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The question: Let $f(x) = e^{-x}\sin(x)$. Calculate $f^{(2001)}(0)$.

Note that in the question, $f^{(n)}(x)$. means the $n$-th derivative of $f$.

I calculated the first six derivatives on zero and got:

$f^{(0)}(0) = 0$.

$f^{(1)}(0) = 1$.

$f^{(2)}(0) = -2$.

$f^{(3)}(0) = 2$.

$f^{(4)}(0) = 0$.

$f^{(5)}(0) = -4$.

$f^{(6)}(0) = 8$.

So, the pattern isn't pretty clear. Also, I thought that if $n$ is even, then:

$f^{(n)}(x) = -e^{-x}\cos(x) - f(x) + f^{(1)}(x) - f^{(2)}(x) + ... + f^{(n-3)}(x) - f^{(n-2)}(x) - f^{(n-1)}(x)$.

and $f^{(n - 1)}(x) = e^{-x}\cos(x) + f(x) - f^{(1)}(x) + f^{(2)}(x) + ... + f^{(n-4)}(x) - f^{(n-3)}(x) - f^{(n-2)}(x)$.

So by substituting $f^{(n - 1)}(x)$ in the first equation, we get:

$f^{(n)}(x) = -2(e^{-x}\cos(x) + f(x) - f^{(1)}(x) + f^{(2)}(x) + ... + f^{(n-4)}(x) - f^{(n-3)}(x))$

I tried to the same thing with $n$ being odd and tried to perform an induction to get a formula for the value of $f^{(n)}(x)$ but couldn't find a way to do so. Any ideas?

Ulivai
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  • Do you know Euler's formula relating sine, cosine, and exponential? – Daniel Fischer Jul 29 '20 at 14:15
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    Hint: If $f(x)=e^{-x}\sin x$ then $f''(x)=-2e^{-x}\cos x$. If $g(x)=e^{-x}\cos x$ then $g''(x)=2e^{-x}\sin x$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jul 29 '20 at 14:16
  • Ow sure, i forgot the we get back to the function with double derivation (as used to integrate by parts). Thanks, now i solved it – Ulivai Jul 29 '20 at 14:29
  • calculate f'(x),f''(x),f''(x),f''''(x), ... and you will see a pattern – miracle173 Jul 29 '20 at 14:30
  • @miracle173 The OP went up to the 6th derivative and he found no pattern. And even if some pattern would emerge, how certain can you be that it is correct? – imranfat Jul 29 '20 at 14:37
  • @imranfat the OP calculated $f^{(k)}(0)$ and not $f^{(k)}(x)$. I am pretty sure he will recognize the pattern. It is easy to proof the pattern. Try it out and you will see! – miracle173 Jul 29 '20 at 14:43
  • Even without complex numbers, define $C:=e^{-x}\cos x,,S:=e^{-x}\sin x$ so $C^\prime=-C-S,,S^\prime:=C-S$ and $C^{\prime\prime}=2S,,S^{\prime\prime}=-2C$. Hence $S^{(4)}=-2C^{\prime\prime}=-4S$, and $S^{(2001)}=(-4)^{500}S^\prime=2^{1000}e^{-x}(\cos x-\sin x)$. – J.G. Jul 29 '20 at 15:02

1 Answers1

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As suggested in the comments, by Euler's formula we have that

$$f(x) = \Im \left(e^{-x+ix}\right) \implies f^n(0)=\Im \left((-1+i)^n\right)$$

then we can use that

  • $\arg (z^n)=n\arg(z)$
  • $|z^n|=|z|^n$
user
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