Replacing the calculator with a computer

Let us consider a basic calculator with buttons to add, subtract, multiply, divide, and take square roots. Using such a simple thing is certainly familiar for any reader of these notes. Indeed, a familiarity with a graphing calculator is expected. Julia makes these familiar tasks just as easy, offering numerous conveniences along the way. In this section we describe how.

The following image is the calculator that Google presents upon searching for "calculator."

Screenshot of a calculator provided by the Google search engine

This calculator should have a familiar appearance with a keypad of numbers, a set of buttons for arithmetic operations, a set of buttons for some common mathematical functions, a degree/radian switch, and buttons for interacting with the calculator: Ans, AC (also CE), and =.

The goal here is to see the counterparts within Julia to these features.

For an illustration of really basic calculator, have some fun watching this video:


Performing a simple computation on the calculator typically involves hitting buttons in a sequence, such as "1", "+", "2", "=" to compute 3 from adding 1 + 2. In Julia, the process is not so different. Instead of pressing buttons, the various values are typed in. So, we would have:

1 + 2

Sending an expression to Julia's interpreter - the equivalent of pressing the "=" key on a calculator - is done at the command line by pressing the Enter or Return key, and in IJulia using the "play" icon, or the keyboard shortcut Shift-Enter. If the current expression is complete, then Julia evaluates it and shows any output. If the expression is not complete, Julia's response depends on how it is being called. Within IJulia, a message about "premature" end of input is given. If the expression raises an error, this will be noted.

The basic arithmetic operations on a calculator are "+", "-", "×", "÷", and "$xʸ$". These have parallels in Julia through the binary operators: +, -, *, /, and ^:

1 + 2, 2 - 3, 3 * 4, 4 / 5, 5 ^ 6
(3, -1, 12, 0.8, 15625)

On some calculators, there is a distinction between minus signs - the binary minus sign and the unary minus sign to create values such as $-1$.

In Julia, the same symbol, "-", is used for each:

-1 - 2

An expression like $6 - -3$, subtracting minus three from six, must be handled with some care. With the Google calculator, the expression must be entered with accompanying parentheses: $6 -(-3)$. In Julia, parentheses may be used, but are not needed. However, if omitted, a space is required between the two minus signs:

6 - -3

(If no space is included, the value "--" is parsed like a different, undefined, operation.)



For everyday temperatures, the conversion from Celsius to Fahrenheit ($9/5 C + 32$) is well approximated by simply doubling and adding $30$. Compare these values for an average room temperature, $C=20$, and for a relatively chilly day, $C=5$:

For $C=20$:

9 / 5 * 20 + 32

The easy to compute approximate value is:

2 * 20 + 30

The difference is:

(9/5*20 + 32) - (2 * 20 + 30)

For $C=5$, we have the actual value of:

9 / 5 * 5 + 32

and the easy to compute value is simply $40 = 10 + 30$. The difference is

(9 / 5 * 5 + 32) - 40

Add the numbers $1 + 2 + 3 + 4 + 5$.

1 + 2 + 3 + 4 + 5

How small is $1/2/3/4/5/6$? It is about $14/10,000$, as this will show:


Which is bigger $4^3$ or $3^4$? We can check by computing their difference:

4^3 - 3^4

So $3^4$ is bigger.


A right triangle has sides $a=11$ and $b=12$. Find the length of the hypotenuse squared. As $c^2 = a^2 + b^2$ we have:

11^2 + 12^2

Order of operations

The calculator must use some rules to define how it will evaluate its instructions when two or more operations are involved. We know mathematically, that when $1 + 2 \cdot 3$ is to be evaluated the multiplication is done first then the addition.

With the Google Calculator, typing 1 + 2 x 3 = will give the value $7$, but if we evaluate the + sign first, via 1 + 2 = x 3 = the answer will be 9, as that will force the addition of 1+2 before multiplying. The more traditional way of performing that calculation is to use parentheses to force an evaluation. That is, (1 + 2) * 3 = will produce 9 (though one must type it in, and not use a mouse to enter). Except for the most primitive of calculators, there are dedicated buttons for parentheses to group expressions.

