uncool
-
Posts
1329 -
Joined
-
Last visited
-
Days Won
4
Content Type
Profiles
Forums
Events
Posts posted by uncool
-
-
Another way that you can take this: (will latexify later)
log_e(x) + C = log_e(x) + pi*i + C_2 = log_e(x) + log_e(-1) + C_2 = log_e(-x) + C_2
Therefore, the two expressions are equivalent.
=Uncool=
0 -
Typo: Integration by parts is:
integral(u dv) = uv - integral(v du)
Personally, I like translation of coordinates with Jacobian...
doubleintegral(F(x,y)dy dx) = doubleintegral(F(x(a,b),y(a,b))J((x,y),(a,b))db da)
=Uncool=
0 -
Once again you are basing your answer on charcoal (lump). I dont' want that. Imagine a carbon atom. Now set fire to it. How much energy do you need to start the ignition? There are no chain reaction here. Puff and there you have CO2. How much energy is released?
Since you are giving the fire the fuel, the only thing the fire needs is oxygen. However, what happens to the CO2? It gets moved away. This was just answered. The only way for fuel to put out the fire is if you have a limited area, for example, in a pot. You can add extra fuel and cover the pot, making the fire go out faster due to more CO2 production. However, if the pot is uncovered, the fire will actually draw in oxygen, and therefore not be extinguished.
=Uncool=
0 -
Electric potential is going to be positive anyway, so I would say C) Half as large.
0 -
If you assume a perfectly random shuffle, meaning that any of the 52! possibilities are equally likely, then you are describing a geometric random variable. The expected value of a geometric random variable is 1/ the parameter, which in this case is 1/52!. So it would take, on average, 52! times to get back to the original order.
0 -
You want the greatest possible value of AB*AC for this, because Area = AB*AC * sin(A)/2
AB = e cos(a)
AC = e cos(45-a)
AB*AC = e^2*cos(a)cos(45-a)
AB*AC*sin(A)/2 = e^2*cos(a)cos(45-a)sqrt(2)/4
The maximum is when a = 45 - a, or a = 22.5.
e^2*cos(22.5)^2*sqrt(2)/4 = e^2*(cos(45)+1)*sqrt(2)/8
= e^2*(1 + sqrt(2))/8
=Uncool=
0 -
About the looking around then eye contact thing, two things:
First, it's counting the hits and forgetting the misses -that is, you only remember when it happens, not when it doesn't. So you think that it happens more often than it does.
Second, people look for eye contact. When they look around, and they see someone staring at them, the natural urge is to stare back. So your experience isn't very strange at all, Bettina.
-Uncool-
0 -
I'm trying to find some general things about orbits in gravitational systems in multiple dimensions. I've already found that it has to all be in one plane (easy enough), and now I'm assuming the following:
[math]F = \frac{g*m_1*m_2}{d^{#dim-1}}[/math]
[math]F_{x} = F*\frac{x}{d}[/math]
[math]F_{y} = F*\frac{y}{d}[/math]
So:
[math]\frac{d^{2}y}{y*dt^2}=\frac{d^{2}x}{x*dt^2}[/math]
[math]Let |x| = e^{f(t)}[/math]
[math]|y|=e^{g(t)}[/math]
[math]\frac{dx}{dt}=f'(t)e^{f(t)}[/math]
[math]\frac{d^{2}x}{dt^{2}}=f''(t)e^{f(t)}+f'(t)^{2}e^{f(t)}[/math]
[math]\frac{dy}{dt}=g'(t)e^{g(t)}[/math]
[math]\frac{d^{2}y}{dt^{2}}=g''(t)e^{g(t)}+g'(t)^{2}e^{g(t)}[/math]
So:
[math]f''(t)+f'(t)^{2}=g''(t)+g'(t)^{2}[/math]
[math]f''(t)-g''(t)=g'(t)^{2}-f'(t)^{2}[/math]
[math]\frac{(f'(t)-g'(t))'}{f'(t)-g'(t)}=-f'(t)-g'(t)[/math]
[math]ln|f'(t)-g'(t)|=-f(t)-g(t)+C[/math]
[math]f(t)=ln(|x|),g(t)=ln(|x|)[/math]
[math]f'(t)=\frac{x'}{x},g'(t)=\frac{y'}{y}[/math]
[math]|\frac{x'}{x}-\frac{y'}{y}|=C/|xy|[/math]
[math]|x'y-y'x|=C[/math]
Since it is constant, let us assume that x'y is always greater.
[math]x'y-y'x=C[/math]
Can anything happen from there?
-Uncool-
0 -
Why aren't rationals suspect to Cantor's diagonal proof? That is, can anyone give a proof that they aren't suspect? And can you try to prove this without taking the method of proving that they are countable first?
