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import numpy as np
import matplotlib.pyplot as plt
import math
%matplotlib inline

7.12. Integral Test and Income Streams

In this lab, we show that it is possible due to continuous compounding of interest to deposit a fixed amount \(P\) at time \(t=0\), and then withdraw at times \(t=1, 2, 3,...\) increasing amounts \(A+Bt\) (where \(A>0\) and \(B>0\) are constants) “in perpetuity” while always maintaining a positive balance in the account.

7.12.1. Integral Test

The integral test is one of the standard tests for convergence/divergence of an infinite series. The basic idea of the test is that for a continuous function \(f(x)\) defined on the interval \(1\le x < \infty\), the infinite series

\[\sum_{n=1}^{\infty} f(n) \]

and the improper integral

\[ \int_{x=1}^{\infty} f(x)\, dx \]

either both converge or both diverge.

7.12.2. P-Series

The integral test can be used to determine all positive values of \(p\) for which the \(p\) series

\[\sum_{n=1}^{\infty} \frac{1}{n^p} \]

converges or diverges.

For example the case \(p=1\) so we have the harmonic series

\[\sum_{n=1}^{\infty} \frac{1}{n}. \]

The corresponding improper integral is divergent:

\[\int_{x=1}^{\infty} \frac{1}{x}\,dx = \lim_{b\rightarrow \infty} \ln x \mid _1^{b} = \lim_{b\rightarrow \infty} \ln b = \infty . \]

Thus the harmonic series is also divergent, as suggested by the circumscribed rectangles representing the first five terms in the harmonic series as shown in the figure below.

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f = lambda x : 1/x
x=np.arange(1,6,.0001)
y=f(x)
x_left = np.arange(1,6,1)
plt.figure(figsize=(5,3))
plt.gca().set_xticks(np.arange(1,7,1))
plt.gca().set_yticks([.1,.2,.3,.4,.5,.6,.7,.8,.9,1])
plt.plot(x,y,'k',markersize=1)
plt.bar(x_left,1/x_left,width=1,alpha=0.1,align='edge',color='r',edgecolor='k')
plt.title('Divergence of the Harmonic Series')
plt.text(1.5,1,'1')
plt.text(2.3,.5,'1/2')
plt.text(3.3,.33,'1/3')
plt.text(4.3,.25,'1/4')
plt.text(5.3,.2,'1/5')
plt.xlabel('x')
plt.ylabel('y')
plt.ylim((0,1.1)) 
#plt.grid()
plt.savefig('harmonic.png')
plt.show()

On the other hand, a \(p\)-series with \(p=2\) series must converge as shown by a convergent improper integral

\[\int_1^{\infty}\frac{1}{x^2}\,dx = \lim_{b\rightarrow\infty} -x^{-1}\mid_1^b = \lim_{b\rightarrow\infty} -\frac{1}{b}+1=1.\]

in comparison with inscribed rectangles representing the tail of the series

\[\frac{1}{2^2}+\frac{1}{3^2}+...<\int_1^{\infty}\frac{1}{x^2}=1 \Rightarrow\]

\[\sum_{n=1}^{\infty} \frac{1}{n^2}= 1+\frac{1}{2^2}+\frac{1}{3^2}+...<1+\int_1^{\infty}\frac{1}{x^2}=2.\]

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f = lambda x : 1/x**2
x=np.arange(1,6,.0001)
y=f(x)
x_right = np.arange(2,6,1)
plt.figure(figsize=(5,3))
plt.gca().set_xticks(np.arange(1,7,1))
plt.gca().set_yticks([.1,.2,.3,.4,.5,.6,.7,.8,.9,1])
plt.plot(x,y,'k',markersize=1)
plt.bar(x_right-1,1/x_right**2,width=1,alpha=0.1,align='edge',color='r',edgecolor='k')
plt.title('Convergence of a p-Series with p=2')
plt.text(1.,.25,'1/2^2')
plt.text(2.,1/8,'1/3^2')
plt.text(3.,1/16,'1/4^2')
plt.text(4.,1/25,'1/5^2')
plt.xlabel('x')
plt.ylabel('y')
plt.ylim((0,1.1)) 
#plt.grid()
plt.savefig('p2.png')
plt.show()

7.12.3. Income Streams

If \(P\) dollars are invested into an account with interest compounded continuously with annual interest rate \(r=.05\) (i.e 5%), then after \(t\) years the amount \(P\) grows to \(Pe^{rt}=Pe^{.05t}\).

