1244.sagews

f = x^2+2*x+1; f;

x^2 + 2*x + 1

f.plot(x,-20,20)

taylor(f, x, 0, 1)

2*x + 1

taylor(f, x, 0, 2)

x^2 + 2*x + 1

taylor(f, x, 0, 5)

x^2 + 2*x + 1

This is not surprising because f is a parabola. Thus, a Taylor polynomial can accurately plot it within two summations.

taylor(f, x, 1, 1)

4*x

taylor(f, x, 1, 2)

(x - 1)^2 + 4*x

(x-1)^2+4*x-(x^2+2*x+1) = 0. (This means that the second approximation is equal to f(x).

taylor(f, x, 1, 1)

4*x

h(x) = abs(4*x - f)

h(1)

0

h(5)

16

h(10)

81

h(100)

9801

This means that, as x is centered farther away from 1, the Taylor polynomial becomes less accurate.

j(x) = (1+x)^-1; j; j.plot(x,-1, 1)

x |--> 1/(x + 1)

taylor(j, x, 1, 1)

x |--> -1/4*x + 3/4

k(x) = (j+1/4*x-3/4); k

x |--> 1/4*x + 1/(x + 1) - 3/4

k(2)

1/12

k(3/2)

1/40

k(5/4)

1/144

k(9/8)

1/544

k(17/16)

1/2112

k(33/32)

1/8320

taylor(j, x, 1, 2)

x |--> 1/8*(x - 1)^2 - 1/4*x + 3/4

m(x) = 1/8*(x - 1)^2 - 1/4*x + 3/4 - j; m

x |--> 1/8*(x - 1)^2 - 1/4*x - 1/(x + 1) + 3/4

m(2)

1/24

m(3/2)

1/160

m(5/4)

1/1152

m(9/8)

1/8704

m(33/32)

1/532480

taylor(j, x, -9/10, 1)

x |--> -100*x - 80

n(x) = -100*x - 80 - j

n(2-9/10)

-4000/21

n(3/2-9/10)

-1125/8

n(5/4-9/10)

-3125/27

n(9/8-9/10)

-10125/98

n(17/16-9/10)

-36125/372

n(33/32-9/10)

-136125/1448

taylor(j, x, -9/10, 2)

x |--> 10*(10*x + 9)^2 - 100*x - 80

p(x) = 10*(10*x + 9)^2 - 100*x - 80 - j

p(2-9/10)

80000/21

p(3/2-9/10)

16875/8

p(5/4-9/10)

78125/54

p(9/8-9/10)

455625/392

p(17/16-9/10)

3070625/2976

p(33/32-9/10)

22460625/23168

p.plot(x,-1,3)

q(x) = arctan(x)

derivative(q,x,2)

x |--> -2*x/(x^2 + 1)^2

r(x) = -2*x/(x^2 + 1)^2; r; r.plot (x, 0, 1)

x |--> -2*x/(x^2 + 1)^2

K=-1.0

derivative(q,x,6)

x |--> -3840*x^5/(x^2 + 1)^6 + 3840*x^3/(x^2 + 1)^5 - 720*x/(x^2 + 1)^4

s(x) = -3840*x^5/(x^2 + 1)^6 + 3840*x^3/(x^2 + 1)^5 - 720*x/(x^2 +1)^4; s; s.plot(x,0,1)

x |--> -3840*x^5/(x^2 + 1)^6 + 3840*x^3/(x^2 + 1)^5 - 720*x/(x^2 + 1)^4

K=-120

derivative(q,x,11)

x |--> 3715891200*x^10/(x^2 + 1)^11 - 8360755200*x^8/(x^2 + 1)^10 + 6502809600*x^6/(x^2 + 1)^9 - 2032128000*x^4/(x^2 + 1)^8 + 217728000*x^2/(x^2 + 1)^7 - 3628800/(x^2 + 1)^6

t(x) = 3715891200*x^10/(x^2 + 1)^11 - 8360755200*x^8/(x^2 + 1)^10 + 6502809600*x^6/(x^2 + 1)^9 - 2032128000*x^4/(x^2 + 1)^8 + 217728000*x^2/(x^2 + 1)^7 - 3628800/(x^2 + 1)^6; t; t.plot(x,0,1)

x |--> 3715891200*x^10/(x^2 + 1)^11 - 8360755200*x^8/(x^2 + 1)^10 + 6502809600*x^6/(x^2 + 1)^9 - 2032128000*x^4/(x^2 + 1)^8 + 217728000*x^2/(x^2 + 1)^7 - 3628800/(x^2 + 1)^6

K=-4e6

taylor(q, x, 0, 1)

x |--> x

taylor(q, x, 0, 5)

x |--> 1/5*x^5 - 1/3*x^3 + x

taylor(q, x, 0, 10)

x |--> 1/9*x^9 - 1/7*x^7 + 1/5*x^5 - 1/3*x^3 + x

-(.5^2)/2

-0.125000000000000

abs(.5-arctan(.5))

0.0363523909991939

This K value works.

-1/2

-1/2

abs(1-arctan(1))

abs(-1/4*pi + 1)

This K value works.

-120*(.5^6)/720

-0.00260416666666667

abs(1/5*.5^5 - 1/3*.5^3 + .5 - arctan(.5))

0.000935724332527255

This K value works.

-120/720

-1/6

abs(1/5*1^5 - 1/3*1^3 + 1 - arctan(1))

abs(-1/4*pi + 13/15)

This K value works.

-4e6*(.5^11)/39916800

-0.0000489298991903159

abs(1/9*.5^9 - 1/7*.5^7 + 1/5*.5^5 - 1/3*.5^3 + .5 - arctan(.5))

0.0000366667928446973

This K value works.

-4e6/39916800

-0.100208433541767

abs(1/9*1^9 - 1/7*1^7 + 1/5*1^5 - 1/3*1^3 + 1 - arctan(1))

abs(-1/4*pi + 263/315)

This K value works.