It is possible to compute the solutions to the numerical problems in this text without the use of a calculator. In fact, depending upon the problem and your ingenuity, accuracy to within a few percent is often achievable. The benefits of doing arithmetic without a calculator are many:

- increased quantitative reasoning skills,
- increased numerical literacy (understanding what the values of numbers really mean), and
- increased reliability (after some practice!).

The key to this approach lies in the principle of **approximation**. Although you have been trained in your mathematics courses to believe that there is one exactly correct answer to every problem, this is seldom true in physics. With the exceptions of unitless numbers like

every number in physics is a measured quantity, and no measurement is exact. For the purposes of this text, every decimal number can be approximated using relatively simple fractions and their multiples:π("pi", the exact ratio of the circumference to the diameter of a perfect circle, whose value is approximately 3.1415927, or about 22/7) and

e(the base of the natural logrithms, whose value is approximately 2.7182818, or about 19/7),

fraction | decimal equivalent | used to approximate |
---|---|---|

1/11 | .0909090909 | .09 |

1/10 | .1 | .10 |

1/9 | .11111111 | .11 |

1/8 | .125 | .12 or .13 |

1/7 | .14285714 | .14 |

1/6 | .16666666 | .17 |

1/5 | .2 | .20 |

1/4 | .25 | .25 |

1/3 | .33333333 | .33 |

1/2 | .5 | .50 |

For example, suppose that you need to compute the quotient 381 / 42.7. Your first step is to express these numbers in a modified version of scientific notation:

.381 * 10You then notice that^{3}/ (.427 * 10^{2}).

.381 is close to .375, which is 3/8, and thatSo you approximate this computation (after dividing out factors of 10) by (3/8 * 10) / (3/7). Further simplifying, our approximate answer is then 7/8 * 10, or 8.75..427 is almost exactly 3/7.

It is important to quantify the error made in such an approximation. We define the **absolute error** as the absolute value of the difference between the exact answer and the approximate answer:

ΔxIn the example above, the "exact" answer (to 8 digits of accuracy) is 8.9227166. This means that for our approximation, the absolute error was .1727166. We can also define the_{abs}= | x_{exact}- x_{approximate}|.

ΔxFor our approximation, the relative error was approximately 1.94 %. Since the numerical problems in this text allow up to a 10% error in your answer, this level of accuracy is more than enough for our purposes._{rel}= 100 % * Δx_{abs}/ x_{exact}.

Computation of the absolute and relative errors are very useful in evaluating experimental data. Suppose you conduct an experiment eight times and obtain the following data points:

9.61350, 9.90505, 9.50567, 9.67618, 10.02710, 9.74638, 9.74732, 9.94688.You did not get the same result each time because there was

0.15751, 0.13404, 0.26534, 0.09483, 0.25609, 0.02463, 0.02369, 0.17587.The maximum of these represents the absolute error in the experiment: all data values are then consistent with

9.77101 + or - 0.26534that is, the range

(9.50567, 10.03635).The relative error is computed in the same way, using the average as the exact value, to obtain

9.77101 + or - 2.71558 %.Notice that the absolute error implies that you actually have no significant digits of precision, but the relative error gives a more accurate description of the quality of the results.

You can also compute squares and roots and other powers without a calculator, but you first must know the following perfect squares:

and the approximate square root of two:

2^{1/2} = 1.414 = 10/7

These facts serve as guideposts for your computations. Suppose we need to square 37. Before beginning, it is important to have an intuitive feel for what sort of answer you *should* get. Since 37 is between 30 and 40,
37^{2} is between 30^{2} and 40^{2}:

30This may seem like a pretty wide margin, but if you get something outside of this range, you know you have made a mistake. This is an example of the sorts of benefits which accrue from doing your own arithmetic: if you make a mistake on your calculator, and do not have any idea of the sort of answer you should end up with, you have no way to know you made a mistake. This is the single most common cause of student errors on tests (other than not studying!).^{2}< 37^{2}< 40^{2}, or3

^{2}* 10^{2}< 37^{2}< 4^{2}* 10^{2}, or900 < 37

^{2}< 1600.

To actually, perform the computation, we use the **binomial expansion**:

(a + b)and write^{2}= a^{2}+ 2 a * b + b^{2}

37Note that this computation was exact: no approximation was involved.^{2}= (30 + 7)^{2}= 30^{2}+ 2 * 30 * 7 + 7^{2}= 900 + 2 * 210 + 49 = 900 + 420 + 49 = 1369.

Binomial expansions have other uses in approximation as well. For cubes, use

(a + b)and for fourth powers,^{3}= a^{3}+ 3 a^{2}* b + 3 a * b^{2}+ b^{3},

(a + b)is helpful. A special use for the latter equation occurs when you have occasion to subtract^{4}= a^{4}+ 4 a^{3}* b + 6 a^{2}* b^{2}+ 4 a * b^{3}+ b^{4}

xwhere x is just a little larger than y. Using the binomial expansion above, we have^{4}- y^{4},

xAs long as c is much smaller than y, all of the terms after the first are at least an order of magnitude smaller, so^{4}- y^{4}= (y + c)^{4}- y^{4}= y

^{4}+ 4 y^{3}* c + 6 y^{2}* c^{2}+ 4 y * c^{3}+ c^{4}- y^{4}= 4 y

^{3}* c + 6 y^{2}* c^{2}+ 4 y * c^{3}+ c^{4}.

