from math 2220 class 13pi.math.cornell.edu/~back/m222_f14/slides/sep26_v2c.pdfsep. 26, 2014 from...

From Math2220 Class 13

Prelim andHW

AbsoluteExtrema

1 VariableTaylor Series

MultivariableTaylorPolynomials

MultivariableTaylorPictures

Higher PartialDerivatives

A Counterex-ample

Higher Partialsin PolarCordinates

GradientEstimate fromGraph

Method ofCharacteristics

From Math 2220 Class 13

Dr. Allen Back

Sep. 26, 2014

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Prelim and HW

Your next homework on 3.2 and 3.3 will not be due soonerthan Mon 10 days from now.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Prelim and HW

No homework quiz today.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Prelim and HW

Prelim 1 next Tues at 7:30 pm in Malott 251 will include 3.1and the local part of 3.3 but not 3.2. (Chapter 2 as well.)We can spend as much of Monday on your review questions asyou like.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Prelim and HW

The exam has 6 questions and is not very intricatecomputationally. Question 6 is a True/False question (not toohard) with the following directions:

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Prelim and HW

Problem 6) (15 Points) For the questions below, please justwrite the complete word TRUE or FALSE as your answer. Forthis question ONLY, no explanations are expected.(If a statement is sometimes true but not always, your answerwould be FALSE.)

You will receive 3 points for a correct answer, 1 point for ablank and −1 point for a wrong answer.

(Thus all 5 parts blank would give you 5 out of 15 while all fivewrong will lower your total on the rest of the exam by 5.)

(The exam is 11 pages with space below the questions to writeyour answers.)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Prelim and HW

In addition to my usual Monday office hours (Evan has sometoo), I will have office hours

Fri 9/26 2:15-3:15

Tues 9/30 1-2:30

Also I’m willing to talk with anyone who likes rightafter class today though I do have an appt withEvan for 2-2:15.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Closed Set contains all its boundary points.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Bounded Set lies inside some single ball of some radius.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Compact Set both closed and bounded.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Theorem A continuous functions whose domain is a compactset always attains (i.e. “has”) an absolute maximum and anabsolute minimum.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Counterexamples when not compact:

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Procedure for finding an absolute extremum for a cont. fcn. ona compact domain:

1 Above theorem guarantees existence since the domain iscompact.

2 Look for critical points (in the interior of the domain.)

3 Investigate the boundary either using lower dimensionalcalculus or Lagrange multipliers.

4 Compare values at all candidates to find the absolute maxand min.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Find the absolute max and absolute min of

f (x , y) = xy(2− x − y)

on the square |x | ≤ 1, |y | ≤ 1.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Absolute Extrema

Find the absolute max and absolute min of

f (x , y) = xy

on (and inside) the triangle with vertices (2, 0), (10, 0), and(4, 1).

Prelim andHW

AbsoluteExtrema

A Counterex-ample

1 Variable Taylor Series

Multivariable Taylor approximation follows (using the chainrule) easily from the 1-variable case, so we’ll start by reviewinghow the results there work out.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Taylor Series of f (x) about x = a:

Σ∞n=0

f n(a)

n!(x − a)n.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Partial sums (the k + 1’st) give the k ’th Taylor Polynomial.

Pk(x) = Σkn=0

f n(a)

n!(x − a)n.

(Pk is of degree k but has k + 1 terms because we start with0.)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

So P1(x) is the linear approximation to f (x) at x = a.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Examples:

(a)ex about x = 0 (b)ex about x = 2

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Where does the formula for Pk come from heuristically?

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Where does the formula for Pk come from heuristically?The k’th Taylor polynomial Pk(x) is the unique polynomial ofdegree k whose value, derivative, 2nd derivative, . . . k’thderivative all agree with those of f (x) at x = a.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Where does the formula for Pk come from heuristically?Or integration by parts: u = f ′(x), v ′ = 1, v = (x − b)

f (b) = f (a) +

af ′(x) dx

f (b) = f (a) +(

(x − b)f ′(x)∣∣ba−

a(x − b)f ′′(x) dx

)f (b) = f (a) + (b − a)f ′(a)+

a(b − x)f ′′(x) dx

The last term is one form of the “remainder” in approximatingf (b) by the first Taylor polynomial (aka linear approximant)P1(b).

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Where does the formula for Pk come from heuristically?

