How to convert $(tx^2+c)^\frac{1}{2}$ into finite polynomial form?

Question

$(tx^2+c)^\frac{1}{2}$ is a member of bigger sum.

I want to convert whole sum into generic finite polynomial form.

How to do it?

I tried to find such a and b, where $(ax+b)^2$ would be equal to $tx^2+c$ to replace $(tx^2+c)^\frac{1}{2}$ with $((ax+b)^2)^\frac{1}{2}$ $=ax+b$ :

$tx^2+c=(ax+b)(ax+b)$

$(t-a^2)x^2+-2abx-(b^2-c)=0$

$D=(-2ab)^2+4(t-a^2)(b^2-c)$

$x=\frac{2ab+D^\frac{1}{2}}{2(t-a^2)}$

than I made substitution for x to get relationship between a and b, I always do have particular t and c. I can treat it as sum of squares. For $t=4$ and $c=2048$ I obtained this relationship for example:

$\frac{16b^2a+32768a^2+64b^2+16ab(8192a^2+16b^2-32768)^\frac{1}{2}}{4a^4-32a^2+64}=0$

Even if this my attempt is theoretically incorrect on your opinion, please ignore my attempt and tell how to do it in correct way on your opinion. The question is how to convert whole parent sum into finite polynomial form.

Answer 1

Well, as many said before, this is not a polynomial (as a function of $x$). Check that $$tx^2+c=a^2x+2abx+b^2$$ implies that $2ab$ has to be zero, so at least one of $a$ and $b$ has to be zero; this leaves as with $b$ or with $ax$. And it is clear that $tx^2+c$ is not $b^2$ nor $(ax)^2$.

About your procedure, even if you give $t$ and $c$ specific values (say $2$ and $3$), there's no point in solving for $x$, since you're not looking for the condition $$2x^2+3=a^2x+2abx+b^2$$ to hold for some specific value/s of $x$, but you want[ed] this to hold for every $x\in\mathbb R$.

Answer 2

You seem not understand that what you want is impossible. Proof below :

Suppose that what you expect be possible. Then a polynomial would exist on the form : $$(tx^2+c)^{1/2}=a_0+a_1x+a_2x^2+...+a_nx^n$$

$$tx^2+c=(a_0+a_1x+a_2x^2+...+a_nx^n)^2$$ Expanding leads to : $$tx^2+c=a_0^2+2a_0a_1x+(a_1^2+2a_0a_2)x^2+...+a_n^2x^{2n}$$ Two equal polynomials are necessary on the same degree. Thus $2n=2$ $$n=1$$ $$c+tx^2=(a_0+a_1x)^2=a_0^2+2a_0a_1x+a_1x^2$$ $$a_0^2=c$$ $$2a_0a_1=0 \quad\implies\quad a_1=0 \quad\text{if } c\neq 0.$$ $$a_1^2=t$$ $a_1^2=t$ is contradictory with $a_1=0$ , except if $c=0$ or $t=0$.

Thus the only case where $(tx^2+c)^{1/2}$ can be equal to a not constant polynomial is when $c=0$ and the polynomial is $t^{1/2}x$.

Conclusion : Definitively $(tx^2+c)^{1/2}$cannot be equal to a polynomial except in the trivial cases $c=0$ or $t=0$.

NOTE :

Don't confuse polynomial with infinite series. $(tx^2+c)^{1/2}$ can be expended to an infinite series : $$(tx^2+c)^{1/2}=c^{1/2}+\frac{t}{2c^{1/2}}x^2-\frac{t^2}{8c^{3/2}}x^4+...+\left(\begin{matrix}1/2\\k \end{matrix} \right)\frac{t^k}{c^{k-\frac12}}x^{2k}+...$$ in $|x|<\frac{c}{t}$ , with binomial coefficients $\left(\begin{matrix}1/2\\k \end{matrix} \right)$ and $k=0\text{ to } \infty.$

Infinite series are not polynomials. For example some transcendantal functions, which of course are not polynomials, are defined by infinite series.

So, if your question was not for polynomial, but for infinite series, my answer should be that it is possible : the infinite series is given above.

Instead of converting what you call "the whole parent sum" into a sum of polynomials, may be you could convert it into a sum of infinite series. Since your question doesn't give the context of the original problem it is not possible to say if it is or not a good method.