Pronouns: under/construction (idk what they are either); I like math and am pretty good at it, I am also very tired and I will be blaming math for that.
forall z contained in the circle C and functions "f" analytic over the closure of C
You do need to find the constant for the outer integral, but this should work.
The proof is pretty short, and is done using the Inverse function theorem.
Recommended preliminaries:
basic complex analysis is recomended
The most advanced thing that is really used is Cauchy's integral formula.
The inverse function theorem is as follows:
To begin The proof, let C be a circle in the complex plane. We then apply Cauchy's integral formula over the image of C under f to the above function:
Now, the integral over the image of a path under an analytic function equals the integral of the path through a modified integrand. This is essentially pre-applying chain rule to the path integral.
This allows us to rewrite our above expression:
From here we can trivially find the inverse to be the primitive of this function, yielding the formula:
QED
some notes:
We didn't actually need to use the inverse function theorem. As it turns out, this trick can be used for the case f^-1(z)=f^-1(z), yeilding
which, as far as I can tell, is still a further depressed version of the Lagrange inversion theorem. However, there is one strength the formula proven above: it lends itself to a very strong description of nonstandard inverse functions. When interpreted as a line integral, the primitive naturally keeps the structure of the inverse in mind, in a "Riemannian way," as far as I can tell. The latter formula doesn't immediately communicate such a property to me, and the Lagrange inversion theorem technically outputs a power series.
Thanks for reading, hope I didn't waste your time!
Yell at me if Iām being stupid plz. Also sorry if this is a bit trivial.
Sorry, I worded that weird. Itās not a special case. Itās actually just saying the inverse is equal to the inverse. If you use Cauchyās integral formula, over f(C), the you can reach that expression.
Yes, it is the one integral :D
I actually just found that version in my textbook (Conwayās functions of one complex variable), but the theorem is only listed āproposition 3.7,ā and stated without proof. Itās not even like āthe proof is an exercise, it just straight up gives it.
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forall z contained in the circle C and functions "f" analytic over the closure of C
You do need to find the constant for the outer integral, but this should work.
The proof is pretty short, and is done using the Inverse function theorem.
Recommended preliminaries:
basic complex analysis is recomended
The most advanced thing that is really used is Cauchy's integral formula.
The inverse function theorem is as follows:
To begin The proof, let C be a circle in the complex plane. We then apply Cauchy's integral formula over the image of C under f to the above function:
Now, the integral over the image of a path under an analytic function equals the integral of the path through a modified integrand. This is essentially pre-applying chain rule to the path integral.
This allows us to rewrite our above expression:
From here we can trivially find the inverse to be the primitive of this function, yielding the formula:
QED
some notes:
We didn't actually need to use the inverse function theorem. As it turns out, this trick can be used for the case f^-1(z)=f^-1(z), yeilding
which, as far as I can tell, is still a further depressed version of the Lagrange inversion theorem. However, there is one strength the formula proven above: it lends itself to a very strong description of nonstandard inverse functions. When interpreted as a line integral, the primitive naturally keeps the structure of the inverse in mind, in a "Riemannian way," as far as I can tell. The latter formula doesn't immediately communicate such a property to me, and the Lagrange inversion theorem technically outputs a power series.
Thanks for reading, hope I didn't waste your time!
Yell at me if Iām being stupid plz. Also sorry if this is a bit trivial.
In Magic: the Gathering, if you play with the original deck building rules it is possible to guarantee a win as soon as you first receive priority (ofc, as long as your opponent isnāt doing everything in their power to stop you), all it requires is to say no to gambling.
Original deck building rules:
A deck can have any number of copies of a card
40 card minimum
No restricted/banned list
We are allowing new cards to be used, not just Limited Edition Alpha.
Deck list (Thassaās Contract):
Elvish spirit guide x3
Manamorphose x2
Savage summoning x1
Thassaās oracle x1
Pact of negation x1
Contract from below x32
Game plan:
Before the game starts, deny ante. Contract from below states āremove this card from your deck if youāre not playing for ante.ā As far as I can tell, there are no stages in which you have to replace this card, and the rules are clear that ante canāt even be used during any kind of tournament play, and states you canāt if it breaks local gambling laws.
