How to write a function for Hermite polynomials?

Question

For 2 numbers x and n entered by the user, my code needs to find Hn(x) defined recursively by the following formulas:

I am trying to implement a recursive version and and iterative version of that function. But I think I am getting the wrong concept of it, since my code doesn't compile due to errors on H(n) and H[n]:

#include "pch.h"
#include <iostream>

int H(int n, int x)   //function for recursion
{
if (n < 0) return -1;
else if (n == 0) return 1;
else if (n == 1) return 2 * x;
return 2 * x * H(n) * x - 2 * n * H(n - 1) * x;
}

int H1(int n, int x) //function for Iterator
{
int *H1 = new int[n + 1];

H[0] * x = 1;
H[1] * x = 2 * x;

for (int i = 0; i <= n; i++)
{
    H[i] * x = 2 * x * H[n] * x - 2 * n * H[n - 1] * x;

}
return H1(n) * x;
}

int main()
{
int n, x;
std::cout << "Enter the number n: ";
std::cin >> n;
std::cout << "Enter the number x: ";
std::cin >> x;

std::cout << "Rec = " << H[n] * x std::endl;
std::cout << "Iter = " << H1[n] * x std::endl;
}

It is confusing, I apologize for that as I am completely new to functions.

I already managed to do this with fibonacci sequence. And there I used only one parameter for function f(x) that is f(int n){... }, but here I am a bit confused with two parameters in function H(int n, int x) , where n is the index of H while x is an integer.

This `H(n)` is not how you call a function that takes two parameters and this `H1[n]` is also not how you call a function. Start by giving distinct names to variables and functions, your code is highly confusing — 463035818_is_not_an_ai, Dec 05 '18 at 17:27
`int *H1 = new int[n + 1];` -- 1) Memory leak. 2) Confusing naming variables the same as function names. — PaulMcKenzie, Dec 05 '18 at 17:27
https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list — 463035818_is_not_an_ai, Dec 05 '18 at 17:28
@user463035818, Yes it is confusing, I apologize for that as I am completely new to functions. — Techtirium, Dec 05 '18 at 17:29
rome wasnt build in one day. Start with a function that prints its parameter on the screen or something as simple as that — 463035818_is_not_an_ai, Dec 05 '18 at 17:30
I already did that with fibonacci sequence , And there I used one parameter for `function f(x)` that is `f(int n){ }`, but here I am a bit confused with two parameters in function `H(int n, int x)` , where `n` is the `index` of `H` while `x` is an `integer`.@ user463035818 — Techtirium, Dec 05 '18 at 17:33
then you should already know that `H[x]` is not calling a function called `H`. There are just too many syntax errors in your code and its not clear what the question is ("please fix my code" isnt really one) — 463035818_is_not_an_ai, Dec 05 '18 at 17:38
yes your are getting the concept wrong. If you have (maths notation) `f(n,x)` then this is not the same as `f(n) * x` — 463035818_is_not_an_ai, Dec 05 '18 at 17:47
Note that your conditions aren't consistent with the Tex in the image posted. Your conditions begin at `n = 1` while the Tex begins with `n = 0`. Please be accurate here. — TrebledJ, Dec 05 '18 at 17:53
@Techtirium I removed your transcription of the formula, since it was more confusing than anything else. WIth the picture in the text, it is much more clearer. I also took over some information of your comments to clarify the question. I hope it's ok like that. — Christophe, Dec 05 '18 at 18:24

Christophe · Accepted Answer · 2018-12-05T18:04:30.247

Yes, you need to translate your matematically indexed function into a function with 2 parameters.

The recursive version is almost ok, except for some shifts in the indexes:

int H(int n, int x)   // recursive version
{
    if (n <= 0)      
        return -1;
    else if (n == 1) 
        return 1;
    else if (n == 2) 
        return 2 * x;
    else 
        return 2 * x * H(n-1, x) - 2 * n * H(n - 2, x); // shift n+1, n n-1 to n, n-1 n-2 
}

Your iterative version needs rework, since you should write it as a loop, if possible without cashing the values that you no longer need. For example:

int Hi(int n, int x) //iterative version
{
    if (n <= 0)      
        return -1;
    else if (n == 1) 
        return 1;

    int am2 = 1;      // start for for n-2
    int am1 = 2*x;    // start for n-1
    if (n == 2) 
        return am1;

    int am; 
    for (int i=3; i<=n; i++) {
        am = 2*x*am1 -  2*i*am2;   // calculate Hn from Hn-1 and Hn-2
        //prepare next interation
        am2=am1; 
        am1=am;
    }
    return am; 
}

Online demo

YSC · Answer 2 · 2018-12-05T18:04:16.850

0

You wrote:

int H(int n, int x)   //function for recursion
{
    if (n < 0) return -1;
    else if (n == 0) return 1;
    else if (n == 1) return 2 * x;
    return 2 * x * H(n) * x - 2 * n * H(n - 1) * x;
}

You're not far from a working program. Drop that H1 function. Let's see:

int H(int n, int x)
{
    switch(n)
    {
    // H_0(x) = 1 
    case 0: return 1;
    // H_1(x) = 2x
    case 1: return 2 * x;
    // H_{n+1}(x) = 2x H_n(x) - 2n H_{n - 1}(x)
    default:
        return 2*x*H(n-1, x) - 2*(n-1)*H(n-2, x);
    }
}

The trick part is realizing than the n in H_{n+1}(x) = 2x H_n(x) - 2n H_{n - 1}(x) and in return 2*x*H(n-1, x) - 2*(n-1)*H(n-2, x);are not the same, they differ by one.

Now, you only need to handle user I/O and calling your H function with user input.

edited Dec 05 '18 at 18:04

answered Dec 05 '18 at 17:51

YSC

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`2*(n-1)*H(n-2, x)` :-) – TrebledJ Dec 05 '18 at 17:55
Originally, it's `H_{n+1}(x) = 2x H_n(x) - 2n H_{n - 1}(x)`. But we're shifting `n` down by 1... so that leads to `H_{n}(x) = 2x H_(n - 1)(x) - 2(n - 1) H_(n - 2)(x)`. – TrebledJ Dec 05 '18 at 18:03

How to write a function for Hermite polynomials?

2 Answers2