How to find first four terms in Maclaurin Series

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Notice to differentiate $f(x) = x^3 \sin (x^2)$ you must use both the chain rule and the product rule. It follows that

$$ f'(x) = 3 x^ 2 \sin (x^2) + 2 x^4 \cos (x^2) $$

But, notice that since $\sin x = \sum \frac{ (-1)^n x^{2n+1 } }{(2n+1)!} $, then

$$ x^3 \sin (x^2) = x^3 \sum_{n \geq 0} \frac{ (-1)^n x^{4n+2} }{(2n+1)!} = \sum_{n \geq 0 } \frac{ (-1)^n x^{4n+5}}{(2n+1)!} $$

Thus, for example, the first and second term are $$ x^5, \frac{ - x^9 }{3!} $$

and so on.

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raasilinin
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Updated on April 04, 2020

Comments

  • raasilinin
    raasilinin over 3 years

    Find the first four non-zero terms of the Maclaurin series for $x^3 \sin(x^2)$.

    I found the derivatives and got as following:

    first derivative = $2x^3 \cos(x^2)$

    second derivative = $-12x^3 \sin(x^2)$

    third derivative = $-72x^3 \cos(x^2)$

    when I used formula, I am substituting $x=0$ and getting all the first four terms as 0. Please help to solve this.

    • Admin
      Admin over 5 years
      Your derivatives are very incorrect. Do you know the product rule? And the chain rule?
    • Clayton
      Clayton over 5 years
      Also, to save yourself some trouble, you don't have to take any derivatives if you know the Taylor expansion for $\sin(x)$.
    • raasilinin
      raasilinin over 5 years
      But, Maclaurin series need to have derivatives.