Review¶

[5]:

%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import sympy as sy

Problem: Summations¶

Use the summation formulas below to evaluate the given summations.

\[\sum_{i = 1}^n i = \frac{n^{2}}{2} + \frac{n}{2}\]

\[\sum_{i = 1}^n i^2 = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}\]

\[\sum_{i = 1}^n i^3 = \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}\]

$\sum_{i = 1}^{10} i^2 - i$
$\sum_{i = 1}^{20} 4i^3 + i^2 - 6$
$\sum_{i = 4}^{10} 2i^2$

[6]:

i = sy.Symbol('i')

[7]:

sy.summation(i**2 - i, (i, 1, 10))

[7]:

$\displaystyle 330$

[11]:

sy.summation(4*i**3 + i**2 - 6, (i, 1, n))

[11]:

$\displaystyle n^{4} + \frac{7 n^{3}}{3} + \frac{3 n^{2}}{2} - \frac{35 n}{6}$

[9]:

n = sy.Symbol('n')

[12]:

sy.summation(2*i**2, (i, 4, 10))

[12]:

$\displaystyle 742$

Problem 2: Mean, Variance, Standard Deviation¶

Below is a sample of historic data relating to NYPD arrests. We count the 20 most frequent violations, and ask you to use these values to compute the mean number of incidents, variance in incident counts, and standard deviation of incident counts.

[13]:

nyc_arrests = pd.read_json('https://data.cityofnewyork.us/resource/8h9b-rp9u.json')

[14]:

nyc_arrests.head()

[14]:

	arrest_key	arrest_date	pd_cd	pd_desc	ky_cd	ofns_desc	law_code	law_cat_cd	arrest_boro	arrest_precinct	jurisdiction_code	age_group	perp_sex	perp_race	x_coord_cd	y_coord_cd	latitude	longitude
0	173130602	2017-12-31T00:00:00.000	566	MARIJUANA, POSSESSION	678.0	MISCELLANEOUS PENAL LAW	PL 2210500	V	Q	105	0	25-44	M	BLACK	1063056	207463	40.735772	-73.715638
1	173114463	2017-12-31T00:00:00.000	478	THEFT OF SERVICES, UNCLASSIFIED	343.0	OTHER OFFENSES RELATED TO THEFT	PL 1651503	M	Q	114	0	25-44	M	ASIAN / PACIFIC ISLANDER	1009113	219613	40.769437	-73.910241
2	173113513	2017-12-31T00:00:00.000	849	NY STATE LAWS,UNCLASSIFIED VIOLATION	677.0	OTHER STATE LAWS	LOC000000V	V	K	73	1	18-24	M	BLACK	1010719	186857	40.679525	-73.904572
3	173113423	2017-12-31T00:00:00.000	101	ASSAULT 3	344.0	ASSAULT 3 & RELATED OFFENSES	PL 1200001	M	M	18	0	25-44	M	WHITE	987831	217446	40.763523	-73.987074
4	173113421	2017-12-31T00:00:00.000	101	ASSAULT 3	344.0	ASSAULT 3 & RELATED OFFENSES	PL 1200001	M	M	18	0	45-64	M	BLACK	987073	216078	40.759768	-73.989811

[16]:

arrests = nyc_arrests['pd_desc'].value_counts().nlargest(20).to_frame()

[18]:

np.mean(arrests)

[18]:

pd_desc    38.25
dtype: float64

[19]:

np.var(arrests)

[19]:

pd_desc    1031.2875
dtype: float64

[20]:

np.std(arrests)

[20]:

pd_desc    32.113665
dtype: float64

[15]:

nyc_arrests['pd_desc'].value_counts().nlargest(20).to_frame()

[15]:

	pd_desc
ASSAULT 3	142
LARCENY,PETIT FROM OPEN AREAS,UNCLASSIFIED	89
TRAFFIC,UNCLASSIFIED MISDEMEAN	74
MARIJUANA, POSSESSION 4 & 5	67
INTOXICATED DRIVING,ALCOHOL	53
ASSAULT 2,1,UNCLASSIFIED	47
ROBBERY,UNCLASSIFIED,OPEN AREAS	37
THEFT OF SERVICES, UNCLASSIFIED	32
LARCENY,GRAND FROM OPEN AREAS,UNCLASSIFIED	29
CONTROLLED SUBSTANCE, POSSESSION 7	25
PUBLIC ADMINISTRATION,UNCLASSIFIED FELONY	23
OBSTR BREATH/CIRCUL	20
MENACING,UNCLASSIFIED	19
RESISTING ARREST	16
WEAPONS, POSSESSION, ETC	16
FORGERY,ETC.,UNCLASSIFIED-FELONY	16
CONTROLLED SUBSTANCE,INTENT TO SELL 3	16
WEAPONS POSSESSION 3	15
TRAFFIC,UNCLASSIFIED MISDEMEANOR	15
WEAPONS POSSESSION 1 & 2	14

Problem: Root Mean Squared Error¶

Suppose we build a model to predict the price of an apartment using the square footage and bedrooms as follows:

\[c(x,y) = 10x + 1.2y + 200\]

where $x$ represents square footage and $y$ the number of bedrooms. The Root Mean Squared Error is defined by:

\[RMSE = \sqrt{\frac{\sum_{i = 1}^n (\hat{y} - y_i)^2 }{n}}\]

essentially the square root of summed squared errors between real and predicted cost.

