Approximating the inverse of an exponent equation

Question

Let $$m=a_{1}n^{\alpha}+a_{2}n^{\beta}$$ where $1>\alpha>\beta>0$ and $a_{1},a_{2}$ are positive constants, and we want to understand $n$ as a function of $m$ , the first order is clearly $$n=a_{1}^{-\frac{1}{\alpha}}m^{\frac{1}{\alpha}}$$ but how can we fine the sceond order and the error magnitude, for large $m$. Any help would be much appreciated.

Answer 1

One may solve to get

$$n=a_1^{-1/\alpha}(m-a_2n^\beta)^{1/\alpha}$$

and with an initial estimate of $n=0$, we get using fixed-point iteration the first few expansions can be turned into asymptotic expansions with the generalized binomial expansion theorem:

\begin{align} n&\approx a_1^{-1/\alpha}m^{1/\alpha}\\\\ n&\approx a_1^{-1/\alpha}(m-a_2a_1^{-\beta/\alpha}m^{\beta/\alpha})^{1/\alpha}\\ &\approx a_1^{-1/\alpha}m^{1/\alpha}-\alpha^{-1}a_2a_1^{-(\beta+1)/\alpha}m^{(\beta-\alpha)/\alpha} \end{align}

Since the sign changes, the error of the first approximation can be bounded by

$$|n-a_1^{-1/\alpha}m^{1/\alpha}|\le\alpha^{-1}a_2a_1^{-(\beta+1)/\alpha}m^{(\beta-\alpha)/\alpha}$$

for large $m$. Repeated iterations can give higher order terms but quickly become very messy to expand.

Answer 2

Consider that you look for the zero of function $$f(n)=a_{1}n^{\alpha}+a_{2}n^{\beta}-m$$ and that the first order is given by $$n_1=a_{1}^{-\frac{1}{\alpha}}m^{\frac{1}{\alpha}}$$ use one iteration of Newton method to get $$n_2=n_1-\frac{a_1 n_1^{\alpha }+a_2 n_1^{\beta }-m}{ a_1\alpha n_1^{\alpha -1}+a_2 \beta n_1^{\beta -1}}$$

Edit

I am not sure that $n_1$ could be consider as the best first order approximation since we could have $a_2 \gg a_1$. I would prefer to write $$(a_{1}+a_{2})n^{\beta} < m <(a_{1}+a_{2})n^{\alpha}$$ This gives two bounds $n_1$ and $n_2$ which bracket the solution. Trying to interpolate linearly, write for example

$$a_{1}n_1^{\alpha}+a_{2}n_1^{\beta}=A +B\, n_1$$ $$a_{1}n_2^{\alpha}+a_{2}n_2^{\beta}=A +B\, n_2$$ Express $A$ and $B$ (this is simple) and generate the first estimate solving $$A +B\, n =m \implies n_0=\frac {m- A} B $$

You could prefer to interpolate in a logarithmic scale and write $$\log\left(a_{1}n_1^{\alpha}+a_{2}n_1^{\beta}\right)=C +D\,\log(n_1)$$ $$\log\left(a_{1}n_2^{\alpha}+a_{2}n_2^{\beta}\right)=C +D\,\log(n_2)$$ and solve $$C+D\log(n)=\log(m)\implies n_0=\frac {\log(m)-C} D$$ Now, use one iteration of Newton method to generate a better one.

Example

Using $a_1=123$, $a_2=456$, $\alpha=0.754$, $\beta=0.543$, $m=13579$, we should get $n_1=65.65$ and $n_2=333.71$.

The linear interpolation gives $A=4071.04$ and $B=49.3023$ then $n_0=192.850$.

The logarithmic interpolation gives $C=6.23919$ and $D=0.635108$ then $n_0=174.147$.

Now, using Newton method on a linear basis would give as first interate $n_1=174.806$. Doing the same on a logarithmic basis would give as first interate $n_1=174.100$.

The exact solution is $n=175.100$