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#1 Re: This is Cool » I disagree with » 2006-04-02 14:43:40


0.9999... = 1-k
Doubling, 1.99999...(8) = 2-2k
If k is not zero, there is a number 2-k between 2-2k and 2
Halfing, there is a number 1 - k/2 between 1-k and 1

Therefore, if 0.9999... isn't 1, then it isn't the largest real number less than one either, which leads me to ask, if 0.999... isn't the largest real number less than 1, then what is?

#2 Re: This is Cool » Sum of the reciprocals of the squares of the prime numbers » 2006-03-02 11:17:52


Maybe the solution lies in the infinite pi-notation format of the reimann zeta function
After all, problems are a lot cooler to solve by hand than to approximate with computer tongue

#3 Re: This is Cool » Time may have a geometry » 2006-02-06 04:28:00


There is a law that says it is impossible to accurately observe anything you are not independent of. For all you know, we could just be part of a huge matrix of data that is constantly being changed by means of various complex functions. We cannot observe our own universe accurately past a certain point.

#4 Re: This is Cool » Time may have a geometry » 2006-02-05 14:36:45


I believe the universe is a 256-dimensional space/time hyper-space that vaguely resembles the structure of a 256-dimensional hyperbola rotated around a 256-dimensional hypersphere perpendicular to the axis of the hyperbola. A warping of the space time continuum actually just shifts the foci in the 32nd, 64th, 96th, 128th, 160th, and 192nd dimensions.

Do I win?

(Where's my name in scientific American?)

#5 Re: This is Cool » 2520 A wonderfull little number » 2006-02-05 14:34:13


100110100000 in binary, 9D8 in hex, 10230 in 7..... that's all I care about

IMHO 65536 is way cooler than 2520.
65536 = 2^2^2^2. That's 2 to it's own power 2^2 times.

#6 Re: Help Me ! » Quotation and Chain Rule HELP » 2006-02-05 14:16:45


0.03 seconds using the mathematica online integrator.

#7 Re: Help Me ! » Permutation » 2006-01-28 13:09:45


you have 6 digits.
So, there are 6 choices for the hundred thousands digit, 5 choices (all the numbers except whatyou already used) for the ten thousands digit, 4 choices (all the numbers except the two you used) for the thousands digit, 3 for the hundreds, 2 for the tens, and 1 (whatever is left over) for the one's digit.

This makes 6*5*4*3*2*1 = 720

However, since there are two 1's and two 5's, we need to divide by 2*1 = 2 twice
(if you have two one's, it doesn't matter which 1 goes in which place. Since there are two ways to order two things, we divide by two)

720/2/2 =


#8 Re: This is Cool » my gosh... » 2006-01-28 13:06:02


Isn't one of the bikini calculus girls an MIT graduate student or something?

But Geez... I downloaded one just to see if it's any good... and it doesnt teach jack

#9 Re: Help Me ! » Recursions » 2006-01-28 03:22:04


Thanks. I had a typo on part iii though:
I meant such that ƒ(x+1) = sin( ƒ(x) )

#10 Re: Help Me ! » volume of cone proving » 2006-01-26 11:16:36


This works just like calculus, except you don't actually use an integral.

The strategy is to find the ratio of the volumes of a cone and it's respective cylinder, since once you find that, that remains constant no matter how you scale them.

Let the radius of the base = r, and without loss of generality (you'll see why), assume that r is an integer. Let the height = r also (to spare us of complexity)
We can approximate the volume of the cone by adding the volumes of r discs, where each disk has an integral radius from 1 to r.

(Ex, the first disk has radius 1, the next has 2, the next has 3 etc... the last has radius r)
Let the height of each disk be 1.
The volume is therefore:
summation of k as k ranges from 1 to r of

times the sum of the first r perfect squares
, by the formula for the sum of the first r perfect squares.

The volume of a cylinder with the same base and height is

The ratio of the area of the cone to the cylinder is therefore

Note that if we had used a height other than r, the height would've canceled out anyway once we divided by the volume of the cylinder.

Obviously, the more disks we have (the bigger r is), the more accurate the ratio will be.
As r gets very very big,

get's very very small, and
get's very very small.

At this point, note that since r is getting very very big, the assumption that it is an integer get's less and less important, because we are taking an infinte number of "samples".

So effectively, the ratio is

, and the volume of the cone is
the volume of the cylinder.

Volume of a cylinder is Area of base x Height, so the volume of a cone is 1/3 x Area of Base x Height

Again, these kinds of approximations work just like calculus and are the foundation of calculus, but you should be able to understand it without calculus.

PS - what is the latex tag on this forum? I'm sick of writing without latex...
PPS - Thanks John

#11 Re: Help Me ! » Scaling y=1/x » 2006-01-26 10:51:52


Picture y = -1/x as scaling the curve by a scale factor of -1.
The fact that you can also rotate the curve to get -1/x is a sidenote.

