[QUOTE=gangstadiddle;39895420]is it possible to find a point of tangency without the tangent line equation. i have a problem that shows a tangent line that passes through a point at (3pi/2, 0) for a function of sin(x), and that's all that it gives.[/QUOTE]
Assuming you can use differentiation; d(sin(x))/dx and solve for your x value, (3pi/2). If Sin(x) is your function though, you won't have it equal 0, but -1 if your domain is limited to 0 and 2pi.
Are you trying to find the sinusoidal function that yields a tangent line that passes through (3pi/2, 0)? Is that the question you're looking for?
(Possibly naive) question: can any iteratively achieved quantity (e.g. Fibonacci) be written as an explicit function of n (that is, an expression solely of n).
I don't know a lot about this. Also, let's exclude stuff like "the sum of the previous two prime numbers", as that obviously would be the holy grail of math.
Yes, I'm pretty sure it's called a closed form. For example, the triangular numbers:
[img]http://upload.wikimedia.org/math/4/9/5/4950ae2d99ac2e83c67b1562e44addfd.png[/img]
The two right side expressions are the closed form or explicit solutions.
You can also find one using the floor function of the Golden ratio to the nth power over the square root of five (for n greater than or equal to 0). As for a formal proof; we're doing that in Structures next Monday. We just briefly hit on it today for recursive sequences before moving on to the towers of Hanoi.
[QUOTE=Number-41;39903242](Possibly naive) question: can any iteratively achieved quantity (e.g. Fibonacci) be written as an explicit function of n (that is, an expression solely of n).
I don't know a lot about this. Also, let's exclude stuff like "the sum of the previous two prime numbers", as that obviously would be the holy grail of math.[/QUOTE]
You have to be very careful what you mean by 'an explicit function of n'... Anything can be written as a single function simply by defining it to be so... But that is somewhat cheating!
It also of course depends what you mean by any iteratively achieved quantity...
If you are allowing [i]any[/i] recursively defined function, the answer is no - basically anything that's not [i]really[/i] nice will probably fail... For a trivial case, just take the definition that x[sub]n[/sub] = f(n) for any f which cannot be expressed in closed form!
However, if the recursion has a lot of nice properties (e.g. linearity) then there will be a closed form expression.
Also, I know a lot of recurrence relations can be thought of as discrete versions of differential equations, so I imagine in principle these are similar questions to whether a closed form solution exists for various differential equations (and again, the answer is certainly not except in some 'nice' cases)
Are there any theorems on this?
[QUOTE=Yahnich;39904404]so i'm using the arclength function to decide at what angle it's faster to walk parabolically instead of using two angled lines
wtf am i doing with my life and why am i perfecting the path to my school[/QUOTE]
It will always be faster to use 2 angled lines
[QUOTE=Yahnich;39904495]nope[/QUOTE]
could you paste a pic? i cant imagine how a curve would be shorter than straight lines :(
[QUOTE=Yahnich;39904507]don't forget the parabola can be very straight, albeit not quite
[editline]13th March 2013[/editline]
[IMG]http://i.imgur.com/JJ4itwO.png[/IMG][/QUOTE]
yeah but wouldn't you be able to 'approximate the curve' with 2 straight lines, which would be shorter?
[QUOTE=Yahnich;39904677]i can't walk like that[/QUOTE]
ohhh right I see, ty
Use variational calculus & Lagrange multipliers to find the optimal path.
Gogogo
wouldn't it be faster to run into a brachistochrone cyclodial curve rather than running along the straight line
Presumably you need to have some kind of function which determines how fast you can move depending on the curvature...
If you just want the shortest length it should be quite straightforward (e.g. imagine threading a piece of string along the route and pulling it taut - that's the route you should take!)
[QUOTE=Number-41;39904343]Are there any theorems on this?[/QUOTE]
It's not really an area I know a lot about, but there are certainly good methods for the 'nice' recurrences e.g. [url]http://delta.cs.cinvestav.mx/~debrup/ho1.pdf[/url] for ones like x[sub] n[/sub] = a[sub] 1[/sub]x[sub] n-1[/sub] + ... + a[sub] k[/sub]x[sub] n-k[/sub]
(If you know how to solve linear homogeneous ODE's with constant coefficients you'll notice that this is an extremely similar method!)
[QUOTE=Yahnich;39904404]so i'm using the arclength function to decide at what angle it's faster to walk parabolically instead of using two angled lines
wtf am i doing with my life and why am i perfecting the path to my school[/QUOTE]
You're using many hours up front to save up a few seconds every time you do one thing?
