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Watno

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Re: Maths thread.
« Reply #800 on: March 12, 2017, 07:34:12 am »
+1

It is not possible to regulate exaxctly what sets exists. Any sufficiently useful axiom system will have some statements it can't decide.
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silverspawn

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Re: Maths thread.
« Reply #801 on: March 12, 2017, 08:43:41 am »
0

I mean, sure you have statements you can't decide. That just means that there are sets that you can't prove exist. You could define "a set exists iff it can be constructed by our axioms" which is how the lecture I read makes it sound like, and have sets where you aren't sure.

liopoil

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Re: Maths thread.
« Reply #802 on: March 12, 2017, 05:08:45 pm »
+1

No, the point is so that you know some sets that DO exist. There will always be sets that might or might not exist, e.g. continuum hypothesis.

EDIT: It also might be undecidable whether a set can be constructed from our axioms or not, so you will still have sets that might or might not exist
« Last Edit: March 12, 2017, 05:11:41 pm by liopoil »
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Re: Maths thread.
« Reply #803 on: March 15, 2017, 08:40:18 am »
0

Okay, so the claim of the axioms is not "these are the sets that exist," but "these are some sets that exist" and then the axiom of foundation says "these are some sets that don't exist," and every set that can't be constructed from the axiom and doesn't contradict the axiom of foundation is up for debate?

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Re: Maths thread.
« Reply #804 on: March 16, 2017, 02:16:13 pm »
0

I've just seen this pop up and will definitely have stuff to contribute.  Can't right now though. :)
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Re: Maths thread.
« Reply #805 on: March 16, 2017, 02:27:10 pm »
+1

I would say statements about the existence of sets are not different from any other statements with regard to their relation to axioms (except that nearly every statement is a statement about the existence of sets in a way).

Note that there is no statement like "this sets exists" in the language of set theory, only "a set with this property exists". (For example "the empty set exists" is not a set-theoretic statement. "There is a set x such that no set y is an element x" is.)
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Re: Maths thread.
« Reply #806 on: March 16, 2017, 06:13:51 pm »
+3

So I finished the set theory lecture a while ago (and read a bit on wikipedia). I got a few questions left, if anyone wants to help. (I got the sense that there are actually a few different models which work slightly differently, so this is about ZFC).

First (this was not really explored in the lecture, I just tried to gather it from thinking & the internet), do I understand it correctly that

ℕ =: ω = {0, 1, 2, ...}

aka the natural numbers and the smallest infinite ordinal. Then the next "bigger" ordinals are

ω+ = ω ∪ {ω} = {0, 1, 2, ..., ω}
ω++ = {0, 1, 2, ..., ω, ω+}
...

and there exists

α := ω ∪ {ω, ω+, ω++, ...} = {0, 1, 2, ..., ω, ω+, ω++, ...}

and then

α+ = α ∪ {α} = {0, 1, 2, ..., ω, ω+, ω++, ... α}
α++ = {0, 1, 2, ..., ω, ω+, ω++, ... α, α+}
...

and there exists

β := {0, 1, 2, ..., ω, ω+, ω++, ... α, α+, α++, ...}

and

γ := {0, 1, 2, ..., ω, ω+, ω++, ... α, α+, α++, ..., β, β+, β++, ...}

and then you can play that game forever and can keep getting ordinals that way, and the ordinal which is the union of all such ordinals is ω1, which is kind of like ℕ x ℕ and therefore still countable in the same way (but the first non-recursive ordinal), so it's still ω1 ~ ℕ and then there exist ω1+, ω1++, ω1+++, ω1 ∪ {ω1, ω1+, ω1++, ...} := Ψ and Ψ+, Ψ++, ω1 ∪ {Ψ, Ψ+, Ψ++}, and this can be done arbitrarily often again, until we get something which is kind of like ℕ x ℕ x ℕ which is still countable, let's call it Φ and then building from Φ you could construct something like ℕ x ℕ x ℕ x ℕ which would still be countable, and if you do that arbitrarily often, then you have the union of all such ordinals, which is also the union of all countable ordinals and is kind of like ℕ x ℕ x ℕ x ℕ ... and called ω0 and the first uncountable ordinal. Is that all correct?

