Three easy pieces(etudes for calculus and complex

variables)

Dr. Rick KreminskiColorado State University - Pueblo

• Visualizing the chain rule for functions from → and →

• Accelerating the convergence of Vieta’s infinite product formula for π

• Taylor’s theorem with remainder

Where is ‘c’…?

erroraxn

afax

afax

afaxafafxf n

n

)(!

)(...)(

!3

)()(

!2

)())(()()(

)(32

1)1(

)()!1(

)()(

nn

axn

cferrorLagrange

Visualizing the chain rule for functions from from → and →

Usual, inputs/outputs, way that functions are introduced (e.g. in beginning algebra, and possibly discrete math)…

…but not used again in calculus

Chain rule for functions from →

Inputs/outputs for a less trivial function: f(x)=2x+3

Fixed point clearly visible

x=.25“Expansion factor” is exactly 2

-4

-2

0

2

4

6

-4

-2

0

2

4

6fx3 2 x

Chain rule for functions from → f(x)=x3

--Observe expansion and contraction, of varying amounts x=.1--Estimate expansionfactor at x=2? i.e. --Expansion factor at 2 for f related to Contraction factor at 8 for f-1 ; i.e.

This ‘explains’ why

-7.5

-5

-2.5

0

2.5

5

7.5

-7.5

-5.

-2.5

0

2.5

5.

7.5

fx x3

?12)2( aboutf

)2(/1)8()( 1 ff

)(/1))(()( 1 afaff

Chain rule for functions from →

Depicting a composition:

-7.5

-5

-2.5

0

2.5

5

7.5

fx x3 gx x13

Chain rule for functions from →

f(x)=x2 for three domainsf isn’t 1-1f inverts order (f is not strictly increasing)(f’ < 0)

0

1

2

3

4

5

0

1

2

3

4

5

fx x2

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-0.2

0

0.2

0.4

0.6

0.8

1

fx x2

0

10

20

30

40

50

60

-10

0

10

20

30

40

50

60

fx x2

Chain rule for functions from →

x=.25

0

10

20

30

40

50

60

-10

0

10

20

30

40

50

60

fx x2

?16)8( f

Chain rule for functions from →• f(2) is 8• Expansion factor due to f(x)=x3 for x=2

was estimated to be 12• Expansion factor due to g(x)=x2 for

x=8 was estimated to be 16• Thus, overall expansion factor due to

g ○ f must be about 1216, i.e. about 192.

• Thus, (g ○ f) (a) = f (a) g ∙ (f(a))• And for three functions composed, (h ○g ○f) (a) = f (a) g ∙ (f(a)) h ∙ (g(f(a))) 0

10

20

30

40

50

60

gx x2

fx x3

0

10

20

30

40

50

60

Chain rule for functions from →Periodic function Generic cubic Usual image

-10

-5

0

5

10

-10

-5

0

5

10

fx Sinx

-15

-10

-5

0

5

10

15

20

-15

-10

-5

0

5

10

15

20

25

fx10 5 x x3

-4 -2 2 4x

-5

5

10

15

20

25

y10 5 x x3

On which intervals is f increasing and decreasing?

Chain rule for functions from →• Domain D: 13 points centered at a=1+i; radius .01. • Image f(D) under f(z) = z3 + 4z2 +3 – 10i• image g(f(D)) where g(z) = -iz4 + 2 - 3i. Note: f(a)=1 and

g(1)=2-4i. Estimate f (a) ?

0.99 0.995 1.005 1.01

0.99

0.995

1.005

1.01

the domain D

1

2

3

45

6

7

8

9

1011

12

13

0.8 0.85 0.9 0.95 1.05 1.1 1.15

-0.2

-0.1

0.1

0.2the image fD

12

3

4

5

6

7

89

10

11

12

13

1.4 1.6 1.8 2.2 2.4 2.6 2.8

-4.75

-4.5

-4.25

-3.75

-3.5

the image gfD1

2

3

4

5

6

7

8

9

101112

13

Chain rule for functions from →

• f (a) = 3a2+8a=8+14i; f (a) has modulus 260 (about 16.12) and argument about 60.25.

• g (f(a)) = -4 i (f(a))3 = -4i , i.e. modulus 4 and argument 90. Oval f(D) appears to have been rotated by -90 and expanded by factor of 4.

• Combining, overall, net result is rotation by about -30 and expansion by about 64. Thus, again quite plausible that

(g ○f) (a) = f (a) g ∙ (f(a))• Considering the middle graph as domain and leftmost

graph as codomain, again appears that (at least locally) (f -1 ) (f(a)) = 1/f (a)

Chain rule for functions from →Image of D under nonconformal map h: R2 R2 h(x,y) = (x3 + 4y2 - 3 x y2 + 4x2 + 3, 8 x y + 3 x2 y – y3 –

10). [f(x,y) = (x3 - 4y2 - 3 x y2 + 4x2 + 3, 8 x y + 3 x2 y – y3 –

10)]

8.925 8.95 8.975 9.025 9.05 9.075

-0.15

-0.1

-0.05

0.05

0.1

0.15

1

2

3

4

5

6

7

89

10

11

12

13

Accelerating the convergence of Vieta’s infinite product formula for π• Can be quickly “derived” in precalculus… double and

half-angle formulas and sin u/u1 as u 0

Vieta’s infinite product formula for π

Let pn be product of first n terms. Viete computed p16 . pn is So

So error goes like c/4n. So error goes like c/10.6 n So pn is accurate to about .6 n digits. So Viete had about 10 digits accuracy for π

Vieta’s infinite product formula for πSo we know the precise form of the error; so we can cancel out terms… perfect Richardson extrapolation…

Vieta’s infinite product formula for π

Let

But then

Vieta’s infinite product formula for π

Example:Vieta computed p10, p11, and p12 (since he computed p16). Had he accelerated convergence, he could have stopped with p12 and obtained the following:

Taylor’s theorem with remainder Where is ‘c’…?

• If n=0, Mean Value Theorem• If n=0, do a few examples, give problems

asking students to find c (or cs)• For larger n, what do you say?

erroraxn

afax

afax

afaxafafxf n

n

)(!

)(...)(

!3

)()(

!2

)())(()()(

)(32

1)1(

)()!1(

)()(

nn

axn

cferrorLagrange

Taylor’s theorem with remainder Where is ‘c’…?

cn for f(x)=ex on [1,2]

Taylor’s theorem with remainder Where is ‘c’…?

cn for f(x)=ex on [1,2] 1/(cn -1)

Taylor’s theorem with remainder Where is ‘c’…?

cn for f(x)=sin(x) on [1,1.5] 1/(cn -1)

Taylor’s theorem with remainder Where is ‘c’…?

cn for f(x)=sin(x) on [0,.3] 1/cn

Taylor’s theorem with remainder Where is ‘c’…?

1/cn for f(x)=sin(x) on [0,2]

Taylor’s theorem with remainder Where is ‘c’…?

1/cn for f(x)=exp(x3) on [0,.125]

Taylor’s theorem with remainder Where is ‘c’…?

For f(x)=1/(1-x) on [0,b], can compute cn exactly. Here, cn -ln(1-b)/n.

For

cn 1/cn

Thank you for your attention!

Questions?• Visualizing the chain rule (for functions over

and ) and more, International Journal of Mathematical Education in Science and Technology, 2009

• Pi to thousands of digits from Vieta’s formula, Mathematics Magazine, 2008

• Taylor’s theorem: the elusive ‘c’ is not so elusive – College Mathematics Journal, 2010