0
$\begingroup$

Using Newton's Method, find the coordinates accurate to 6 decimal points of the point on the function y=e^x that is closest to the origin.

  • 0
    Newton's Method looks for the zeros of a function. You might start telling us about your attempts to write down that function. – 2017-01-23
  • 0
    @avs There is a Newton's method for finding the zeros of a function, and also a Newton's method for minimizing a smooth function $f$. (The second Newton's method works by using the first Newton's method to solve $\nabla f(x) = 0$.) – 2017-01-23
  • 1
    @littleO, yes, so, essentially, there is only one Newton's method.:) And I was trying to encourage ConMan to write down the suitable $f$. – 2017-01-24

3 Answers 3

1

Hint

We want the $x$ that yields the minimum value of $D(x)$, the distance between $(0,0)$ and $(x, e^x)$. This is the value for which $D'(x)=0$ and $D''(x)>0$.

Here we would approximate $D'(x) = 0$ using Newton's Method as given by: $$ x_{n + 1} = x_n - \frac{D'(x_n)}{D''(x_n)} $$

Now we just need an expression for $D(x)$ and a first approximation, $x_1$.

  • 0
    Should be fixed now @user251257 – 2017-01-23
1

The squared distance of $(x,e^x)$ from the origin is given by $x^2+e^{2x}$, whose derivative is given by $2x+2e^{2x}$. It follows that the minimal distance is achieved at the only real root of $e^{2x}+x$ (that is a convex function). By choosing $x_0=0$ as starting point, Newton's method $$ x_{n+1}=x_n-\frac{e^{2x_n}+x_n}{2e^{2x_n}+1} $$ gives $x\approx -0.426303$ with just $4$ iterations. The minimal distance is so $\approx\color{red}{0.779767}$.
Here we have $x=-\frac{1}{2}W(2)$ where $W$ is Lambert's function.

-1

Newton's Method will approximate the solution of $ f(x) = 0 $. Draw a diagram, consider your function, and apply the formula: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ until you get your answer