(It's pretty weird. I like it. If it actually works that way.)
This is a pretty good summary, yes.  You have to be a bit careful with the first uncountable ordinal, in that pretty much any union you can physically write down is going to be smaller than that ("union of all countables" works fine of course).  Other than that, yes, you have the idea.

More general picture:


It is important to remember that set theory is not trying to be a 100% precise description of the mathematical world.  Maybe that is what the original set theorists wanted, but the reality is that that's not feasible.  Instead, set theory is trying to write down a decent approximation of what we believe the mathematical world to be.  This, silver, is where your mild confusion arises, I believe.

As logicians, we write down some number of axioms that we believe should hold in mathematics (Foundation is a great example - we take it as an axiom because, as mathematicians, we do not want to deal with sets that might be infinitely nested or horribly ugly in other ways.  It is a statement of our belief that mathematics should be well-founded on a nice clean basis without infinite recursions.  But you don't have to take Foundation - indeed, I believe - though don't fully understand - that computer scientists often like to use set theory without Foundation, because it suits their theoretical basis somehow.).    But we do not expect - indeed we know that it is impossible - to write down a set of axioms that covers absolutely every facet of possibility.   

At this point you need some grounding in Logic/Model Theory,  The idea is: we have written down some axioms.  But since our axioms are not complete (and indeed they cannot be), there are bound to be lots of different structures that satisfy those axioms.  In other words, there are lots of "models of ZFC"  (or at least we believe that there should be).  Here, "model of ZFC" means "collection of things/sets that satisfies all the axioms".  There might well be lots of different models of ZFC, and we want to believe that the "true" mathematical world is one of those models - but we don't know which one. 

So, to your questions, if you ask "does a set with *some weird property* exist?", if it's not proven or disproven by the axioms, the answer will be:
"Some models of ZFC contains such a set, and some do not."  We haven't decided yet, as mathematicians, which of the various models of ZFC is the "right" one - if we ever do, that would be tantamount to adding an additional axiom.  (For instance we might take GCH as an axiom - though this is unlikely to happen any time soon.).

This goes into the scope of some slightly deeper theory (can send you some notes on that if you like, silver).  Foundation is once again a good example.  It is proven that the axiom of Foundation is independent of the other axioms (excluding Choice).  How do we prove this? "Easy". We write down two structures.  One of them satisfies all of the axioms including Foundation (still ignoring Choice), and one satisfies all of the axioms except it DOESN'T satisfy Foundation. 
So, if we decided we weren't sure about the Axiom of Foundation any more, then we would have at least two possible models of Set Theory - and we'd have to accept both as equally valid (though one contains some infinitely nested sets, and the other doesn't).

No idea whether this makes things clearer.  Anyhoo, general abstract chat over, wanted to reply to some quotes:
Second, if this is all grounded in this set language and can all be done formally, why aren't all mathematical theorems just provable or disprovable through some software which handles First-order logic?
Even if computers were any good at producing proofs (they're not) - there's the issue that not every mathematical idea can be expressed in first-order - even when you use all the power of set theory.

Third, is this in any way applicable beyond providing a generally better understanding of maths? (Not saying this wouldn't be enough reason to study it.)Depending on how you view it, it's more an exercise in its own right - though the understanding of infinite cardinals is VERY useful mathematically.

And fourth, I'm a bit confused about the whole principle of axioms. If the idea is that they give you ways to construct new sets, and your universe always contains exactly all sets which can be constructed this way, doesn't the axiom of foundation (which was the only one introduced which said something doesn't exist) just assert something that must be true anyway (since otherwise it is inconsistent with the other axioms), but for which no-one found a proof (and presumably there is none?) If so, that seems more like a theory that everyone beliefs in rather than an axiom to me, because it's asserting something rather than defining it, and because introducing it doesn't change the universe.
I've alluded to this above.  The bit in italics is probably your biggest stumbling block.  The important thing is that the axioms lay out some rules for what sets SHOULD look like (we should have all of these sets, but not these ones, etc), but it doesn't lay out every POSSIBLE rule.  The axioms leave room for some uncertainty about which things are sets and which aren't.
And 4.5, isn't the power set axiom implied by the replacement scheme? It seemed as if otherwise the axioms were meant to have as little redundancy as possible, so that seems odd.
I don't think this is true.
Responses in bold.

