So in this program i am trying to print out the standard deviation of a set of numbers the user enters in. The formula to calculate standard deviation is correct (or as correct as it needs to be) so that is not the problem, but when i run the program everything goes well until the console prints out the results. It prints that totalStandardDeviation = nan
what exactly does than mean? is nan the same as nil? has it lost the value somehow and not been able to find it? thanks for any help you may provide.
#include <iostream>
#include <cmath>
using namespace std;
double returnStandardDeviation(double x, int counter);
double total;
int userInput = 1;
int counter = 0;
double x;
double x1;
double x2;
double standardDeviation;
double totalStandardDeviation;
int main(int argc, const char * argv[])
{
cout << "Please enter a list of numbers. When done enter the number 0 \n";
cin >> userInput;
while (userInput != 0) // As long as the user does not enter 0, program accepts more data
{
counter++;
x = userInput;
returnStandardDeviation( x, counter); // send perameters to function
cout << "Please enter next number \n";
cin >> userInput;
}
cout << "The standard deviation of your "
<< counter
<< " numbers is : "
<< returnStandardDeviation(x, counter);
return 0;
}
double returnStandardDeviation(double x, int counter)
{
x1 += pow(x,2);
x2 += x;
totalStandardDeviation = 0;
totalStandardDeviation += (sqrt(counter * x1 - pow(x2,2))) / (counter * (counter - 1));
return totalStandardDeviation;
}
NaN stands for "Not a number".
NaN can e.g. be the result of:
- Dividing by zero
- Taking the square root of a negative number
In your function, both of these could happen. Division by zero e.g. when counter is <= 1; and x1 and x2 are uninitialized (+= adds the value on the right to their current value - which was never set, and is therefore random gibberish), which can easily lead to your function trying to take the square root of some value < 0.
This expression
counter * x1 - pow(x2,2)
can very easily yield a negative number. You then proceed to take its square root. This would result in a nan.
Next, this one
counter * (counter - 1)
yields 0 when counter is 1. Dividing by zero gives nan.
Your formula is wrong. You are either dividing by zero or taking the square root of a negative number.
Check your formula!
Additional info:
NaN is "Not a number". It is an IEEE floating point value that signals an invalid results, like log(-1), or sqrt(-4).
Additionally, know that Positive Infinity and Negative Infinity are also floating point values.
Related
I want to get square root of a number upto 9 precision points so I did something like below but I am not getting correct precision. Here e is the precision which is greater than 10^9 then also ans is upto 5 precision points. What am I doing wrong here??
#include <iostream>
using namespace std;
long double squareRoot(long double n)
{
long double x = n;
long double y = 1;
long double e = 0.00000000000001;
while (x - y > e)
{
x = (x + y) / 2;
y = n / x;
}
cout << x << "\n";
return x;
}
int main()
{
int arr[] = {2,3,4,5,6};
int size = sizeof(arr)/sizeof(arr[0]);
long double ans = 0.0;
for(int i=0; i<size; i++)
{
ans += squareRoot(arr[i]);
}
cout << ans << "\n";
return 0;
}
The output is
1.41421
1.73205
2
2.23607
2.44949
9.83182
What should I do to get precision upto 9 points??
There are two places at which precision plays a role:
precision of the value itself
precision of the output stream
You can only get output in desired precision if both value and stream are precise enough.
In your case, the calculated value doesn't seem to be a problem, however, default stream precision is only five digits, i. e. no matter how precise your double value actually is, the stream will stop after five digits, rounding the last one appropriately. So you'll need to increase stream precision up to the desired nine digits:
std::cout << std::setprecision(9);
// or alternatively:
std::cout.precision(9);
Precision is kept until a new one is set, in contrast to e. g. std::setw, which only applies for next value.
try this
cout << setprecision(10) << x << "\n";
cout << setprecision(10) << ans << "\n";
I need help figuring out where my code went wrong. I want to reset the values for the loop so that it isn't compiling because my output right now is using past input values in current calculation whereas I want the output to be different every time as though it is the firs time running the code. The code works fine when I don't use the while loop, but then I have to rerun the program each time. I want the output to prompt a new input every time, but not use past inputs in the new calculations. I know I'm not explaining it very well, but I'm just sort of lost. Anything helps!
