Consider the nine angles x = pi/18, 3pi/18, 5pi/18, 7pi/18, 9pi/18, 11pi/18, 13pi/18, 15pi/18 and 17pi/18
These all satisfy the equation cot(9x) = 0. The idea is to find a polynomial whose roots are the tangents of these angles (excluding 9pi/18 = pi/2 for which the tangent is infinite).
Let t = tanx. Iterating tan(3x) = ( 3t - t^3 )/( 1 - 3t^2 ), we can find an expression for tan(9x) in terms of t. The denominator of tan(9x) (or the numerator of cot(9x)) comes out as
9t^8 - 84t^6 + 126t^4 - 36t^2 + 1
The roots of this polynomial are the values of tan(x). Only even powers of t are present because the values of tan(x) come in opposite sign pairs (e.g. tan(pi/18) = -tan(17pi/18)). Let y = t^2, then
9y^4 - 84y^3 + 126y^2 - 36y + 1 = 0
has roots y1 = tan^2(pi/18), y2 = tan^2(3pi/18), y3 = tan^2(5pi/18), y4 = tan^2(7pi/18). Note that tan^4(3pi/18) = (1/sqrt3)^4 = 1/9. Therefore, using some properties of roots of polynomials, we end with
tan^4(pi/18) + tan^4(5pi/18) + tan^4(7pi/18) =
tan^4(pi/18) + tan^4(3pi/18) + tan^4(5pi/18) + tan^4(7pi/18) - 1/9 =
y1^2 + y2^2 + y3^2 + y4^2 - 1/9 =
(y1 + y2 + y3 + y4)^2 - 2(y1y2 + y1y3 + ...) - 1/9 =
(-84/9)^2 - 2(126/9) - 1/9 =
59
I knew it was similar to a question almost two years ago about sums of fourth powers of degree-3 algebraic conjugates.
(Well, not exactly. And I read NxCy's post first.)
Anyway, with a method of solution similar to what I did back then, it's possible to do without going past degree 3:
First of all, tan4(5pi/18)=tan4(13pi/18), so it is the same as evaluating tan4(pi/18)+tan4(7pi/18)+tan4(13pi/18).
Let a=tan(pi/18), b=tan(7pi/18) and c=tan(13pi/18). As NxCy stated above, if tan(x)=t, then tan(3x)= (3t - t3)/(1 - 3t2). Now, no matter whether we put x=pi/18 or 7pi/18 or 13pi/18, that gives us tan(3x)=1/sqrt(3), and so a,b,c satisfy the polynomial t3 - sqrt(3)t2 - 3t + sqrt(3)/3 = 0.
Since (t-a)(t-b)(t-c) = t3 - (a+b+c)t2 + (ab+bc+ac)t - abc which must be the same as t3 - sqrt(3)t2 - 3t + sqrt(3)/3, this gives a+b+c=sqrt(3) and ab+bc+ac=-3. So a2+b2+c2 = (a+b+c)2-2(ab+bc+ac) = 3-2(-3) = 9.
Finally, if we define f(n)=an+bn+cn, then f(n) is the solution to a recursive formula with characteristic polynomial t3 - sqrt(3)t2 - 3t + sqrt(3)/3 with the initial conditions f(0)=3, f(1)=sqrt(3), f(2)=9. That is: f(n) = sqrt(3)f(n-1) + 3f(n-2) - (sqrt(3)/3)f(n-3). This formula gives f(3) = sqrt(3)*9 + 3*sqrt(3) - sqrt(3) = 11*sqrt(3), and f(4) = sqrt(3)*11*sqrt(3) + 3*9 - (sqrt(3)/3)*sqrt(3) = 33 + 27 - 1 = 59. So the answer is 59.
Note that technically a,b,c aren't degree 3; the polynomial I gave doesn't have rational coefficients. Actually they're degree 6 and the conjugates include all tan(n*pi/18) with n=1,5,7,11,13,17.
Maybe not exactly a maths problem, but here's a cool little trick I saw in a Youtube video recently.
