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 May 10th, 2014, 10:44 AM #1 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 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 May 10th, 2014, 11:32 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra 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. May 10th, 2014, 12:59 PM   #3
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 Originally Posted by v8archie 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. May 10th, 2014, 04:59 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2666 Math Focus: Mainly analysis and algebra Shouldn't you write $$(1 -D)y = f(x) \Longrightarrow y = -\frac{1}{D-1}f(x)$$ and then proceed as in the blog? May 11th, 2014, 04:54 AM   #5
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 Originally Posted by v8archie 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 Tags diff, equation, form, series, solutions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post solrob Applied Math 1 October 14th, 2013 11:47 AM gen_shao Calculus 3 July 9th, 2013 12:52 PM Grayham1990 Complex Analysis 2 March 24th, 2012 05:57 PM zgonda Calculus 1 November 27th, 2010 04:23 PM cmmcnamara Differential Equations 0 February 25th, 2010 08:41 AM

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