递归 function 指数 function

Question

I need implement recursive function exponential function (e^x) with help row Taylor: e^x = 1 + x + x2/2. + x3/3. +.:. But i can't understand what i do wrong I have next code:

function fact(n){
    return n * fact(n - 1);
}

function myPow(x, n){
    return x * myPow(x, n - 1);
}

function expon(x ,n){
    if(n == 1){
        return expon(x, n - 1) * x;
    }
    else{
        return expon(x, n - 1) + (myPow(x, n)/fact(n));
    }

}

console.log(expon(1, 10));

Answer 1

Your factorial function has no base case.

function fact(n) {
  if (n == 1)
    return 1;
  if (n < 1)
    return 0;
  return n * fact(n - 1);
}

A similar change will be needed for myPow . Although since powers are funny, I think if n == 0 return 1.

Answer 2

Your code should look like this:

function fact(n){
  if (n == 1)
    return 1;
  return n * fact(n - 1);
}

function myPow(x, n){
  if(n == 1)
    return n;
  return x * myPow(x, n - 1);
}

function expon(x ,n){
  if(n == 1){
    return 1;
  }
  else{
    return expon(x, n - 1) + (myPow(x, n)/fact(n));
  }

console.log(expon(1, 10));

Answer 3

This looks like an assignment so I won't debug your code but give you some hints.

You seem to not understand how recursion ends with giving a result in the end.

You have to provide a simple case where you return a value with a non recursive call.

Sometimes you don't write any if (bad!), and sometimes there is an if , but both cases use a recursive call.

Answer 4

This will solve your problem:

function fact(n) {
  if (n == 1) return 1;
  return n * fact(n - 1);
}

function myPow(x, n) {
  if (n == 0) return 1;
  return x * myPow(x, n - 1);
}

function expon(x, n) {
  if (n == 0) return 1;
  return expon(x, n - 1) + myPow(x, n) / fact(n);
}
console.log(expon(1, 10));