Linear relationship at small x

[watertreader]

in general, how do we show linear relationship over a small x of the equation?

Is the equation being able to be Taylor expand show itself as linear? likewise can we consider the series expansion at about x=0 (maclaurin series) to be linear too?


If not, is there any examples countering the above notion? appreciate if anyone can point out to how to substantiate an equation to be linear at small x?

thanks

 

[Mark44]

You can use a linear function to approximate a given function by ignoring all terms of degree two and higher in the Taylor or Maclaurin series.

For example, the Maclaurin series for e^x is 1 + x + x^2/2! + ... The linear approximation, for small x (x close to 0) is y = 1 + x.

 

[watertreader]

just another query...if we have done a Maclaurin expansion on an equation ...the result is a0 + a2x^2 + a3 x^3 +...

does it still qualify to say that the equation is still linear? since the only terms in the expansion up to x terms are only a0

thankls

 

[Tedjn]

In this one might say the best linear (more precisely affine) approximation at x = 0 is just a0, a horizontal line. This is because the derivative of the function at 0 is zero.

 

[Redbelly98]

just another query...if we have done a Maclaurin expansion on an equation ...the result is a0 + a2x^2 + a3 x^3 +...

does it still qualify to say that the equation is still linear? since the only terms in the expansion up to x terms are only a0

thankls

No, that is not linear if you are including the x² and x³ terms.

 

[watertreader]

In this one might say the best linear (more precisely affine) approximation at x = 0 is just a0, a horizontal line. This is because the derivative of the function at 0 is zero.

so if we take an approximation to a0, even if we made a small change to x, we will still obtain a straight line or a constant

Would this suggest that at small x, the function itself is not a function of x?

thanks

 

[Mark44]

For small x, y = a₀ is a constant function of x.