Quadratic Approximations When linear approximations are insufficient, we must turn to quadratic
approximations . For example, in economics and finance, models often rely
on logarithmic or quadratic scales to reduce complexity. Understanding
these models requires reducing computations to quadratic approximations.
Quadratic approximations are just an extension of linear approximations.
We're merely taking an extra step in detail. The general formula:
f ( x ) β f ( x 0 ) + f β² ( x 0 ) β
( x β x 0 ) + f β² β² ( x 0 ) 2 ( x β x 0 ) 2 f(x) \approx f(x_0) + f'(x_0) \cdot (x-x_0) + \dfrac{f''(x_0)}{2}(x-x_0)^2 f ( x ) β f ( x 0 β ) + f β² ( x 0 β ) β
( x β x 0 β ) + 2 f β²β² ( x 0 β ) β ( x β x 0 β ) 2 Where is that ( 1 / 2 ) f β² β² ( 0 ) {(1/2)f''(0)} ( 1/2 ) f β²β² ( 0 )
f ( x ) = a + b x + c x 2 f(x) = a + bx + cx^2 f ( x ) = a + b x + c x 2 Differentiating the function above:
f β² ( x ) = b + 2 c x f'(x) = b + 2cx f β² ( x ) = b + 2 c x Differentiating a second time:
f β² β² ( x ) = 2 c f''(x) = 2c f β²β² ( x ) = 2 c Next, we want to determine a , {a,} a , b , {b,} b , c {c} c
f ( 0 ) = a + b ( 0 ) + c ( 0 ) 2 = a f β² ( 0 ) = b + 2 c ( 0 ) = b f β² β² ( 0 ) = 2 c \begin{aligned} f(0) = a + b(0) + c(0)^2 = a \\ f'(0) = b + 2c(0) = b \\ f''(0) = 2c \end{aligned} f ( 0 ) = a + b ( 0 ) + c ( 0 ) 2 = a f β² ( 0 ) = b + 2 c ( 0 ) = b f β²β² ( 0 ) = 2 c β Notice that the final approximation, f β² β² ( 0 ) {f''(0)} f β²β² ( 0 ) 2 c . {2c.} 2 c . c {c} c 2. {2.} 2.
f β² β² ( 0 ) = 2 c f β² β² ( 0 ) 2 = c \begin{aligned} f''(0) &= 2c \\[1em] \dfrac{f''(0)}{2} &= c \end{aligned} f β²β² ( 0 ) 2 f β²β² ( 0 ) β β = 2 c = c β We now have our c {c} c c = f β² β² ( 0 ) 2 . {c = \dfrac{f''(0)}{2}.} c = 2 f β²β² ( 0 ) β . 1 / 2 {1/2} 1/2
f ( x ) β f ( x 0 ) + f β² ( x 0 ) β
( x β x 0 ) + f β² β² ( x 0 ) 2 ( x β x 0 ) 2 f(x) \approx f(x_0) + f'(x_0) \cdot (x - x_0) + \dfrac{f''(x_0)}{2}(x - x_0)^2 f ( x ) β f ( x 0 β ) + f β² ( x 0 β ) β
( x β x 0 β ) + 2 f β²β² ( x 0 β ) β ( x β x 0 β ) 2 Notice that this really is just the linear approximation formula, with the
additional term of a second derivative and a factor of 2. {2.} 2. f ( x ) = ln β‘ ( 1 + x ) . {f(x) = \ln(1+x).} f ( x ) = ln ( 1 + x ) . ln β‘ ( 1 + x ) β x . {\ln(1+x) \approx x.} ln ( 1 + x ) β x .
f ( x ) = ln β‘ ( 1 + x ) f β² ( x ) = 1 1 + x f β² β² ( x ) = β 1 ( 1 + x ) 2 \begin{aligned} &f(x) = \ln(1+x) \\[1em] &f'(x) = \dfrac{1}{1+x} \\[1em] &f''(x) = -\dfrac{1}{(1+x)^2} \end{aligned} β f ( x ) = ln ( 1 + x ) f β² ( x ) = 1 + x 1 β f β²β² ( x ) = β ( 1 + x ) 2 1 β β Then, we evaluate the functions at 0. {0.} 0.