You should look into programming and algorithms and all that as a carrier path, it's right up your alley :v:
Hey, I'm doing discrete mathematics right now and we're working on this:
[quote]We recursively define the following function f, from binary strings to natural numbers, as follows. f(λ) is 3, and if w is any string, f(w0) = f(w) and f(w1) = -f(w). What is the value of f(w) for a general string w?[/quote]
My book doesn't cover this, so could someone tell me keywords to search on google? I have no clue what to search for.
I find it quite fascinating how a problem I work on (I just finished doing limits on Khanacademy) seems to take me ages to do it initially, with my notepad having scribbles and scrawls everywhere. Then they slowly take less time until I can do them very quickly and seem quite satisfying to do.
Mathematics is getting more fascinating.
[QUOTE=Sobotnik;39907533]I find it quite fascinating how a problem I work on (I just finished doing limits on Khanacademy) seems to take me ages to do it initially, with my notepad having scribbles and scrawls everywhere. Then they slowly take less time until I can do them very quickly and seem quite satisfying to do.
Mathematics is getting more fascinating.[/QUOTE]
Yeah, that has to be the best part about learning math for me, it feels so satisfying once you get something down.
[QUOTE=Sobotnik;39907533]I find it quite fascinating how a problem I work on (I just finished doing limits on Khanacademy) seems to take me ages to do it initially, with my notepad having scribbles and scrawls everywhere. Then they slowly take less time until I can do them very quickly and seem quite satisfying to do.
Mathematics is getting more fascinating.[/QUOTE]
I've done a lot (and I mean a lot) of math over the course of two years during my studies. After those two years, looking back at the problems that used to take 1+ page to solve correctly, I realized most of those we would now just admit the proof in one line because it had become trivial.
So when I tried to help some people that were a few years before me in their studies with math, things that I had no problem doing at the time had actually become difficult: I had to look for how I was supposed to write the proof other than "trivially, ..." :v:
That why I was also a very bad teacher and I couldn't help people with "basic" math very well. I got pretty upset when they asked me "but why does this work?"
WELL UUUH IT WORKS. Having to explain something that had become an evidence was pretty taxing. :v:
[editline]14th March 2013[/editline]
I don't mean that I'm very good at maths right now, but I sure was a lot (a [i]lot[/i]) better after those two years than before.
It is quite amazing how a bit of maths you find completely incomprehensible at first can become incredibly easy after just working through it a few times, it's one of the things I love about it!
[QUOTE=DoctorSalt;39907401]Hey, I'm doing discrete mathematics right now and we're working on this:
[quote]We recursively define the following function f, from binary strings to natural numbers, as follows. f(λ) is 3, and if w is any string, f(w0) = f(w) and f(w1) = -f(w). What is the value of f(w) for a general string w?
[/quote]
My book doesn't cover this, so could someone tell me keywords to search on google? I have no clue what to search for.[/QUOTE]
Are you having trouble with what the problem actually [i]means[/i] or just doing it?
To actually do it: [sp]From the definitions you see that if you have a string with a zero at the end you can ignore it, and if you have a 1 at the end you change the sign of the answer, so you can systematically add/remove zeros/ones from w to transform it into λ... Then you need to keep track of how many times you added/removed 1's (the 0's don't make a difference)[/sp]
[QUOTE=Joey90;39909651]It is quite amazing how a bit of maths you find completely incomprehensible at first can become incredibly easy after just working through it a few times, it's one of the things I love about it!
Are you having trouble with what the problem actually [i]means[/i] or just doing it?
To actually do it: [sp]From the definitions you see that if you have a string with a zero at the end you can ignore it, and if you have a 1 at the end you change the sign of the answer, so you can systematically add/remove zeros/ones from w to transform it into λ... Then you need to keep track of how many times you added/removed 1's (the 0's don't make a difference)[/sp][/QUOTE]
We covered it a lil further, but the book doesn't explain the lambda and omega notation and how strings are defined recursively. I think I got it now (thanks)
Can someone please tell me how proofing for strong induction differs from mathematical induction? I know that mathematical induction involves proving that if a statement holds for n it holds for n+1. And I think strong induction is proving for all cases between n and m?
Also, what about structural induction?
My professor failed to explain any of these things in class, I went to her office hours and she said I should know these things already.
I managed to finish that assignment; the [B]most difficult[/B] assignment I've [B]ever[/B] had -- It felt amazing.
One of those questions:
[QUOTE][IMG]http://i.imgur.com/2yNRJ4S.png[/IMG][/QUOTE]
How I solved it:
Proof: [Direct Proof]
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?Let\; x^n = x^n[/IMG][/QUOTE]
Where any [I]x[/I], to any [I]n[/I], will equal another number.
Add zero to both sides, by cancelling summations.
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?x^n +\left (\sum_{y=0}^{(n-1)}x^y \right ) -\left (\sum_{y=0}^{(n-1)}x^y \right ) = x^n[/IMG][/QUOTE]
Subtract 1 from both sides.
