transgalactic Posted January 26, 2009 Share Posted January 26, 2009 Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 Please learn to use this forum's [math] capability. That way I can quote your work and help you along the way. You didn't get the right answer because you solved the wrong problem. The problem at hand is to show [math]\frac{d^n}{dx^n}\left(x^{n-1}e^{1/x}\right) = (-1)^n\frac{e^{1/x}}{x^{n+1}}[/math] Define some terms to make this a bit easier to understand: Let [math]f_n(x) \equiv x^{n-1}e^{1/x}[/math] [math]g_n(x) \equiv (-1)^n\frac{e^{1/x}}{x^{n+1}}[/math] With these definitions, the problem is [math]\frac{d^n}{dx^n}f_n(x) = g_n(x)[/math] Hint: Forget about the right-hand side for a bit. First find a way to express [math]\frac{d^{n+1}}{dx^{n+1}}f_{n+1}(x)[/math]. Second hint: [math]f_{n+1}(x) = xf_n(x)[/math] Link to comment Share on other sites More sharing options...
transgalactic Posted January 27, 2009 Author Share Posted January 27, 2009 i did like you said http://img517.imageshack.us/img517/7919/66574613oo1.gif still i am not getting the k+1 expression i used program called math type how to covert it into [math] code?? Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 Re LaTeX: See this Quick LaTeX Tutorial thread. Apparently MathType can export LaTex. See http://www.dessci.com/en/support/mathtype/workswith/t/latexee.htm. You didn't do anything close to what I suggested you do. I said to forget about the right-hand side for a bit. The problem is that you are ignoring that the functions being differentiated are themselves a series. Hint: Given only that [math]f_{n+1}(x) = xf_n(x)[/math], what is [math]\frac{d^{\,k}}{dx^k}f_{n+1}(x)[/math]? Link to comment Share on other sites More sharing options...
transgalactic Posted January 27, 2009 Author Share Posted January 27, 2009 you showed the my function on the left side is not f(x) but x*f(x) so i need do a derivative on x*f(x) that what you meant that what i did i my last document ?? Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 You are ignoring that the functions being differentiated on the left-hand side are different for each n. In particular, this statement is wrong: "so when we do another derivative we should get the n=k+1" Please type your posts! I had to type your response by hand because there is no way to quote or cut-and-paste text from an image. Link to comment Share on other sites More sharing options...
transgalactic Posted January 27, 2009 Author Share Posted January 27, 2009 (edited) so my mistake is by defferentiating the "k" expression instead of the k+1 expression ?? what is the k+1 expression?? if i will input n=k+1 into the given expression i would get [math] {{d^{k + 1} } \over {dx^{k + 1} }}(x^k e^{{1 \over x}} ) = ( - 1)^{k + 1} {{e^{{1 \over x}} } \over {x^{k + 2} }} [/math] obviously that the expression i should get as a result so what expression should i differentiate Edited January 27, 2009 by transgalactic Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 No. Why don't you at least try doing what I suggested in posts #2 and 4? Link to comment Share on other sites More sharing options...
transgalactic Posted January 27, 2009 Author Share Posted January 27, 2009 ok [math]f_n (x) \equiv x^{n - 1} e^{1/x} [/math] \cr [math]g_n (x) \equiv ( - 1)^n {{e^{1/x} } \over {x^{n + 1} }}[/math] \cr [math]{{d^n } \over {dx^n }}f_n (x) = g_n (x)[/math] \cr our{\rm{ n + 1 expression is:}} \cr [math]{{d^{n + 1} } \over {dx^{n + 1} }}f_{n + 1} (x)[/math] \cr [math]f_{n + 1} (x) \equiv x^n e^{1/x} = xx^{n - 1} e^{1/x} = x*f(x)[/math] \cr [math]{{d^{n + 1} } \over {dx^{n + 1} }}x*f(x)[/math] \cr {\rm{i dont know what to do now??}} \cr} Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 What you do now is find a way to express [math]\frac{d^{n+1}}{dx^{n+1}}f_{n+1}(x)[/math] in terms of [math]f_n(x)[/math] Hint: [math]f_{n+1}(x) = x f_n(x)[/math] Differentiating, [math]\frac{d}{dx}f_{n+1}(x) = x\frac{d}{dx}f_n(x) + f_n(x)[/math] What happens when you differentiate again, and again, and ... n+1 times? Can you see a pattern arising? Link to comment Share on other sites More sharing options...
transgalactic Posted January 27, 2009 Author Share Posted January 27, 2009 i see that after the n+1 differentiations there will be x*f'(x) in the start i cant see how would look he rest of the expression? Merged post follows: Consecutive posts mergedeven if we will find some pattern that matches the n+1 case its only a pattern its not a mathematical proof Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 Try it one step at a time. What is the second derivative, the third, ... You should see a pattern. Then prove that the pattern is general. Link to comment Share on other sites More sharing options...
transgalactic Posted January 27, 2009 Author Share Posted January 27, 2009 after i have done the first derivative where i substitute the [math] f_{n+1}(x) = x f_n(x) [/math] for the next derivative Link to comment Share on other sites More sharing options...