Or integration by parts: u = f ′′(x), v ′ = (b − x), v = −(b−x)2

f (b) = f (a) + (b − a)f ′(a)+

a(b − x)f ′′(x) dx

)f (b) = f (a) + (b − a)f ′(a)+

(−(b − x)2

2f ′′(x)

∣∣∣∣ba

−∫ b

−(b − x)2

2f ′′′(x) dx

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Where does the formula for Pk come from heuristically?Or integration by parts:

f (b) = f (a) + (b − a)f ′(a)+

(−(b − x)2

2f ′′(x)

∣∣∣∣ba

−∫ b

−(b − x)2

2f ′′′(x) dx

)f (b) = f (a) + (b − a)f ′(a)+

(b − a)2

2f ′′(a)

(b − x)2

2f ′′′(x) dx

Prelim andHW

AbsoluteExtrema

A Counterex-ample

The above “integral formula” for the remainder can be turnedinto something simpler to remember but sufficient for mostapplications:The next Taylor term except with the higher derivativeevaluated at some unknown point c between a and b.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Taylor’s Theorem with RemainderIf f (x), f ”(x), . . . f (k)(x) all exist and are continuous on [a, b]and f (k+1)(x) exists on (a, b), then there is a c ∈ (a, b) so that

f (b) = Pk(b) + Rk(b)

where Pk is the k’th Taylor polynomial of f about x = a and

Rk(b) =f k+1(c)

(k + 1)!(b − a)k+1.

Please remember this!

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Example: Accuracy of sin x byP1(x) (or P3(x)) (about x = 0)for |x | < .1? Or x < .01?

Prelim andHW

AbsoluteExtrema

A Counterex-ample

f (x) = 11+x+x2 about x = 1.

Approximation properties of different Taylor polynomials nearx = 1.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Taylor polynomials of f (x) = 11+x+x2 about x = 1.

Typically higher order approximate well for a greater distance.

(Bad behavior beyond the radius of convergence.)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Show that the Maclaurin series of sin x converges to sin x forall x .

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Show that the Maclaurin series of sin x converges to sin x forall x .This is different than saying the radius of convergence is ∞.Some functions, e.g.

f (x) =

{0 if x ≤ 0

e−1x if x > 0

have a Taylor series which always converges, but not to f (x)even near 0. In this case, every derivative is 0 at x = 0 and theMaclaurin series is identically 0!

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Show that the Maclaurin series of sin x converges to sin x forall x .The point here is that all derivatives are bounded functionstaking values between −1 and +1. So you can show theremainder Rn(x) goes to 0 as n→∞.Taylor polynomials of f (x) = 1

1+x+x2 about x = 1.Typically higher order approximate well for a greater distance.

(Bad behavior beyond the radius of convergence.)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

It is more advanced than our course, but there is an amazingconnection between when Taylor series converge to a functionand differentiability of f(x) for x viewed as a complex variable.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

It is more advanced than our course, but there is an amazingconnection between when Taylor series converge to a functionand differentiability of f(x) for x viewed as a complex variable.For example, consider the MacLaurin series of f (x) = 1

1+x2 .

This fcn. is complex differentiable when x 6= ±i = ±√−1. And

if one identifies the complex number a + bi with the point(a, b) ∈ R2, the nearest bad point of ±i ∼ (0,±1) is distance 1from 0 = 0 + 0i ∼ (0, 0). So from this advanced point of view,the radius of convergence is 1!

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Similarly, consider the above picture for f (x) = 11+x+x2 about

x = 1. This fcn. is bad when 1 + x + x2 = 0, or x = −1±i√

The distance from 1 + 0i ∼ (1, 0) to −(12 ,√

32 ) is

√3, so from

this advanced point of view, that is the radius of convergencefor this Taylor series.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Multivariable Taylor Polynomials

Given f : U ⊂ Rn → R, to find the formula for the k’th Taylorpolynomial Pk of a C k function about the pointp0 = (a1, a2, . . . , an) ∈ U, we consider the 1-variable functiong(t) defined by

g(t) = f (p0 + th)

where h = (h1, h2, . . . , hn) ∈ Rn.If we are interested in relating the Taylor polynomial Pk to thebehavior of f at the point p = (x1, x2, . . . , xn) ∈ U we mightchoose

h = (x1 − a1, x2 − a2, . . . , xn − an)

as long as the entire line segment between p0 and p lies in U.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

g(t) = f (p0 + th)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Problem: Find the first and second Taylor polynomials of

f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Prelim andHW

AbsoluteExtrema

A Counterex-ample

f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Estimate the error in approximating f by P1 for |x − π

2| < .1

and |y − π| < .1.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Estimate the error in approximating f by P1 for |x − π

2| < .01

and |y − π| < .01.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

f (x , y) = cos (x + 3y) + sin x

about (x , y) = (π

2, π).

Find an ε so that the error in approximating f by P1 for

|x − π

2| < ε and |y − π| < ε is at most .01.

