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?
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.
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?)
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)
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
The volume of a cylinder with the same base and height is
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
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) )?
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
We use slang a lot. Chavs* speak in slang all the time, even when they are not online.
God bless urban dictionary
PS: No offense for the derogatory urban dictionary definition, I don't write those things
(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.
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:
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