# finite number of discontinuety points

• Sep 16th 2011, 12:16 PM
transgalactic
finite number of discontinuety points
i have a function
f(x)= 1 for x=1/n
f(x)=0 for otherwise

n is a natural number
prove that f(x) is defferentiable [0,1]
?

first i tried to prove that there are finite number of discontenuety points

n>1/a so 0<1/n<a
then the book says that there at most n-1 discontinuty ponts
dont know how they got it from the last innequality

?
• Sep 16th 2011, 02:26 PM
Plato
Re: finite number of discontinuety points
Quote:

Originally Posted by transgalactic
i have a function
f(x)= 1 for x=1/n
f(x)=0 for otherwise
n is a natural number
prove that f(x) is differentiable [0,1]
?
first i tried to prove that there are finite number of discontenuety points
n>1/a so 0<1/n<a
then the book says that there at most n-1 discontinuty ponts
dont know how they got it from the last innequality

I cannot figure out if you are translating these questions or if you are just unfortunately in a very poor course.
I don't think I have ever seen a poorer written question.
There is absolutely no part of that question that is correct.
Even the definition of the function is nonsense.

Moreover, if a function is differentiable at a point, then it must be continuous at that point. Do you see why it is all nonsense?
• Sep 16th 2011, 04:43 PM
HallsofIvy
Re: finite number of discontinuety points
Quote:

Originally Posted by transgalactic
i have a function
f(x)= 1 for x=1/n
f(x)=0 for otherwise

n is a natural number
prove that f(x) is defferentiable [0,1]
?

first i tried to prove that there are finite number of discontenuety points

n>1/a so 0<1/n<a
then the book says that there at most n-1 discontinuty ponts
dont know how they got it from the last innequality

?

The function you have defined is 1 at x= 1, 1/2, 1/3, 1/4, 1/5, 1/6, etc. 0 everywhere else. Since for any number that is NOT of the form 1, 1/2, 1/3, etc., there exist an interval around it such none of those are in the interval. As a result, the function is continuous and differentiable (with derivative 0) at all x except 1, 1/2, 1/3, 1/4, etc. It is not continuous at those points so there is an infinite number of points at which it is not continuous.

But, since you say "there are at most n-1 discontinuity point", which only makes sense if n is a fixed number, did you mean to define the function $\displaystyle f_n$ as $\displaystyle f_n(x)= 1$ if x= 1/n, $\displaystyle f(x)= 0$ otherwise, for a specific n? That is, $\displaystyle f_2$ is 0 everywhere except x= 1/2 where it is 1, $\displaystyle f_3(x)$ is 0 everywhere except at x= 1/3 where it is 1, etc? That also doesn't quite make sense- any such function differentiable everywhere except at the single point $\displaystyle x= 1/n$.
• Sep 16th 2011, 04:52 PM
Plato
Re: finite number of discontinuety points
Quote:

Originally Posted by HallsofIvy
The function you have defined is 1 at x= 1, 1/2, 1/3, 1/4, 1/5, 1/6, etc. 0 everywhere else. Since for any number that is NOT of the form 1, 1/2, 1/3, etc., there exist an interval around it such none of those are in the interval. As a result, the function is continuous and differentiable (with derivative 0) at all x except 1, 1/2, 1/3, 1/4, etc. It is not continuous at those points so there is an infinite number of points at which it is not continuous.

Even under your reading of this question, how can $\displaystyle f$ be differentiable on $\displaystyle [0,1]~?$
• Sep 16th 2011, 08:04 PM
chisigma
Re: finite number of discontinuety points
Quote:

Originally Posted by Plato
Even under your reading of this question, how can $\displaystyle f$ be differentiable on $\displaystyle [0,1]~?$

May be that the 'true' question is: is f differentiable in x=0?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 16th 2011, 10:12 PM
transgalactic
Re: finite number of discontinuety points
sorry i ment
n-1 discontinuety point in [a,1]

we need to prove defferentiability in [a,1]

0<a<1

cant imagine why there are at most n-1 discontinuety points
• Sep 17th 2011, 04:21 AM
HallsofIvy
Re: finite number of discontinuety points
You appear to be writing the function incorrectly. You do NOT mean "f(1/n)= 1, for all n, f(x)= 0 otherwise" nor do you mean $\displaystyle f_n(1/n)= 1$, f(x)= 0 otherwise" for a specific n. You seem to be saying "$\displaystyle f_n(1/i)= 1$ for all [tex]i\le n[/itex], f(x)= 0 otherwise..

If that is correct, look at some examples: $\displaystyle f_2(x)= 0$ for all x except 1/2- one point of discontinuity.

$\displaystyle f_3(x)= 0$ for all x except 1/2 and 1/3- two points of discontinuity.

$\displaystyle f_4(x)= 0$ for all x except 1/2, 1/3, and 1/4- three points of discontinuity.