You will have at least 2 elvish spirit guides and a manamorphose in hand.
As soon as you receive priority, exile both elvish spirit guides for mana, to cast manamorphose, adding 2 blue mana and drawing the last remaining card in the deck.
At this stage you are guaranteed to have elvish spirit guide, Thassaās oracle, pact of negation, savage summoning, and a scrap manamorphose in hand. Exile the remaining spirit guide to cast savage summoning, which allows you to cast Thassaās oracle at instant speed. Which wins the game on the spot!
Iād love to say that pact of negation gives counter spell backup, but it doesnāt really matter since savage summoning is uncounterable and causes the Thoracle to be uncounterable. Itās really there to give you a 7/8 chance to play manamorphose with counter spell backup or deal with any funny business.
You lose if your opponent forces you to draw a card, at any point really, or if your opponent counters the manamorphose. But this deck beats chancellor of the dross x40, since you can respond to the turn 0 triggers, and even works against most other turn 0 wins.
I actually made a version that was sooo close to being uncounterable, but it evades me. I have what Iāve got so far below:
Deck list (doesnāt work ;-;):
Elvish spirit guide x2
Savage summoning x2
Krosan wayfarer x1
Thoracle x1
Lotus Vale x1
Contract from below x33
Game plan (although itās all for naught):
Deny ante.
Your starting hand is whatever in the list isnāt contract from below (I donāt wanna type that all out).
Exile an elvish spirit guides to cast savage summoning, and then exile the other to cast Krosan wayfarer, both uncounterable.
Sacrifice the wayfarer to play lotus vale. In response to the sacrifice trigger, tap lotus vale to add 1 green mana and 2 blue mana. Cast the other copy of savage summoning and then cast Thoracle and then WIN!
ā¦
And then remember that lotus vale can only add 1 color at a timeā¦
You rewind play to before you made the mistake, choose blue instead, pass priority, your opponent plays a turn 1 ragavan, passes the turn. You go to draw and lose on the spot, still with the taste of victory on your lips, although itās turning sour now. Oh, what a fool you were, to think you deserve such a glorious victory. At least you werenāt playing for ante. Thatās the only comfort you have, as you stare into the taunting figure of Ragavan. Itās a hollow comfort. They have the full art foil. It just adds insult to injury.
If anyone finds an uncounterable ver PLEASE TELL ME. Iām so sadā¦ itās 1 mana off.
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
ā Live Streamingā Interactive Chatā Private Showsā HD Qualityā Free Actions
Free to watch ā¢ No registration required ā¢ HD streaming
Have you ever wondered what a "half composition" of a function would be?
If we say a function "f" is composed with itself "k" times, we can intuitively interpret that as fāfā...k times (ie, the kth composition of f), so long as k is an integer; but what about when k isn't an integer?
We could interpret the "a/b th composition of f," as some function "g", such that, when g is composed "b" times with itself, it is equal to f composed "a" times with itself. While this interpretation is sound it still begs a few questions: is there an intuitive way to think of "real" compositions, what about complex?
These questions are interesting, however, I feel they are dwarfed by one you could be considering right now:
If we have some smooth function f, can we find functions that are these "exotic composites" of f? Can we do so for any smooth function?
Recommended preliminaries:
A good understanding of taylor expansions is very necessary.
Familiarity with ODEs would be helpful, but they are not used directly.
I had initially planned to include alt text for formulas but after much trial and error, I have given up. If anyone finds this interesting but was unable to engage with the content properly due to the lack of alt text I would be (genuinely) happy to converse about the subject via dm.
We will be breaking this into two steps:
Determining the a first order approximation of the desired function.
Finding higher order derivatives of such a function function.
We will be using the following notation to denote multiple compositions, as well as preserving the following homomorphism:
Theorem 1:
given a smooth function f:A->A, that fixes for some value x_f in A (ie, f(x_f)=x_f), then:
proof: If we repeatedly apply chain rule we can form the following product:
Clearly, this product can't be made to work for all values of s naturally, as that would inherently require some non-integer composition, but what is well understood is extending the product of a constant.
