The integral $$ I_n = \int_0^1 x^n e^x\,\mathrm{d}x \quad n=0,1,2,\cdots $$ suggests a recursion relation obtained by integration by parts: $$ I_n = \left[ x^n e^x \right]_0^1 - n\int_0^1x^{n-1}e^x\,\mathrm{d}x = e - nI_{n-1} $$ terminating with $I_0 = e-1$. However, this algorithm, applied "forwards" for increasing $n$ is numerically unstable since small errors (such as floating point rounding errors) are magnified at each step: if the error in $I_n$ is $\epsilon_n$ such that the estimated value of $I_n$ is $I_n' + \epsilon_n$ then $$ \epsilon_n = I_n' - I_n = (e-nI_{n-1}') - (e - nI_{n-1}) = n(I_{n-1} - I_{n-1}') = -n\epsilon_{n-1}, $$ and hence $|\epsilon_n| = n!\epsilon_{0}$. Even if the error in $\epsilon_0$ is small, that in $\epsilon_n$ is larger by a factor $n!$ which can be huge.
The numerically stable solution, in this case, is to apply the recursion backwards for decreasing $n$:
$$ I_{n-1} = \frac{1}{n}(e - I_n) \quad \Rightarrow \epsilon_{n-1} = -\frac{\epsilon_n}{n}. $$ That is, errors in $I_n$ are reduced on each step of the recursion. One can even start the algorithm at $I'_N = 0$ and providing enough steps are taken between $N$ and the desired $n$ it will converge on the correct $I_n$.
import numpy as np
import matplotlib.pyplot as plt
def Iforward(n):
if n == 0:
return np.e - 1
return np.e - n * Iforward(n-1)
def Ibackward(n):
if n >= 99:
return 0
return (np.e - Ibackward(n+1)) / (n+1)
N = 35
Iforward = [np.e - 1]
for n in range(1, N+1):
Iforward.append(np.e - n * Iforward[n-1])
Ibackward = [0] * (N+1)
for n in range(N-1,-1,-1):
Ibackward[n] = (np.e - Ibackward[n+1]) / (n+1)
n = range(N+1)
plt.plot(n, Iforward, label='Forward algorithm')
plt.plot(n, Ibackward, label='Backward algorithm')
plt.ylim(-0.5, 2)
plt.xlabel('$n$')
plt.ylabel('$I(n)$')
plt.legend()
plt.show()
The figure below shows the forwards algorithm becoming extremely unstable for $n>16$ and fluctuating between very large positive and negative values; conversely, the backwards algorithm is well-behaved.
