Hello, and happy new year first of all.

What I am trying to do is make a program that according to the values the user inputs then displays a polynomial.
The forumla for the polynomial is this:

``````float par(float x) //H timh ths synarthshs sto shmeio Xo
{
float tel=trap, timx=1;
int i;
for(i=0;i<n-1;i++)
{
timx=timx*(x-trap[i]);
tel=tel+trap[i+2]*timx;

}
return tel;
}``````

Where then the user is asked to input a value for x and then the result of the polynomial is calculated. It all works fine- the result is correct and all- but I can't seem to get the program to display the polynomial before the result is calculated.
It should be like this:
f(x0)+f[x0,x1](x-x0)+f[x0,x1,x2](x-x0)(x-x1)...+f[x0,...,xn](x-x0)...(x-xn)

where f[...]=trap[j]

Thanks for the help people.

It shouldn't be too difficult. You could either create another function to display the polynomial or add a couple of cout statements in your current function. In either case, you'll probably need two for loops, an outer one that will iterate over f(x0), f[x0, x1], ..., f[x0, ... xn] and …

Sorry but I'm not very good at explaining...

Then why don't you post an exact example of what you want the output to look like? It's always easier to show than explain, although both are important.

I don't think I can use cout

But, how will you print anything without using cout?

part of my problem is that I'm not sure how I can let x stay as a character since it is a part of the formula and the function

You can't do it like …

## All 11 Replies

It shouldn't be too difficult. You could either create another function to display the polynomial or add a couple of cout statements in your current function. In either case, you'll probably need two for loops, an outer one that will iterate over f(x0), f[x0, x1], ..., f[x0, ... xn] and an inner one that will iterate over x0, x1, ..., xk, for each f[x0, ..., xk] (it looks like you store x0, x1, ..., xk as trap, trap, ..., trap[k], so, you already have everything you need). By the way, I believe you also need to use two loops in order to compute the value of the polynomial (which implies that your current code doesn't calculate the result correctly).

Also, I have a question too: Do you need this for Mrs. Gousidou's exercise?

EDIT:

Ok, I see what you do now. You don't need two loops to compute the result. But if you want to print the polynomial, you'll need two loops.

It shouldn't be too difficult. You could either create another function to display the polynomial or add a couple of cout statements in your current function. In either case, you'll probably need two for loops, an outer one that will iterate over f(x0), f[x0, x1], ..., f[x0, ... xn] and an inner one that will iterate over x0, x1, ..., xk, for each f[x0, ..., xk] (it looks like you store x0, x1, ..., xk as trap, trap, ..., trap[k], so, you already have everything you need). By the way, I believe you also need to use two loops in order to compute the value of the polynomial (which implies that your current code doesn't calculate the result correctly).

Thanks for the help, but also part of my problem is that I'm not sure how I can let x stay as a character since it is a part of the formula and the function... I'm not very experienced in programming, so in my terms this is what I mean:

``xval=xval*([b]"x"[/b]-table[i]);``

I don't think I can use cout because I need to have it looped. Sorry but I'm not very good at explaining...

Also, I have a question too: Do you need this for Mrs. Gousidou's exercise?

Hmm nah not sure who that Mrs. Gousidou is, it's a project I started a few days ago by myself and I still haven't finished it. Tough stuff.

Sorry but I'm not very good at explaining...

Then why don't you post an exact example of what you want the output to look like? It's always easier to show than explain, although both are important.

Well I want the polynomial to be displayed like this:

f(x0)+f[x0,x1](x-x0)+f[x0,x1,x2](x-x0)(x-x1)...+f[x0,...,xn](x-x0)...(x-xn)
eg.

2+3(x-4)-1(x-4)(x-5)+7(x-4)(x-5)(x-2)

I know that I need one loop to change x0,x1,...xn inside the parenthesis. And I also need another loop to change the number of the parenthesis each time, but I'm not sure how to do that since the parenthesis contains x, which should be just a character.
Get it now?

I don't think I can use cout

But, how will you print anything without using cout?

part of my problem is that I'm not sure how I can let x stay as a character since it is a part of the formula and the function

You can't do it like this. You'll have to use two loops:

``````for i = 0 -> N
{
if ( /* something involving i */ ) print "+"

print "("
print f[x0, ..., xi]
print ")"

for j = ??? -> ???
{
print "(x-"
print xj
print ")"
}
}``````

I don't know how it will work if I break this line

``xval=xval*(x-table[i]);``

with cout.

Well, as I said above, you can't do it like this. Leave that as it is and write
another function to print the polynomial, using the pseudocode I posted
above. It shouldn't be too difficult, unless the code you posted isn't yours.

Thanks it worked indeed... I need more exercising in C++...
I might need more help later but for now I'm covered, thanks!

Alright, I have one more question... can anyone give me a hint of how I can convert reverse polish notation into "normal" mathematical notation whn using C++?
Thanks.Alright, I have one more question... can anyone give me a hint of how I can convert reverse polish notation into "normal" mathematical notation whn using C++?
Thanks.

Recommend you search the site as this is a common problem to solve with C++. If you have any questions after doing the search, then start another thread given there is no continuity between the polynomial thread and the reverse polish logic other than you are the poster for both questions.

HINT: stacks are commonly used.

Yeah got carried away sorry. I'll make a search and be a good boy.

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