4 ybecomes an excellent approximation for the original difference.^{3}* c

We can also compute approximate square roots by utilizing the following observation: if in the binomial expansion,

b < < a(read "b is much less than a"), then as an approximation we can ignore the b

(a + b)Now suppose that we need the square root of x, and that we can write x as the product of two numbers which are very close together, differing only by a small number, which we shall call 2b:^{2}= a^{2}+ 2 a * b.

x = a * (a + 2b).Then

x = aand therefore^{2}+ 2 a * b= (a + b)

^{2},

x^{1/2}= a + b.

As an example, suppose we need the square root of 13,421. Before we start, we note that this is 1.3421 * 10^{4}, so its square root will be 1.3421^{1/2} * 100, or between 100 and 200. Now it is too hard to factor 13,421, but it is very easy to factor 13,400:

13,400 = 134 * 100.

Using our technique, we write this as

100 * (100 + 34),so with a = 100 and 34 = 2b, we see that the square root of 13,421 should be close to 117. In fact, 117

100% * | 115.84904 - 117 | / 115.84904 = .993 %This is very good, but in general your results will be directly related to your ability to find two factors which are very nearly equal. This is more important as the numbers get smaller. As in all of these approximation techniques, practice improves your performance greatly!

Of course, if graphs are available you should make use of them:

- a "3-4-5" right triangle has an approximately 53 degree angle,
- a "30-60-90" triangle has sides whose lengths are in the ratio 1 : 3
^{1/2}: 2 and - an isoceles triangle has sides whose lengths are in the ratio 1 : 2
^{1/2}: 1.

from which, with our basic knowledge of the trigonmetric functions, we can build the following table:

Note that 2

θsin θcos θtan θ0 0 1 0 30 .5 3 ^{1/2}/ 21 / 3 ^{1/2}37 .6 .8 .75 45 2 ^{1/2}/ 22 ^{1/2}/ 21 53 .8 .6 4 / 3 60 3 ^{1/2}/ 2.5 3 ^{1/2}90 1 0 (infinity)

Suppose that you need the sine of 35 degrees. It turns out that within a 10% margin of error,

Sin x = xfor angles from 0 to 42 degrees (about .75 radians). Therefore the sine of 35 degrees is

35 * π / 180The last step is an example of=35 * 22 / (7 * 180)= 11 / 18

= something between 5/9 (.556) and 6/9 (.667)

= .556 + (.667 - .556) / 2

= .61

The sine of 35 degrees is approximately .57357644, so the error in this computation is about 7%. With this approximation, as well as with the others shown below, the error grows with the argument: **these approximations are much better for small values**. This is because they are the leading term(s) of an **infinite series** of terms, whose subsequent terms constitute smaller and smaller corrections to the leading term.

cos θ = sin (π/2 - θ),this means that the cosine of 55 degrees is also approximately .61 (using our approximation above). If you need the cosine of 35 degrees, you can always use the identity

costo find that^{2}θ + sin^{2}θ = 1

cos 35 = (1 - (.6 + .01) * .61)(using some of our previous techniques as well). Cos 35 is actually approximately .81915204, so our error is about 8%.^{1/2}= (1 - .427)^{1/2}= .573

^{1/2}= 57.3

^{1/2}/ 10= .75

For angles less than 42 degrees, you can also use the approximation

cos θ = 1 - xwhich in this example yields^{2}/ 2,

cos 35 = 1 - (11/18)This is of course much better because we did not rely on an approximation for sin 35.^{2}/ 2= 1 - 121 / (324 * 2)= 1 - 121/648

= 1 - 1/5.4

= 1 - (something between 1/5 and 1/6)

= 1 - .18

= .82.

Tangents can be computed using the identity

tan θ = sin θ / cos θ.In all cases, it is important to know the behaviors of these functions:

sin -θ = - sin θas well as their domains and ranges. The identities for the sum or difference of two angles are also handy:

cos θ = cos θ

tan -θ = - tan θ

sin (θ ± φ) = sin (θ) cos (φ) ± cos (θ) sin (φ)and

cos (θ ± φ) = cos (θ) cos (φ) -+ sin (θ) sin (φ).

Finally, it will be useful to know that for values of x up to .9,

tanto within 10% accuracy.^{-1}x = x - x^{3}/ 3

The absolute error in u is 5, and its relative error is 5%; the absolute error in v is 3, and its relative error is 20%.
Consider the quantity f = u v^{2}: it can range from 13,680 to 34,020. The relative error in f is then 47%!
It is clear that the error has grown considerably.

For our purposes, the **root-mean-square** (or **RMS**) error will provide a good way to approximate the propagated error in such a calculation.
Suppose that A, B and C are exact quantities (ie., constants) and u and v are measured quantities. If

f1 = A u + B v + C,the relative error in f1 is

f1For_{rel}= √ (A^{2}u_{rel}^{2}+ B^{2}v_{rel}^{2}).

f2 = A uthe relative error in f2 is^{ B}v^{ C},

f2So the error in a function which is computed from uncertain quantities has a form which depends on the type of computation you are doing. The latter formula (where A is a, B is 1 and C is 2), gives a relative error for our previous example of 40%, commensurate with the 47% error computed by hand._{ = A √ (B2 urel2 + C2 vrel2). }

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©2012, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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