"
an infinite number of peptide hormones involved in the digestive system
The Math Geek Says: That's not possible. There'd have to be no upper bound on the number of atoms in the hormone. In a person who weighs, say, 100 kg, there are no more than 6.022*1028 atoms (that's the number if they were all hydrogen atoms); and there can't be more atoms in the hormone than there are in the entire body.
Now, granted, the only upper bound I can place on the number of hormones (without knowing more about their structure) is (6.022*1028)!, which certainly qualifies as Lots. But it's still finite. :-)" from ![[livejournal.com profile]](https://www.dreamwidth.org/img/external/lj-userinfo.gif)
metageek 's comment at least a year ago.
Possible hormones in the digestive system are peptide hormones so they are composed of any of 20 amino acids, in a chain of length 3 to 30 amino acids long. So a mathematical expression for that would be:
20^3 +20^4 +20^5 + ... + 20^30, where the ... is terms of 20^n for n=6 to 29. Is there any compact way to express this? Is there a way of doing this in a shorter way?
![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
The Math Geek Says: That's not possible. There'd have to be no upper bound on the number of atoms in the hormone. In a person who weighs, say, 100 kg, there are no more than 6.022*1028 atoms (that's the number if they were all hydrogen atoms); and there can't be more atoms in the hormone than there are in the entire body.
Now, granted, the only upper bound I can place on the number of hormones (without knowing more about their structure) is (6.022*1028)!, which certainly qualifies as Lots. But it's still finite. :-)" from![[livejournal.com profile]](https://www.dreamwidth.org/img/external/lj-userinfo.gif)
metageek 's comment at least a year ago.
Possible hormones in the digestive system are peptide hormones so they are composed of any of 20 amino acids, in a chain of length 3 to 30 amino acids long. So a mathematical expression for that would be:
20^3 +20^4 +20^5 + ... + 20^30, where the ... is terms of 20^n for n=6 to 29. Is there any compact way to express this? Is there a way of doing this in a shorter way?