When I assembled a recurrence formula including the following matrix with jupyter notebook, the definition of the recurrence formula itself seems to be working, but I took out one element of the matrix and made a graph to see the change of that element. When I try, I get the following error:
Specifically, the initial state is a matrix called add, and it is a recurrence formula that multiplies it by a matrix called UE and continues to add add.
It appears in a 1x2 matrix, but I want to draw a graph of the element (F (t) [0]) in the first row of that element at the tth.
add = np.array ([[1,2]]). T
UE = np.array ([[1, 1], [1, 1]])
def F (t):
if t == 0:
return add
else: else:
return np.dot (UE, F (t1)) + add
print (F (10))
# [[1022]
# [1025]]
print (F (10) [0])
# [1022]
t = np.arange (0, 1000)
plt.plot (t, F (t) [0])
 
ValueError Traceback (most recent call last)
<ipythoninput43c7b4a9e0b37b>in<module>
>1 plt.plot (t, F (t) [0])
<ipythoninput4043c0249f5b44>in F (t)
1 def F (t):
>2 if t == 0:
3 return add
Four
5 else:
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any () or a.all ()

Answer # 1

Answer # 2
The error says that you can't perform comparison operations on multiple arrays.
t = np.arange (0, 1000) plt.plot (t, F (t) [0])
This is probably because the argument t of the function F requires an integer as a definition, but the array created by np.arrange () is inserted as it is.
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The reason for the error is that the vector value was input to the function F that expects the scalar value to be input, as TakaiY answered.
To achieve what the questioner wants to do, apply numpy.vectorize, which transforms a scalar>scalar function into a vector>vector function. Change the last part as follows.
In addition, since the original function F is a scalar → a vector of two elements, vectorize is applied after making it a scalar → scalar function by shaping it so that only the first element is extracted using the lambda function. I will.
Also, if t is extended to 1000, repeated nesting will exceed the limit of Python, so I set it to 100. I think this improvement goes beyond the premise of the question of using recurrence formulas, so it is not supported.