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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?

After all, problems are a lot cooler to solve by hand than to approximate with computer

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?)

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.

0.03 seconds using the mathematica online integrator.

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 =

permutationsIsn'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

Thanks. I had a typo on part iii though:

I meant such that (x+1) = sin( (x) )

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,

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

The fact that you can also rotate the curve to get -1/x is a sidenote.

**God**- 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 (x) such that (x+1) = sin( (x) )?

-thanks

Here are my functions >:)

y = 1

y = |x| / x

y = sqrt(x^2)

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.

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

Translation:

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

*see: http://www.urbandictionary.com/define.php?term=chavs

God bless urban dictionary

-----

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

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

The parametric square = awesome

(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.

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.

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

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:

Derivative =0

x^2 = 0, x = 0

or

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

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

is log natural log or base 10?

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?