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?x^n +\left (\sum_{y=0}^{(n-1)}x^y \right ) -\left (\sum_{y=0}^{(n-1)}x^y \right ) -1 = x^n -1[/IMG][/QUOTE]
Expanding the LHS with some rearranging will result in:
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?1 + x - 1 + x^2 -x + ... + x^{n-1}-x^{n-2} + x^n - x^{n-1} -1 = x^n -1[/IMG][/QUOTE]
Tidy it up the LHS by cancelling the 1's, and grouping.
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?(x - 1) + (x^2 -x) + ... + (x^{n-1}-x^{n-2}) + (x^n - x^{n-1}) = x^n -1[/IMG][/QUOTE]
Factor out (x-1) from the LHS terms.
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?1(x - 1) + x(x -1) + ... + x^{n-2}(x-1) + x^{n-1}(x - 1) = x^n -1[/IMG][/QUOTE]
Divide both sides by (x-1).
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?1 + x + ... + x^{n-2} + x^{n-1} = \frac{x^n -1}{x-1}[/IMG][/QUOTE]
[QUOTE][IMG]http://latex.codecogs.com/gif.latex?Where \; x\neq 1, as\; the \;RHS \;would\; be \;undefined.[/IMG][/QUOTE]
[B]QED[/B]
Not as much English as my other proofs; thoughts? critiques?
[QUOTE=Mr_Razzums;39919237]Can someone please tell me how proofing for strong induction differs from mathematical induction? I know that mathematical induction involves proving that if a statement holds for n it holds for n+1. And I think strong induction is proving for all cases between n and m?
Also, what about structural induction?
My professor failed to explain any of these things in class, I went to her office hours and she said I should know these things already.[/QUOTE]
Proving for strong induction doesn't really differ.
Basic induction is
1) prove that it's true for 0 or 1 or whatever rank you start at
2) a- suppose that it's true for a given rank n
2) b- show that it is then true for the next rank n+1
And logically, it means that it's true for every rank >=0 (or 1). Because it's true for 1, so it's true for 1+1=2, so it's true for 2+1=3 etc.
Strong induction is sometimes needeed for some proofs. For step 2)b- you will sometimes need the assumption to be true not only for the previous rank, but for all previous ranks.
So this is how you do it:
1) prove that it's true for 0 or 1 or whatever rank you start at
2) a- suppose that it's true for [b]every rank between 0 (or 1) and a given rank n[/b]
2) b- show that it is then true for the next rank n+1 (and at this step you will have to use the fact that it's true for every rank before n+1)
And it works the same: it's true for 0, so it's true for 1. It's true for 0 and 1, so it's true for 2. It's true for [0..7] so it's true for 8, etc.
So really it's the same, it's just that sometimes you need more than just the previous rank.
I know I also sometimes used induction with two steps: prove that it's true for 0 and 1, and suppose that it's true for n-1 and n, then show that it's true for n+1.
I'm sorry if you don't understand some things; not only am I bad at explaining things, I also never did maths in english and I don't really know the terms I should use... I hope it helps anyway.
[editline]15th March 2013[/editline]
Can't help you about structural induction, I read what wikipedia had to say on the subject and I can say I never heard of it :v:
[QUOTE=Mr_Razzums;39919237]Can someone please tell me how proofing for strong induction differs from mathematical induction? I know that mathematical induction involves proving that if a statement holds for n it holds for n+1. And I think strong induction is proving for all cases between n and m?
Also, what about structural induction?
My professor failed to explain any of these things in class, I went to her office hours and she said I should know these things already.[/QUOTE]
As Zanpa says:
(Weak) Induction requires you to prove that if a statement holds for n, it holds for n+1.
Strong Induction requires you to prove that if a statement holds for m,...,n it holds for n+1.
And they both require you to prove a base case m (usually 0 or 1)
You can easily see that these are equivalent: Clearly Strong => Weak, since if you know it's true for m,...,n it's certainly true for n. Conversely, if your statement is P(n), consider the statement Q(n) = 'P(k) is true for k=m,...,n' then weak induction on Q is the same as strong induction on P.
These are then both equivalent to the Well Ordering Principle (that the natural numbers are well ordered)
Structural Induction is used on some recursively defined structure and works in the same sort of way - you prove it for the minimal cases, and then prove that if it's true for substructures of something then it's true for the structure itself.
[QUOTE=Bradyns;39920032]I managed to finish that assignment; the [B]most difficult[/B] assignment I've [B]ever[/B] had -- It felt amazing.
Not as much English as my other proofs; thoughts? critiques?[/QUOTE]
Very nice, there are only 2 things I'd critique:
Firstly, the first few lines are pretty unnecessary, you could start with a line that says
[img]http://latex.codecogs.com/gif.latex?\left%20(\sum_{y=0}^{(n-1)}x^y%20\right%20)%20-\left%20(\sum_{y=0}^{(n-1)}x^y%20\right%20)%20+%20x^n%20-1%20=%20x^n%20-1[/img]
and proceed from there.