There are models of ZF(C) without the power set axiom, which are actually some of the more populary axiom systems known as ZF(C)^{-}.
Unless you're using quite an unusual terminology, ZF^- is typically ZF without Foundation - removing Powerset is not a familiar idea to me.

In each individual model, yes it does. However, there are models where it does, and models where it doesn't.

Consider for example the theory consisting only of the axiom "Empty sets exists". The model consisting only of a single set is a model of that, but so is any model of ZFC.
This is a good way to put it.

Oh yes, I was assuming that. Isn't that the point of the axiom system, to regulate exactly which sets exist?
It gives you some of the rules, but it CANNOT regulate it entirely.  There will always be some things which may or may not exist as sets - at that point it is up to your philosophy to decide which

Okay, so the claim of the axioms is not "these are the sets that exist," but "these are some sets that exist" and then the axiom of foundation says "these are some sets that don't exist," and every set that can't be constructed from the axiom and doesn't contradict the axiom of foundation is up for debate?
This seems like a good description.

Note that there is no statement like "this sets exists" in the language of set theory, only "a set with this property exists". (For example "the empty set exists" is not a set-theoretic statement. "There is a set x such that no set y is an element x" is.)
Hmm.  In  the presence of the Extensionality Axiom, this distinction is pretty much negligible.  The Empty Set is unique by extensionality, as are many of the other sets given by axioms like Union/Pairs/Comprehension.




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Re: Maths thread.
« Reply #807 on: March 17, 2017, 02:34:09 pm »
0

Thanks! That was very helpful.

SirPeebles

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Re: Maths thread.
« Reply #808 on: March 19, 2017, 10:30:07 am »
+4

An example that I've found useful for understanding the distinctions around axioms and models is to think about group theory. The definition of a group is essentially a list of axioms. A model of the group axioms is any collection of things which satisfy the group axioms -- in other words, a group! There are lots of statements which are undecidable from the group axioms. For instance, the statement "there is a nonidentity element which is its own inverse" is an undecidable statement, since there are some models where it is true (Z mod 2) and some where it is false (Z mod 3).

Similarly for rings. The definition of a ring lays out the axioms, and each ring is a model of those axioms. The statements "multiplication is commutative" and "each nonzero element has a multiplicative inverse" are independent of the ring axioms. If we choose to make these new axioms, then the models we are left with are called fields.

In a real analysis class, you probably laid out the axioms for the real numbers, namely that they form a complete ordered field. Any two such fields are isomorphic, but there are still different models. Two of the most common are Dedekind Cuts and equivalence classes of Cauchy sequences.

And of course no discussion of axioms is complete without mentioning Euclid's postulates. For a long time people wonder if his fifth postulate, known as the parallel postulate, could be deduced from the first four. This was finally settled a few centuries ago by producing alternative models of the first four axioms: flat, spherical, and hyperbolic geometry. The parallel postulate holds in flat but fails in the others, demonstrating that it is independent of the other axioms.
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silverspawn

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Re: Maths thread.
« Reply #809 on: May 01, 2017, 05:24:59 am »
0

I need help calculating the length of an arc (or rather showing that it has no finite length). I don't think this is supposed to be hard, but I can't figure it out

the function is α: [0,1] -> ℝ^2, α(t) = (t, t*cos(pi/t)) for t > 0 and (0,0) for t = 0

I researched and found out how to bring it back to the integral



But I can't figure out how to show that isn't finite, either.

pacovf

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Re: Maths thread.
« Reply #810 on: May 01, 2017, 08:46:50 am »
+2

Off the top of my head, I would try to calculate (or put a lower bound) on the length of the arc in between two points where the cosinus is 1 (or zero, whichever is more convenient), and go from there. Does that work?
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Titandrake

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Re: Maths thread.
« Reply #811 on: May 02, 2017, 03:48:50 am »
+2

Off the top of my head, I would try to calculate (or put a lower bound) on the length of the arc in between two points where the cosinus is 1 (or zero, whichever is more convenient), and go from there. Does that work?

Solution follows.


cos(pi/t) = 1 when t = 1/2, 1/4, 1/6, ...
cos(pi/t) = -1 when t = 1, 1/3, 1/5, ...

