Header-only compile-time template library for representing algebraic expressions, deriving differential and integral of the expression, and evaluating them by values.
An algebraic expression is composition of variables and constants connected with arithmetic operations and functions. Before we evaluate expression with values, the expression is just a representation retaining arithmetic structures among variables and constants.
These characteristic of algebraic expression well suits with expression template pattern in C++ generic coding. Using expression template pattern and its relevants, we can construct relations of variables and constants, then evaluate them at exact timing we want.
Download all header files in (REPOSITORY)/src/difint, then include difint.h
- Variables
A varible is a placeholder to get input value, and defined as follows.
DIFINT_VAR(C2);
DIFINT_VAR_4(x, x2, x3, x4);
std::cout << C2 << ", " << x << ", " << x2 << ", " << x3 << ", " << x4 << std::endl;Output Result
C2, x, x2, x3, x4
Since variables should be distinguished eachother, each variable has its own constexpr key value in it. The key value is derived by encoding its name (above example :'C2, x, x2, x3, x4'), and when we need the string name, the key value is decoded to string. By this reason, the length of variable name is limited by at most 10 characters.
- Constants
A constant object needs to be initialized by number. The constants can be initialized explicitely as follows.
di::C<double> Num(3.5);
di::C<double> N1(1);
di::C<double> N2(2);
di::C<double> N3(3);However, there will be few cases of the explicit initialize in a common usage. Rather we can omit the explicit definition of the constants because of the implicit type conversion from default (double compatible) types to constants as follows.
auto y = x4 + 3 + x1.e(2);Expressions are generated by arithmetic operations and functions. The expression is made, but not evaluated by actual values.
DIFINT_VAR_6(C1, C2, x, x2, x3, x4);
auto y1 = x.e(C1) + ((x + x2 - x3)* C1 * C2.e(x)).e(C1); // expression is generated
std::cout << y1 << std::endl;
auto y2 = x4.e(y1);
std::cout << y2 << std::endl;Output Result
x^C1+((x+x2-x3)*C1*C2^x)^C1
x4^(x^C1+((x+x2-x3)*C1*C2^x)^C1)
Using derivative member function, partial derivative can be derived. The derivation is generated by type estimation in compile-time.
- An example
DIFINT_VAR_4(x1, x2, x3, x4);
auto y = 2 * x1.e(2) + 3 * x2.e(2) + 4 * x3.e(2);
std::cout << y.derivative(x1) << std::endl;
std::cout << y.derivative(x2) << std::endl;
std::cout << y.derivative(x3) << std::endl;
std::cout << y.derivative(x4) << std::endl;Output Result
2.000000*2.000000*x1^(2.000000-1.000000)*1.000000
3.000000*2.000000*x2^(2.000000-1.000000)*1.000000
4.000000*2.000000*x3^(2.000000-1.000000)*1.000000
0.000000
The Result is dirty, we need reduction ...
- Another example
auto expn = x.e(x2.e(x3.e(x4.e(x.e(x2.e(x3.e(x4.e(x.e(x)))))))));
std::cout << expn << std::endl;
std::cout << expn.derivative(x) << std::endl;Output Result
C4^x^x2^x3^x4^x^x2^x3^x4^x^x
C4^x^x2^x3^x4^x^x2^x3^x4^x^x*x^x2^x3^x4^x^x2^x3^x4^x^x*(x2^x3^x4^x^x2^x3^x4^x^x*x3^x4^x^x2^x3^x4^x^x*x4^x^x2^x3^x4^x^x*x^x2^x3^x4^x^x*(x2^x3^x4^x^x*x3^x4^x^x*x4^x^x*x^x*(1.000000*lnx+1.000000*x/x)*lnx4*lnx3*lnx2*lnx+1.000000*x2^x3^x4^x^x/x)*lnx4*lnx3*lnx2*lnx+1.000000*x2^x3^x4^x^x2^x3^x4^x^x/x)*lnC4
more examples on simple derivatives
more examples on complicated derivatives
By input the numeric values into variables, we can evaluate the expression. Inputing values to variables and evaluation of expression is runtime features.
- Evaluation Example
DIFINT_VAR(x);
auto y1 = x + 3;
auto y2 = 3 - x;
auto y3 = 3 * x;
x = 4;
std::cout << y1.eval() << std::endl;
std::cout << y2.eval() << std::endl;
std::cout << y3.eval() << std::endl;Output Result
7
-1
12
- Evaluation Exception Example When there is at least one variables which are not input with numeric value, the expression cannot be evaluated, then throw except.
DIFINT_VAR_2(x, x2);
auto y4 = 3 / x + x2;
x = 4;
double resultVal = y3.eval(); //throw exceptionOutput Result (need to be decorated)
Exception : x2
- Plotting Graph Example
DIFINT_VAR(x);
auto y = x.e(2);
for (int idx = 0; idx < 100; ++idx)
{
x = idx / 10.0;
std::cout << "(" << x.eval() << "," << y.eval() << "," << y.derivative(x).eval() << ")" << std::endl;
}Output Result
(0,0,0)
(0.1,0.01,0.2)
(0.2,0.04,0.4)
(0.3,0.09,0.6)
(0.4,0.16,0.8)
(0.5,0.25,1)
...
(9.5,90.25,19)
(9.6,92.16,19.2)
(9.7,94.09,19.4)
(9.8,96.04,19.6)
(9.9,98.01,19.8)
Moved to Issues.