Jump to content

polynomials and numerical method ... help ?


Recommended Posts

Posted (edited)

 

 

 

 

 

 

 

 

 

f(x) = polynomial ??

 

f(x) = transcendental ??

 

f(x) = polynomial of degree one ?? therefore linear ?

 

f(x) = polynomial greater than degree one ?? therefore non linear ??

 

 

also all the above mentioned stuff about a linear change against a non linear change ???

 

Yes x is the independent variable and y the dependent variable.

 

So can I assume you are now responding directly to my posts?

 

If so the rest of what you posted is quite irrelevent.

 

Please don't try to go beyond a simple answer to a simple question.

Especially as you have only answered the first one.

 

To help I will examine a few functions (of x) in the light of my definition of linear.

 

Here are a few functions.

 

y = x

y = x2

y = sin(x)

y = x + 10

 

 

Calculate y for x = 1 and for x = 5 that is for x1=1 and for x2=5. or x2 = 5x1

 

y = x So that y(1) = 1; y(5) = 5

y = x2 So that y(1) = 1; y(5) = 25

y = sin(x) So that y(1) = 0.84; y(5) = -0.96

y = x + 10 So that y(1) = 11; y(5) = 20

 

Now my definition of linear is that

 

y(5x) = 5y(x)

 

For which of these functions is this true?

 

Can you now answer my questions 2 and 3 in post49?

Edited by studiot
Posted (edited)

In y = f(x) do you understand what is meant by the independent variable and the dependent variable; please state which is which?

For the function is y = 5 does my definition make the function linear or nonlinear?

Is this true for any y = a constant?

 

 

x is independent ,y is dependent

y=5 is linear and for any value of a yes it is linear

for any value of 'a', yes it is linear

 

 

let me rearrange few things one more time ....

 

 

 

f(x) = polynomial ??

 

f(x) = transcendental ??

 

f(x) = polynomial of degree one ?? therefore linear ? Solution of Linear Equations? Direct Methods ? http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/12-LinEqs_Direct.pdf

 

 

f(x) = polynomial of degree one ?? therefore linear ? Solution of Linear Equations? Indirect Methods ? http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/13-LinEqs_Indirect.pdf

 

f(x) = polynomial greater than degree one ?? therefore non linear ?? solving nonlinear equations ? Fixed Point ?http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_1_FixedPoint.pdf

 

f(x) = polynomial greater than degree one ?? therefore non linear ?? solving nonlinear equations ? Bisection ? http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_2_Bisection.pdf

 

f(x) = polynomial greater than degree one ?? therefore non linear ?? solving nonlinear equations ? Newton ? http://www.ce.utexas.edu/prof/mckinney/ce311k/Overheads/14-NonlinearEquations_3_Newton.pdf

 

 

 

is this right so far ??

 

its better than getting stuck .... atleast things are going to get better in 2016 ....

Edited by bimbo36
Posted

I am sorry, I am clearly wasting my time.

 

We have a saying in English.

 

You can lead a horse to water.

But you can't make him drink.

 

I feel exactly this way seeing your last post.

Posted

hello studiot , thanks for all the help , support and questions...

 

why i dont work with problems right now is because , i was a bit weak in algebra in general ...

 

i was also looking for small problems to begin working with examples ...

 

thank god , i found this excellent beginner book on algebra ....

 

 

Peter Selby, Steve Slavin-Practical Algebra_ A Self-Teaching Guide-John Wiley & Sons (1991)

 

i am also ... going to buy two more books from an online store ...

 

 

How to Ace Calculus: The Streetwise Guide

 

 

with the help of those two books .. i am going to work on examples of algebra and calculus ....

 

 

later , after i get those books... i will start working on differential equations too ... right now i am not touching that subject ....

 

 

i am glad i got a better overall picture of this subject ...

 

now its time to practise from basics.... with the help of those books ....

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.