What's "z0" in here?

Question

0.00/5 (No votes)

See more:

Python

from scipy.optimize import fsolve


def f(z):
	x= z[0]
	y= z[1]
	return [x+2*y, x**2+y**2-1]
	
z0 = [0,1]
z = fsolve(f,z0)
print(z)
print(f(z))

and plz tell me the difference between print(z) & print(f(z))

What I have tried:

Plz help me

Python

from scipy.optimize import fsolve


def f(z):
	x= z[0]
	y= z[1]
	return [x+2*y, x**2+y**2-1]
	
z0 = [0,1]
z = fsolve(f,z0)
print(z)
print(f(z))

and plz tell me the difference between print(z) & print(f(z))

Posted 21-Aug-17 17:04pm

Member 13360036

Updated 22-Aug-17 2:52am

markkuk

v3

Add a Solution

Comments

PIEBALDconsult 21-Aug-17 23:16pm

z0 is an array?
What's fsolve?

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

markkuk · Answer 1 · 2017-08-21T20:00:00

Solution 1

From the documentation of fsolve()[^]: z0 is the initial guess for the root (zero point) of function f(z). print(z) outputs the root found by fsolve() and print(f(z)) outputs the value of the function at z (should be zero).

Posted 21-Aug-17 20:00pm

markkuk