If we choose to evaluate this at x=x_f a peculiarity arises. Compositions of fixed points do not change, that is f(f(x_f))=f(x_f)=x_f. This allows the product to be simplified as follows:
we can then allow s to vary beyond integer values. QED
Now we could generate an approximation of our desired function nearby the fix point, however, this is fairly pointless.
one could notice that by repeatedly differentiating the initial product, you can actually continue to get higher order approximations, by applying the definition of the geometric series whenever nessesary. However, this endeavor is Extremely computationally intense; and no obvious pattern immediately discern itself (at least for me, I would invite anyone with blatant disregard for their own mental well being to try to find one directly).
Instead, we will be constructing an alternate method of improving our approximation.
Theorem 2:
For a given smooth function f:A->A, with some fix point x_f, and for some integer m>0, the following holds:
where B_{m,k} denote bell polynomial coefficients. (Just to be clear, the "mathcal B" defined for shorthand, it is not related to the statement directly.)
proof: Fa di Bruno's theorem is a generalization of the chain rule to higher order derivatives. It states the following:
where B_{m,k} are bell polynomial coefficients with the derivatives of g as "inputs".
However, rather than let this define the mth derivative of a composition of different functions, we instead use it to take the mth derivative of the s+1th composition of f with itself:
Note that the final term of this sum contains the mth derivative of the sth composition of f, making this expression recursively defined. We rectify this by subtracting this problem term from both sides (also, we begin to evaluate at the fix point, as it allows some slight clean up of the expression).
While this does get all unknowns on the same side of the expression, this is still unsolved.
To begin to solve this we will attempt to factor the "left side." There is a linear operator T_s called the "forward operator" defined so that T_sF(s)=F(s+1). We will simply take it for granted that this operator commutes with differential operators, but there are many proofs of this fact. This allows us to modify the above to a more neat form (the "right side" is unchanged):
our motivation for this simplification will be apparent momentarily. First, we will be applying the inverse of T_s to both sides (ie, the "backwards operator" T_s^-1F(s)=F(s-1)). We can directly apply it to the right by replacing all s with s-1.
(Here we use a slight abuse of notation, as the "1" on the left should in reality be the identity operator "I," however it is not of consequence.)
Note that the expression on the "left" consists of a linear operator L=(1-[f(x_f)]^mT_s^-1) next to an expression. Now, if we where to naively attempt to invert this operator we would have the reciprocal expression in the form 1/(1-L). This is where we use the definition of the geometric series.
We say this series approaches the desired operator so long as it converges (the convergence condition is noted below it). If this is unfamiliar, I would recommend really trying to grasp it (it's a rather interesting concept; try to apply this to (1-L)g(s) and see what happens I would also recommend replacing the main in the sum with L^l).
If we apply this inverse operator to both sides we can reach a formula explicitly for the derivative, assuming it converges.
Since m is finite we can exchange these sums without issue. I'm going to avoid specifics, but we can arrange to reach the final formula. introducing the shorthand.
Substitute the expression on the left side into the convergence condition found before to reach the above convergence condition. note that for any chosen m this expression only references derivatives less than m, which allows us to recursively apply this without issue.
To reach the expression in the beginning simply replace m with m+1 in this formula. QED
This formula allows us to find all derivatives of the desired function so long as the expression converges and the function fixes. The convergence condition is a little rough to determine, however from what I've found it is sufficient that |f'(x_f)|<1 (that proof was more arduous than the one above, so I have omitted it, more for my sake than yours); which is rather restrictive, however, if this condition is not fulfilled, you may simply invert the function on a nearby interval and the derivative must necessarily have a reciprocal magnitude by the inverse function theorem; leaving out only the case where |f'(x_f)|=1
Using theorems 1 and 2, we can obviously create a taylor expansion. Letting the following hold (yes I'm restating the short hand again; sue me):
"Most" differentiable functions defined on f:C->C, have at least one fix point, so as long as you aren't "unlucky" this should work.
A few notes I think are important:
is this the best way to do this?
Probably not. I was looking at math overflow for information, and I heard vague talk about a professor that had solutions to this very problem that worked for |f'(x_f)|=1, but I couldn't find anything substantial at the time. From the discussion I saw I was getting the vibe it was never published so idk.
I couldn't even begin to consider how my formula fairs in terms of convergence speed.
