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proving equality of a function

aruwin

Member
Jul 4, 2012
121
I have the solution to the question, but I don't understand the first step so I am just gonna paste the first step here.I need people to explain to me.

Question:
For the function f(x) = (logx)/x, prove the following equality.

fn (1) = (-1)(n-1) n![1+ 1/2 +1/3 +...+1/n]

First step of the solution(The bolded ones are what I don't understand):
Let y(x) = 1 - x and g(x) = f(y(x)) = log(1 - x) / (1 - x). Then:

g(y(x)) = log(1 - (1 - x)) / (1 - (1 - x)) = log(x) / x = f(x)

Thus:

f'(x) = g'(y(x)) * y'(x) ... (chain rule)
= g'(y(x)) * -1
= -g'(y(x))

f''(x) = -g''(y(x)) * y'(x)
= -g''(y(x)) * -1
= g''(y(x))

So on, by induction, we can see that:

f^(n)(x) = (-1)^n * g^(n)(y(x))
===> f^(n)(1) = (-1)^n * g^(n)(y(1)) = (-1)^n * g^(n)(0)

Ok, first off, why and how do we know to put y(x) = 1 - x?
Secondly, the chain rule part. Where does he get y(x) in that chain rule?
I thought f'(x)=g'(y(x))
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
When you write [tex]f^n[/tex] do you mean the nth derivative? That is more commonly represented with (): [tex]f^{(n)}[/tex]. Just [tex]f^n[/tex] usually means the nth iteration of f but that can't be what you mean here: f(1)= 0 so [tex]f^2(1)[/tex] is not defined.

I'm not sure what there is about "y= 1- x" not to understand! They are defining a new variable in terms of the old If y= 1- x, then x= 1- y so that f(x)= ln(x)/x= ln(1- y)/(1- y)= g(y). f and g are really the same function- the only difference is that we are thinking of g as a function of y, f as a function of x. They are defining g(y) to be ln(1- y)/(1- y)= f(y). The derivative of g with respect to x is, by the chain rule, [tex]\frac{dg}{dy}= \frac{df}{dy}\frac{dy}{dx}. Of course, with y= 1- x, dy/dx= -1 so we have dg/dy= -df/dy, or g'= -f', as they say. Differentiating again, the same thing happens so that we multiply by -1 again: g''= (-1)(-1)f''= f''. It should be easy to see that as we continue differentiating, we just keep multiplying by -1 so the nth derivative of g is [itex](-1)^n[/itex] times the nth derivative of f.

Secondly, the chain rule part. Where does he get y(x) in that chain rule?
I thought f'(x)=g'(y(x))
No, that is precisely what the chain rule says: with the derivative of f with respect to x and the the derivative of g with respect to y, we must have f'(x)= g'(y) (dy/dx).
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I have the solution to the question, but I don't understand the first step so I am just gonna paste the first step here.I need people to explain to me.

Question:
For the function f(x) = (logx)/x, prove the following equality.

fn (1) = (-1)(n-1) n![1+ 1/2 +1/3 +...+1/n]

First step of the solution(The bolded ones are what I don't understand):
Let y(x) = 1 - x and g(x) = f(y(x)) = log(1 - x) / (1 - x). Then:

g(y(x)) = log(1 - (1 - x)) / (1 - (1 - x)) = log(x) / x = f(x)

Thus:

f'(x) = g'(y(x)) * y'(x) ... (chain rule)
= g'(y(x)) * -1
= -g'(y(x))

f''(x) = -g''(y(x)) * y'(x)
= -g''(y(x)) * -1
= g''(y(x))

So on, by induction, we can see that:

f^(n)(x) = (-1)^n * g^(n)(y(x))
===> f^(n)(1) = (-1)^n * g^(n)(y(1)) = (-1)^n * g^(n)(0)

Ok, first off, why and how do we know to put y(x) = 1 - x?
Well I would suggest that \(y=1-x\) is an interesting change of variable to consider since we are obviously looking at the coefficients in the Taylor series expansion of \(f(x)\) about \(x=1\).

Trial and error would then have been used to find the given relation and that would then be transformed into a proof with no given reason for the choice other than it works.

What you are seeing is a streamlined proof with the scaffolding used to construct it removed.

Secondly, the chain rule part. Where does he get y(x) in that chain rule?
I thought f'(x)=g'(y(x))
No, the prime denotes differentiation with respect to \(x\), so the chain rule is:

\[\frac{d}{dx}g(y)=\frac{d}{dy}g(y)\frac{dy}{dx}\]

CB
 
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