Question

The Linearization is the Best Linear Approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c(m\) and \(c\) constants). If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions i. \(E(a)=0\) ii. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \((x-a)\) .

Short answer

Considering the two constraints, i.e, \(E(a)=0\) and \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\), it has been demonstrated that \(g(x)\) must necessarily be equal to the linearization \(L(x) = f(a) + f'(a)(x - a)\) at \(x=a\). Therefore, the linearization, under these conditions, gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison to \((x-a)\), which proves the statement.

Step-by-step solution

Understanding the Constraints

Our task is to prove that under these constraints, the function \(g(x)\) actually coincides with the linearization of function \(f(x)\), denoted as \(L(x)\), at \(x=a\). The two constraints imposed on \(g(x)\) are: i) \(E(a)=0\) which means that the error between \(f(a)\) and \(g(a)\) is zero; and ii) \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) which means the ratio of the error to \(x-a\) tends to zero as \(x\) tends to \(a\).

Apply the First Constraint

The first condition states that \(E(a)=0\). This means that the error between the function \(f(x)\) and its approximation \(g(x)\) at \(x=a\) is zero. Or in other words, \(f(a)=g(a)\). Recalling that \(g(x)=m(x-a)+c\), when we substitute \(x=a\), we get \(f(a)=m(a-a)+c \Rightarrow f(a)=c\). Thus, the constant \(c\) is actually \(f(a)\). Let's substitute \(c=f(a)\) in \(g(x)\), we get: \(g(x)=m(x-a)+f(a)\).

Apply the Second Constraint

The second condition states that \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\). Expressing \(E(x)\) as \(f(x) - g(x)\), we get \(\lim _{x \rightarrow a} \frac{f(x) - g(x)}{x-a}=0\). This can be rearranged as \(\lim _{x \rightarrow a} \frac{f(x) - m(x-a)-f(a)}{x-a}=0\). Given that the limit of a function can be split into addends, we can reformulate to \(\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a}-\lim _{x \rightarrow a}m=0\). The first limit is by definition is the derivative of \(f\) at \(a\), i.e. \(f'(a)\). Therefore we get \(f'(a)-m=0\), solving for \(m\), we get \(m=f'(a)\). So, if we substitute \(m=f'(a)\) into the function \(g(x)\), we obtain \(g(x)=f'(a)(x-a)+f(a)\).

Completion of Proof

The final result, \(g(x)=f'(a)(x-a)+f(a)\), can be rearranged to \(g(x)=f(a) + f'(a)(x-a)\). This is actually the definition of the linearization \(L(x)\) of \(f(x)\) at \(x=a\). So under these conditions, \(g(x)\) is equivalent to \(L(x)\). This completes the proof.