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Have you tried anything yet?
What are some of the biggest obstacles to solving this equation in its current state?
What have you done so far?
I think the best strategy would be to multiply x²+x+1 by other polys and see if you can get another polynomial out that only has 3 nonzero terms. One thing you know about these polys that they must have a constant term of 1. Do you see why?
Also, you could try making a multiplication grid:
x²
x
1 1
1 ax ...
Basically, you want to see what kinds of polynomials you can get out of this multiplication.
First of all, Rolle's Theorem allows for a very easy proof of the MVT. The MVT is essential to many proofs in analysis, for example, relating the second derivative definition of concave up/down with a secant definition. There are many, many instances when it is extremely difficult, if not impossible, to calculate the derivative of a differentiable function, but an application of MVT may be able to give you all you need.
Do you want a single most likely value? Or do you want a most likely value with error values on x and y? Cause the first one is waaay easier.
So, the graph was like a step functon then, because you can't really graph a summation so it's smooth like a polynomial or something.
How were you able to graph that sum?
It seems like, if you have your n integers arranged in a row, you will have to choose some starting number, and then pick a number that borders on it for the second number in the series. Then pick a number than is a neigbor on the side of that group. It just continues to build up on the original number picked, like a one-dimesional pearl building on a piece of sand. How's that for a math metaphore!
If I understand you right, you want to try to find patterns in the decimals of pi. I would think the best way to do this is to start from the beginning (of pi). Granted, some of the power series for calculating pi converge slowly, and are therefor impractical. However, the Chudnovsky algorithm that I gave above converges very quickly, giving 14 decimal places of pi per term of the series.
An excellent site describing the methods for finding pi throughout history is the one given below.
http://www.cecm.sfu.ca/organics/papers/borwein/index.html
Unfortunately, this seemingly simple and straight forward expansion for arcsine is very, very, very slow. I tried it on my TI-86 and it took forever. It is simply not practical to compute pi by using 2*arcsin(1).
The image at the bottom of the page shows the Chudnovsky algorithm, which is the fastest way, according to MathWorld, to get lots of digits of pi. The Chudnovsky's are very smart. Their algorithm is complicated. It is based on harder math than arcsine.
Thanks Patrick!
Okay, so I was just wondering how fast the convergence on that series Krassi poster is.
I also figured that the sigma notation for that super slow converging arcsine series is:
The x term will just drop out if you are looking for the arsine of 1.
I'm sorry, but could someone just show in quotes how to put Latex into this board. It should be simple, I know, but I can't get it to work. It's not just: "\[_code_\]".
As far as that first integral, i would just replace (sin x)^3 with (1-(cos x)^2)(sin x) and sub in with u=cos x
du=-sinx dx
Here's a good question:
Given the number of different ways of representing date and time, how many times are there that the digits in the date/time string will be consecutive in either increasing or decreasing order?
One thing we can note is that the sine of 90 degrees is 1. Then, in radians, the sine of pi/2 is 1.
The arcsine (sin-¹) of 1 is 90 degrees, or in radians it is pi/2. So then we have:
pi=2*sin-¹ (1)
If we can find a way to calculate sin-¹ as accurately as we want, that will let us calculate pi as accurately as we want.
It turns out that there is a way to do this:
sin-¹(x)=x+1/2*(x^3)/3+(1*3)/(2*4)*(x^5)/5+(1*3*5)/(2*4*6)*(x^7)/7+...
However, this is a very sloooooooooooowwwww way to get pi. I know, because I put a program for it on my calc, and it took forever just to get the first 2 digits of pi correctly. There are lots of faster ways than this.
Good Work! The formula that you came up with is a good one. The only problem with it is the cos(θ) part. Since we do have caclulators and use them all the time, we kind of take for granted the fact that we can always have a perfectly accurate cosine function. The problem is, our calculators actually only give acurate trigonometric functions for the 10 digits that show on our calculators; or 11 digits, or 12, or 13...anyway, the calculator is not accurate for an infinite number of digits, so what this problem really comes down to is:
How can we get infinitly accurate trig functions, or more truthfully, functions as accurate as we want to make them?
Before proceding on, I think it would be useful to know how much math you've had, especially experience with: radians, trigonometric functions, inverse trigonometric functions, and sigma notation.
P.S. It is interesting to note that a few years ago, I myself pondered the same question about pi.
Ok, thanks!
Here's the mathworld website:
http://mathworld.wolfram.com/Infinitesimal.html
hmm..i don't know why, but my infinity sign came out as a '?'
I think they often call 1/? infintessimal, try checking it out on mathworld (google mathworld, it's the first result).
The way I look at it:
f(x)=1/x
limit f(x) as 1->? is 0
This is especially true if we are applying 1/? in a prarical setting.
Hmm...
I would almost just write off the initial '364' as having no control over the value of the string, but then again, all the resulting numbers have 8 digits.
Is the some way you could check what the numeric values of other strings is?
Where did this problem come from?
Ok I admit it, this is more of a ponder than a discovery.
First, by ^n(x), I mean the function iterated on x, n times.
Now, if P(x) is a polynomial, is P^∞(x) a polynomial?
Is it continous?
Is it a step (also called discrete, I think) function?
Is it just a fractal set?
Hmmm....
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