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The result of calculation deviates after some digits [duplicate]

This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 3 years ago.
Result of calculation shown below should be equal to 0,007306897 (I referred from 2 books) . But when i check result from Watch 1, i see that result is equal to 0,007306882. I split the process into some parts. And the problem is occurring when c is calculating.
///Declarations
double sigma = 1.00000000;
double a,b,e,c;
a = (1 / Math.Sqrt(2 * Math.PI)); //calculated properly
c = -(i * i + j * j) / 2.00000000 * (sigma * sigma); //i and j are equal to -2
e = Math.E; //calculated properly
b = Math.Pow(e, c);
result=a * b;

Unfortunately the double type is not highly accurate. The link below shows that some numbers like 1.05 can not be stored accurately by the double type.
http://www.binaryconvert.com/convert_double.html
For normal usage it is usually accurate enough, if you need accuracy to the level you describe you probably need to use something like binary coded decimal. That will mean the normal math library won't work. You could try asking on some physics sites to see what they use when modelling complex systems to maintain accuracy.

Correlate two floats using c# [closed]

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I know im an idiot and I am sure this is simple math. But I cannot seem to wrap my head around it here is my situation
When X = 3, I need Y = 0, and when X = 0 I need y = 1;.
I am trying to fill a progress bar based upon how low X is.
The value to fill the progress bar (Y) must be between 0 and 1.
Math?

// "Single" is just like "float"
Single y = (3.0f - x) / 3.0f;
So that
x=3 -> y=0.00
x=2 -> y=0.33
x=1 -> y=0.66
x=0 -> y=1.00
Alternatively:
// different points of view are better
Single y = -(x - 3.0f) / 3.0f;

As I know best from my high school:
y = a*x + b
You must solve equations:
0 = a*3 + b and
1 = a*0 + b
a = -b/3; b =1
So your equation is: y=-1/3*x+1
private float GetProgressValue(float x)
{
return x/-3f + 1f;
}

How to restrict calculated value to four decimal places [duplicate]

This question already has answers here:
How do you round a number to two decimal places in C#?
(15 answers)
Closed 5 years ago.
I want to limit decimal place till 4 for below code, how to do that,
var a = Convert.ToDecimal( 80794992640) / (1024 * 1024);

I guess you can round it if you want to fix it to 4 digits:
var a = Math.Round(Convert.ToDecimal( 80794992640) / (1024 * 1024),4 );
but if your concern is to restrict it in the display then you can just apply the restriction in the ToString method:
a.ToString("0.####");
the latter method will keep the precision for the calculations but cut the precision only for display

This will round it to 4 decimal places:
var a = Math.Round(Convert.ToDecimal( 80794992640) / (1024 * 1024), 4);

c# - outputting a value of infinity [closed]

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I have part of the calculation that is modified to explain what issue I'm having. I'm converting a working Java code into a c# application.
A double variable is declared doing the following calculation and the following input values being passed along for example:
double input1 = 45
double input2 = 42
double input3 = 1
double a = (input1 - input2 * Math.Pow(input3, 2.0)) / (1 - Math.Pow(input3, 2.0));
This outputs as an Infinite in c# but Java shows the value. With infinite being outputted, the other calculations below are affected.
double b = input1 + input2 - 2 * a (becomes -Infinite)
double c = b + a (gives you NaN)
a,b, and c are outputted into individual text boxes and each either show as Infinite or NaN.
Textbox1.Text = a.ToString("F") (Infinite)
Textbox2.Text = b.ToString("F") (-Infinite)
Textbox3.Text = c.ToString("F") (NaN)
I know that this is due to IEEE standards, but is there actually a way that variable a shows the value instead of showing infinite so that it doesn't affect the calculations or output the results into a text box in 2 decimal places?

Your code is returning a value 0 for the divide part which results into infinity value. which is true here there is nothing wrong in the code. You need to check the calculation.
double input1 = 45;
double input2 = 42;
double input3 = 1;
double a = (input1 - input2 * Math.Pow(input3, 2.0))
/ (1 - Math.Pow(input3, 2.0));
Console.WriteLine((input1 - input2 * Math.Pow(input3, 2.0)));
Console.WriteLine((1 - Math.Pow(input3, 2.0)));
Console.WriteLine(a);
check the output on console.

I pasted into a Java test program the code you posted for calculating a, and made minimal changes to turn it into compilable Java. I added ; to the first three declarations, and changed Math.Pow to Math.pow:
public class Test {
public static void main(String[] args) {
double input1 = 45;
double input2 = 42;
double input3 = 1;
double a = (input1 - input2 * Math.pow(input3, 2.0))
/ (1 - Math.pow(input3, 2.0));
System.out.println(a);
}
}
The program prints, as expected, "Infinity".
If your Java program is getting something different, there is some error in how you are converting it to C#.

Math.Round with Mid point Rounding [closed]

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This sample code is producing unexpected result
decimal s = 463.815M;
decimal a = Math.Round(s, 2, MidpointRounding.AwayFromZero);
decimal b = Math.Round(s, 2, MidpointRounding.ToEven);
decimal t = 4.685M;
decimal c = Math.Round(t, 2, MidpointRounding.AwayFromZero);
decimal d = Math.Round(t, 2, MidpointRounding.ToEven);
Console.WriteLine(a);
Console.WriteLine(b);
Console.WriteLine(c);
Console.WriteLine(d);
Console.Read();
It produces
463.82
463.82
4.69
4.68
I was expecting a and c to have incremented by 1 which c did but to my surprise a didn't. Can anyone explain the reason for this please?
[update]
a and c are expected to have same results as:
a has .815 and c also has .685 i.e. 5 at the end.
a and c both are using MidpointRounding.AwayFromZero

This is the expected result, because 0.815 fraction is rounded up to 0.82. The same exact thing happens when you round to even, because 2 is even.
The result would be different if you used 0.825 as a fraction:
decimal s = 463.825M;
decimal a = Math.Round(s, 2, MidpointRounding.AwayFromZero);
decimal b = Math.Round(s, 2, MidpointRounding.ToEven);
Now the code prints
463.83
463.82
to illustrate the difference between MidpointRounding.AwayFromZero and MidpointRounding.ToEven.