Two people work together to perform a card trick. The first person chooses five cards randomly from a standard deck. After analysing the cards, he chooses one of the five cards to keep for himself, arranges the remaining four cards in an order of his choice and passes them to the second person. The second person, after observing the four cards they received, correctly announces the identity of the fifth card that was kept by the first person.
How is the trick performed?
Since we're in the mood, more hardcore trigonometry. Let us derive a result first found by a particular irrelevant mathematician. Warning: this has LOTS of calculations
Let x=2pi/17
Set
cos x + cos 4x = a
cos 2x + cos 8x = b
cos 3x + cos 5x = c
cos 6x + cos 7x = d
a + b = e
c + d = f
Evaluate cos x.
Hint: find a quadratic equation for e and f, then use these values to find quadratic equations for the other sums until you get the desired result.
So if we look at this from the perspective of information theory. The passed hand contains the identities of the four cards and their order. There are 5 choose 4 ways of choosing which cards to pass, and 4! ways of choosing how to pass them.
The receiver of the cards only has partial access to this information though. (which is 6.9 bits of information). He or she obviously has access to the 4! ordering. But this is only 4.6 bits of information, and you need 5.6 bits to specify 1 of 48 remaining cards.
You can add the final bit by handing the cards either face up or face down, which is very magic-trick-like, in that there's a trick. And that gives you the exact number of bits required to specify 1 of 48 unique cards. I don't know if they do this though. But here's how I'd run an encoding:
First, define any unambiguous ordering to the cards. (0=ace of clubs, 1=two of clubs, etc up to 51=king of spades)
Second, choose any card as the secret card and find the number of this card (after removing the other 4 cards from the ordering.) Call this number n.
Third, find the modulo 24 of n, call this m.
Four, construct the mth permutation of the remaining four cards from sorted order.
Fifth, if n >= 24 then hand the cards face up, otherwise hand them face down.
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
The latest Numberphile video brought up an interesting pattern in the prime numbers. Or, rather, in the list of sums of two successive primes. This list is:
5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, 78, 84, 90, 100, 112, 120, 128, 138, 144, 152, 162, 172, 186, 198, 204, 210, 216, 222, 240, 258, and so on.
Curiously, the segment 12, 18, 24, 30, 36, 42 consists of successive multiples of 6.
Then there are multiples of 6 at semi-regular intervals until we reach another surprisingly long segment of consecutive multiples of 6: 198, 204, 210, 216, 222.
(The next two numbers, 240 and 258, are also multiples of 6, but not the next consecutive ones after 222.)
I wonder if this is just coincidence.
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
Warp wrote:
The latest Numberphile video brought up an interesting pattern in the prime numbers. Or, rather, in the list of sums of two successive primes. This list is:
5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, 78, 84, 90, 100, 112, 120, 128, 138, 144, 152, 162, 172, 186, 198, 204, 210, 216, 222, 240, 258, and so on.
Curiously, the segment 12, 18, 24, 30, 36, 42 consists of successive multiples of 6.
Then there are multiples of 6 at semi-regular intervals until we reach another surprisingly long segment of consecutive multiples of 6: 198, 204, 210, 216, 222.
(The next two numbers, 240 and 258, are also multiples of 6, but not the next consecutive ones after 222.)
I wonder if this is just coincidence.
For p > 3 every prime is equal to 6k +- 1, and for small enough numbers a good portion of numbers that can be written this way are in fact primes. So these sequences just mean that 7 consecutive or so of such numbers happen to be primes.
For reference, the next higher (than six consecutive multiples) sequence can be seen starting with the 4004237th prime: 67944059, 67944073, 67944083, 67944091, 67944101, 67944109, 67944119, 67944127, 67944137, 67944143.
This gives a sum sequence of 135888132, 135888156, 135888174, 135888192, 135888210, 135888228, 135888246, 135888264, 135888280.
This is sequence of seven multiples of 18. (Their differences are 24, 18, 18, 18, 18, 18, 18, 16.)
Though there are a lot of six-consecutive ones before that. The multiples that come up are of 6 (only once), 12, 18, 24, 30, and 36.
