• Mathematician Chat v. 3.999...
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[QUOTE=sltungle;47575112]I accidentally drew the most aesthetically pleasing '2' in my entire life today and there's no way it's ever happening again.[/QUOTE] [img]http://i.imgur.com/Y3ubb2z.png[/img]
[QUOTE=JohnnyMo1;47575312][img]http://i.imgur.com/Y3ubb2z.png[/img][/QUOTE] Silly bugger should have just added [C1cos(32) - C1cos(23)].
What kind of pen do you guys use to math?
[QUOTE=JohanGS;47575513]What kind of pen do you guys use to math?[/QUOTE] Pilot G2.
[media]http://www.youtube.com/watch?v=DgmuGqeRTto[/media] It's interesting but goddamn it's so exhausting to hear him stumble over every single sentence...
Much easier to read his blog. Terry Tao is awesome though. Wish I was as smart as that dude, or Charles Fefferman. [editline]22nd April 2015[/editline] If I was Fefferman I'd be a full professor by now. If I was Terry Tao, by next year.
[QUOTE=JohanGS;47575513]What kind of pen do you guys use to math?[/QUOTE] Mitsubishi Uniball. My handwriting is atrocious without it. The clicking thing the lid does when you turn it also helps keep my hands distracted when I'm thinking.
I think I'm just fucking around and this is probably meaningless, but it's pretty weird considerating that if you can set a value for it you can essentially calculate the square root of two: Let's start with this equation: [IMG]http://latex.codecogs.com/gif.latex?x%3D1-x[/IMG] The solution is x=1/2, right? Then: [IMG]http://latex.codecogs.com/gif.latex?1/2%3Dx%3D1-x[/IMG] Now, since x=1-x, what if we replace the x on the left side of the equation by the left side itself? [IMG]http://latex.codecogs.com/gif.latex?x%3D1-%281-x%29[/IMG] Now, what if we do it and infinity of times? [IMG]http://latex.codecogs.com/gif.latex?x%3D1-%281-%281-...%5CRightarrow%201/2%3D1-%281-%281-...[/IMG] Ok, here's where the mistake or nonsense is, but what if we calculate the square root of x? [IMG]http://latex.codecogs.com/gif.latex?%5Csqrt%7Bx%7D%3D%5Csqrt%7B1-x%7D%5CRightarrow%20%5Csqrt%7Bx%7D%3D%5Csqrt%7B1-%5Csqrt%7B1-...%7D%7D[/IMG] Then: [IMG]http://latex.codecogs.com/gif.latex?%5Csqrt%7B1/2%7D%3D%5Csqrt%7B1-%5Csqrt%7B1-...%7D%7D[/IMG] It looks meaningless, and probably it is, but if the left side is a rational number, then this expresion: [IMG]http://latex.codecogs.com/gif.latex?%5Csqrt%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Csqrt%7B1-...%7D%7D%7D[/IMG] Would make sense. [editline]23rd April 2015[/editline] Hell, if the left side was rational the square root of 2 would be rational too, so never mind
[QUOTE=Cosa8888;47586163]I think I'm just fucking around and this is probably meaningless, but it's pretty weird considering that if you can set a value for it you can essentially calculate the square root of two: It looks meaningless, and probably it is, but if the left side is a rational number, then this expresion: [IMG]http://latex.codecogs.com/gif.latex?%5Csqrt%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Csqrt%7B1-...%7D%7D%7D[/IMG] Would make sense.[/QUOTE] I'm not really sure how it would make sense even if it were rational.
Now that I think of it, you can do something like this with equation that has the form x=c±kx, where k and c are any rational numbers
Something in your steps is definitely not valid, because sqrt(1 - sqrt(1 - sqrt(... etc. is calculable (and does not equal 1/sqrt(2)).
Interesting, what could it be? I'd like to know what was my mistake [editline]23rd April 2015[/editline] And by the way, what is it equal to?
Calculating sqrt(1 - sqrt(1 - sqrt(1 - ... etc. is the same as finding x when x = sqrt(1 - x). Square both sides and you get a quadratic which is easily solved by the quadratic formula. I was practically speechless when I saw this trick for a nested radical for the first time. It's so damn slick.
Yeah, I thought about that when I saw that 1-1/2=1/2. It reminded me of that problem on a newspaper that [URL="http://en.wikipedia.org/wiki/Srinivasa_Ramanujan"]Ramanujan[/URL] solved with a continued fraction. It went like this: [QUOTE]"Imagine that you are on a street with houses marked 1 through n. There is a house in between (x) such that the sum of the house numbers to the left of it equals the sum of the house numbers to its right. If n is between 50 and 500, what are n and x?"[/QUOTE] I looked up how he did it but I didn't quite understood it, but it involved nested radicals somewhere.
Sitting in the maths lectures for my engineering course, waving our arms and chanting "To first order, to first order"
[QUOTE=r0b0tsquid;47670375]Sitting in the maths lectures for my engineering course, waving our arms and chanting "To first order, to first order"[/QUOTE] Ahh, makes me pine for my physics classes. Leading order is the best order.
I have an overdetermined system of equations. From the exponential decay formula I have values for I and t, how do I find I_0 and lambda using linear algebra (least square, QR, whatevs)?
Why is it called linear algebra? What other kinds of algebra are there?
Non-linear and abstract are those I can read in a year or two.
