A Guide to Calculus [WIP]

What is a function?
    -A function, in examples denoted as f(x), is a series of operations one does on an input.
    -The first part, the 'f' in f(x), is the name of the function, and the second part, 'x', is the name of the input.
    -If f(x) = 5x, then the output will be five times the input (f(1) = 5, f(2) = 10).  This is because the number placed in between the brackets is the input, and so within the function, x = whatever was inside f().  This does not apply outside of the function (in f(3)+7x, where f(x) = 5x, the x next to the 7 doesn't equal 3.)

What is a Limit?
    -A limit is the value a function approaches but can never reach.
      (image from wikipedia) is how it is denoted.  For the sake of this guide, I will write it as:         lim(x->p)f(x) = L.  x=The input of a function.  p=Number x approaches.  L=The number f(x) gets closer to as x gets closer to p.
    - p or L may equal ∞.  This is when a number can get as big as it wants, but it will never quite reach a certain value.  Think of it like a hyperbola, the function goes on forever, and it keeps getting closer to the axes, but it never quite reaches it.  Another example is: Joe wants to go to Starbucks.  However, he can only move half the remaining distance at a time.  In the beginning, he has 10 miles to go.  Then, he has 5 miles to go.  After that, 2.5, and then 1.25, followed by 0.625, and infinitely on, never reaching 0 miles to go.
    - Sometimes, a function may have two limits.  Then the limits are referred to as left-hand and right-hand, with left-hand approaching from the left (duh), and right-hand from the right.  To show mathematically which hand you're talking about, do lim(x->p^-) for left-handed, and lim(x->p⁺) for right-handed

Example 1: lim(x->0)f(x)=0, when f(x) = ((x+1)/x)^-1.  This is because you can't have something divided by 0, which is what would happen if you made x = 0.  As x gets closer to 0, however, the values decrease.  Theoretically, if you could divide by 0, the answer would be 0/1 in the end, which equals 0, so L=0.

Example 2: lim(x->0)f(x)=1, where f(x) = (sin(x))/x.  If you examine the outputs, you will see a trend, as x goes down, f(x) goes up.

How to find the Limit

    -To find the limit, you'll probably have to use a bit more logic than you're used to.  For instance, you must find the lim(x->5)f(x), and f(x) = (x^2-25)/(x^2+x-30).  If you substitute x for 5, the number x is supposed to be approaching, you'll get 0/0, which isn't a valid value.  However, if you factor the polynomials to get ((x+5)(x-5))/((x-5)(x+6)), you'll notice that you can simplify it to (x+5)/(x+6), which is (5+5)/(5+6), aka 10/11.  Therefore, the limit is 10/11!
    -Also, if you think about it, if you know which 'hand' the function's limit is (left-handed or right-handed), you can also solve with absolute values in the function.  Given lim(x->2^-)f(x) and f(x) = (5+x)(x(1-(2/x)))/|x-2|, we know that the limit is left-handed (the ^-).  That means that the values are approaching 2 from the left, so they must be less than 2.  If x<2, x-2<0.  Thus, x-2 is a negative number.  Now, |x-2| is in the function.  For the sake of finding the limit, temporarily change |x-2| to -(x-2).  If we look at the top half and simplify it, to just one multiplication, we get (5+x)(x-2).  Since -(x-2) is in the denominator, we can factor it out to get (5+x)/-1, which equals -(5+x).  Since x is approaching 2, we substitute it in for x to get -7.

What is a Derivative?
    - (image from Paul's Online Notes).  This is the function^[L] for finding a derivative.
    -A derivative is the rate of change of a function.
    -A derivative of a function is denoted with an ' (f'(x) = derivative of x)
    -It can also be shown using df/dx|x=?, which is equivalent to derivative of function f in terms of x.  In df/dx, the bold letters are the names of the function and the function's input, respectively.  The reason for this is that df/dx matches with the integral's ^[In] dx notation.
    -From Paul's Online Notes, this is how you will mostly see the notations written:
 (left side is ' notation and right side is df/dx notation).  For the simplicity of this guide, if you ever see me use df/dx notation like that, I will say (df/dx(=?)), ? being what x equals.  Of course, sometimes you don't know what x is.  In this case, instead of df/dx|x=?, you will only do df/dx.  

Example 1:
f'(1)=?, f(x)=2x.  Plug it in to the derivative equation.  f'(x)=lim(h->0)((f(1+h)-f(1))/h), f(1+h)=2(1+h), f(1) = 2.  f'(x)=lim(h->0)((2+2h-2)/h) = f'(x)=lim(h->0)(2h/h) = f'(x)=lim(h->0)(h), plug 0 in (since h is approaching 0), and you've got f'(x) = 0!  Why?  Well, remember that 2x will always go in a continuous line diagonally, so there isn't a 'change' in the outputs.  A good rule of thumb is that, in f(x)=Cx^e, f'(x)=C(e-1)x^(e-2), with C being coefficient and e being exponent.  It won't work unless the function can be split into monomials of the form Cx^e.

Example 2:
Using the rule we established in example 1, f'(x)=C(e-1)x^(e-2), let's try f(x) = 5x^4.  f'(x)=5(4-1)x^(4-2) = 15x^2!  Much faster this way.  Tip:  If e = 1, the output will be 0.  If e = 2, the output will be 1.  Otherwise, it will be in terms of x.