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
You can add the final bit by handing the cards either face up or face down, which is very magic-trick-like, in that there's a trick. And that gives you the exact number of bits required to specify 1 of 48 unique cards.
There is a way to always correctly determine the final card solely from the four cards that were passed and their order. There's no additional trickery involved (handing the cards face up vs face down or providing additional information by other means). I guess there are probably multiple ways to do it, but the solution I saw is quite elegant.
You can add the final bit by handing the cards either face up or face down, which is very magic-trick-like, in that there's a trick. And that gives you the exact number of bits required to specify 1 of 48 unique cards.
There is a way to always correctly determine the final card solely from the four cards that were passed and their order. There's no additional trickery involved (handing the cards face up vs face down or providing additional information by other means). I guess there are probably multiple ways to do it, but the solution I saw is quite elegant.
(This is probably why mathematicians like setting up "prisoner problems", as opposed to "magic trick problems". Honesty can be enforced in the former case. :) )
This took me a while to figure out, but here's one way that isn't complicated and works with the playing cards specifically, while eradicating any form of trickery or deception (as in, no communication is allowed except to the extent spelled out by the given rules):
Out of 5 cards, there are at least two cards which are the same suit. Taking the ranks as the cycle A-2-3-4-5-6-7-8-9-10-J-Q-K-A-2-3-... , of two cards of the same suit, one of the cards is 6 or less steps ahead of the other in the cycle. So the strategy could be as follows:
First person: Out of the 5 cards given, take any two cards of the same suit. Keep the card which is n<=6 steps ahead of the other, and place the other card 1st in the set of four. Order the remaining three cards based on contract bridge: ranks from highest to lowest (A-K-Q-J-10-9-...-2) and if ranks are tied, suits from highest to lowest (spades-hearts-diamonds-clubs). Then arrange these cards in the 2nd to 4th slots indicating n as a lexicographic order:
low-mid-high: n=1
low-high-mid: n=2
mid-low-high: n=3
mid-high-low: n=4
high-low-mid: n=5
high-mid-low: n=6
Pass these cards to the second person.
Second person: Given the four cards, the card in the 1st slot indicates that the card kept by the first person is the same suit and 1-6 steps ahead in the cycle. The number of steps ahead is determined by the arrangements of cards in the 2nd to 4th slots above. Using this, the second person can determine the card held by the first person.
This is probably not optimal, in the sense that N=52 is not the highest number of cards for which there is a strategy that works. The optimal number is probably somewhat higher, but the strategy would be too complicated for a normal person to remember and it would take computation to figure out; it's similar to solving a covering problem.
Nice! That's exactly the intended solution. Apologies for my framing of the problem. I'll have to avoid words like "trick" and "magic" in the future :P
If anyone's interested I saw it at around 26:25 in this video
https://youtu.be/TCZ3YwbcDaw
Oh, I should have known it was a Mathologer video.
Apparently it's extremely well-known as "Fitch Cheney's Five-Card Trick". Furthermore I found out that not only is it possible to do for a deck of N=124 cards (the most obvious upper bound ensuring that, for N cards, there are enough ordered 4-card combinations to encode all unordered 5-card combinations), but it's easily proven AND there is an ingenious strategy that is reasonable to perform. It goes something like this (not a complete description, just brief):
Suppose the cards are numbered 0, 1, ... , 123 and the first person is given five cards n0, n1, n2, n3, n4 (assume it's in increasing order). Compute their sum mod 5 to be j, and keep the card nj. Then, given only what the other four cards are, there are exactly 24 possible values for the fifth card (by deleting the values of the four cards from the ordering and taking every fifth number); encode nj using the four cards and pass them to the second person. Then the second person just deduces what the fifth card is, from the four cards and their order.
Let x=2pi/17
Set
cos x + cos 4x = a
cos 2x + cos 8x = b
cos 3x + cos 5x = c
cos 6x + cos 7x = d
a + b = e
c + d = f
Evaluate cos x.