[QUOTE=cathal6606;47679698]Why is it called linear algebra? What other kinds of algebra are there?[/QUOTE] Abstract algebra (which really contains linear algebra), multilinear algebra, homological algebra, categorical algebra, universal algebra. There's a lot of algebras.
calculus question, not a homework question, just wondering about how is this how you might derive the following: if f(x) = (a * b) and g(x) = x + c and h(c) = x + (b*d) h'(x) would be x' + (b*d)' g'(x) would be x' + c' f'(x) would be a'b + ab' so h'(g(f(x))) would be: g'(f(x)) + (b'd + bd') and g'(x) would be f'(x) + c' and f'(x) would be a'b + ba' so g'(x) is: ((a'b + ab') + c') + (b'd + bd') or am I wrong
[QUOTE=proboardslol;47692011] and h(c) = x + (b*d) [/quote] Is that supposed to be c or x? [QUOTE=proboardslol;47692011] so g'(x) is: ((a'b + ab') + c') + (b'd + bd') or am I wrong[/QUOTE] You don't have to go through all that to find g'(x) since you've already defined it in terms of x and c(x): g(x) = x + c g'(x) = 1 + c' What you've found is different, so there should be a mistake somewhere. What the exact problem is depends on the definition of h.
I feel stupid. I have an abstract algebra/number theory question. Okay, it's about permutation, disjoint cycles, etc. So we're given [b]S[/b]=(1,3,6,7,2)(2,5)(2,4,6,5) and [b]T[/b]=(1,3)(1,4,7,5)(2,6,3). Now, the book was pretty straight forward on some things, but this study guide given to me by my professor asks us to compute [b]S[/b]. I don't understand how this computation works. I know how to compute the product of two permutation when they're in the two-line notation, but these particular disjointed cycles are throwing me off. How do I put [b]S[/b] in two-line notation or a non-disjoint cycle? I used WolframAlpha and it gave me the answer, but I need to know what is the process. WolframAlpha gives: [b]S[/b]=(1,3,5,4,6,7,2), but how does 3 go to 5 and how does 5 go to 4?
[QUOTE=Ausare;47719677]I feel stupid. I have an abstract algebra/number theory question. Okay, it's about permutation, disjoint cycles, etc. So we're given [b]S[/b]=(1,3,6,7,2)(2,5)(2,4,6,5) and [b]T[/b]=(1,3)(1,4,7,5)(2,6,3). Now, the book was pretty straight forward on some things, but this study guide given to me by my professor asks us to compute [b]S[/b]. I don't understand how this computation works. I know how to compute the product of two permutation when they're in the two-line notation, but these particular disjointed cycles are throwing me off. How do I put [b]S[/b] in two-line notation or a non-disjoint cycle? I used WolframAlpha and it gave me the answer, but I need to know what is the process. WolframAlpha gives: [b]S[/b]=(1,3,5,4,6,7,2), but how does 3 go to 5 and how does 5 go to 4?[/QUOTE] There are 2 opposite conventions for composing permutations - either right-to-left or left-to-right I learned right-to-left as it matches function application (if you apply 'f' first, then 'g', you write it as g.f), doing it this way gives the answer [b]S[/b] = (1,3,6)(2,4,7). e.g. for 6, the last cycle takes 6 to 5, the middle cycle takes 5 to 2, then the first cycle takes 2 to 1. It looks like WolframAlpha is using the left-to-right strategy so you get a different result. e.g. for 3, the first cycle takes 3 to 6, the middle cycle does nothing to 6, then the last cycle takes 6 to 5. You should work out which convention you're supposed to be using and calculate accordingly!
[QUOTE=Ausare;47719677]I feel stupid. I have an abstract algebra/number theory question. Okay, it's about permutation, disjoint cycles, etc. So we're given [b]S[/b]=(1,3,6,7,2)(2,5)(2,4,6,5) and [b]T[/b]=(1,3)(1,4,7,5)(2,6,3). Now, the book was pretty straight forward on some things, but this study guide given to me by my professor asks us to compute [b]S[/b]. I don't understand how this computation works. I know how to compute the product of two permutation when they're in the two-line notation, but these particular disjointed cycles are throwing me off. How do I put [b]S[/b] in two-line notation or a non-disjoint cycle? I used WolframAlpha and it gave me the answer, but I need to know what is the process. WolframAlpha gives: [b]S[/b]=(1,3,5,4,6,7,2), but how does 3 go to 5 and how does 5 go to 4?[/QUOTE] Tried to draw out the process: [IMG]http://i.imgur.com/QEjT3VY.jpg[/IMG] I've drawn the separate cycles of [B]S[/B] and what they switch around. Then it's easy to follow the "path" of a single number. You don't have to draw it like this but it helps if you're having trouble. If you want to do it in your head, you could follow this kind of thinking: "Well the 1st one changes, 1 to 3, 2nd does nothing to 3 and 3rd also nothing to 3. So in the final solution, 1 switches to 3. Next up, 3 switches to 6 in the 1st one, 2nd one does nothing to 6 and 3rd one switches 6 to 5. So 3 goes to 5. Then, 1st one does nothing to 5 (working with 5 because we ended on 5), 2nd one switches 5 to 2 and 3rd 2 to 4. So 5 switches to 4." and so on, until you loop back to what you started from.
So, I'm taking GCSE Maths and I need help with A* questions such as surds, vectors and other difficult questions. I've been able to get A's on test papers but I want to make sure I get an A/A* on my paper.
Thanks joey and block that helped me out a lot. Apparently other students were confused on that as well. Sadly, the study guide was mostly useless for preparing for the exam. I'll take the course again some other semester. It's unfortunate that the same professor is the only one who teaches it.
Learning about Taylor series.. interesting stuff!
[url=http://vixra.org/abs/1503.0193]Saw this gem on vixra today.[/url] My favourite part: [img]http://i.imgur.com/xMrNvoA.png[/img]
[QUOTE=Fourier;47763935]Learning about Taylor series.. interesting stuff![/QUOTE] I missed lectures on that back when I was doing calculus so I ended up just ignoring it. You're post made me go look it up and learn it. I seriously overestimated it, its not nearly as difficult as I thought, I guess the formula intimidated me.
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