Linear approximation of Non linear system by Taylor series

[mac1]

I have a equation which represents a nonlinear system.I need to linearize it to obtain a linear system.I have studied various notes and asked my teachers but they are unable to explain how the solution has been obtained.I have the solution but I want to know how it has been done.Please could someone explain it to me.

Image file eqn1- The highlighted text shows the simple way of linearisation using Taylor series.

View attachment 2166

Image file eqn2- The FIRST highlighted equation shows the equation to be linearized and the SECOND highlighted equation is the linearized form.

View attachment 2167

So,please help me to find the method by which eqn 5.6 has been obtained from eqn 5.5b.

Thanks.

 

[HallsofIvy]

First point: to get a linear approximation, you don't really need to know about "Taylor series". You just need to know that tangent line to y= f(x), at $$\left(\overline{x}, f(\overline{x})\right)$$ is given by $$y= f'(\overline{x})(x- \overline{x})+ f(\overline{x})$$.

In this particular case, $$y= f(x)= \frac{5.68x}{1+ 4.68x}$$, we can use the "quotient rule" to differentiate:
$$\frac{dy}{dx}= \frac{5.68(1+ 4.68x)- 5.68x(4.68)}{(1+ 4.68x)^2}$$

The two "5.68(4.68x)" terms in the numerator cancel leaving
$$\frac{dy}{dx}= \frac{5.68}{(1+ 4.68x)^2}$$
So the linear approximation to the function, at $$\overline{x}$$ is
$$y= \frac{5.68}{(1+ 4.68\overline{x})^2}(x- \overline{x})+ \frac{5.68\overline{x}}{1+ 4.68\overline{x}}$$

 

[mac1]

First point: to get a linear approximation, you don't really need to know about "Taylor series". You just need to know that tangent line to y= f(x), at $$\left(\overline{x}, f(\overline{x})\right)$$ is given by $$y= f'(\overline{x})(x- \overline{x})+ f(\overline{x})$$.

In this particular case, $$y= f(x)= \frac{5.68x}{1+ 4.68x}$$, we can use the "quotient rule" to differentiate:
$$\frac{dy}{dx}= \frac{5.68(1+ 4.68x)- 5.68x(4.68)}{(1+ 4.68x)^2}$$

The two "5.68(4.68x)" terms in the numerator cancel leaving
$$\frac{dy}{dx}= \frac{5.68}{(1+ 4.68x)^2}$$
So the linear approximation to the function, at $$\overline{x}$$ is
$$y= \frac{5.68}{(1+ 4.68\overline{x})^2}(x- \overline{x})+ \frac{5.68\overline{x}}{1+ 4.68\overline{x}}$$

Thanks for the reply sir.I have understood how linearisation was done for the first equation 5.1.
But I want to know how eqn 5.6 was obtained from eqn 5.5b.I used the same concept and applied it to 5.5b but I am getting two terms more than the eqn 5.6 : 1/Mn(yn-1*Vn - ynVn). consider yn as yn bar.