1. ## Taylor series

I have to expand the f(x) thanks to the taylor series, and then solve it
How do i expand the first squareroot(senx)?
There is no actual formula for this... I know how to expand sen^2x but how about sen^1/2 x??
Thanks to everyone

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2. ## Re: Taylor series

Originally Posted by domy7997
I have to expand the f(x) thanks to the taylor series, and then solve it
How do i expand the first squareroot(senx)?
There is no actual formula for this... I know how to expand sen^2x but how about sen^1/2 x??
Thanks to everyone

Sent from my Redmi Note 4X using Tapatalk
Use the chain rule:
$\displaystyle \dfrac{d}{dx} sin \left ( \sqrt{x} \right ) = cos \left ( \sqrt{x} \right ) \cdot \dfrac{1}{2 \sqrt{x}}$

etc.

-Dan

3. ## Re: Taylor series

Originally Posted by topsquark
Use the chain rule:
$\displaystyle \dfrac{d}{dx} sin \left ( \sqrt{x} \right ) = cos \left ( \sqrt{x} \right ) \cdot \dfrac{1}{2 \sqrt{x}}$

etc.

-Dan
I believe you can also just write the series for $\sin(x) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{2k+1}$

and just replace $x \to \sqrt{x}$ yielding

$\Large \sin(\sqrt{x}) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{\frac{2k+1}{2}}$

4. ## Re: Taylor series

Originally Posted by romsek
I believe you can also just write the series for $\sin(x) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{2k+1}$

and just replace $x \to \sqrt{x}$ yielding

$\Large \sin(\sqrt{x}) = \sum \limits_{k=0}^\infty~\dfrac{(-1)^k}{(2k+1)!}x^{\frac{2k+1}{2}}$
From the app i don't know how to write the math correctly sry!

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5. ## Re: Taylor series

Originally Posted by domy7997
$f(x) =\sum \limits_{k=0}^\infty ~f^{(k)}(a)\dfrac{(x-a)^k}{k!}$
$f^{(k)}(a) = \left . \dfrac{d^kf}{dx^k}(x) \right |_{x=a}$