Why do the coefficients of a equation, expressed in terms of a small parameter epsilon, have to be 0?

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I have trouble understanding this very elementary example of perturbation theory, especially the part marked in red ("It is because epsilon is variable that we can conclude that the coefficient of each power of epsilon in (7.1.5) is separately equal to zero"). Please help!

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The given conclusion is valid because $$a\epsilon^2 + b\epsilon + c = \mathcal O(\epsilon^3) \qquad (\epsilon\to0)$$ Means that a quadratic polynomial must "shrink as fast as a cubic", wich is only possible if $a=b=c=0$. To see that you must consider the definition of $\mathcal O$. It says the following:

There exists an $\epsilon_0 > 0$ and a constant $C \ge 0$ such that for all $0 < \epsilon < \epsilon_0$ $$|a \epsilon^2 + b\epsilon + c| \le C\epsilon^3$$

Now for this to hold it is already necessary that $a=b=c=0$, wich leads to the conclusion.

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Yes. Since $\epsilon$ is an "arbitrary small quantity" (a variable), then the coefficient of each term must vanish.

Let's consider a usual quadratic equation: $ax^2+bx+c=0$. If you consider the coefficients to be fixed, then $x$ can only be some special numbers. But if we need a relation that always vanishes (for all $x$'s), then $a$, $b$ and $c$ must be zero.

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the underlying idea is that you seek a function $F(x,y):\mathbb{R}^2 \rightarrow \mathbb{R}$ such that

(i) $F(x,0)$ - the un-perturbed equation has an easily-identifiable root ($\xi$,say),

(ii) there is a small number $\eta$ such that $F(x,\eta)=0$ is the perturbed equation $f(x)=0$: you wish to solve.

the implicit function theorem provides that in some neighbourhood of $(\xi,0)$ there is a function $g:U \rightarrow \mathbb{R}$ satisfying: $$ F(g(\epsilon),\epsilon) = 0 $$ the text you cite explains a technique for obtaining the coefficients of a power-series expansion of $g$. the significance of the (red-pencilled) remark about $\epsilon$ being variable derives from the fact that a power series is identically zero if and only if each coefficient is zero.

this allows you to find $g$, from which you can then read off the root corresponding to the particular value of $\epsilon$ which corresponds to your perturbed equation (the value referred to above as $\eta$).