In Julia, the entire expression is typed in before being evaluated, so the usual conventions of mathematics related to the order of operations may be used. These are colloquially summarized by the acronym PEMDAS.

PEMDAS. This acronym stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. The order indicates which operation has higher precedence, or should happen first. This isn't exactly the case, as "M" and "D" have the same precedence, as do "A" and "S". In the case of two operations with equal precedence, associativity is used to decide which to do. For the operations +, -, *, / the associativity is left to right, as in the left one is done first, then the right. However, ^ has right associativity, so 4^3^2 is 4^(3^2) and not (4^3)^2. (Be warned that some calculators - and spread sheets, such as Excel - will treat this expression with left associativity.)

With rules of precedence, an expression like the following has a clear interpretation to Julia without the need for parentheses:

1 + 2 - 3 * 4 / 5 ^ 6

Working through PEMDAS we see that ^ is first, then * and then / (this due to associativity and * being the leftmost expression of the two) and finally + and then -, again by associativity rules. So we should have the same value with:

(1 + 2) - ((3 * 4) / (5 ^ 6))

If different parentheses are used, the answer will likely be different. For example, the following forces the operations to be -, then *, then +. The result of that is then divided by 5^6:

(1 + ((2 - 3) * 4)) / (5 ^ 6)



The percentage error in $x$ if $y$ is the correct value is $(x-y)/y \cdot 100$. Compute this if $x=100$ and $y=98.6$.

(100 - 98.6) / 98.6 * 100

The marginal cost of producing one unit can be computed by finding the cost for $n+1$ units and subtracting the cost for $n$ units. If the cost of $n$ units is $n^2 + 10$, find the marginal cost when $n=100$.

(101^2 + 10) - (100^2 + 10)

The average cost per unit is the total cost divided by the number of units. Again, if the cost of $n$ units is $n^2 + 10$, find the average cost for $n=100$ units.

(100^2 + 10) / 100

The slope of the line through two points is $m=(y_1 - y_0) / (x_1 - x_0)$. For the two points $(1,2)$ and $(3,4)$ find the slope of the line through them.

(4 - 2) / (3 - 1)

Two ways to write division - and they are not the same

The expression $a + b / c + d$ is equivalent to $a + (b/c) + d$ due to the order of operations. It will generally have a different answer than $(a + b) / (c + d)$.

How would the following be expressed, were it written inline:

\[ ~ \frac{1 + 2}{3 + 4}? ~ \]

It would have to be computed through $(1 + 2) / (3 + 4)$. This is because unlike /, the implied order of operation in the mathematical notation with the horizontal division symbol (the vinicula) is to compute the top and the bottom and then divide. That is, the vinicula is a grouping notation like parentheses, only implicitly so. Thus the above expression really represents the more verbose:

\[ ~ \frac{(1 + 2)}{(3 + 4)}. ~ \]

Which lends itself readily to the translation:

(1 + 2) / (3 + 4)

To emphasize, this is not the same as the value without the parentheses:

1 + 2 / 3 + 4

Infix, postfix, and prefix notation

The factorial button on the Google Button creates an expression like 14! that is then evaluated. The operator, !, appears after the value (14) that it is applied to. This is called postfix notation. When a unary minus sign is used, as in -14, the minus sign occurs before the value it operates on. This uses prefix notation. These concepts can be extended to binary operations, where a third possibility is provided: infix notation, where the operator is between the two values. The infix notation is common for our familiar mathematical operations. We write 14 + 2 and not + 14 2 or 14 2 +. (Though if we had an old reverse-Polish notation calculator, we would enter 14 2 +!) In Julia, there are several infix operators, such as +, -, ... and others that we may be unfamiliar with. These mirror the familiar notation from most math texts.


The Google calculator has two built in constants, e and π. Julia provides these as well, though not quite as easily. First, π is just pi:

π = 3.1415926535897...

Whereas, e is is not simply the character e, but rather a unicode character typed in as \euler[tab].