f ( 0 ) = 0 f β² ( 0 ) = 1 f β² β² ( 0 ) = β 1 f(0) = 0 \\ f'(0) = 1 \\ f''(0) = -1 f ( 0 ) = 0 f β² ( 0 ) = 1 f β²β² ( 0 ) = β 1 Applying our formula, we get:
ln β‘ ( 1 + x ) β 0 + 1 β
( x β 0 ) + β 1 2 ( x β 0 ) 2 β x β x 2 2 \begin{aligned} \ln(1+x) &\approx 0 + 1 \cdot (x - 0) + \dfrac{-1}{2}(x - 0)^2 \\[1em] &\approx x - \dfrac{x^2}{2} \end{aligned} ln ( 1 + x ) β β 0 + 1 β
( x β 0 ) + 2 β 1 β ( x β 0 ) 2 β x β 2 x 2 β β Thus, we have the quadratic approximation
ln β‘ ( 1 + x ) β x β x 2 2 \ln(1+x) \approx x - \dfrac{x^2}{2} ln ( 1 + x ) β x β 2 x 2 β Applying the formula above to ln β‘ ( 1.1 ) , {\ln(1.1),} ln ( 1.1 ) ,
ln β‘ ( 1.1 ) = ln β‘ ( 1 + 1 10 ) β 1 10 β 1 2 ( 1 10 ) 2 β 0.095 \begin{aligned} \ln(1.1) &= \ln \left(1 + \dfrac{1}{10}\right) \\[1em] &\approx \dfrac{1}{10} - \dfrac{1}{2}\left( \dfrac{1}{10} \right)^2 &\approx 0.095 \end{aligned} ln ( 1.1 ) β = ln ( 1 + 10 1 β ) β 10 1 β β 2 1 β ( 10 1 β ) 2 β β 0.095 β Recall that ln β‘ ( 1.1 ) = 0.095310179804325. {\ln(1.1) = 0.095310179804325.} ln ( 1.1 ) = 0.095310179804325.
Next, let's consider the quadratic approximation for f ( x ) = ( 1 + x ) r . {f(x) = (1+x)^r.} f ( x ) = ( 1 + x ) r .
f ( x ) = ( 1 + x ) r f β² ( x ) = r ( 1 + x ) r β 1 f β² β² ( x ) = r ( r β 1 ) ( x + 1 ) r β 2 \begin{aligned} &f(x) = (1+x)^r \\ &f'(x) = r(1+x)^{r-1} \\ &f''(x) = r(r-1)(x+1)^{r-2} \end{aligned} β f ( x ) = ( 1 + x ) r f β² ( x ) = r ( 1 + x ) r β 1 f β²β² ( x ) = r ( r β 1 ) ( x + 1 ) r β 2 β Evaluating at x = 0 , {x = 0,} x = 0 ,
f ( 0 ) = ( 1 + 0 ) = 1 f β² ( 0 ) = r ( 1 + 0 ) r β 1 = r f β² β² ( 0 ) = r ( r β 1 ) ( 0 + 1 ) r β 2 = r ( r β 1 ) \begin{aligned} &f(0) = (1+0) = 1 \\ &f'(0) = r(1+0)^{r-1} = r \\ &f''(0) = r(r-1)(0+1)^{r-2} = r(r-1) \end{aligned} β f ( 0 ) = ( 1 + 0 ) = 1 f β² ( 0 ) = r ( 1 + 0 ) r β 1 = r f β²β² ( 0 ) = r ( r β 1 ) ( 0 + 1 ) r β 2 = r ( r β 1 ) β Plugging in these findings to our quadratic approximation formula, we have:
( 1 + x ) r β 1 + r ( x β 0 ) + r ( r β 1 ) 2 ( x β 0 ) 2 ( 1 + x ) r β 1 + r x + r ( r β 1 ) 2 x 2 \begin{aligned} (1 + x)^r &\approx 1 + r(x-0) + \dfrac{r(r-1)}{2}(x-0)^2 \\[1em] (1 + x)^r &\approx 1 + rx + \dfrac{r(r-1)}{2}x^2 \\[1em] \end{aligned} ( 1 + x ) r ( 1 + x ) r β β 1 + r ( x β 0 ) + 2 r ( r β 1 ) β ( x β 0 ) 2 β 1 + r x + 2 r ( r β 1 ) β x 2 β Hence, we have the quadratic approximation:
( 1 + x ) r β 1 + r x + r ( r β 1 ) 2 x 2 (1 + x)^r \approx 1 + rx + \dfrac{r(r-1)}{2}x^2 ( 1 + x ) r β 1 + r x + 2 r ( r β 1 ) β x 2 Let's examine the other functions we saw for linear approximation. To do
