Post on 21-Dec-2015
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 →
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 →• 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
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’…?
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