Question

Alchemista or anyone who knows C++ , i have a question (related to math too)

Answer 1

Anyway I'm just making an easy converter, but i was wondering about this right.. when converting from Far to Celsius, why is it that when i do this:
cout << value << " Farenheit is " << value*5/9-32*5/9 << " Celsius." << endl;
the program works, but..

Answer 2

when I did:
cout << value << " Farenheit is " << (value-32)*(5/9) << " Celsius." << endl; it didn't ?

Answer 3

if this is any help when i type in 90 farenheit it is showing me 0 celsius but when i use the first method i do get 33 celsius

Answer 4

You need to be aware of the various data types in C++

Answer 5

When you perform division on two integers it will perform integer division. The remainder will simply be chopped off instead of being included in the quotient.

Answer 6

(value - 32.0) * (5.0/9.0)

Answer 7

ohh okay i see.. so i have to use .0 after decimals?

Answer 8

i mean after integers..

Answer 9

when .0 is added to a literal the compiler will interpret it as a floating point type

Answer 10

oh okay thanks =)!

Answer 11

this is what my stuff looks like so far..

Answer 12

#include <iostream>

using namespace std;

int main () {
char choice;
float value;
cout << "Select choice:" << endl
<< "1) Kilometers -> Meters" << endl
<< "2) Meters -> Kilometers" << endl
<< "3) Feet -> Meters" << endl
<< "4) Meters -> Feet" << endl
<< "5) Farenheit -> Celsius" << endl
<< "6) Exit" << endl;
for ( ;; ) {
cout << "Select: " << flush;
cin >> choice;
switch ( choice ) {
case '1':
cout << "Kilometers=" << flush;
cin >> value;
cout << value << " kilometers is " << value*1000.0 << " meters." << endl;
break;
case '2':
cout << "Meters=" << flush;
cin >> value;
cout << value << " meters is " << value/1000.0 << " kilometers." << endl;
break;
case '3':
cout<< "Feet=" << flush;
cin >> value;
cout << value << " feet is " << value*0.3048 << " meters." << endl;
break;
case '4':
cout<< "Meters=" << flush;
cin >> value;
cout << value << " meters is " << value/0.3048 << " feet." << endl;
break;
case '5':
cout<< "Farenheit=" << flush;
cin >> value;
cout << value << " Farenheit is " << (value-32.0)*(5.0/9.0) << " Celsius." << endl;
break;
case '6':
return 0;
default:
cout << "Unknown choice!" << endl;
}
}

}

Answer 13

that's if u're interested.. i mean nothing serious.. Ohhh can u differentiate too?

Answer 14

What do you mean? Like numerically? Or symbolically?

Answer 15

both but id say numerically would be good enough..

Answer 16

C++ is a general purpose programming language and its very low level (close to the machine). The idea of a mathematical function doesn't exist so you can't really symbolically differentiate.

Answer 17

i see.. i read that before in a thread on some programming site..

Answer 18

You could write all the code to do this, including all the various rules (chain rule, product rule, power rule, etc) and then develop a way to encode symbolic mathematical expressions but this gets very complicated.

Answer 19

okay so it is possible i might do it eventually as a project or something.. now how about numerically?

Answer 20

You can also just numerically differentiate, and that's very simple. Recall the difference quotient:
\[\frac{f(x + h) - f(x)}{h}\]

To take the numerical approximation of the derivative at x, just use a very small h.

Answer 21

yes..

Answer 22

Why ".." is there something you don't understand?

Answer 23

no im just thinking of putting that into code.. can you give me an example?

Answer 24

ohh wait a second.. hmm if i did h=0, then it would give me an error wouldnt it?

Answer 25

err

Answer 26

thats wrong one second

Answer 27

float h = 0.001;
int x = 5;
int approx = ((x + h)*(x + h) - x*x)/h;

Answer 28

hmm but what if i wanted a derivative of some function.. suppose:
x^2+x+1 would that be possible? because u are saying that x is an integer already..

Answer 29

like f(x) = x^2+x+1
f'(5)-?

Answer 30

its simpler just to define a function, but im assuming you haven't gotten that far yet.

Answer 31

okay anyway i gtg will continue tomorrow thanks for the help, bye

Answer 32

typedef float FUNC_REALTOREAL ( float input );
float myPolynomial ( float input ) {
return input * input + input + 1;
}
float derivative ( FUNC_REALTOREAL *realToRealFunc, float x, float h ) {
return ( realToRealFunc ( x + h ) - realToRealFunc ( x ) ) / h;
}
int main () {
int result = derivative ( myPolynomial, 5, 0.001 );
}

Answer 33

Some of the above might not be accessible until you learn a bit more C/C++.

Answer 34

ALCHEMISTA I NEED YOUR HELP!!! go to a new thread please!