so, we must gather all of the terms we need:
f {f} f f β² {f'} f β² f β² β² {f''} f β²β² f ( 0 ) {f(0)} f ( 0 ) f β² ( 0 ) {f'(0)} f β² ( 0 ) f β² β² ( 0 ) {f''(0)} f β²β² ( 0 ) sin β‘ x {\sin x} sin x cos β‘ x {\cos x} cos x β sin β‘ x {- \sin x} β sin x sin β‘ 0 = 0 {\sin 0 = 0} sin 0 = 0 cos β‘ 0 = 1 {\cos 0 = 1} cos 0 = 1 β sin β‘ 0 = 0 {- \sin 0 = 0} β sin 0 = 0 cos β‘ x {\cos x} cos x β sin β‘ x {- \sin x} β sin x β cos β‘ x {- \cos x} β cos x cos β‘ 0 = 1 {\cos 0 = 1} cos 0 = 1 β sin β‘ 0 = 0 {- \sin 0 = 0} β sin 0 = 0 β cos β‘ 0 = β 1 {- \cos 0 = -1} β cos 0 = β 1 e x {e^x} e x e x {e^x} e x e x {e^x} e x e 0 = 1 {e^0 = 1} e 0 = 1 e 0 = 1 {e^0 = 1} e 0 = 1 e 0 = 1 {e^0 = 1} e 0 = 1
Applying the information above, we have the following quadratic
approximations:
sin β‘ x β x cos β‘ x β 1 β 1 2 x 2 e x β 1 + x + 1 2 x 2 Β β£ x β 0 \large \left. \begin{aligned} &\sin x \approx x \\[1em] &\cos x \approx 1 - \dfrac{1}{2}x^2 \\[1em] &e^x \approx 1 + x + \dfrac{1}{2}x^2 \end{aligned} ~\right\vert_{x \approx 0} β sin x β x cos x β 1 β 2 1 β x 2 e x β 1 + x + 2 1 β x 2 β Β β£ β£ β x β 0 β Recall that, where a k = ( 1 + 1 k ) 2 , {a_k = \left( 1 + \dfrac{1}{k} \right)^2,} a k β = ( 1 + k 1 β ) 2 ,
lim β‘ k β β a k = e \lim\limits_{k \to \infty}a_k = e k β β lim β a k β = e We determined by first taking the logarithm of both sides:
ln β‘ ( a k ) = k β
ln β‘ ( 1 + 1 k ) 2 \begin{aligned} \ln(a_k) &= k \cdot \ln \left( 1 + \dfrac{1}{k} \right)^2 \\[1em] \end{aligned} ln ( a k β ) β = k β
ln ( 1 + k 1 β ) 2 β Then, applying the limit:
ln β‘ ( a k ) = lim β‘ k β β k β
ln β‘ ( 1 + 1 k ) 2 = e \begin{aligned} \ln(a_k) &= \lim\limits_{k \to \infty} k \cdot \ln \left( 1 + \dfrac{1}{k} \right)^2 \\[1em] &= e \end{aligned} ln ( a k β ) β = k β β lim β k β
ln ( 1 + k 1 β ) 2 = e β We can achieve the same result with quadratic approximation by first
applying linear approximation. First, we know that:
ln β‘ ( 1 + x ) β x Β Β whereΒ Β x = 1 k β 0 \ln(1 + x) \approx x ~\text{ where }~ x = \dfrac{1}{k} \approx 0 ln ( 1 + x ) β x Β Β whereΒ Β x = k 1 β β 0 It follows then that:
ln β‘ a k = k ( 1 + 1 k ) β k β
1 k \begin{aligned} \ln a_k &= k \left( 1 + \dfrac{1}{k} \right) \\[1em] &\approx k \cdot \dfrac{1}{k} \end{aligned} ln a k β β = k ( 1 + k 1 β ) β k β
k 1 β β Thus we have:
ln β‘ a k = k β
ln β‘ ( 1 + 1 k ) β k β
1 k \begin{aligned} \ln a_k &= k \cdot \ln \left( 1 + \dfrac{1}{k} \right) \\[1em] & \approx k \cdot \dfrac{1}{k} \end{aligned} ln a k β β = k β
ln ( 1 + k 1 β ) β k β
k 1 β β The value k ( 1 / k ) {k(1/k)} k ( 1/ k ) 1. {1.} 1.
