I'm trying to compute a series using C++.
The series is:
(for those wondering)
My code is the following:
#include <iostream>
#include <fstream>
#include <cmath> // exp
#include <iomanip> //setprecision, setw
#include <limits> //numeric_limits (http://en.cppreference.com/w/cpp/types/numeric_limits)
long double SminOneCenter(long double gamma)
{
using std::endl; using std::cout;
long double result=0.0l;
for (long double k = 1; k < 1000 ; k++)
{
if(isinf(pow(1.0l+pow(gamma,k),6.0l/4.0l)))
{
cout << "infinity for reached for gamma equals: " << gamma << "value of k: " << k ;
cout << "maximum allowed: " << std::numeric_limits<long double>::max()<< endl;
break;
}
// CAS PAIR: -1^n = 1
if ((int)k%2 == 0)
{
result += pow(4.0l*pow(gamma,k),3.0l/4.0l) /(pow(1+pow(gamma,k)),6.0l/4.0l);
}
// CAS IMPAIR:-1^n = -1
else if ((int)k%2!=0)
{
result -= pow(4.0l*pow(gamma,k),3.0l/4.0l) /(pow(1+pow(gamma,k)),6.0l/4.0l);
//if (!isinf(pow(k,2.0l)*zeta/2.0l))
}
// cout << result << endl;
}
return 1.0l + 2.0l*result;
}
Output will be, for instance with gamma = 1.7 :
infinity reached for gamma equals: 1.7 value of k: 892
The maximum value a long double can represent, as provided by the STL numeric_limits, is: 1.18973e+4932.
However (1+1.7^892)= 2.19.... × 10^308 which is way lower than 10^4932, so it shouldn't be considered as infinity.
Provided my code is not wrong (but it very well might be), could anyone tell me why the discussed code evals to infinity when it should not?
You need to use powl rather than pow if you want to supply long double arguments.
Currently you are hitting the numeric_limits<double>::max() in your pow calls.
As an alternative, consider using std::pow which has appropriate overloads.
Reference http://en.cppreference.com/w/c/numeric/math/pow
Be gentle ... I'm 5 weeks into studying C++. I've dug and dug and cannot figure out why Visual Studio Express (and online compilers) are throwing errors about this.
Note that I've included all my declarations for the sake of clarity -- most are used in different section of code. The line that gets the errors is this one: newsharePrice = perchangeEnd * profitLoss << '\n';
The error I get is c2296, left operand has type double. I have no idea why it doesn't like this ... I multiply other doubles just fine.
double numberShares,
sharePrice,
profitLoss,
profitGain,
commissionPaid,
commissionCost,
sharesCost,
totalshareCost,
newtotalshareCost,
newcommissionCost,
newsharePrice;
double perChange;
double perchangeEnd;
const int minVALUE = 1;
const int maxVALUE = 100;
int seed = time(0);
srand (seed);
perChange = (rand() % (maxVALUE - minVALUE + 1)) + minVALUE;
cout << perChange << '\n';
perchangeEnd = perChange / 100;
int flip = rand() % 2 + 1;
if (flip == 1)
profitLoss = 1;
else
profitLoss = -1;
newsharePrice = perchangeEnd * profitLoss << '\n';
newsharePrice = newsharePrice + sharePrice;
cout << newsharePrice << '\n';
newtotalshareCost = numberShares * newsharePrice;
cout << "You've now paid " << newtotalshareCost << " for your shares." << '\n';
newcommissionCost = newtotalshareCost * commissionRate;
cout << "The new amount of commission for this is " << newcommissionCost << " ." << '/n';
Well, just read the problematic line:
newsharePrice = perchangeEnd * profitLoss << '\n';
// ▲▲▲▲▲▲▲▲
That << '\n' is not part of the multiplication; a copy-pasta fail from your cout lines?
In this context, the compiler has no choice but to assume you're trying to perform a bitwise left-shift operation, which cannot be performed on doubles; only on integers.
While the compilation error is now fixed, your domain error is still there (today is Friday, isn't it?). Why would share price fluctuation affect your commission in any way? You already hace your position. You also measure your number of shares with floating-point precision. While in some cases you might have uneven number of shares, this happens quite seldom. Are you really account for this or just incorrectly use double? Most systmes would count number of shares as integer. Also, you can have negative position, which after all calctulations will give negative commission! Brokers would not agree to that ;). The last, but not the least, in US commission is rarely expressed as percentage of transaction value. It is usually charged in a form of cents per share (or fixed transaction cost for most retail brokers).
To clarify, this is a lab for class. I'm just learning all the fundamentals right now. I am trying to figure out how to print the total using a float variable AND a double variable (I thought we can just choose one or the other) while using loop mechanisms (while, do-while or for). I decided to go with the for loop. Any suggestions would be helpful. What I have now prints every single fraction until it gets to the very last one. I tried different variations but so far I got nothing.