Posting the solution to this one. The trick is to solve a tower of quadratic equations. Let us first find a quadratic equation for e and f.
Notice that we can write cos nx = (enx+e(17-n)x)/2. From this, e+f is simply the sum of 16 17-th roots of unity, excluding 1. Since when we sum all roots of unity, we must obtain 0, we can conclude
e + f = -1/2
Now, we turn our attention to the product ef = (a+b)(c+d). So, in order to calculate it, we need to evaluate ac, ad, bc and bd. When we do this, we get terms of the form cos(nx)cos(mx). We can change it to a sum of cosines applying the relation cos(nx)cos(mx) = (cos(n+m)x + cos(n-m)x)/2
Our hope is that a miracle occurs, and when we sum all these terms together, they add up to something we can evaluate. Indeed, this miracle does occur.
2ac = cos 2x + cos 4x + cos 6x + cos 4x + cos x + cos 7x + cos x + cos 9x = 2a + b + d
2ad = cos 5x + cos 7x + cos 6x + cos 8x + cos 10x + cos 2x + cos 3x + cos 11x = 2d + b + c
2bc = cos 5x + cos x + cos 7x + cos 3x + cos 5x + cos 11x + cos 3x + cos 13x = 2c + a + d
2bd = cos 4x + cos 8x + cos 9x + cos 5x + cos 14x + cos 2x + cos x + cos 15x = 2b + a + c
So, ef = ac + ad + bc + bd = 2(a+b+c+d) = 2(e+f) = -1
Therefore, e and f are solutions of the equation y² -y/2 -1, y = (-1 +- sqrt(17))/4
To find which is which, we can compute everything numerically. Notice that e is positive, so we must have
e = (-1 + sqrt(17))/4, f = (-1 - sqrt(17))/4
Now we can evaluate a and b. We see that a + b = e and
2ab = cos x + cos 3x + cos 7x + cos 9x + cos 2x + cos 6x + cos 12x + cos 4x = a + b + c + d = e + f = -1/2
So, a and b are solutions of y² -ey -1/4. In a similar way, we can find cd = -1/4, so that c and d are roots of y² -fy -1/4. Solving the quadratic equations, and computing the expressions numerically to identify the correct sums we obtain, after long calculations:
8a = -1 + sqrt(17) + sqrt(34-2sqrt(17))
8b = -1 + sqrt(17) - sqrt(34-2sqrt(17))
8c = -1 - sqrt(17) + sqrt(34+2sqrt(17))
8d = -1 - sqrt(17) - sqrt(34+2sqrt(17))
Finally, since cos x + cos 4x = a, and cos x*cos 4x = (cos 3x + cos 5x)/2 = c/2. Therefore, cos x and cos 4x are both roots of y² -ay +c/2. Solving this, we pick the + sign for cos x, since it's clearly the larger one, and after writing all the square roots, we obtain the amazing formula
16*cos(2pi/17) = -1 + sqrt(17) + sqrt(34-2sqrt(17))+2sqrt(17+3sqrt(17)-sqrt(34-2sqrt(17))-2sqrt(34+2sqrt(17)))
The formula, since it writes cos(2pi/17) in terms of square roots, implies that the regular heptadecagon is constructible, a result first obtained by Gauss.
Notice that the problem could be solved because of the miraculous grouping of the cosine sums.
Perhaps this could be a subject of another question. How does one "know" one has to group the cosines that way to solve the problem? The answer can be found in Gauss' Disquisitiones Arithmeticae, for anyone who is curious...
For a project I would need to solve this:
Given x and a, solve y:
tan(y) * ((1 - a*sin(y)) / (1 + a*sin(y)) ^ (a/2) = x
If it's not possible with a closed-form expression, then an approximation (eg. using Newton's method) would be fine too.
For a project I would need to solve this:
Given x and a, solve y:
tan(y) * ((1 - a*sin(y)) / (1 + a*sin(y)) ^ (a/2) = x
If it's not possible with a closed-form expression, then an approximation (eg. using Newton's method) would be fine too.
Approximation is pretty much the only way I see.