This is my problem:
An approximate value of pi can be calculated using the series given
below:
pi = 4 · [ 1 – 1/3 + 1/5 – 1/7 + 1/9 ... + (–1 ^ n)/(2n + 1) ]
Write a C++ program to calculate the approximate value of pi using
this series. The program takes an input n that determines the number
of terms in the approximation of the value of pi and outputs the
approximation. Include a loop that allows the user to repeat this
calculation for new values n until the user says she or he wants to
end the program.
#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
int n;
double sum=0;
cout << "Enter the number of terms to approximate (or zero to quit):\n";
cin >> n;
if (n == 0)
{
return 0;
}
while (n != 0)
{
{ for(int i=0;i<n;i++)
{
if (i%2==0)
{
sum += 1.0/(2*i+1);
}
else
{
sum += -1.0/(2*i+1);
}
}
cout.setf(ios::showpoint);
cout.precision(3);
cout << "The approximation is " << sum*4 << " using " << n << " terms." << endl;
}
cout << "Enter the number of terms to approximate (or zero to quit):\n";
cin >> n;
}
return 0;
}
This is my output:
This is what the output should be:
You do not reset sum before entering the for loop. Just add
sum=0;
before the for line.
Notice that 2.67 = 6.67 - 4.00.
You want your program to compute a sum for several values of n.
The sum must be initialized to 0 at the beginning of each calculation, inside the while loop.
Actually, it should even be declared there.
C++ doesn't require you to define variables at the start of a function, so it is perfectly legal to write:
while (n != 0)
{
double sum = 0.0;
this would solve your problem. Alternatively, if you want to keep the declaration of sum at the top of the function, just change your code to
while (n != 0)
{
sum = 0.0;
I am trying to write a calculator in C++ that does the basic functions of /, *, -, or + and shows the answer to two decimal places (with 0.01 precision).
For example 100.1 * 100.1 should print the result as 10020.01 but instead I get -4e-171. From my understanding this is from overflow, but that's why I chose long double in the first place!
#include <iostream>
#include <iomanip>
using namespace std;
long double getUserInput()
{
cout << "Please enter a number: \n";
long double x;
cin >> x;
return x;
}
char getMathematicalOperation()
{
cout << "Please enter which operator you want "
"(add +, subtract -, multiply *, or divide /): \n";
char o;
cin >> o;
return o;
}
long double calculateResult(long double nX, char o, long double nY)
{
// note: we use the == operator to compare two values to see if they are equal
// we need to use if statements here because there's no direct way
// to convert chOperation into the appropriate operator
if (o == '+') // if user chose addition
return nX + nY; // execute this line
if (o == '-') // if user chose subtraction
return nX - nY; // execute this line
if (o == '*') // if user chose multiplication
return nX * nY; // execute this line
if (o == '/') // if user chose division
return nX / nY; // execute this line
return -1; // default "error" value in case user passed in an invalid chOperation
}
void printResult(long double x)
{
cout << "The answer is: " << setprecision(0.01) << x << "\n";
}
long double calc()
{
// Get first number from user
long double nInput1 = getUserInput();
// Get mathematical operations from user
char o = getMathematicalOperation();
// Get second number from user
long double nInput2 = getUserInput();
// Calculate result and store in temporary variable (for readability/debug-ability)
long double nResult = calculateResult(nInput1, o, nInput2);
// Print result
printResult(nResult);
return 0;
}
setprecision tells it how many decimal places you want as an int so you're actually setting it to setprecision(0) since 0.01 get truncated. In your case you want it set to 2. You should also use std::fixed or you'll get scientific numbers.
void printResult(long double x)
{
cout << "The answer is: " << std::fixed << setprecision(2) << x << "\n";
}
working example
It is not due to overflow you get the strange result. Doubles can easily hold numbers in the range you are showing.
Try to print the result without setprecision.
EDIT:
After trying
long double x = 100.1;
cout << x << endl;
I see that it doesn't work on my Windows system.
So I searched a little and found:
print long double on windows
maybe that is the explanation.
