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Differential Calculus

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16 Mar 2008 21:41:00 IST

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Qn intended for all except sboosy:

For what n is the nth derivative of sin(x³) non-zero?

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Anand Hegde

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16 Mar 2008 22:28:32 IST

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3 is not right???

sandeep ramesh

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16 Mar 2008 22:29:37 IST

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if its right, then why hasnt bhattji told so?

gokul subramanian

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16 Mar 2008 22:31:05 IST

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rohit there r more solutions isnt tht obvious

sandeep ramesh

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16 Mar 2008 22:33:37 IST

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well then you can as well say that its once in 3 cycles da i.e. for all 3n its non 0

sandeep ramesh

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16 Mar 2008 22:50:40 IST

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what sir? is it right?

jeesucks

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16 Mar 2008 22:54:12 IST

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answer is 3rd derivative

shridhar ramachandran

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16 Mar 2008 23:06:22 IST

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i think the answer is every $(6k -3)$ th derivative where $k$ = 1,2,3...
i.e. 3rd, 9th, 15th and so on..

sandeep ramesh

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16 Mar 2008 23:07:44 IST

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WHY NOT 3n??? bcos again that cos(x^3) or whatever will undergo a cycle

shridhar ramachandran

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16 Mar 2008 23:09:30 IST

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i didnt do it like that da.. i did using series expansion..
anyway, lets wait for bhattji to post

sandeep ramesh

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16 Mar 2008 23:10:54 IST

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Sirji plz tell us whether its right or not

jeesucks

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16 Mar 2008 23:12:26 IST

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Hari Shankar

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16 Mar 2008 23:13:09 IST

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taylor series expansion is right. it contains only the powers of the form 6k-3.
So that means the other derivatives are zero at x =0

abhishek gupta

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20 Mar 2008 19:12:07 IST

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sir the answer is n=2..........

Hari Shankar

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20 Mar 2008 21:23:40 IST

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See the sine series is

sinx = x-x³/3! + x⁵/5! - ...

So sin(x³) = x³ - x⁹/3! + x¹⁵/5! - ...

Compare this with Taylor series for f(x) =sin(x³)

f(x) = f(0) + x f'(0) + x²/2! f"(0) + x³/3! f"'(0) + ...

This tells us that fⁿ(0) is non-zero if the coefficient of xⁿ is non-zero i.e. for n = 3(2k-1) = 6k-3

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