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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on May 27, 2009, 02:25:17 pm

Title: hard integral
Post by: /0 on May 27, 2009, 02:25:17 pm: What is the 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999th derivative of sin(x)
Title: Re: hard integral
Post by: kamil9876 on May 27, 2009, 02:55:43 pm: -sinx
Title: Re: hard integral
Post by: evaporade on May 27, 2009, 03:17:10 pm: -cosx
Title: Re: hard integral
Post by: kamil9876 on May 27, 2009, 04:23:33 pm: yes, it's the derivative of my answer :P forgot how to count :(
Title: Re: hard integral
Post by: pHysiX on May 27, 2009, 04:25:23 pm: -cos(x) =]
Title: Re: hard integral
Post by: 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)

Consider first of all numbers of the form $d = \sum_{0}^n 10^{n} = 1 + 10 + 100 + \dots$ (numbers whose digits are all 1, with $n \geq 1$ ).

Considering modulo 4, we find $d \equiv \left(100 + 1000 + \dots \right) + 11 \pmod{4} \equiv \left(0 + 0 + \dots \right) + 3 \pmod{4} \equiv 3 \pmod{4}$ .

Therefore we can say that numbers of the form $f = \sum_{0}^n 9 \cdot 10^{n}$ are $3$ modulo $4$ . (since: $3 \cdot 9 \pmod{4} \equiv 27 \pmod{4} \equiv 3 \pmod{4}$ )

Therefore we know that we wish to differentiate $\sin(x)$ $p$ times, where $p$ is a number which is $3$ modulo 4. This means that the sought derivative is equal to the 3rd derivative of $\sin(x)$ , which is $-\cos(x)$ .
Title: Re: hard integral
Post by: kamil9876 on May 30, 2009, 12:59:53 am: 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:

$10^n-1$
$=5^n*2^n-1$
$=5^n*2^{n-2}*4-1$
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 :-[
Title: Re: hard integral
Post by: enwiabe on May 30, 2009, 03:23:09 am: Did you mean 5^n instead of 5^2, kamil?
Title: Re: hard integral
Post by: kamil9876 on May 30, 2009, 10:53:11 am: yep sorry :P there u go, another mistake :P