# Talk:MyOpenMath/Solutions/Big-O

## Dropping a higher order term

In my proof that the complicated formula for nλ reduces to s sinθ I chose the big-O notation over the Taylor expansion. Halfway through I realized that I was going to unnecessary high order because S=R+O-ε to some power. That lack of forethought caused me to include terms which would only confuse the reader. Here I place what I think is the extra terms that were deleted from the essay:

### Copied from deleted version of resource page

In order to ensure that the first order calculation is sufficient in the MyOpenMath version of this question, the software verifies that the first order solution is within 10% of the true answer. It is convenient to define $R={\sqrt {L^{2}+y^{2}}},$ so that:

$r_{1}={\sqrt {L^{2}+(y-d/2)^{2}}}={\sqrt {R^{2}-yd+d^{2}/4}}$ $r_{2}={\sqrt {L^{2}+(y+d/2)^{2}}}={\sqrt {R^{2}+yd+d^{2}/4}}$ Note from the figure that $y/R=\sin \theta$ , and that the two paths are effectively parallel when $d< . The exact formula for the path difference is:

$\Delta R=r_{2}-r_{1}={\sqrt {T^{2}+yd}}-{\sqrt {T^{2}+yd}},$ where,

$T={\sqrt {L^{2}+y^{2}+{\frac {d^{2}}{4}}}}={\sqrt {R^{2}+{\frac {d^{2}}{4}}}}$ Apparently, if the two paths are nearly parallel, we should be able to show that:

$\Delta R\approx {\frac {yd}{R}}=d\sin \theta$ .

To verify this we can perform a Taylor series of $\Delta R=\Delta R(y),$ or equivalently use this expansion for small $\epsilon$ :

$(1+\epsilon )^{1/2}=1+{\frac {\epsilon }{2}}-{\frac {\epsilon ^{2}}{8}}+{\frac {\epsilon ^{3}}{16}}+{\mathcal {O}}(\epsilon ^{4})$ The first four terms on the RHS refer to the zeroth, first, second, and third order terms, respectively. The last (fifth-order) term will be briefly discussed but not calculated.

Replacing $\epsilon$ by $-\epsilon$ into the aforementioned expression, we see that the zeroth, second, and fourth order terms cancel when we subtract:

$(1+\epsilon )^{1/2}-(1-\epsilon )^{1/2}=$ $1+{\frac {\epsilon }{2}}-{\frac {\epsilon ^{2}}{8}}+{\frac {\epsilon ^{3}}{16}}+{\mathcal {O}}(\epsilon ^{4})$ $-\left(1-{\frac {\epsilon }{2}}-{\frac {\epsilon ^{2}}{8}}-{\frac {\epsilon ^{3}}{16}}+{\mathcal {O}}(\epsilon ^{4}\right))$ $\therefore \quad (1+\epsilon )^{1/2}-(1-\epsilon )^{1/2}=\epsilon +{\frac {\epsilon ^{3}}{8}}+{\mathcal {O}}(\epsilon ^{5})$ I replaced ${\mathcal {O}}^{4}$ by ${\mathcal {O}}^{5}$ because it is obvious that the fourth order terms also cancel due to the subtraction. In order to obtain a useful value for our small parameter $\epsilon$ , we divide our expression for $\Delta R$ by $T$ :

${\frac {\Delta R}{T}}={\frac {{\sqrt {T^{2}+yd}}-{\sqrt {T^{2}+yd}}}{T}}={\sqrt {1+{\frac {yd}{T^{2}}}}}=(1+\epsilon )^{1/2}$ where we have defined the small parameter,

$\epsilon ={\frac {yd}{T^{2}}}={\frac {yd}{R^{2}+d^{2}/4}}={\frac {yd}{R^{2}}}\left[1+\left({\frac {d}{2R}}\right)^{2}+{\mathcal {O}}\left({\frac {d}{2R}}\right)^{4}\right]$ 