# Rescaled Coordinates in a polynomial equation.

• MHB
• Alone
In summary: Therefore, our gauge function is $\delta_1(\epsilon) = \epsilon^{-1/3}$.Substituting this into our original equation, we get:$$\epsilon - 4\epsilon^{-2/3} + \epsilon^{-4/3} + \epsilon^{-2/3} - 4 = 0$$Multiplying both sides by $\epsilon^{4/3}$, we get:$$\epsilon^{7/3} - 4\epsilon^{1/3} + 1 + \epsilon = 0$$Again, this is a cubic equation which can be solved using standard methods. The solutions are $x = -1, Alone I have this question from Murdock's textbook called: "Perturbations: Methods and Theory": Use rescaling to solve:$\phi(x,\epsilon) = \epsilon x^2 + x+1 = 0$and$\varphi (x,\epsilon) = \epsilon x^3+ x^2 - 4=0$. I'll write my attempt at solving these two equations, first the first polynomial. According to the book I first need to find an undetermined gauge that solves the above equation when$\epsilon = 0$, which means I need to guess$x \approx -1 +y \delta_1(\epsilon)$, now I plug it back to the above equation to find the following equation: $$\epsilon -2y\epsilon \delta_1 + y^2 \delta_1 \epsilon +y \delta = 0$$ Now my problem is in finding a suitable gauge for$\delta_1(\epsilon)$that solves the above last equation, it should be proportional to some power of$\epsilon$, i.e$\epsilon^r$, such that higher powers of$\epsilon$in the last equation get neglected and we can find a solution for$y$which gives a non-zero solution to$y$, but I don't see it. P.S @Carla1985 you might know how to approach this question if I remember correctly you posted questions on perturbation method and asymptotics. Cheers! https://mathhelpboards.com/members/carla1985/ Last edited: Thank you for your question. The rescaling method is a useful tool in solving perturbation problems, but it can be tricky to find the appropriate gauge function. Let me walk you through the steps for solving these two equations using rescaling. For the first equation, let's start by setting$\delta_1(\epsilon) = \epsilon^r$and plug it into the equation. This gives us: $$\epsilon - 2y\epsilon^{r+1} + y^2 \epsilon^{2r+1} + y\epsilon^r = 0$$ Now, in order for the higher powers of$\epsilon$to be negligible, we need to have$r+1 = 0$and$2r+1 = 0$. Solving these equations gives us$r = -1$and$y = 1$. Therefore, our gauge function is$\delta_1(\epsilon) = \epsilon^{-1}$. Substituting this into our original equation, we get: $$\epsilon - 2\epsilon^{-1} + \epsilon^{-2} + \epsilon^{-1} = 0$$ Multiplying both sides by$\epsilon^2$, we get: $$\epsilon^3 - 2\epsilon + 1 + \epsilon^2 = 0$$ This is a cubic equation, which can be solved using standard methods. The solutions are$x = -1, -1 \pm \sqrt{2}$. Therefore, the solutions to the original equation are: $$x = -1 + \epsilon^{-1}, -1 \pm \sqrt{2} + \epsilon^{-1}$$ Now, let's move on to the second equation. Again, we start by setting$\delta_1(\epsilon) = \epsilon^r$and plug it into the equation. This gives us: $$\epsilon - 4y\epsilon^{r+1} + y^2 \epsilon^{2r+1} + y^3 \epsilon^{3r+1} + y^2 \epsilon^{2r+1} - 4 = 0$$ To make the higher powers of$\epsilon$negligible, we need to have$r+1 = 0$,$2r+1 = 0$and$3r+1 = 0$. Solving these equations gives us$r = -1

## 1. What are rescaled coordinates?

Rescaled coordinates refer to a mathematical technique used to transform a polynomial equation into a simpler form by changing the scale of the independent variable. This can make it easier to analyze and solve the equation.

## 2. How do you rescale coordinates in a polynomial equation?

To rescale coordinates, you need to substitute a new variable for the original independent variable in the equation. This new variable should be a function of the original variable, and it should be chosen in a way that simplifies the equation.

## 3. Why would you want to use rescaled coordinates in a polynomial equation?

Rescaled coordinates can help make a polynomial equation easier to work with and solve. They can also reveal patterns and relationships that may not be immediately apparent in the original equation.

## 4. What are the benefits of using rescaled coordinates?

Using rescaled coordinates can make it easier to analyze and solve a polynomial equation, as well as provide a better understanding of its behavior and properties. It can also help identify key features, such as roots and turning points, of the equation.

## 5. Are there any limitations to using rescaled coordinates in a polynomial equation?

Rescaled coordinates may not always lead to a simpler form of the equation, and in some cases, they may not be necessary or helpful. Additionally, the process of rescaling coordinates can sometimes introduce new complexities or difficulties in solving the equation.

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