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For example the equation

$$ n2^n = 8 $$

Is true for $n=2$ which can be guessed, but more complicated examples would require some sort of approach. Also with trigonometric functions,

$$ x\sin(x) + B\cos(x) = A $$

I read that solutions of these kinds of equations can not be expressed in closed form, why is this the case?

vladimirm
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  • It depends on which function you consider. For $n2^n$, if you use $f(n)=2^n$, then the first $n$ is outside of $f$. But if you consider $g(n)=n2^n$, then everything is inside $g$. – Kim Jong Un Sep 10 '14 at 12:55
  • I meant the standard functions such as exponential, trigonometric, etc. But don't know how they are collectively called. – vladimirm Sep 10 '14 at 12:59
  • As far as I know the reason is simple the expressions involved are highly non linear and even not polinomial. – rlartiga Sep 10 '14 at 13:00
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    Because most of the ones you mention coupled with their argument outside the function give rise to transcendental equations and for such equations there's no algebraic solution (except for the exponential which can be solved via the Lambert function). –  Sep 10 '14 at 13:07
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    Your question is an example of why the inverse function theorem is so important. By taking inverse functions, we easily make new and useful functions out of old ones. Examples include: square roots; logarithms; inverse trig functions; et cetera. – Lee Mosher Sep 10 '14 at 13:26

2 Answers2

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It depends on what you mean by "closed form". If you allow the natural logarithm (the inverse of $y = e^x$) you can solve equations like $5^x = 7$ by writing $x = \dfrac{\ln 7}{\ln 5}$.

In the case of expressions like $n2^n = 8$ there is a function called the Lambert $W$-function which is defined as the inverse of $y = xe^x$, $x \ge -1$. That is $x = W(y)$ is equivalent to $y = xe^x$. Then $$8 = n2^n = n e^{\ln 2} = \frac{n \ln 2 e^{n \ln 2}}{\ln 2}$$ so that $n \ln 2 e^{n \ln 2} = 8 \ln 2$, implying $n\ln 2 = W(8 \ln 2)$ and thus $n = \dfrac{W(8 \ln 2)}{\ln 2}$.

This "solves" $8 = n2^n$ in much the same way that $x=(\ln 7)/(\ln 5)$ solves $5^x = 7$.

Umberto P.
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First off, you need to realize that solving equations is the process of determining the inverse function. If I have $n2^n=8$, what I'm really doing is starting with the function $f(n)=n2^n$, and that statement $f(n)=8$ means I need to find $f^{-1}(8)$. When I can find a functional expression for $f^{-1}$, then I have a closed form solution, i.e. a function that I can write down that gives me the inverse of $f$ across all of $f$'s domain.

The reason finding closed form solutions is so hard is that the process of functional inversion doesn't distribute over simple algebra.

For example, look at derivation. Derivation takes a function and returns you a different function, just like inversion does, and it behaves well with arithmetic. Using $d$ to represent the derivation operator, $d(f+g) = df + dg$ and $d(f\times g) = f\times dg + df\times g$. This works no matter how complex $f$ and $g$ are.

However, inversion doesn't play well with others. Using $I$ to represent the inversion operator, there's no general formula for $I(f+g)$ or $I(f \times g)$. That's why things get hairy even before complex functions come into play. For instance, while there are simple algebraic solutions for low order polynomials (e.g. the quadratic equation), there are entire classes of specialized functions created just to express closed form solutions for higher order polynomials.

As an exercise: With $f(x)=2x$ and $g(x)=3x$, try finding a simple formula for the inverse of $f$ + $g$ written in terms of $f, g, f^{-1},$ and $g^{-1}$. Now try that formula using $f(x)=x^2$ and $g(x)=x^2$.

Zimul8r
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