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 April 4th, 2017, 03:28 PM #1 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 Power Series When you take the derivative of this why does n increment by +1? What does equating const x^0 means? What does equating const x^n means? Differential Equations & Linear Algebra (9780136054252), Pg. 641, Ex. 1 :: Homework Help and Answers :: Slader April 4th, 2017, 03:41 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 1. It doesn't. As the term for which n = 0 is a constant, its derivative is zero and can be omitted. 2 and 3. Think about it. What is the power series for the function that is zero everywhere? April 4th, 2017, 05:33 PM   #3
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 Originally Posted by skipjack 2 and 3. Think about it. What is the power series for the function that is zero everywhere?
The power series is a base zero "a0" which equals 0. I still don't know, could you give me the reason why too? April 4th, 2017, 09:48 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 As $y\,^\prime - y = 0$, the power series for $y\,^\prime - y$ is also just 0. Hence the $n$th term of that power series is zero. April 5th, 2017, 05:33 AM #5 Member   Joined: Feb 2014 Posts: 91 Thanks: 1 That makes sense, it is homogeneous. I asked about equating const x^0 & x^n because there seems to be inconsistency in how to get the power series with the solutions I was provided. I'm really stuck on question 5. I'm confused because this guy did the problem and even got a 5 star rating yet he did not use the same method he used for the previous question number 1. This mainly starts around equating x^0 where he doesn't equate a0, He ignores it. I really wanted to figure this out on my own which is why I only for meaning of equating x^0 & x^n Differential Equations & Linear Algebra (9780136054252), Pg. 641, Ex. 5 :: Homework Help and Answers :: Slader The question I'm really stuck on. April 5th, 2017, 07:47 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 The solutions do not contain "equating const x^n". One solution contains "Equating coeff. of ($x^n$) terms to 0; the other contains "equating $x^n$ terms to 0". Those are slightly different wordings, but the method is essentially the same. Note that $x^0 = 1$. April 5th, 2017, 11:23 AM   #7
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 Originally Posted by skipjack The solutions do not contain "equating const x^n". One solution contains "Equating coeff. of ($x^n$) terms to 0; the other contains "equating $x^n$ terms to 0". Those are slightly different wordings, but the method is essentially the same. Note that $x^0 = 1$.
Thank you, I know what it is, I know how to do, now I just need to learn why so that I don't just do work that I don't understand why I am doing it. Tags power, series 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 Diehardwalnut Math Books 0 October 13th, 2014 06:47 PM m.nikolov Real Analysis 2 January 16th, 2013 02:10 PM g0bearmon Real Analysis 2 May 22nd, 2012 12:10 PM aaron-math Calculus 4 November 28th, 2011 10:53 AM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

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