2. Is this even right?
Uhhh, I think; but idk I guess. I don't see much why it isn't correct, and like, I did just go over the proof, so if you think I'm wrong just like... idk what to say (ā _ ā ) sry. The formula is disastrously hard to test so I just haven't tested it; if anyone wants to test it feel free. Just know if it isn't working you're actually wrong and just need more terms (įµ ā¢ į“ ā¢) yeah, yeah...
If you do find out I'm wrong I actually would like to know.
3. How long did it take to find this solution?
... I don't wanna talk about that... then why did I bring it up??
One issue here I can see is that you assume that if you have a formula for f*s(x_f) and its derivatives at integer values of s that can be extended to the reals, that it does extend to the reals. Unless you prove continuity with respect to s the non integer values could be unrelated to the integer values. Assuming you do (you probably can), you need to prove that f*s(x_f) is analytic as a function of s (you cannot prove this by the formula for integer values being analytic when extended to reals) in order to say that the analytic continuation of the integer values is equal to the true function.
You probably can do so, but youāre missing important details.
I left out the details on the first point because I thought it might be unnecessary, but the proof does seem incomplete without it:
Letās consider a point āpā in A a very small distance ārā away from the fix point. Now, if we apply f to p and x_f, x_f clearly does not move, however f(p) is not p, ie it āmoves.ā The real question is āby how much?ā That is totally dependent on the distance r. As we let the distance r grow very small, it is clear that this point p is moved closer to x_f+fā(x_f)r. This is because our first order Taylor expansion about x_f is: x_f-(x-x_f)fā(x_f). This kinda feel like where not actually getting anywhere, but the point is if we substitute this approximation into itself repeatedly we at first can say we seem to get x_f-(x-x_f)[fā(x_f)]^s. Now why is this any better evidence? For linear functions, composition can already be extended, and in exactly this fashion (you can check from here that the homomorphism is held, which does mean that the expression is valid; but uniqueness is uncertain).
The proof that I placed up there is essentially the veiled version of this logic: composing an approximation with certain properties with itself that are certainly preserved.
Tbh, I thought the second point didnāt have much merit, however, I double checked and, it is very important to consider. The entire formulation does not explicitly reference the differentiability of f*s(x) with respect to s anywhere, and almost none of the intermediate steps actually require analyticity in terms of s, only that f*s(x) exists. That is, all but commuting the forward operator and the differentiation operator.
The quickest proof that the forward operator and derivative commute uses the identity e^{partial_s}=T_s, which clearly commutes with other partial derivatives, unless the chosen function is not smooth.
But I do think that this is the only point of issue, as nothing else obviously requires f*s(x) be analytic in terms of s, it all more acts as a proof that it must be (which can be see via and inductive argument).
I think this might be due to a uniqueness issue. If we assume f*s(x) must be analytic, the formulation continues without issue and leads to a result without obvious contradiction. This result should have all the desired properties regardless of the fact that it was constrained. It is possible that there is an f*s(x) that is not analytic, but those answers are not considered.
I think the point you were trying to make was slightly different, but itās all about deciding what versions of our function we are considering. I should probably prove that a valid analytic function exists; but thatāll take me more time. I might post it later.
If this doesnāt seem like enough, or if anyone wants to take a stab at the proof just lmk!
Have you ever wondered what a "half composition" of a function would be?
If we say a function "f" is composed with itself "k" times, we can intuitively interpret that as fāfā...k times (ie, the kth composition of f), so long as k is an integer; but what about when k isn't an integer?
We could interpret the "a/b th composition of f," as some function "g", such that, when g is composed "b" times with itself, it is equal to f composed "a" times with itself. While this interpretation is sound it still begs a few questions: is there an intuitive way to think of "real" compositions, what about complex?
These questions are interesting, however, I feel they are dwarfed by one you could be considering right now:
If we have some smooth function f, can we find functions that are these "exotic composites" of f? Can we do so for any smooth function?
Recommended preliminaries:
A good understanding of taylor expansions is very necessary.
Familiarity with ODEs would be helpful, but they are not used directly.
I had initially planned to include alt text for formulas but after much trial and error, I have given up. If anyone finds this interesting but was unable to engage with the content properly due to the lack of alt text I would be (genuinely) happy to converse about the subject via dm.
