A years-long explanation on differentials.

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Charlie Root 2026-03-30 21:00:49 -03:00
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@ -3008,7 +3008,7 @@ Veja também a Seção~\ref{sec:sine-cosine} pra compreender o ``co'' em
\subsection{Differentials} \subsection{Differentials}
If $y = f(x)$, where $f$ is a differentiable function, then the If $y = f(x)$, where $f$ is a differentiable function, then the
differential $dx$ is an independent variable; that is, $dx$ can be given differential $dx$ is an independent variable, that is, $dx$ can be given
the value of any real number. The differential $dy$ is then defined in the value of any real number. The differential $dy$ is then defined in
terms of $dx$ by the equation terms of $dx$ by the equation
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@ -3026,22 +3026,36 @@ x$. Then, the change in $y$ is
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$$\Delta y = f(x + \Delta x) - f(x).$$ $$\Delta y = f(x + \Delta x) - f(x).$$
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Certainly, because $f(x + \Delta x)$ is the height of the point $Q$; Certainly, because $f(x + \Delta x)$ is the height of the point $Q$.
then, if we subtract the height of the point $P$, we get the change in Then, if we subtract the height of the point $P$, we get the change in
$y$ as we moved from $P$ to $Q$. $y$ as we moved from $P$ to $Q$.
Now, $dy$ expresses the change in the linearization of $f(x)$, which for Now, $dy$ expresses the change in the linearization of $f(x)$, which for
a concave down curve with positive slope, would be larger than $\Delta a concave down curve with positive slope, would be larger than $\Delta
y$. y$.
In a concave down graph with positive slope, for example, let $L(x)$ be In a concave down graph with positive slope, for example, let $L(x)$
a tangent to the curve at $P$. So $\Delta y < dy$. The value of $\Delta be a tangent to the curve at $P$. So $\Delta y < dy$. The value of
y$ is simply $f(x + \Delta x) - f(x)$, so the value of $dy$ should be $\Delta y$ is simply $f(x + \Delta x) - f(x)$, so the value of $dy$
something like $L(x + \Delta x) - L(x)$. However, the value of $dy$ is should be something like $L(x + \Delta x) - L(x)$, embora a definição
$f'(x) dx$. So it must follow that seja $dy = f'(x) \dx$. Isso significa que tem que ser verdadeiro que
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$$L(x + \Delta x) - L(x) = f'(x) \dx.$$ $$L(x + \Delta x) - L(x) = f'(x) \dx.$$
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E de fato é---hoje podemos explicar por quê. Declare $\Delta x =
dx$. Comparemos $\Delta f$ com $dy$. Quando nos movemos numa
quantidade $\Delta x$, a variação na curva $f(x)$ é $\Delta f = f(x +
\Delta x) - f(x)$. Já a variação na reta $L(x)$ é
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\begin{align*}
L(x + \Delta x) - L(x) &= L(x + dx) - L(x)\\
&= f(a) + f'(a)(x + dx - a) - (f(a) + f'(a)(x - a))\\
&= \cancel{f(a)} + \cancel{f'(a)x} + f'(a)\dx - \cancel{f'(a)a} - \cancel{f(a)} - \cancel{f'(a)x} + \cancel{f'(a)a}\\
&= f'(a) \dx,
\end{align*}
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como desejado.
%% \subsection{Newton's method} %% \subsection{Newton's method}