You can replace sin(y)=z to get z/sqrt(1-z^2) * (1-az)/(1+az)^(a/2) = x and we need to solve it where |z|<=1
Assuming x is positive, we can take natural log of both sides:
ln(z) - (1/2)ln(1-z^2) + ln(1-az) - (a/2)ln(1+az) = ln(x)
Then if you want to use Newton's Method (assuming a good enough first guess):
next value = z - f(z)/f'(z)
where f(z) = ln(z) - (1/2)ln(1-z^2) + ln(1-az) - (a/2)ln(1+az) - ln(x)
and f'(z) = 1/z + z/(1-z^2) - a/(1-az) - (a^2/2) 1/(1+az)
How does one "know" one has to group the cosines that way to solve the problem? The answer can be found in Gauss' Disquisitiones Arithmeticae, for anyone who is curious...
Maybe I should answer this, here. The trick to find the value of cos(2pi/17) is actually a pretty straightforward application of concepts from number theory.
The idea is to look at F17, the finite field with 17 elements (it is indeed a field because 17 is a prime number). It turns out that this field is isomorphic to the 17-th roots of unity exp(2pi*n/17), for n ranging from 0 to 16, where multiplication in the roots of unity corresponds to addition in F17.
To solve the problem of determining the roots of unity algebraically, we only need to look at what would correspond to multiplication in F17. It turns out that F17 has another important property, its [ur=https://en.wikipedia.org/wiki/Unit_(ring_theory)#Group_of_unitsl]group of units[/url] has a primitive root, meaning that we can write any element as a power of a generator element. There are several possible choices, but we can take 3 as primitive unit.
If we write down 3, 3², 3³, 3⁴,..., we get a particular ordering of the elements of the units in F17.
3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4, 12, 2, 6, 1
If we map these to the roots of unity as n -> exp(2pi*n/17), notice that elements which are a distance of 8 apart are complex conjugates of each other, so if we sum them we obtain something proportional to the cosine. If we let x=2pi/17
S(1,8) = 2cos(3x)
S(2,8) = 2cos(8x)
S(3,8) = 2cos(7x)
S(4,8) = 2cos(4x)
S(5,8) = 2cos(5x)
S(6,8) = 2cos(2x)
S(7,8) = 2cos(6x)
S(8,8) = 2cos(x)
Similarly, we can compute sums of all elements a distance of four apart. Since there are 16 elements, there are four sums, each containing four terms. Alternatively, one could compute them by summing the sums of distance 8:
S(1,4)= 2(cos(3x)+cos(5x))
S(2,4)= 2(cos(2x)+cos(8x))
S(3,4)= 2(cos(6x)+cos(7x))
S(4,4)= 2(cos(x)+cos(4x))
We can do the same for distance 2:
S(1,2)=2(cos(3x)+cos(5x)+cos(6x)+cos(7x))
S(2,2)=2(cos(x)+cos(2x)+cos(4x)+cos(8x))
Now it should be easy to identify these as the variables a, b, c, d, e, f that helped us solve the problem. The pattern here is that any symmetric combination of all S(n,8) can be written in terms of S(m,4), and so on. Therefore, this method gives us a way to create a tower of quadratic equations to arrive at the value of each unit.
There is nothing that stops you from doing this for primes other than 17. For a prime p, you could construct sums of any value that divides p-1. However, if you want to find constructible polygons, you can only use quadratic equations. So, constructible polygons with a prime number of sides should have the form p=2n+1. We can refine this further by noticing that, if n has any odd factors, it's possible to factor p. Therefore, n should be a power of 2. From this, we can infer that regular polygons whose number of sides is a Fermat prime are constructible.
If one proves that if the previous method fails, there's no way to find a compass and straightedge construction, as was done by Wantzel, then we have found all possible regular constructible polygons!
Is it possible to have two functions f(x) and g(x) such that
lim(x->a) f(x) = 0,
lim(x->a) g(x) = 0
lim(x->a) f(x)g(x) = 0
and f(x) is not just the constant 0 (nor reduces to the constant zero, ie. eg. "f(x)=x-x" would be too boring of an answer)?