However, when the accompanying package, CalculusWithJulia, is loaded, the character e will refer to the Euler constant (as it brings in the Base.MathConstants package):

using CalculusWithJulia
ℯ = 2.7182818284590...

In the sequel, we will just use e for this constant (though more commonly the exp function), with the reminder that base Julia alone does not reserve this symbol.

Mathematically these are irrational values with decimal expansions that do not repeat. Julia represents these values internally with additional accuracy beyond that which is displayed. Math constants can be used as though they were numbers, such is done with this expression:


Numeric literals

For some special cases, Julia implements multiplication without a multiplication symbol. This is when the value on the left is a number, as in 2pi, which has an equivalent value to 2*pi. However the two are not equivalent, in that multiplication with numeric literals does not have the same precedence as regular multiplication - it is higher. This has practical importance when used in division or powers. For instance, these two are not the same:

1/2pi, 1/2*pi
(0.15915494309189535, 1.5707963267948966)

Why? Because the first 2pi is performed before division, as multiplication with numeric literals has higher precedence than regular multiplication, which is at the same level as division.

To confuse things even more, consider


Is this the same as 2 * (pi^2) * pi or (2pi)^(2pi)?. The former would be the case is powers had higher precedence than literal multiplication, the latter would be the case were it the reverse. In fact, the correct answer is 2 * (pi^(2*pi)):

2pi^2pi, 2 * (pi/2) * pi, (2pi)^(2pi), 2 * (pi^(2pi))
(2658.978166443007, 9.869604401089358, 103540.92043427199, 2658.97816644300

This follows usual mathematical convention, but is a source of potential confusion. It can be best to be explicit about multiplication, save for the simplest of cases.


On the Google calculator, the square root button has a single purpose: for the current value find a square root if possible, and if not signal an error (such as what happens if the value is negative). For more general powers, the $x^y$ key can be used.

In Julia, functions are used to perform the actions that a specialized button may do on the calculator. Julia provides many standard mathematical functions - more than there could be buttons on a calculator - and allows the user to easily define their own functions. For example, Julia provides the same set of functions as on Google's calculator, though with different names. For logarithms, $\ln$ becomes log and $\log$ is log10 (computer programs almost exclusively reserve log for the natural log); for factorials, $x!$, there is factorial; for powers $\sqrt{}$ becomes sqrt, $EXP$ becomes exp, and $x^y$ is computed with the infix operator ^. For the trigonometric functions, the basic names are similar: sin, cos, tan. These expect radians. For angles in degrees, the convenience functions sind, cosd, and tand are provided. On the calculator, inverse functions like $\sin^{-1}(x)$ are done by combining $Inv$ with $\sin$. With Julia, the function name is asin, an abbreviation for "arcsine." (Which is a good thing, as the notation using a power of $-1$ is often a source of confusion and is not supported by Julia.) Similarly, there are asind, acos, acosd, atan, and atand functions available to the Julia user.

The following table summarizes the above:


$+$, $-$, $\times$, $\div$

+, -, *, /



$\sqrt{}, \sqrt[3]{}$

sqrt, cbrt



$\ln$, $\log$

log, log10

$\sin, \cos, \tan, \sec, \csc, \cot$

sin, cos, tan, sec, csc, cot

In degrees, not radians

sind, cosd, tand, secd, cscd, cotd

$\sin^{-1}, \cos^{-1}, \tan^{-1}$

asin, acos, atan



Using a function is very straightforward. A function is called using parentheses, in a manner visually similar to how a function is called mathematically. So if we consider the sqrt function, we have:

sqrt(4), sqrt(5)
(2.0, 2.23606797749979)

The function is referred to by name (sqrt) and called with parentheses. Any arguments are passed into the function using commas to separate values, should there be more than one. When there are numerous values for a function, the arguments may need to be given in a specific order or may possibly be specified with keywords.

