Our next step is proving this formula for Radon inversion and the goal is to justify

this formula here which allows us to recover the original density f from Radon data Rf.

So we can input Rf which would be a sinogram and then we have to concatenate some operations

so a one dimensional Fourier transform, some filtering, an inverse Fourier transform again

and then we get back the original density.

But before we can prove this we have to prove an auxiliary lemma first which gives us a

few symmetry properties of the Ruanot transform and of its one dimensional Fourier transform.

This is nothing too complicated, this is just plugging in the definitions.

So for example Rf of s and minus theta vector, this is defined as the integral from minus

infinity to infinity of f of s theta vector plus t, sorry, this is minus theta vector

and then of course we have minus theta vector perpendicular as well dt and well what could

we do now?

We can now shift this negative sign on the scalar variable in front so minus theta vector

plus, sorry, minus t times theta vector perpendicular dt.

Now what do we do, we define t prime as minus t then a few things change.

First we get a minus sign because of dt prime being minus dt then we integrate from plus

infinity to minus infinity because of the sign change in the substitution.

Then we get f of minus s theta vector plus t theta vector perpendicular and then well

accounting for the fact that we can swap integration limits with an additional minus sign we can

combine those two things and we get integral from minus infinity to infinity over f minus

s theta vector plus t theta vector perpendicular, sorry t prime here and then we can just call

this dt again t here so t is t prime now just as a matter of notation and as you can see

this is the same as the Radon transform of f in minus s and theta vector.

Okay so that was quite easy and this will be similar for the Fourier transform of the

Radon transform.

I will spare the details you can do this as an exercise so first part is that, two part

as an exercise.

It basically works exactly like the computation I just showed you so with these symmetry properties

we can now try to prove this very important theorem.

So we want to prove theorem 4.6 now, proof of theorem 4.6 so we start with f of x and

what can we do, how can we recover this function?

We apply Fourier transformation twice so this x is in our two so this is the inverse Fourier

transform of the Fourier transform e to the power of 2 pi i x transposed xi d xi and what

we can now do is convert this into polar coordinates.

So how do we do this?

This is a two dimensional integral with respect to the variable xi so now we can write this

as an integral with respect to radius and angle and the angle will be between 0 and

1 where 1 means 360 degrees and the radius will be from 0 to plus infinity and now we

just have to insert, let me make this a bit more explicit so now we write xi as, which

is in R2 of course, we can write this as R times cosine of 2 pi, let's call it theta,

so good notation in this context, and R sine 2 pi theta and of course this is the same

thing as R times theta vector because this is exactly what the definition of theta vector

was so we can put that in here e to the power of 2 pi i x transposed R theta vector.

Now we integrate with respect to R and theta, not theta vector, theta which is the angle

but in order to account for this transformation, so you might recall from your calculus class

that you have to include the determinant of the Jacobi matrix here and this is the absolute

value of R or just R because R is actually positive but it doesn't hurt to do this.

Okay and now we can use the central slice theorem which said that f hat of R theta vector

is the same as the one dimensional Fourier transform of the Radon transform in R theta

vector, so that is here same thing as, so this doesn't change, this is one dimensional