I need help calculating the total for:
1/1 + 1/2 + 1/3 + 1/4 +....... 1/99999999 + 1/100000000
This is what I have so far:
#include <iostream>
using namespace std;
int main()
{
float answer = 0;
int num;
for (int den = 1; den <= 100000000; ++den)
{
num = 1;
cout << num << "/" << den;
if (den == 100000000)
cout << " = " << endl;
else
cout << " + ";
answer += ( (float)num ) / ( (float)den );
}
cout << answer << endl;
}
Thanks!
At any point in time, your program has to be using either a float or a double, so I guess you're wanting to reuse your calculation code for each of those types in turn. Here's an example of how to do the calculation twice using a template - first for float, then for double:
#include <iostream>
template <typename T>
void calculate()
{
T answer = 0;
for (int den = 1; den <= 100000000; ++den)
answer += T(1) / T(den);
std::cout << answer << '\n';
}
int main()
{
calculate<float>();
calculate<double>();
}
My school give me an assignment to calculate pi.
The result should be :
Question 4
Accuracy set at : 1000
term pi
1 4
100 3.13159
200 3.13659
300 3.13826
400 ...
... ...
The result in my program :
term pi
1 4
100 3
200 3
300 3
400 ...
... ...
I guess that when I do (4 / denominator), the result will lose the decimal number although I have changed some declarations of data type from int to double.
(Some websites tell me to do this.)
Maybe I do it wrongly.
How can I deal with this problem?
The following is my program.
#include <iostream>
using namespace std;
class Four
{
private:
int inputedAccuracy;
double pi;
int denominator;
int doneTermCounter;
double oneTerm;
int negativeController;
public:
double question4()
{
cout << "Accuracy set at : " ;
cin >> inputedAccuracy;
cout << endl;
pi = 0.0;
denominator = 1.0;
doneTermCounter = 0;
negativeController = 1;
cout << "Term" << " " << "pi" << endl;
cout << "1 " << " " << "4" << endl;
for (inputedAccuracy; inputedAccuracy > 0; inputedAccuracy -= 100)
{
for (int doneTerm = 0; doneTerm < 100; doneTerm++)
{
pi = pi + (negativeController * 4 / denominator);
negativeController *= -1;
denominator += 2;
doneTermCounter++;
}
if (doneTermCounter >= 10000)
cout << doneTermCounter << " " << pi << endl;
else
if (doneTermCounter >= 1000)
cout << doneTermCounter << " " << pi << endl;
else
cout << doneTermCounter << " " << pi << endl;
}
return 0.0;
}
};
Thank you for your attention!
pi = pi + (negativeController * 4 / denominator);
The (negativeController * 4 / denominator) expression results in an int because both negativeController and denominator are int. In other words, you're doing an integer division here which explains why you don't get the expected result.
Declare either (or both) of them as double to force a floating-point division.
You should change :-
int denominator; to
double denominator;
See here
pi = pi + (negativeController * 4 / denominator);
In this line, you have an integer division (because both operands of / are of type int), meaning that the fractional part of the division's result is discarded.
To use floating point division, at least one operand/side needs to be of type float or (long) double. The easiest way to achieve this would in this case be a change of 4 (a literal of type int) to 4.0 (a literal of type double):
Then, when calculating the result of *, negativeController will also be converted to double (usual arithmetic conversions), yielding a double as the left-hand side operand of / which in turn causes denominator (the rhs) to also be converted into a double and so on.
I think changing negativeController and denominator to int would do the trick as the sub-expression is being evaluated on integers thus loosing precision.
What I mean is the following:
double d1 =555;
double d2=55.343
I want to be able to tell that d1 is an integer while d2 is not. Is there an easy way to do it in c/c++?
Use std::modf:
double intpart;
modf(value, &intpart) == 0.0
Don't convert to int! The number 1.0e+300 is an integer too you know.
Edit: As Pete Kirkham points out, passing 0 as the second argument is not guaranteed by the standard to work, requiring the use of a dummy variable and, unfortunately, making the code a lot less elegant.
Assuming a c99 and IEEE-754 compliant environment,
(trunc(x) == x)
is another solution, and will (on most platforms) have slightly better performance than modf because it needs only to produce the integer part. Both are completely acceptable.
Note that trunc produces a double-precision result, so you don't need to worry about out of range type conversions as you would with (int)x.
Edit: as #pavon points out in a comment, you may need to add another check, depending on whether or not you care about infinity, and what result you want to get if x is infinite.
Assuming you have the cmath <math.h> library, you can check the number against it's floor. If the number might be negative, make sure you get the absolute first.
bool double_is_int(double trouble) {
double absolute = abs( trouble );
return absolute == floor(absolute);
}
avakar was almost right - use modf, but the detail was off.
modf returns the fractional part, so the test should be that the result of modf is 0.0.
modf takes two arguments, the second of which should be a pointer of the same type as the first argument. Passing NULL or 0 causes a segmentation fault in the g++ runtime. The standard does not specify that passing 0 is safe; it might be that it happens to work on avakar's machine but don't do it.
