# Problems with Spivak Calculus Chapter 6 Problem 6

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• Oct 3rd 2012, 07:07 PM
patrickmanning
Problems with Spivak Calculus Chapter 6 Problem 6
I have the solutions book and see the answers, but they make no sense to me.
Any/All help is appreciated.
a) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... but continuous at all other points.
b) Find a function that is discontinuous at 1, 1/2, 1/4, 1/4... and at 0 but continuous at all other points.

Problem 7 doesn't seem to have enough explanation in the solution either. How is f being continuous 0 used?
7) Suppose f satisfies f(x+y) = f(x) + f(y) and f is continuous at 0. Prove it is continuous at A for all values of A.
• Oct 3rd 2012, 09:46 PM
MaxJasper
Re: Problems with Spivak Calculus Chapter 6 Problem 6
• Oct 4th 2012, 08:19 AM
patrickmanning
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Well that certainly didn't help at all. Thanks, anyway.
• Oct 4th 2012, 08:38 AM
emakarov
Re: Problems with Spivak Calculus Chapter 6 Problem 6
For 6a, you can also consider

$f(x)=\begin{cases}\frac{1}{k}& \frac{1}{2^k}
• Oct 4th 2012, 09:48 AM
johnsomeone
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Problem #7: Ignoring issues of the domain, $f$ continuous at $a$ means: $\lim_{x\to a} f(x) = f(a)$.

Now observe that, always, $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$.
• Oct 4th 2012, 10:09 AM
emakarov
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Quote:

Originally Posted by johnsomeone
Problem #7: Ignoring issues of the domain, $f$ continuous at $a$ means: $\lim_{x\to a} f(x) = f(a)$.

Now observe that, always, $\lim_{x\to a} f(x) = \lim_{h\to 0} f(a + h)$.

I think that one also has to prove that f(0) = 0.
• Oct 4th 2012, 10:58 AM
patrickmanning
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Thanks, I managed something with 7, but I still don't understand 6. Is there any way to show how you construct an answer?
• Oct 4th 2012, 11:45 AM
emakarov
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Quote:

Originally Posted by patrickmanning
I still don't understand 6. Is there any way to show how you construct an answer?

Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase.
• Oct 4th 2012, 12:22 PM
patrickmanning
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Quote:

Originally Posted by emakarov
Can you construct a function that is discontinuous at x = 0? For example, f(x) = 0 for x <= 0 and f(x) = 1 for x > 0. If this is clear, then what is difficult about problem 6? I suggested a function whose graph looks like a staircase.

In your example in this post, f(x) is really discontinuous at x = 0? It seems I don't even understand continuity properly. At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right?
Wouldn't that mean in your first example, f(x) was also discontinuous at x 0?
• Oct 4th 2012, 12:45 PM
emakarov
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Quote:

Originally Posted by patrickmanning
At the point where there is a jump, that point is discontinuous? Because it is different when approached from left and right sides, right?

Yes. Would you agree that a person's spacial position as a function of time is a continuous function? If f(t) = 0 for t <= 0 and f(1) = 1 for t > 0 described my spacial position, then I would be at position 0 before time 0 and then suddenly move to position 1 (say, 1 mile from position 0). If you know how to construct such transportation, I would certainly like to hear!

Formally, f(t) is not continuous at 0 because any interval containing t = 0 is mapped by f to a set that contains some $y_1$ and $y_2$ such that $|y_1 - y_2| = 1$. In other words, however small you make an interval around 0, the image of that interval contains numbers that are sufficiently far apart, i.e., not arbitrarily close. It does not matter how precisely you approximate t = 0: if you change t just a little bit, f(t) can change by 1.
• Oct 4th 2012, 01:30 PM
patrickmanning
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Thank you. I hate to be a bother, but can you continue helping by answering my question about your first example? Isn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k?
• Oct 4th 2012, 01:36 PM
emakarov
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Quote:

Originally Posted by patrickmanning
sn't f(x) discontinuous at x = 0 in your example with f(x) = 1/k?

As defined in post #4, f(x) = 0 for $x\le 0$. As far as x > 0, f(x) decreases as x decreases. In fact, f(x) tends to 0 as x tends to 0 from the right. Since $\lim_{x\to0}f(x)=f(0)$, f is continuous at 0.
• Oct 4th 2012, 03:21 PM
patrickmanning
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Thank you. I understand well now. Your example actually is not discontinuous at 1/3 then. Changing it a bit to:
f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that.
• Oct 4th 2012, 03:27 PM
emakarov
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Quote:

Originally Posted by patrickmanning
Your example actually is not discontinuous at 1/3 then. Changing it a bit to:
f(x) = 1/b , 1/a < x <= 1/b where a = b+1 and b = 1, 2, 3... will fix that.

I though that "1, 1/2, 1/4, 1/4..." from post #1 was supposed to mean "1, 1/2, 1/4, 1/8...," but you are right if it should be "1, 1/2, 1/3, 1/4..."
• Oct 4th 2012, 03:38 PM
patrickmanning
Re: Problems with Spivak Calculus Chapter 6 Problem 6
Haha, you are right. I didn't realise I typed 1/4 twice. Thank you so much for everything. I understand this all now.
For 6b, you can just change it to "otherwise, x = -1" and now x = 0 will be discontinuous as from the left, f(x) approaches -1 but from the right, f(x) still approaches 0.
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