Linearization of a non linear equation

So then the problem becomes finding the best fitting a. In summary, the poster is trying to linearize the given equation by taking the natural log of both sides and using basic log laws. However, this may not be the best approach as it may not lead to a linear form. It may be more effective to use algebraic rearrangement to approximate the equation to a linear form for estimating the constants a and b using regression analysis tools such as Microsoft Excel's 'LINEST'.
  • #1
red98
2
0

Homework Statement



I'm sure this is easy but I've been looking at it for an hour and can't get anywhere. I have an equation that I need a linear form of.

Homework Equations



y = a*b*(x1*x22-(x3*x44/c))/(1+b*x1)

That's the equation I have to write a linear form of.

The Attempt at a Solution



I'm struggling to make a start, first thought was to try multiply the LHS by the 1+a*x1 that the RHS is divided by. Then take the natural log of both sides.

Assuming basic log laws I ended up separating it into:

2*ln(y) + ln(b) + ln(x1) = ln(a) + ln(b) + ln(x1) + 2*ln(x2) - ln(a) + ln(b) + ln(c) + ln(x3) + 4*ln(x4)

I know this is wrong since it cancels to

2*ln(y) = ln (b) + 2*ln(x2) + ln(c) + ln(x3) + 4*ln(x4)

And that would mean the constant a vanishes as does x1. I guess I'm approaching the linearization wrong, that or my interpretation of log laws is flawed?
 
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  • #2
red98 said:
EDIT: realized that might not be the clearest form to interpret, this is the equation i was trying to write above:
http://g.imagehost.org/0914/equation1.png [Broken]

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Assuming basic log laws I ended up separating it into:

2*ln(y) + ln(b) + ln(x1) = ln(a) + ln(b) + ln(x1) + 2*ln(x2) - ln(a) + ln(b) + ln(c) + ln(x3) + 4*ln(x4)

That's not right. For instance, [itex]\ln(y(1+bx_1))=ln(y)+ln(1+bx_1)[/itex]. But you can't separate [itex]ln(1+bx_1)[/itex] any further.


That said, I'm a bit confused by what you mean by "linearizing" the equation. When I hear that expression I think of finding a linear approximation of [itex]y=f\left(x_1,x_2,x_3,x_4\right)[/itex]. But this necessarily involves calculus, and you placed this in the Precalc forum. Can you explain further what you mean by linearizing an equation?
 
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  • #3
Tom Mattson said:
That's not right. For instance, [itex]\ln(y(1+bx_1))=ln(y)+ln(1+bx_1)[/itex]. But you can't separate [itex]ln(1+bx_1)[/itex] any further.That said, I'm a bit confused by what you mean by "linearizing" the equation. When I hear that expression I think of finding a linear approximation of [itex]y=f\left(x_1,x_2,x_3,x_4\right)[/itex]. But this necessarily involves calculus, and you placed this in the Precalc forum. Can you explain further what you mean by linearizing an equation?

Ok, thanks for your prompt reply, I'd better explain the context of the problem since I guess I'm using the wrong term but I'm fairly certain that the solution isn't calculus based.

I've realized my attempt at taking the natural log of each side was wrong, I forgot my basic log laws :blushing:

That equation is supposed to be rearranged into a form where the constants a and b can be estimated using a function like Microsoft Excels 'LINEST' or similar to do regression analysis. So I thought this meant some sort of rearrangement so it could be approximated to a form such as: [itex]y=mx+c[/itex] or [itex]1/y=m/x+1/c[/itex]. My initial thought was to try and linearize it by the logs. Now I'm thinking it's probably alegraic rearrangement (without logarithms) but I'm not getting anywhere.
 
  • #4
Sorry, I don't know how LINEST works. :redface: Also I'm at a loss as to how you could estimate [itex]a[/itex] and [itex]b[/itex] separately, given that they occur in the combination [itex]ab[/itex].

If you want to use natural logs then I would try the following.

[tex]\ln(y)=\ln(ab)+\ln\left(x_1x_2^2-\frac{x_3x_4^4}{c}\right)-\ln\left(1+bx_1\right)[/itex]

Now define new variables [itex]u_1=x_1x_2-x_3x_4^4/c[/itex] and [itex]u_2=1+bx_1[/itex].
 
  • #5
See page 4 in <http://en.wikipedia.org/wiki/Gradient_descent>. [Broken] I think by linearization the poster might mean something like estimating a in y = e**ax using least squares. Take the log base e of both sides, then subject it to linear least squares.
 
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1. What is linearization of a non linear equation?

The linearization of a non linear equation is the process of approximating a non linear function with a linear function. This allows for easier analysis and calculation of the function's properties.

2. Why is linearization of a non linear equation important?

Linearization is important because it allows us to understand the behavior of non linear systems by approximating them with simpler linear systems. This simplification is useful in many fields such as physics, economics, and engineering.

3. How is linearization of a non linear equation done?

Linearization is typically done by finding the tangent line to the non linear function at a specific point, known as the linearization point. The equation of this tangent line is the linear approximation of the non linear function.

4. What is the purpose of finding the linearization point?

The linearization point is used to find the best linear approximation of the non linear function. By choosing a point close to the non linear function's curve, the linear approximation will be more accurate.

5. Can linearization be applied to all non linear equations?

No, not all non linear equations can be linearized. Some functions are inherently non linear and cannot be approximated with a linear function. Additionally, linearization may only be valid for a certain range of values for the non linear function.

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