i tried to prove this equation in differential way..

[transgalactic]

[D H]

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]

[transgalactic]

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??

[D H]

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]?

[transgalactic]

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

??

[D H]

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.

[transgalactic]

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

[D H]

No.

 

Why don't you at least try doing what I suggested in posts #2 and 4?

[transgalactic]

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}

[D H]

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?

[transgalactic]

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 merged

even if we will find some pattern that matches the n+1 case

its only a pattern

its not a mathematical proof

[D H]

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.

[transgalactic]

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

[D H]

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 f_n+1(x).

[transgalactic]

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]

[D H]

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.

[transgalactic]

& [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.

[D H]

You made a mistake in deriving the third derivative. That led you to the wrong pattern in the n^th derivative of f_n+1(x).

 

Seeing tentative patterns and then proving them to be true is how a lot of the recursive-type relations were developed.

[transgalactic]

where is the mistake in the third derivative

??

[D H]

What is [math]\frac{d}{dx}\left(\frac{d}{dx}f_n(x)\right)[/math]?

[transgalactic]

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?

[D H]

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.

[transgalactic]

so if i wrote it in a correct way

where is the mistake then?

[D H]

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]

[transgalactic]

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]