CODEWITHSUNDEEP
pythonmath
Cawen Tv
Cawen Tv

Reputation: 41

Why my code is giving the error name 'sin' is not defined?

I'm trying to implement a code for the following function: f(x) = sen(2πx), x ∈ [0, 2], to get the graph of it. But I'm getting the return that the sine is not defined. I'm not quite understanding what I would have to correct. I would be grateful if someone can help me

Code I used:

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

def f(x): return sin*(2*np.pi*x)  
  
x = np.linspace(0,2)  
plt.plot(x, f(x))    
plt.grid()  
plt.show()

This was the only code I thought of to solve this issue, because I thought it would print correctly and without errors

Upvotes: 1

Views: 2782

Answers (3)

Stef
Stef

Reputation: 15525

There are two distinct ways to get access to function sin from module numpy:

Either:

from numpy import sin

sin([3,4,5])
# array([ 0.14112001, -0.7568025 , -0.95892427])

Or:

import numpy as np

np.sin([3,4,5])
# array([ 0.14112001, -0.7568025 , -0.95892427])

You got an error because you tried a mix of both.

Upvotes: 0

islam abdelmoumen
islam abdelmoumen

Reputation: 664

The problem is that you are trying to use the sin function without import it. You can find the function in the numpy package:

import numpy as np
import matplotlib.pyplot as plt

def f(x):
    return np.sin(2*np.pi*x)  

x = np.linspace(0,2)  
plt.plot(x, f(x))    
plt.grid()  
plt.show()

Upvotes: 1

Luckk
Luckk

Reputation: 538

Because the sine function is defined inside numpy, so you should explicitly specify the library where the function is defined:

def f(x): return np.sin*(2*np.pi*x)

Upvotes: 1

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