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# Math 152: Intro to Mathematical Software
### 2017-01-30
### Kiran Kedlaya; University of California, San Diego
### adapted from lectures by William Stein, University of Washington

## **Lecture 9: Symbolic Calculus (part 2)**

Math 152: Intro to Mathematical Software

2017-01-30

Kiran Kedlaya

Symbolic Calculus (part 2)

%md
Please fill out the week 4 feedback survey! Thanks.

(URL removed)

Topics

  1. finding roots: symbolic, numerical

  2. numerical approximation of a symbolic expression

  3. more about 2d plotting

  4. more about 3d plots

Finding Roots: symbolic

You can use the solve command to solve for zeroes of a function.



x^2+3 = 5
Error in lines 1-1 Traceback (most recent call last): File "/projects/sage/sage-7.5/local/lib/python2.7/site-packages/smc_sagews/sage_server.py", line 976, in execute exec compile(block+'\n', '', 'single') in namespace, locals File "<string>", line 1 SyntaxError: can't assign to operator 

x^2 + 3 == 5
x^2 + 3 == 5 
eqn = x^2 + 3 == 5
show(eqn)
x2+3=5\displaystyle x^{2} + 3 = 5
eqn.add_to_both_sides(-3)
x^2 == 2 
pi == pi
pi == pi 
bool(pi == pi)
True 
solve(x^4 + 2*x + 3 == 0, x)
[x == -1/2*sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3)) - 1/2*sqrt(-(2*I*sqrt(15) + 2)^(1/3) - 4/(2*I*sqrt(15) + 2)^(1/3) + 4/sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3))), x == -1/2*sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3)) + 1/2*sqrt(-(2*I*sqrt(15) + 2)^(1/3) - 4/(2*I*sqrt(15) + 2)^(1/3) + 4/sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3))), x == 1/2*sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3)) - 1/2*sqrt(-(2*I*sqrt(15) + 2)^(1/3) - 4/(2*I*sqrt(15) + 2)^(1/3) - 4/sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3))), x == 1/2*sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3)) + 1/2*sqrt(-(2*I*sqrt(15) + 2)^(1/3) - 4/(2*I*sqrt(15) + 2)^(1/3) - 4/sqrt(((2*I*sqrt(15) + 2)^(2/3) + 4)/(2*I*sqrt(15) + 2)^(1/3)))] 
show(solve(x^4 + 2*x + 3 == 0, x)[0])
x=−12 (2i 15+2)23+4(2i 15+2)13−12 −(2i 15+2)13−4(2i 15+2)13+4(2i 15+2)23+4(2i 15+2)13\displaystyle x = -\frac{1}{2} \, \sqrt{\frac{{\left(2 i \, \sqrt{15} + 2\right)}^{\frac{2}{3}} + 4}{{\left(2 i \, \sqrt{15} + 2\right)}^{\frac{1}{3}}}} - \frac{1}{2} \, \sqrt{-{\left(2 i \, \sqrt{15} + 2\right)}^{\frac{1}{3}} - \frac{4}{{\left(2 i \, \sqrt{15} + 2\right)}^{\frac{1}{3}}} + \frac{4}{\sqrt{\frac{{\left(2 i \, \sqrt{15} + 2\right)}^{\frac{2}{3}} + 4}{{\left(2 i \, \sqrt{15} + 2\right)}^{\frac{1}{3}}}}}}
solve(sqrt(x) == 2, x)
[x == 4] 
solve(sin(x) == 0, x)
[x == 0] 
solve(e^(3*x) == 5, x)
[x == log(1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3)), x == log(-1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3)), x == 1/3*log(5)] 
v = solve(e^(3*x) == 5, x, solution_dict=True); v
[{x: log(1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3))}, {x: log(-1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3))}, {x: 1/3*log(5)}] 
%var x, y
solve([x^2 == y^2, x^3 == y^3 + 5], [x,y])
[[x == 1.35720887245841, y == -1.35720887245841], [x == (-0.6786044041487247 + 1.175377306225595*I), y == (0.6786044041487285 - 1.175377306225602*I)], [x == (-0.6786044041487247 - 1.175377306225595*I), y == (0.6786044041487285 + 1.175377306225602*I)]] 
g = implicit_plot(x^3 == y^3 + 5, (x, -3, 3), (y,-3,3))
g += implicit_plot(x^2 == y^2, (x, -3, 3), (y,-3,3),color='red')
g

