Here's a quick visual proof of the first result: if f''x(x) is negative for all x, the function is concave down, which means it's less than or equal to its tangent line at any a. If f'(a)≠0, that tangent line intersects the x-axis, meaning that there's somewhere that f(x) is not positive. Therefore, it's negative everywhere in order to have a sign pattern.
@@minamagdy4126 It can be vastly slower if your system doesn't have a barrel shifter and you need to shift right the maximum number of times. But, you could go the other way, shift left, and get the sign from the carry flag, if your system has that. All that being said, most systems have a sign flag you can check directly, no bit arithmetic needed. Though, you should also check the zero flag, or else zero will be regarded as positive.
You're right, but all n before k make their term equal to zero. It's just better written as starting at k because that's when the terms start being non-zero.
Pet peeve of mine: needless proof by contradiction. Proof by contradiction should be reserved for when you can't construct the thing that makes your statement true. But you did construct it! It's precisely the x_0 that you defined. You could have just done a direct proof. Still, a very neat result. Yet another way that being analytic is actually a very rigid condition.
If we want to be rigorous, we should write in the definition that sign(n-thderivative…) = sign(a_n) and not just equal to a_n. Since a_n is a number either equal to 1 or -1 we can’t say that the sign(something) equals a number.
Pretty cool. Suppose we have a function f from reals to reals, which is infinitely differentiable, and such that there is a sequence of nested non-empty intervals A_k \subseteq A_{k+1} such that the union of all the A_k is the full real line, such that the k-th derivative of f, f^{(k)} , has a sign sequence when restricted to the interval A_k . This is a slightly weaker condition than “f has a sign sequence on the whole real line”, but has a similar flavor. What possible sign sequences are there for f restricted to A_0 ? Is it still just those 4? If we use that construction for x_0 , perhaps we can obtain something about how quickly the A_k can grow, given a particular sign sequence? If we know the sign sequence for f on A_0 , then we get, I guess a limit on where the boundary of A_0 has to be relative to the boundary of A_2 or something like that? uh, x_0 = a - f(a)/f’(a) , so f(x_0) , if x_0 is in A_0 , must have sign a_0 , while f’’(c) , for c between a and x_0 , if c is in A_2 , must have sign a_2 , And, if x_0 is in A_0 , then as A_2 contains A_0 , and A_0 contains both a and x_0 , it must contain c, so c is in A_2 . So if x_0 is in A_0 , a_0=a_2 ? So, if a_2=-a_0 , then x_0 is not in A_0 , and so a-f(a)/f’(a) is outside A_0 ? Ah, well, of course it can’t be just those 4, because we can take e^x plus some polynomial with large coefficients, and then the first finitely many derivatives on some interval will fail to match the pattern, but then after the polynomial has been differentiated into oblivion, we are left with e^x , which has a sign pattern on the whole line. But, is any sign pattern for such an f, *eventually* the same as one of those 4? Or, another question: What constraints on a function can we obtain from knowing its sign sequence on an interval?
if f(x) is infinitely differentiable on R, then it has the same sign pattern on R or the derivatives have zeros. Thus the union of A_k cannot fill the whole real line but leaves at least one hole. If you allow such holes, you can have other sequences. For example with f(x)=cos(x).
@@deinauge7894 what about f(x) = e^x - 1000 x^2 though? On the interval (0,1) it has the sign pattern (-1,-1,1,1,1,…) but f’’ has the sign pattern (1,1,1,…) on all of R , right? Like, this f does have zeros, at like, approximately sqrt(1/1000) or so, and f’ has a zero at like, somewhere between 1/1000 and e/1000 . So, I was thinking if the zeros are further and further out for higher derivatives of the function?
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Here's a quick visual proof of the first result: if f''x(x) is negative for all x, the function is concave down, which means it's less than or equal to its tangent line at any a. If f'(a)≠0, that tangent line intersects the x-axis, meaning that there's somewhere that f(x) is not positive. Therefore, it's negative everywhere in order to have a sign pattern.
Regarding the definition of sgn(x), I once came across sgn(x)=(x>0)-(x
I've used this idea before, but I believe I used (00)-1, because that was slightly faster given the architecture.
