Question:
$$(3^\frac{2}{7}-2^\frac{3}{7}) * 112 = ?$$
How should i approach such question ? First look, i tried to make power fractions equal for both the individual expressions. Like the following:
$$3^\frac{2}{7}-2^\frac{3}{7}=9^\frac{1}{7}-8^\frac{1}{7}$$
Now this looks like :
$$(a^x-b^x) or (a^x-(a-1)^x)$$
or may be :
$$a^x*(1-(\frac{a-1}{a})^x)$$
or instead of doing this, should i go with approximation by considering $9^\frac{1}{7}$ close to $9^\frac{1}{8}$ and then taking square root 3 times ?
or is there any specific reason to multiply this expression by 112 ?
Need some help figuring out the solution.
I do not want to use any calc for this and am trying to figure out a few possible approaches which can give me an approximate result with atleast 2 decimal point accuracy if possible.
If my life depended on it, I would probably compute the seventh roots of $8$ and $9$ in base $2$ with the root algorithm used in written computation. Omitting details (the algorithms is actually a dichotomic search), the crucial step in the algorithm is to compute
$$(2x+1)^7,$$ where $x$ is an approximation. You can do this using the Binomial expansion (and noting that multiplying by a power of $2$ is a mere shift). The coefficients are $1,7,21,35,35,21,7,1$ (decimal).
But I guess that the computation via $y^7=y\,y^2(y^2)^2$ will be faster/simpler.
Next you convert to base $10$ and multiply by $112$. A first step is to determine the number of significant bits required. Count two of three hours of work.
Alternatively, the seventh roots can be evaluated by Newton-Raphson, but this will take the evaluation of sixths powers and divisions.
Another interesting alternative is the secant method, starting from the inequalities
$$1.3^7=6.2748517<8<9<1.4^7=10.5413504.$$
As the curve is pretty flat, the next estimates are obtained by linear interpolation. Optionally, when a new approximation is obtained, it can be truncated to a smaller number of decimals to ease the evaluation of the seventh power.
For example, the next approximation of $\sqrt[7]8$ is $1.340434755\cdots$, and you can use $1.34$ instead.
If you have a fast way to compute the square roots, by some magic, you can use the fixed-point iteration
$$x\leftarrow \sqrt{\sqrt{\sqrt{yx}}},$$ the solution of which is $x=y^{1/7}$.