2. Newton's Method for Solving Equations

The graph is as follows:

Try x₀ = 1.5

Then

`x_1=x_0-(f(x_0))/(f’(x_0))`

` = 1.5-(f(1.5))/(f’(1.5))`

Now f(1.5) = 2(1.5)² − 1.5 − 2 = 1

f '(x) = 4x − 1 and f '(1.5) = 6 − 1 = 5

So

`x_1=1.5-1/5=1.3`

So `1.3` is a better approximation.

Continuing the process,

`x_2=x_1-(f(x_1))/(f’(x_1))`

` = 1.3- (f(1.3))/(f’(1.3))`

` = 1.3-0.08/4.2`

`=1.2809524`

We can continue for as many steps as necessary to give the required accuracy.

Check: Using some Computer Algebra Systems, (eg Mathcad) we also need to give an initial guess (say, x = 2) and the result is:

root(2x² − x − 2, x) = 1.2807764064044

The software just keeps applying the algorithm until the decimal digits don't change any more. It then returns the answer.

Let y = 1− t² + 2^t

The graph of `y(t)`:

There appear to be 2 roots, one near t = −1 and the other near t = 3. However, if we look more carefully in the region near t = 3 (by zooming in), we see that there is more than one root there.

By simple substitution, we can see that one of the roots is exactly t = 3:

1 − (3)² + 2³ = 0

Now for the root near t = 3.4.

We will use Newton's Method to approximate the root. We need to differentiate y = 1− t² + 2^t. Since we have t as an exponent in our expression, we need to use logarithms when differentiating. [See how to differentiate logarithms in Derivative of the Logarithmic Function].

Let's differentiate 2^t by itself first.

Let h = 2^t.

Take natural log of both sides:

`ln\ h=t\ ln\ 2`

`1/h (dh)/dt=ln\ 2`

`(dh)/(dt)=h\ ln\ 2=2^t\ ln\ 2`

So

`(dy)/(dt)=f'(t)=-2t+2^t\ ln\ 2`

So for Newton's Method in this example, we would have:

`(f(t))/(f'(t))=(1-t^2+2^t)/(-2t+2^t ln\ 2)`

Initial guess: t₀ = 3.4

`t_1=t_0-(f(t_0))/(f'(t_0))`

`t_1=3.4-(f(t))/(f'(t)) -` ` (1-3.4^2+2^3.4)/(-2(3.4)+2^3.4 ln\ 2) ` `= 3.407615918`

We can write this more conveniently (for later steps) showing the substitution as:

`t_1=3.4-[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.4)` `=3.407615918`

Now, doing another step of Newton's Method:

`t_2=t_1-(f(t_1))/(f'(t_1))`

`t_2` `=3.407615918-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.407615918)` `=3.407450596`

And doing another step:

`t_3` `=3.407450596-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.407450596)` `=3.407450522`

And another:

`t_4` `=3.407450522-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=3.407450522)` `=3.407450505`

We can conclude that correct to 7 decimal places, t = 3.4074505.

Using Graphs Instead

Using a computer algebra system, we can zoom into the root and we can see (from where the graph cuts the y-axis) that t is indeed close to `3.40745`.

Now for the negative case. Let t₀ = −1 be our initial guess.

`t_1` `=-1-[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=-1)` `=-1.213076633`

`t_2` `=-1.213076633-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=-1.213076633)` `=-1.198322474`

`t_3` `=-1.198322474-` `[(1-t^2+2^t)/(-2t+2^t ln\ 2)]_(t=-1.198322474)` `=-1.198250199`

We could continue until we obtained the required accuracy.

Comparing this to the zoomed in graph, we can see that the solution is t = −1.198250197, correct to 9 decimal places.

Conclusion

So the solutions for 1− t² + 2^t = 0 are

t = −1.19825,

t = 3, or

t = 3.40745,

correct to 5 decimal places.

We'll learn in the Curve Drawing section that a cubic curve can have either:

• One real root;
• Two real roots (one of them repeated); or
• Three distinct roots

The best thing to do is to sketch the curve, but let's assume that's too hard for now. We can determine where at least one of the roots occurs by substituting values of `x` until we see a change in the sign for `y` (which would indicate the curve must have passed through the `x`-axis in that interval).

Starting at `x=0`, we find that `y=2`, which is positive.

Next, try `x=1`, and we get `y=1`, which is still positive. (Of course, the curve could have dipped below the `x`-axis between `0` and `1`, but without a sketch, we don't know.)

Try `x=2` and we get `y=0`. This is a lucky guess, and gives us one of the roots.

Trying more positive values (`x=3,4,5...`) gives us larger and larger values for `y` (which are `17,70,177,...`), so we are clearly moving away from the root.

We continue to stumble around in the dark, and assume maybe there is a root near `x=1`. Let's try the algorithm with `x_0=1` and see what happens.

We'll need the derivative:

`dy/dx = 9x^2-18x+5`

So

`x_1=x_0-(f(x_0))/(f’(x_0))`

` = 1-(f(1))/(f’(1))`

` = 1-1/(-4)`

` = 1.25`

Another step:

`x_2=x_1-(f(x_1))/(f’(x_1))`

` = 1.25-(f(1.25))/(f’(1.25))`

` = 1.26363636363`

Continuing on we get:

x₃ = 1.26376260461638

x₄ = 1.26376261582597

x₅ = 1.26376261582597

x₆ = 1.26376261582597

Since there is no more change to our value, we can assume we've found the second root, at `x=1.26376261582597`.

You can use the interactive graph applet to find the third root.

Of course, we can easily solve this one by inspection.

But I've included it because it illustrates a case when Newton's Method is not very efficient.

Newton's Method works well when the slope of the line is reasonably steep, but in cases where it's quite flat near the root (similar to Example 2 above), it does not converge very quickly.

(You can investigate this example in the interactive graph applet.)

In fact, even after 12 steps of the method, we are still not very close to the root.

Let's start at `x=-4`.

The derivative:

`dy/dx = 2x`

So

`x_1=x_0-(f(x_0))/(f’(x_0))`

` = -4-(f(-4))/(f’(-4))`

` = -4-16/(-8)`

` = -2`

Step 2:

`x_2=x_1-(f(x_1))/(f’(x_1))`

` = -2-(f(-2))/(f’(-2))`

` = -2-4/(-4)`

` = -1`

Continuing on we get:

x₃ = -0.5

x₄ = -0.25

x₅ = -0.125

x₆ = -0.0625

x₇ = -0.03125

x₈ = -0.015625

x₉ = -0.0078125

x₁₀ = -0.00390625

x₁₁ = -0.001953125

x₁₂ = -0.0009765625

Generally, however, Newton's Method is a simple and neat way to find roots of equations. It is commonly used (in conjunction with other efficiency algorithms) by computers and calculators.