I wonder whether, using bit arithmetic, getting the sign bit is faster than evaluating x < 0
you're thinking of the Iverson bracket: [p] is 1 if p is true and 0 if p is false. sgn(x) = [x>0]-[x
@@minamagdy4126 It can be vastly slower if your system doesn't have a barrel shifter and you need to shift right the maximum number of times. But, you could go the other way, shift left, and get the sign from the carry flag, if your system has that. All that being said, most systems have a sign flag you can check directly, no bit arithmetic needed. Though, you should also check the zero flag, or else zero will be regarded as positive.
@@rubenverg I've not heard this name before. Thanks, that is interesting! Also happy to see any mention of APL.
"My institution says its ok for us to do a talk if its ok with your institution" 😂
Such sign patterns of derivatives are used in Budan-Fourier theorem for root counting.
I think the sum should start at n=k at 15:53
You're right, but all n before k make their term equal to zero. It's just better written as starting at k because that's when the terms start being non-zero.
At 19:20, I expected you to use the fact that 1/[(2^n)*(n!)] < 1/(2^n) = 1/2 + 1/4 + ... = 1. I didn't even notice that this was related to e^x.
Pet peeve of mine: needless proof by contradiction. Proof by contradiction should be reserved for when you can't construct the thing that makes your statement true. But you did construct it! It's precisely the x_0 that you defined. You could have just done a direct proof. Still, a very neat result. Yet another way that being analytic is actually a very rigid condition.
I thought it was neat that he _didn't_ need to assume f analytic, only that it is infinitely differentiable.
If we want to be rigorous, we should write in the definition that sign(n-thderivative…) = sign(a_n) and not just equal to a_n. Since a_n is a number either equal to 1 or -1 we can’t say that the sign(something) equals a number.
as far as i'm concerned doesn't the sign function just output 1 or -1 (or 0) based on the sign of the input?
Pretty cool.
Suppose we have a function f from reals to reals, which is infinitely differentiable, and such that there is a sequence of nested non-empty intervals A_k \subseteq A_{k+1} such that the union of all the A_k is the full real line, such that the k-th derivative of f, f^{(k)} , has a sign sequence when restricted to the interval A_k .
This is a slightly weaker condition than “f has a sign sequence on the whole real line”, but has a similar flavor.
What possible sign sequences are there for f restricted to A_0 ? Is it still just those 4?
If we use that construction for x_0 , perhaps we can obtain something about how quickly the A_k can grow, given a particular sign sequence?
If we know the sign sequence for f on A_0 , then we get, I guess a limit on where the boundary of A_0 has to be relative to the boundary of A_2 or something like that?
uh, x_0 = a - f(a)/f’(a) , so f(x_0) , if x_0 is in A_0 , must have sign a_0 , while f’’(c) , for c between a and x_0 , if c is in A_2 , must have sign a_2 ,
And, if x_0 is in A_0 , then as A_2 contains A_0 , and A_0 contains both a and x_0 , it must contain c, so c is in A_2 .
So if x_0 is in A_0 , a_0=a_2 ?
So, if a_2=-a_0 , then x_0 is not in A_0 , and so a-f(a)/f’(a) is outside A_0 ?
Ah, well, of course it can’t be just those 4, because we can take e^x plus some polynomial with large coefficients, and then the first finitely many derivatives on some interval will fail to match the pattern, but then after the polynomial has been differentiated into oblivion, we are left with e^x , which has a sign pattern on the whole line.
But, is any sign pattern for such an f, *eventually* the same as one of those 4?
Or, another question: What constraints on a function can we obtain from knowing its sign sequence on an interval?
if f(x) is infinitely differentiable on R, then it has the same sign pattern on R or the derivatives have zeros. Thus the union of A_k cannot fill the whole real line but leaves at least one hole.
If you allow such holes, you can have other sequences. For example with f(x)=cos(x).
@@deinauge7894 what about f(x) = e^x - 1000 x^2 though?
On the interval (0,1) it has the sign pattern (-1,-1,1,1,1,…) but f’’ has the sign pattern (1,1,1,…) on all of R , right?
Like, this f does have zeros, at like, approximately sqrt(1/1000) or so,
and f’ has a zero at like, somewhere between 1/1000 and e/1000 .
So, I was thinking if the zeros are further and further out for higher derivatives of the function?
f(x)=e^(positive_number*x) also has a sign pattern a_n = 1
The sign patterns are limited to four different possibilities, the given functions are only examples that achieve those.
21:49
You are one of the most dedicated people I’ve ever seen
@@JOSHUVASRINATH I know, I’m just expressing my admiration
I thought this was tacitly known, but I like that there is the proof
I thought all this was already kind of pretty obvious or already known but cool review I guess.