Yes. Combine some terms and you're there.

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]

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?

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]

No! You are just guessing, and not very well.

 

Try taking the fourth derivative.

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 January 29, 200916 yr by transgalactic

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]

if you say that my 4th derivative is correct

then i think that the pattern is

 

[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^k }}{{dx^k }}f_n (x)[/math]

 

 

what to do now??

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]

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?

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.

but how should i go from the n=k case to the n=k+1 case

by derivation of n=k expression??

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.

so you are saying that because our pattern is true

the n=k+1 case is true then n=k is true

 

but our pattern is only a presumption

we cant say that its a fact

I asked you to prove it. So do that: Prove it.

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 January 30, 200916 yr by transgalactic
Consecutive posts merged.

I've already given more help here than I feel comfortable giving, transgalactic. The next step is yours.

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]

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 g_n 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 -e^1/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 e^1/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.

 

(e^1/x/xⁿ⁺¹)' = (e^1/x)' (1/xⁿ⁺¹) + (e^1/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 (e^1/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 g_n(x). Which you did not do right in the preceding couple of posts. So please try again to find (e^1/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 nx^n-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 January 31, 200916 yr by Martin

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?

That is still not right, transgalactic. What is the quotient rule for differentiation?

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??

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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]

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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

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[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 February 1, 200916 yr by transgalactic
Consecutive posts merged.