Thread: 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

?

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?

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 as if x= 1/n, otherwise, for a specific n? That is, is 0 everywhere except x= 1/2 where it is 1, 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 .

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 be differentiable on

Originally Posted by Plato

Even under your reading of this question, how can be differentiable on

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

Kind regards

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

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 , f(x)= 0 otherwise" for a specific n. You seem to be saying " for all [tex]i\le n[/itex], f(x)= 0 otherwise..

If that is correct, look at some examples: for all x except 1/2- one point of discontinuity.

for all x except 1/2 and 1/3- two points of discontinuity.

for all x except 1/2, 1/3, and 1/4- three points of discontinuity.