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I was looking for a solution to the following equation:

$$x^4 - x^3 = C$$

where $x$ is real and $C$ is a parameter.

Wolfram Alpha returned a very complex answer. I am looking for a simpler answer, which is approximately correct when $C$ is sufficiently large (i.e., it approaches the optimal solution when $C\to\infty$).

How can I find such a solution for this equation in particular, and for other polynomial equations in general?

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4 Answers 4

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When $C$ is sufficiently large, then $|x|>>1$ holds because following graph.

enter image description here

So $C=x^4-x^3=x^3(x-1)\sim x^4$. Asymptotic answer will be $x=\pm^4\sqrt C$.

+) Or more accurate answer.

$(x-\frac14)^4=x^4-x^3+\frac38x^2-\frac1{16}x+\frac1{256}\sim x^4-x^3$, so we can approximate $x^4-x^3$ as $(x-\frac14)^4$.

So the answer is $x=\frac14\pm^4\sqrt C$.

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    $\begingroup$ Thanks! This gives the first-order term. Is there a way to add a second-order term? $\endgroup$ Jan 19 at 4:33
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    $\begingroup$ Yes, this can be done. Write $D^4 = C$. Then the two solutions become $x = D +1/4 + (3/32)/D$ and $x = -D + 1/4 - (3/32)/D$. $\endgroup$
    – M. Wind
    Jan 19 at 5:30
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Consider that $$x^4-x^3=x^4-x^3+O\left(\frac{1}{x^4}\right)\qquad\qquad \color{red}{\large (!!)}$$ and use series reversion and, in the real domain, you should get (using $d=\sqrt[4]{C}$) $$x=d+\frac{1}{4}+\frac{3}{32 d}+\frac{1}{32 d^2}+\frac{15}{2048 d^3}-\frac{77}{65536 d^5}-\frac{3}{4096 d^6}+O\left(\frac{1}{d^7}\right)$$ Trying for a small number $C=100$, this truncated formula will give $$x=\frac{16588752+66155123 \sqrt{10}}{65536000}=\color{red}{3.4452761}77$$ while the exact solution is $\color{red}{3.445276105}$.

Using this formula $$x^4-x^3-C=\frac{1989}{2097152 C}+O\left(\frac{1}{C^{3/2}}\right)\sim \frac 1 {1000 C}$$

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  • $\begingroup$ Series reversion looks like a very useful and general tool, but I could not understand how exactly you use it in this case. I have $y = x^4-x^3$, so $a_1=a_2=0, a_3=-1, a_4=1, a_5=...=0$, so all coefficients $A_i$ are divided by 0. $\endgroup$ Jan 22 at 20:12
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How in general ...?

Factor out the big part. In this case, the power of $x$ of largest degree. \begin{align*} C &= x^4 - x^3 \\ &= x^4 \left( 1 - \frac{1}{x} \right) \text{.} \end{align*} Here, we see the approximate solutions $x \in \{\sqrt[4]{C}, \mathrm{i}\sqrt[4]{C}, -\sqrt[4]{C}, -\mathrm{i}\sqrt[4]{C} \}$. (If you do not want complex approximate solutions, ignore the complex approximate solutions; that still leaves two real approximate solutions.) This approximation improves "as fast as" $-1/x$ becomes negligible in your application.

More generally, take the polynomial $P(x)$, of degree $n$, written in the form $$ P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0 \text{.} $$ Then $$ a_n x^n \left( \frac{a_{n-1}}{a_n x} + \frac{a_{n-2}}{a_n x^2} + \cdots + \frac{a_1}{a_n x^{n-1}} \right) = -a_0 \text{.} $$ Approximate solutions are the $n$ roots of $$ x^n = \frac{-a_0}{a_n} \text{.} $$ (And again, ignore the complex approximate solutions if they are not of use to you. If so, if $n$ is even, you get positive and negative $n^{\text{th}}$ roots and if $n$ is odd, only the root whose sign matches that of $-a_0/a_n$.) If we are interested in large $x$, this approximation improves at least as fast as $1/x$, although large coefficients in the parenthesized term can delay this a bit (until the power of $x$ in the relevant term is larger than the coefficient, so starts to overwhelm it).

