# Basic Operations Solutions¶

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from sympy import *
init_printing()


For each exercise, fill in the function according to its docstring.

## Substitution¶

Write a function that takes a list of expressions, a variable, and a point, and evaluates each expression in the list at that point.

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def evaluate(exprs, x, x0):
"""
Evaluate each expression in exprs at the point x = x0.

>>> x, y = symbols('x y')
>>> exprs = [x**2, cos(x), x*y]
>>> evaluate(exprs, x, 1)
[1, cos(1), y]
>>> evaluate(exprs, y, 0)
[x**2, cos(x), 0]
"""
return [expr.subs(x, x0) for expr in exprs]

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x, y = symbols('x y')

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exprs = [x**2, cos(x), x*y]

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evaluate(exprs, x, 1)

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evaluate(exprs, y, 0)


Write a function that computes

$${\underbrace{x^{{}^{.^{.^{.^x}}}}}_\text{n copies of x}}$$

That is, x**(x**(...x)), with n copies of x. In Knuth up-arrow notation, $x\uparrow\uparrow n$.

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def uparrow(x, n):
"""
Computes x**(x**(...x)), with n copies of x.

>>> x = symbols('x')
>>> uparrow(x, 3)
x**(x**x)
>>> uparrow(x, 1)
x
>>> uparrow(x**x, 3)
(x**x)**((x**x)**(x**x))
"""
expr = x
for i in range(n - 1):
expr = x**expr
return expr

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x = symbols('x')

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uparrow(x, 3)

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uparrow(x, 1)

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uparrow(x**x, 3)


Write a function that takes a function and nests it within itself n times.

For example, if we started with $x^x$, and $n=3$, we would end up with

$$\left(\left(x^{x}\right)^{\left(x^{x}\right)}\right)^\left({\left(x^{x}\right)^{\left(x^{x}\right)}}\right)$$
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def nest(expr, x, n):
"""
Nests expr into itself (in the variable x) n times.

>>> x, y = symbols('x y')
>>> nest(x**x, x, 3)
((x**x)**(x**x))**((x**x)**(x**x))
>>> nest(sin(x)*cos(y), x, 2)
sin(sin(x)*cos(y))*cos(y)
>>> nest(sin(x)*cos(y), y, 2)
sin(x)*cos(sin(x)*cos(y))
>>> nest(x**2, x, 1)
x**2
"""
origexpr = expr
for i in range(n - 1):
expr = expr.subs(x, origexpr)
return expr

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x, y = symbols('x y')

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nest(x**x, x, 3)

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nest(sin(x)*cos(y), x, 2)

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nest(sin(x)*cos(y), y, 2)

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nest(x**2, x, 1)


Write a function that replaces all trig functions in terms of $\sin(x)$ and $\cos(x)$. You can assume that the trig function will be of the form $t(x)$, where $x$ is the variable. The trig functions implemented in SymPy are

$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$$$\cot(x) = \frac{\cos(x)}{\sin(x)}$$$$\sec(x) = \frac{1}{\cos(x)}$$$$\csc(x) = \frac{1}{\sin(x)}$$
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def trig_rewrite(expr, x):
"""
Rewrite all trig functions t(x) in terms of sin(x) and cos(x)

>>> x, y = symbols('x y')
>>> trig_rewrite(tan(x), x)
sin(x)/cos(x)
>>> trig_rewrite(tan(x) + cos(x)*sec(x), x)
sin(x)/cos(x) + 1
>>> trig_rewrite(cot(x) + sin(x)*csc(x), x)
1 + cos(x)/sin(x)
>>> trig_rewrite(tan(x)*tan(y), x)
sin(x)*tan(y)/cos(x)
>>> trig_rewrite(tan(x)*tan(y), y)
sin(y)*tan(x)/cos(y)
"""
return expr.subs([(tan(x), sin(x)/cos(x)), (sec(x), 1/cos(x)), (csc(x), 1/sin(x)), (cot(x), cos(x)/sin(x))])

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x, y = symbols('x y')

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trig_rewrite(tan(x), x)

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trig_rewrite(tan(x) + cos(x)*sec(x), x)

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trig_rewrite(cot(x) + sin(x)*csc(x), x)

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trig_rewrite(tan(x)*tan(y), x)

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trig_rewrite(tan(x)*tan(y), y)


Suppose we are working with the series expansion of a function at $x=0$, like $a_0 + a_1x + a_2x^2 + \cdots$. In this expansion, the terms with higher powers are less significant to the calculation. For example, if we had

$$1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24} + \frac{x^{5}}{120} + \frac{x^{6}}{720} + \frac{x^{7}}{5040} + \frac{x^{8}}{40320} + \frac{x^{9}}{362880}$$

We might only care about the terms with powers less than 5

$$1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24}$$

We will see later that this is can be done automatically using the O class, but it can also be done using subs.

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def series_reduce(expr, x, p):
"""
Remove all powers of x in expr with power greater than p.

You may assume that there are no powers of x greater than 10.

Bonus: which functions are represented by the series expansions below (you
can use expr.series(x, 0, 10) to check if you are right)?

>>> x, y = symbols('x y')
>>> series_reduce(1 - x**2/2 + x**4/24 - x**6/720 + x**8/40320, x, 5)
x**4/24 - x**2/2 + 1
>>> series_reduce(1 + x + x**2 + x**3 + x**4 + x**5 + x**6 + x**7 + x**8 + x**9 + x**10, x, 0)
1
>>> series_reduce(x*y + x**3*y**3/3 + 2*x**5*y**5/15 + 17*x**7*y**7/315 + 62*x**9*y**9/2835, x, 5)
2*x**5*y**5/15 + x**3*y**3/3 + x*y
"""
return expr.subs([(x**i, 0) for i in range(11) if  i > p])

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x, y = symbols('x y')

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series_reduce(1 - x**2/2 + x**4/24 - x**6/720 + x**8/40320, x, 5)

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series_reduce(1 + x + x**2 + x**3 + x**4 + x**5 + x**6 + x**7 + x**8 + x**9 + x**10, x, 0)

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series_reduce(x*y + x**3*y**3/3 + 2*x**5*y**5/15 + 17*x**7*y**7/315 + 62*x**9*y**9/2835, x, 5)


## Evalf¶

At the 762nd place in the decimal expansion of $\pi$, there is 999999 (see the below comic ref: http://www.qwantz.com/index.php?comic=1013). Note T-Rex is counting the digits like

    π: 3 . 1 4 1 5 9

digit:     1 2 3 4 5
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from IPython.core.display import Image
Image(filename='../imgs/comic2-1040.png')


Write a function that takes a symbolic expression (like pi), and determines the first place where 999999 appears.

Tip: Use the string representation of the number. You can take the part after the . by using split('.', 2)[1].

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'1.2345'.split('.', 2)[1]


And remember that Python starts counting at 0, so if you use str.find, you'll need to add one.

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str(1.2345).split('.', 2)[1].find('345')

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str(1.2345).split('.', 2)[1].find('345') + 1


And remember that str.find returns -1 if it cannot find the string.

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'abcd'.find('def')

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def find_999999(expr, limit=100000):
"""
Find the first place in the decimal expr where 999999 appears.

Only checks up to limit digits.

Returns False when 999999 does not appear.

>>> find_999999(pi)
762
>>> find_999999(E)
False
>>> find_999999(E, 1000000) # This one will take a few seconds to compute
384340
"""
found = str(expr.evalf(limit)).split('.', 2)[1].find('999999') + 1
if found == 0:
return False
return found

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find_999999(pi)

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find_999999(E)

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find_999999(E, 1000000)

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