Draw straight lines between t=1, t=1/2, t=1/3, and so on. The sum of those line lengths is a lower bound for the length of the curve. The line between t=1/k and t=1/(k+1) is at least 1/k + 1/(k+1) long, so the length of the curve is lower bounded by the harmonic series, which diverges.
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silverspawn

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Re: Maths thread.
« Reply #812 on: May 02, 2017, 05:01:09 am »
0

I get the solution except for the line between t=1/k and t=1/(k+1) being at least 1/k + 1/(k+1) long. Isn't it just/also at least 2 long, since it's sqrt(2 + [1/k - 1/(k+1)] > sqrt(2) = 2?

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Re: Maths thread.
« Reply #813 on: May 02, 2017, 01:06:24 pm »
+1

I get the solution except for the line between t=1/k and t=1/(k+1) being at least 1/k + 1/(k+1) long. Isn't it just/also at least 2 long, since it's sqrt(2 + [1/k - 1/(k+1)] > sqrt(2) = 2?

No, because the function is t*cos(pi/t), not just cos(pi/t).

More detail: Consider just t=1 and t=1/2. (1,-1) and (1/2,1/2) are two points on the graph. Draw a right triangle between the two points. Clearly the length of the hypotenuse is a lower bound for the  arc length between those two points. The length of a leg of the triangle is also a lower bound. The height of the triangle is 1/2+1. Now consider t=1/2 and t=1/3 giving (1/2,1/2) and (1/3,-1/3). Draw the triangle between the two points. The height of the triangle is 1/2+1/3. Etc. The sum of the heights of all such triangles is a lower bound on the arc length. That sum is at least as large as the harmonic series which diverges.

Edited wording.
« Last Edit: May 02, 2017, 01:08:48 pm by Polk5440 »
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silverspawn

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Re: Maths thread.
« Reply #814 on: May 02, 2017, 02:38:19 pm »
0

I get the solution except for the line between t=1/k and t=1/(k+1) being at least 1/k + 1/(k+1) long. Isn't it just/also at least 2 long, since it's sqrt(2 + [1/k - 1/(k+1)] > sqrt(2) = 2?

No, because the function is t*cos(pi/t), not just cos(pi/t).

-.- right. Okay, thanks.

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Re: Maths thread.
« Reply #815 on: May 05, 2017, 02:12:09 pm »
0

I have a second problem (the same stupid lecture). I need to prove something, but it really appears to be false. That probably means I messed up, but idk how, so...

Let (E^2, d) be an Euclidean space and [P, Q] and [P', Q'] two line segments of equal length. I am supposed to prove that there are exactly two possible isometries T :: E^2 -> E^2 which map P onto P' and Q onto Q'; T(P) = P' and T(Q) = Q'. Also an isometry is defined as a function which preserves distance, so d(P, Q) = d(T[p], T[Q]).

Except this isn't true. I think.

Let f be a function which just shifts all points by (P' - P), so that P lands exactly on P' and all other points land wherever.

Let g be a rotation with fixed point P' that rotates exactly far enough so that f(g(Q)) = Q'.

Let h be a reflection around the axis that goes through P' and through the middle point of f(Q) and Q', so that h(f(Q)) = Q'

Then g ∘ f and h ∘ f both map P onto P' and Q onto Q' (f already maps P onto P' and g and h both don't change the point P').

Similarly, by moving Q to Q' first and then doing the same maneuvers to the point where P lands to P' without changing Q', I get two more functions.

All of them preserve distance, and they don't seem to be identical. You can get 4 more by first swapping/mirroring and then shifting I believe (and there are probably infinity other ways), but well just having 4 proves that there aren't just 2. Am I wrong or is the hypothesis wrong?

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Re: Maths thread.
« Reply #816 on: May 05, 2017, 02:18:34 pm »
0

Oh, I also made a graphic! This shows P and Q and P' and Q' and f and g and h.

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Re: Maths thread.
« Reply #817 on: May 05, 2017, 03:58:25 pm »
0

Similarly, by moving Q to Q' first and then doing the same maneuvers to the point where P lands to P' without changing Q', I get two more functions.
Have you checked wether these two are actually different from the 2 before?
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Re: Maths thread.
« Reply #818 on: May 05, 2017, 09:55:31 pm »
+1

I have a second problem (the same stupid lecture). I need to prove something, but it really appears to be false. That probably means I messed up, but idk how, so...