#12 Help Me ! » Recursions » 2006-01-26 10:12:46

Replies: 3

Say we define a sequence such that:
a_0 = 1
and for each successive term;
a_n+1 = sin(a_n)

i. The infinite limit of the sequence a_n is 0, right?
ii. Does this infinite series have a finite sum? If so, what is it?
iii. Is there any explicit function ƒ(x) such that ƒ(x+1) = sin( ƒ(x) )?


#13 Re: Help Me ! » Is there a definition of an integer? » 2006-01-21 12:24:42


Here are my functions >:)
y = 1
y = |x| / x
y = sqrt(x^2)

#14 Re: Help Me ! » inflection points » 2006-01-11 10:57:18


Wikipedia article: Inflection point

a point on a curve at which the curvature changes sign. The curve changes from being concave upwards (positive curvature) to concave downwards (negative curvature), or vice versa. If one imagines driving a vehicle along the curve, it is a point at which the steering-wheel is momentarily 'straight', being turned from left to right or vice versa.

#15 Re: This is Cool » school » 2006-01-11 10:35:22


yeh we use slang a lot chavs use it all da time we dont need 2 be online

We use slang a lot. Chavs* speak in slang all the time, even when they are not online.


God bless urban dictionary smile

PS: No offense for the derogatory urban dictionary definition, I don't write those things

#16 Re: This is Cool » Trigonometric function with half-square graph? » 2006-01-11 10:30:48


I'm not really getting a square... at least not with the explicit equations.

The parametric square = awesome

#17 Re: Help Me ! » Long term behaviour of sequences » 2006-01-11 09:58:04


(n^4 + 2n^2) / (4n^4 + 7n)

Rick, I think he's dealing with the sequence a_n = (n^4 + 2n^2) / (4n^4 + 7n), not the sum of the infinite series.

So if that is the case, your method is correct, by dividing the top and bottom by n^4, you find that as n -> infinity, the value of a_n approaches 1/4.

If you are in fact dealing with the sum of the infinite series, keep in mind that as you are approaching a_n -> 1/4, you're adding a whole bunch of (more specifically, an infinite ammount of) 1/4's.

So the sequence converges but the series does not.

#18 Re: Help Me ! » inflection points » 2006-01-11 09:52:46


The first derivative doesn't have to be 0 to have a PoI.

For example, the function f(x) = x^3 - 3x has inflection point when x = 0 because the following two conditions are met when x = 0:

f''(x) = 6x, which is 0 when x = 0
f'''(x) = 6, is not equal to 0 when x = 0

Note that although f'(x) = 3x^2 - 3 and f'(0) = -3, the function still changes concavity.

#19 Re: Help Me ! » inflection points » 2006-01-11 09:27:51


Inflection point is when the slope changes from increasing to decreasing or vice versa via 0. In other words, f''(x) = 0 and f'''(x) is not equal to zero. Another way of putting it is that the function changes concavity.

#20 Re: Help Me ! » A cubic... » 2006-01-08 09:41:47


I guess you might as well just use Newton's recursion for these things...

#21 Re: Help Me ! » weird equation » 2006-01-08 09:34:08


By the rational roots theorem, the only possible rational solutions are -9 -3 -1 1 3 and 9
By the rule of signs, there is one sign change, so there is a positive root, so start with positive numbers.
After finding that 3 is a root, we know that we can factor out (x-3)

Use synthetic division to find out the remaining quadratic:

#22 Re: Help Me ! » derivative.. » 2006-01-08 09:20:49


Derivative =0

x^2 = 0, x = 0


2xlnxe - lnx = 0
lnx(2xe-1) = 0

lnx = 0 or 2ex - 1 - 0
x = 1 or x = 1/(2e)

So, horizontal tangents occur at
x = 1
x = 1/(2e)
x = 0

Double check that the second derivative is never equal to 0 for each of those points... when it does, it isn't an extrema

#23 Re: Help Me ! » weird equation » 2006-01-08 09:12:44


so basically x^3 - 6x - 9 = 0

We can factor out (x-3) from here, leaving
(x-3)(x^2+3*x+3) = 0

x-3 = 0, so x=3 is a solution

x^2 + 3x + 3 = 0
This can be solved easily using the quadratic formula

#24 Re: Help Me ! » Diffrentiation » 2006-01-08 05:30:07


is log natural log or base 10?

#25 Re: Help Me ! » Diffrentiation » 2006-01-08 05:25:02


y=3x^6 + sin2x +e^2x + e^4x log e^2x

Use power rule for 3x^6 -> 18x^5
Use chain rule for next two terms:
e^2x -> 2e^2x; sin2x -> 2cos2x

so so far that's dy/dx = 18 x^5 + 2 cos2x + 2 e^2x + ...

... can you put parenthases in the last term so that I know what exactly it says?

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