How would you find the maclaurin series for this
sin(x^5)
thanks.
recall that the Maclaurin series for sine is:
$\displaystyle \sin x = \sum_{n = 0}^{\infty} \frac {(-1)^n}{(2n + 1)!} x^{2n + 1} = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} +...$ for ALL $\displaystyle x$
so for $\displaystyle \sin \left( x^5 \right)$, just replace $\displaystyle x$ with $\displaystyle x^5$ in the series above and simplify
A McLaurin series is a Taylor series for x = 0. So we have:
$\displaystyle f(x) \approx f(0) + \frac{1}{1!}f^{\prime}(0)x +
\frac{1}{2!}f^{\prime \prime}(0)x^2 + ~ ...$
$\displaystyle f(x) = sin(x^5)$
So
$\displaystyle f(0) = 0$
$\displaystyle f^{\prime}(x) = cos(x^5) \cdot 4x^5$ ==> $\displaystyle f^{\prime}(0) = 0$
$\displaystyle f^{\prime \prime}(x) = 20x^3 cos(x^5) - 25x^8sin(x^5)$ ==> $\displaystyle f^{\prime \prime}(0) = 0$
$\displaystyle f^{\prime \prime \prime}(x) = -125x^{12}cos(x^5) + 60x^2 cos(x^5) - 300x^7sin(x^5)$ ==> $\displaystyle f^{\prime \prime \prime}(0) = 0$
etc.
I'll give you a hint. After a lot of hard work on your part, you will find that $\displaystyle f^{(6)}(0)$ is the first non-zero term in the series.
(This one's just nasty to do by hand!)
-Dan
I don't like telling people to just memorize stuff, but yes, you should have the Maclaurin series for functions like sine, cosine, e^x, 1/(1 - x) ... memorized.
the Maclaurin series is just the Taylor series centered at zero. topsquark showed you how to derive it, but in this case, i think the memorization route is easiest. unless your professor requires you to derive it
The answer in the book is summation from n = 1 to infinity of (-1)^n times x^(5n)
I see the (-1)^n since the values are changing from positive to negative every time, but where does the x^(5n) come from? Also how did you find your results... 1 - x + x^2 - x^3 ... I don't quite understand. Thank you.
he treated $\displaystyle \frac {1}{1 + x}$ as $\displaystyle \frac {1}{1 - (-x)}$ and then used the formula i gave you.
where did the $\displaystyle x^{5n}$ come from? did you actually look at the series? $\displaystyle \sum_{n = 0}^{\infty}(-1)^n\left( x^5 \right)^n = \sum_{n = 0}^{\infty}(-1)^nx^{5n}$