Can You Find a Root of f(x) with Only f(x_1)==y and f ' (x)?

[soandos]

Is it possible to find a root of f(x), given just f(x_1)==y and f ' (x)?
if so, how would one go about it?
If this is in the wrong forum can a mod please move it?
thanks.

 

[Feldoh]

Yes, by the fundamental theorem of calculus.

 

[soandos]

Elaborate please.

 

[Feldoh]

\int_{x_1}^{x} f'(t) dt = f(x)-f(x_1)

So effectively you have found f(x).

 

[soandos]

so solve that for x?

 

[Feldoh]

Why do you think you would solve for x? How would you normally find the roots of an equation once you know what the equation is?

 

[soandos]

Im sorry, I am confused.
I think to solve for x, as that was the original variable.
I generally solve equations by setting them to zero and then using inverses.
I was asking for a different method.

 

[Feldoh]

Oh I see what your saying -- sorry I misunderstood.

You're saying solve for x such that f(x) = 0 right?

 

[soandos]

yes.

 

[HallsofIvy]

Is it possible to find a root of f(x), given just f(x_1)==y and f ' (x)?
if so, how would one go about it?
If this is in the wrong forum can a mod please move it?
thanks.

Feldoh assumed you meant you were given the value of f at the single point x_1 and the derivative for all x. His point was that you can then integrate the derivative, using that one point to determine the constant of integration, to recover the function itself:
f(x)= \int_{x_1}^x f&#039;(x)dx+ y[/itex]<br /> Set that equal to 0 and solve. <br /> <br /> Probably the best way to solve that equation, given the information you have, would be the Newton-Raphson Algorithm.<br /> <br /> Starting with any convenient value of x, perhaps x_1, Do <br /> x_{n+1}= x_n- \frac{f(x_n)}{f&amp;#039;(x_n}= x_n-\frac{\int_{x_1}^{x_n} f&amp;#039;(x)dx+ y}{f&amp;#039;(x)}<br /> repeatedly until two succesive values are within your error tolerance of each other.

 

[soandos]

But is there a way to get exact roots using that information?

 

[Office_Shredder]

In general you can't invert functions/find roots. For many useful specific cases, you can. Whether you can find exact roots depends on what your function f'(x) is

 

[soandos]

here was my thinking.
a root is determined by f(x)=0
in general, f(x_1)=y, so f(x)-y = 0
can the derivative of f(x) tell us how much x had to change.
If this was simple slope, it would be easy, distance to change/slope.
cant seem to do it with the derivative.
I also know that there is a way to solve for the roots of an n degree polynomial exactly, but i don't quite get it. thing is, it shows that its always possible even when the function has no inverse.
yes i realize that broadening it to f(x) is not quite the same thing
wondering if its possible though.

 

[HallsofIvy]

here was my thinking.
a root is determined by f(x)=0
in general, f(x_1)=y, so f(x)-y = 0
can the derivative of f(x) tell us how much x had to change.
If this was simple slope, it would be easy, distance to change/slope.
cant seem to do it with the derivative.
I also know that there is a way to solve for the roots of an n degree polynomial exactly, but i don't quite get it.

Then you know something no mathematician knows! It was proved, by Galois in the nineteenth century, that there cannot exist a formula for solutions of polynomials of degree five or greater.

thing is, it shows that its always possible even when the function has no inverse.
yes i realize that broadening it to f(x) is not quite the same thing
wondering if its possible though.

Thing is, you are completely wrong.

 

[soandos]

not what he said. what Galois said was that it was not possible for it to be done with arithmetic operations, so anything that involves calculus is possible.
if you are interestind in how, here is the link
http://hal.ccsd.cnrs.fr/docs/00/00/32/74/PDF/The-roots-of-any-polynomial-equation.pdf

 

[Count Iblis]

Yes, it is easy to derive a series expansion or write down the differential equation the inverse function has to satisfy.

 

[soandos]

How is that done?