Suppose \(f(t)=200+10t\) dollars are invested into an account with the same interest rate at times \(t=1, 2, 3\) (in years) and the total including interest is computed at \(t=3\).

The amount \(\$ 210\) invested at \(t=1\) grows to \(A_1=\$ 210 e^{.05(3-1)}=210e^{.05(2).}\)

The amount \(\$ 220\) invested at \(t=2\) grows to \(A_2=\$ 220 e^{.05(3-2)}=220e^{.05(1).}\)

The amount \(\$ 230\) invested at \(t=3\) grows to \(A_3=\$ 230 e^{.05(3-3)}=230e^{.05(0)}=230.\)

Note that

\[A_1+A_2+A_3 < \int_1^3 (200+10t)e^{.05(3-t)}.\]

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f = lambda t : (200+10*t)*math.e**(.05*(3-t))
a = 0; b = 3; N = 3
n = 10 # Use n*N+1 points to plot the function smoothly
#right endpoint rule
t = np.linspace(a,b,N+1)
y = f(t)
T = np.linspace(a,b,n*N+1)
Y = f(T)

plt.figure(figsize=(5,5))
plt.gca().set_xticks([1,2,3])
plt.plot(T,Y,'r', linestyle =':')
t_right = t[1:] # right endpoints
y_right = y[1:]
plt.plot(t_right,y_right,'k.',markersize=10)
plt.bar(t_right,y_right,width=-(b-a)/N,alpha=0.1,align='edge',color='k',edgecolor='k')
plt.title('Right Rectangle Sum, N = {}'.format(N))
plt.xlabel('Time t investment of 200+10t dollars with interest')
plt.ylabel('Value of investment at t=3')
plt.text(1,232.5,"f(t)=(200+10t)e^{.05(3-t)}",color='r')
plt.text(.3,232,"A_1")
plt.text(1.3,231,"A_2")
plt.text(2.3,230,"A_3")
plt.ylim((225, 233)) 
#plt.grid()
plt.savefig('3 period.png')
plt.show()

7.12.4. Exercises

Exercises

1. Suppose \(f(t) = 10,000+500 t\) dollars are invested into an account (with the same interest rate) at times \(t=1, 2, 3,\) and \(4\) (\(t\) in years). At time \(t=4\), find the total value of the investment with interest \(A_1+A_2+A_3+A_4.\)

2. Show geometrically that the sum \(10,500 e^{.05(3)} + 11,000 e^{.05(2)} + 11,500 e^{.05} + 12,000 \) is less than \(I_4=\int_0^4 (10,000+500 t) e^{.05 (4-t)} dt\).

3. Explain why

\[ I_4=\int_0^4 (10,000+500 t) e^{.05 (4-t)} dt = e^{.05 (4)} \int_0^4 (10,000+500 t) e^{-.05 t} dt = e^{.05(4)}P_4. \]

4. Explain why an amount \(P_4=\int_0^4 (10,000+500 t) e^{-.05 t} dt\) invested at time \(t=0\) will allow withdrawals whose total value with interest is \(A_1+A_2+A_3+A_4\).

5. Extend the reasoning in exercises 1-4 to show that \(P_N = \int_0^N (10,000+500 t) e^{-.05t } dt \) deposited at \(t=0\) is sufficient to allow \(f(t) = 10,000+500 t\) to be withdrawn from the account at \(t=1,2,...,N\). Use integration by parts to find \(P_{\infty} = \int_0^{\infty} (10,000+500 t) e^{-.05 t} dt \). (If this finite amount \(P_{\infty}\) is deposited at \(t=0\), then \(f(t)=10,000 + 500 t\) can be withdrawn at \(t=1, 2, 3,...\) (“in perpetuity”.)