calculatorcalculuscubicsquadraticsroots
For example, many calculators come equipped with the quadratic formula, and so you can give them $A,B,C$ coefficients, and they will show you out the solution using the formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
My calculator however finds $x$ when we have $x^3$ too, and this time it asks for $A,B,C,D$.
I was wondering how exactly it did this? Is there a cubic power "formula"? My calculator even calculates irrational roots so I'm not sure how exactly it can do this without a formula.
This is what I would do to calculate $e^x$ with perfect 8-digit accuracy.
Take $\lfloor x\rfloor$ and $b = x - \lfloor x\rfloor$, the whole and fractional parts respectively. ($e^x = e^{\lfloor x \rfloor + (x - \lfloor x\rfloor)} = e^{\lfloor x \rfloor} e^{x - \lfloor x \rfloor}$)
Use the $(((b/4 + 1)b/3 + 1)b/2 + 1)b + 1$ pattern to calculate $e^{x - \lfloor x\rfloor}$. (If the fractional part of the exponent is .4 or less, only terms up to $b/7$ are needed for 8-digit accuracy. Worst-case scenario ($b \rightarrow 1$) terms up to $b/10$ are needed.) This method is easy to implement on a simple calculator, especially ones with a memory slot to quickly reinsert $b$.
When that is finished, multiply by $e$ ($\approx 2.71828183$ by memorization) and press the equals button $\lfloor x\rfloor$ times to repeat multiply. The result is $e^x$.
I've analyzed the number of terms required for full accuracy. Here is the chart (click to view full image):
Basically, if terms up to $b/t$ are needed to fully calculate $e^x$ up to 8-digit precision (7 digits after the decimal point), accounting for rounding (the $\frac{1}{2}$ term), the relation between $x$ and $t$ is given by $\sqrt[t]{\frac{1}{2}10^{-7}t!} = x$.
If a polynomial is given numerically, (coefficients $a_0..a_n$ are given), the resultant method can be used to get at a numerical value of the discriminant. The coefficients of the polynomial and its derivative are put in a (n+2)- squared Sylvester matrix. Then the determinant is the discriminant wanted. Where writing out the discriminant of a matrix containing symbols is prohibitive, the discriminant can be calculated swiftly numerically using existing matrix packages in $O(N^3)$ time.
It is very nicely explained in
http://www2.math.uu.se/~svante/papers/sjN5.pdf
This is example 4.7
If $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, then Theorem 3.3 yields
\begin{align} \begin{array}{| ccccccc |} & a & b & c & d & e & 0 & 0 &\\ & 0 & a & b & c & d & e & 0 \\ & 0 & 0 & a & b & c & d & e \\ & 4a& 3b& 2c& d & 0 & 0 & 0 \\ & 0 & 4a& 3b& 2c& d& 0 & 0 \\ & 0 & 0 & 4a& 3b& 2c& d & 0 \\ & 0 & 0 & 0 & 4a& 3b& 2c& d \\ \end{array} \end{align}
= $b^2c^2d^2 - 4b^2c^3 e - 4b^3d^3 + 18b^3cde - 27b^4e^2 - 4ac^3d^2 + 16ac^4e + 18abcd^3 - 80abc^2de - 6ab^2d^2e + 144ab^2ce^2 - 27a^2d^4 + 144a^2cd^2e - 128a^2c^2e^2 - 192a^2bde^2 + 256a^3e^3$
Best Answer
There are formulas for the cubic and quartic. However it is impossible to find a solution to every quintic function. This took many years to figure out.
I would strongly recommend understanding how the bisection method works for functions. This provides a "numerical solution" instead of a formula which provides a "analytic solution".
There are may numerical techniques that can approximate values that can not be solved directly. Even quintic equations which sometimes don't have analytic solutions, can be solved with numerical solutions.
Edit:
Here is an example of bisection method.This quintic is not solvable: $f(x) = x^5 - x - 1$. But notice $f(1) = - 1$ and $f(2) = 29$ are opposite signs.
Running through the algorithm by hand you will notice that
top - bottomis getting smaller and smaller buttopandbottomare of opposite signs the whole time and surround a root.