Let's just take rationals between 0 and 1 in binary to simplify things this way.
Cantor's diagonal proof for the real numbers goes as follows:
Proven: The real numbers between 0 and 1 are uncountable (they have no one-to-one correspondence with the integers).
Proof:
Let us assume the real numbers between 0 and 1 are countable. Then write them down in any order in a table. Put them in binary for simplicity.
Then, take the 1st number after the decimal place in the first number, the second digit after the decimal place in the second, etc. and create a number such that the digits are switched from the number created by the earlier method. This number must be different from all numbers in the table because for the nth number, the nth digit is different. Even if you add this number, there will be another number for the new table which will be different than every element.
For example:
Table:
.0
Element not in the table: .1
Table:
.00
.10
Element not in table: .11
.000
.100
.110
Element not in table: 0.111
etc.
-Uncool-
0 -
OK, matt, you are correct. I made a mistake (I divided two logs, and accidentally put the division inside). Shows wat happens when you skip a step...
A few line break problems with what I had, too, though. The ln[n] is always decreasing... is supposed to be on a separate line.
-Uncool-
0 -
Matter and antimatter don't quite make 0. they make two (or more) photons, usually gamma rays. These are extremely high high energy light beams. However, the energy is equal to the energy made by the masses of the matter and antimatter.
-Uncool-
0 -
Someone tell me if my math is correct?
[math] n^n[/math] acts like [math]n!e^n[/math], as follows:
[math] \frac{(n+1)^{n+1}}{n^n} = (n+1)\frac{(n+1)^n}{n^n}[/math]
[math]=(n+1)\frac{n^n+n*n^{n-1}+\frac{n(n-1)}{2}*n^{n-2}+...}{n^n}[/math]
[math]=(n+1)\frac{n^n*(1+1+\frac{1}{2}+...)+n^{n-1}*(-\frac{1}{2}-\frac{3}{6}-\frac{6}{24}-\frac{10}{120}-...)}{n^n}[/math]
The last terms can be dropped out because they contain a [math]\frac{1}{n}[/math] term.
[math]=(n+1)\frac{n^n*e}{n^n}=e*(n+1)[/math]
so after a while, the two act similarly.
In this case, similarly means that if you take the ln of both and divide them, the quotient will approach one, as shown below:
n [math]n!*e^n[/math] [math]n^n[/math] Ratio 1 2.718281828 1 2.718281828 2 14.7781122 4 3.694528049 3 120.5132215 27 4.46345265 4 1310.355601 256 5.118576566 5 17809.57909 3125 5.69906531 6 290468.7313 46656 6.225752986 7 5527031.118 823543 6.711284193 8 120192226 16777216 7.164014938 9 2940447096 387420489 7.58980792 10 79929639076 10000000000 7.992963908 11 2.38998E+12 2.85312E+11 8.376748609 12 7.79598E+13 8.9161E+12 8.743710992 13 2.75492E+15 3.02875E+14 9.095885843 14 1.04841E+17 1.1112E+16 9.434928193 15 4.27481E+18 4.37894E+17 9.762205686 16 1.85922E+20 1.84467E+19 10.07886392 17 8.59161E+21 8.2724E+20 10.38587387 18 4.2038E+23 3.93464E+22 10.684067 19 2.17115E+25 1.97842E+24 10.97416181 20 1.18036E+27 1.04858E+26 11.25678423 21 6.73795E+28 5.84259E+27 11.53248351 22 4.02944E+30 3.41428E+29 11.80174472 23 2.51923E+32 2.08805E+31 12.06499882 24 1.64351E+34 1.33374E+33 12.32263074 25 1.11688E+36 8.88178E+34 12.57498598 26 7.89361E+37 6.15612E+36 12.82237603 27 5.7934E+39 4.43426E+38 13.06508291 28 4.40947E+41 3.31455E+40 13.30336289 29 3.47599E+43 2.56769E+42 13.53744971 30 2.83462E+45 2.05891E+44 13.76755723 31 2.38864E+47 1.70692E+46 13.99388172 32 2.07776E+49 1.4615E+48 14.21660385 33 1.86382E+51 1.2911E+50 14.43589032 34 1.72257E+53 1.17566E+52 14.65189539 35 1.63885E+55 1.10251E+54 14.86476209 36 1.60375E+57 1.06387E+56 