The other thing is that you should assume that [img]http://latex.codecogs.com/gif.latex?x\neq%201[/img] [i]before[/i] you do the division, rather than just doing it and then working out what the problem is. (I would say something like 'Divide both sides by [img]http://latex.codecogs.com/gif.latex?x%20-%201[/img] (nonzero since [img]http://latex.codecogs.com/gif.latex?x\neq%201[/img])')
Also as a purely stylistic point, sums are usually indexed by i, j, k, l, m or n... To me x and y are usually real variables (of course this doesn't change the proof at all, it's completely arbitrary)
[QUOTE=Joey90;39920677]Firstly, the first few lines are pretty unnecessary, you could start with a line that says
[IMG]http://latex.codecogs.com/gif.latex?\left (\sum_{y=0}^{(n-1)}x^y \right ) -\left (\sum_{y=0}^{(n-1)}x^y \right ) + x^n -1 = x^n -1[/IMG]
and proceed from there.[/QUOTE]
I wanted to make sure for my first assignment, that I was over the top, rather than fell short.. Considering that my lecturer emphasised [I][U][B]USE COMPLETE SENTENCES[/B][/U][/I], even though there aren't any in this proof.
Essentially, I'd rather more descriptive, then less.
[QUOTE=Joey90;39920677]The other thing is that you should assume that [IMG]http://latex.codecogs.com/gif.latex?x\neq 1[/IMG] [I]before[/I] you do the division, rather than just doing it and then working out what the problem is. (I would say something like 'Divide both sides by [IMG]http://latex.codecogs.com/gif.latex?x - 1[/IMG] (nonzero since [IMG]http://latex.codecogs.com/gif.latex?x\neq 1[/IMG])')[/QUOTE]
When I handed it in, I actually wrote it before the division that x cannot equal one.. I was going off memory. (I was posting exclusively from memory [SUB][SUB][SUB][SUB]whilst inebriated.[/SUB][/SUB][/SUB][/SUB]). :v:
[QUOTE=Joey90;39920677]Also as a purely stylistic point, sums are usually indexed by i, j, k, l, m or n... To me x and y are usually real variables (of course this doesn't change the proof at all, it's completely arbitrary)[/QUOTE]
YES! I actually put in the variable in the assignment as m... So it was [Sigma (x^m) from m=0 to (n-1)], and the original axiom was [a^n = y]. Once again, working from memory.
[QUOTE=Bradyns;39920032]Not as much English as my other proofs; thoughts? critiques?[/QUOTE]
Were you not allowed to use induction? That would speed it up to a couple of lines.
[QUOTE=ThisIsTheOne;39942448]Were you not allowed to use induction? That would speed it up to a couple of lines.[/QUOTE]
I could have...
How would you do this problem by induction?
[QUOTE=Bradyns;39943187]I could have...
How would you do this problem by induction?[/QUOTE]
For n=1 the property is true (1=(x-1)/(x-1))
Let the property be true for n, let's prove it's also true for n+1 (ie 1+x+...+x^(n)=(x^(n+1)-1)/(x-1))
We have 1+x+...+x^n-1=(x^(n)-1)/(x-1) by hypothesis
Add x^n to both sides:
<=> 1+x+...+x^n-1+x^n=(x^(n)-1)/(x-1)+x^(n)
Put x^n over the same denominator:
<=> 1+x+...+x^n-1+x^n=((x^(n)-1)+(x-1)*x^(n))/(x-1)
The two x^n from the right side cancel out, leaving only:
<=> 1+x+...+x^n-1+x^n=(x^(n+1)-1)/(x-1)
Which proves the relation for n+1, therefore for all n.
Is there a set of disks with disjoint interiors whose union is a larger disk?
[QUOTE=Krinkels;39946323]Is there a set of disks with disjoint interiors whose union is a larger disk?[/QUOTE]
Are these closed discs or open discs? In R[sup]n[/sup] or some more general metric space?
If they're closed it's not very interesting, you can just take the trivial set of discs being merely the points of a bigger disc.
If they're open it's harder to decide:
It's [i]not[/i] possible in R[sup]n[/sup] - a union of more than one open discs is not connected (just take one disc and the union of all the others) whereas a larger disc clearly is connected, thus they can't make up a larger disc.
In a general metric space it can be possible though - e.g. with a discrete space the union of the singleton discs (radius <1) is the whole space (disc radius >1)
I can't think of an obvious condition for when it's possible or not... If all singletons are discs then it always is possible, alternatively if all discs are connected then it's [i]not[/i] possible, but there are spaces that satisfy neither of these.
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