final_exam.ipynb

Kernel: Python 3 (Ubuntu Linux)

Final Exam

Technology for Calculus

Due Tuesday, 12/11/18 at 5PM

Problem 1

Using SymPy, find the third derivative of $$f(x) = \frac{1}{(4x + \sqrt{x^{2} + 9})^{2}}.$$

from sympy import *
init_printing()
x = Symbol('x')
diff(1/(4*x + sqrt(x**2 + 9))**2, x, 3)

$$- \frac{6 \left(\frac{x \left(\frac{x^{2}}{x^{2} + 9} - 1\right)}{\left(x^{2} + 9\right)^{\frac{3}{2}}} + \frac{3 \left(\frac{x}{\sqrt{x^{2} + 9}} + 4\right) \left(\frac{x^{2}}{x^{2} + 9} - 1\right)}{\left(4 x + \sqrt{x^{2} + 9}\right) \sqrt{x^{2} + 9}} + \frac{4 \left(\frac{x}{\sqrt{x^{2} + 9}} + 4\right)^{3}}{\left(4 x + \sqrt{x^{2} + 9}\right)^{2}}\right)}{\left(4 x + \sqrt{x^{2} + 9}\right)^{3}}$$

Problem 2

Using SymPy, compute $$\int_{-\infty}^{\infty}\frac{1}{4 x^{2} + 3 x + 5}\,dx.$$

expr = 1 / (4*x**2 + 3*x + 5)
expr.integrate((x,-oo,oo))

$$\frac{2 \sqrt{71} \pi}{71}$$

Problem 3

Using the midpoint rule, estimate $\displaystyle\int_{2}^{3}\sin(x^{2})\,dx$ by computing $M_{10}$. You may copy relevant code from the "Quadrature in Python" notes on Blackboard. If you do so, be sure to use NumPy's version of the sine function instead of SymPy's version of the sine function.

import numpy as np

# define a midpoints function since we'll need it a couple times
midpoint = lambda x: ( x[1:] + x[:-1] ) / 2

def midpoint_rule(f, lower, upper, n):
    x_vals = np.linspace(lower, upper, n + 1)

    # define Delta X
    delta_x = ( upper - lower ) / n

    # define midpoints
    midpoints = midpoint(x_vals)

    # compute approximation
    # width * height
    areas = delta_x * f(midpoints)
    return areas.sum()

f = lambda x: np.sin(x**2) # define our function for this integral

midpoint_rule(f, 2, 3, 10)

$$-0.030010607817955004$$

Problem 4

Let $$a_{k} = \frac{1}{\sqrt[k]{k!}} = (k!)^{-1/k}.$$ Using SymPy, find $\lim\limits_{k\to\infty}a_{k}$.

k = Symbol('k')
expr = (factorial(k)**(-1/k))
limit(expr, k , oo)

$$0$$

Problem 5

Find the first five nonzero terms of the Taylor series for $$\sin(\sqrt{x})$$ centered at $a = \pi^{2}$.

series(sin(sqrt(x)), x0 = pi**2, n =6)

$$- \frac{x - \pi^{2}}{2 \pi} + \frac{\left(x - \pi^{2}\right)^{2}}{8 \pi^{3}} + \left(- \frac{1}{16 \pi^{5}} + \frac{1}{48 \pi^{3}}\right) \left(x - \pi^{2}\right)^{3} + \left(- \frac{1}{64 \pi^{5}} + \frac{5}{128 \pi^{7}}\right) \left(x - \pi^{2}\right)^{4} + \left(x - \pi^{2}\right)^{5} \left(- \frac{7}{256 \pi^{9}} - \frac{1}{3840 \pi^{5}} + \frac{3}{256 \pi^{7}}\right) + O\left(\left(x - \pi^{2}\right)^{6}; x\rightarrow \pi^{2}\right)$$