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May 10th, 2014, 10:44 AM   #1
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Solutions for diff equation in form of series

Given the following diff equation: , being D = d/dx, the "implicit" solution is: , so, for "explicit" the solution is necessary to expand the fraction 1/(1-D) by identity: , but this infinity series is true only for |x|<1, for |x|>1 is necessary utilize the following series: . Happens that D isn't a number for I say that |D| is less or greater than 1. So, how can I interprate this form of solution correctly?

PS, this ideia from Operational Calculus:
https://es.wikipedia.org/wiki/Transf...hist.C3.B3rica
Heaviside’s Operational Calculus | Rip's Applied Mathematics Blog
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May 10th, 2014, 11:32 AM   #2
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D is an operator. You can't leave it with nothing to operate on.

If I were solving that with a series, I'd say $Dy(x) = y(x) - f(x)$ and then crreate a Taylor series. Of course, this relies on having an initial condition.

Then $y(x) = y(0) + Dy(0)x + \cdots = y(0) + (y(0) - f(0)x + \cdots$

We can get $D^2y(x)$ and higher orders by differentiating the equation for $Dy(x)$.

Edit: having browsed the blog, I think I've seen something similar before. I would rather understand why it works than the approach seen there, where it seems to for my functions, so lets say it's right.

Last edited by v8archie; May 10th, 2014 at 11:49 AM.
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May 10th, 2014, 12:59 PM   #3
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Quote:
Originally Posted by v8archie View Post
D is an operator. You can't leave it with nothing to operate on.

If I were solving that with a series, I'd say $Dy(x) = y(x) - f(x)$ and then crreate a Taylor series. Of course, this relies on having an initial condition.

Then $y(x) = y(0) + Dy(0)x + \cdots = y(0) + (y(0) - f(0)x + \cdots$

We can get $D^2y(x)$ and higher orders by differentiating the equation for $Dy(x)$.

Edit: having browsed the blog, I think I've seen something similar before. I would rather understand why it works than the approach seen there, where it seems to for my functions, so lets say it's right.
In the first pages (13-17) has more about this...
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May 10th, 2014, 04:59 PM   #4
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Shouldn't you write
$$(1 -D)y = f(x) \Longrightarrow y = -\frac{1}{D-1}f(x)$$
and then proceed as in the blog?
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May 11th, 2014, 04:54 AM   #5
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Quote:
Originally Posted by v8archie View Post
Shouldn't you write
$$(1 -D)y = f(x) \Longrightarrow y = -\frac{1}{D-1}f(x)$$
and then proceed as in the blog?
Now the ideia is expand the fraction $\displaystyle \frac{1}{1-D}$ with $\displaystyle \frac{1}{1-D}=\sum_{0}^{\infty}D^n \Delta n$ or with $\displaystyle \frac{1}{1-D} = -\sum_{-\infty}^{-1} D^n \Delta n$...

PS: look this article from page 13 until 17.
http://www.latp.univ-mrs.fr/~chaabi/...20,2006%29.pdf
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