We will be breaking this into two steps:
Determining the a first order approximation of the desired function.
Finding higher order derivatives of such a function function.
We will be using the following notation to denote multiple compositions, as well as preserving the following homomorphism:
Theorem 1:
given a smooth function f:A->A, that fixes for some value x_f in A (ie, f(x_f)=x_f), then:
proof: If we repeatedly apply chain rule we can form the following product:
Clearly, this product can't be made to work for all values of s naturally, as that would inherently require some non-integer composition, but what is well understood is extending the product of a constant.
If we choose to evaluate this at x=x_f a peculiarity arises. Compositions of fixed points do not change, that is f(f(x_f))=f(x_f)=x_f. This allows the product to be simplified as follows:
we can then allow s to vary beyond integer values. QED
Now we could generate an approximation of our desired function nearby the fix point, however, this is fairly pointless.
one could notice that by repeatedly differentiating the initial product, you can actually continue to get higher order approximations, by applying the definition of the geometric series whenever nessesary. However, this endeavor is Extremely computationally intense; and no obvious pattern immediately discern itself (at least for me, I would invite anyone with blatant disregard for their own mental well being to try to find one directly).
Instead, we will be constructing an alternate method of improving our approximation.
Theorem 2:
For a given smooth function f:A->A, with some fix point x_f, and for some integer m>0, the following holds:
where B_{m,k} denote bell polynomial coefficients. (Just to be clear, the "mathcal B" defined for shorthand, it is not related to the statement directly.)
proof: Fa di Bruno's theorem is a generalization of the chain rule to higher order derivatives. It states the following:
where B_{m,k} are bell polynomial coefficients with the derivatives of g as "inputs".
However, rather than let this define the mth derivative of a composition of different functions, we instead use it to take the mth derivative of the s+1th composition of f with itself:
Note that the final term of this sum contains the mth derivative of the sth composition of f, making this expression recursively defined. We rectify this by subtracting this problem term from both sides (also, we begin to evaluate at the fix point, as it allows some slight clean up of the expression).
While this does get all unknowns on the same side of the expression, this is still unsolved.
To begin to solve this we will attempt to factor the "left side." There is a linear operator T_s called the "forward operator" defined so that T_sF(s)=F(s+1). We will simply take it for granted that this operator commutes with differential operators, but there are many proofs of this fact. This allows us to modify the above to a more neat form (the "right side" is unchanged):
our motivation for this simplification will be apparent momentarily. First, we will be applying the inverse of T_s to both sides (ie, the "backwards operator" T_s^-1F(s)=F(s-1)). We can directly apply it to the right by replacing all s with s-1.
(Here we use a slight abuse of notation, as the "1" on the left should in reality be the identity operator "I," however it is not of consequence.)
Note that the expression on the "left" consists of a linear operator L=(1-[f(x_f)]^mT_s^-1) next to an expression. Now, if we where to naively attempt to invert this operator we would have the reciprocal expression in the form 1/(1-L). This is where we use the definition of the geometric series.
We say this series approaches the desired operator so long as it converges (the convergence condition is noted below it). If this is unfamiliar, I would recommend really trying to grasp it (it's a rather interesting concept; try to apply this to (1-L)g(s) and see what happens I would also recommend replacing the main in the sum with L^l).
If we apply this inverse operator to both sides we can reach a formula explicitly for the derivative, assuming it converges.
Since m is finite we can exchange these sums without issue. I'm going to avoid specifics, but we can arrange to reach the final formula. introducing the shorthand.
Substitute the expression on the left side into the convergence condition found before to reach the above convergence condition. note that for any chosen m this expression only references derivatives less than m, which allows us to recursively apply this without issue.
To reach the expression in the beginning simply replace m with m+1 in this formula. QED
This formula allows us to find all derivatives of the desired function so long as the expression converges and the function fixes. The convergence condition is a little rough to determine, however from what I've found it is sufficient that |f'(x_f)|<1 (that proof was more arduous than the one above, so I have omitted it, more for my sake than yours); which is rather restrictive, however, if this condition is not fulfilled, you may simply invert the function on a nearby interval and the derivative must necessarily have a reciprocal magnitude by the inverse function theorem; leaving out only the case where |f'(x_f)|=1
Using theorems 1 and 2, we can obviously create a taylor expansion. Letting the following hold (yes I'm restating the short hand again; sue me):
"Most" differentiable functions defined on f:C->C, have at least one fix point, so as long as you aren't "unlucky" this should work.
