# Using Sympy for Analytical Maths

### Saturday, August 25th 2018

### This is an example of how i would use sympy to evaluate a set of analytical questions, for example:

Find $\forall a \in [0,12,24,30,99]$ and $b=100$:

$\int_a^b \frac{\alpha x^{3} - \sin(2x)}{\sqrt{\beta e^x}} dx$

Where $\alpha = 2.341$, $\beta = e^x$

First we need to import the library (and start the printing for nice easy to read stuff):

```
import sympy as sp
sp.init_printing()
```

Now create the eqn above so we can evaluate it

`a, b, alpha, beta, x = sp.symbols('a b \\alpha \\beta x')`

```
expr = (alpha*x**3 - sp.sin(2*x)) * (1/sp.sqrt(beta*sp.E**x))
expr
```

$\frac{1}{\sqrt{\beta e^{x}}} \left(\alpha x^{3} - \sin{\left (2 x \right )}\right)$

This like the integrand we hav above. Now theres a few things we can do, if we know $\alpha$ and $\beta$, which we do, we can sub them in. then we will have a more simple expression:

```
expr2 = expr.subs({alpha: 2.341, beta:sp.E**x})
expr2
```

$\frac{1}{\sqrt{e^{2 x}}} \left(2.341 x^{3} - \sin{\left (2 x \right )}\right)$

If we wanted we could now evaluate this using the `.subs()`

command again (this time with a value for x such as `.subs(x,3)`

), but we're asked to evalue the integral:

```
integral = sp.integrate(expr2, (x,a,b))
integral
```

$\frac{2.341 a^{3}}{\sqrt{e^{2 a}}} + \frac{7.023 a^{2}}{\sqrt{e^{2 a}}} + \frac{14.046 a}{\sqrt{e^{2 a}}} - \frac{2.341 b^{3}}{\sqrt{e^{2 b}}} - \frac{7.023 b^{2}}{\sqrt{e^{2 b}}} - \frac{14.046 b}{\sqrt{e^{2 b}}} + \frac{0.2}{\sqrt{e^{2 b}}} \sin{\left (2 b \right )} + \frac{0.4}{\sqrt{e^{2 b}}} \cos{\left (2 b \right )} - \frac{14.046}{\sqrt{e^{2 b}}} - \frac{0.2}{\sqrt{e^{2 a}}} \sin{\left (2 a \right )} - \frac{0.4}{\sqrt{e^{2 a}}} \cos{\left (2 a \right )} + \frac{14.046}{\sqrt{e^{2 a}}}$

Now we need to calculate all the values of this expression for every combination of $a \in [0,12,24,30,99]$ and $b=100$. We can do this by starting with lambdifying the expression into a callable function:

```
my_callable = sp.lambdify((a, b), integral)
my_callable(1,2) # this will sub in a=1 and b=2 to the above.
```

$1.6786017689393038$

Now we just loop over the varable s for a and we can print out the values:

```
for a in [0,12,24,30,99]:
print("a =",a)
print(my_callable(a,100))
print()
```

```
a = 0
13.645999999999999
a = 12
0.03219056961943131
a = 24
1.3876938438535285e-06
a = 30
6.546926902186657e-09
a = 99
1.470461229758446e-37
```

This decreasing sequence is expected as were taking essentially taking area slices of the below plot, moving further left each time as a starting point.

```
%matplotlib inline
import matplotlib.pyplot as plt
X = [x for x in range(0,101)]
expr2_callable = sp.lambdify(x, expr2)
plt.plot(X,[expr2_callable(x) for in_x in X])
```

`[<matplotlib.lines.Line2D at 0x24a166e8668>]`