Complex Derivatives TBA
    

What is an Integral?
    -An integral is the antiderivative^[D] of a function.
    -There are two kinds of integrals, the definite integral and the indefinite integral.
    -The indefinite integral returns a function^[L] and the definite integral returns a value.
    -∫*function*dx is the symbol for an indefinite integral.  'dx' is not a variable, but rather what denotes the ending of the ∫.  Treat it as a ), where ∫ is a (.
    -The definite integral has the same symbol as the indefinite integral, but it has a number above and below it, usually called 'b' and 'a', respectively.
    -Integrals are used to find the area in between the x-axis of a grid and a line.
    -'B' and 'A' are used as the limits, as a line goes on forever, so it has an infinite area.  So ⁵∫₄f(x)dx would mean 'the area of a line, but only the bit that is in between the x-coordinates of 4 and 5.  'B' is the upper limit, 'A' is the lower limit.
    -∫f(x)dx returns the function F(x), which is the antiderivative of f(x).  If f(x) = x, F(x) = ∫ x dx (which is (x/2)^2, by the way.)  The function F(x) will return a value equivalent to ^x∫₀, which leads us to the next bullet point, the fundamental theory of calculus.
    -  ^b∫_a f(x) dx = F(b) - F(a), where F(x) is the ∫ of f(x).
    -What this means is that the indefinite integral finds function that can be used to find the area between its input and x=0, and the definite integral finds the area between two points.  (This makes sense.  If you have a line overlapping another line, and it is 5 units big and the smaller line is only 3 units big, you can find the length of the non-overlapping part of the bigger line by subtracting the smaller (3) from the larger (5), in this case resulting in 2 units.  Its simple Segment Addition Postulate.

Indefinite Integrals

How the operation ∫f(x)dx works:
    -Step 1: If f(x) is a polynomial, you must look at the individual monomials while doing your calculations.  For example, if    f(x) = x^2 + x, you must look at it as x^2, do the math to it, and then look at the lonely x.
    -Step 2: All monomials are in terms of x, where x is the input of f(x).  Other variables can be treated as constants.  This means that if f(x) = 3x^2 + 4x + 3, you must regard it as 3x^2 + 4x^1 + 3x^0.  (Refresher: Anything to the 0th power = 1).
    -Step 3: Once the function is divided into monomials, take the coefficients (in 4x, '4' is the coefficient) out of the equation temporarily (remember, 4 = 4x^0).  Note: In this case, other variables should be considered part of the coefficient as well, and should be taken out, as per step 2 (so in 4yx, '4y' is to be considered the coefficient).
    -Step 4: Let e=x's exponent (in x^4, e=4, in x, e=1 (x = x^1), and in the case of x^0, e=0).  Add one to it, and make the old exponent the new exponent (so x^4 becomes x^5).
    -Step 5: Divide your new number by e+1 (so if you started with x^4, this step and step 4 make it (x^5)/5).
    -Step 6: Multiply the coefficient back in (so if it was 4x in the beginning, it should now be (4x^2)/2.  If it was just 4 in the beginning, it should now be 4x, because 4((x^(0+1))/(0+1)) = 4x.
    -Step 7: Do the same for every monomial, and then add them back together.

Example:
    f(x)=3x^2+5x+4 -> 3x^2, 5x, and 4 -> 3(x^2), 5(x), and 4(x^0) -> 3((x^3)/3), 5((x^2)/2), and 4(x) -> x^3,5(x^2)/2, and 4x -> x^3 + 5(x^2)/2 + 4x, which can also be written as (2x^3+5x^2+8x)/2

The result I got, (2x^3+5x^2+8x)/2, is not equivalent to f(x), but to F(x) (Remember, F(x) = ∫f(x)dx).

Definite Integrals

    -Step 1: Perform the indefinite integral operation on f(x), assuming f(x) is the name of your function (the function can be named anything, but is generally called f(x).
    -Step 2: Remembering the fundamental rule of calculus (^b∫_a f(x) dx = F(b) - F(a)), we now must plug in the values a and b into our new function, F(x), the result of ∫f(x)dx.  
    -Step 3: Subtract F(a) from F(b)

Example:
    ⁵∫₂ f(x) dx -> ∫f(x)dx (to save time, f(x) will be the same as in the previous example, and so ∫f(x)dx equates to (2x^3+5x^2+8x)/2).  F(b) - F(a) = F(5)-F(2) -> F(5) = (2(5)^3+5(5)^2+8(5))/2 = (2(125)+5(25)+40)/2 = (250 + 125 + 80)/2 = 455/2 = 227.5.  F(2) = (2(2)^3+5(2)^2+8(2))/2 = (2(8)+5(4)+16)/2 = (16+20+16)/2 = 52/2 = 27 -> 227.5-27 = 200.5.  Thus, ⁵∫₂ f(x) dx = 200.5.


Multiple Integrations


    -While you may think ∫∫ will mean 'the integral of an integral', you are wrong.  (If it were, an easy way to calculate it would be my below formula, where m = how many integrations (∫∫? ∫∫∫?), C_n = coefficient of X^e in the 'n'th monomial (if n=2, in 5x^2 + 6x, we would be looking at 6x), and e_n = exponent of X in the 'n'th monomial.  I made it myself (although somebody probably came up with it before I did), so you better use it...  XD)
   
    -Will continue multiple integration later.  It's some complex stuff.  Have fun with my formula in the meantime!

Complex Integrals TBA

Note: A good resource for finding integrals, in my opinion, integrals.wolfram.com/index.jsp?

More circles!

TBA

End of Calculus!