Use the formula to find the RMSE of our models predictions below.

<tr style="text-align: right;">      <th></th>      <th>square footage</th>      <th>bedrooms</th>      <th>real_cost</th>      <th>predicted_cost</th>    </tr>  </thead>  <tbody>    <tr>      <th>0</th>      <td>400</td>      <td>1</td>      <td>4300.76</td>      <td>4201.2</td>    </tr>    <tr>      <th>1</th>      <td>500</td>      <td>1</td>      <td>5301.01</td>      <td>5201.2</td>    </tr>    <tr>      <th>2</th>      <td>600</td>      <td>3</td>      <td>6302.63</td>      <td>6203.6</td>    </tr>    <tr>      <th>3</th>      <td>700</td>      <td>1</td>      <td>7300.91</td>      <td>7201.2</td>    </tr>    <tr>      <th>4</th>      <td>800</td>      <td>1</td>      <td>8301.07</td>      <td>8201.2</td>    </tr>    <tr>      <th>5</th>      <td>900</td>      <td>3</td>      <td>9303.85</td>      <td>9203.6</td>    </tr>    <tr>      <th>6</th>      <td>1000</td>      <td>2</td>      <td>10302.35</td>      <td>10202.4</td>    </tr>  </tbody></table>

[21]:

(4300.76-4201.2)**2 + (5301.01 - 5201.2)**2

[21]:

9912.19360000008

[27]:

arrests['predictions'] = arrests['pd_desc'] + np.random.randint(1, 20, 20)

[30]:

np.mean((arrests['pd_desc'] - arrests['predictions'])**2)

[30]:

157.0

[31]:

np.sqrt(157)

[31]:

12.529964086141668

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Problem: Area under a curve.¶

Evaluate the definite integrals below and represent the solution visually as the area under the curve $f(x)$.

$\int_{1}^3 x^3 - \frac{1}{x} dx$
$\int_{\pi}^{4\pi} 2 \cos(x) dx$
$\int_4^9 1.04(5)^x dx$

[32]:

from scipy.integrate import quad

[37]:

def f(x): return x**3 - 1/x
def g(x): return 2*np.cos(x)
def h(x): return 1.04*5**x

[34]:

quad(f, 1, 3)

[34]:

(18.90138771133189, 2.0984755834487654e-13)

[36]:

quad(g, np.pi, 4*np.pi)

[36]:

(-1.5809616649023117e-15, 1.306472773737261e-13)

[38]:

quad(h, 4, 9)

[38]:

(1261682.718116748, 1.4007492034248646e-08)

Problem: Area between curves¶

Given the functions:

\[f(x) = 1 + x + e^{x^2 - 2x} \quad g(x) = x^4 - 6.5x^2 + 6x + 2\]

define the regions R and S shown below.

<img src = images/a4p4.png />
</center>

Prove that the lines intersect at $x = 1$.
Set up definite integrals to represent the areas $R$ and $S$
Evaluate the integrals using technology.

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Problem: Volumes and Revolution¶

Find the volume of the solid generated by rotating the region bounded by $y = x$, $x = 0$, and $y = (x-1)^2 + 1$. Sketch an image of this region or try to use Python to visualize.
Find the volume of the solid formed by rotating the region R from previous problem about the $x$-axis. Sketch an image of this region.

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Problem: Gini Index¶

The World Bank provides access to data about world GINI Indicies here. Take a look around at a country of your choice. What does the GINI Index say about this country?

The United States Census gathers and provides data related to Income and Poverty in the United States. Visit their site here, and explore the data available. Download one data table and discuss the information your found and what it says about income and poverty in the United States.

[ ]:

# def c(x, y): return 10*x + 1.2*y + 200

# x = np.arange(400, 1100, 100)

# y = np.random.randint(1, 4, 7)

# zhat = np.round(c(x, y), 2)

# z = zhat + np.round(np.random.normal(100, size = 7), 2)

# df = pd.DataFrame({'square footage': x, 'bedrooms': y, 'real_cost': z, 'predicted_cost': zhat})
# df.to_html()

[ ]:

# def f(x): return 1 + x + np.e**(x**2 - 2*x)
# def g(x): return x**4 - 6.5*x**2 + 6*x + 2
# x = np.linspace(0, 2, 1000)
# plt.plot(x, f(x), label = '$f(x)$')
# plt.plot(x, g(x), label = '$g(x)$')
# plt.legend()
# plt.fill_between(x, f(x), g(x), color = 'grey', alpha = 0.1)
# plt.text(0.4, 2.5, 'R', fontsize = 30)
# plt.text(1.4, 2.2, 'S', fontsize = 30)
# plt.savefig('images/a4p4.png')