# B Decomposition of a function into even and odd parts

#### PainterGuy

What do you assume about your functions? Are they necessarily smooth? Is one necessarily dominating the other? Studying some finite collection of examples is not very useful outside of potentially giving us some (mis-leading) intuition.
Thank you but I have already been told that and agree with those replies. In my last posting I was only trying to clarify a point from @mfb 's post.

#### olgerm

Gold Member
If function f is real:
$f_{even}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*f(x_1))))$
$f_{odd}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Im(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*f(x_1))))$
or
$f_{even}(x)=IFT( Re(FT(f)))(x)$
$f_{odd}(x)=IFT( Im(FT(f)))(x)$
FT fouruer transform. IFT is inverst fourier transform.

i am not sure if it also always works if f is complex function.

Last edited:

#### mfb

Mentor
Let $g(x)=i f(x)$. Then $\hat g(\omega) = i \hat f(\omega)$ due to the linearity of the Fourier transform. What was the real part is now the imaginary part and vice versa (+sign), so these formulas don't work any more.

#### olgerm

Gold Member
For complex functions it should be:
$f_{even}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Re(f(x_1)))))+\sqrt{-1}*\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Im(f(x_1)))))$
$f_{odd}(x)=\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Im(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Re(f(x_1)))))+\sqrt{-1}*\int_{-\infty}^{\infty}(d\omega*e^{i*2\pi*\omega*x}* Re(\int_{-\infty}^{\infty}(dx_1*e^{-i*2\pi*\omega*x_1}*Re(f(x_1)))))$
or
$f_{even}(x)=IFT( Re(FT(Re(f))))(x)+\sqrt{-1}*IFT( Re(FT(Im(f))))(x)$
$f_{odd} (x)=IFT( Im(FT(Re(f))))(x)+\sqrt{-1}*IFT( Im(FT(Im(f))))(x)$
FT fouruer transform. IFT is inverst fourier transform.

• weirdoguy