Let $f$ be a function such that $f^\prime(x)=\sin(x^2)$ and $f(0)=0$.What are the first three nonzero terms of the Maclaurin series for $f\\$ ?
I've tried:$f^{\prime\prime}(x)=2x\cos(x^2)$ so $f^{\prime\prime}(0)=0$; $f^{\prime\prime\prime}(x)=2\cos(x^2)-4x^2\sin(x^2)$ so $f^{\prime\prime\prime}(0)=2$.
Hence the first nonzero term is $\frac{f^{\prime\prime\prime}(0)}{3!}x^3=\frac{x^3}{3}$.But it seemed difficult for me to find the next nonzero term, so I checked the answer and found it was $\frac{x^3}{3}-\frac{x^7}{42}+\frac{x^{11}}{1320}$. Do I have to calculate the 7th and 11th derivative of $f$ in order to find $f^{(7)}(0)$ and $f^{(11)}(0)$? Put another way, is there any quick way to find $f^{(n)}(0)$?
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Aaron Lee
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1Note that $f’’’(x)=2\cos(x^2) -4x^2f’(x)$. Yes you need to compute the derivatives, but try and be smart about it. :) – Dr. Tony Tony Chopper May 04 '23 at 14:30
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1Hint: If you're familiar with the Maclaurin series of $\sin(x)$ you can easily find the one for $\sin(x^2)$ – jjagmath May 04 '23 at 14:32
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@jjagmath But it's said that $f^\prime(x)=\sin(x^2)$ instead of $f(x)$, can I still compute that way? – Aaron Lee May 04 '23 at 14:36
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1@Dr.TonyTonyChopper Thanks, I think I get the point. – Aaron Lee May 04 '23 at 14:39
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2If $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3+\cdots$ then $f'(x) = a_1 +2 a_2 x + 3a_3 x^2+\cdots$, so it's pretty easy to see the relation between the coefficients of $f$ and $f'$ – jjagmath May 04 '23 at 14:39
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@jjagmath Thanks, I've tried and it works. – Aaron Lee May 04 '23 at 14:42
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This exercise follows by uniqueness of the Maclaurin expansion. In other words, if two power series appear to be a Taylor expansion of the same function in a region, then they are in fact the same power series, coefficient for coefficient. Thus, if we have the Maclaurin series of a function $g$, then we can derive the Maclaurin series for functions such as $2g, g+x^2$ and so on simply by modifying the Maclaurin series of $g$ appropriately.
In this case, the Maclaurin series for $\sin(x)$ is $$ \sin(x) = x-\frac{x^3}{3!}+\frac{x^5}{5~!} - \frac{x^7}{7!} + \ldots $$ By substituting $x^2$ in place of $x$, $$ \sin(x^2) = x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!} - \frac{x^{14}}{7!}+\ldots $$ is the Maclaurin series for $\sin(x^2)$, by uniqueness. Now, if $f'(x) = \sin(x^2)$, then we can find $f(x)$ by integrating the Maclaurin series of $\sin(x^2)$ term by term. This will give us the Maclaurin series of $f$ by uniqueness.
Doing this, $$ f(x) = \frac{x^3}{3}-\frac{x^7}{7 \times 3!} + \frac{x^{11}}{11 \times 5!} - \ldots $$ which gives the answers $\frac 13,-\frac{1}{42},\frac{1}{1320}$ for the powers $x^3,x^7,x^{11}$.
Obviously, this also works in finding $f^{(n)}(0)$ because the coefficient of $x^n$ in the Maclaurin series of $f(x)$ is $\frac{f^{n}(0)}{n!}$. In this case, we know that the coefficient of $x^{4k+3}$ is $\frac{(-1)^k}{(4k+3)(2k+1)!}$. Hence, $$ f^{(4k+3)}(0) = \frac{(-1)^k(4k+2)!}{(2k+1)!} $$
Sarvesh Ravichandran Iyer
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