I have an exact equation:

a=√1+x2−√(1+x2)2−2x2cos2θ\begin{equation}

a = \sqrt{1 + x^2 – \sqrt{(1 + x^2)^2 – 2 x^2 \cos^2\theta}}\tag{1}

\end{equation}

and its approximation:

a=xcos2θ√1+x2\begin{equation}

a = \dfrac{x \cos^2\theta}{\sqrt{1 + x^2}}\tag{2}

\end{equation}

I’m doing most theoretical calculations using equation (2)(2) and to check the outcomes I’m doing numerical calculations of (1)(1) in Mathematica.

For example, finding dadx|x=0\left.\frac{da}{dx}\right|_{x = 0} dividing by dadθ|θ=0\left.\frac{da}{d\theta}\right|_{\theta = 0} and integrating:

A∫0dadx|θ=0dadθ|θ=0dx\begin{equation}

\int\limits_{0}^{A}\frac{\left.\frac{da}{dx}\right|_{\theta = 0}}{\left.\frac{da}{d\theta}\right|_{\theta = 0}}dx

\end{equation}

And I want to check that this (or any other) procedure doesn’t differ much from the same for precise and approximate function.

How can I verify that the use of the approximate formula doesn’t make the calculations inaccurate?

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A slightly more accurate approximation might be had by using xcosθ√1+x2\dfrac{x\cos\theta}{\sqrt{1+x^2}} 11−(xcosθ2(1+x2))2\dfrac1{1-\left(\frac{x\cos\theta}{2(1+x^2)}\right)^2}…

– J. M.♦

Nov 25 ’12 at 9:45

1

x|cosθ|√1+x2\dfrac{x\lvert\cos\theta\rvert}{\sqrt{1+x^2}} seems to be pretty good to me.

– Rahul

Nov 25 ’12 at 11:01

Looks more like a math question.

– Jens

Nov 25 ’12 at 21:40

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2 Answers

2

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I don’t understand how you get your approximate expression. I’d do a series expansion of your expression and check how well it approximates it :

exact2[x_, y_] = Sqrt[1 + x^2] Sqrt[1 – Sqrt[1 – 2 (y/(1 + x^2))^2]];

Simplify[exact2[x, x Cos[\[Theta]]], Assumptions -> {x \[Element] Reals, \[Theta] \[Element] Reals}] ==

Simplify[exact[x, \[Theta]], Assumptions -> {x \[Element] Reals, \[Theta] \[Element] Reals}]

(* True *)

appr[x_, \[Theta]_] = Simplify[Normal[Series[exact2[x, y], {y, 0, 2}]],

Assumptions -> {x \[Element] Reals}] /. y -> (x Cos[\[Theta]])

(* (x Cos[\[Theta]])/Sqrt[1 + x^2] *)

You can check how well the approximation is valid for xcos(θ)≈0x \cos(\theta) \approx 0 :

RegionPlot[Abs[exact[x, \[Theta]] – appr[x, \[Theta]]] < 0.001, {x, -5, 5}, {\[Theta], -\[Pi], \[Pi]} , FrameTicks -> {Automatic, {-\[Pi]/2, \[Pi]/2}}, FrameLabel -> {“x”, “\[Theta]”}, PlotPoints -> 50]]

exact[x_, \[Theta]_] :=

Sqrt[1 + x^2 – Sqrt[(1 + x^2)^2 – 2 x^2 Cos[\[Theta]]^2]]

approximate[x_, \[Theta]_] := x Cos[\[Theta]]^2/Sqrt[1 + x^2]

I’m not sure what kind of verification you want, or what your values are, but you could make a plot to get an idea (now with relative error as @J.M suggested)

absoluteError[x_, \[Theta]_] :=

exact[x, \[Theta]] – approximate[x, \[Theta]]

relativeError[x_, \[Theta]_] :=

1 – approximate[x, \[Theta]]/exact[x, \[Theta]] // Abs

Plot3D[absoluteError[x, \[Theta]], {x, -0.2, 0.2}, {\[Theta], -Pi, Pi},

AxesLabel -> (Style[#, 20] & /@ {“x”, “\[Theta]”})]

Plot3D[relativeError[x, \[Theta]], {x, -0.02, 0.2}, {\[Theta], -Pi,

Pi}, AxesLabel -> (Style[#, 20] & /@ {“x”, “\[Theta]”}),

PlotRange -> {0., 0.2}]

I’ve updated the question. I think this is now much easier to understand.

– m0nhawk

Nov 25 ’12 at 9:10

3

One might also consider looking at the relative error in addition to the absolute error…

– J. M.♦

Nov 25 ’12 at 9:37