A few notes I think are important:
is this the best way to do this?
Probably not. I was looking at math overflow for information, and I heard vague talk about a professor that had solutions to this very problem that worked for |f'(x_f)|=1, but I couldn't find anything substantial at the time. From the discussion I saw I was getting the vibe it was never published so idk.
I couldn't even begin to consider how my formula fairs in terms of convergence speed.
2. Is this even right?
Uhhh, I think; but idk I guess. I don't see much why it isn't correct, and like, I did just go over the proof, so if you think I'm wrong just like... idk what to say (ā _ ā ) sry. The formula is disastrously hard to test so I just haven't tested it; if anyone wants to test it feel free. Just know if it isn't working you're actually wrong and just need more terms (įµ ā¢ į“ ā¢) yeah, yeah...
If you do find out I'm wrong I actually would like to know.
3. How long did it take to find this solution?
... I don't wanna talk about that... then why did I bring it up??
Extension of the Fibonacci #s via Composition of Functions
There are plenty of ways to derive the extension of the Fibonacci numbers to the reals. But one way is a personal favorite for me.
We start by defining the following function (1):
And we let the following notation denote repeated compositions (f:A->A) (2):
We can (by inspection, Iām not proving this) see the first output of repeated compositions of F output the Fibonacci numbers (starting with input of x_0=(1,0)). (3):
We want to determine the Jacobian of the composition of F with itself s times. The Jacobian of a composition of functions can be determined using the chain rule repeated. In this case we have (4):
However, note that for this specific F, the Jacobian is constant, thus (5):
This matrix has an Eigendecomposition, and thus can have its exponent extended to the reals (6):
Since its higher order derivatives are zero and F(0)=0, we can simply say F(x)=Jx. Thus if you had the patience to multiply the matrices (or access to a good enough algebra software) you can substitute the above into (3) to reach (7):
My general thought process in this derivation is currently my banner image at time of posting this. This one started because I got stranded somewhere, and I had chalk.
If I made a mistake remember to kill me with hammers. But I seriously do want to know, I checked the matrix product a while back with an algebra software to see if it was right, but I might have changed something since then.
That wraps up my first ārealā mathblr post. Toodles!
So from the action of (1) and following to (5) should not that matrix be ((1 1), (1 0)), since acting F on the vector in (1), one would want the x1 to carry over. Though I may be misunderstanding how you arrived at that matrix.
Either way, I was able to reproduce the very left hand matrix in (6), but I have the columns swapped. I'm pretty sure that's because of the differing matrices, I continued on with the one I mentioned before.
But so from there, deriving the very right hand matrix in (6), I'm not entirely sure how you derived it if not from the shortcut for getting the inverse of a 2x2 or row reduction. I assume some of my confusion stems from some trickery of the golden ratio, always able to rewrite it into slightly more convenient forms, but regardless of that, it looks as if you calculated the determinant of the left hand matrix to be 2+phi, but I calculated 1-2phi. Well, really I got 2phi-1, but due to the column swapping I know it differs by a minus sign.
But so 1-2phi is sqrt5, which is what you have in the denominator in (7), which is not equal to 2+phi, so I don't know if I'm misunderstanding something or there was a typo at some point or something. At the end, using your matrix for F in (5) I got the same as your (7), if I understand the result is basically just the upper right hand of the matrix representing the Jacobian of Fs? So as far as checking your matric calculator, checked!
But using the different matrix that I suggested I have basically the same thing, but a minus sign on the second term, so [phi^s-(-phi)^(-s)]/sqrt5.
I was curious, so I plotted these, at lest the best a real number plotter would do with that -phi, and both your solution and the one I got lie on the curves for cosh(s*ln(phi)) and sinh(s*ln(phi)) That makes sense of course since that essentially looks like both solutions, just depending on the fractional representation of whatever decimal is being input for s.
