Finding qth real root of a real number

In this post we will see how to find n^th roots of a positive real number. We will use the Newton-Raphson method to deduce an iterative formula, and see its convergence.

The task is to find the real q^th root of a real number. Let there be a non-negative real number C, of which the real q^th root is to be found. Let the square root of C be x. We represent this with $x=\sqrt[q]{C}$ .

Now:
$x=\sqrt[q]{C}$
$\Rightarrow x^{q}=C$
$\Rightarrow x^{q}-C=0$

So we are now to find the roots of the q^th order function $f(x)=x^{q}-C$

Finding qth Roots

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To find the root of $f(x)=x^{q}-C$ we will use the Newton-Raphson method of finding roots. The Newton-Raphson is an iterative method, which starts with a guess root $x_{n}$ and then finds the next guess $x_{n+1}$ by drawing tangent at the first guess point $(x_{n},f(x_{n}))$ and taking the x ordinate of the intersection of the tangent and X axis as $x_{n+1}$ . It continues this process till the exact root is found or some root is found within tolerable error. Taking $x_{0}$ as the first guess we can find the next guess by Newton-Raphson method from the below expression.

$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$

For more detail, check this link in Wikipedia, about Newton’s method or check any standard text book.

We will use this method to get roots of real number. In our case the function is:

$f(x_{n})=x^{q}-C$
And the first derivative is:
$f'(x_{n})=qx^{q-1}$

Putting these values in the Newton-Raphson’s formula we get:

$x_{n+1}=x_{n}-\frac{x_{n}^{q}-C}{qx_{n}^{q-1}}=\frac{qx_{n}^{q}-x_{n}^{q}+C}{qx_{n}^{q-1}}=\frac{(q-1)x_{n}^{q}+C}{qx_{n}^{q-1}}$
$\Rightarrow x_{n+1}=\frac{(q-1)x_{n}^{q}+C}{qx_{n}^{q-1}}$

Putting q=2,3,4 …. q in the last equation we can find square, cube, fourth ….. q^th root of the real number C.

Finding q^th root of a real number:
$x_{n+1}=\frac{(q-1)x_{n}^{q}+C}{qx_{n}^{q-1}}$

Convergence

Now we talk about if this method converges to the root. As per Newton-Raphson’s convergence condition, the method will converge to its root if

$\left|\frac{f(x_{n})f''(x_{n})}{(f'(x_{n}))^{2}}\right|<1$

Putting the values for our case we get

$\left|\frac{q(q-1)x_{n}^{q-2}(x_{n}^{q}-C)}{q^{2}x_{n}^{2q-2}}\right|=\left|\frac{(q-1)}{q}\times\frac{x_{n}^{q}-C}{x_{n}^{q}}\right|<1$
as $(x_{n}^{q}-C)<x_{n}^{q}$ and $(q-1)<q$

Thus this process converges to its root. If C is negative then the roots of the equation is imaginary, and this process will fail to compute the roots. The initial guess is very important for convergence or the results might oscillate infinitely back and forth the negative and positive side of X axis.

Square Root

To find the square root of a real number with this method, read this post Evaluating square root of a real number

4 thoughts on “Finding qth real root of a real number”

Pingback: Evaluating square root of a real number | Phoxis
jay.me says:

April 27, 2010 at 9:09 pm

Hi Arjun, I dont get any of that =(
I find Isaac Newtons religious views very interesting, he was anti-trinitarian =)

Reply
1. phoxis says:
  
  April 28, 2010 at 4:52 pm
  
  now what was that, i didn’t either get what was “anti-trinitarian”
  
  Reply
jay.me says:

April 28, 2010 at 6:22 pm

Follow the links from the one you posted and get here…..http://en.wikipedia.org/wiki/Isaac_Newton's_religious_views

Newton was anti-trinitarian, he did not believe in the doctrine of the trinity, he believed in a monotheistic God.

“…Newton held a Christian faith that, had it been made public, would not have been considered orthodox by mainstream Christianity…”

“Though he is better known for his love of science, the Bible was Sir Isaac Newton’s greatest passion. He devoted more time to the study of Scripture than to science, and he said, “I have a fundamental belief in the Bible as the Word of God, written by those who were inspired. I study the Bible daily.”

cool.

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Finding qth real root of a real number

Finding qth Roots

Convergence

Square Root

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