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    How to Approximate Like A Boss

    1. Find out what 2.0015, 8.62/3, e-0.15, and 1/1200 are approximately without using a calculator.

    We use a really simple method that works well for polynomials and describe how we can determine how well our approximation does relative to the actual answer.

    Table Of Content

    1. Prologue – Brook Taylor the mathemagician

    Let’s see an example

    2. Now let’s go through some actual problems

    2.0015

    8.62/3

    e-0.16

    1/1200

    1. Prologue – Brook Taylor the mathemagician

    In the 17th century, Brook Taylor stole a bunch of math from even more badass mathemagicians before him and published a method under which any smooth analytical functions can be expressed as polynomials. In the end, no one remembered anything about Taylor (not even his first name) except for the famous “Taylor expansion” for which he is named after. Worry not, we do not need to dwell upon him any further, all we need is the following equation.

    f(a + b) = f(a) + b \times \left.\dfrac{\partial f(x)}{\partial x}\right|_{x = a} + \mathbb{O}(b^2)

    Or more succinctly, we define the first order taylor expansion of a function f(x) as

    f(a + b) \approx f(a) + b f'(a)

    Let’s see an example

    Let f(x) = x^2, and we want to approximate f(x + \epsilon) with a first order taylor expansion. Then by dumbly plugging the necessary variables into the above expression (a = x, b = \epsilon), we find that

    f(x + \epsilon) = (x + \epsilon)^2 \approx f(x) + \epsilon f'(x) = x^2 + 2 \epsilon x

    Since (x + \epsilon)^2 = x^2 + 2 \epsilon x + \epsilon^2, then for $latex \|\epsilon\| < 1$, we can see that the approximation is pretty reasonably close.

    2. Now let’s go through some actual problems

    The year was 2012, and let me tell you, I had one hell of a physics professor, Seamus Davis, who always expected absolute perfection from all of his students in his Waves course. Here, absolute perfection can be defined as accurate to within two significant figures. Needless to say, the students loved him.

    We were allowed to bring in a crude and unruly 100+ function non-graphing (just barbaric) calculator to the exams, and I, always being the dumbass, forgot to bring mine to the final exam. Fortunate for me, I didn’t have to do any of these dumb approximations because Seamus Davis isn’t a dick. He let me borrow one of his calculators.

    Anyways, let’s get to some actual real life problems that someone asked on reddit.

    The general rule is to pick the epsilon such that \mathbb{O}(\epsilon^2 f''(\xi)) will be reasonably small, where \xi is some parameter that yields the largest derivatives within the general vicinity around x.

    2.0015

    Let f(x) = x^5, find f(2 + 0.001)

    f(x + \epsilon) \approx x^5 + 5\epsilon x^4 = 32 + 0.005 \times 16 = 32.08

    Compare this to the actual answer 32.08008004001, we see that we’re pretty damn close.

    8.62/3

    Let f(x) = x^{\frac{2}{3}}, find f(8 + 0.6)

    f(x + \epsilon) \approx x^{\frac{2}{3}} + \frac{2}{3} \epsilon x^{-\frac{1}{3}} = 4 + \frac{2}{3} 0.6 \frac{1}{2} = 4.2

    Considering that 8.62/3 is 4.1975798, we can see that we are again pretty darn close.

    e-0.16

    Let f(x) = e^x, find f(0 + -0.16)

    f(x + \epsilon) \approx e^{x} + \epsilon e^x= 1 - 0.16 = 0.84

    However, the actual solution to e-0.16 is 0.8521437, which means that we’re only accurate to a single sig fig. Why were the other approximations so much seemingly better? From wikipedia, we see that the error can be approximated as \frac{\epsilon^2}{2} f''(x) = \frac{0.16^2}{2} e^0 \approx 0.012, which we see is the difference between the actual solution and our approximation.

    There are a few things we can do to refine our approximation.

    First, we can use the approximation that $latex e^{-x} = 1 – x\ \forall\ \|x\| << 1$ in order to make a better approximation using f(-0.16) = f(-0.10 + -0.06) = 0.90 - 0.06*0.90 = 0.846. Alternatively, as seen above, we can also simply add in the error term.

    1/1200

    Let f(x) = 1/x, find f(1000 + 200)

    f(x + \epsilon) \approx \frac{1}{x} - \epsilon \frac{1}{x^2}= 0.001 - 200\times 0.000001 = 0.0008

    This also has an approximate error of \frac{f''(x)}{2} \epsilon^2 = \frac{-2}{-2x^3}\epsilon^2 = \frac{200^2}{1000^3}

    While the actual solution is 0.0008\overline{33}

    Posted By Lee

    Woosh