Ask Experts Questions for FREE Help !
Ask
    vickia's Avatar
    vickia Posts: 7, Reputation: 1
    New Member
     
    #1

    Nov 5, 2011, 03:32 PM
    Help understanding how to construct a nonlinear function
    Can you construct a nonlinear function which has a fixed point in [0,1]
    I'm more interested on how you got the answer than the answer itself

    Thank you
    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
    Senior Member
     
    #2

    Nov 5, 2011, 09:24 PM
    Do you mean that the function passes through the point (0,1)? In other words, for the function f(x), you're saying that f(0)=1?

    Or do you mean that the function is constant on the interval [0,1]?

    Or do you mean something else entirely?

    If it's the first case, there are of course many answers. A nonlinear function is basically anything other than a constant or a polynomial of degree 1. Quadratic and cubic polynomials, for example, are nonlinear, as are exponential and logarithmic functions, trigonometric functions, hyperbolic functions, etc. Basically, if it's not explicitly the equation of a line, then it's nonlinear.

    You could take any nonlinear function, f(x); compute f(0); then add or subtract whatever constant c is necessary to make the answer 1. f'(x)=f(x)+c. f'(0)=1.

    For example if you choose the nonlinear function , you can evaluate it to find f(0)=0. To get that to be 1, it's necessary to add 1, so we could say . Now we know we have a function, f'(x), which is nonlinear and passes through the point (0,1).
    vickia's Avatar
    vickia Posts: 7, Reputation: 1
    New Member
     
    #3

    Nov 6, 2011, 05:16 AM
    I meant that the function has a unique fixed point in the interval [0,1]
    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
    Senior Member
     
    #4

    Nov 6, 2011, 01:20 PM
    Hmmm... I'm still not following. Sorry.

    The very definition of a function is that it maps any value of the abcissa (usually denoted with x) to one and only one ordinate (usually denoted with y). It doesn't change... ever.

    For a given function, that means all points over any interval are fixed at all times. passes through the point (2,4). If you check it tomorrow, next week, or 10 billion years from now, it's still going to pass through (2,4). Clearly you must mean something else.

    Perhaps you're talking about a function of more than one variable? Or perhaps it's a family of functions F(x) which are parametrized by some parameter (let's call it a). For example, , and you want that whole family of functions to pass through a particular point on the interval [0,1], regardless of the value of a? In this example, we have a winner because that family of functions passes through (0,0) for any value of a. You could change the value of the exponent, add any constant, or replace x by (x-b), where b is some value between 0 and 1. All of those polynomial functions would still pass through a fixed point within the desired interval, regardless of the value of a.

    There are countless other functions besides polynomials, as well. For example will always pass through (0,1), regardless of the value of a. will always pass through (0,0). will always pass through the point (1,0).

    Is that what you mean?

Not your question? Ask your question View similar questions

 

Question Tools Search this Question
Search this Question:

Advanced Search

Add your answer here.


Check out some similar questions!

Nonlinear inequality [ 1 Answers ]

express the solution using interval notation and graph the solution set (x+2)(x-3)<0 && (x+2)(x-1)(x-3)<0 -

How do you measure the strength of a nonlinear relationship? [ 2 Answers ]

The correlation coefficient r measures the strength of a linear relationship between ordered pairs (xi,yi). Suppose you are testing the strength of a non-linear relationship. How would you modify the formula for the calculation of r? To save you time, here is one representation of the formula...

Nonlinear Systems [ 1 Answers ]

Solve the given Nonlinear System (confused how) y=4x-x^2+8 y=x^2-2x I'm not sure what to do, my substitution is not giving me the correct answer. I need help bad.

Nonlinear Systems of Equation [ 2 Answers ]

Nonlinear Systems of Equation: Problem: y = x^2 3x = y +2 Any assistance you can give me will greatly appreciated. Thank you!:)


View more questions Search