# Open neighbourhoods and equating coefficients of polynomials

• MHB
• Carla1985
In summary: Your Name]In summary, the conversation discusses the concept of linear transformations and their properties. The equation given by the supervisor is used to illustrate that for the coefficients of $x$ to be equal, the coefficients $a$ and $\bar{a}$ must also be equal. This holds true even without knowing the exact form of the mapping $\lambda$. The same principles can also be applied to equations with multiple variables.
Carla1985
Hi all,

I am trying to understand some examples given to me by my supervisor but am struggling with some bits. The part I don't understand is: if the equation
$$ax+b\lambda=\bar{a}x-\bar{d}y$$
holds for any $x,y\in V$, an open neighbourhood of the origin, and $\lambda$ is a mapping from $V$ to $R^2$, then the coefficients of $x$ must be equal, so $a=\bar{a}$. I don't see how I can state this for definite without yet knowing what $\lambda$ is. Can someone please explain why this is the case? Thanks.

Also if I have
$$ax+bx\lambda +c\lambda+d=\bar{a}x+\bar{b}xy+\bar{c}y$$
can I apply the same principles?

Carla

Hi Carla,

It seems like your supervisor is discussing linear transformations and their properties. Let me try to break down the equation for you to provide a better understanding.

Firstly, the equation
$$ax+b\lambda=\bar{a}x-\bar{d}y$$
is representing a linear transformation in the form of an equation. Here, $x$ and $y$ are variables in a vector space $V$ and $\lambda$ is a mapping from $V$ to $R^2$. This means that for any given values of $x$ and $y$ in an open neighbourhood of the origin, the equation must hold true.

Now, for the coefficients of $x$ to be equal, we can compare the terms $ax$ and $\bar{a}x$ on both sides of the equation. In order for these terms to be equal, the coefficients $a$ and $\bar{a}$ must be equal. This is because they are both multiplied by the same variable $x$.

So, even without knowing the exact form of the mapping $\lambda$, we can still say that $a=\bar{a}$ based on the equation given to us. This is a fundamental property of linear transformations.

Moving on to your second question, the same principles can be applied. In this case, we have multiple variables and terms on both sides of the equation. However, the same logic applies. In order for the terms containing $x$ to be equal, the coefficients $a$ and $\bar{a}$ must be equal. Similarly, for the terms containing $y$ to be equal, the coefficients $b$ and $\bar{b}$ must be equal.

I hope this explanation helps you understand the concept better. Let me know if you have any further questions.

## 1. What are open neighbourhoods?

Open neighbourhoods are subsets of a topological space that contain a point and are open sets. In simpler terms, they are sets of points that are close to a given point and have no boundaries.

## 2. How are open neighbourhoods used in mathematics?

In mathematics, open neighbourhoods are used to define the concept of continuity and convergence in topological spaces. They also play a crucial role in the definition of limits, differentiation, and integration in calculus.

## 3. What is the importance of equating coefficients of polynomials?

Equating coefficients of polynomials is important in solving equations and determining the properties of polynomials. It allows us to find the roots or solutions of polynomial equations and also helps in proving theorems about polynomials.

## 4. How do you equate coefficients of polynomials?

To equate coefficients of polynomials, we set the polynomials equal to each other and compare the coefficients of the same degree terms on both sides. We can then solve for the unknown coefficients by using algebraic manipulation.

## 5. Can equating coefficients of polynomials be used in real-world applications?

Yes, equating coefficients of polynomials can be used in various real-world applications such as signal processing, data analysis, and modeling physical systems. It is also used in computer graphics and cryptography.

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