I am currently trying to find the roots of this function below.
$$ f_s(c,s,\alpha,\rho, \gamma, us) =(\alpha s^{\rho-1} ((1-\alpha)c^\rho + \alpha s^\rho)^{((1-\gamma)/\rho) – 1}) – us $$
, where us is one whole variable.
I want to find the roots of this function, given fixed values of s,alpha,rho,gamma and us. The issue is, I have typed this function into wolfram trying different fixed values, but the result is always that it yields no roots.
s = 40, alpha = 0.5, rho = 0.5, gamma = 2, us = 3
s = 20, alpha = 0.2, rho = 0.5, gamma = 2, us = 2
Is there something I'm missing or does this equation truly have no roots? I've attached some more images below that might be of help, thanks a lot!
Wolfram Alpha Giving the Origin of that function above, and us is just a new variable I made up to subtract from it.
EDIT 1 : Thank you for all your suggestions; I have tried graphing it, and for sure there are no roots. This is supposed to work with either of the fixed values I chose, so I would conclude that this function really has no roots and I was down the wrong track. If anyone else has any insights, that would be great but if not, thanks for your help everyone and I'll close this thread.
Best Answer
Let's rename $us$ to $u$, since it's more common to work with single-character variable names in math (and it makes it easier to read). The equation is $$\alpha s^{\rho-1}\big[(1-\alpha)c^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1}=u.$$ Now, for every quintuple $(\alpha,s,c,\rho,\gamma)$, there is (exactly) one value of $u$ for which the equation is satisfied. This means that, for at least some values of $(\alpha,s,\rho,\gamma,u)$, there is some value $c$ that satisfies the equation: pick some real $c_0$ and some parameters $(\alpha,s,\rho,\gamma)$, set $$u_0=\alpha s^{\rho-1}\big[(1-\alpha)c_0^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1},$$ and then $c=c_0$ will be a root of the equation with parameters $(\alpha,s,\rho,\gamma,u_0)$.
This may not be that satisfying, since it doesn't tell you much about when a solution might exist. Let's try to solve the system. We can first divide by $(\alpha s^{\rho-1})$ to get $$\big[(1-\alpha)c^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1}=\frac{u}{\alpha s^{\rho-1}}.$$ Now, we can take each side to the power of $\frac{1-\gamma-\rho}{\rho}$ to get $$(1-\alpha)c^\rho+\alpha s^\rho = \left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}.$$ This gives $$c^\rho=\frac1{1-\alpha}\left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}-s^\rho\frac{\alpha}{1-\alpha}.$$ So, depending on your value of $\rho$, this might be solvable (if $\rho$ is an odd integer, there is a root somewhere, while otherwise you probably want the right side to be positive). In general, though, if $$\left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}\geq \alpha s^\rho,$$ then there is a (positive real) solution for $c$.
Edit (to explain the remark on how the solvability depends on $\rho$): Say $\rho$ is some arbitrary real number. Where is $c^\rho$ defined? A priori, it's hard to say how to define exponentation exactly -- it's easy when $\rho$ is an integer and we can just multiply $c$ together a bunch of times, but harder for arbitrary $\rho$. It turns out (and Desmos, for example, will give you a graph of this) that you can define $c^\rho$ for any real $\rho$, as long as $c$ is positive (for an example of why this constraint is reasonable, think about $(-1)^{1/2}$; it turns out we can define exponentiation $x^y$ for all reals, but we lose some nice properties, like the fact that $x^y$ is a real number). These values are always positive, so there is only a solution to $c^\rho=x$ in $c$ for arbitrary $\rho$ when $x$ is positive.
If we're in the special case where $\rho$ is an integer, then we can define $c^\rho$ for all real $c$, and then try to solve for $c$. If $\rho$ is even, $c^\rho\geq 0$ for all $c$, and so we can't solve $c^\rho=x$ for negative $x$. However, if $\rho$ is odd, we can!