Superposition Theorem

Discussion in 'Homework Help' started by steeve_wai, Sep 19, 2007.

  1. steeve_wai

    Thread Starter Active Member

    Sep 13, 2007
    47
    0
    why can this theorem be applied only to linear functions such as voltage and current,and not power which is non-linear.please explain with equations.thank you.:confused:
     
  2. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    its not ' non linear functions' that they r not applicable to IIRC (correct me if i m wrong).
    its not applicable to non linear circuits.
     
  3. Eduard Munteanu

    Active Member

    Sep 1, 2007
    86
    0
    If you model the current entering/exiting a node as a function, then you can say that f(\sum_{k=1}^n{I_k}) = \sum_{k=1}^n{f(I_k)} iff f is linear, where I_k is the current contributed by source k. This is easy to prove if one writes f(x) = ax + b. It's also easy to extend it to voltages.

    How to prove it the other way eludes me at the moment.

    f(E) = {1 \over R} E^2, which is for power, certainly doesn't obey the above relationship.
     
  4. BlackBox

    Member

    Apr 22, 2007
    20
    0
    Sorry to appoint it Eduard, but you are wrong:  f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" alt=" f(x_1+x_2)=f(x_1)+f(x_2)[\tex]
    given x in a vector field

    Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals).

    The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation:

    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" />" alt=" f(ax)= a \cdot() f (x)[\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" alt=" f(x_1+x_2)=f(x_1)+f(x_2)[\tex]
    given x in a vector field

    Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals).

    The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation:

    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" />" />" alt=" f(x) = a x + b[\tex] is not a linear function if you want to speak of the superposition theorem.

    A linear function is a function between vector fields (a scalar field can be a vector field too).
    A function is linear if it follows the following, simple, rules:
    given x in a vector field

    Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals).

    The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation:

    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" alt=" f(x_1+x_2)=f(x_1)+f(x_2)[\tex]
    given x in a vector field

    Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals).

    The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation:

    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" />" alt=" f(ax)= a \cdot() f (x)[\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" alt=" f(x_1+x_2)=f(x_1)+f(x_2)[\tex]
    given x in a vector field

    Coming to your question: the sum is a linear operation, and the result of the combination of linear operators is linear, so given that the basic components of a linear circuit (resistor, inductance, capacitor) are linear, any combination of these will be linear; so the transfer function of your circuit will always be linear, because it is a combination of linear functions (and functionals).

    The power instead is a nonlinear function, because it involves the square of a value (voltage or current), so it's results are not a vector field, simple demonstation:

    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" alt=" f(x) = x^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" alt=" f(x_1 + x_2) = (x_1 + x_2)^2 [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" alt=" f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex]
    Q.E.D.

    It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...

    Holla" />" alt=" f(x_1) + f(x_2) = x_1^2 + x_2^2 [\tex]
     f(x_1 + x_2) \neq f(x_1) + f(x_2) [\tex] <br />
Q.E.D.<br />
<br />
It's not easy to understand if you are not confortable with vector fields, ask if you have some problems...<br />
<br />
Holla" />" />" />" />" />" />
     
Loading...