Finding the Function for 3/2(g(x))^2: Proving 3/2(g(x))^2 = g''(g)

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i want to find the function that satisfies the below. been trying a lot, but all my tactics have proven useless.

f:[0,=>] -> R

f(x)= [0]\int[/x]( 1/(1+t^3)^0.5)

show that 3/2(g(x))^2 = g''(g)
 
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monsmatglad said:
i want to find the function that satisfies the below. been trying a lot, but all my tactics have proven useless.

f:[0,=>] -> R

f(x)= [0]\int[/x]( 1/(1+t^3)^0.5)

show that 3/2(g(x))^2 = g''(g)
How does g(x) enter the picture? How is it related to f(x)? What does g''(g) mean?
 


Mark44 said:
How does g(x) enter the picture? How is it related to f(x)? What does g''(g) mean?

Sorry for not being precise. g''(g) was supposed to be g''(x), the second derivative of g. The function g is not really mentioned any further in the assignment, but I'm pretty sure it's the inverse function of f.

Mons
 


monsmatglad said:
i want to find the function that satisfies the below. been trying a lot, but all my tactics have proven useless.

f:[0,=>] -> R

f(x)= [0]\int[/x]( 1/(1+t^3)^0.5)

show that 3/2(g(x))^2 = g''(g)

I'd first make sure it's right Mon. How about I write it as:

Let:

y=f(x)=\int_0^x \frac{1}{\sqrt{1+t^3}}dt

Then if:

x=f^{-1}(y)

show:

3/2(f^{-1}(y))^2=\frac{d^2 f^{-1}}{dy^2}

I dont' know how to do that right off-hand analytically but I do know how to check it numerically. I'll leave this Mathematica code for you to interpret if you're interested:

Code: 
f[x_] := NIntegrate[1/Sqrt[1 + t^3], 
    {t, 0, x}]; 
mytable = Table[{f[x], x}, 
    {x, 0, 1, 0.01}]; 
myeqn = Table[x^n, {n, 0, 10}]
myfit = Fit[mytable, myeqn, x]
p1 = Plot[(3/2)*myfit^2, {x, 0, 1}, 
   PlotStyle -> Red]
p2 = Plot[Evaluate[D[myfit, {x, 2}]], 
   {x, 0, 1}, PlotStyle -> Blue]
Show[{p1, p2}]

And the reason for doing that is to give me some assurance that the problem is even written correctly and now that I know it's probably correct (at least in the range I checked it), it's worth working on to try and prove it analytically.
 
Last edited:


Ok, I see it. How about you start with just differentiating f(x) and recall that the derivative of the inverse is:

\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}

Now what can you think to do?
 


jackmell said:
Ok, I see it. How about you start with just differentiating f(x) and recall that the derivative of the inverse is:

\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}

Now what can you think to do?

Hm. I've been trying to diferentiate and integrate g(x), but the numbers don't seem to add up? did they for you? please show me the way=).
 


monsmatglad said:
Hm. I've been trying to diferentiate and integrate g(x), but the numbers don't seem to add up? did they for you? please show me the way=).

Here's what jackmell is saying in different notation. Since f and g are inverses, f(g(x))=x. If you differentiate that using the chain rule, you get f'(g(x))*g'(x)=1, right? Can you apply that to this problem? What's f'(g(x))?
 


hm, that's actually what I've been doing, but with no luck. i get that g'(f(x))=(1+x^3)^0.5
and f'(g(x)) is 1/((1+t^3)^0.5)
 


monsmatglad said:
hm, that's actually what I've been doing, but with no luck. i get that g'(f(x))=(1+x^3)^0.5
and f'(g(x)) is 1/((1+t^3)^0.5)

That's somewhat messed up. f'(g(x)) doesn't have a t in it. How can what it is equal to be a function of t? Can you run through how you got f'(g(x)) again?
 
  • #10


sorry. f'(x) must be 1/(1+x^3)^0.5. and according to the rule which i don't know what is called, the derivative of g(f(x)) is 1/f'(x). this is what i have been using, but with no success. i don't get the answers to fit together like they should.
 
