Integration adding an 'e' ??

[ae_jester]

Heres the best answer I got.

By the way this attachment is a word document NOT A ZIP FILE.

Just rename the file to myanswer.doc after u download it

Clint

[Helger]

Thanks for your answer jester. I did those funky equations with the formula editor that came with word 2000. In the menu 'insert' hit 'object...' and somewhere down the line in the dialogue that will pop up, you can find the MS formula editor.

Hmmm - sorry I'm still a bit unsure about why that e^-rt pops up there. (I know about what exponential functions will do.)
Speaking of the e: Is it something like lim(original funct's denominator) = e ? Is that it? It's simply the limes of the geometrical Reihe?(what's it called in English? row, series?)

Let's see if I got it:
e^-rt is merely what happens to the /(1+r)^t thing for sufficiently big t. So the function right of the integration sign is the same as sum above and the integration sign tells me that it all happens in continuous time.

Close enough?

thanks,

Helger

[noble]

e^-t represents exponential decay
e^t represents exponential growth

from the first equation....
if you analyze what is happening p is a constant and
U*C(t) represents an initial value that varies with time.

When t=0, e^0 = 1 so that term is simply U. As time increases
the value of e^-t decreases exponentially....exponential decay.

The first equation simply states that as time increases so does
the p*t term. Since, through simple integration, int(e^-pt)dt =
-(1/p)e^-pt, you can see that the denominator term gets large
so the total term with respect to time gets smaller and smaller.

The second equation is just the opposite, they're saying that
the function P(t) exponentially grows as time increases. When
t=0, P(t) = P(0), the initial value. And as time goes on, the e^pt
term gets larger and larger. Since you know, the initial value,
and time, you can figure out P(t) for any value of time.

I hope this helps.

[Helger]

Thanks. One is decaying exponentially and one is increasing exponentially. I can work with that.

Is there any mathematical reason or necessity why they chose 'e' in that equation? Or did they simply throw it in for convenience? I mean if I replace 'e' by any number >1, I would get growth/decay at a slower or faster rate.

I'm a bit thick with that one.

thx,

Helger

[noble]

e is a constant ~2.7 something.

it's an exponential constant that's related to logarithms

[Helger]

It's defined as limit of 1+1/n)^n for n--> infinity. Or as the limit of Sum(1/k!) over k = 0 to n for n-->infinity.

e is very convenient since the derivative of e^x = e^x so it will save you a lot of calculus.

Is that conveniency the only reason it is in there or is there any other? Could I replace it by a random number >1 here if I wanted to? Say replace it with 3?

Helger

[Helger]

Oh yes: and why would the r(or actually rho) from the original sum make it into the exponent of e together with t?

Before: Sum(U/(1+r)^t)

After U*e^(-r*t)

What?????

sigh,

Helger

[Helger]

I have updated the 'troublesome_equations.zip' file.

It is a Word-document - anyone interested just rename it.

[Helger]

Thanks for all your help again. I finally could contact the professor and he told me it is just ok to know what I know about the problem. No need to understand why the integration is done the way it's done.
The result for me is: first equation: example with discrete time. second equation: example with continuous time. no questioning about why or how. Wow - I'm glad I never considered joining the maths department.

Helger