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i tried to prove this equation in differential way..


transgalactic

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is that the correct pattern?

 

[math]

\frac{{d^3 }}{{dx^3 }}f_{n + 1} (x) = x\frac{{d^3 }}{{dx^3 }}f_n (x) + 3\frac{{d^2 }}{{dx^2 }}f_n (x) \\

[/math]

[math]

\frac{{d^n }}{{dx^n }}f_{n + 1} (x) = x\frac{{d^n }}{{dx^n }}f_n (x) + 3\frac{{d^{n - 1} }}{{dx^{n - 1} }}f_n (x) \\

[/math]

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Why three? There isn't a three in the first and second derivatives. Can you see a pattern?

 

[math]

\begin{aligned}

\frac {d}{dx}f_{n+1}(x) &= x\frac {d}{dx}f_{n}(x) + f_n(x) \\

\frac {d^2}{dx^2}f_{n+1}(x) &= x\frac {d^2}{dx^2}f_{n}(x) + 2\frac{d}{dx}f_{n}(x) \\

\frac {d^3}{dx^3}f_{n+1}(x) &= x\frac {d^3}{dx^3}f_{n}(x) + 3\frac {d^2}{dx^2}f_{n}(x)

\end{aligned}[/math]

 

 

In particular, what would you expect [math]\frac {d^{k+1}}{dx^{k+1}}f_{n+1}(x)[/math] to be? Can you prove that this is the case?

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its an obvious result form

[math]

\begin{aligned}

\frac {d}{dx}f_{n+1}(x) &= x\frac {d}{dx}f_{n}(x) + f_n(x) \\

\frac {d^2}{dx^2}f_{n+1}(x) &= x\frac {d^2}{dx^2}f_{n}(x) + 2\frac{d}{dx}f_{n}(x) \\

\frac {d^3}{dx^3}f_{n+1}(x) &= x\frac {d^3}{dx^3}f_{n}(x) + 3\frac {d^2}{dx^2}f_{n}(x)

\end{aligned}

[/math]

the 4th derivative is

[math]

\frac{{d^4 }}{{dx^4 }}f_{n + 1} (x) = \frac{{d^3 }}{{dx^3 }}f_n (x) + x\frac{{d^4 }}{{dx^4 }}f_n (x) + 3\frac{{d^3 }}{{dx^3 }}f_n (x) = x\frac{{d^4 }}{{dx^4 }}f_n (x) + 4\frac{{d^3 }}{{dx^3 }}f_n (x) \\

[/math]

 

so i cant see why my pattern is wrong??

Edited by transgalactic
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Look at what you wrote in post #29:

i think this is the k+1 case

correct?

[math]\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_{n + 1} (x) = x\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_n (x) + (k + 1)\frac{{d^2 }}{{dx^2 }}f_n (x)[/math]

 

The fourth derivative (k=3) should then be

 

[math]\frac{d^4}{dx^4}f_{n + 1} (x) = x\frac{d^4}{dx^4}f_n (x) + 4\frac{d^2 }{dx^2}f_n (x)[/math]

 

this disagrees with the known value of the fourth derivative, which you derived correctly in post #31:

 

[math]\frac{d^4}{dx^4}f_{n + 1} (x) = x\frac{d^4}{dx^4}f_n (x) + 4\frac{d^3 }{dx^3}f_n (x)[/math]

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Finally!

 

You need to prove this is indeed a correct expression (recursion again). Then use this expression to aid in the proof of the original relationship,

 

[math]\begin{aligned}

f_n(x) &\equiv x^{n-1}e^{1/x} \\

g_n(x) &\equiv (-1)^n\frac{e^{1/x}}{x^{n+1}} \\

\frac{d^n}{dx^n}f_n(x) &= g_n(x)\end{aligned}[/math]

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i cant understand how this pattern helps

i got my base case

[math]

\frac{d^n}{dx^n}f_n(x) = g_n(x)

[/math]

 

and i need to input n=k+1 in the base case

 

and prove that this k+1 equation is correct using the n=k equation

 

how this pattern helps me in doing that?

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The problem at hand involves proving that

 

[math]\frac{d^n}{dx^n}f_n(x) = g_n(x)[/math]

 

This is obviously proved recursively. Show that it is true for some simple case (i.e., n=1), assume it is true for some given n>1, and show that this assumption means the expression is true for the successor case (i.e., n+1),

 

[math]\frac{d^{n+1}}{dx^{n+1}}f_{n+1}(x) = g_{n+1}(x)[/math]

 

What does the relationship I finally pulled out of you in post #33 help you? Hint: Look at the left-hand side of the above and the left-hand side of the equation in post 33.

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Since

 

[math]\frac {d^{k+1}}{dx^{k+1}} f_{n+1}(x) =

\frac {d^{k+1}}{dx^{k+1}} f_n(x) + (k + 1)\frac {d^k}{dx^k} f_n (x)[/math]

 

for all k>=0, it is certainly valid for k=n so long as n>=0.

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i need to prove it by derivation of n=k case?


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[math]

 

{\rm{ }}\frac{{d^n }}{{dx^n }}f_n (x) = g_n (x) \\

[/math]

[math]

 

\frac{{d^{n + 1} }}{{dx^{n + 1} }}f_n (x) = \frac{{d^{} }}{{dx^{} }}g_n (x) = ( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} + (n + 1)x^n e^{1/x} {\rm{] }} \\

[/math]

 

what to next?

