The Grainger College of Engineering
Physics Van

Decimal Expansion of E

Most recent answer: 11/25/2014

Q:
If a decay rate constant is 0.05, such that e^-0.05t gives the amount remaining after time t, why can it not be said that 0.95^t (1-0.05 = 0.95 being that amount not decaying)also gives the amount remaining? e^0.05 = 0.951 229 being slightly greater than 0.95.
- Chris Oldman (age 65)
Cheltenham, Gloucestershire, England
A:

Hello Chris,

The complete Taylor series expansion of ex is:

1 + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}+ \cdots = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots\! = \sum_{n=0}^\infty \frac{x^n}{n!}.

Your value of 1 - .05 = .95 is only the first two terms in the series.    If  you add a few more terms you can get as close as you want to the true value.  For example adding the next two terms gets you .9512291667  .

LeeH

p.s. Your expression e^(0.05t) needs some unit, e.g. 0.05t/second or 0.05t/year. mw


(published on 11/25/2014)

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