  • #11


monsmatglad said:
sorry. f'(x) must be 1/(1+x^3)^0.5. and according to the rule which i don't know what is called, the derivative of g(f(x)) is 1/f'(x). this is what i have been using, but with no success. i don't get the answers to fit together like they should.

You have to be more careful about the arguments of the functions. You are being too sloppy. Yes, f'(x) is 1/(1+x^3)^(1/2). No, the derivative of g(f(x)) is NOT 1/f'(x). g(f(x))=x. I'd say the derivative of x is 1.
 
  • #12


ok. but what do you get when you derivate g(f(x)) in this problem? I see I've been thinking the wrong things, but i really believe i just need to see the correlation, and then i will understand.
 
  • #13


monsmatglad said:
ok. but what do you get when you derivate g(f(x)) in this problem? I see I've been thinking the wrong things, but i really believe i just need to see the correlation, and then i will understand.

You start with the expression I mentioned before. Since g and f are inverses, f(g(x))=x, correct? Can you differentiate both sides of that expression? If you can, you can find g'(x) in a useful form.
 
  • #14


hm, yes. f(g(x)) = x. f'(g(x))*g'(x)=1. but even though I've been using the wrong definitions, i think what i now get is sort of similar to what I've tried.
 
  • #15


monsmatglad said:
hm, yes. f(g(x)) = x. f'(g(x))*g'(x)=1. but even though I've been using the wrong definitions, i think what i now get is sort of similar to what I've tried.

That's right. If that's similar to what you've tried, then you were on the right track. Now solve for g'(x) and use your expression for f'.
 
  • #16


yes. but this is really where my tactics failed. i get thet g'(x)= (1+x^3)^0.5 and when i integrate and derivate this, the results don't fit like the are supposed to.
 
  • #17


monsmatglad said:
yes. but this is really where my tactics failed. i get thet g'(x)= (1+x^3)^0.5 and when i integrate and derivate this, the results don't fit like the are supposed to.

That's because you did something wrong in solving for g'(x). What is f'(g(x))?
 
  • #18


hm, f'(g(x)) is 1/(1+x^3)
 
  • #19


monsmatglad said:
hm, f'(g(x)) is 1/(1+x^3)

That is f'(x) (well, it is if you stick the 0.5 power in). That is NOT f'(g(x)).
 
  • #20


hm, i guess this is what made me fail. and i really don't know what f'(g(x)) is.
 
  • #21


monsmatglad said:
hm, i guess this is what made me fail. and i really don't know what f'(g(x)) is.

It's easier than you think. f'(x)=1/(1+x^3)^(1/2). Substitute 'g(x)' for 'x'. That's all.
 
  • #22


so f'(g(x)) is 1/(1 + g(x)^3 )^(1/2). ?
 
  • #23


monsmatglad said:
so f'(g(x)) is 1/(1 + g(x)^3 )^(1/2). ?

Sure it is. Getting the arguments of functions right isn't that hard. So now what's g'(x)?
 
  • #24


(1+g(x)^3)^0.5 i guess =)
 
  • #25


monsmatglad said:
(1+g(x)^3)^0.5 i guess =)

Yes, it is. Can you finish it off? What's g''(x)?
 
  • #26


(3/2)*g(x)^2*(1+g(x)^3)^-0.5
 
  • #27


monsmatglad said:
(3/2)*g(x)^2*(1+g(x)^3)^-0.5

You are missing something. You didn't finish with the chain rule. Where's the g'(x) on the right side?
 
  • #28


g'(x)*(3/2)*g(x)^2*(1+g(x)^3)^-0.5?
 
  • #29


yes, that must be it
 
  • #30


monsmatglad said:
g'(x)*(3/2)*g(x)^2*(1+g(x)^3)^-0.5?

I hope that '?' doesn't mean you don't know why there should be a g'(x).
 
  • #31


i get it. thanks so much. really appreciate the patience you've showed me.
 

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