D H Posted January 27, 2009 Share Posted January 27, 2009 I already showed you the first derivative, a simple application of the chain rule: [math]\frac{d}{dx}f_{n+1}(x) = x\frac{d}{dx}f_n(x) + f_n(x)[/math] Differentiate the above to get the second derivative of fn+1(x). Link to comment Share on other sites More sharing options...
transgalactic Posted January 28, 2009 Author Share Posted January 28, 2009 is this the correct second derivative? [math]{d \over {dx}}f_{n + 2} (x) = x{d \over {dx}}f_{n + 1} (x) + f_{n + 1} (x) = x(f_n (x) + xf'_n (x)) + xf_n (x)[/math] Link to comment Share on other sites More sharing options...
D H Posted January 28, 2009 Share Posted January 28, 2009 No. Second derivative means [math]\frac {d^2}{dx^2}[/math]. The equation in post number 14 is the first derivative, so differentiating again will yield the second derivative. Link to comment Share on other sites More sharing options...
transgalactic Posted January 28, 2009 Author Share Posted January 28, 2009 (edited) & [math]{d \over {dx}}f_{n + 1} (x) = x{d \over {dx}}f_n (x) + f_n (x)[/math] \cr [math] {{d^2 } \over {dx^2 }}f_{n + 1} (x) = {d \over {dx}}f_n (x) + x{{d^2 } \over {dx^2 }}f_n (x) + {d \over {dx}}f_n (x) = x{{d^2 } \over {dx^2 }}f_n (x) + 2{d \over {dx}}f_n (x) [/math] [math] \begin{array}{l} \frac{{d^3 }}{{dx^3 }}f_{n + 1} (x) = \frac{d}{{dx}}[x\frac{{d^2 }}{{dx^2 }}f_n (x) + 2\frac{d}{{dx}}f_n (x)] = x\frac{{d^3 }}{{dx^3 }}f_n (x) + 3\frac{d}{{dx}}f_n (x) \\ \\ \frac{{d^n }}{{dx^n }}f_{n + 1} (x) = x\frac{{d^n }}{{dx^n }}f_n (x) + n\frac{d}{{dx}}f_n (x) \\ \end{array} [/math] am i correct? Edited January 28, 2009 by transgalactic Consecutive posts merged. Link to comment Share on other sites More sharing options...
D H Posted January 28, 2009 Share Posted January 28, 2009 You made a mistake in deriving the third derivative. That led you to the wrong pattern in the nth derivative of fn+1(x). Seeing tentative patterns and then proving them to be true is how a lot of the recursive-type relations were developed. Link to comment Share on other sites More sharing options...
transgalactic Posted January 28, 2009 Author Share Posted January 28, 2009 where is the mistake in the third derivative ?? Link to comment Share on other sites More sharing options...
D H Posted January 28, 2009 Share Posted January 28, 2009 What is [math]\frac{d}{dx}\left(\frac{d}{dx}f_n(x)\right)[/math]? Link to comment Share on other sites More sharing options...
transgalactic Posted January 28, 2009 Author Share Posted January 28, 2009 its the second derivative obviously its not [math] \frac{{d^2 }}{{dx^2 }}f_{n } (x) [/math] if you are pointing at this place. so what is it? Link to comment Share on other sites More sharing options...
D H Posted January 28, 2009 Share Posted January 28, 2009 What is [math]\frac{d}{dx}\left(\frac{d}{dx}f_n(x)\right)[/math']? its the second derivative obviously its not [math] \frac{{d^2 }}{{dx^2 }}f_{n } (x) [/math] if you are pointing at this place. so what is it? Good lord. Obviously that is exactly what it is. Link to comment Share on other sites More sharing options...
transgalactic Posted January 28, 2009 Author Share Posted January 28, 2009 so if i wrote it in a correct way where is the mistake then? Link to comment Share on other sites More sharing options...
D H Posted January 28, 2009 Share Posted January 28, 2009 Your mistake is here: [math]\frac{{d^3 }}{{dx^3 }}f_{n + 1} (x) = \frac{d}{{dx}}[x\frac{{d^2 }}{{dx^2 }}f_n (x) + 2\frac{d}{{dx}}f_n (x)] = x\frac{{d^3 }}{{dx^3 }}f_n (x) + 3\frac{d}{{dx}}f_n (x)[/math] which led you to make this mistake here: [math]\frac{{d^n }}{{dx^n }}f_{n + 1} (x) = x\frac{{d^n }}{{dx^n }}f_n (x) + n\frac{d}{{dx}}f_n (x)[/math] Link to comment Share on other sites More sharing options...
transgalactic Posted January 29, 2009 Author Share Posted January 29, 2009 is this the correct third derivative [math]\frac{{d^3 }}{{dx^3 }}f_{n + 1} (x) = \frac{d}{{dx}}[x\frac{{d^2 }}{{dx^2 }}f_n (x) + 2\frac{d}{{dx}}f_n (x)] = \frac{{d^2 }}{{dx^2 }}f_n (x) + x\frac{{d^3 }}{{dx^3 }}f_n (x) + 2\frac{{d^2 }}{{dx^2 }}f_n (x) [/math] Link to comment Share on other sites More sharing options...
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