Edited by transgalactic
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ok but its a shot in the dark

i dont know what to do after the derivative.

from the given derivative of n i find the same derivative of n+1

but it differs the expression we need to get fron the n case

??

[math]

\frac{{d^{n + 1} }}{{dx^{n + 1} }}f_{n + 1} (x) = ( - 1)^{n + 1} {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 2} + (n + 2)x^{n + 1} e^{1/x} {\rm{]}}

[/math]

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Hello trans,

in case DH is busy with other stuff, I can try to help you. I think you made an error differentiating g_n

[math]

\frac{{d^{} }}{{dx^{} }}g_n (x) = ( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} + (n + 1)x^n e^{1/x} {\rm{] }} \\

[/math]

 

That's way wrong, as if you don't know how to use the product rule or quotient rule---either one works here let's focus on the product rule.

 

For starters let's recall what gn is. Remember what DH said in his first post on this thread:

... 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]

...

 

[math]g_n = (-1)^n e^{1/x} / x^{n+1}[/math]

 

do you know the product rule for derivatives?

 

(p(x)q(x))' = p'q + pq'

 

so set the (-1)n aside for a moment and just look at [math] e^{1/x} / x^{n+1}[/math]

 

that is the product of[math] e^{1/x}[/math] multiplied by [math] 1/ x^{n+1}[/math]

 

The derivative of the first is -e1/x/x2. do you have any trouble with that?

 

The derivative of the second factor is -(n+1)/xn+2

 

So when I apply the product rule, (p(x)q(x))' = p'q + pq', to what you have namely e1/x/xn+1

I get something totally different from you. It makes me worry that you don't know how to use the product rule for differentiating.

 

(e1/x/xn+1)' = (e1/x)' (1/xn+1) + (e1/x)(1/xn+1)'

 

I don't want to completely do it for you so why don't you finish the calculation. Put in what is (e1/x)' and put in what is (1/xn+1)'.

 

At this point I think the whole thing depends on your being able to correctly differentiate gn(x). Which you did not do right in the preceding couple of posts. So please try again to find (e1/x/xn+1)', like I suggested above.

 

======================

 

BTW are you OK with the prime notation? f' means the same as df/dx, just easier on the eyes.

And are you OK with negative exponents: 1/xn = x-n ?

And the derivative of xn is nxn-1

So naturally the derivative of x-n is -nx-n-1 = -n/x-(n+1)

in other words: (1/xn)' = -n/x-(n+1)

If you have any problem with these facts please say. They are basic to what I'm trying to tell you.

Edited by Martin
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you are correct i forgot the denominator part

 

[math]

 

\frac{{d^{} }}{{dx^{} }}g_n (x) = \frac{{( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} + (n + 1)x^n e^{1/x} {\rm{]}}}}{{x^{2n + 2} }}{\rm{ }}

 

[/math]

 

what to do now??

how to get to the n+1 expression?

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sorry forgot the minus

 

[math]

\frac{{d^{} }}{{dx^{} }}g_n (x) = \frac{{( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} - (n + 1)x^n e^{1/x} {\rm{]}}}}{{x^{2n + 2} }}{\rm{ }}

[/math]

 

what to do next??


Merged post follows:

Consecutive posts merged

we assume that f_n correct so the f_n+1 is correct too(except the derivative part)

 

but when we take deriavative n+1 we get a different equations the we supposed to get

??

[math]

 

\frac{{d^n }}{{dx^n }}f_{n + 1} (x) = ( - 1)^{n + 1} \frac{{e^{1/x} }}{{x^{n + 2} }}

[/math]

[math]

\frac{{d^{n + 1} }}{{dx^{n + 1} }}f_{n + 1} (x) = ( - 1)^{n + 1} \frac{{ - \frac{1}{{x^2 }}e^{1/x} x^{n + 2} - (n + 2)x^{n + 1} e^{1/x} }}{{x^{2n + 4} }} [/math]


Merged post follows:

Consecutive posts merged

how to interpret this prove into d/dx symbols??

 

 

Note that the commutator

[math]

[D,x]=Dx-xD=1

[/math]

and so

[math]

[D^n,x]=D^{n-1}[D,x]+D^{n-2}[D,x]D+\cdots+[D,x]D^{n-1}=nD^{n-1}.

[/math]

Thus we have


Merged post follows:

Consecutive posts merged

[math]

D^{n+1}x^n\exp(x^{-1})

&=[D^{n+1},x]x^{n-1}\exp(x^{-1})+xD^{n+1}x^{n-1}\exp(x^{-1})\\

&=(n+1)D^nx^{n-1}\exp(x^{-1})+xDD^nx^{n-1}\exp(x^{-1})\\

&=(-1)^n(n+1)x^{-n-1}\exp(x^{-1})+xD(-1)^nx^{-n-1}\exp(x^{-1})

 

[/math]

by the inductive hypothesis. Now do the differentiation and you get

[math]D^{n + 1} x^n \exp (x^{ - 1} ) = ( - 1)^n (n + 1)x^{ - n - 1} \exp (x^{ - 1} )\quad + ( - 1)^n x(( - n - 1)x^{ - n - 2} \exp (x^{ - 1} ) - x^{ - n - 3} \exp (x^{ - 1} )) = ( - 1)^{n + 1} x^{ - n - 2} \exp (x^{ - 1} )[/math] \\

\end{array}

 

 

which finishes the induction step.

Edited by transgalactic
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