The next big statement that we will have to derive and prove will be the central slice

theorem.

And, well, it's probably easiest when I just write down what it means.

So we take the Radon transform of f, our f of s and theta, and now we integrate it with

respect to this Fourier-type term.

And this is the same as taking the function whose Radon transform we've just obtained,

taking the Fourier transform of it and evaluating it in r times theta vector.

So there is geometrical structure in there, but for now it's just a formula that we want

to prove.

Okay, the easiest way is to start with the left-hand side.

So we have minus infinity to infinity of rf, s theta vector, e to the minus 2 pi, isr,

ds.

Now we insert the definition of the Radon transform, so we get another integral in here,

where we have f of s theta vector plus t theta vector perpendicular, d, sorry, e to the minus

2 pi, oh no, that's outside of this bracket, dt, e to the minus 2 pi, isr, ds.

And now we call this object here, which is in r2, we call this x, and note that x transposed

theta vector, well this is s theta vector plus t theta vector perpendicular transposed

theta.

Well, those two terms here are perpendicular to each other, so that's zero, and what is

remaining is s times theta vector transposed theta vector, and this is 1 because theta

is a unitary vector.

Okay, now we remove those brackets and put everything next to each other, f of s theta

plus t theta vector perpendicular, e to the minus 2 pi, i, and now we include what we

just computed, this is the same thing as writing s theta vector plus t theta vector perpendicular

transposed theta vector times r.

So we've just replaced s by this longer expression here, dt ds, and now we interpret this as

a two dimensional integral, well it is a two dimensional integral, with respect to dt and

ds, and we write this x as phi of s and t.

And then we can write this as phi of, so f of phi of s t, e to the minus 2 pi i phi of

s t transposed theta vector times r, so phi is a map from r2 to r2.

We can also compute the Jacobian, and well, let's remember that, I need a bit more space

for that.

Let's move that here, so a short note in here, so phi of s t is s theta vector plus t theta

vector perpendicular, and by definition of those quantities, this is s, sorry, s times

cosine 2 pi theta sine of 2 pi theta, now this is not a vector anymore, this is the

angle theta plus t, and that is sine of 2 pi theta, well let's make the brackets explicit

here, and minus cosine of 2 pi theta.

Now we can compute the Jacobian of this map from r2 to r2, so we take s and t, and this

is mapped to that.

The Jacobian matrix looks like this, cosine 2 pi theta sine 2 pi theta sine 2 pi theta

and minus cosine of 2 pi theta, and note that the determinant of d phi of s t, so that's

minus cosine squared minus sine squared, this is minus 1, so the absolute value of the determinant

is 1, so this means we can insert this term without doing any harm, because it's 1, d

phi of s t, d s and t.

And now you can apply integration by substitution, and this is exactly, well let's be a bit more

verbose, so let's call all of this g of phi of s t, so this is g of phi of s t times determinant

of the Jacobian of phi of s t d s t, and by the integration by substitution rule, we can

rewrite this, well phi is a bijective mapping from r2 to r2, so the integration domain does

not change, then this is g of, you can just call it x y, the determinant term drops, and

we have d of x y, and now we can write down what this is, this is the integral over r2