00:02
Okay, so for a, we have g of s equals sum over s k times p k, which is equal to p0 plus s p1 plus s squared p2 plus s2k times pk plus da da da.
00:31
And now we're gonna take the derivative of g with respect to s, the kth derivative, which is 0 plus 0 plus k times k minus 1 times pk plus k plus 1 times k.
01:08
S p k plus 1 and now notice that this is k factorial so we're gonna divide both sides by k factorial and get dk g s which is equal to p k plus k plus 1 k k tatata 1 over k factorial times s pk plus 1 plus tatata.
01:45
So, we have 1 over k dk g of s.
01:54
S equals 0 is equal to pk.
01:58
Now, for b, we have g equals sum over sk times pk, which implies that the derivative of g with respect to s is equal to d s to k times pk and infinite and k times s k minus 1 times pk.
02:52
Now we're gonna name it 1, so this means that d g s of s equals 1 is equal to sum over k times pk, which is the expected value of x...