15.07462336 37 1.61299E+59 1.05551E+58 15.28160301 38 1.66614E+61 1.07591E+60 15.48581657 39 1.76632E+63 1.12595E+62 15.68737206 40 1.92054E+65 1.20893E+64 15.88637066 41 2.14044E+67 1.33088E+66 16.08290725 42 2.44369E+69 1.50131E+68 16.27707102 43 2.85634E+71 1.73438E+70 16.46894589 44 3.4163E+73 2.05077E+72 16.65861095 45 4.17892E+75 2.48064E+74 16.84614083 46 5.22536E+77 3.06803E+76 17.03160605 47 6.67588E+79 3.87792E+78 17.21507334 48 8.71052E+81 5.00702E+80 17.39660591 49 1.1602E+84 6.60097E+82 17.57626371 50 1.57688E+86 8.88178E+84 17.75410364 51 2.18607E+88 1.21921E+87 17.9301798 52 3.09002E+90 1.70677E+89 18.10454364 53 4.45176E+92 2.43568E+91 18.27724417 54 6.53462E+94 3.54212E+93 18.4483281 55 9.76961E+96 5.24745E+95 18.61784 56 1.4872E+99 7.91643E+97 18.78582241 57 2.3042E+101 1.2158E+100 18.952316 58 3.6329E+103 1.9003E+102 19.11735968 59 5.8264E+105 3.0218E+104 19.28099067 60 9.5026E+107 4.8874E+106 19.44324464 61 1.5757E+110 8.0375E+108 19.60415578 62 2.6555E+112 1.3436E+111 19.76375689 63 4.5477E+114 2.2827E+113 19.92207945 64 7.9116E+116 3.9402E+115 20.0791537 65 1.3979E+119 6.9083E+117 20.23500872 66 2.5079E+121 1.23E+120 20.38967246 67 4.5675E+123 2.2234E+122 20.54317183 68 8.4427E+125 4.0795E+124 20.69553274 69 1.5835E+128 7.596E+126 20.84678014 70 3.0131E+130 1.435E+129 20.9969381 71 5.8153E+132 2.7501E+131 21.14602983 72 1.1381E+135 5.3449E+133 21.29407772 73 2.2585E+137 1.0533E+136 21.44110339 74 4.543E+139 2.1045E+138 21.58712774 75 9.2619E+141 4.2618E+140 21.73217094 76 1.9134E+144 8.7465E+142 21.87625251 77 4.0049E+146 1.8188E+145 22.01939133 78 8.4914E+148 3.8316E+147 22.16160566 79 1.8235E+151 8.176E+149 22.30291319 80 3.9654E+153 1.7668E+152 22.44333104 81 8.731E+155 3.8662E+154 22.58287582 82 1.9461E+158 8.5652E+156 22.7215636 83 4.3908E+160 1.9208E+159 22.85940999 84 1.0026E+163 4.3597E+161 22.99643011 85 2.3165E+165 1.0014E+164 23.13263866 86 5.4154E+167 2.3274E+166 23.26804987 87 1.2807E+170 5.4724E+168 23.40267759 88 3.0635E+172 1.3016E+171 23.53653527 89 7.4114E+174 3.1312E+173 23.66963596 90 1.8132E+177 7.6177E+175 23.80199238 91 4.4851E+179 1.874E+178 23.93361686 92 1.1216E+182 4.661E+180 24.06452141 93 2.8355E+184 1.172E+183 24.19471773 94 7.2453E+186 2.9786E+185 24.32421718 95 1.871E+189 7.6514E+187 24.45303083 96 4.8825E+191 1.9863E+190 24.58116948 97 1.2874E+194 5.2102E+192 24.70864361 98 3.4295E+196 1.3809E+195 24.83546345 99 9.2291E+198 3.6973E+197 24.961639 100 2.5087E+201 1E+200 25.08717995 101 6.8876E+203 2.7319E+202 25.2120958 102 1.9097E+206 7.5373E+204 25.33639579 103 5.3468E+208 2.1001E+207 25.46008894 104 1.5115E+211 5.9084E+209 25.58318404 105 4.3143E+213 1.6783E+212 25.7056897 106 1.2431E+216 4.8131E+214 25.82761431 107 3.6156E+218 1.3934E+217 25.94896604 108 1.0615E+221 4.0716E+219 26.0697529 109 3.145E+223 1.2008E+222 26.18998271 110 9.404E+225 3.5743E+224 26.3096631 111 2.8374E+228 1.0736E+227 26.42880153 112 8.6385E+230 3.254E+229 26.54740531 113 2.6535E+233 9.9509E+231 26.66548157 114 8.2227E+235 3.0701E+234 26.78303728 115 2.5704E+238 9.5555E+236 26.90007926 116 8.1051E+240 3E+239 27.01661421 117 2.5777E+243 9.5005E+241 27.13264864 118 8.2683E+245 3.0344E+244 27.24818895 119 2.6746E+248 9.7744E+246 27.36324141 120 8.7243E+250 3.175E+249 27.47781213 121 2.8695E+253 1.04E+252 27.59190713 122 9.5162E+255 3.4348E+254 27.70553227 123 3.1817E+258 1.1437E+257 27.81869332 124 1.0725E+261 3.8396E+259 27.93139591 125 3.6441E+263 1.2994E+262 28.04364558 126 1.2481E+266 4.4329E+264 28.15544773 127 4.3087E+268 1.5243E+267 28.26680769 128 1.4992E+271 5.2829E+269 28.37773065 129 5.257E+273 