You could also use fmod(a,b) which calculates the a modulo b passing 1.0. This also should give the fractional part.
#include<cmath>
#include<iostream>
int main ()
{
double d1 = 555;
double d2 = 55.343;
double int_part1;
double int_part2;
using namespace std;
cout << boolalpha;
cout << d1 << " " << modf ( d1, &int_part1 ) << endl;
cout << d1 << " " << ( modf ( d1, &int_part1 ) == 0.0 ) << endl;
cout << d2 << " " << modf ( d2, &int_part2 ) << endl;
cout << d1 << " " << ( modf ( d2, &int_part2 ) == 0.0 ) << endl;
cout << d2 << " " << modf ( d2, &int_part2 ) << endl;
cout << d1 << " " << ( modf ( d2, &int_part2 ) == 0.0 ) << endl;
cout << d1 << " " << fmod ( d1, 1.0 ) << endl;
cout << d1 << " " << ( fmod ( d1, 1.0 ) == 0 ) << endl;
cout << d2 << " " << fmod ( d2, 1.0 ) << endl;
cout << d2 << " " << ( fmod ( d2, 1.0 ) == 0 ) << endl;
cout.flush();
modf ( d1, 0 ); // segfault
}
int iHaveNoFraction(double d){
return d == trunc(d);
}
Now, it wouldn't be C if it didn't have about 40 years of language revisions...
In C, == returns int but in C++ it returns bool. At least on my Linux distro (Ubuntu) you need to either declare double trunc(double); or you could compile with -std=c99, or declare the level macro, all in order to get <math.h> to declare it.
How about
if (abs(d1 - (round(d1))) < 0.000000001) {
printf "Integer\n"; /* Can not use "==" since we are concerned about precision */
}
Fixed up to work using rounding to reflect bug Anna found
Alternate solutions:
if ((d1 - floor(d1) < 0.000000001) || (d1 - floor(d1) > 0.9999999999)) {
/* Better store floor value in a temp variable to speed up */
printf "Integer\n"; /* Can not use "==" since we are concerned about precision */
}
Theres also another one with taking floor, subtracting 0.5 and taking abs() of that and comparing to 0.499999999 but I figure it won't be a major performance improvement.
Just compare ceil and floor value of d
return floor(d)==ceil(d);
So, for d1=555, the above statement will return 555==555, i.e, true, so it's an integer.
And for d2=555.6, the above statement will return 555==556, i.e, false, so it's a double.
How about this?
if ((d1 - (int)d1) == 0)
// integer
#define _EPSILON_ 0.000001
bool close_to_int(double &d)
{
double integer,
fraction = modf(d, &integer);
if(fraction < _EPSILON_)
{
d = integer;
return true;
}
if((1.0 - fraction) < _EPSILON_)
{
d = integer + 1;
return true;
}
return false;
}
This looks at both side of the integer value and sets the value of d if it is within the limits of an integer value.
try:
bool isInteger(double d, double delta)
{
double absd = abs(d);
if( absd - floor(absd) > 0.5 )
return (ceil(absd) - absd) < delta;
return (d - floor(absd)) < delta;
}
#include <math.h>
#include <limits>
int main()
{
double x, y, n;
x = SOME_VAL;
y = modf( x, &n ); // splits a floating-point value into fractional and integer parts
if ( abs(y) < std::numeric_limits<double>::epsilon() )
{
// no floating part
}
}
In many calculations you know that your floating point results will have a small numerical error that can result from a number of multiplications.
So what you may really want to find is the question is this number within say 1e-5 of an integer value. In that case I think this works better:
bool isInteger( double value )
{
double flr = floor( value + 1e-5 );
double diff = value - flr;
return diff < 1e-5;
}
I faced a similar questions.
As I needed to round the double anyway, that's what I find working:
double d = 2.000000001;
int i = std::round(d);
std::fabs(d-i) < 10 * std::numeric_limits<double>::epsilon()
modf uses std::nearbyint(num) that why you should use nearbyint which return a double without decimal and may be faster.
#include <iostream>
#include <cmath>
int main() {
double number = 55.12;
if (!(number - std::nearbyint(number))) {
std::cout << "Is integer!";
} else {
std::cout << "Has decimal!";
}
return 0;
}
A sample code snipped that does it:
if ( ABS( ((int) d1) - (d1)) )< 0.000000001)
cout <<"Integer" << endl;
else
cout <<"Flaot" << endl;
EDIT: Changed it to reflect correct code.
Below you have the code for testing d1 and d2 keeping it very simple. The only thing you have to test is whether the variable value is equal to the same value converted to an int type. If this is not the case then it is not an integer.
#include<iostream>
using namespace std;
int main()
{
void checkType(double x);
double d1 = 555;
double d2 = 55.343;
checkType(d1);
checkType(d2);
system("Pause");
return 0;
}
void checkType(double x)
{
if(x != (int)x)
{
cout<< x << " is not an integer "<< endl;
}
else
{
cout << x << " is an integer " << endl;
}
};