1
$\begingroup$

In the preface of Foundations of Differential Calculus there's a section that says:

Thus, if the quantity $x$ is given an increment $\omega$, so that it becomes $x + \omega$, its square $x^2$ becomes $x^2 + 2x\omega + \omega^2$, and it takes the increment $2x\omega + \omega^2$. Hence, the increment of $x$ itself, which is $\omega$, has the ratio to the increment of the square, which is $2x\omega+\omega^2$, as $1$ to $2x+\omega$. This ratio reduces to $1$ to $2x$, at least when $\omega$ vanishes.

$f(x) = x^2$

$f(x+w) = (x+\omega)^2 = x^2 + 2x\omega + \omega^2$

$(x+\omega)^2 - x^2 = 2x\omega+\omega^2$

And then the part I don't understand where $\omega = 2x\omega+\omega^2$ goes to "as $1$ to $2x+\omega$".

$\endgroup$
  • $\begingroup$ maybe it is better to do $\frac{(x+\omega)^2-x^2}{\omega}=2x+\omega$ $\endgroup$ – janmarqz Feb 20 '16 at 18:31
2
$\begingroup$

The paragraph says that the increment of $x$ (which means $(x + \omega) - x$, i.e. $\omega$) and the corresponding increment of $f$ from $f(x) = x^2$ to $f(x + \omega) = x^2 + 2 \omega x + \omega ^2$ (which means $f(x + \omega) - f(x) = 2 \omega x + \omega ^2$) have the ratio

$$\frac {(x + \omega) - x} {f(x + \omega) - f(x)} = \frac {\omega} {2 \omega x + \omega ^2} = \frac 1 {2 x + \omega}$$

which, when $\omega \to 0$, is exactly $\frac 1 {2x}$.

This prepares you mentally for later considering the limit of the reversed fraction:

$$\lim \limits _{\omega \to 0} \frac {f(x + \omega) - f(x)} {(x + \omega) - x} = 2x ,$$

which is an illustration of the concept of "derivative".

$\endgroup$
  • $\begingroup$ I get it now, thanks. The language was confusing me. $\endgroup$ – Elohgee Feb 21 '16 at 0:06
0
$\begingroup$

The text shaded in khaki has numerator and denominator interchanged in its last part. Instead it should read as follows:

Thus, if the quantity $x$ is given an increment $ω$, so that it becomes $x + ω$. In this way the square $x^2$ becomes $x^2 + 2xω + ω^2$, so that the square takes the increment $2xω + ω^2$. Hence, the ratio of the increment of the square, which is $2xω+ω^2$, to the increment of $x$ itself, which is $ω$, is $2x+ω$. This ratio reduces to $2x$ when $ω$ is "negligible", or "goes to $0$".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.