I guess I really only ended up having the one question to ask about how you arrived at your matrix representation for JF. Anyway, I hope I'm not being too forward in typing all this, this is a really fun problem to think through!
I looked at the photo of the blackboard I did this on and I totally got the Jacobian flipped when translating it over in (5) ;-; and I straight up am inncorrect in (7) T_T.
Regardless my work that reaches (6) uses the correct matrix, so if yours is different I would have to imagine that either I made a mistake or that when you made your Eigendecomposition, you chose different columns to have the chosen eigenvalues.
Also in (6) to invert the matrix I augmented it with the identity matrix and did row operations (I had limited internet and donāt generally memorize shortcuts), and then I simplified (using golden ratio magic) till I thought it looked good enough. I think in the process of simplifying I lost a ^{-1} exponent in the numerator of the top right entry of the far right matrix.
For (7), the known closed form solution is a difference in the numerator (which you have), and I donāt have access to the program that I made do my algebra, but I remember at least trying to check it.
In future I will not be posting math at; I think it was ~1:00?
Thanks for catching me, working with actual examples is a weakness of mine.
Anya is live and ready to show you everything. Watch her strip, dance, and perform exclusive shows just for you. Interact in real-time and make your fantasies come true.
ā Live Streamingā Interactive Chatā Private Showsā HD Qualityā Free Actions
Free to watch ā¢ No registration required ā¢ HD streaming
Has a game ever genuinely changed how you think about something?
Feed your dashboard by answering my question, blogger.
I used to play factory sims, primarily satisfactory. While I think the game is wonderful, I didnāt like it (I want it clear that I think itās a good game, not my cup of tea). But I would play it anyway. I would put hours into that game, and I felt each hour was wasted. After so many hours is when I realized that if Iām not having fun, that isnāt an absence of some QoL feature, that means I shouldnāt be playing the game.
That kind of started a wave (or perhaps was caused by some wave) of me realizing that I should do what I enjoy. It was such a bizarre thought: that I was allowed to enjoy myself. I even still wonder why I ever thought otherwise, or why it often feels like this thought still looms over me.
Iām pretty much over playing games I donāt enjoy, but Iāve found it leads to me to play games very little, and I think itās just a little sad in that sense.
Water is actually super dangerous, consider the facts! (NOT PROFESSIONAL ADVICE)
Water is an acid so strong that itās commonly referred to as āthe universal solvent!ā It has a PH of 7, the highest PH of ANY ACID! Thatās more than TWICE the PH of hydrochloric acid!
Water is used in all kind of TOXIC products. Antifreeze is largely made up of water, as well as many types of chemical drain cleaners! That still isnāt all that much to go off of, for all we know water isnāt even doing the harm in those products! But thereās more:
After putting all of this to the side we still have chemistry 101 safety to think about. Your stomach is filled with acid used to dissolve food. When Diluting an acid it is paramount to add the acid to the water; āyou know that you otter š¦¦ add acid to water.ā When water is added to an acid, it generates large amounts of heat and since the acid typically has a lower boiling point than water, can begin to boil! I donāt need to explain why having a boiling pot of acid in your belly is a bad thing! Thus if you insist on keeping the bad habit of drinking water, basic chemistry seems to suggests that you should extract the acid from your stomach (in a controlled manner, else you can damage your esophagus) and then drink the water, and then, slowly add the acid back to your stomach.
Water is used EVERYWHERE in Nuclear power plants! Itās essentially the power source (ie, they actually allow for the energy to be transferred)! It is also are used as a āneutron moderator,ā in which it is able to absorb neutrons and become radioactive (nuclear power plant water is extensively tested before it is allowed to be used for anything else, please donāt defund nuclear power). If a Hydrogen atom that makes up the water absorbs an electron it can become deuterium, which is completely stable, but sounds scary. But if the exact same hydrogen atom where to absorb another neutron, then it would become tritium a kinda radioactive isotope of hydrogen. Tritium is generally safe outside of the body, but if ingested can wreak havoc on the body for as many as 14 days!
But this is all chemistry, not real science like āsafety engineering.ā
The CDC claims that out of all hospitalized burn treatments from 2001-2006, somewhere between 33-58% of those burns where Scalds, ie, vapor based heat transfer, ie, VERY LIKELY WATER BASED. Now they donāt actually ever say that it was water, but, like what else would it be?
