all the world’s a polynomial chris harrow [email protected] [email protected]
TRANSCRIPT
All the World’s a Polynomial
Historically, students struggle to understand the utility and origins of Taylor Series. This session makes use of local linearity and statistical regressions to explain tangent lines in a way that is useful to all AP Calculus students before extending the approach to create Taylor Series for AP Calculus BC. This introduction is understandable by both pre-calculus and calculus students. The session will conclude with a student project around a famous Euler problem and techniques for using series to connect circular and hyperbolic trigonometry.
How can you compute ? e3
Perhaps a graph of near could help.
e3 =e1/3
y=ex x =13
But isn’t linear ...So make it linear by zooming in (LOCAL LINEARITY), and pick some ordered pairs from the resulting “line”.
y=ex
Compute a linear “equivalent” to .
But this equation is very close to .
So, near , , making .
y=ex
e3 ≈13+1
y=x+1
x =0 ex ≈x+1
How close was the estimate?
and
So the percentage error is
e3 ≈1.395614
3≈1.33333
e3 −43
e3≈0.044625 ≈4.46%
Analyzing error
Look at the residuals for . ex ≈x+1
That looks quadratic!Compute an equation for the linear residuals and use that to enhance your approximation.
But this equation is very close to .
So, , making .
ylinresid =12
x2
ex − x+1( ) ≈12
x2 ex ≈12
x2 + x+1
Improving the estimate
e3 ≈12
13
⎛⎝⎜
⎞⎠⎟
2
+13+1=
2518
≈1.38889
And the percentage error is
e3 −2518
e3≈0.0048176 ≈0.48%
Analyzing error again
Look at the residuals for . ex ≈12
x2 + x+1
And that looks cubic!Compute an equation for the cubic residuals and use that to enhance your approximation.
And this equation is very close to .
So, , making .
yquadresid =16
x3
ex −12
x2 + x+1⎛⎝⎜
⎞⎠⎟≈
16
x3 ex ≈16
x3 +12
x2 + x+1
A faster way.Compute a cubic regression on the original data. This gives the same result, but faster.
I prefer the build up rather than the “black box.”
Quartic Regressionsex ≈
124
x4 +16
x3 +12
x2 + x+1
e3 ≈124
13
⎛⎝⎜
⎞⎠⎟
4
+16
13
⎛⎝⎜
⎞⎠⎟
3
+12
13
⎛⎝⎜
⎞⎠⎟
2
+13
⎛⎝⎜
⎞⎠⎟+1
≈27131944
≈1.395576
e3 −27131944e3
≈0.000036292 ≈0.0036%
What about sine & cosine?
That’s suspicious
These regressions suggest
ex ≈124
x4 +16
x3 +12
x2 + x+1
sin x( ) ≈−16
x3 + x
cos x( ) ≈124
x4 −12
x2 +1
Connections
If you can evaluate , then
eix =...+124
ix( )4 +16
ix( )3 +12
ix( )2 + ix( )+1
=...+124
x4 −i16
x3 −12
x2 + ix+1
= ...+124
x4 −12
x2 +1⎛⎝⎜
⎞⎠⎟+ i⋅ ...−
16
x3 + x⎛⎝⎜
⎞⎠⎟
=cos x( )+ i⋅sin x( )
e3
Another Strange Result
Euler
You have one series for sine:
sin x( ) =...+1
120x5 −
16
x3 + x
=x−x3
3!+
x5
5!−...
But what if you thought of sine as a polynomial via its factors? Then,
sin x( ) =Ax⋅ x−π( ) x+π( ) x−2π( ) x+2π( )...
Euler 2
sin x( ) =Ax⋅ x−π( ) x+π( ) x−2π( ) x+2π( )...
=Aµx⋅ 1−xπ
⎛⎝⎜
⎞⎠⎟
1+xπ
⎛⎝⎜
⎞⎠⎟
1−x
2π⎛⎝⎜
⎞⎠⎟
1+x
2π⎛⎝⎜
⎞⎠⎟...
=Aµx⋅ 1−x2
π 2
⎛
⎝⎜⎞
⎠⎟1−
x2
2π( )2⎛
⎝⎜
⎞
⎠⎟...
But what if you thought of sine as a polynomial via its factors? Then,
Euler 3
x−
x3
3!+
x5
5!−...=Aµx⋅ 1−
x2
π 2
⎛
⎝⎜⎞
⎠⎟1−
x2
2π( )2⎛
⎝⎜
⎞
⎠⎟...
Two polynomials representing the same curve must be equivalent, so
Comparing linear terms gives . Aµ=1
Euler 4
−x3
3!= −
x3
π 2−x3
2π( )2 −
x3
3π( )2 − ...
−x3
6= −
x3
π 21
12+1
22+1
32+ ...
⎛⎝⎜
⎞⎠⎟
π 2
6=1
12+1
22+1
32+ ...
Comparing cubic terms gives ...
QED