I asked this question many years ago, and one of the answers suggested was:
f(x) = exp(-x-4),
g(x) = x2
a = 0
I was satisfied with the answer back then. However, I recently realized that f(x) isn't actually smooth (at least in the sense that it has a discontinuity at the critical point we are interested in, ie. x = 0.)
I suppose I have to update my question a bit:
Is it possible to have two functions f(x) and g(x) such that
f(a) = 0,
g(a) = 0
lim(x->a) f(x)g(x) ≠ 1
with the conditions that f(x) cannot be the null function, and both functions must be smooth (or analytic, I'm not sure about the difference) around a? Ie. there cannot be any discontinuity in either function at x=a.
(I'm using the term "smooth function" meaning that the functions are continuous and all of their derivatives are also continuous at x=a. I'm not sure if the requirement that they be analytic is also required.)
Also, no case trickery (ie. "f(x) = { (some function) if x ≠ 0, 0 if x = 0 }".)
For reference, Warp's previous post on this question.
Analytic is actually important to this question. (A function f(x) is analytic at x=c if its Taylor series at x=c is equal to f(x) on some open interval containing c. It's stronger than just being "smooth", or infinitely differentiable.)
It turns out (see this article) that if f(x) and g(x) are analytic everywhere and not the zero function, with the limit as x--->0 of both functions being 0, then the limit of f^g must be 1.
If "analytic" is replaced with "infinitely differentiable", then the statement would be false. Ex:
f(x) = exp(-1/x^4) if x≠0, 0 if x=0,
g(x) = x^2
Both f(x) and g(x) are infinitely differentiable (including at x=0), limit as x--->0 of both functions is 0 and limit of f^g is 0. Note that f(x) is not analytic at x=0: its Taylor series is 0 at x=0 (all derivatives are 0 at x=0) but f(x) is not the zero function on any open interval containing 0.
However, there aren't any convenient ways to express functions that are infinitely differentiable but not analytic without using case notation. Rather than attempt to express such a function using some form of trickery that satisfies no one, I will just leave this question as I have already answered it.
It turns out (see this article) that if f(x) and g(x) are analytic everywhere and not the zero function, with the limit as x--->0 of both functions being 0, then the limit of f^g must be 1.
Dammit, I knew this to be true for some large family of functions, but for something like 30 years I couldn't prove (or disprove) it, or find the proper constraints or terminology, because I'm not a mathematical expert. Turns out it's a result known at least as far back as the 1970's.
When I was in high school (what now feels like an eternity ago), in calculus class the teacher presented us with the classical problem of finding the limit of xx when x approaches 0 (from the positive side). Since back then I was really into math, I found the proof that the limit is 1 to be marvelously interesting. (I have presented the question here too, years ago.)
I don't remember exactly when, but at some point I noticed that the same seemed to be true not only for xx, but in fact for a whole lot of different kinds of functions f(x) and g(x), for the limit of f(x)g(x) when both approach 0. I tried to come up with a proof of this, using similar techniques as with the proof of xx, but it wasn't as simple (back then I thought I had a solid proof, but it turned out later to be flawed and thus invalid. I don't think the result can be proven in the exact same manner as the more simplistic case.)
At university I presented the hypothesis to a maths professor who was immediately very dismissive of it, and in fact acted in a bit of a rude manner. He later apologized for his dismissive attitude and considered the hypothesis more seriously, and presented a counter-example. I don't remember what it was exactly, but it was a function that had a discontinuity at x=0, and he simply used a case definition (ie. like "f(x) { (that function) when x≠0, 0 when x=0 }").
I wasn't satisfied with that counter-example because it just felt like "cheating". Back then I didn't know the proper mathematical terminology for what I was trying to express. Asking around I came to the conclusion that what I was looking for were so-called "smooth" functions. Although, apparently, that isn't enough of a restriction either, and what I was actually looking for were analytic functions.
