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April 28, 2024, 10:15:34 am
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Author Topic: hard integral  (Read 1314 times)  Share 

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/0

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hard integral
« on: May 27, 2009, 02:25:17 pm »
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What is the 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999th derivative of sin(x)
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kamil9876

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Re: hard integral
« Reply #1 on: May 27, 2009, 02:55:43 pm »
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-sinx
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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."

evaporade

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Re: hard integral
« Reply #2 on: May 27, 2009, 03:17:10 pm »
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-cosx
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kamil9876

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Re: hard integral
« Reply #3 on: May 27, 2009, 04:23:33 pm »
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yes, it's the derivative of my answer :P forgot how to count :(
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pHysiX

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Re: hard integral
« Reply #4 on: May 27, 2009, 04:25:23 pm »
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-cos(x) =]
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dcc

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Re: hard integral
« Reply #5 on: May 30, 2009, 12:45:14 am »
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Quote from: /0 on May 27, 2009, 02:25:17 pm
What is the (large number divisible by 3)th derivative of sin(x)

Consider first of all numbers of the form (numbers whose digits are all 1, with ).

Considering modulo 4, we find .

Therefore  we can say that numbers of the form are modulo .  (since: )

Therefore we know that we wish to differentiate times, where is a number which is modulo 4.  This means that the sought derivative is equal to the 3rd derivative of , which is .
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kamil9876

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Re: hard integral
« Reply #6 on: May 30, 2009, 12:59:53 am »
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Quote from: dcc on May 30, 2009, 12:45:14 am
Quote from: /0 on May 27, 2009, 02:25:17 pm
What is the (large number divisible by 3)th derivative of sin(x)



I think it's much more efficient to look at it as a number of the form:




and so that is a number before a multiple of 4. so 3 in mod 4.
But hey, you got it right unlike me who counted sin(x) as the 1st derivative of sin(x) rather than the 0th  :-[
« Last Edit: May 30, 2009, 10:53:34 am by kamil9876 »
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enwiabe

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Re: hard integral
« Reply #7 on: May 30, 2009, 03:23:09 am »
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Did you mean 5^n instead of 5^2, kamil?
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kamil9876

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Re: hard integral
« Reply #8 on: May 30, 2009, 10:53:11 am »
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yep sorry :P there u go, another mistake :P
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