is actually much neater because by splitting them up, you can see the pattern much clearer :P

Now how to apply limits and get rid of the term?
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kamil9876

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Re: TT's Maths Thread
« Reply #82 on: November 15, 2009, 10:18:30 pm »
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so you can do just like in my example find the product to n=k. Find this as a function of k which you can do since heaps of denominators cancel etc. Once you do you will basically get the quotient of two polynomials, so you just take their limit to infinity.
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."

TrueTears

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Re: TT's Maths Thread
« Reply #83 on: November 15, 2009, 10:29:54 pm »
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so you can do just like in my example find the product to n=k. Find this as a function of k which you can do since heaps of denominators cancel etc. Once you do you will basically get the quotient of two polynomials, so you just take their limit to infinity.
Ahh I see and that would be more tedious I guess, which is why this pattern method works a bit better  ;)
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kamil9876

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Re: TT's Maths Thread
« Reply #84 on: November 15, 2009, 10:42:28 pm »
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It actually turns out quite nicely.

the fact that in the first product, the denominator of the ith term cancels with numerator of (i+2)th term gives:


the fact that in the second product the numerator of the ith term cancels with the denominator of the (i+1)th term leaves only the denominator of first term, and numerator of last term hence:



Hence the product of these is:




whose limit as approaches infinity is:

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."

TrueTears

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Re: TT's Maths Thread
« Reply #85 on: November 15, 2009, 10:56:42 pm »
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Thanks kamil.

What exactly did you do though lol, did you just n = k, k+1, k+2 etc?
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Re: TT's Maths Thread
« Reply #86 on: November 15, 2009, 11:01:00 pm »
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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."

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Re: TT's Maths Thread
« Reply #87 on: November 15, 2009, 11:07:02 pm »
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ahhh thanks for that, that's just like telescoping but for products!
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Re: TT's Maths Thread
« Reply #88 on: November 15, 2009, 11:39:12 pm »
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Let be the integer closest to . Find .
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Re: TT's Maths Thread
« Reply #89 on: November 16, 2009, 01:24:15 am »
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sketch:

we need to know what the values of f(n) are first of all.

iff:





How many values of satisfy this will depend on , notice that the difference is an indication of how many values of solve this. In fact it is the number of values of n that solve this, since the difference is itself an integer(since it's a difference of two numbers with the same decimal part). After doing the messy algebra, you get an expression for the number of times appears, now you can figure out which numbers and how many of them you are exactly summing, and then see if you can try to find the sum.
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."