f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Forum for the GRE subject test in mathematics.
lawliet
Posts: 2
Joined: Mon Apr 15, 2013 11:15 am

f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Postby lawliet » Mon Apr 15, 2013 11:20 am

I have difficulties to deal with these kinds of questions. So, thank you for answering!

There is a missing condition: f has the 12th derivative. So f(x)=1/2*x^2+x does not satisfy the condition. Sorry!
Last edited by lawliet on Mon Apr 15, 2013 8:25 pm, edited 1 time in total.

tarheel
Posts: 10
Joined: Wed Sep 26, 2012 4:37 pm

Re: f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Postby tarheel » Mon Apr 15, 2013 5:10 pm

lawliet wrote:I have difficulties to deal with these kinds of questions. So, thank you for answering!


If this is a multiple choice question and there is only one correct answer, then it's simple. Let f(x)=\frac{1}{2}x^2+x, then g(x)=\frac{1}{2}x^{20}+x^{10}. So g^{(11)}(x)=\frac{1}{2}\cdot 20 \cdot 19...\cdot 10 x^9. So g^{(11)}(0)=0.

lawliet
Posts: 2
Joined: Mon Apr 15, 2013 11:15 am

Re: f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Postby lawliet » Mon Apr 15, 2013 8:26 pm

tarheel wrote:
lawliet wrote:I have difficulties to deal with these kinds of questions. So, thank you for answering!


If this is a multiple choice question and there is only one correct answer, then it's simple. Let f(x)=\frac{1}{2}x^2+x, then g(x)=\frac{1}{2}x^{20}+x^{10}. So g^{(11)}(x)=\frac{1}{2}\cdot 20 \cdot 19...\cdot 10 x^9. So g^{(11)}(0)=0.

Sorry, there is a missing condition that f(x) has the 12th derivative. So, do you have other methods?

waiting512
Posts: 61
Joined: Sat Dec 10, 2011 10:41 pm

Re: f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Postby waiting512 » Mon Apr 15, 2013 9:51 pm

His example does have a 12th derivative.

2help
Posts: 12
Joined: Fri Mar 22, 2013 7:29 pm

Re: f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Postby 2help » Mon Apr 15, 2013 11:02 pm

If you try to find out the pattern in which the 10th derivative of g=f(x^10) expands , you will find that the only term that does not become 0 when you calculate g^(11)(0) is the first term which is 10!f'(x^10). However, when you take the derivative for the 11th time , there is a x^9 comes out of the composite part of it , which finally makes it 0. so the answer is 0.

tarheel
Posts: 10
Joined: Wed Sep 26, 2012 4:37 pm

Re: f'(0)=f''(0)=1, g=f(x^10), the 11th derivative of g at x=0?

Postby tarheel » Tue Apr 16, 2013 10:19 am

lawliet wrote:
tarheel wrote:
If this is a multiple choice question and there is only one correct answer, then it's simple. Let f(x)=\frac{1}{2}x^2+x, then g(x)=\frac{1}{2}x^{20}+x^{10}. So g^{(11)}(x)=\frac{1}{2}\cdot 20 \cdot 19...\cdot 10 x^9. So g^{(11)}(0)=0.

Sorry, there is a missing condition that f(x) has the 12th derivative. So, do you have other methods?


Yeah, waiting512 is right. If f(x)=\frac{1}{2}x^2+x, then f^{(12)}=0,\, \forall x, which exists. Its derivative being zero doesn't mean the derivative doesn't exist.




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