Solution of Equations by the Differential Calculus

Excerpts

Solution of Equations by the Differential Calculus
Let $a + h$ be the real value, in which $h$ will be a small quantity. It follows that $\phi(a + h) = 0$, or, which is nearly true, $\phi a + \phi' a\, h = 0$.
Solution of Equations by the Differential Calculus
For example, let $x^{2} + x - 4 = 0$ be the equation. Here $\phi x = x^{2} + x - 4$, and $\phi(x + h) = (x + h)^{2} + x + h - 4 = x^{2} + x - 4 + (2x + 1)h + h^{2}$; so that $\phi' x = 2x + 1$.
Solution of Equations by the Differential Calculus
A near value of $x$ is $1.57$; let this be $a$. Then $\phi a = .0349$, and $\phi' a = 4.14$. Hence $-\dfrac{\phi a}{\phi' a} = -.00843$.
Solution of Equations by the Differential Calculus
If we proceed in the same way with $1.5616$, we shall find a still nearer value of $x$, viz., $1.561553$.
Solution of Equations by the Differential Calculus
We have here chosen an equation of the second degree, in order that the student may be able to verify the result in the common way; it is, however, obvious that the same method may be applied to equations of higher degrees, and even to those which are not to be treated by common algebraical method, such as $\tan x = ax$.

Solution of Equations by the Differential Calculus
\phi x = 0
The equation whose root x is sought is a function of x set equal to zero.
Solution of Equations by the Differential Calculus
\phi(a + h) = 0
Since a + h is the real root, the function vanishes at a + h.
Solution of Equations by the Differential Calculus
\phi a + \phi' a\, h = 0
Replacing phi(a + h) by its first-order expansion gives a nearly true linear relation for the small correction h.
Solution of Equations by the Differential Calculus
a - \dfrac{\phi a}{\phi' a}
The value a - phi(a)/phi'(a) is a nearer approximation to the root x than a, since h is nearly -phi(a)/phi'(a).
Solution of Equations by the Differential Calculus
\phi' x = 2x + 1
For phi x = x^2 + x - 4, the derivative phi' x equals 2x + 1.
Solution of Equations by the Differential Calculus
\phi(x + h) = (x + h)^{2} + x + h - 4 = x^{2} + x - 4 + (2x + 1)h + h^{2}
For phi x = x^2 + x - 4, expanding phi(x + h) gives the original value plus (2x + 1)h plus h^2.
Solution of Equations by the Differential Calculus
x^{2} + x - 4 = 0
The example equation of the second degree whose root is sought by the method.
Solution of Equations by the Differential Calculus
\tan x = ax
An equation not treatable by common algebra, to which the same method of successive approximation applies.

No exercises in this chapter.