I am looking for a solution:
A= {0,1,2,3,4};
F(x) = 3x - 1 (mod5)
Could you help me to find the inverse. I am struggling with this as it seems to be not to be onto or 1to1.
Thank you for your help.
x = 2y + 2, where y = F(x)
-> 3x - 1 = 3(2y+2) - 1 = 6y + 5 = y (mod 5)
edit: if you want this to be evaluated for the list of principal values mod 5 [0,1,2,3,4], just evaluate 2y+2 for each of these, and what you get is [2,4,1,3,0]. Which, if you plug back into 3x-1, you get [0,1,2,3,4] as expected.
Related
I want to calculate the sum of the series as below
Lim
X->1 (2/3 - x/3 -(x^2)/3 +(x^3)*2/3 -..). I am not sure whether we have a formula for finding the sum of this kind of series. Tried a lot but couldn't find any. Any help is appreciated.
This seems to be more maths than computing.
It factorises as (1 + x^3 + x^6 + ...)(2 - x - x^2)/3
If x = 1-d (where d is small), then to first order in d, the (2 - x - x^2) term becomes (2 - (1-d) - (1-2d)) = 3d
And the (1 + x^3 + x^6 + ...) term is a geometric progression, with sum 1/(1-x^3), or here 1/(1-(1-d)^3), and the denominator to first order in d is (1 - (1-3d)) = 3d
Hence the whole thing is (1/3d) (3d) / 3 = 1/3
But we can also verify computationally with a value close to 1 (Python code here):
x = 0.999999
s = 0
f = (2 - x - x*x) / 3.
x3 = x ** 3
s_prev = None
while s != s_prev:
s_prev = s
s += f
f *= x3
print(s)
gives:
0.33333355556918565
I want to create a polynomial with given coefficients. This seems very simple but what I have found till now did not appear to be the thing I desired.
For example in such an environment;
n = 11
K = GF(4,'a')
R = PolynomialRing(GF(4,'a'),"x")
x = R.gen()
a = K.gen()
v = [1,a,0,0,1,1,1,a,a,0,1]
Given a list/vector v of length n (I will set this n and v at the begining), I want to get the polynomial v(x) as v[i]*x^i.
(Actually after that I am going to build the quotient ring GF(4,'a')[x] /< x^n-v(x) > after getting this v(x) from above) then I will say;
S = R.quotient(x^n-v(x), 'y')
y = S.gen()
But I couldn't write it.
This is a frequently asked question in many places so it is better to leave it here as an answer although the answer I have is so simple:
I just wrote R(v) and it gave me the polynomial:
sage
n = 11
K = GF(4,'a')
R = PolynomialRing(GF(4,'a'),"x")
x = R.gen()
a = K.gen()
v = [1,a,0,0,1,1,1,a,a,0,1]
R(v)
x^10 + a*x^8 + a*x^7 + x^6 + x^5 + x^4 + a*x + 1
Basically (that is, ignoring the specifics of your polynomial ring) you have a list/vector v of length n and you require a polynomial which is the sum of all v[i]*x^i. Note that this sum equals the matrix product V.X where V is a one row matrix (essentially equal to the vector v) and X is a column matrix consisting of powers of x. In Maxima you could write
v: [1,a,0,0,1,1,1,a,a,0,1]$
n: length(v)$
V: matrix(v)$
X: genmatrix(lambda([i,j], x^(i-1)), n, 1)$
V.X;
The output is
x^10+ax^8+ax^7+x^6+x^5+x^4+a*x+1
This is a fairly simple question. I need need an equation to determine whether two 2 dimensional lines collide with each other. If they do I also need to know the X and Y position of the collision.
Put them both in general form. If A and B are the same then they're parallel. Otherwise, create two simultaneous equations and solve for x and y.
Let A and B represented by this parametric form : y = mx + b
Where m is the slope of the line
Now in the case of parallelism of A and B their slope should be equal
Else they will collide with each other at point T(x,y)
For finding the coordinates of point T you have to solve an easy equation:
A: y = mx + b
B: y = Mx + B
y(A) = y(B) means : mx + b = Mx + B which yields to x = (B - b)/(m - M) and by putting
the x to the line A we find y = ((m*(B - b))/(m - M)) + b
so : T : ((B - b)/(m - M) , ((m*(B - b))/(m - M)) + b)
I have this recursive function:
f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
f(1) = 2
f(2) = 8
I know from experience that explicit form of it would be:
f(n) = 3 ^ n - 1 // pow(3, n) - 1
I wanna know if there's any way to prove that. I googled a bit, yet didn't find anything simple to understand. I already know that generation functions probably solve it, they're too complex, I'd rather not get into them. I'm looking for a simpler way.
P.S.
If it helps I remember something like this solved it:
f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
// consider f(n) = x ^ n
x ^ n = 2 * x ^ (n-1) + 3 * x ^ (n-2) + 4
And then you somehow computed x that lead to explicit form of the recursive formula, yet I can't quite remember
f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
f(n+1) = 2 * f(n) + 3 * f(n-1) + 4
f(n+1)-f(n) = 2 * f(n) - 2 * f(n-1) + 3 * f(n-1) - 3 * f(n-2)
f(n+1) = 3 * f(n) + f(n-1) - 3 * f(n-2)
Now the 4 is gone.
