Garry Saini - 3 years ago 119

Python Question

`>>> import math`

#defining first function

>>> def f(a):

return a-math.sin(a)-math.pi/2

#defining second fuction

>>> def df(a):

return 1-math.cos(a)

#defining third function which uses above functions

>>> def alpha(a):

return a-f(a)/df(a)

How to write a code in which alpha(a) takes a starting value of a=2, and the solution of alpha(2) will become the input the next time. For eg: let's suppose alpha(2) comes to 2.39 , hence the next value would be alpha(2.39) and go on {upto 50 iterations}. Can somebody please help me a bit. thanks in advance.

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Answer Source

You can let the program **iterate with a for loop**, and

```
temp = 2 # set temp to the initial value
for _ in range(50): # a for loop that will iterate 50 times
temp = alpha(temp) # call alpha with the result in temp
# and store the result back in temp
print(temp) # print the result (optional)
```

`print(temp)`

will print the intermediate results. It is not required. It only demonstrates how the `temp`

variable is updated throughout the process.

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