[MUSIC] Our next and final goal, is to define the binomial theorem for arbitrary exponents. Suppose we have an expression 1. + q and want to raise it to some rational power, alpha. Alpha is a rational number. Not necessarily integer, so how to define this? Well analysis suggest us that, we need to define it using the exponential function. So we set by definition? This to be equal to the exp ( alpha.logarithm(1 + Q)). This is the definition of an arbitrary, rational power. An exercise which is not completely obvious. Show that if alpha is integer or if alpha is in here. This definition gives us the same thing as the usual definition. So this is (1 + q) to the power of n which is (1 + q)...(1+q) n times for n positive Or. 1 / 1 + q N times for n negative. This is not completely obvious. To show this, you need to establish some properties of the exponential function. Now comes the binomial theorem for arbitrary exponentials. So let us compute this as a formal power series. Theorem. Let alpha be rational. In this case, (1 + q) to the power alpha = as a formal power series to 1 + alpha.q. plus (alpha(Alpha- 1)/2.q squared + alpha(alpha -1)(alpha- 2)/3 factorial or let me put a factorial here q to the 3 +... And this can be written as the sum as a usual binomial for, but this sum will be infinite for all n greater than or equal to =. Something called (alpha choose n) q to the power n where this thing, the binomial coefficient for a probably non-integer alpha. (alpha chose n) = alpha(alpha- 1)..., and we need to take n factors here so the last one will be (alpha- n + 1) / n factorial. So if alpha is integer. here we recover the usual definition of the binomial coefficient. And if not, we get something new. Let us prove the binomial theorem. Of course we need to start with the definition of the left hand side, (1 + q ) to the power alpha. (1+q) to the power alpha is. Exponent of alpha logarithm (1+q). And now, let us compute the logarithmic derivative of this expression. So we take the logarithm of. The (exponential(alpha, logarithm(1+q))) And take a 0, so the logarithmic derivative of this expression is. As it was stated in the exercise, the (exp (alpha logarithm (1+q))) prime / the exp(alpha logarithm(1 + q)). So this is just ((1 + q) to the power alpha)) prime / (1 + q) to the power alpha. On the other hand, this expression, under the duration sign, is just the logarithm of (1 + q) alpha. Because the logarithm and the exponential are mutually inverse. So, on the other hand. The left-hand side, well this expression is, it is equal to (alpha logarithm (1 + q)) prime. Which is just alpha divided by 1 + q. As simple as that. So,we see that ((1 + q) to the power alpha) / (1 + q) to the power alpha = alpha/ 1 + q. From this relation we can compute (1 + q) to the power alpha as a formal power series. Computing its coefficients one by one. So suppose that (1 + q) to the power alpha is. Well this is a formal series starting with 1, because the value in q = 0 is obviously 1. So this is 1 + a1q + a2q squared +... And ((1 + q) to the power of alpha) prime is the derivative of this series, so this is a1 +2a2q +3a3q squared +... And we see that we get the relation. Alpha(1 +a1 q+ a2 qsquared+...) = (a1 + 2a2q + 3a3q squared+...)(1+q). So this is exactly this relation. So from this, we find the recurrent relation on the coefficient of this series, into this follows Kak + (K- 1) ak-1 = alpha.ak-1. So, knowing ak, ak-1, we can compute the next one. Let's do that, a. Okay is. Alpha- (k- 1) / k times ak- 1. And ak- 1 can be expressed by ak- 2 by using the same formula. So this = (alpha- (K- 1))( alpha- (K- 2) /k (k- 1) ak -2. Etc, etc. And finally, we get the expression (alpha- (k- 1)) (alpha- (k- 2)...(alpha- 1)a/ k factorial. And this is exactly this expression. (alpha choose K). The binomial theorem for arbitrary exponents is proved. [MUSIC]