Or at most 10−6.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Multivariable Taylor Pictures

Graph of f (x , y) = 1x+y2 for .5 ≤ x , y ≤ 1.5

Prelim andHW

AbsoluteExtrema

A Counterex-ample

P2 about (1, 1) for f (x , y) = 1x+y2 for .5 ≤ x , y ≤ 1.5

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Both superimposed for .5 ≤ x , y ≤ 1.5

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Error in using P2 for f for .5 ≤ x , y ≤ 1.5

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Error in using P2 for f for .75 ≤ x , y ≤ 1.25

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Error in using P2 for f for 0 ≤ x , y ≤ 3

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Higher Partial Derivatives

A typical application of mixed partials arises out of yourhomework problem 3.1 #22 where you show that the waveequation

∂x2=∂2w

becomes after a change of coordinates

x = u + v , y + u − v

∂u∂v= 0

Prelim andHW

AbsoluteExtrema

A Counterex-ample

∂x2=∂2w

becomes after a change of coordinates

x = u + v , y + u − v

∂u∂v= 0

Prelim andHW

AbsoluteExtrema

A Counterex-ample

∂x2=∂2w

x = u + v , y + u − v∂2w

∂u∂v= 0

implying the general solution

w = f (u) + g(v) = f (x − y

2) + g(

for any C 2 functions f and g .

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

When f is not C 2, fxy need not equal fyx even when both exist.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

An example illustrating this is

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This function

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This functionis clearly C 2 for (x , y) 6= (0, 0) since it is a rational function(quotient of polynomials) with nonzero denominator.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This functionsatisfies f (x , y) = −f (y , x) implying

fxy = fyx

whenever x = y . (A good chain rule exercise . . . )

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This functionsatisfies fx(0, 0) = 0 and fy (0, 0) = 0 since f vanishes along theaxes.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This function

satisfies fx =y(x2 − y2)

x2 + y2+

x2 + y2− 2x2y(x2 − y2)

(x2 + y2)2.

implying

fx(0, y) = −y (the last two terms vanish)

and∂2f

∂y∂x(0, 0) = −1.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A Counterexample

f (x , y) =

{xy(x2−y2)

x2+y2 if (x , y) 6= (0, 0)

0 if (x , y) = (0, 0)

This function

fx(0, y) = −y (the last two terms vanish)

and∂2f

∂y∂x(0, 0) = −1.

By fxy (x , x) = −fyx(x , x) (mentioned above) or directly we see

∂x∂y(0, 0) = 1

and the mixed partials really differ!

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Higher Partials in Polar Cordinates

Let x = r cos θ, y = r sin θ. Given f (x , y), express

in terms of partials of f with respect to x and y .

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Really there are two functions here; namely f (x , y) and

g(r , θ) = f (r cos θ, r sin θ).

But many applied fields routinely use the same letter f for bothfunctions.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

The key step is realizing that the task (via the chain rule) of

relating∂f

∂rto

∂xand

∂yis just like the task of doing the

analagous thing with expressions like

(The right hand side “operator notation” means partially

differentiate the function∂f

∂xof (x , y) with respect to r .)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

x = r cos θ y = r sin θ

∂r+∂f

∂xcos θ +

∂ysin θ

))∂x

))∂y

(∂2f

)cos θ +

(∂2f

∂y∂x

)sin θ

Prelim andHW

AbsoluteExtrema

A Counterex-ample

A similar computation with∂2f

∂θ2would show that the laplacian

∂x2+∂2f

is given in polar coordinates by

∂r2+

(∂2f

∂θ2

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Can you estimate the value of the gradient?

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Method of Characteristics

Problem 33 on page 146 from your homework asks you to show

using the chain rule that if z = f (y

x) for a 1-variable

differentiable function f , then

∂x+ y

∂y= 0.

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Problem 33 on page 146 from your homework asks you to show

using the chain rule that if z = f (y

x) for a 1-variable

differentiable function f , then

∂x+ y

∂y= 0.

You don’t need it to do this homework problem, but a naturalquestion is

Does x∂z

∂x+ y

∂y= 0 imply z = f (

for some 1-variable differentiable function f ?

Prelim andHW

AbsoluteExtrema

A Counterex-ample

The answer is yes, and we can easily show it using what welearned in section 2.6!(i.e. essentially chain rule ideas applied to curves.)

Prelim andHW

AbsoluteExtrema

A Counterex-ample

The beautiful idea for the equation

a(x , y)∂z

∂x+ b(x , y)

∂y= c(x , y , z)

is to consider curves (x(t), y(t) called characteristics) satisfying

dt= a(x(t), y(t))

dt= b(x(t), y(t)).

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Here for

∂x+ y

∂y= 0

we havedx

with solutions x(t) = x0et and y(t) = y0et .

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Here for

∂x+ y

∂y= 0

we havedx

with solutions x(t) = x0et and y(t) = y0et .The key observation is that for a solution z(x , y)

dt[z(x(t), y(t))] =

∂xx+

Prelim andHW

AbsoluteExtrema

A Counterex-ample

Here for

∂x+ y

∂y= 0

with solutions x(t) = x0et and y(t) = y0et .The key observation is that for a solution z(x , y)

dt[z(x(t), y(t))] =

∂xx+

So z(x , y) is constant along linesy

x= k for any k and thus

z(x , y) = z(1,y

x) = f (

as we wanted to show.

from math 2220 class 13pi.math.cornell.edu/~back/m222_f14/slides/sep26_v2c.pdfsep. 26, 2014 from...

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