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I have a composite function problem. Given $f(x) =e^{-x}$ Find $f^{215} (x)$ I use graphing calculator and I realise $f^4(x$) is the same as $f^2(x)$. With this is mind I'd say $f^{215}$ is the same as $f^3$.

But how could I show it.

arthur anderson
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  • I've added one (of several, not particularly standard) way to indicate that superscripted numbers on function names represent repeated composition (not repeated differentiation). Does this capture your intent? – Eric Towers Oct 11 '19 at 15:32
  • @EricTowers Sorry but I think that your interpretation is really uncorrect. You should rollback and maybe ask to the OP about your ideed before to edit the question in such way. – user Oct 11 '19 at 15:39
  • @gimusi : The first signal that composition was intended was "composite" in the title and in the body of the question. The second signal was "But this is a composite function problem. Can we use differentiation to solve it." as the first comment to your answer. OP seems very clear that composition, not differentiation, is intended. – Eric Towers Oct 11 '19 at 15:46
  • @gimusi : The third signal is OP's use of "f{f(x)}, f{f(f(f(x))} and so on" in further comments. – Eric Towers Oct 11 '19 at 15:48
  • @EricTowers In teh original question the OP said "I use graphing calculator and I realise f4(x) is the same as f2(x)". Is that true for composition? – user Oct 11 '19 at 15:49
  • @gimusi : Of course not. Does this mean the OP used the composition he said he was using with his calculator? – Eric Towers Oct 11 '19 at 15:51
  • @EricTowers I think the OP can clarify that in a moment. – user Oct 11 '19 at 15:55
  • @Eric Towers. I got this problem from a friend of mine. And my first thought in this question was applying the differentiation approach. But my friend used the composite function approach and he could not solve the problem. The aim of asking this question is to see whether this question should be solved with differentiation approach or composite function approach. – arthur anderson Oct 11 '19 at 16:00
  • @arthuranderson : There is nothing in your question indicating that there are two methods under discussion or that differentiation is even a candidate. How did you expect others to guess that from what you wrote? – Eric Towers Oct 11 '19 at 16:03
  • @Eric Towers. My intent is to use composite function approach but it seems this question require differentiation approach to make sense. – arthur anderson Oct 11 '19 at 16:06
  • Since there does not seem to be any coherent way to convert what is actually written in your Question to what you meant, I have rolled back my edits. – Eric Towers Oct 11 '19 at 16:08

1 Answers1

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We have that

  • $f'(x)=\frac{d}{dx}e^{g(x)}=e^{g(x)}\cdot g'(x)=-f(x)$
  • $f''(x)=f(x)$
  • $f'''(x)=f'(x)=-f(x)$
  • $\dots$

therefore your guess is correct.

To prove it rigorously, from $f'(x)=-f(x)$, we have that for any $n\in N$

$$f^{2n-1}(x)=-f(x)\implies f^{2n}(x)=-\frac{d}{dx}f^{2n-1}(x)=-\frac{d}{dx}f(x) =-f'(x)=f(x)$$

$$\implies f^{2n+1}(x)=\frac{d}{dx}f^{2n}(x)=\frac{d}{dx}f(x)=-f(x)$$

and we are looking for $n$ from $1$ to $107$.

user
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