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I want to find $f^{(80)}(0)$, where $f(x) = \displaystyle \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$.

My attempt:

Let $g(x) = \displaystyle \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Now $f(x)$ can be rewritten as $$ f(x) = g{\left(x^{20}\right)} - g(0). $$

Taking the derivative $80$ times, $$ f^{(80)}(x) = \sum_{i = 1}^{80}{a_i \times g^{(i)}{\left(x^{20}\right)}}, $$ where $\{a\}$ is a sequence of integers. I tried to find the sum, but little progress was made.

I noticed that if I plugged $x = 0$ into the equation, all of the "$x$-linked" terms would be canceled. hence, only the first few terms would be significant, which is $\displaystyle \sum_{i = 1}^{4}{a_i \times g^{(i)}{\left(x^{20}\right)}}$.

Another observation is that $a_i$ will become $0$, i.e. $g^{(i)}$ will vanish after 20 times the derivative was taken. Thus, the final answer would be $a_4 \times g^{(4)}{\left(x^{20}\right)}$.

I only managed to find $a_1 = 20!$ for the first 20 derivatives and failed to derive explicit formulas for the other three.

I would like to know how to solve this problem appropriately, thanks in advance.

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    $\begingroup$ Have you tried finding a closed form for $f(x)$? Substituting $u=\sqrt t$ and by-parts would help. $\endgroup$
    – Tavish
    Dec 19, 2020 at 18:05
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    $\begingroup$ The function $f$ is not differentiable at the origin as it is not defined to the left of the origin. $\endgroup$
    – hunter
    Dec 19, 2020 at 18:10
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    $\begingroup$ @hunter Why do you think $f$ is not defined there? Actually, $f$ is an even function. $\endgroup$
    – VIVID
    Dec 19, 2020 at 18:22
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    $\begingroup$ @hunter You might be confusing $f(x)$ with $f(t)$. $\endgroup$
    – Tavish
    Dec 19, 2020 at 18:26
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    $\begingroup$ Think the power series way. $\endgroup$
    – metamorphy
    Dec 19, 2020 at 18:27

3 Answers 3

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Some simple rules allow us to find the first derivative

Namely the fundamental theorem of calculus and the chain rule

Which gives

$$ \frac{df(x)}{dx} = \frac{df(x)}{d(x^{20})}\cdot \frac{d(x^{20})}{dx} = x^{10} \cdot \sin x^{10}\cdot 20x^{19}= 20x^{29}\sin x^{10}$$

We know that deriving this another 79 times gives the 80th derivative of $f$. To find the 79th derivative of this derivative we use

The taylor expansion of $\sin u$ at $u=0$ and substitute $u=x^{10}$

Which gives

$$20x^{29}\sin x^{10} = 20x^{29}\sum_{k=0}^{\infty} \frac{(-1)^k (x^{10})^{1 + 2 k}}{(1 + 2 k)!}$$

Which allows us to use the following fact

the 79th derivative of this function at zero will be $79!$ multiplied by the coefficient of $x^{79}$ - this is valid because all other terms have either vanished or are stil multiplied by $x=0$

We can find the corresponding $k$ as follows

$$10(1+2k) = 79-29 \iff k = 2$$

Now we apply arithmatic to find the solution

$$f^{(80)}(0) = \frac{20\cdot 79!}{5!} = \frac{79!}{6}$$

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    $\begingroup$ Finding the derivative by Taylor's series is very new to me. Thank you for your detailed answer. $\endgroup$
    – KM02
    Dec 19, 2020 at 20:03
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We have $f'(x) = 20x^{19}x^{10}\sin(x^{10}) = 20x^{29}\sin(x^{10})$. What do the first few terms of the Taylor series of $\sin$ tell you?

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  • $\begingroup$ Thanks for the suggestion of using the Taylor series, I will try it. $\endgroup$
    – KM02
    Dec 19, 2020 at 20:04
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Let $$F(t) = \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$$ Then, we have $f(x) = F(x^{20})-F(0).$

Differentiating once you get $$ f'(x) = 20x^{19}F'(x^{20}) = 20x^{19}(\sqrt{x^{20}}\sin\sqrt{x^{20}}) \\ f'(x) = 20x^{29}\sin(x^{10})$$ I hope continuing is not difficult.

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  • $\begingroup$ Something's wrong. At the end you have $x^{19}\sqrt{x^{20}}=x^{39}$ $\endgroup$
    – saulspatz
    Dec 19, 2020 at 18:09
  • $\begingroup$ @saulspatz Oh yes, there should be an absolute value. $\endgroup$
    – VIVID
    Dec 19, 2020 at 18:17
  • $\begingroup$ @VIVID No there shouldn't, it's something else :) $\endgroup$
    – witten_s
    Dec 19, 2020 at 18:18
  • $\begingroup$ I see now, haha :D $\endgroup$
    – VIVID
    Dec 19, 2020 at 18:23

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