Can Mathematical Induction Verify This Complex Differential Equation?

[transgalactic]

i need to prove that this equation is true
$\frac{d^n}{dx^n}\left(x^{n-1}e^{1/x}\right) = (-1)^n\frac{e^{1/x}}{x^{n+1}}$

i tried by induction
$f_n(x) \equiv x^{n-1}e^{1/x}$

$g_n(x) \equiv (-1)^n\frac{e^{1/x}}{x^{n+1}}$
this is the pattern if i keep differentiating the base case.
$\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_{n + 1} (x) = x\frac{{d^{k + 1} }}{{dx^{k + 1} }}f_n (x) + (k + 1)\frac{{d^k }}{{dx^k }}f_n (x)$

then i tried to differentiate the "n" equation inorder to get to the n+1 equation
$\frac{{d^{n + 1} }}{{dx^{n + 1} }}f_n (x) = \frac{{d^{} }}{{dx^{} }}g_n (x) = ( - 1)^n {\rm{[}}e^{1/x} ( - \frac{1}{{x^2 }}{\rm{)}}x^{n + 1} + (n + 1)x^n e^{1/x} {\rm{] }} \\$

but it didnt work
??

 

[Mark44]

Base case (n = 1); d/dx $e^{1/x} = - e^{1/x}/x^2$
So the equation holds for n = 1.

Induction step (n = k)
Asssume that $d^k/dx^k(x^{k -1}e^{1/x}) = (-1)^k e^{1/x}/x^{k + 1}$

Now show that the equation holds for n = k + 1. Note that the (k + 1)st derivative is the derivative of the kth derivative.

 

[transgalactic]

i did that in the first post
it doesn't give me the n+1 expression as a result
??

 

[Mark44]

Where did you find this equation?

After some thought, it doesn't seem to be true. I showed that when n = 1 it's true, namely that
$$d/dx(x^0 e^{1/x}) = d/dx(e^{1/x}) = x^{-2}e^{1/x}$$
But when n = 2, you have
$$d^2/dx^2(x e^{1/x}) = d/dx (x^{-2}e^{1/x})$$
$$= e^{1/x}(1/x^4 + 2/x^3)$$
and this doesn't fit the pattern.

Things will only get worse for higher derivatives, since you'll get an additional term for each additional derivative.

 

[marcus]

You have made a mistake taking the derivative. Incorrect application of the product rule or the quotient rule.

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

Set the (-1)ⁿ aside and try again to take the derivative of $\frac{e^{1/x}}{x^{n+1}}$

This is the product of two pieces $e^{1/x}$ and $\frac{1}{x^{n+1}}$

The derivative of the first piece is $-\frac{e^{1/x}}{x^{2}}$
The derivative of the second piece is $-(n+1)\frac{1}{x^{n+2}}$

How about you apply the product rule and tell us what the derivative of g_n is?

If you can just get the derivative of e^1/x/xⁿ⁺¹ right, then I think the rest will go OK.
BTW the quotient rule is just a poorly disguised version of the product rule. You really only need one rule and the product rule will work in either case. Correctly applied they both give the same answer. You must learn at least to apply the product rule correctly

 

[transgalactic]

is this correct??
what to do next??
$g_n '(x) = ( - 1)^n \frac{{ - \frac{1}{{x^2 }}e^{1/x} x^{n + 1} - (n + 1)x^n e^{1/x} }}{{x^{2n + 2} }}$

 

[marcus]

is this correct??
what to do next??
$g_n '(x) = ( - 1)^n \frac{{ - \frac{1}{{x^2 }}e^{1/x} x^{n + 1} - (n + 1)x^n e^{1/x} }}{{x^{2n + 2} }}$

well, let's see. I will apply the product rule and see if what you have is right.

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

((e^1/x)( 1/xⁿ⁺¹))' = (-e^1/x/x²) (1/xⁿ⁺¹) + (e^1/x)(-(n+1)/xⁿ⁺²)

= -e^1/x/xⁿ⁺³ - (n+1) e^1/x/xⁿ⁺²

= -(n + 1 + 1/x) e^1/x/xⁿ⁺²

You can put back in the factor of (-1)ⁿ.

Now is this what you got? I can't read your "itex" text, it is too small, please use plain "tex". I will copy your answer but with "tex":
==quote==
is this correct??

$$g_n '(x) = ( - 1)^n \frac{{ - \frac{1}{{x^2 }}e^{1/x} x^{n + 1} - (n + 1)x^n e^{1/x} }}{{x^{2n + 2} }}$$
==endquote==

Yes! It looks right! You just didn't simplify it, by canceling stuff.

OK now we know that (g_n(x))' = (n + 1 + 1/x) g_n+1(x)

Make sure you understand that much.

 

[marcus]

OK now we know that (g_n(x))' = (n + 1 + 1/x) g_n+1(x).

From here it should be a straight shot.
For convenience let's write D for d/dx.
Now just successively differentiate f_n+1

Df_n+1 = f_n + xDf_n
D²f_n+1 = 2Df_n + xD²f_n
D³f_n+1 = 3D²f_n + xD³f_n
D⁴f_n+1 = 4D³f_n + xD⁴f_n
Dⁿ⁺¹f_n+1 = (n+1)Dⁿf_n + xDⁿ⁺¹f_n

I think you have done that. Make sure you understand so far. Do it again, I would suggest, for good measure. It is now easy to finish solving. Try it. You don't have to depend, I think. Any questions about what's been done?