Is the assumption consider cheating in mathematics? Given an equation
$g(p_0,v_0)=g(p_1,v_0+h)$
I am require to find the value of $p_1$ which satisfy the equation
Rearrange the equation and expand it using taylor series
$g(p_1,v_0+h)-g(p_0,v_0)=0$
$g_p(p_1-p_0)+g_v(v_0+h-v_0)+\frac{1}{2}[g_{pp}(p_1-p_0)^2+2g_{pv}(p_1-p_0)(h)+g_{vv}h^2]+...=0$
In order to solve it, I start by assumption $p_1=p_0+a+b+c+...$
Which is then assume the coefficient of $h^n=0$ where $n=1,2,3,....$
$g_p(a)+g_vh=0h$
which found
$a=-h\frac{g_v}{g_p}$
substitute $a$ to second order term of taylor expansion and group all term of $h^2$
$g_p(b)+\frac{1}{2}[g_{pp}(-h\frac{g_v}{g_p})^2+2g_{pv}(-h\frac{g_v}{g_p})(h)+g_{vv}h^2]=0h^2$
$b=-\frac{h^2}{2}[g_{pp}(\frac{g_v^2}{g_p^3})-2g_{pv}(\frac{g_v}{g_p^2})+\frac{g_{vv}}{g_p}]$
Repeat the same method as above to compute c and so on
In this method, I will need to solve for a to get b, and solve for b to get c and so on but
is the direct assumption of $p_1=p_0+a+b+c+...$ and input of a,b and c into 1st,2nd and 3rd order term of taylor series to derive the value and $h^n=0$ consider "cheating"? As one may claim how you know the assumption is true?
 A: It's not cheating, such an assumption is called an ansatz, and it's quite common. When using one you need to verify that your final solution is indeed valid and note that if so there may be other solutions. But other than that it's a perfectly good technique and sometimes even the only reasonably effective one. 
However, when using an ansatz if you conclude that there is no solution you can not also conclude the original problem has no solution. 
To make it more rigorous, let $S$ be the set of valid solutions and $A$ be the set of solutions (valid or not) that satisfy your ansatz. Then your assumption is $\exists x : (x \in S \wedge x\in A)$. This may be valid or not. Using it we find some concrete solution $s \in A$. You verify this solution (or abort the proof if the verification doesn't hold), finding $s \in S$. Now we can prove our initial assumption was justified with $s$.
This isn't circular logic because we found a concrete object $s$ that breaks the cycle - that object exists regardless of any assumptions. Critically though, we must verify $s \in S$ independently of our assumption, otherwise it would be circular.