Some more examples:

exp(2), log(10), sqrt(100), 10^(1/2)
(7.38905609893065, 2.302585092994046, 10.0, 3.1622776601683795)

<script> // XXX This seems to be no longer needed, as 2^(-1) is defined when -1 is a numeric literal // // Julia's design embraces polymorphism, a term to indicate that the // same function may have different implementations and the one // ultimately called depends on the number and types of the // arguments. Polymorphism is also commonly referred to as multiple // dispatch. This is a great convenience for the user who only needs to // remember one function name for related uses, that may differ only in // technicalities. For example, there are about $200$ methods // implemented for the generic function +. This may be expected from // mathematical analogy: the details of the operation of addition are // different for integers, than rational numbers, than polynomials, // though all can be added. // // // // The power operator, ^, provides a concrete example. For integer // bases there is a different implementation than there is for real // bases. When the base is an integer, the answer may or may not be an // integer: // // julia; // 2^(-1), 2^2 // // // However, for floating point bases the answer is always a floating point number // // julia; // 2.0^(-1), 2.0^2 // </script>

Multiple arguments

For the logarithm, we mentioned that log is the natural log and log10 implements the logarithm base 10. As well there is log2. However, in general there is no logb for any base b. Instead, the basic log function can take two arguments. When it does, the first is the base, and the second the value to take the logarithm of. This avoids forcing the user to remember that $\log_b(x) = \log(x)/\log(b)$.

So we have all these different, but related, uses to find logarithms:

log(e), log(2, e), log(10, e), log(e, 2)
(1, 1.4426950408889634, 0.43429448190325176, 0.6931471805599453)

In Julia, the "generic" function log not only has different implementations for different types of arguments (real or complex), but also has a different implementation depending on the number of arguments.



A right triangle has sides $a=11$ and $b=12$. Find the length of the hypotenuse. As $c^2 = a^2 + b^2$ we have:

sqrt(11^2 + 12^2)

A formula from statistics to compute the variance of binomial random variable for parameters $p$ and $n$ is $\sqrt{p (1-p)/10}$. Compute this value for $p=1/4$ and $n=10$.

sqrt((1/4 * (1 - 1/4)) / 10)

Find the distance between the points $(-3, -4)$ and $(5,6)$. Using the distance formula $\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$, we have:

sqrt((5 - -3)^2 + (6 - -4)^2)

The formula to compute the resistance of two resistors in parallel is given by: $1/(1/r_1 + 1/r_2)$. Suppose the resistance is $10$ in one resistor and $20$ in the other. What is the resistance in parallel?

1 / (1/10 + 1/20)


Not all computations on a calculator are valid. For example, the Google calculator will display Error as the output of $0/0$ or $\sqrt{-1}$. These are also errors mathematically, though the second is not if the complex numbers are considered.

In Julia, there is a richer set of error types. The value 0/0 will in fact not be an error, but rather a value NaN. This is a special floating point value indicating "not a number" and is the result for various operations. The output of $\sqrt{-1}$ (computed via sqrt(-1)) will indicate a domain error:

ERROR: DomainError with -1.0:
sqrt will only return a complex result if called with a complex argument. Try sqrt(Complex(x)).

For integer or real-valued inputs, the sqrt function expects non-negative values, so that the output will always be a real number.

There are other types of errors. Overflow is a common one on most calculators. The value of $1000!$ is actually very large (over 2500 digits large). On the Google calculator it returns Infinity, a slight stretch. For factorial(1000) Julia returns an OverflowError. This means that the answer is too large to be represented as a regular integer.

ERROR: OverflowError: 1000 is too large to look up in the table; consider using `factorial(big(1000))` instead

How Julia handles overflow is a study in tradeoffs. For integer operations that demand high performance, Julia does not check for overflow. So, for example, if we are not careful strange answers can be had. Consider the difference here between powers of 2:

2^62, 2^63
(4611686018427387904, -9223372036854775808)

On a machine with $64$-bit integers, the first of these two values is correct, the second, clearly wrong, as the answer given is negative. This is due to overflow. The cost of checking is considered too high, so no error is thrown. The user is expected to have a sense that they need to be careful when their values are quite large. (Or the user can use floating point numbers, which though not exact, can represent much bigger values.)


