I'm trying to write some code that'll calculate the value of sin(0.75) using the Taylor expansion, and print each iteration until the absolute difference between the value calculated using the expansion, and the value calculated using Fortran's intrinsic sin function is less than 1E-6. Here is my code:
program taylor
implicit none
real :: x = 0.75
do while (x - sin(0.75) < 10**(-6))
print *, x
x = x - ((x**3)/6) + ((x**5)/120) - ((x**7)/5040)
end do
end program taylor
However, this doesn't print anything out? Why is this?
It looks too obvious to most people so no-one even wanted to answer, but it should be said explicitly
The condition x - sin(0.75) < 10**(-6) is obviously not true when x very different from sin(0.75), so the do while loop is never entered.
Also, as IanH commented 10**(-6) will give 0 because the result of the power of two integers is again an integer. The literal real number 10^-6 should be expressed as 1e-6.
If you change it to x - sin(0.75) > 1e-6 the loop will proceed, but it will run forever, because your iteration is wrong. Taylor series works differently, you should compute
y = 0
y = y + x**1/1!
y = y - x**3/3!
y = y + x**5/5!
y = y - x**7/7!
...
and so on, which is a very different kind of loop.
Try this one:
program taylor
implicit none
real :: x = 0.75
real :: y, fact
integer :: sgn, i
fact = 1
sgn = 1
y = 0
do i = 1, 10, 2
y = y + sgn * x**i / fact
fact = fact*(i+1)*(i+2)
sgn = -sgn
end do
print *, y, sin(x)
end program taylor
Related
I'm attempting to write a FORTRAN 90 program that calculates Pi using random numbers. These are the steps I know I need to undertake:
Create a randomly placed point on a 2D surface within the range [−1, 1] for x and y, using call random_number(x).
calculate how far away the point is from the origin, i'll need to do both of these steps for N points.
for each N value work out the total amount of points that are less than 1 away from origin. Calculate pi with A=4pir^2
use a do loop to calculate pi as a function of N and output it to a data file. then plot it in a graphing package.
This is what I have:
program pi
implicit none
integer :: count, n, i
real :: r, x, y
count = 0
CALL RANDOM_SEED
DO i = 1, n
CALL RANDOM_NUMBER(x)
CALL RANDOM_NUMBER(y)
IF (x*x + y*Y <1.0) count = count + 1
END DO
r = 4 * REAL(count)/n
print *, r
end program pi
I know i've missed out printing the results to the data file, i'm not sure on how to implement this.
This program gives me a nice value for pi (3.149..), but how can I implement step 4, so that it outputs values for pi as a function of N?
Thanks.
Here is an attempt to further #meowgoesthedog effort...
Program pi
implicit none
integer :: count, n, i
real :: r, x, y
count = 0
Integer, parameter :: Slice_o_Pie = 8
Integer :: Don_McLean
Logical :: Purr = .FALSE.
OPEN(NEWUNIT=Don_McLean, FILE='American.Pie')
CALL RANDOM_SEED
DO i = 1, n
CALL RANDOM_NUMBER(x)
CALL RANDOM_NUMBER(y)
IF (x*x + y*Y <1.0) count = count + 1
Purr = .FALSE.
IF(MODULO(I, Slice_o_Pie) == 0) Purr = .TRUE.
IF (Purr) THEN
r = 4 * REAL(count)/i
print *, i, r
WRITE(LUN,*) 'I=',I,'Pi=',Pi
END IF
END DO
CLOSE(Don_McLean)
end program pi
Simply put the final calculation step inside the outer loop, and replace n with i. Also maybe add a condition to limit the number of results printed, e.g. i % 100 = 0 to print every 100 iterations.
program pi
implicit none
integer :: count, n, i
real :: r, x, y
count = 0
CALL RANDOM_SEED
DO i = 1, n
CALL RANDOM_NUMBER(x)
CALL RANDOM_NUMBER(y)
IF (x*x + y*Y <1.0) count = count + 1
IF ([condition])
r = 4 * REAL(count)/i
print *, i, r
END IF
END DO
end program pi
I have already posted a question regarding this problem, and have implemented what i've learned from the answers. I'm now at a point where the answers that are printed out on the screen are very close, but incorrect. Here is the code I now have:
program taylor
implicit none
integer :: k = 0
real :: y = 0.75
real :: x = 0.75
do while (abs(y - sin(0.75)) > 1E-6)
k = k + 1
y = y + ((y * (-x * x)) /( 2 * k * (2 * k + 1 )))
print *, y
end do
end program taylor
I can't seem to spot the error here, why is this not working? The first answer it prints is correct, but then it seems to get progressively lower, instead of closer to the true value. (The do while loop is to ensure the program stops when the absolute value between the calculation and the intrinsic sin function is less than 1E-6). I can see that the program is constantly reducing the final output, and the Taylor series is suppose to alternate between - and +, so how can I write that in my program?
Thanks.
The taylor series uses factorials and power of x.
taylor_sin(x) = sum[0->n] ( [(-1 ^ n) / (2n + 1)!] * x**(2n + 1) )
(Sorry for my poor keyboard math skills...)
program taylor
implicit none
integer :: n = 0
real :: fac = 1
real :: y = 0.75
real :: x = 0.75
real :: px = 1
integer :: sgn = -1
integer :: s = 1
do while (abs(y - sin(0.75)) > 1E-6)
n = n + 1
fac = fac * (2 * n) * ((2 * n) + 1)
px = px * (x * x)
s = s * sgn
y = y + ((s /fac) * (px * x))
print *, y
end do
end program taylor
Computing the factorial fac in a loop is rather trivial.
To compute x**(2n + 1), compute and keep x**(2n) and add an extra multiplication by x.
I use an integer to compute -1**n since using a real may lose precision over time.
[EDIT] silly me... you can save that extra multiplication by x by setting px to 0.75 before the loop.
I'm attempting to write a program that calculates the value of sin(0.75) using the Taylor series, until the absolute difference between the calculated value and FORTRAN'S intrinsic sin function is less then 1E-6, printing each iteration to the screen. (it also needs to print the absolute difference at each step, but I haven't gotten round to implementing that yet!). Here's my code:
program taylor
implicit none
real :: x = 0.75
if (abs(x - sin(0.75)) > 10.00**(-7)) then
print *, x
x = x - ((x**3)/6)
x = x + ((x**5)/120)
x = x - ((x**7)/5040)
end if
end program taylor
This just prints out 0.75 to the terminal, why is this? How can I fix it so that it does what I specified above?
Thanks, any help is appreciated.
So i'm implementing a heuristic algorithm, and i've come across this function.
I have an array of 1 to n (0 to n-1 on C, w/e). I want to choose a number of elements i'll copy to another array. Given a parameter y, (0 < y <= 1), i want to have a distribution of numbers whose average is (y * n). That means that whenever i call this function, it gives me a number, between 0 and n, and the average of these numbers is y*n.
According to the author, "l" is a random number: 0 < l < n . On my test code its currently generating 0 <= l <= n. And i had the right code, but i'm messing with this for hours now, and i'm lazy to code it back.
So i coded the first part of the function, for y <= 0.5
I set y to 0.2, and n to 100. That means it had to return a number between 0 and 99, with average 20.
And the results aren't between 0 and n, but some floats. And the bigger n is, smaller this float is.
This is the C test code. "x" is the "l" parameter.
//hate how code tag works, it's not even working now
int n = 100;
float y = 0.2;
float n_copy;
for(int i = 0 ; i < 20 ; i++)
{
float x = (float) (rand()/(float)RAND_MAX); // 0 <= x <= 1
x = x * n; // 0 <= x <= n
float p1 = (1 - y) / (n*y);
float p2 = (1 - ( x / n ));
float exp = (1 - (2*y)) / y;
p2 = pow(p2, exp);
n_copy = p1 * p2;
printf("%.5f\n", n_copy);
}
And here are some results (5 decimals truncated):
0.03354
0.00484
0.00003
0.00029
0.00020
0.00028
0.00263
0.01619
0.00032
0.00000
0.03598
0.03975
0.00704
0.00176
0.00001
0.01333
0.03396
0.02795
0.00005
0.00860
The article is:
http://www.scribd.com/doc/3097936/cAS-The-Cunning-Ant-System
pages 6 and 7.
or search "cAS: cunning ant system" on google.
So what am i doing wrong? i don't believe the author is wrong, because there are more than 5 papers describing this same function.
all my internets to whoever helps me. This is important to my work.
Thanks :)
You may misunderstand what is expected of you.
Given a (properly normalized) PDF, and wanting to throw a random distribution consistent with it, you form the Cumulative Probability Distribution (CDF) by integrating the PDF, then invert the CDF, and use a uniform random predicate as the argument of the inverted function.
A little more detail.
f_s(l) is the PDF, and has been normalized on [0,n).
Now you integrate it to form the CDF
g_s(l') = \int_0^{l'} dl f_s(l)
Note that this is a definite integral to an unspecified endpoint which I have called l'. The CDF is accordingly a function of l'. Assuming we have the normalization right, g_s(N) = 1.0. If this is not so we apply a simple coefficient to fix it.
Next invert the CDF and call the result G^{-1}(x). For this you'll probably want to choose a particular value of gamma.
Then throw uniform random number on [0,n), and use those as the argument, x, to G^{-1}. The result should lie between [0,1), and should be distributed according to f_s.
Like Justin said, you can use a computer algebra system for the math.
dmckee is actually correct, but I thought that I would elaborate more and try to explain away some of the confusion here. I could definitely fail. f_s(l), the function you have in your pretty formula above, is the probability distribution function. It tells you, for a given input l between 0 and n, the probability that l is the segment length. The sum (integral) for all values between 0 and n should be equal to 1.