ln β‘ ( ( 1 + 1 k ) 2 ) β 1 \ln\left(\left(1 + \dfrac{1}{k}\right)^2\right) \approx 1 ln ( ( 1 + k 1 β ) 2 ) β 1 Question: What is the rate of convergence for ln β‘ a k ? {\ln a_k?} ln a k β ?
ln β‘ a k = k β
ln β‘ ( 1 + 1 k ) \ln a_k = k \cdot \ln(1 + \dfrac{1}{k}) ln a k β = k β
ln ( 1 + k 1 β ) approach 1 ? {1?} 1 ?
ln β‘ a k β 1 ln β‘ a k β 1 β 1 β 1 ln β‘ a k β 1 β HowΒ bigΒ isΒ this? β 0 \ln a_k \to 1 \\[1em] \ln a_k - 1 \to 1 - 1 \\[1em] \underbrace{\ln a_k - 1}_{\text{How big is this?}} \to 0 ln a k β β 1 ln a k β β 1 β 1 β 1 HowΒ bigΒ isΒ this? ln a k β β 1 β β β 0 The answer lies in quadratic approximation.
Again, all of the approximations above depend on the assumption that x {x} x 0. {0.} 0.
Function Approximation Formula sin β‘ x {\sin x} sin x x {x} x cos β‘ x {\cos x} cos x 1 β 1 2 x 2 {1 - \dfrac{1}{2}x^2} 1 β 2 1 β x 2 e x {e^x} e x 1 + x + 1 2 x 2 {1 + x + \dfrac{1}{2}x^2} 1 + x + 2 1 β x 2 ln β‘ ( 1 + x ) {\ln(1 + x)} ln ( 1 + x ) x β 1 2 x 2 {x - \dfrac{1}{2}x^2} x β 2 1 β x 2 ( 1 + x ) 2 {(1 + x)^2} ( 1 + x ) 2 1 + r x + r ( r β 1 ) 2 x 2 {1 + rx + \dfrac{r(r-1)}{2}x^2} 1 + r x + 2 r ( r β 1 ) β x 2
Quadratic approximations can get very nasty. Accordingly, we should only
resort to them if we truly aren't satisfied with a linear approximation.
Consider the following problem:
Problem. Find the the quadratic approximation for the function
f ( x ) = e β 3 x ( 1 + x ) β 1 / 2 {f(x) = e^{-3x}(1+x)^{-1/2}} f ( x ) = e β 3 x ( 1 + x ) β 1/2 x {x} x 0. {0.} 0.
Carrying out the approximation with the approximation formulas above, we
have:
e β 3 x ( 1 + x ) β 1 / 2 β ( 1 + ( β 3 x ) + ( β 3 x ) 2 2 ) ( 1 β 1 2 x + 1 2 ( β 1 2 ) ( β 3 2 ) x 2 ) β 1 β 3 x β 1 2 x + 3 2 x 2 + 9 2 x 2 + 3 8 x 2 \begin{aligned} e^{-3x}(1+x)^{-1/2} &\approx \left(1 + (-3x) + \dfrac{(-3x)^2}{2}\right) \left(1 - \dfrac{1}{2}x + \dfrac{1}{2}\left(- \dfrac{1}{2}\right)\left(- \dfrac{3}{2}\right)x^2\right)\\[1em] & \approx 1 - 3x - \dfrac{1}{2}x + \dfrac{3}{2}x^2 + \dfrac{9}{2}x^2 + \dfrac{3}{8}x^2 \end{aligned} e β 3 x ( 1 + x ) β 1/2 β β ( 1 + ( β 3 x ) + 2 ( β 3 x ) 2 β ) ( 1 β 2 1 β x + 2 1 β ( β 2 1 β ) ( β 2 3 β ) x 2 ) β 1 β 3 x β 2 1 β x + 2 3 β x 2 + 2 9 β x 2 + 8 3 β x 2 β If we applied the approximation formulas in their entirety, we'd notice
that there are some missing terms. These terms were dropped. Why? Because
these higher order terms get so small that they're negligible. Simplifying
the aproximation above, we have:
e β 3 x ( 1 + x ) β 1 / 2 β 1 β 7 2 x + 51 8 x 2 e^{-3x}(1+x)^{-1/2} \approx 1 - \dfrac{7}{2}x + \dfrac{51}{8}x^2 e β 3 x ( 1 + x ) β 1/2 β 1 β 2 7 β x + 8 51 β x 2 Notice that the term 1 β 7 2 x {1 - \dfrac{7}{2}x} 1 β 2 7 β x 51 8 x 2 {\dfrac{51}{8}x^2} 8 51 β x 2