Did Homer Simpson disprove Fermat's Theorem?

Fermat's theorem states there are no solutions over the integers to $a^n + b^n = c^n$ when $n > 2$. In the photo accompanying the linked article, we see:

\[ ~ 3987^{12} + 4365^{12} - 4472^{12}. ~ \]

If you were to do this on most calculators, the answer would be $0$. Were this true, it would show that there is at least one solution to $a^{12} + b^{12} = c^{12}$ over the integers - hence Fermat would be wrong. So is it $0$?

Well, let's try something with Julia to see. Being clever, we check if $(3987^{12} + 4365^{12})^{1/12} = 4472$:

(3987^12 + 4365^12)^(1/12)

Not even close. Case closed. But wait? This number to be found must be at least as big as $3987$ and we got $28$. Doh! Something can't be right. Well, maybe integer powers are being an issue. (The largest $64$-bit integer is less than $10^{19}$ and we can see that $(4\cdot 10^3)^{12}$ is bigger than $10^{36})$. Trying again using floating point values for the base, we see:

(3987.0^12 + 4365.0^12)^(1/12)

Ahh, we see something really close to $4472$, but not exactly. Why do most calculators get this last part wrong? It isn't that they don't use floating point, but rather the difference between the two numbers:

(3987.0^12 + 4365.0^12)^(1/12) - 4472

is less than $10^{-8}$ so on display with $8$ digits may be rounded to $0$.

Moral: with Julia and with calculators, we still have to be mindful not to blindly accept an answer.



Compute $22/7$ with Julia.


Compute $\sqrt{220}$ with Julia.


Compute $2^8$ with Julia.


Compute the value of

\[ ~ \frac{9 - 5 \cdot (3-4)}{6 - 2}. ~ \]


Compute the following using Julia:

\[ ~ \frac{(.25 - .2)^2}{(1/4)^2 + (1/3)^2} ~ \]


Compute the decimal representation of the following using Julia:

\[ ~ 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} ~ \]


Compute the following using Julia:

\[ ~ \frac{3 - 2^2}{4 - 2\cdot3} ~ \]


Compute the following using Julia:

\[ ~ (1/2) \cdot 32 \cdot 3^2 + 100 \cdot 3 - 20 ~ \]


Wich of the following is a valid Julia expression for

\[ ~ \frac{3 - 2}{4 - 1} ~ \]

that uses the least number of parentheses?


Wich of the following is a valid Julia expression for

\[ ~ \frac{3\cdot2}{4} ~ \]

that uses the least number of parentheses?


Which of the following is a valid Julia expression for

\[ ~ 2^{4 - 2} ~ \]

that uses the least number of parentheses?


In the U.S. version of the Office, the opening credits include a calculator calculation. The key sequence shown is 9653 + which produces 11532. What value was added to?


We saw that 1 / 2 / 3 / 4 / 5 / 6 is about $14$ divided by $10,000$. But what would be a more familiar expression representing it:


One of these three expressions will produce a different answer, select that one:


One of these three expressions will produce a different answer, select that one:


One of these three expressions will produce a different answer, select that one:


What is the value of $\sin(\pi/10)$?


What is the value of $\sin(52^\circ)$?


Is $\sin^{-1}(\sin(3\pi/2))$ equal to $3\pi/2$? (The "arc" functions do no use power notation, but instead a prefix of a.)


What is the value of round(3.5000)


What is the value of sqrt(32 - 12)


Which is greater $e^\pi$ or $\pi^e$?


What is the value of $\pi - (x - \sin(x)/\cos(x))$ when $x=3$?


Factorials in Julia are computed with the function factorial, not the postfix operator !, as with math notation. What is $10!$?


Will -2^2 produce 4 (which is a unary - evaluated before ^) or -4 (which is a unary - evaluated after ^)?


A twitter post from popular mechanics generated some attention.

What is the answer?

Does this expression return the correct answer using proper order of operations?


Why or why not: