Lecture # 10 - Derivatives of Functions of One Variable (cont.)

Concave and Convex Functions

We saw before that •

f 0 (x) 0 f (x) is a decreasing function ≤ ⇐⇒

Wecanusethesamedefintion for the second derivative: •

f 00 (x) 0 f 0 (x) is a decreasing function ≤ ⇐⇒

Definition: a function f ( ) whose first derivative is decreasing(and f (x) 0)iscalleda • · 00 ≤ concave function

In turn, •

f 00 (x) 0 f 0 (x) is an increasing function ≥ ⇐⇒

Definition: a function f ( ) whose first derivative is increasing (and f (x) 0)iscalleda • · 00 ≥ convex function

Notes: •

— If f 00 (x) < 0, then it is a strictly concave function

— If f 00 (x) > 0, then it is a strictly convex function

Graphs • Importance: • — Concave functions have a maximum

— Convex functions have a minimum.

1 Economic Application: Elasticities

Economists are often interested in analyzing how demand reacts when price changes • However, looking at ∆Qd as ∆P =1may not be good enough: • — Change in quantity demanded for coffee when price changes by 1 euro may be huge — Change in quantity demanded for cars coffee when price changes by 1 euro may be insignificant

Economist look at relative changes, i.e., at the percentage change: what is ∆% in Qd as • ∆% in P

We can write as the following ratio • ∆%Qd ∆%P

Now the rate of change can be written as ∆%x = ∆x . Then: • x d d ∆Q d ∆%Q d ∆Q P = Q = ∆%P ∆P ∆P Qd P ·

Further, we are interested in the changes in demand when ∆P is very small = Take the • ⇒ limit ∆Qd P ∆Qd P dQd P lim = lim = ∆P 0 ∆P · Qd ∆P 0 ∆P · Qd dP · Qd µ → ¶ µ → ¶ So we have finally arrived to the definition of price elasticity of demand • dQd P = dP · Qd

d dQd P P Example 1 Consider Q = 100 2P. Then = dP Qd = 2 100 2P . So the elasticity − · − · − varies with price

d 2 dQd P 3 P Example 2 Consider Q = 500P − . Then = dP d = (500) ( 2) P − 2 = 2. · Q − · 500P − − In general, if Q = AP b, then = b

2 We can also have income elasticity:of demand, or price elasticity of supply. In general, given • a function y = f (x) , the elasticity is: dy x Elxf (x)= dx · y

3 Linear Approximation

Sometimes, when we have a complicated function, we can use a simpler function (as an • approximation) to work with it.

Consider a function f (x) that is differentiable at point x = a • — By defintion of tangent: y f (a) f 0 (a)= − x a − — Solving for y :

y = f (a)+f 0 (a)(x a) − Then the linear approximation to f (x) around x = a is: •

f (x) f (a)+f 0 (a)(x a) ≈ −

3 1 2 3 2 Example 3 Suppose f (x)=x (so f 0 (x)= 2 x )

— The linear approximation to f (x) around x =1is:

f (x) f (1) + f 0 (1) (x 1) ≈ − 3 (1) + (x 1) ≈ 2 − µ ¶ 3 1 x ≈ 2 − 2 Then, the approximate value for f (x) for x =1.05 is 1.075 (exact value is 1.07592983) Notice that the derivative of 3 x 1 is 3 ,sameasf (x) at x =1 ∗ 2 − 2 2 0 — The linear approximation to f (x) around x =4is:

f (x) f (4) + f 0 (4) (x 4) ≈ − (8) + (3) (x 4) ≈ − 3x 4 ≈ − Then, the approximate value for f (x) for x =3.95 is 7.85 (exact value is 7.85047) Notice that the derivative of 3x 4 is 3,sameasf (x) at x =4 ∗ − 0 4 Polynomial Aproximations

Linear approximation is sometimes NOT ’close enough’ = Need a better one: • ⇒ Why not a ”quadratic” aproximation as follows: • 1 2 f (x) f (a)+f 0 (a)(x a)+ f 00 (a)(x a) ≈ − 2 −

3 1 1 2 3 2 3 2 Example 4 Suppose again x (so f 0 (x)= 2 x AND f 00 (x)= 4 x− )

— The linear approximation to f (x) around x =1is:

1 2 f (x) f (1) + f 0 (1) (x 1) + f 00 (1) (x 1) ≈ − 2 − 3 1 3 (1) + (x 1) + x2 2x +1 ≈ 2 − 2 4 − µ ¶ µ ¶ 3 1 3 3 3 ¡ ¢ x + x2 x + ≈ 2 − 2 8 − 4 8 3 3 1 x2 + x ≈ 8 4 − 8 Then, the approximate value for f (x) for x =1.05 is 1.0759375 (exact value is 1.07592983) Notice that the first derivative of 3 x2 + 3 x 1 evaluated at x =1is 3 ,sameas ∗ 8 4 − 8 2 f 0 (x) at x =1 Notice that the SECOND derivative of 3 x2 + 3 x 1 is 3 ,sameasf (x) at x =1 ∗ 8 4 − 8 4 00 — The linear approximation to f (x) around x =4is:

1 2 f (x) f (4) + f 0 (4) (x 4) + f 00 (4) (x 4) ≈ − 2 − 1 3 (8) + (3) (x 4) + x2 8x +16 ≈ − 2 8 − µ ¶ 3 3 ¡ ¢ 3x 4+ x2 x +3 ≈ − 16 − 2 3 3 x2 + x 1 ≈ 16 2 − Then, the approximate value for f (x) for x =3.95 is 7.85046 (exact value is 7.85047) Notice that the first derivative of 3 x2 + 3 x 1 evaluated at x =4is 3,sameas ∗ 16 2 − f 0 (x) at x =4 Notice that the SECOND derivative of 3 x2 + 3 x 1 is 3 ,sameasf (x) at x =1 ∗ 16 2 − 8 00 5 Inthesameway,wecanfind more accurate aooriximation by using higher order derivatives: • General formula: nth-order Taylor polynomial about x = a: • f (a) f (a) f 0 (a) f n (a) f (x) f (a)+ 0 (x a)+ 00 (x a)2 + 00 (x a)3 + ... + (x a)n ≈ 1! − 2! − 3! − n! −

where n is a positive integer

Notice that a Taylor polinomial is an approximation. The difference between the real • function and the Taylor polynomial is called the remainder after n terms

Lagrange form of the remainder: (which will be the n +1term) • 1 (n+1) n+1 Rn+1 (x)= f (c)(x a) (n +1)! − where c is a value between a and x.

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