The graph at the top of page 7 confuses this point. It plots l vs. f_s(l), but you have to watch out for the stray factors it puts on the side. You notice that the values on the bottom go from 0 to 1, but there is a factor of x n on the side, which means that the l values actually go from 0 to n. Also, on the y-axis there is a x 1/n which means these values don't actually go up to about 3, they go to 3/n.
So what do you do now? Well, you need to solve for the cumulative distribution function by integrating the probability distribution function over l which actually turns out to be not too bad (I did it with the Wolfram Mathematica Online Integrator by using x for l and using only the equation for y <= .5). That however was using an indefinite integral and you are really integration along x from 0 to l. If we set the resulting equation equal to some variable (z for instance), the goal now is to solve for l as a function of z. z here is a random number between 0 and 1. You can try using a symbolic solver for this part if you would like (I would). Then you have not only achieved your goal of being able to pick random ls from this distribution, you have also achieved nirvana.
A little more work done
I'll help a little bit more. I tried doing what I said about for y <= .5, but the symbolic algebra system I was using wasn't able to do the inversion (some other system might be able to). However, then I decided to try using the equation for .5 < y <= 1. This turns out to be much easier. If I change l to x in f_s(l) I get
y / n / (1 - y) * (x / n)^((2 * y - 1) / (1 - y))
Integrating this over x from 0 to l I got (using Mathematica's Online Integrator):
(l / n)^(y / (1 - y))
It doesn't get much nicer than that with this sort of thing. If I set this equal to z and solve for l I get:
l = n * z^(1 / y - 1) for .5 < y <= 1
One quick check is for y = 1. In this case, we get l = n no matter what z is. So far so good. Now, you just generate z (a random number between 0 and 1) and you get an l that is distributed as you desired for .5 < y <= 1. But wait, looking at the graph on page 7 you notice that the probability distribution function is symmetric. That means that we can use the above result to find the value for 0 < y <= .5. We just change l -> n-l and y -> 1-y and get
n - l = n * z^(1 / (1 - y) - 1)
l = n * (1 - z^(1 / (1 - y) - 1)) for 0 < y <= .5
Anyway, that should solve your problem unless I made some error somewhere. Good luck.
Given that for any values l, y, n as described, the terms you call p1 and p2 are both in [0,1) and exp is in [1,..) making pow(p2, exp) also in [0,1) thus I don't see how you'd ever get an output with the range [0,n)
This is something that's I've wanted to know recently, mostly out of curiousity. I'm in the mood to learn some old coding styles, and FORTRAN seems like a good place to start.
I guess I should help you guys out by providing a good starting point.
So how would this C procedure be written in FORTRAN?
int foo ( int x , int y )
{
int tempX = x ;
x += y / 2 ;
y -= tempX * 3 ; // tempX holds x's original value.
return x * y ;
}
I know the entire function could be a single line:
return ( x + ( y / 2 ) ) * ( y - ( x * 3 ) ) ;
But the point of me asking this question is to see how those four statements would be written individually in FORTRAN, not neccessarily combined into a single statement.
Don't blame me - you said old coding styles:
C234567
SUBROUTINE FOO(I,J,K)
C SAVE THE ORIGINAL VALUES
IORIG = I
JORIG = J
C PERFORM THE COMPUTATION
I = I + J/2
J = J - IORIG*3
K = I * J
C RESTORE VALUES
I = IORIG
J = JORIG
END SUBROUTINE FOO
I shudder as I write this, but
all variables are implicitly integers, since they start with letters between I and N
FORTRAN passes by reference, so reset I and J at the end to give the same effect as the original function (i.e. pass-by-value, so x and y were unchanged in the calling routine)
Comments start in column 1, actual statements start in column 7
Please, please, please never write new code like this unless as a joke.
Your function might look like this in Fortran
integer function foo(m, n)
integer i
i = m
m = m + n/2
n = n - i*3
foo = m*n
end function foo
You should be able to get started with any tutorial on Fortran. Some thing like this might help http://www.cs.mtu.edu/~shene/COURSES/cs201/NOTES/fortran.html
cheers
See Functions and Subroutines:
INTEGER FUNCTION foo(i, j)
...
foo = 42
END
then later:
k = foo(1, 2)
Where do you learn FORTRAN from? Just take a look at the wikibooks!
Derived from the example, I'd say:
function func(x, y) result(r)
integer, intent(in) :: x, y
integer :: r
integer :: tempX
tempX = x
x = x / 2
y = y - tempX * 3
r = x * y
end function foo
Similar to above, but with a main program to illustrate how it would be called.
C2345678
program testfoo
implicit none
integer r, foo
r = foo(4,5)
print *, 'result = ', r
end
integer function foo(x,y)
integer x, y
integer tx, ty
tx = x + y / 2
ty = y - x * 3
foo = tx * ty
return
end
Note that this is Fortran 77, which is what I learned 23 years ago.
True old style in applying the rule IJKLMN for integer
C2345678
FUNCTION IFOO(I,J)
II = I + J/2
JJ = J - I*3
IFOO = II*JJ
END