2011-03-16, 09:55 AM
What you're saying is that the rate of change of the bacteria is proportional to its current population (that's how you get exponential growth). Therefore, you claim:
dx/dt = kx
dx/dt -kx = 0
(dx/dt)(e^(-kt)) +x*(-k*e^(-kt)) = 0
∫ (dx/dt)(e^(-kt)) +x*(-k*e^(-kt)) dt = C
x*(e^(-kt)) = C
x = C*e^(kt)
x = C*(e^k)^t
Since you specified that the bacteria is doubling in population every unit of time, e^k = 2. Therefore, you have:
x = C*2^t
You also specified that the initial condition that there is one unit of bacterial population. Therefore:
1 = C*2^(0) = C
C = 1
Thus, your model states:
x = 1*2^t
If your initial condition is 0, then:
x = C*2^t
0 = C*2^(0) = C
C = 0
x = 0*2^t = 0
Obviously, in an initial population of 0, there will be nothing going on, because doubling 0 will result in 0 and so on. You haven't proven anything at all regarding 0^0.
dx/dt = kx
dx/dt -kx = 0
(dx/dt)(e^(-kt)) +x*(-k*e^(-kt)) = 0
∫ (dx/dt)(e^(-kt)) +x*(-k*e^(-kt)) dt = C
x*(e^(-kt)) = C
x = C*e^(kt)
x = C*(e^k)^t
Since you specified that the bacteria is doubling in population every unit of time, e^k = 2. Therefore, you have:
x = C*2^t
You also specified that the initial condition that there is one unit of bacterial population. Therefore:
1 = C*2^(0) = C
C = 1
Thus, your model states:
x = 1*2^t
If your initial condition is 0, then:
x = C*2^t
0 = C*2^(0) = C
C = 0
x = 0*2^t = 0
Obviously, in an initial population of 0, there will be nothing going on, because doubling 0 will result in 0 and so on. You haven't proven anything at all regarding 0^0.
shouri Wrote:The 0^0 argument is like the why .999... = 1 argument (sort of). In that people tend to argue craploads for multiple sides and no one like to agree with the other side. 0^x = 0 for any nonzero, thus people say that 0 ^ 0 should be 0. But x ^ 0 = 1 for any non zero x, thus people say that 0^0 should be 1.Actually, the .999... = 1 argument is completely different from the 0^0 argument. By using summations, fractions, and various other methods, there actually exists valid proof that .999... = 1. For 0^0=0, all arguments supporting 0^0=0 are incomplete, including the 0^x argument.