Let (E^2, d) be an Euclidean space and [P, Q] and [P', Q'] two line segments of equal length. I am supposed to prove that there are exactly two possible isometries T :: E^2 -> E^2 which map P onto P' and Q onto Q'; T(P) = P' and T(Q) = Q'. Also an isometry is defined as a function which preserves distance, so d(P, Q) = d(T[p], T[Q]).

Except this isn't true. I think.

Let f be a function which just shifts all points by (P' - P), so that P lands exactly on P' and all other points land wherever.

Let g be a rotation with fixed point P' that rotates exactly far enough so that f(g(Q)) = Q'.

Let h be a reflection around the axis that goes through P' and through the middle point of f(Q) and Q', so that h(f(Q)) = Q'

Then g ∘ f and h ∘ f both map P onto P' and Q onto Q' (f already maps P onto P' and g and h both don't change the point P').

Similarly, by moving Q to Q' first and then doing the same maneuvers to the point where P lands to P' without changing Q', I get two more functions.

All of them preserve distance, and they don't seem to be identical. You can get 4 more by first swapping/mirroring and then shifting I believe (and there are probably infinity other ways), but well just having 4 proves that there aren't just 2. Am I wrong or is the hypothesis wrong?

They may not seem identical, but they are.
It might help to think of a concrete example, here's a spoilered one.
You're translating (f), rotating (g), and reflecting(h).  Take a piece of paper, move it around, flip it over rotate it.  If you want to move 2 holes onto 2 nails you can do it in lots of ways, but there are only 2 different end results.
Anyways, I don't think that will directly help you prove it, but it should improve your intuition about it.
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Re: Maths thread.
« Reply #819 on: May 05, 2017, 11:12:21 pm »
0

I like theorel's post. You should get only one solution without using reflections, and only one using a single reflection. I think I wouldn't try to prove that all those transformations you mentioned are the same, so much as prove that they can't be different, but that depends on how you think about these things.

If you want a hint: you don't need to know what 2D isometries look like, beyond the fact that they conserve distance.
« Last Edit: May 05, 2017, 11:28:21 pm by pacovf »
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Re: Maths thread.
« Reply #820 on: May 06, 2017, 05:56:55 am »
0

Similarly, by moving Q to Q' first and then doing the same maneuvers to the point where P lands to P' without changing Q', I get two more functions.
Have you checked wether these two are actually different from the 2 before?

I had checked whether g ∘ f and h ∘ f were different. I only briefly checked whether g_ ∘ f_ and h_ ∘ f_ are different and must have made a mistake there... looking again I see they aren't. too bad, I was hoping to be right ;_;

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Re: Maths thread.
« Reply #821 on: May 07, 2017, 10:55:43 am »
+1

I like theorel's post. You should get only one solution without using reflections, and only one using a single reflection. I think I wouldn't try to prove that all those transformations you mentioned are the same, so much as prove that they can't be different, but that depends on how you think about these things.

Okay, so you just look at a third point H, and because distance of H' to P' and Q' has to be the same as of H to P and Q, you already only have two options of where to put it, so you can either put it to the "left" or "right" of P' and Q'. Now all that's left to show is that a function can't choose different sides for different points, which probably also works through H'.

Would have admittedly saved me a lot of time if I hadn't tried to disprove it before trying to prove it.

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Re: Maths thread.
« Reply #822 on: May 09, 2017, 01:19:03 pm »
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For a somewhat ridiculous extra credit question (I don't think the teacher realized what he was doing), I need to solve this integral:

∫(cos(θ) + 2sin(θ))*sqrt((cos(θ)+2sin(θ))^2-4.75) dθ

It's used for a definite integral that can be calculated numerically (and gives the correct answer), but I'm curious if there's any way to figure out the indefinite integral.  No online calculators have been able to find the answer.  It's probably not an elementary function, but I want to see if anybody can figure it out.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

Polk5440

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Re: Maths thread.
« Reply #823 on: May 09, 2017, 02:09:54 pm »
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.... I want to see if anybody can figure it out.

Do I get the credit? :P
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sudgy

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Re: Maths thread.
« Reply #824 on: May 09, 2017, 02:38:12 pm »
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.... I want to see if anybody can figure it out.

Do I get the credit? :P

It was an offhand remark and he had no idea how hard it would be.  The class isn't focused on integrals (he usually says setting up the integral is enough), so this isn't really necessary.  I don't even know if I'll get extra credit.  I was just curious if anybody could figure it out.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm
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