1.8453E+272 28.48822172 130 1.8577E+276 6.4958E+274 28.59828591 131 6.6152E+278 2.3043E+277 28.70792812 132 2.3736E+281 8.2368E+279 28.81715318 133 8.5814E+283 2.9667E+282 28.92596581 134 3.1258E+286 1.0766E+285 29.03437064 135 1.1471E+289 3.936E+287 29.14237222 136 4.2405E+291 1.4497E+290 29.24997503 137 1.5792E+294 5.3792E+292 29.35718345 138 5.9239E+296 2.0105E+295 29.46400178 139 2.2383E+299 7.5693E+297 29.57043425 140 8.518E+301 2.8703E+300 29.67648501 141 3.2648E+304 1.0962E+303 29.78215814 142 1.2602E+307 4.2164E+305 29.88745764
-Uncool-
0 -
[math]
L = \lim_{n \to \infty} n!^\frac{1}{n}
ln[L] = ln[\lim_{n \to \infty} n!^\frac{1}{n}] = \lim_{n \to \infty} ln[n!^\frac{1}{n}] = \lim_{n \to \infty} \frac{ln[n!]}{n}[/math]
Ratio of one to the next:
[math]ln[n]\frac{n-1}{n}
ln[n][/math] is always increasing, and will increase infinitely.
[math]\frac{n-1}{n}[/math] is always decreasing, and will decrease towards 1.
Therefore, the ratio will increase beyond one, and remain beyond one.
Therefore, ln[L] does not exixt.
Therefore, L does not exist.
-Uncool-
P.S. that was my first real LATEX thing. yay!
0 -
Hey guys, how do you do all those symbols (sum and product, etc.)?
-Uncool-
0 -
I think I have the answer as well.
[hide]F projects to itself, as the two are on exactly the same plane. from H to M goes up one, over 1/2, and over the other way 1/2. From M to F goes up one, over 1/2, and over the other way 1/2. So the three are collinear, so F is on the plane, so its projection is itself.[/hide]
-Uncool-
0 -
From where are you getting r1, Johnny?
Or do you mean to try to find r1 from the coefficients without doing what I did?
-Uncool-
0 -
Define "fault"
-Uncool-
0 -
Yes
-Uncool-
0 -
Benson: No, No
Wiggle: Define have anything to do with...
-Uncool-
0 -
Johnny, I challenge you to prove your statement (that is, that there exists a moment x etc.)
-Uncool-
0 -
This maze seems relatively easy, actually - just let yourself wander along a path, and go in a different direction if you find yourself in the same place, as there are no dead ends. Then erase the loops, and you have a short path.
-Uncool-
0 -
Imaginary numbers are the square roots of negative real numbers.
Complex numbers are the sum of real numbers and imaginary numbers.
-Uncool-
0 -
OK, to remember it, just remember that you want to eliminate the middle terms. First the A term, then the B term.
To eliminate the A term: Let [math]Y = Z + A/3[/math]
Let this leave you with [math]Y^3 + PY + Q = 0[/math]
To eliminate the P term: Let [math]Y = X-P/(3X)[/math]
Let this leave you with [math]X^3 + M - N/X^3 = 0[/math]
Find X^3 because this is a quadratic equation. Both will give the same final answer.
Backsolve to find Z.
Use this solution to reduce the cubic to a quadratic.
Solve the quadratic for the other two roots.
So pretty much all you have to do is remember how to eliminate each middle term.
-Uncool-
0 -
Pie is a type of good food.
Pi, on the other hand, is the ratio of the length of the circumference of a circle to the length of the diameter. It is approximately 3.14159265.
What else would you like to know?
-Uncool-
0
modulo
in Mathematics
Posted
I believe he means that o(a mod b)|h for all a^h = 1 (mod b)
o means order of a mod b - that is, the lowest number o>0 such that a^o = 1 (mod b)
It's a simple proof: Assume we have an h such that o(a mod b) does not divide h. Then by division algorithm, we have an r: d = r + xh, 0 <= r < h. a^d = a^r*a^xh = a^r * (a^h)^x = a^r*1^x = a^r, so a^r = 1. But then, as r < h, that means r must be 0. Therefore, d = xh, so h|d.
=Uncool=