So I tried
long double x = 100.1;
cout << (double)x << endl;
which worked fine.
2nd EDIT:
Also see this link provided by Raphael
http://oldwiki.mingw.org/index.php/long%20double
The default floating point presentation switches automatically between presentation like 314.15 and 3.1e2, depending on the size of the number and the maximum number of digits it can use. With this presentation the precision is the maximum number of digits. By default it's 6.
You can either increase the maximum number of digits so that your result can be presented like 314.15, or you can force such fixed point notation by using the std::fixed manipulator. With std::fixed the precision is the number of decimals.
However, with std::fixed very large and very small numbers may be pretty unreadable.
The setprecision() manipulator specifies the number of digits after the decimal point. So, if you want 100.01 to be printed, use setprecision(2).
When you use setprecision(0.01), the value 0.01 is being converted to int, which will have a value of 0.
It wouldn't have hurt if you had actually read the documentation for setprecision() - that clearly specifies an int argument, not a floating point one.
I am learning C++ language, but much confused about a little thing, that is
below I have putted some code of making a square of an integer. But I am not understanding the treatment of x in raiseToPow function. Here double x is an argument taking the value from calling function and passing to x in for loop. So how it makes square of that integer passing from x. Please guide me.
#include <iostream>
using namespace std;
double raiseToPow(double x, int power){
double result;
result = 1.0;
for (int i = 1; i <= power; i++)
{
result *= x;
}
return (result);
}
main()
{
double x;
int i;
cout << "Enter the Number: ";
cin >> x;
cout << "Enter the Power: ";
cin >> i;
cout << x << "Raise to power " << i << " is equal to " << raiseToPow(x , i);
}
double x is the base of the exponentiation xpower.
The function uses the definition xpower is 1 multiplied by x power times. Thus 23 is 1*2*2*2 and 45 is 1*4*4*4*4*4. So double result is set to be 1, then the for loop executes power times, each time multiplying result by x and storing that back in result.
Even though this is C++, not C, the logic is the same.
You pass in a number(x) and its exponent(power). The value of x is multiplied 'power' times.
For example, if you want to know the value of 2 raised to the power of 3, you would pass 2, 3 to the function raiseToPow
The statement
result*=x
is the same as
result = x*result.
So, the first time through the loop, result would just be the value of x (2, in this example). The second time through the loop, result*=x would multiply the value of x(which is 2) times the value stored in result (which, for now, is also 2). The third and final time through the loop, result *=x would multiply x times the value stored in result (which now is 4).
The value of 2^3=8, and that is what the value stored in the variable result is, so the value 8 is returned from the call raiseToPow(2,3).
I need help on how to get nth root of some number.
User enters number n and number he wants root of. I need to solve this without cmath lib and with divide and conquer method.
Here's my code that doesn't work yet:
#include<iostream>
using namespace std;
float pow(float a,float c){
if (a == 0)
return 0;
else if(a == 1)
return 1;
else{
float p = pow(a,(c/2));
if(c%2)
return p*p*a;
else
return p*p;
}
}
int main(){
float a,b;
float c;
cout << "Enter positive number:(base)" << endl;
do{
cin >> a;
}while (a < 0);
cout << "Enter number: (root)" << endl;
cin >> b;
c = 1/b;
cout << "Result:"<<pow(a,c) << endl;
system("pause");
return 0;
}
Any ideas on how to approach this problem would be more than useful.
Let me tell you how you can use divide and conquer for finding square root. The nth root would be similar.
For a given number x, you need to search for it's square root between 0 and x. Divide it by 2 = x2. If the x2 * x2 < x then your search space moves to x2 -> x or else it will be 0 -> x2. If x2 * x2 matches x then your square root is x2. Similar technique for nth root.
For those not doing numerical experiments: use the <cmath> functions sqrt and cbrt (cube-root) to construct the any root that is factorable by 2 and 3. For example, the 4th root is sqrt(sqrt(x)) and the 6th root is sqrt(cbrt(x)). If you need something for general use you can construct a recursive function which calls sqrt and cbrt appropriately.
I'm guessing this will give a faster and more accurate answer than pow, if that matters. If it doesn't, just use pow.