C code homework help (syntax error)

[JewishBacon]

this is my code that i am working on, but i cant seem to get it to work. *warning math terms inbound* the end goal is to take the integral of x *ln(x) from [0,1] using simpsons method which should be 1/4. basically im taking the area under a curve by using a formula for those of you that dont know what an integral or simpsons rule is.  ANYWAY this program returns fx to be zero, and since the sum is started at zero i get zero for my answer, the odd part is the 3 terms that make up fx each have a nonzero value... i think my issue is a syntax error, but i am new to C and cant really see where the error is

 

#include <stdio.h>#include <math.h> int main(){double g,x,count,sum,fx,t1,t2,t3,r,m;sum=0;r=10;m=logf(6);        for (count=1;count<=r;count++)        {        x=count/r;        t1=x*logf(x);        g=x+1;        t2=4*((g/2)*logf (g));        t3=1*logf (1);        fx=(1/6)*(t1+t2+t3);        sum=sum+fx;        printf ("sum =%lf  x=%lf  fx= %lf t1=%lf   t2=%lf   t3=%lf \n",sum,x,fx, t1,t2,t3);        } printf ("the requested integral is = ___%lf___\n",sum);return 0;}

Edited April 8, 2014 by alpenwasser
added code tags

[fizzlesticks]

The problem is that (1/6) is an integer division that results in 0. To make it a floating point division change it to (1.0/6.0)

[aidenrelkoff]

I take Pre-calc accelerated Math 10, Have a vast knowlage of HTML, Minecraft plugin coding and Java script and im lost here...

[fizzlesticks]

Also logf(1) is zero which probably isn't what you're expecting.

[JewishBacon]

The problem is that (1/6) is an integer division that results in 0. To make it a floating point division change it to (1.0/6.0)

ya thats getting me a value at least, but its still not what im going for, do you see anything else? it might be a math error though

[Ciccioo]

The problem is that (1/6) is an integer division that results in 0. To make it a floating point division change it to (1.0/6.0)

yep

or just

fx = (t1 + t2 + t3) / 6;

ya thats getting me a value at least, but its still not what im going for, do you see anything else? it might be a math error though

no? that 0 is multiplying t1+t2+t3 so it screws up everything

[JewishBacon]

Also logf(1) is zero which probably isn't what you're expecting.

i would like it to be 0, because i know it is, however c is spitting out 1.21 for t3....

why you do this c?!

[JewishBacon]

yep

or just

fx = (t1 + t2 + t3) / 6;

no? that 0 is multiplying t1+t2+t3 so it screws up everything

i have changed the code to this

 

#include <stdio.h>

#include <math.h>

 

int main()

{

double g,x,count,sum,fx,t1,t2,t3,r,m;

sum=0;

r=10;

m=logf(6);

        for (count=1;count<=r;count++)

        {

        x=count/r;

        t1=x*logf(x);

        g=x+1;

        t2=4*((g/2)*logf (g));

        t3=1*logf(1);

        fx=(1.0/6.0)*(t1+t2+t3);

        sum=sum+fx;

        printf ("sum =%lf  x=%lf  fx= %d t1=%lf   t2=%f   t3=%lf \n",sum,x,fx, t1,t2,t3);

        }

 

printf ("the requested integral is = ___%lf___\n",sum);

return 0;

}

[Ciccioo]

i would like it to be 0, because i know it is, however c is spitting out 1.21 for t3....

why you do this c?!

are you sure? it gives 0 to me

[JewishBacon]

 

are you sure? it gives 0 to me

 

 

Use "fg" to return to Pico

 

[4]+  Stopped                 pico -z simpson.c

 

sum =-0.003429  x=0.100000  fx= 286331157 t1=-0.003429   t2=-0.230259   t3=0.209682

sum =0.015851  x=0.200000  fx= 1145324610 t1=0.019281   t2=-0.321888   t3=0.437572

sum =0.069344  x=0.300000  fx= 1861152496 t1=0.053493   t2=-0.361192   t3=0.682147

sum =0.165278  x=0.400000  fx= -858993460 t1=0.095934   t2=-0.366516   t3=0.942122

sum =0.310248  x=0.500000  fx= -536870912 t1=0.144970   t2=-0.346574   t3=1.216395

sum =0.509835  x=0.600000  fx= 1073741824 t1=0.199586   t2=-0.306495   t3=1.504012

sum =0.768912  x=0.700000  fx= -733723580 t1=0.259077   t2=-0.249672   t3=1.804136

sum =1.091831  x=0.800000  fx= -1968526677 t1=0.322919   t2=-0.178515   t3=2.116032

sum =1.482535  x=0.900000  fx= -1281779303 t1=0.390703   t2=-0.094824   t3=2.439045

sum =1.944633  x=1.000000  fx= 0 t1=0.462098   t2=0.000000   t3=2.772589

the requested integral is = ___1.944633___

 

