Question

Hi,

Could someone help me with the quadratic approximation of x ln x? I've checked the solutions but still don't really understand. it says to substitute 1 + h, but once I do, I'm not certain how to differentiate that, nor the other steps. It's from problem set 3 (2A-12E). Thanks!

Answer 1

The quadratic approximation is
\[ f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{1}{2} f''(x_0) (x - x_0)^2 \]
For your problem, we want to approximate the function
\[ f(x) = x \ln(x) \]
in the vicinity of x=1. That means we want to use \(x_0 = 1 \) in the approximation formula (which is the first 3 terms in a Taylor Series expansion).

From the formula, you see we need \(x_0 \) (good, we have it, as 1), we need
\( f(x_0) = f(1) \), which we can do.
and we need the first and second derivatives of f(x), evaluated at x=1

Answer 2

we can work through the formula and get the result. (we would need to use the product rule to find the derivative of x lnx )

In the posted answer, they took a slightly different approach (apparently to avoid taking the derivative of a product )

they said, let x be 1+ h
then
\[ f(x) = x \ln(x) \rightarrow f(x)= (1+h) \ln(1+h) \]
now they used the quadratic approximation formula to approximate *only* the ln part
in other words, evaluate about \( x_0=1\)
we get
\[ \ln(x) \approx \ln(1) + \frac{1}{x}\bigg|_{x=1} (x-1) + -\frac{1}{2}\frac{1}{x^2}\bigg|_{x=1}(x-1)^2 \\
\ln(x) \approx (x-1) - \frac{(x-1)^2}{2}
\]
or, if we use x= 1+h,
\[ \ln(x) \approx h - h^2/2 \]

using this approximation in
\[ f(x) = x \ln(x) \approx (1+h)(h - h^2/2) \]
multiply this out to get
\[ x \ln(x) \approx h - h^2/2 +h^2 - h^3/2 \]
Combine like terms, but only keep powers of 2 (h^3 is higher order than quadratic):
\[ x \ln(x) \approx h +h^2/2 \]
or , putting back (x-1) for h
\[ x \ln(x) \approx (x-1) +\frac{(x-1)^2}{2} \]

Answer 3

Looking at the pdf that went along with Section 25
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-a-approximation-and-curve-sketching/session-25-introduction-to-quadratic-appoximation/MIT18_01SCF10_Ses25a.pdf

I see another reason they said let x= 1+ h

in the pdf they show
\[ \ln(1+x) \approx x + \frac{x^2}{2} \]
pattern match that to immediately get
\[ \ln(1+h) \approx h + \frac{h^2}{2} \]
so just use that result in
\[ x \ln x \approx (1+h) \left( h + \frac{h^2}{2} \right) \]
expand, combine like terms, and drop all terms with exponents greater than 2.
and then replace h with (x-1) to get their answer.