Iām not even going to begin to cover how a moist environment can promote bacteria growth, or how nearly 100% of boat incidents are on or near water, or even how water vapor high in the air can cause lighting bolts! And donāt even get me started on Tsunamis!
I think the message is clear, just DONāT DRINK WATER. If you feel the urge to drink water, remember, based on the above information, itās probably safer just to drink coffee or juice or something, rather than take the risk and drink water.
TLDR;
According to basic chemistry, drinking water might give you a tummy ache and is used in things that are bad for you. Also sometimes things happen to make water bad for you. So never ever drink water. Drink like tea or something, idk just not water.
Extension of the Fibonacci #s via Composition of Functions
There are plenty of ways to derive the extension of the Fibonacci numbers to the reals. But one way is a personal favorite for me.
We start by defining the following function (1):
And we let the following notation denote repeated compositions (f:A->A) (2):
We can (by inspection, Iām not proving this) see the first output of repeated compositions of F output the Fibonacci numbers (starting with input of x_0=(1,0)). (3):
We want to determine the Jacobian of the composition of F with itself s times. The Jacobian of a composition of functions can be determined using the chain rule repeated. In this case we have (4):
However, note that for this specific F, the Jacobian is constant, thus (5):
This matrix has an Eigendecomposition, and thus can have its exponent extended to the reals (6):
Since its higher order derivatives are zero and F(0)=0, we can simply say F(x)=Jx. Thus if you had the patience to multiply the matrices (or access to a good enough algebra software) you can substitute the above into (3) to reach (7):
My general thought process in this derivation is currently my banner image at time of posting this. This one started because I got stranded somewhere, and I had chalk.
If I made a mistake remember to kill me with hammers. But I seriously do want to know, I checked the matrix product a while back with an algebra software to see if it was right, but I might have changed something since then.
That wraps up my first ārealā mathblr post. Toodles!
I donāt know all that much math, the highest level course Iāve formally taken so far was an ODE course, but my school doesnāt have anything higher so Iām just getting credits for subjects I already know this semester.
Iāve been reading analysis books, and I think it is the favorite so far (subject to change). I feel a bit out of depth, but I think Iām getting enough out of it anyway.
I havenāt interacted with much else. I tried to look into some topology but the book I got was very boring. Iāve been reading a lot more, so If anyone has any (math) book recommendations Iād love to hear them.
ā¦..not even six hours later i got an offer of a well paying full time long-term job with free room and board in queens in nyc, allowing me independence and a way to escape an abusive situation and an unhealthy environment
likes charge reblogs cast, folks, this is the good luck post
the last time I reblogged this post right before I got a great job, in a permanent work-from-home position, with benefits, retirement, and a salary literally 3x what I was making before, doing something I really like.Ā
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Here's a fun little math puzzle, not too hard. Classify all mereomorphic functions f such that for some point p f misses p except possibly at p in which case f(p)=p
I gave it a go bc Iāve been studying complex analysis and Iām not all that confident (please let me know if I made mistakes):
Iām assuming āmisses pā means there dne t in C that is not equal to p such that f(t)=p. With that definition we can state f(z)=p is a condition we disallow.
Since p is technically an entire function f(z)-p is meromorphic. Any meromorphic function can be expressed as a ratio of entire functions, which in turn, may be factored using the Weirstreiss factorization theorem. If the function in the numerator has a root f(z)-p must have a root thus breaking the condition. Skipping a few steps using the weirstreiss factorization, we should be able to reach f(z)=p+(1/g(z)), where g is an entire function.
If we want to specifically make the exception for p to be a fix point of f, it is first necessary that g does not have a root at p and second we must have some factor that goes to zero in the numerator iff z=p. f(z)=p+(z/p - 1)^m / g(z), should be sufficient, assuming m is a natural number (including zero) and g is an entire function without a root at p.
I stand in the wind and I let the moments wash over me. Cold and biting, hiding me from my own warmth.
As I stand dormant, my emotions are greying. Resigned to my torment, I feel Iām fraying.
I seek no shelter, there is none that I see, spare the grass that stands here with me. Itās insufficientā Iām insufficient, insufficient to stand in this field.