 

 

 

this is what is returned for this code 

 

 

#include <stdio.h>

#include <math.h>

 

int main()

{

double g,x,count,sum,fx,t1,t2,t3,r,m;

sum=0;

r=10.0;

m=logf(6.0);

        for (count=1;count<=r;count++)

        {

        x=count/r;

        t1=x*logf(x);

        g=x+1.0;

        t2=4.0*((g/2)*logf (g));

        t3=1.0*logf(1.0);

        fx=(1.0/6.0)*(t1+t2+t3);

        sum=sum+fx;

        printf ("sum =%lf  x=%lf  fx= %d t1=%lf   t2=%f   t3=%lf \n",sum,x,fx, t1,t2,t3);

        }

 

printf ("the requested integral is = ___%lf___\n",sum);

return 0;

}

 

 

[fizzlesticks]

i would like it to be 0, because i know it is, however c is spitting out 1.21 for t3....

why you do this c?!

In that case just get rid of it, 1*logf(1) is always 0 and t3 won't have any effect on the rest of the problem.

[fizzlesticks]

 

 

sum =-0.003429  x=0.100000  fx= 286331157 t1=-0.003429   t2=-0.230259   t3=0.209682

sum =0.015851  x=0.200000  fx= 1145324610 t1=0.019281   t2=-0.321888   t3=0.437572

sum =0.069344  x=0.300000  fx= 1861152496 t1=0.053493   t2=-0.361192   t3=0.682147

sum =0.165278  x=0.400000  fx= -858993460 t1=0.095934   t2=-0.366516   t3=0.942122

sum =0.310248  x=0.500000  fx= -536870912 t1=0.144970   t2=-0.346574   t3=1.216395

sum =0.509835  x=0.600000  fx= 1073741824 t1=0.199586   t2=-0.306495   t3=1.504012

sum =0.768912  x=0.700000  fx= -733723580 t1=0.259077   t2=-0.249672   t3=1.804136

sum =1.091831  x=0.800000  fx= -1968526677 t1=0.322919   t2=-0.178515   t3=2.116032

sum =1.482535  x=0.900000  fx= -1281779303 t1=0.390703   t2=-0.094824   t3=2.439045

sum =1.944633  x=1.000000  fx= 0 t1=0.462098   t2=0.000000   t3=2.772589

the requested integral is = ___1.944633___

Well that just don't make no damn sense. The end result is the same as mine where t3 is 0. But whatever value you get for t3 you get it shouldn't be changing.. try a different compiler maybe?

[Ciccioo]

i can't really say why, but the error is the warning given by

printf ("sum =%lf  x=%lf  fx= %d t1=%lf   t2=%f   t3=%lf \n",sum,x,fx, t1,t2,t3);

if you put a %lf after fx=, it works

printf prints t3 wrong

must investigate!

 

edit:

when i say "it works", i mean "it displays the real value of t3"

the result is always the same

 

edit again:

it seems like printf says "hey, i'm expecting an int but you're giving me a double. ah, whatever, i'll just print a shitrandom int, and i will place your double in the next substitution that, luckily, is expecting a double"

 

so, those wrong values of t3 are actually values of t2: all the printf parameters are shifted starting from fx

[JewishBacon]

i can't really say why, but the error is the warning given by

printf ("sum =%lf  x=%lf  fx= %d t1=%lf   t2=%f   t3=%lf \n",sum,x,fx, t1,t2,t3);

if you put a %lf after fx=, it works

printf prints t3 wrong

must investigate!

 

edit:

when i say "it works", i mean "it displays the real value of t3"

the result is always the same

ok well that seems to work for getting t3 to be correct, but this answer is bugging me 

 

ass i make the value for "r" higher, the end value for sum should approach .25, however it just gets larger

[JewishBacon]

the most current version of the code is as follows 
 

 

 

#include <stdio.h>

#include <math.h>

 

int main()

{

double g,x,count,sum,fx,t1,t2,t3,r,m;

sum=0;

r=10.0;

m=logf(6.0);

        for (count=1;count<=r;count++)

        {

        x=count/r;

        t1=x*logf(x);

        g=x+1.0;

        t2=4.0*((g/2)*logf (g));

        t3=1.0*log(1.0);

        fx=(1.0/6.0)*(t1+t2+t3);

        sum=sum+fx;

        printf ("sum =%lf  x=%lf  fx= %lf t1=%lf   t2=%f   t3=%lf \n",sum,x,fx, t1,t2,t3);

        }

 

printf ("the requested integral is = ___%lf___\n",sum);

return 0;

}

 

 

returns this 

 

