If Statement when x is near a value

Question

I am trying to solve the numerical equation:

sin^2(x)tan(x) = 0.499999

Using a while loop in C. However I was only able to get program to print an answer if I rounded to 0.5. This led me to thinking, is there a way of writing:

For(x=0;x<=360;x=x+0.001)
{ y=f(x)
If(y **is near x**(e.g. Within 1 percent) )
Do something etc.
}

Is there a way of telling the computer to execute a task if the value is sufficiently near. Such as in this if statement? Thank you.

Google *epsilon* in the context. For example, `if(fabs(x - y) < EPSILON) { ... }` — Weather Vane, Mar 08 '17 at 21:23
@WeatherVane: If `x` is 0.0002 and y is 0.0001, your test is true but the difference is 50%, not 1% as requested in the question. — Jonathan Leffler, Mar 08 '17 at 21:26
@JonathanLeffler that's true, my comment was not an answer but a suggested avenue of approach, an *absolute* approach, not a *proportional* one as in your answer. — Weather Vane, Mar 08 '17 at 21:28

score 4 · Accepted Answer · answered Mar 08 '17 at 21:25

4

Use a relative difference, such as:

#include <math.h>

static inline double reldiff(double x, double y)
{
    return fabs(x - y) / fmax(fabs(x), fabs(y));
}

Now your test becomes:

if (reldiff(x, y) < 0.01)   // 1% as requested

answered Mar 08 '17 at 21:25

Jonathan Leffler

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score 0 · Answer 2 · answered Mar 08 '17 at 21:28

0

Well, a simple solution would be.

if (y >= x-x/100 && y <= x+x/100)
{
   [execute code...]
}

Essentially what it does is check if y is between [x minus 1% of x] and [x plus 1% of x]. Should work alright.

answered Mar 08 '17 at 21:28

stelioslogothetis

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If Statement when x is near a value

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