1

The core problem was simple:

Determine the interior angles of triangle ABC given side $a = 12.34cm$ and hypotenuse $c = 35.32cm$.

Simple. Clearly, the solution is:

$\sin A = \frac{12.34cm}{35.32cm}; A \approx 20.45º$

$\cos B = \frac{12.34cm}{35.32cm}; B \approx 69.55º$

Now, his argument was that if we round $A$ and $B$ to the same number of significant figures as our input (4 significant figures), in some cases, $A + B \neq 90.00º$ due to "rounding error". As a result, you should calculate $A$, then calculate $B = 90º - A$ "to ensure $A + B = 90º$".

He claims he's graded assignments in which this happened, but I'm skeptical. What would be the point of sig figs if they only work part of the time? I'm trying to find a way to formally disprove his statement. I understand the intuition behind how significant figures would prevent this from happening.

I have a feeling that this would generalize to a proof that for any $a$, $b$, and $z$ such that

$z = a + b$

the following statement (with errors $\epsilon$ and $\omega$ determined by the number of significant figures) will hold:

$z + (\epsilon + \omega) = (x + \epsilon) + (y + \omega)$

Edit to clarify:

Clearly $\arcsin x + \arccos x = 1$ exactly.

The professor is saying that, given inputs with, say, 4 significant figures, calculating $\arcsin x$ would give you, say, $59.34º$. Then, calculating $\arccos x$ could give you something like $30.67º$ (obviously this doesn't work, but he claims some such $x$ exists). Thus, he recommended that we instead just calculate $\arcsin x$ and then subtract it from 90º to calculate $\arccos x$.

Steven Petryk
  • 113
  • 1
    If you round to whole numbers first, $44.5+45.5 \approx 45 + 46 \neq 90$. – Paul May 20 '16 at 13:58
  • $44.5 + 45.5 \approx 90.0$, since the number of significant figures needs to match for both the addends and sum. – Steven Petryk May 20 '16 at 13:59
  • 2
    Say $a=1/8=0.125$, $b=7/8=0.675$, so $a+b=1$. If we round these to two significant figures, we have $a \approx 0.13$, $b \approx 0.68$, so $a+b \approx 1.01$. The rounded figures don't add exactly to $1$. (If you prefer, you can write $a=1.0/8.0$, $b=7.0/8.0$ if you're assuming these correspond to some measurements that are accurate to two significant figures.) – kccu May 20 '16 at 14:00
  • In this case, $a + b \approx 1.0$, since you'd need to express the answer with two significant figures. – Steven Petryk May 20 '16 at 14:01
  • 1
    The point of sig figs is a limitation of measurement resolution. If you're just doing math it doesn't matter. Once you apply it to the real world, sig figs have to be accounted for. – Paul May 20 '16 at 14:02
  • But significant figures rules are still based on provable mathematical principles, and he's wrong to say that they can't be relied upon. – Steven Petryk May 20 '16 at 14:03
  • @StevenPetryk "For quantities created from measured quantities by addition and subtraction, the last significant decimal place (hundreds, tens, ones, tenths, and so forth) in the calculated result should be the same as the leftmost or largest decimal place of the last significant figure out of all the measured quantities in the terms of the sum." From https://en.wikipedia.org/wiki/Significant_figures – kccu May 20 '16 at 14:04
  • @StevenPetryk what probable mathematical principles? – Stella Biderman May 20 '16 at 14:04
  • I could also give a different example: $a=1.0/8.0=0.125 \approx 0.13$ and $b=5.0/8.0 =0.625 \approx 0.63$. Then $a+b=6.0/8.0=0.75$, but $0.13+0.63=0.76$. The point is, rounding errors do happen, even when using significant figures. This grader or professor is correct. – kccu May 20 '16 at 14:06
  • Why is it surprising that doing a calculation with the wrong numbers gives you the wrong answer? – preferred_anon May 20 '16 at 14:07
  • @StevenPetryk The phenomenon you describe in the question happens all the time (on a more blatant way when more than two numbers are summed, but here you are actually calculating trig functions, which is far worse...). Due to the real-word interest of this, there are in fact a disciplines that study it systematically. –  May 20 '16 at 14:09
  • @kccu, in this case, you're doing intermediate rounding, which is going to introduce imprecision early on. – Steven Petryk May 20 '16 at 14:09
  • @StellaBiderman, see http://math.stackexchange.com/questions/208133/what-is-the-proof-of-the-rules-of-significant-figures – Steven Petryk May 20 '16 at 14:09
  • Did the prof actually explicitly say what is mentioned in your title or are you paraphrasing? I'm wondering if what he meant was that rounding too much too soon (like in the examples @kccu is providing) can lead to errors. Note that this is different from rounding the final answers to the same number of sig figs while keeping more digits in intermediate calculations. –  May 20 '16 at 14:09
  • I made sure to ask a few questions to probe as to whether that's really what he was saying. – Steven Petryk May 20 '16 at 14:11
  • @StevenPetryk But the point is that you are also rounding before adding the results. If you don't round before adding, then mathematically the result must be exactly $90$. – kccu May 20 '16 at 14:14
  • @kccu that's what I'm saying! I'm saying that he said even if you round at the very end, they could still get you an answer not equal to 90 (or 90.0 or 90.00 or 90.000, etc). i.e., you could get 90.01. See my edit for clarity. – Steven Petryk May 20 '16 at 14:17
  • @StevenPetryk In that case, the $\arcsin$ and $\arccos$ would have to both be rational and end in a $5$ (so that both results get rounded up and produce an error in the sum). Because of the nature of the $\arcsin$ and $\arccos$ functions, I doubt that any rational inputs would give such a result. – kccu May 20 '16 at 14:35

0 Answers0