# derivative of a function using cosine transform

If we have a periodic function y(x) the derivative of function can be found using Fourier transform as

$derivative = y'(x)$

taking Fourier transform

$F(derivative) = -kiy(x)$ i is the complex number

$derivative = F^{-1}(-kiy(x))$

Now if the function is real non-periodic function with Neumann boundary condition, can its derivative be found by taking cosine transform?

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With the Fourier transform the function does not have to be periodic, that requirement is only given for the Fourier series approximation for a function. –  peterm Mar 6 '13 at 9:33
@peterm, But cant we express any continuous function with Fourier series? –  alekhine Mar 6 '13 at 9:52
If the function is not periodic and you wish to represent it over all of $\mathbb{R}$ using a Fourier series, you have to take the period as being infinite and then you recover the Fourier transform definition, if you are only interested in representing the function over some finite interval then a Fourier series with a period equal to the length of that interval would suffice (see this question math.stackexchange.com/questions/219725/…) –  Graham Hesketh Aug 5 '13 at 9:59

so to reduce the Fourier transform representation of the derivative to the cosine transform only we would require: $$\int_{-\infty}^{\infty}\left(2\pi\xi\right)\hat{F}(\xi)\sin(2\pi x \xi){d\xi}=0$$ and as $\sin$ is an odd function this integral vanishes when $\left(2\pi\xi\right)\hat{F}(\xi)$ is an even function which occurs when: $$\xi\hat{F}(\xi)=-\xi\hat{F}(-\xi),$$ $$\hat{F}(\xi)=-\hat{F}(-\xi),\tag{2a}$$ that is when $\hat{F}(\xi),$ is odd, and we see from $(1)$ that this requires $$f(x)=-f(-x)\tag{2b}$$ i.e. it requires $f(x)$ to be odd. But if $(2)$ holds then $(1)$ can be written: \begin{aligned}\hat{F}(\xi)&=-i\int_{-\infty}^{\infty}f(x)\sin(2\pi x \xi){dx},\tag{3}\\ f(x)&=i\int_{-\infty}^{\infty}\hat{F}(\xi)\sin(2\pi x \xi){d\xi},\end{aligned}
So to summarise, assuming $(1)$ exists and $(2)$ holds we may define the derivative by the following cosine transform: $$\dfrac{d}{dx}f(x)=i2\pi\int_{-\infty}^{\infty}\xi\hat{F}(\xi)\cos(2\pi x \xi){d\xi}$$ which we see would also result from applying differentiation under the integral sign to $(3)$. With that said, you have not specified what it is that you want to cosine transform in order to find the derivative; if this equation holds: $$\dfrac{d}{dx}f(x)=\int_{-\infty}^{\infty}\left(i2\pi\xi\right)\hat{F}(\xi)e^{i2\pi x \xi}{d\xi},$$ then we could simply define: $$\hat{F}(\xi)=\hat{G}(\xi)\cos(2\pi x \xi)e^{-i2\pi x \xi},$$ to obtain a cosine transform representation of the derivative but this is being a bit pedantic; some clarification of what it is you hope to cosine transform would help to clear that up.