I just intuitively knew, all these decades, that there had to be a concrete family of functions for which that hypothesis is true, and it turns out that I was right, all these years. I wish I had known about that result a long time ago.
Btw, a somewhat related question:
I saw someone in another forum claim that this equality does not hold, and can't be assumed:
lim(x->c)(f(x)g(x)) = lim(x->c)(f(x))lim(x->c)(g(x))
Is that the case?
Btw, a somewhat related question:
I saw someone in another forum claim that this equality does not hold, and can't be assumed:
lim(x->c)(f(x)g(x)) = lim(x->c)(f(x))lim(x->c)(g(x))
Is that the case?
Yes, and I think we can use the same example.
f(x) = exp(-1/x^2)
g(x) = x^2
Then lim(x->0) [ f(x)^g(x) ] = lim(x->0) [ exp(-1) ] = 1/e
but
[ lim(x->0) f(x) ] ^ [ lim(x->0) g(x) ] = 0^0 is undefined
Yes, and I think we can use the same example.
f(x) = exp(-1/x^2)
g(x) = x^2
Then lim(x->0) [ f(x)^g(x) ] = lim(x->0) [ exp(-1) ] = 1/e
but
[ lim(x->0) f(x) ] ^ [ lim(x->0) g(x) ] = 0^0 is undefined
I don't think that's a valid argument because you could use the exact same argument to argue that the rule
lim(a/b) = lim(a) / lim(b)
is invalid (even though it's one of the basic laws of limits).
Yes, and I think we can use the same example.
f(x) = exp(-1/x^2)
g(x) = x^2
Then lim(x->0) [ f(x)^g(x) ] = lim(x->0) [ exp(-1) ] = 1/e
but
[ lim(x->0) f(x) ] ^ [ lim(x->0) g(x) ] = 0^0 is undefined
I don't think that's a valid argument because you could use the exact same argument to argue that the rule
lim(a/b) = lim(a) / lim(b)
is invalid (even though it's one of the basic laws of limits).
It's been a while since I've studied analysis, but that rule you mentioned has some assumptions behind it. I believe the simplest version goes: if lim(a) = A and lim(b) = B where B is not equal to 0 then lim(a/b) = A/B. It can probably be extended to cases where lim(a) or lim(b) = infinity, but you have to be careful with indeterminate forms.
As for one function to the power of another, we have
lim(x->c) f(x)^g(x) = lim(x->c) exp[ g(x)*ln(f(x)) ] = exp[ lim(x->c) g(x)*ln(f(x)) ]
where I've probably assumed f is positive and used the fact that exp is a continuous function. If the limits of g(x) and ln(f(x)) are nice then we can use the limit product rule to get the limit to be exp[ g(c)*ln(f(c)) ], which then unravels to get f(c)^g(c) as you wanted, but I think it's clear that there are several things that can go wrong, for example the limit of f at c may be 0, in which case you would end up with ln(0) and (at least the simplest version of) the limit product rule no longer applies.
So I think the answer is 'sometimes yes, but not in general'. If anyone has a more clear-cut answer that'd be nice.
Yes, I think that the "limit laws" do state that lim(A(x)/B(x)) = lim(A(x))/lim(B(x)) only if lim(B(x)) is not 0. In other words, the law is conditional.
However, the post I saw stated that lim(A(x)^B(x)) = lim(A(x))^lim(B(x)) can never be assumed, regardless of any conditionals, which I don't think is true. The only situation where it can't be assumed, I think, is when both limits are zero.
Warp, I think the disconnect you're having is that you're assuming tacitly that both limits will approach 0 at the same rate. But that is not necessarily the case.
Take a really stupid example where f(x) = e^x and g(x) = 1/x
Now it's clear that limit x->0 f(x)^g(x) is just e. However, the two limit case is different. To make this a bit more clear, I'm going to do a change of variables:
limit x->0 f(x)^(limit y->0 g(y)). Now it's clear that if we have a free variable k that relates x and y (y = kx), then we can make the two stage limit equal any complex number (given k complex), because the limit in that case will be e^k.
(In this case I also selected the indeterminate form 1^inf)
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