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Power function

For any real number base x, the powers of x is defined as: $x^0 = 1, x^1 = x, x^2 = x⋅x, x^3 = x⋅x⋅x,$ etc. The exception is $0^0$, which is considered indeterminate.

Powers are also called exponents.

Fractional exponents is defined in terms of roots:

Power functions have the general form $y = x^p$, for any real value of p, for $x \gt 0$.

$f(x)=x^2$: quadratic equation forms a curved parabola in the XY-plane. In this case, the curve is said to be concave up, i.e., it “holds water.” Similarly, the graph of $–x^2$ is concave down; it “spills water".

$f(x)=x^{-1}$. This is the graph of a hyperbola, which has two branches, one in the first quadrant and the other in the third. As we move to the right, the graph approaches the X-axis as a horizontal asymptote.

The hyperbola is one of the three kinds of conic section, formed by the intersection of a plane and a double cone. The other conic sections are the parabola and the ellipse.

Exponential function

The var $x$ is an exponent. $f(x) = 2^x$.

The inverse of the exponential function $y = b^x$ is, by definition, the logarithm function $y = log_b (x)$.

Lambert W function Solve a hard equations: $2^x = x$.

$$ x = 2^x $$ $$ x = e^{\ln(2^x)} = e^{x \ln 2} $$

To further on, we needs knowlege of Lambert W function.


The Lambert W function (also called the omega function or product logarithm, is a set of functions, namely the branches of the inverse relation of the function $f(z) = ze^z$ where $z$ is any complex number. In other words $$ z=f^{-1}(ze^{z})=W(ze^{z}) $$ which can also be expressed as $$ f(W) = We^W $$ Wolfram.

By substituting the above equation in $z'=ze^{z}$, we get the defining equation for the W function (and for the W relation in general): $$ z'=W(z')e^{W(z')} $$ for any complex number $z'$.

Since the function $f$ is not injective, the relation W is multivalued (except at 0).


\begin{aligned} 2^{t}&=5t\\ 1&={\frac {5t}{2^{t}}}\\ 1&=5t\,e^{-t\ln 2}\\ {\frac {1}{5}}&=t\,e^{-t\ln 2}\\ {\frac {-\ln 2}{5}}&=(-t\ln 2)\,e^{(-t\ln 2)}\\ W\left({\frac {-\ln 2}{5}}\right)&=-t\ln 2\\ t&=-{\frac {W\left({\frac {-\ln 2}{5}}\right)}{\ln 2}} \end{aligned}

Another example: \begin{aligned} x^{x}&=z\\ \Rightarrow x\ln x&=\ln z\\ \Rightarrow e^{\ln x}\cdot \ln x&=\ln z\\ \Rightarrow \ln x&=W(\ln z)\\ \Rightarrow x&=e^{W(\ln z)}\\ \end{aligned}

or, equivalently, $x={\frac {\ln z}{W(\ln z)}}$ since: $ \ln z=W(\ln z)e^{W(\ln z)}$ by definition.

Quadratic formula

For quadratic equation $0 = ax^2 + bx + c$, the quadratic formula is: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Formula derivation


For equation $0 = ax^2 + bx + c$. $$ 0 = x^2 + \frac{b}{a} x + \frac{c}{a} \quad (\forall a \neq 0)$$ Using binomial theorem, $ (x+a)^2 = x^2 + 2xa + a^2$, so: $$ 0 = x^2 + \frac{b}{a} x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a} $$ $$ (x + \frac{b}{2a})^2 = (\frac{b}{2a})^2 - \frac{c}{a} $$ $$ x + \frac{b}{2a} = \pm \sqrt{ (\frac{b}{2a})^2 - \frac{c}{a} } $$ In this step, $(\frac{b}{2a})^2 - \frac{c}{a} = \frac{b^2 - 4ac}{(2a)^2} \geq 0 \Rightarrow b^2-4ac \geq 0$ $$ x + \frac{b}{2a} = \pm \sqrt{ (\frac{b^2}{(2a)^2}) - \frac{4ac}{(2a)^2} } $$ $$ x = -\frac{b}{2a} \pm \frac{1}{2a} \sqrt{b^2 - 4ac} $$ $$ x = -\frac{b \pm \sqrt{b^2 - 4ac}}{2a} $$

Binomial Theorem

Wikipedia: binomial theorem.

$$ (a+b)^n = \sum_{i=0}^{n} \binom{n}{i} a^i b^{n-i} $$ where $\binom{n}{i}$ denotes the total number of different combinations of $i$ items chosen from within $n$ items. $$ \binom{n}{i} = \frac{n!}{i!(n-i)!} $$

See wikipedia: binomial expansion visualization for geometry interpretation of binomial expansion. Here is a general approach for solving problems like this. Before getting into the details, let's first consider a simpler version -- if x+y+z=0, proving that x^3+y^3+z^3=3xyz. Now, some people happen to remember the identity that factors x^3+y^3+z^3−3xyz, which makes that problem easy. But the point is that even for that case, the amount of brute-force work it takes to multiply everything out and prove that identity is not negligible, and one can imagine how much work it would be to do this for seventh powers. So that's the motivation of why we need a better approach. Now, if we have symmetric expressions for three variables x,y,z, it is interesting to look at them as roots of a third-degree polynomial. In particular, if x+y+z=0, then these guys are the roots of a polynomial t^3+at+b for some a,b -- and it is not hard to see that a=xy+yz+zx, b=−xyz -- this follows from the definition of a polynomial with particular roots. The nice thing is that this will be the only brute force work that will be needed. With this approach, the simpler version of the problem is a one-line solution: take t^3+at+b=0 for t=x,y,z, add them up, and since any constant times the sum of the three variables vanishes anyway, we get x^3+y^3+z^3+3b=0. Finally turning to the more complex problem given: note that x^2+y^2+z^2=(x+y+z)^2−2(xy+yz+zx), so x^2+y^2+z^2=−2a. As for the higher powers, consider the following: t^5+at^3+bt^2=0 so summing across x,y,z and reusing the results for second and third powers already known yields x^5+y^5+z^5=−a(−3b)−b(−2a)=5ab Finally, t^6=(at+b)2, so t^7=t(at+b)^2=a^2t^3+2abt^2+b^2t Again doing the sum yields x7+y7+z7=a2(−3b)+2ab(−2a)=−7a2b Therefore, we just need to prove that (−2a/2)(5ab/5)=−7a^2b/7, which is obvious.