(Spivak) - a function with strange behaviour.

[kioria]

1) Find a function, f(x) which is discontinuous at 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4} ..., but continuous at any other points.

Solution (I have come across, probably wrong and a half):
f(x) = { 1 for all real x; 0 for 1/x where x is natural numbers.

Can anyone tell me the answer to this?

 

[ReyChiquito]

The function you have founded have indeed that property, there is an infinite set of functions that can do the job...

 

[kioria]

The function you have founded have indeed that property, there is an infinite set of functions that can do the job...

Excellent! Thank you.

 

[Nexus[Free-DC]]

Does the function need to be defined everywhere? If not, you can construct an elegant solution as follows:

f(x)=\frac{(x-1)(x-1/2)(x-1/3)...}{(x-1)(x-1/2)(x-1/3)...}

This function is equal to 1 except at the points \frac{1}{n}, where it is undefined.

 

[matt grime]

Part of the definition of a function is its domain, and it needs to be defined on its domain.

If you mean, say, is 1/x a function from R to R? No: you've not defined what it is at zero, until you do it is at best a function from R\{0} to R.

This is a big problem that is not taught properly when it first arises and causes many unnecessary problems.

The one you gave has the nice property that one may define f at the points in question so that it is 1, and is continuous at all those points.

 

[Gokul43201]

discontinuity

see attachment for a family of solutions

 

[kioria]

Thanks for all the help :)

 

[kuengb]

Kioria, I just want to add that the function you gave is also discontinuous at x=0, not only at x=1/n. But you can change it like this to kill that bug:

f(x) = { 0 for all real x; x for x=1/n , n any natural number

 

[matt grime]

If you're going to get picky then the original definition doesn't define a function.

 

[kioria]

Part of the definition of a function is its domain, and it needs to be defined on its domain.

If you mean, say, is 1/x a function from R to R? No: you've not defined what it is at zero, until you do it is at best a function from R\{0} to R.

This is a big problem that is not taught properly when it first arises and causes many unnecessary problems.

The one you gave has the nice property that one may define f at the points in question so that it is 1, and is continuous at all those points.

Interesting. Good stuff.

 

[fourier jr]

The Gauss Transformation is a function that is like that. It looks like this:

G(x) = 1/x + [ 1/x ] for x in the interval (0,1]
G(x) = 0 @ x=0

[] means floor, aka least integer function.

 

[stefanfuglsang]

Another suggestion

\Gamma( \frac {-1}{x} ), for x > 0