Question about Sawtoothwave Function

[solve]

Homework Statement

For a sawtooth wave function with a period of P where the first branch of the function is given for [a, a+P) we can say that :

f(x)=some expression in x for [a, a+P)

f(x+nP)=f(x)

Homework Equations

The Attempt at a Solution

Why is 'a+P' not included in the x-interval? Wouldn't f(x) equal 1 if x was 1? What would happen if 'a+P' was included in the interval?

Thank You.

 

[micromass]

We have that f(a+P)=f(a). So knowing the value of f(a) means that we also know the value of f(a+P). So there is no need to include a+P.

 

[solve]

We have that f(a+P)=f(a). So knowing the value of f(a) means that we also know the value of f(a+P). So there is no need to include a+P.

The graph in my textbook has a straight line branch between x=0 and x=1 that repeats itself indefinitely. The highest Y equals 1. For [0,1) the output from f is given as f(x)=x.

So if x is 0.5 then f(x) is 0.5. What is f(x) if x=1?

 

[SammyS]

If f(0) = 0, and the function is periodic with period 1, then f(1) = 0 also.

 

[solve]

If f(0) = 0, and the function is periodic with period 1, then f(1) = 0 also.

From the book:

"f(x+1)=f(x) for [0,1)

For example, f(1.5)=f(0.5+1)=f(0.5)=0.5"

Period here is 1. y=1 when x=1 on the graph. Does it mean that x=1 is undefined or something?

 

[micromass]

f(1)=f(0)=0.

 

[SammyS]

From the book:

"f(x+1)=f(x) for [0,1)

For example, f(1.5)=f(0.5+1)=f(0.5)=0.5"

Period here is 1. y=1 when x=1 on the graph. Does it mean that x=1 is undefined or something?

It may look like y=1 when x=1, however y only gets very close to 1 when x gets very close to 1.

f(0.999995) = 0.999995 , f(0.99999999993) = 0.99999999993 , f(1) = 0

 

[solve]

f(1)=f(0)=0.

f(1)=0 if you go by definition, f(1)=1 if you go by the picture of the graph. Too confusing.

 

[solve]

It may look like y=1 when x=1, however y only gets very close to 1 when x gets very close to 1.

f(0.999995) = 0.999995 , f(0.99999999993) = 0.99999999993 , f(1) = 0

So what would happen if I were to include 1 in the interval?

 

[micromass]

f(1)=0 if you go by definition, f(1)=1 if you go by the picture of the graph. Too confusing.

The picture of the graph is wrong. On several points. For example, that line straight down doesn't happen, you just have a discontinuity

So what would happen if I were to include 1 in the interval?

You can't. If you include 1, then f(1) would have multiple definitions, this is not good.