Taylor Expansion of algebraic equivalent equation yields different results

Question

I am applying the perturbation method to find an approximate solution. However, I don't think that this information is of importance to this question. It's more an algebraic question.

I have the following equation:

$$\frac{1}{(p-\delta)^4}+ a +c(p-\delta)=0\tag{1}$$

$\delta$ is a small variation (perturbation) around a known point $p$. A truncated first order Taylor expansion approximation yields:

$$a+\delta \left(\frac{4}{p^5}-c\right)+c p+\frac{1}{p^4}\tag{2}=0$$

Now, I could have started a bit differently: I could have multiplied both sides of equation 1 by $(p-\delta)^4$ without changing its meaning. I do this to emphasis that it is a quintic equation:

$$c (p-\delta)^5+a (p-\delta)^4+1 \tag{3}=0$$

However, now, when I do the truncated first order Taylor expansion approximation, I get:

$$\delta \left(-4 a p^3-5 c p^4\right)+a p^4+c p^5+1=0 \tag{4}$$

This is not the same result as before, although eq. 1 and 3 yield the same solution.

This is evident, if I solve both equation 2 and equation 4 for $\delta$:

From Equation 2, I get:

$$\delta = \frac{p \left(c p^5+a p^4+1\right)}{c p^5-4} \tag{5}$$

From Equation 4, I get:

$$\delta = \frac{c p^5+a p^4+1}{p^3 (4 a+5 c p)}\tag{6}$$

Assigning numeric values ($\{a\to 1,c\to 1,p\to 1\}$):

• Equation 5 yields $$\delta = -1$$
• Equation 6 yields $$\delta = \frac{1}{3}$$

To me both equation 1 and equation 3 are the same, hence the final $\delta$ should be the same too. I don't quite understand what I did wrong, and would highly appreciate your help.

EDIT: Maybe in other words: If eq.1 and eq.3 are the same, then shouldn't their Taylor expansion be the same too? The thing is: At the end I am interested in how much the perturbation $\delta$ changes the known exact solution(of a simpler system) $p$. However, it seems that depending on which equation (1 or 3) I take, I end up with a different perturbation $\delta$. As this calculation is done for a real physical system, I need to be sure about the perturbation to predict failure modes. Hence, I would be very glad, if you could help me spot my thinking mistake.

For those of you, who like to work with Mathamatica:

Solve[Normal[
   Series[1/(p - \[Delta])^4 + a + c*(p - \[Delta]), {\[Delta], 0, 
     1}]] == 0, \[Delta]]

Solve [Normal[
   Series[1 + a*(p - \[Delta])^4 + c*(p - \[Delta])^5, {\[Delta], 0, 
     1}]] == 0, \[Delta]]

Answer 1

Let $f(\delta) = \frac{1}{(p-\delta)^4} + a + c(p-\delta)$. Assume that $p\ne 0$.

You want to solve the equation $f(\delta) = 0$.

Method 1: You solve $f(0) + f'(0) \delta = 0$.

Method 2: You solve $g(0) + g'(0) \delta = 0$ where $g(\delta) = f(\delta)(p-\delta)^4$.

Note that $g(0) = f(0)p^4$ and $g'(0) = f'(0)p^4 - 4p^3f(0)$.

Thus, in Method 2, you actually solve $f(0)p^4 + (f'(0)p^4 - 4p^3f(0))\delta = 0$, or $f(0) + f'(0)\delta - \frac{4}{p}f(0)\delta = 0$.

Unless $f(0) = 0$, the solutions in Method 1 and Method 2 are different.