Can we do better if there is useful structure? Sure, but such methods depend on the structure and you have given no hint what kind of structure to expect.

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  • $\begingroup$ So I can just ignore all the terms that are not of the maximum degree? $\endgroup$ Jan 19 at 5:35
  • $\begingroup$ @ErelSegal-Halevi : That is one general method for approximation. You don't state accuracy requirements, simplicity requirements, or polynomial structure promises, so you got the simplest method for general polynomials having accuracy that improves as $x$ increases. $\endgroup$ Jan 19 at 16:24
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Let $C\to t$ and $Q^4=t$, then $Q=\left\{-\sqrt[4]{t},-i \sqrt[4]{t},i \sqrt[4]{t},\sqrt[4]{t}\right\}$.

First approximation $x=\dfrac{2 Q (2 Q-1)}{4 Q-3}$.

More exact approximation $x=\dfrac{Q \left(1024 Q^5-2816 Q^4+3168 Q^3-1776 Q^2+481 Q-48\right)}{4 (2 Q-1)^2 (4 Q-3) \left(16 Q^2-20 Q+9\right)}$.

More-more exact approximation $x=\frac{Q \left(4398046511104 Q^{21}-51677046505472 Q^{20}+291095703453696 Q^{19}-1044845284032512 Q^{18}+2679340185681920 Q^{17}-5216549166120960 Q^{16}+7999588518592512 Q^{15}-9892945964564480 Q^{14}+10019888069345280 Q^{13}-8393581748289536 Q^{12}+5847885318586368 Q^{11}-3395429015715840 Q^{10}+1640930164996096 Q^9-657039601832960 Q^8+216160980335232 Q^7-57679294645056 Q^6+12246042446563 Q^5-2011062174240 Q^4+244701755352 Q^3-20572244352 Q^2+1050589440 Q-23887872\right)}{16 (2 Q-1)^2 (4 Q-3) \left(16 Q^2-20 Q+9\right) \left(1024 Q^5-2816 Q^4+3168 Q^3-1776 Q^2+481 Q-48\right)^2 \left(1024 Q^6-3584 Q^5+5472 Q^4-4656 Q^3+2341 Q^2-660 Q+81\right)}$.

Verify code for Wolfram CAS:

t=RandomInteger[{-1000000,1000000}];
P=x^4-x^3-t;
Print["Equation: ",P,"=0"];
Print["Solution by CAS:"];
Print[NSolve[P==0,x,16]];
Q={-t^(1/4),-I t^(1/4),I t^(1/4),t^(1/4)};
Print["Solution by formula:"];
Print["x = ",N[(Q (-23887872+1050589440 Q-20572244352 Q^2+244701755352 Q^3-2011062174240 Q^4+12246042446563 Q^5-57679294645056 Q^6+216160980335232 Q^7-657039601832960 Q^8+1640930164996096 Q^9-3395429015715840 Q^10+5847885318586368 Q^11-8393581748289536 Q^12+10019888069345280 Q^13-9892945964564480 Q^14+7999588518592512 Q^15-5216549166120960 Q^16+2679340185681920 Q^17-1044845284032512 Q^18+291095703453696 Q^19-51677046505472 Q^20+4398046511104 Q^21))/(16 (-1+2 Q)^2 (-3+4 Q) (9-20 Q+16 Q^2) (-48+481 Q-1776 Q^2+3168 Q^3-2816 Q^4+1024 Q^5)^2 (81-660 Q+2341 Q^2-4656 Q^3+5472 Q^4-3584 Q^5+1024 Q^6)),16]//Sort]
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  • $\begingroup$ How did you get the first approximation? $\endgroup$ Jan 22 at 20:16
  • $\begingroup$ @ErelSegal-Halevi, see videos "Solving Polynomial Equations" by Norman Wildberger, I use same technique. By induction can get full form of recursions in numerator and denominator, as this. And for odd-degree polynomial can get rational approximation of one real root without radicals. $\endgroup$ Jan 23 at 5:47

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