Exponent question

[Joseph Aldred]

I am new to C and I am writing a program for a class I am taking. I need to
use a exponent but I can not figure how to do it. I figure I need to use
exp() or pow() but I can not figure out how to do it. Any help that anyone
can give me would be appreciated greatly.

Thanks,
Joseph

 

[Frane Roje]

You can use pow()
pow(base, exponent)

HTH

--
Frane Roje

Have a nice day

Remove (*dele*te) from email to reply

 

[Joseph Aldred]

Thank you for the help.

 

[Joseph Aldred]

Ok I am still having problems so I must be doing something wrong. What I am
supposed to be doing is writing a problem to solve the polynomial

3x^3 - 5x^2 + 6
for x=2.55

What I came up with for a program is:

#include <stdio.h>
#include <math.h>

main ()
{
double a ;

a = 2.55

printf ("3 * pow(%f,3) - 5 * pow(%f,2) + 6\n", a);

return 0;
}


or I also tried this

#include <stdio.h>
#include <math.h>

main ()
{
float a = 2.55;

float b = 3 * pow(%f,3) - 5 * pow(%f,2) +6, a, a;

printf ("3 * pow(%f,3) - 5 * pow(%f,2) + 6 = ", a, a);

return 0;
}


On the second one I keep getting parse error before '%' token

Does anyone have any suggetions as to what I am doing wrong?

I appreciate any help anyone is willing to offer.

Thank you,
Joseph

 

[Ben Pfaff]

Ok I am still having problems so I must be doing something wrong. What I am
supposed to be doing is writing a problem to solve the polynomial

3x^3 - 5x^2 + 6
for x=2.55

I think you mean that you want to "evaluate", not "solve", the
polynomial.

What I came up with for a program is:

#include <stdio.h>
#include <math.h>

main ()

You should write
int main(void)
explicitly.

{
double a ;

a = 2.55

Missing semicolon.

printf ("3 * pow(%f,3) - 5 * pow(%f,2) + 6\n", a);

This is misguided. What is inside the parentheses will be output
verbatim, except that %f will be expanded to the value of the
floating-point arguments (you supplied one too few of them, by
the way) and \n will be expanded to a new-line character.

return 0;
}


or I also tried this

#include <stdio.h>
#include <math.h>

main ()
{
float a = 2.55;

float b = 3 * pow(%f,3) - 5 * pow(%f,2) +6, a, a;

This is also misguided. % is only used in this way in the format
string argument to printf() and similar functions. Here you want
to write
float b = 3.0 * pow(a, 3) - 5.0 * pow(a, 2) + 6.0;
or
float b = 3.0 * a * a * a - 5.0 * a * a + 6.0;
Also, I'd recommend using double instead of float.

printf ("3 * pow(%f,3) - 5 * pow(%f,2) + 6 = ", a, a);

This will print the polynomial, but you omitted printing the
result.

return 0;
}

Maybe you should buy a C textbook?

 

[Joseph Aldred]

Thank you Ben,

I have a book called "Programming in ANSI C". by Stephen G Kochan. It is
the textbook for the class I am taking. I am not sure that I like it very
well, my instructor seems pretty critical of the way the material is
presented. I look in the index and there is only one place for exponents
and all that has to do with is scientific notation.

But thank you for your help. I am so glad that there are people here that
can give help to us newbies.

J

 

[Malcolm]

I look in the index and there is only one place for exponents
and all that has to do with is scientific notation.

We use exponents in two places. Firstly, where numbers are so large or so
small that decimal point notation is not human-readable. For instance
100,000,000,000 is best written as 1 * 10 ^11. The second place is where the
program logic calls for a value to be raised to the power of another value.
For instance, say we have a virus multiplying in a large population, and
each infected person infects another five on average before dying. How many
people will be infected after N rounds? The answer is 5^N, or, in C,
pow(5.0, N).
For mathematical reasons which are too involved to go into here the
expression e^x has special characteristics, so it has its own function,
exp(). The expression x^0.5, or sqrt(x) is also special and has its own
function - here it is easy for a non-mathematician to see the value of this.

 

[Martin Ambuhl]

Ok I am still having problems so I must be doing something wrong. What I am
supposed to be doing is writing a problem to solve the polynomial

3x^3 - 5x^2 + 6
for x=2.55

What I came up with for a program is:

[misuse of printf() as if it were a Basic interpreter ignored, since the
relevant errors otherwise are repeated below.]

or I also tried this

#include <stdio.h>
#include <math.h>

main ()

main returns an int. You should say so
int main(void)

{
float a = 2.55;

float b = 3 * pow(%f,3) - 5 * pow(%f,2) +6, a, a;

float b; /* but you really want doubles; whatever */
b = 3*pow(a,3) - 5*pow(a,2) + 6; /* pow() takes an argument.
This isn't a lambda expression. */
But, obviously, this is avoids the function call:
b = 3*a*a*a - 5*a*a +6;
And better is
b = 6 + a*a*(3*a - 5);

printf ("3 * pow(%f,3) - 5 * pow(%f,2) + 6 = ", a, a);

^^ ^
%f\n , b
return 0;

}


On the second one I keep getting parse error before '%' token

Does anyone have any suggetions as to what I am doing wrong?

Using a format specifier instead of an argument.

 

[CBFalconer]

I am new to C and I am writing a program for a class I am taking.
I need to use a exponent but I can not figure how to do it. I
figure I need to use exp() or pow() but I can not figure out how
to do it. Any help that anyone can give me would be appreciated
greatly.

Your answer is somewhere among the following references.

--
Some useful references:
<http://www.ungerhu.com/jxh/clc.welcome.txt>
<http://www.eskimo.com/~scs/C-faq/top.html>
<http://benpfaff.org/writings/clc/off-topic.html>
<http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/> (C99)

 

[CBFalconer]

Ok I am still having problems so I must be doing something wrong.
What I am supposed to be doing is writing a problem to solve the
polynomial

3x^3 - 5x^2 + 6
for x=2.55

This is a different question. Start by applying some algebra:

y = 3 * x^3 - 5 * x^2 + 6
= ((((3 * x) - 5) * x) + 0) * x + 6

Do you see a pattern?

--
Some useful references:
<http://www.ungerhu.com/jxh/clc.welcome.txt>
<http://www.eskimo.com/~scs/C-faq/top.html>
<http://benpfaff.org/writings/clc/off-topic.html>
<http://anubis.dkuug.dk/jtc1/sc22/wg14/www/docs/n869/> (C99)