Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$?

Question

I have tried many times to evaluate $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ as polynomial for some values of integer $n$ which are greater than $1$ for the solution of the titled equation for looking why exactly $c$ always closed to integer which is $-2*10^n$ looking to the behavior of polynomial coefficients but I can't expand that, For numerical evidence I have used such that an example for $n=2$ is montioned here, Now my question here is :

Why $c$ closed to $-2\times10^n$ in $(1-c^2)^3+(c^3+10^nc^2-1)^3+(10^n c^2-1)^3=n$ for $n >1$ ? And does this gives any evidence about unknown representation numbers as sum of cubic like $n=390,732,\cdots $, because What I have tried the decimal expansion of $c$ increases with the digits $9$ ?

Addendum:This part is the motivation of this question, Now I should deduce this question for Unknown representations numbers as $390$, Could we have an integer $c$ for $n=390$ in the titled equation since we have increasing decimal expansion of $c$ with $999999999\cdots$?

Answer 1

This is a very interesting post from a numerical point of view.

Why is the solution so close to $-2\times 10^n$ ? @John Omielan gave good explantions; what I noticed is that we have the same behavior setting the rhs equal to $0$ instead of $n$.

Working with illimited precision, I was able to solve for $n=2000$ to $10000$ places without any difficulty and the result contains $6000$ $9$'s.

What I found interesting is to let $c=-2\times 10^n+\epsilon$ and to perform one single step of Newton method. Hoping no mistake, I obtained $$\epsilon=\frac {10^{-n}}{12} \, \frac {160\times 10^{6n}-48\times 10^{4n}+12\times10^{2n}+n+1} { 16\times 10^{7n}+40\times 10^{4n}-8\times10^{2n}+10^n+1}\sim 10^{-2n}$$ which is almost exactly what is observed.

Edit

We can make things simpler ignoring the $n$ in the rhs and considering the equation $$(1-c^2)^3+(c^3+A c^2-1)^3+(A c^2-1)^3=0$$ with a solution written as $c=-2A+\epsilon$.

Expanding, we face a polynomial of degree $9$ in $\epsilon$ which can be considered as an expansion to $O(\epsilon^{10})$. Now, using series reversion we end with $$\epsilon=\frac{5}{6 A^2}-\frac{1}{4 A^4}+\frac{25}{36 A^5}+O\left(\frac{1}{A^6}\right)$$ Making $A=10^n$ gives the previous result. Moreover, this also explains the appearance of other sequence in the exact solution as shown below $$\left( \begin{array}{cc} n & c_n - \lceil c_n \rceil \\ 2 & -0.99991666909722222222222222222222222222222222222222222222222 \\ 3 & -0.99999916666691597222222222222222222222222222222222222222222 \\ 4 & -0.99999999166666669165972222222222222222222222222222222222222 \\ 5 & -0.99999999991666666666916659722222222222222222222222222222222 \\ 6 & -0.99999999999916666666666691666597222222222222222222222222222 \\ 7 & -0.99999999999999166666666666669166665972222222222222222222222 \\ 8 & -0.99999999999999991666666666666666916666659722222222222222222 \\ 9 & -0.99999999999999999916666666666666666691666666597222222222222 \\ 10 & -0.99999999999999999999166666666666666666669166666665972222222 \\ 11 & -0.99999999999999999999991666666666666666666666916666666659722 \\ 12 & -0.99999999999999999999999916666666666666666666666691666666667 \\ 13 & -0.99999999999999999999999999166666666666666666666666669166667 \\ 14 & -0.99999999999999999999999999991666666666666666666666666666917 \\ 15 & -0.99999999999999999999999999999916666666666666666666666666667 \end{array} \right)$$

However, the same can be observed with much smaller values than $10$; considering $A=2^n$, the following table reports the estimated and exact values of $\epsilon$ (the rhs being included in the equation). $$\left( \begin{array}{ccc} n & \epsilon_{\text{estimated}} & \epsilon_{\text{exact}} \\ 1 & 0.214409722222222 & 0.214480559652020 \\ 2 & 0.051784939236111 & 0.051777771960045 \\ 3 & 0.012980990939670 & 0.012980993387867 \\ 4 & 0.003252055909899 & 0.003252057549138 \\ 5 & 0.000813584360811 & 0.000813584401775 \\ 6 & 0.000203436266424 & 0.000203436267195 \\ 7 & 0.000050861719097 & 0.000050861719110 \\ 8 & 0.000012715599976 & 0.000012715599976 \\ 9 & 0.000003178910770 & 0.000003178910770 \\ 10 &0.000000794728370 & 0.000000794728370 \end{array} \right)$$

Answer 2

Your expression is

$$(1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 = n \tag{1}\label{eq1A}$$

for $n \gt 1$. For some real $d$, let

$$c = d \times 10^n \tag{2}\label{eq2A}$$

Note that within each of the $3$ terms on the left side of \eqref{eq1A}, there's a term of $1$ or $-1$, but the other terms are all at least $c^2$. If $d$ in \eqref{eq2A} is near $0$, this gives you other solutions, such as the $2$ that Wolfram Alpha gives of $c \approx \pm 0.134$ in your link.

Assume $d$ is not very close to $0$ from this point on (such as $|d| \gt 1$), so you have $c^2 \gg 1$. To make the analysis easier, for now ignore those terms of $\pm 1$. As such, using this and \eqref{eq2A} in the LHS of \eqref{eq1A}, gives

$$\begin{equation}\begin{aligned} & (1 - c^2)^3 + (c^3 + 10^n c^2 - 1)^3 + (10^n c^2 - 1)^3 \\ & \approx (c^2)^3 + ((c^2)(c + 10^n))^3 + (10^n c^2)^3 \\ & = ((d\times10^n)^2)^3 + (((d\times10^n)^2)(d\times10^n + 10^n))^3 + (10^n (d\times10^n)^2)^3 \\ & = d^6\times 10^{6n} + d^6(d+1)^3\times 10^{9n} + d^6\times10^{9n} \\ & = d^6\times 10^{6n}(1 + 10^{3n}((d+1)^3 + 1)) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Next, you have $d^6\times 10^{6n} \gg n$, so the value within the outside brackets in \eqref{eq3A} should be close to $0$. Also, for $n \gt 1$, you have $10^{3n} \gg 1$, so this means you will then need to have

$$\begin{equation}\begin{aligned} 10^{3n}((d+1)^3 + 1) & \approx 0 \\ (d + 1)^3 + 1 & \approx 0 \\ d + 1 & \approx -1 \\ d & \approx -2 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

With $d = -2$, so $c = -2 \times 10^n$, note in the LHS of \eqref{eq1A} that the first term is negative, but much less in magnitude than either of the next two terms. Also, the second term is negative, with it being just slightly less than $-64 \times 10^{9n}$. Finally, the third terms is positive, just about equal in magnitude to the second term, but just slightly less than $64 \times 10^{9n}$. Thus, the overall result is negative. However, as indicated in \eqref{eq3A} and \eqref{eq4A}, as $d$ increases, the overall result increases, so you would expect a value just a bit greater than $c = -2 \times 10^n$ to be a root of \eqref{eq1A}. In fact, in your Wolfram Alpha result, if you click on the "More Digits" button for the $c = -200$ result, you get that $c \approx -199.99991666909716587$.