Penrose is giving us the following definition for ez, where z is a complex number:
And says that we must use (1) to show the following:ea+b=eaeb
(2)
In order to do this, we have to apply (1) to both sides of (2) and show that they give equivalent quantities. Let's first consider ea+b, which according to (1) expands as follows:
This can be expanded as follows for the first few terms:
Which suggests a general pattern for the expansion up to the nth term, and continuing onwards (using Penrose's hint for the coefficient of the nth term of the expansion of (a+b)n:
Which in turn can be written more concisely as:
Now we look at the right hand side of (2). Using (1):
Which expands as:
Comparing (6) and (7) we see that the overall equality boils down to whether:
The easiest way I've found to show (8) is to re-examine the products on the left hand side in a different way:
This appears to be exactly the operation described by the right hand side of (8), which we derived from the expression for e(a+b). In this way we have demonstrated (2).
Hi! Your detaild derivation is helpful. But I don't understand the last argument you provide. About re-examining the products on the left hand side.
ReplyDeletem = 3
sum(a^i/i!)*sum(b^i/i!) =
(a^1/1! + a^2/2!)*(b^1/1! + b^2/2!) =
= a*b + (a*b^2)/2! + (a^2*b)/2! + (a^2*b^2)/(2!*2!)
which contains 2 extra terms a*b and (a^2*b^2)/(2!*2!)
Could you please explain why left_hand_side = right_hand_side?