As you said the next step is letting f(n) = x ^ n
x^(n+1) = 3 * x^n + x^(n-1) - 3 * x^(n-2)
divide by x^(n-2)
x^3 = 3 * x^2 + x - 3
x^3 - 3 * x^2 - x + 3 = 0
factorise to find x
(x-3)(x-1)(x+1) = 0
x = -1 or 1 or 3
f(n) = A * (-1)^n + B * 1^n + C * 3^n
f(n) = A * (-1)^n + B + C * 3^n
Now find A,B and C using the values you have
f(1) = 2; f(2) = 8; f(3) = 26
f(1) = 2 = -A + B + 3C
f(2) = 8 = A + B + 9C
f(3) = 26 = -A + B + 27C
solving for A,B and C:
f(3)-f(1) = 24 = 24C => C = 1
f(2)-f(1) = 6 = 2A + 6 => A = 0
2 = B + 3 => B = -1
Finally
f(n) = 3^n - 1
Ok, I know you didn't want generating functions (GF from now on) and all the complicated stuff, but my problem turned out to be nonlinear and simple linear methods didn't seem to work. So after a full day of searching, I found the answer and hopefully these findings will be of help to others.
My problem: a[n+1]= a[n]/(1+a[n]) (i.e. not linear (nor polynomial), but also not completely nonlinear - it is a rational difference equation)
if your recurrence is linear (or polynomial), wikihow has step-by-step instructions (with and without GF)
if you want to read something about GF, go to this wiki, but I didn't get it till I started doing examples (see next)
GF usage example on Fibonacci
if the previous example didn't make sense, download GF book and read the simplest GF example (section 1.1, ie a[n+1]= 2 a[n]+1, then 1.2, a[n+1]= 2 a[n]+1, then 1.3 - Fibonacci)
(while I'm on the book topic) templatetypedef mentioned Concrete Mathematics, download here, but I don't know much about it except it has a recurrence, sums, and GF chapter (among others) and a table of simple GFs on page 335
as I dove deeper for nonlinear stuff, I saw this page, using which I failed at z-transforms approach and didn't try linear algebra, but the link to rational difference eqn was the best (see next step)
so as per this page, rational functions are nice because you can transform them into polynomials and use linear methods of step 1. 3. and 4. above, which I wrote out by hand and probably made some mistake, because (see 8)
Mathematica (or even the free WolframAlpha) has a recurrence solver, which with RSolve[{a[n + 1] == a[n]/(1 + a[n]), a[1] == A}, a[n], n] got me a simple {{a[n] -> A/(1 - A + A n)}}. So I guess I'll go back and look for mistake in hand-calculations (they are good for understanding how the whole conversion process works).
Anyways, hope this helps.
In general, there is no algorithm for converting a recursive form into an iterative one. This problem is undecidable. As an example, consider this recursive function definition, which defines the Collatz sequence:
f(1) = 0
f(2n) = 1 + f(n)
f(2n + 1) = 1 + f(6n + 4)
It's not known whether or not this is even a well-defined function or not. Were an algorithm to exist that could convert this into a closed-form, we could decide whether or not it was well-defined.
However, for many common cases, it is possible to convert a recursive definition into an iterative one. The excellent textbook Concrete Mathematics spends much of its pages showing how to do this. One common technique that works quite well when you have a guess of what the answer is is to use induction. As an example for your case, suppose that you believe that your recursive definition does indeed give 3^n - 1. To prove this, try proving that it holds true for the base cases, then show that this knowledge lets you generalize the solution upward. You didn't put a base case in your post, but I'm assuming that
f(0) = 0
f(1) = 2
Given this, let's see whether your hunch is correct. For the specific inputs of 0 and 1, you can verify by inspection that the function does compute 3^n - 1. For the inductive step, let's assume that for all n' < n that f(n) = 3^n - 1. Then we have that
f(n) = 2f(n - 1) + 3f(n - 2) + 4
= 2 * (3^{n-1} - 1) + 3 * (3^{n-2} - 1) + 4
= 2 * 3^{n-1} - 2 + 3^{n-1} - 3 + 4
= 3 * 3^{n-1} - 5 + 4
= 3^n - 1
So we have just proven that this recursive function does indeed produce 3^n - 1.
how can i calculate the polynomial that has the tangent lines (1) y = x where x = 1, and (2) y = 1 where x = 365
I realize this may not be the proper forum but I figured somebody here could answer this in jiffy.
Also, I am not looking for an algorithm to answer this. I'd just like like to see the process.
Thanks.
I guess I should have mentioned that i'm writing an algorithm for scaling the y-axis of flotr graph
The specification of the curve can be expressed as four constraints:
y(1) = 1, y'(1) = 1 => tangent is (y=x) when x=1
y(365) = 1, y'(365) = 0 => tangent is (y=1) when x=365
We therefore need a family of curves with at least four degrees of freedom to match these constraints; the simplest type of polynomial is a cubic,
y = a*x^3 + b*x^2 + c*x + d
y' = 3*a*x^2 + 2*b*x + c
and the constraints give the following equations for the parameters:
a + b + c + d = 1
3*a + 2*b + c = 1
48627125*a + 133225*b + 365*c + d = 1
399675*a + 730*b + c = 0
I'm too old and too lazy to solve these myself, so I googled a linear equation solver to give the answer:
a = 1/132496, b = -731/132496, c = 133955/132496, d = -729/132496
I will post this type of question in mathoverflow.net next time. thanks
my solution in javascript was to adapt the equation of a circle:
var radius = Math.pow((2*Math.pow(365, 2)), 1/2);
var t = 365; //offset
this.tMax = (Math.pow(Math.pow(r, 2) - Math.pow(x, 2), 1/2) - t) * (t / (r - t)) + 1;
the above equation has the above specified asymptotes. it is part of a step polynomial for scaling an axis for a flotr graph.
well, you are missing data (you need another point to determine the polynomial)
a*(x-1)^2+b*(x-1)+c=y-1
a*(x-365)^2+b*(x-365)+c=y-1
you can solve the exact answer for b
but A depends on C (or vv)
and your question is off topic anyways, and you need to revise your algebra