 

sum =-0.003429  x=0.100000  fx= -0.003429 t1=-0.230259   t2=0.209682   t3=0.000000

sum =0.015851  x=0.200000  fx= 0.019281 t1=-0.321888   t2=0.437572   t3=0.000000

sum =0.069344  x=0.300000  fx= 0.053493 t1=-0.361192   t2=0.682147   t3=0.000000

sum =0.165278  x=0.400000  fx= 0.095934 t1=-0.366516   t2=0.942122   t3=0.000000

sum =0.310248  x=0.500000  fx= 0.144970 t1=-0.346574   t2=1.216395   t3=0.000000

sum =0.509835  x=0.600000  fx= 0.199586 t1=-0.306495   t2=1.504012   t3=0.000000

sum =0.768912  x=0.700000  fx= 0.259077 t1=-0.249672   t2=1.804136   t3=0.000000

sum =1.091831  x=0.800000  fx= 0.322919 t1=-0.178515   t2=2.116032   t3=0.000000

sum =1.482535  x=0.900000  fx= 0.390703 t1=-0.094824   t2=2.439045   t3=0.000000

sum =1.944633  x=1.000000  fx= 0.462098 t1=0.000000   t2=2.772589   t3=0.000000

the requested integral is = ___1.944633___

 

[Ciccioo]

ok well that seems to work for getting t3 to be correct, but this answer is bugging me

i edited my answer, it's a little less vague now, at least we can see a pattern in the behaviour

 

and please@JewishBacon, use the code tag

[JewishBacon]

i edited my answer, it's a little less vague now, at least we can see a pattern in the behaviour

 

and please@JewishBacon, use the code tag

 

will do but, how do i do the code tag.

 

how do i change the code to make the printf expect a double?

[Ciccioo]

will do but, how do i do the code tag.

[code][/code]

how do i change the code to make the printf expect a double?

you already did that

the %lf in the printf format string stands for double

 

and for your problem, are you sure about the maths? i don't know simpson's method, i'm looking it up now, but it doesn't look too much like a right implementation to me

[JewishBacon]

[code][/code]

you already did that

the %lf in the printf format string stands for double

 

and for your problem, are you sure about the maths? i don't know simpson's method, i'm looking it up now, but it doesn't look too much like a right implementation to me

 

this is a link to the problem i am trying to solve 

 

https://www.wolframalpha.com/input/?i=integrate+x+*+ln+x+from+0+to+1

[Ciccioo]

this is a link to the problem i am trying to solve 

 

https://www.wolframalpha.com/input/?i=integrate+x+*+ln+x+from+0+to+1

yep

 

but why is simpson's rule implemented as an iterative algorithm?

 

maybe you're looking for something like the trapezoid/rectangle methods

[JewishBacon]

yep

 

but why is simpson's rule implemented as an iterative algorithm?

 

maybe you're looking for something like the trapezoid/rectangle methods

I checked the math and i misunderstood what t1 is supposed to be, i readjusted and this new code is within error for what i should be getting, now i just need to make the code look a little prettier so i can turn it in thanks @Ciccioo

[Ciccioo]

i mean, consider that i just read about is, so bear with me and be patient, but this is my implementation for simpson's rule

#include <stdio.h>#include <math.h>double  f(double x){    if(x == 0)        return 0;    return x * logf(x);}int main(){    double a = 0, b = 1;        double result = (b - a) / 6 * (f(a) + 4 * f((a + b) / 2) + f(b));    printf ("the requested integral is = ___%lf___\n", result);    return 0;}

result: -0.231049

[JewishBacon]

this is my code:

 

how do i do the color code thing

 

@Ciccioo

 

 

 

#include <stdio.h>

#include <math.h>

 

int main()

{

double x,count,fx,t1,t2,t3;

 

 

 

        t1=.000000001*logf(0.000000001);

        t2=4.0*((1.0/2.0)*logf (1.0/2.0));

        t3=1.0*log(1.0);

        fx=(1.0/6.0)*(t1+t2+t3);

 

printf ("the requested integral is = ___%lf___\n",fx);

return 0;

}

 

 

i get the same answer you do  

 

the confusion you are feeling is normal, welcome to engineering school

 

[Ciccioo]

how do i do the color code thing

[code]// put your code hereint somethingfloat something elsefunction a lot of colors[/code]

 

result:

// put your code hereint somethingfloat something elsefunction a lot of colors

[JewishBacon]

#include <stdio.h>

#include <math.h>

 

int main()

{

double x,count,fx,t1,t2,t3;

 

 

 

        t1=.000000001*logf(0.000000001);

        t2=4.0*((1.0/2.0)*logf (1.0/2.0));

        t3=1.0*log(1.0);

        fx=(1.0/6.0)*(t1+t2+t3);

 

printf ("the requested integral is = ___%lf___\n",fx);

return 0;

}

 

[code]