Please allow a minor correction towards the conclusion of the proof of Liouville's Theorem: you say "So RHS of the above inequality becomes arbitrarily large" but what you mean is "So *the denominator* of the RHS of the above inequality becomes arbitrarily large". Very nice lecture. This whole series is fantastic.
In the last of two given exercises, there is an error. Instead of integrating from 0 to π we should integrate from 0 to 2π. This would give us the result 2π instead of π.
Please allow a minor correction towards the conclusion of the proof of Liouville's Theorem: you say "So RHS of the above inequality becomes arbitrarily large" but what you mean is "So *the denominator* of the RHS of the above inequality becomes arbitrarily large". Very nice lecture. This whole series is fantastic.
GOOD
In the last of two given exercises, there is an error. Instead of integrating from 0 to π we should integrate from 0 to 2π. This would give us the result 2π instead of π.
No the equation is correct as it is.[Hint: Use Liouville's theorem]
You are right @Anubhab!