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Definitions

Terminology
We would call this “x raised to the a power”

x is called the base
a is called the exponent

Example:
is called “seven raised to the eleventh power”
the base is 7
the exponent is 11

Integer exponents
An integer exponent just means multiplying a number by itself.


Example:

Fraction exponents
An exponent which is the reciprocal of an integer means taking the root


Example:

Negative exponents
A negative exponent indicates a reciprocal


Example:

NOTICE: This means you don’t ever have to deal with fractions or roots ever again if you don’t want to! You can treat them all like exponents.

Algebra rules

Combining exponents
When two numbers with the same base are multiplied, the exponents may be added


Example:

Exponents in ratios
When exponents in the numerator an denominator have the same base, their exponents may be subracted


NOTE: This is just a combination of the negative exponent rule with the multiplication rule!

Distributing exponents
If you have 2 layers of exponents, they may be combined


Example:

Negative bases
When the base is a negative number, the sign of the expression overall depends on whether the exponent is even or odd.
   if n is an odd integer
   if n is an even integer

Therefore:
   if n is an odd integer
   if n is an even integer

Examples:

A warning about addition
Exponents DO NOT distribute cleanly over addition


You either have to find a way to simplify the base first OR use the definition of the exponent to simplify things.
Examples:

Functions of a certain form can be scaled and translated by modifying the constants contained in the expression. Suppose we have some function of x that has not been shifted or scaled in any way.

$$ y = f(x) $$

We can modify that function to be shifted by (h, k) and scaled by the factor a.

$$y = a*f(x-h)+k$$

• Scaling occurs by simply multiplying the expression by some scale factor. In my examples here, that scale factor is a.
• Translation left and right can be accomplished by replacing x with (x-h), where h is how far the function is shifted to the right.
• Translation up and down is accomplished by simply adding a constant k, where k is how far the function is shifted up.

Here is an interactive example showing how a parabolic function may be shifted and scaled. (If the embedded applet isn’t working, you can access it here.)

And here is the scaling of an exponential function.
(If the embedded applet isn’t working, you can access it here.)

You can also play with examples of function shifting and scaling with the Desmos online graphing calculator.

Standard forms let you learn valuable information about an equation just by looking at what number was put where. There are three common forms of quadratic equations, also known as second-order polynomials.

Note that variables like a, b, and c in these descriptions will change in meaning from form to form.

Standard form

$$y=\text{ax}^2+\text{bx}+c$$

Advantage: Relative mathematical simplicity, sets you up to use the quadratic formula, and easy to find the y-intercept. It is just the constant y=c.

Example: $$y=2 x^2+12 x-14$$

Factored form

$$y=a(x+b)(x+c)$$

Advantage: Easy to find the x-intercepts. They are at x=-b and x=-c. You don’t even need the quadratic formula!

Example $$y = 2(x+7)(x-1)$$

Vertex form

$$y=a(x-h)^2+k$$

Advantage: Easy to find the vertex, which is at (h,k).

Example: $$y=2 (x+3)^2-32$$

This was inspired by the Quora question “What is Pi backwards?” It seemed like a good early post to try converting Mathematica code and LaTeX to WordPress.

The concise answer to the question is “That question is meaningless,” but why is this the case? After all we can write 123 “backwards” as 321, why not π?

One explanation is that π is an irrational number with unlimited digits, and something without an end cannot be reversed. Or something with no end, written backwards, becomes a thing with no beginning. Or one could say that “reversing” any number really doesn’t have much meaning in the first place. Yet, this question gives us a good opportunity to explore what we mean when we describe π as an irrational number.

One way to look at π is as a summation, like this summation equivalent to the arctangent function. $$4*ArcTan(1) = \pi$$

$$4 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}}{2 n-1}=4 \left(1 \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}\text{…}\right)=\pi$$

If you look at that sum, it adds up an infinite terms, and every term is smaller than the previous one. Thus, every term we evaluate brings us closer to the actual value of π. But, we never run out of terms, so we never find the exact value of π. We just get closer and closer forever.

We can approximate π by going out to 100 terms.

$$4 \sum _{n=1}^{100} \frac{(-1)^{n-1}}{2 n-1} = 3.13159$$

This gets us in the neighborhood. (Note that I rounded to 7 digits.) But we can get closer by going to 1000 terms.

$$4 \sum _{n=1}^{1000} \frac{(-1)^{n-1}}{2 n-1} = 3.14059$$

Now π is starting to look like the number we recognize! If I kept adding more terms, then I would need to round to a larger number of digits to see the change. If we could go forever, the summation converges to π itself. Now you probably have some idea of why approximating π to a certain number of digits is such a popular problem to throw at big computers.

This sets us up to see why “π backwards” ends up being meaningless. The number has no end, so there is nowhere to “start” the reverse-order number. Just for fun, though, we can interpret “π backwards” as an excuse to switch the starting and ending place of the summation.

$$4 \sum _{n=\infty }^1 \frac{(-1)^{n-1}}{2 n-1}=\pm 4\left(\frac{1}{2 \infty -1}-\frac{1}{2 \infty -3}+\frac{1}{2 \infty -5}\text{…}\right)$$

Notice that we are starting at infinity, so we will never reach a definite answer to this summation either. But it’s worse than that. We are effectively dividing by infinity in every term, so every term is practically zero. We can’t even figure out if the “first” term is positive or negative! (Not that it really matters when every term is zero.)

$$4 \sum _{n=\infty }^1 \frac{(-1)^{n-1}}{2 n-1}= \pm 4\left(\frac{1}{\infty }-\frac{1}{\infty }+\frac{1}{\infty }\right) =\pm (0-0+0\text{…})$$

So we do have an answer for what “Pi backwards” might look like. Every term is effectively zero, yet infinitesimally larger than the previous term, we can’t even figure out the sign of the terms, and there is an infinite number of these infinitely insignificant terms to add up. We will never get to a term that isn’t zero because by starting at the “infinity” term, we put ourselves infinitely far away the “end” of our summation.

So “Pi backwards” looks like an infinite summation of zeroes. At least looking at π forwards gave us a reasonable approximation of the actual number because it prioritized the largest, most significant terms.

Film is one of the three universal languages, the other two: mathematics and music.

Frank Capra