Ultrafunctions and Transformations

I know I’ve written a whole ream of entries on this subject, but as I said before, it’s proved pretty darned fruitful.

Remember that the ultrafunction is defined thus:

u(f(a),b) = f(f(f(…f(a)…) (applied b times)

Now, consider the transformation t(g(a)) from the set of all functions into the set of all functions. Then, consider the ultrafunction of the transformation of the function:

u(t(f(a)),b) = t(f(t(f(t(f(…t(f(a))…) (as before, b repeats)

Would it be possible to “factor out” the transformation? Well, if that were possible, then we’d have:

t(f(f(f(…f(a)…) = t(f(t(f(t(f(…t(f(a))…)

But, since the transformations of two functions are equal only if they are taken of the same function, then:

f(f(f(…f(a)…) = f(t(f(t(f(t(…t(f(a))…)

Which will obviously only be true for a relatively small subset of the set of all functions.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: