Question

In 2A-12, d, part of the explanation in the answer sheet says that "ln(1+h) ~= h". Isn't this the linear approximation? In 2A-12, e, "ln(1+h) ~= h-(h^2/2)". Can someone please explain why the linear approx. is used in 12d, but the quad. approx is used in 12e, when both are asking for a quadratic approximation?

Answer 1

we start with
\[ \cos x \approx 1 - \frac{x^2}{2} \]
and thus
\[ \ln( \cos x) \approx \ln \left(1 - \frac{x^2}{2}\right) \]

next, we use
\[ \ln(1-y) \approx -y - \frac{y^2}{2} \]
with \( y= \frac{x^2}{2} \)
we get
\[ \ln( \cos x) \approx -\left( \frac{x^2}{2}\right) - \frac{1}{2} \left( \frac{x^2}{2} \right)^2\]

notice that the second term is to the fourth power, and thus we drop it from our quadratic approximation, to get the result
\[ \ln( \cos x) \approx - \frac{x^2}{2} \]
Whoever posted the answer assumed that this process was "obvious to the most casual observer", i.e. "clearly" we need only consider the linear approximation for ln.

Answer 2

Thank you for your quick and helpful reply. To clarify, we know that the linear approximation is enough for ln(1-x^2/2) because it's a squared term (and therefore the quartic term will drop off, so we only need take the linear approximation)? And for 2A-12e, we keep the quadratic approximation, because x=1-h, and h^2 is squared and not quartic (and so will not drop off)?

Answer 3

For 2A-12e

\[ x \ln x, \ x \approx 1 \]
let x= 1+h (and \( h \approx 0 \))
and in terms of h we have
\[ x \ln x \approx (1+h) \ln(1+h) \]
use \[ \ln(1+h) \approx h - \frac{h^2}{2} \]
to get
\[ x \ln x \approx (1+h) \ln(1+h) \approx (1+h) \left( h - \frac{h^2}{2}\right)\]
multiplying out,
\[ x \ln x \approx h - \frac{h^2}{2} +h^2 - \frac{h^3}{2} = h^2 + \frac{h^2}{2} - \frac{h^3}{2}\]
with h near 0, the higher order terms approach zero faster than the linear or quadratic terms, and it's reasonable to drop them. i.e. the quadratic approximation is
\[ h^2 + \frac{h^2}{2} \]
put in terms of x (from x= 1+h, we have h= x-1):
\[ (x-1)^2 + \frac{(x-1)^2}{2} \]

Answer 4

Hijacking this old post, since it concerns the same problem (2A-12e)

I got different answers by trying two different methods that (in my mind) should be equivalent. One is the method posted above, using x=h+1.

The other is finding the quadratic approximations of x and ln(x) near x=1, using the formula

\[f(x) = f(1) + f'(1)(x-1) + \frac{ 1 }{ 2 }f''(1)(x-1)^{2}\]

For ln(x), this gives

\[\ln x = 0 + (x-1) - \frac{ (x-1)^2 }{ 2 }\]

and

\[x = 1 + (x-1) = x\]


Plugging both of these into the original equation gives:

\[x \left[ (x-1)-\frac{ (x-1)^2 }{ 2 } \right]\]

Which does not simplify to the answer given by the other method. Did I make a mistake somewhere?