使用Fortran编写用于定点迭代的程序 [英] Write a program for a fixed point iteration using Fortran

问题描述

大家好，我是编程的新手。鉴于给定定点迭代的编程，我遇到了很大的麻烦。

 P = g（Po）
g（ x）= x-  sin （pi.x），pi与22/7相同。 
 Po =  0 。 95  
 TOL = 10 **  -   6  
 NO =  20  
 TOL  if （Pi-Po）< = 10 **  -   6  
 
步骤
  1 。设置i = 1 
  2 。  while  i< No，执行步 3   -   6  
  3 。设置P = g（Po）
  4 。测试如果 | P-Po | < TOL 
输出（P）; （过程成功完成）
  STOP  
  5 。设置i = i + 1 
  6 。设置P = Po 
  7 输出（没有迭代后方法失败）
   NO，NO; （过程已成功完成）
 停止 

SIMPLY FORTRAN是我能够下载的FORTRAN软件。请问我该怎么办？
为上述问题编写算法。谢谢



我尝试了什么：



我试过了使用视频教程但似乎没有任何工作。 SIMPLY FORTRAN是我能够下载的软件的名称，尝试下载FORTRAN 90或77，但没有。如果您有这样的软件并愿意与我分享。我很高兴。

解决方案

好吧，我后来和其他网站的人交谈过，他们建议我停止使用Fortran并尝试使用QB64，观看视频它看，我成了胜利者。这是使用QBasic的解决方案。享受吧！

 CLS 
i = 1：po = 0.95：tol = 10 ^ -6：no = 20：CONST pi = 4 * ATN1 
 PRINT ,,; 课程代码分配
打印，; 通过候选名称
打印i，po，p，ABS（p-po）
 2如果i< no THEN 
p = po  -  SIN（pi * po）
 PRINT i，
 PRINT po，
 PRINT p，
 PRINT ABS（p  -  po）
 IF ABS（p-po）<那么
 PRINT（p）; 程序成功完成
 END IF 
i = i + 1：po = p 
 GOTO 2 
 ELSE 
 PRINT方法在20次迭代后失败*
 END IF 

Hello everyone, I am a complete novice to programming. I am facing a big trouble writing a programming for a fixed point iteration given

P= g(Po)
 g(x) = x-sin (pi.x), pi as in 22/7.
Po = 0.95
TOL = 10**-6
NO = 20
TOL if (Pi-Po)<=10**-6

Steps
1. Set i=1
2. While i <No,  do step 3-6
3. Set P=g(Po)
4. Test if |P-Po| < TOL
OUTPUT (P); (Procedure completed successfully )
STOP
5. Set i = i+1
6. Set P = Po
7 output (method failed after no iteration)
"NO", NO; (Procedure completed Successfully)
STOP

SIMPLY FORTRAN is the FORTRAN software I was able to download. Please how can I
Write the algorithm for the above problem. Thank you

What I have tried:

I've tried using Video tutorials but nothing seems to be working. SIMPLY FORTRAN is the name of the software I was able to download, tried downloading FORTRAN 90, or 77 but nothing. Please If you have such software and willing to share with me. Please I'd be very glad.

解决方案

Well, i later talked to guys from another site and they recommended i stop using simply Fortran and try QB64, watched videos on it and behold, i became a victor. Here is the solution Using QBasic. ENJOY!

CLS
i = 1: po = 0.95: tol = 10 ^ -6: no = 20: CONST pi = 4 * ATN1
PRINT , , " "; "COURSE CODE ASSIGNMENT"
PRINT , "       "; "By CANDIDATE NAME"
PRINT "i", "po", "p", "ABS(p-po)"
2 IF i < no THEN
    p = po - SIN(pi * po)
    PRINT i,
    PRINT po,
    PRINT p,
    PRINT ABS(p - po)
    IF ABS(p - po) < tol THEN
        PRINT (p); "Procedure completed successfully"
    END IF
    i = i + 1: po = p
    GOTO 2
ELSE
    PRINT "method failed after 20 iterations*"
END IF

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