show(v[0][x])
log⁡(12i⋅5133−12⋅513)\displaystyle \log\left(\frac{1}{2} i \cdot 5^{\frac{1}{3}} \sqrt{3} - \frac{1}{2} \cdot 5^{\frac{1}{3}}\right)
(e^(v[0][x]*3)).simplify_full()
5 
# or use roots:
v = (e^(3*x) - 5).roots()
v
[(log(1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3)), 1), (log(-1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3)), 1), (1/3*log(5), 1)] 
show(v[0][0])
log⁡(12i⋅5133−12⋅513)\displaystyle \log\left(\frac{1}{2} i \cdot 5^{\frac{1}{3}} \sqrt{3} - \frac{1}{2} \cdot 5^{\frac{1}{3}}\right)

Finding A SINGLE Root in an interval: numerical

f(x) = e^(3*x) - 5
plot(f, -2, 2, ymax=1)

f.find_root(0, 1)
Error in lines 1-1 Traceback (most recent call last): File "/projects/sage/sage-7.5/local/lib/python2.7/site-packages/smc_sagews/sage_server.py", line 976, in execute exec compile(block+'\n', '', 'single') in namespace, locals File "", line 1, in <module> NameError: name 'f' is not defined

Numerical approximation of a symbolic expression

alpha = log(1/2*I*5^(1/3)*sqrt(3) - 1/2*5^(1/3))
show(alpha)
log⁡(12i⋅5133−12⋅513)\displaystyle \log\left(\frac{1}{2} i \cdot 5^{\frac{1}{3}} \sqrt{3} - \frac{1}{2} \cdot 5^{\frac{1}{3}}\right)
f = pi/10^30 + pi - e/10^30 - acos(0)
f
500000000000000000000000000001/1000000000000000000000000000000*pi - 1/1000000000000000000000000000000*e 
numerical_approx(f)
1.57079632679490 


numerical_approx(alpha)
0.536479304144700 + 2.09439510239320*I 
numerical_approx(alpha, prec=200)
0.53647930414470012486691977774206254650853378475617257397088 + 2.0943951023931954923084289221863352561314462662500705473166*I 
numerical_approx(alpha, digits=20)
0.53647930414470012487 + 2.0943951023931954923*I 
# This is the number of digits used in computing the result, NOT the number of correct digits in output!
numerical_approx(alpha, digits=3)
0.536 + 2.09*I 
# Interval arithmetic --
# Every displayed digit except last in the output is definitely right:
ComplexIntervalField(20)(alpha)
0.53648? + 2.09440?*I 
N is numerical_approx
True 
alpha.N()
0.536479304144700 + 2.09439510239320*I 
alpha.n()
0.536479304144700 + 2.09439510239320*I 
alpha.numerical_approx()
0.536479304144700 + 2.09439510239320*I

More about 2d plots

You can do much more than just plot(function...). E.g.,

  • a point

  • a bunch of points

  • text

  • "line" through a bunch of points

  • polygon

  • ellipse

  • implicit plot

  • contour plot

  • vector field

point((1,2))

point2d((1,2), pointsize=150, color='red')

point([(random(), random()) for i in range(100)])

point([(x,x^2) for x in range(50)])

print r"adlkjf\nlksjdfljs"
adlkjf\nlksjdfljs 
print "adlkjf\\lksjdfljs"
adlkjf\lksjdfljs 
g =  point((1,2), pointsize=300, color='red')
g += text(r"Get $\int_0^{\pi} \sin(x) dx$ points!", (1,2), alpha=0.5, fontsize=30, fontweight='bold', color='black', rotation=20, zorder=1)
g.show(frame=True, axes=False)

line([(0,0), (1,1), (1,2), (2,0), (3,1)], thickness=3, color='blue')

v = [(0,0), (1,1), (1,2), (2,0), (3,1)]
line(v, thickness=3, color='blue', zorder=-1) + points(v, pointsize=200, color='grey', zorder=1)

%var x, y
pl = implicit_plot(y^2 + cos(y*e^x) == x^3 - 2*x + 3, (x,-2,2), (y,-3,3))

pl.save('test.pdf')


oo
+Infinity 
-oo
-Infinity 



show(plot(x^2, -100, 100), ymax=3, ymin=2)