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December 14th, 2017, 05:19 AM   #1
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Equation solution?

$\displaystyle e\frac{dz}{dt}=z(t+1) \; \; $ where $\displaystyle e-$ [euler constant]
or equation can be written as: $\displaystyle \; eZ'_{t} - Z_{t+1}=0$

Last edited by idontknow; December 14th, 2017 at 05:24 AM.
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December 14th, 2017, 06:16 AM   #2
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$z = ce^t$ for any constant $c$.
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January 13th, 2018, 03:05 AM   #3
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Quote:
Originally Posted by SDK View Post
$z = ce^t$ for any constant $c$.

Hmmm. If $z= ce^t$ then $z'= ce^t$ and $z(t+ 1)= ce^{t+ 1}= ce e^t$ so the equation becomes $ce^t= ce e^t$ so $e= 1$. No, that's not true.

This is a "differential-difference" equation. Typically, one can assign values to an interval of length 1, such as t = 0 to 1, and then solve for other values. For example, if $z(t) = \sin(t)$ for t = 0 to t = 1, then $z(t+ 1)= e \cos(t)$ so that $z(t)= e \cos(t- 1)$ for t= 1 to 2, etc. If, instead, $z(t)= t^2$ for t = 0 to 1, then $z(t+ 1)= 2e t$ so that $z(t)= 2e(t- 1)$ for t = 1 to 2, etc.

Last edited by skipjack; January 18th, 2018 at 07:55 PM.
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January 18th, 2018, 07:34 PM   #4
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Quote:
Originally Posted by Country Boy View Post
Hmmm. If $z= ce^t$ then $z'= ce^t$ and $z(t+ 1)= ce^{t+ 1}= ce e^t$ so the equation becomes $ce^t= ce e^t$ so $e= 1$. No, that's not true.

This is a "differential-difference" equation. Typically, one can assign values to an interval of length 1, such as t = 0 to 1, and then solve for other values. For example, if $z(t) = \sin(t)$ for t = 0 to t = 1, then $z(t+ 1)= e \cos(t)$ so that $z(t)= e \cos(t- 1)$ for t= 1 to 2, etc. If, instead, $z(t)= t^2$ for t = 0 to 1, then $z(t+ 1)= 2e t$ so that $z(t)= 2e(t- 1)$ for t = 1 to 2, etc.
I'm not sure what you mean by "differential-difference" equation. The usual name for this type of equation is a delay differential equation. This is because the function depends on its derivative at a different moment in time given in this case by a linear lag time.

In any case, you are correct that $z = ce^t$ yields $z' = ce^t$ and also $z(t+1) = ce\cdot e^t$. However, the delay equation is $ez' = z(t+1)$ so we note that $ez' = e\cdot ce^t = ce^{t+1} = z(t+1)$ so it satisfies the equation.

Last edited by skipjack; January 18th, 2018 at 07:55 PM.
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