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R-Raise the name of the Fraternity.
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What is the first term in the Fibonacci sequence to contain 1000 digits?
After the first few numbers in the Fib sequence, Fib(n) is very nearly equal to (phi)n / sqrt(5) where phi is the Golden Ratio = [1+sqrt(5)]/2 [The difference is around 0.00003 by Fib(20)] So you want the smallest n such that (phi)n / sqrt(5) ≥ 10999 Taking logs, n*log(phi) - 0.5*log(5) ≥ 999 n*log(phi) ≥ 999 + 0.5*log(5) = 999.349 n ≥ 999.349/log(phi) = 999.349/0.2090 So n = 4781
How to convert degree to radian?
n degrees = n*tau/360 or n*pi/180 radians.
What has the author Geoffrey N Pendleton written?
Geoffrey N. Pendleton has written: 'BATSE data analysis' -- subject(s): Gamma ray bursts, Observations
Does N have rotational symmetry but no line symmetry?
Yes, the capital letter N has rotational symmetry but no lines of symmetry:
How do you calculate the nth term in the Fibonacci sequence?
For low terms, it is probably most economical to to simply list out the sequence until you arrive at the nth term, but for large terms, there actually is a closed-form equation. Let (phi) = (1 + sqrt(5))/2. As an aside, this is the golden ratio. Then the nth term of the Fibonacci sequence is given by F(n) = [(phi)^n - (1 - (phi))^n]/sqrt(5), where we are letting F(0) = 0, F(1) = 1, F(2) = 1, etc. The derivation of this formula involves linear algebra, and, in short, the key is in setting this up as a matrix equation and diagonalizing the matrix. While I shall not run through the process here, I will give you the matrix and vector to start with: Let A be the 2x2 matrix [0 1] [1 1] and x(n) be the column vector in R^2 [F(n)] [F(n+1)] In particular, x(0) = (F(0),F(1)) = (0,1). notice that [A][x(n)] = [F(n+1)] [F(n)+F(n+1)] which is x(n+1). Thus, it follows that x(n) = [A]^n[x(0)] (you can prove this rigorously through induction). Now, F(n) is the first term of the vector x(n), so what one needs to do is diagonalize A, raise A to the appropriate power, multiply it with x(0), and finally take the upper term. The result is what I presented at the outset. Note that with this formula you can show that lim(n-->infinity)[F(n+1)/F(n)] approaches phi.
