eaxdx

Back to Contents

If I have something in the amount P0 and it increases by a fraction x in one standard interval, then after one such interval elapses I will have the amount P1 = P0(1+x), which becomes the starting amount for the next interval. After the elapse of a number of such intervals represented by the letter a, I will have Pa = P0(1+x)a. I have tacitly assumed that I calculate the increase in the amount only at the end of each standard interval. Suppose, instead, that I calculate the increased amount at the end of each of n evenly spaced instants across each standard interval: I then get Pa' = P0(1+x/n)an. Let's focus our attention entirely upon the multiplier of P0 and ask what happens as n tends toward infinity; that is, what limit do we reach, if any, as n increases without bound?

We have

(Eq'n 1)

We can use the binomial theorem to open out the argument to get

(Eq'n 2)

But now the limit process applies only to the coefficient in that expression, so we have in consequence

(Eq'n 3)

and, therefore,

(Eq'n 4)

With that expression in mind we can explore the properties of F(x). We have immediately that F(0) = 1 (because x0 = 1 for all values of x, however small, we can take the limit as x tends to zero and define 00 = 1, just as we defined 0! = 1: for all other values of k the terms zero out).

Next we want to calculate F(x+y). We have

(Eq'n 5)

Expand the n-th term, (x+y)n/n!, by way of the binomial theorem and get

(Eq'n 6)

If we sum those subseries over all values of n, we get the same infinite series that we would have obtained from multiplying together the series and , so we infer that

(Eq'n 7)

Now we have F(0) = 1 = F(x-x) = F(x)F(-x), which entails

(Eq'n 8)

But we also have

(Eq'n 9)

which alternates positive and negative terms. Do we truly have

(Eq'n 10)

as true to mathematics? If so, then the product of the two series must equal one, so let's multiply them together. We get

(Eq'n 11)

If we gather together all of the terms for which n+m = k, then we have the subseries

(Eq'n 12)

The coefficients of xk in that subseries comprise the k-th row of the alternating Pascal's triangle divided by k!, but we know that each of the rows of the alternating Pascal's triangle except the zeroth add up to zero, so we have

(Eq'n 13)

So now we know that the argument x of the function F(x) has the properties of an exponent, the power to which we must raise some base number in order to calculate the value of F(x). We can calculate the value of that base number from Equation 4 by letting x = 1, so we have

(Eq'n 14)

for which

(Eq'n 15)

If we take and integrate it term by term, we get

(Eq'n 16)

Because we can represent any constant as a sum of any other constants, such as C = C'+1/a, we have at last

(Eq'n 17)

Example Application: Diminution of Radioactivity;

An engineer has been asked to design a power system for a space probe that will be sent to a nearby star. The voyage will take 100,000 years: the probe must have enough power to maintain itself during the interstellar crossing and must then be able to generate substantially more power when it reaches its destination. The engineer has chosen to exploit the fact that Plutonium-239 decays into Uranium-235 by spitting out an alpha particle that carries 5.147 Mev of kinetic energy. She designs a system that exploits the energy coming from the decay of the Plutonium during the trans-stellar drift and then uses the accumulated Uranium in a fission reactor at the end of the voyage.

Among the facts that she has at her disposal we find that Pu-239 has a half-life of 24,100 years (a = 2.876 x 10-5 per year), that U-235 has a half-life of 710,000,000 years, and that one kilogram of pure Pu-239 produces energy at a rate of 1.8928 watts due to its decay into U-235. She calculates that over the span of 100,000 years the probe's original load of plutonium will have diminished to e-at = 0.056352425 of its original amount. Thus, at the end of the probe's voyage each kilogram of Plutonium will have become 56.35 grams of Plutonium, 927.86 grams of Uranium, 15.79 grams of Helium, and a minuscule amount of Uranium decay products.

Our engineer needs one other piece of information: she wants to know how much energy each kilogram of Plutonium will release over the course of the interstellar crossing, over the span of one hundred millenia. At the beginning of the voyage each kilogram releases energy at a rate that corresponds to P = 5.9732x107 Joule per year. After an elapse of t years that rate declines by a factor of e-at, so the engineer calculates the total energy output by integrating that declining rate over time; that is,

(Eq'n 18)

With that fact and others in mind, our engineer can proceed to design her probe.

afafafebebeb

Back to Contents