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Weierstrass functions: continuous but not differentiable anywhere
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Re: Weierstrass functions: continuous but not differentiable anywhere
#22Fractal curves like the Koch snowflake are likewise continuous and nowhere differentiable.
John McCarthy (yes, the inventor of Lisp) has a simpler to analyze construction of a Weierstrass-like function from a triangle wave:
http://www-formal.stanford.edu/jmc/weierstrass/weierstrass.h...
Here's the big picture. When you analyze variations in f at the kth scale by setting |dx| = 2^(-2^k), the only term that really contributes is the kth term, 2^(-k) g(2^(2^k) x), since the higher terms completely vanish and the lower terms are insignificant by comparison. Thus a difference quotient df/dx at the kth scale is very nearly the derivative of 2^(-k) g(2^(2^k) x), which is just +/- 2^(-k) 2^(2^k). This has unbounded magnitude as k increases (the exponential term 2^k outgrows the linear term -k). For a differentiable function the rate of variation should settle down at finer scales. Instead of settling down at finer scales, McCarthy's function actually gets progressively more jagged.
If you want to grok the details, step 3 in his proof may be too terse. When bounding max(n
max(n Re: Weierstrass functions: continuous but not differentiable anywhere
#23See also, smooth but analytic nowhere:
Re: Weierstrass functions: continuous but not differentiable anywhere
#24Fractal curves like the Koch snowflake are likewise continuous and nowhere differentiable. John McCarthy (yes, the inventor of Lisp) has a simpler to analyze construction of a Weierstrass-like function from a triangle wave: http://www-formal.stanford.edu/jmc/weierstrass/weierstrass.h... Here's the big picture. When you analyze variations in f at the kth scale by setting |dx| = 2^(-2^k), the only term that really cont…
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Re: Weierstrass functions: continuous but not differentiable anywhere
#25Much simpler construction: http://en.wikipedia.org/wiki/Takagi_function
Re: Weierstrass functions: continuous but not differentiable anywhere
#26The construction of the Weierstrass function ( http://en.wikipedia.org/wiki/Weierstrass_function ) was new to me, even though I suffered through a semester of baby Rudin. To me, the canonical example of a continuous but nowhere differentiable function is: any Brownian motion sample path. Which, like the Weierstrass function, illustrates that functions that look simple may be so weird that our standard intuition does…
any Brownian
Almost any. Although they have probability zero of actually occurring, a Brownian motion can generate a continuous path.
Nitpicky, but this is a comment thread about math.
Re: Weierstrass functions: continuous but not differentiable anywhere
#27Earlier quoted context omitted.
Not sure what you mean by this, since your comment has no context -- there are no limits or sums in the link. However, even the humble Riemann integral would be able to find the area under the graph of a bounded function on a compact interval which is continuous on a set of full Lebesgue measure. Any other function only has a "well-defined area" for some seriously stunted notion of area; you can integrate more functi…
Consider this function: f(x) = sum[n = 0 to infinity]((1/2)^n sin((1/20)^n pi x)) It converges because each successive term is bounded by [-1/2, +1/2]. But if you integrate it you get: f(x) = sum[n = 0 to infinity](10^n / pi cos((1/20)^n pi x)) (Well... you get that if you play fast and loose about swapping the order of the integral and the sum) Which diverges because the input to the cosine function limits to 0 as n…
What you did in the limit in your second expression does not make any sense. You can't obtain the indefinite integral (the expression with cos()) and then take the limit as n -> infinity. This is not meaningful.
You can only do that with a definite integral (under conditions).
Using your reasoning, even a function like exp(x) would not integrate, because the terms in the integral of its series expansion would all go to zero as n -> infinity.
Another way to think about this is that each term in the indefinite integral you wrote has an arbitrary constant of integration, which can depend on n, that you did not include. This constant precludes you from actually finding the limit.
Re: Weierstrass functions: continuous but not differentiable anywhere
#28The construction of the Weierstrass function ( http://en.wikipedia.org/wiki/Weierstrass_function ) was new to me, even though I suffered through a semester of baby Rudin. To me, the canonical example of a continuous but nowhere differentiable function is: any Brownian motion sample path. Which, like the Weierstrass function, illustrates that functions that look simple may be so weird that our standard intuition does…
any Brownian Almost any. Although they have probability zero of actually occurring, a Brownian motion can generate a continuous path. Nitpicky, but this is a comment thread about math.
As long as you're nit picking, I'd suggest you re read carefully what you wrote, because it is wrong.
What you meant to say is that on a set of measure zero, the paths can have points where continuity is not satisfied. But you did not say that.
In any case, the version of Brownian motion I studied was the version that excludes this null set. ;-)
Re: Weierstrass functions: continuous but not differentiable anywhere
#29No, no! We don't want that!
What we want is differentiable everywhere but where the derivative is not Riemann integrable!!!!!! Everyone should have one of those, just to carry around.
We might keep in mind that a function is Riemann integrable iff it is continuous everywhere except on a set of measure 0. So, we want the points where the function is not continuous to be of positive measure!
Much more than you ever wanted to know about continuity, differentiability, and the Riemann integral!
Re: Weierstrass functions: continuous but not differentiable anywhere
#30Earlier quoted context omitted.
any Brownian Almost any. Although they have probability zero of actually occurring, a Brownian motion can generate a continuous path. Nitpicky, but this is a comment thread about math.
As long as you're nit picking, I'd suggest you re read carefully what you wrote, because it is wrong. What you meant to say is that on a set of measure zero, the paths can have points where continuity is not satisfied. But you did not say that. In any case, the version of Brownian motion I studied was the version that excludes this null set. ;-)
Yes I meant to say differentiable, brownian motion is always continuous. But with probability zero, brownian motion can generate a differentiable curve.
Take the typical random walk construction as the limit of a piecewise continuous curve. I.e., 50% chance of being y=x, 50% chance of being y=-x on [0,1], then subdivide. With zero probability, the walk can converge to a straight line x=t, which is certainly differentiable.
It's early in the morning and I had no caffeine.