Derivative of a finite continued fraction

[AvgStudent]

I got this as a challenge question from my professor, but not sure how to start. Let the fraction be be iterated 2018 times, and
[math]f(x)=x-\frac{1}{x-\frac{1}{...-\frac{1}{x-1}}}[/math]Find [imath]f'(0)[/imath]

 

[Deleted member 4993]

I got this as a challenge question from my professor, but not sure how to start. Let the fraction be be iterated 2018 times, and
[math]f(x)=x-\frac{1}{x-\frac{1}{...-\frac{1}{x-1}}}[/math]Find [imath]f'(0)[/imath]

Since this is a challenge question - you need to show more effort !!

I see a recursive relationship here.

 

[AvgStudent]

Since this is a challenge question - you need to show more effort !!

I see a recursive relationship here.

With your clue:
[math]f_{n+1}(x)=x-\frac{1}{f_{n}(x)}[/math]However, I'm unsure about the first iteration. Is it [imath]f_1(x)=x[/imath] or [imath]f_1(x)=x-1[/imath]?

 

[blamocur]

With your clue:
[math]f_{n+1}(x)=x-\frac{1}{f_{n}(x)}[/math]However, I'm unsure about the first iteration. Is it [imath]f_1(x)=x[/imath] or [imath]f_1(x)=x-1[/imath]?

The latter, i.e. [imath]x-1[/imath]

 

[AvgStudent]

Next, I tried to compute the derivative.
[math]f'_{n+1}(x)=\frac{d}{dx}\Bigg(x-\frac{1}{f_{n}(x)}\Bigg)=1+\frac{f'_n(x)}{f_n(x)^2}[/math]Is this correct? If so, how do I evaluate at 0?

 

[blamocur]

Looks good to me.

But: don't be afraid to make mistakes. Move on, even if you are not 100% sure, and see where it gets you. Making mistakes is a very efficient way to learn.

 

[AvgStudent]

So I tried writing out the first few terms for [imath]f_n(0)[/imath].
[math]f_1(0)=0-1=-1\\ f_2(0)=0-\frac{1}{0-1}=1\\ f_3(0)=-1\\ f_4(0)=1\\[/math]I see a pattern. For odd n, then [imath]f_n(0)=-1[/imath]. Whereas for even n, [imath]f_n(0)=1[/imath].

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As far as [imath]f'_n(0)[/imath], I also tried writing out the first few terms.
[math]f'_1(0)=1\\ f'_2(0)=1+\frac{1}{(-1)^2}=2[/math][math]f'_3(0)=1+\frac{2}{1^2}=3\\ f'_4(0)=1+\frac{3}{(-1)^2}=4\\[/math]

 

[blamocur]

Looks to me that you are almost done.

 

[AvgStudent]

I don't know why it took me a while to generalize the nth terms. But
[math]f'_{n+1}(0)=1+\frac{n}{(-1)^{2n}}=1+n\\ f'_{2018}(0)=1+2017=2018?[/math]

 

[blamocur]

Time to change to "AboveAvgStudent" ?

 

[AvgStudent]

Time to change to "AboveAvgStudent" ?

Lol. I'm still on the fence about the answer because the question has 2018 iterations i.e n=2018. So are we looking for [math]f'_{2019}(0) \quad or\quad f'_{2018}(0)?[/math]

 

[blamocur]

It depends on whether [imath]x-1[/imath] is considered 0-th or 1st iteration Seriously, I don't know how to interpret the text of the problem, but I hope your teacher appreciate the solution without nitpicking on exact number of iterations.