NumberCatalog v1 - Common Calculations

The Basic Calculus Tutorial - Part 2

Welcome back to our three-part tutorial on how to learn and perform simple calculus!

⚠️ It is assumed you have already read and understood large number methods.

Reverse of Operations

The method described is known as "reverse of operations," or, in short, as a reverse-algebraic-geometry method for solving algebraic problems. We've written an algebra calculator app for this tutorial. Just copy the calculator into the space provided in your calculator and you can try using it on your own problems or just watch our tutorials to see how it's used for a real world problem. Here, I've asked for an exact solution. Because it is an exact solution, there may be cases when a result that is an algebraic fraction, for example 1÷(3 - x^2) would have a result of 3.

Let's consider the following problem, a linear algebraic geometry problem:

The slope of the tangent line at P to the graph of the function y = f(x) is given by f' = 1 + x + 1÷(2x), which has an equation y' = (3 - x - 2÷(4x)) + f'. We need to find an algebraic way to obtain f' = 1 + x + 1÷(2x) in terms of y', i.e. solve 3x = y'. The inverse of f' = 3 - x - 2÷(4x) is y' = f' · (2x)÷(1 + 4x^2).

Note the change of sign and that y' must equal y', as we derived y = f'. This can be confirmed by a little algebra: if y' = (3 - x - 2÷(4x)), we get

y' - f' = 1 = -y,

y - 2f'x - (3 - x) - 1 = 2y - x, from which we find that the new equation is 3x + 4y - f' = 0. In terms of 1÷x we get x^(-3)(2x) = 4÷(3 - (4x^(-1)^3)) which simplifies to

-3 = (4÷1 + (3÷1 - (4÷1)^(-1) = (1÷x - 2).

Note the change of sign and that y = 4x÷(x^3 + 1) is also the correct tangent to y' = 4 - 4÷(-8x^2), as seen from 3x = y.

How to solve such an algebraic function f'(y') using an exact method of mathematics is an algebra problem, with a similar solution procedure as 1÷(y'). It turns out to be rather tedious to find a closed formula in such cases, so the following two methods will give you some tips on solving this kind of a problem in a shorter manner, with a different, algebraic-looking method that could be useful to a physicist. You should only use one method as described here.

The Way Out

The method that is presented here, known as "the way out," can also be used when a computer can only perform 32-precision, and if you can obtain the value of f(x), but it must be known exactly as a root. It also can be used, but we're not going into details at the moment, so it can be found in a text on algebra.

The first method we use to solve a problem is the way out. We do not really have any formulas to write. For example, I cannot write x = (5 - 6). Here I write "The slope is "5 - 6" because the method is known as "reverse of operations." That means "reverses operations, i.e. it is reverse-algebraic geometry" (in reverse-geometry the slope is found in 40÷3). What you actually get is x = 2÷(2·6), but 5 and 6 are the unknown variables to solve.

The formula that I found for x to find is: 3 = x, or x = 1.

The method described here can also be used when we obtain an algebraic fraction from 2 steps backward in terms of 3 numbers (a root), which will lead you to the exact algebraic equation (as opposed to x^3 = 50, in the following).

What follows is the most straightforward method and it will probably give you a different method from 3 times of the inverse of f'(y') which was given earlier. What the inverse is in 4 steps backwards is explained above, so you can have a look if you don't like to see the derivation or not.

If x = y÷(2·6) = 1, then it's possible that:

So 6x + 6x = 8·10·5x

But it turns out to be more straightforward than y^4 - (x - 4·x + x - 10)·x = (8·x·6)

The 6 in front of x + x = x(1 + 1) gives us that:

Six Steps Backwards

The other problem of the first part of our tutorial can also be solved by 6 steps backwards, in that order (the "butterflies"). In fact, when the steps of backward-derivation of y' and then 3 steps backwards of 1÷y are solved, then it is found that 4·(3-4) = -5 which is y. The steps 5·6 + 7·5 are: -2·(8·5) = -8 and -1·(40·6) = -24, 3 = y, then 4·(3-4) = -5 is 5.

As the next step is "finding 4y", in general, one may say "what is y^2, multiplied by x = (1÷(4-6)·4÷2 + 1÷(20))?"

Let's calculate:

That's not as nice, so the way to do is the one below:

y' - 3 = -3. The solution of 4x - 4 + y is: -3 = x = -4, that's not 3·x, so this is not 1÷y but -3 is 2x. Note: I just solved this with algebra.

The third solution, by multiplying 4 to y, will be used in 7th steps backwards of the original method ("the butterflies"): -5x - 15x + 7 = (3·15x)(x^5)

Since you cannot use negative numbers (and y > 0), there is no y (I've also not seen that problem on this topic so far). If the negative of y > 0, you can multiply with it, like in 7th step backwards: 3÷27x - 4÷9x - 5÷3 + 13÷27

The same thing could also be written as (y · (5 + y·20÷4))÷(-84x). However, we have not used x. Here the last step backward is that 5÷x^7. What I'm finding is:

y + y·3 - (5÷8) - (20·5) - (y·(45·x^4)·(5 - y))

The next 3 steps back of this equation will give 2x: -y, y - x, then x - (2÷78),

so if y = -6 then y' = (9) and y^6 - 15 - (-84x) is x = 12. There you get y^5 - 2 - y = (26x)(x^6), then y' = (64÷2516) and

-5÷5 + (x^8÷764)·(8·(8+8)·5·(6·x)) = x^8 + (49186÷x)

After solving all this equations we will get a value that can be compared with x.

If y' = (2·x)(5 + (2516÷x^6))(6·x)^5 is (7,530). Now the "butterflies" have gone from y' to -x which was -5÷(y'), which is y + 65÷x = -266x, we multiply and obtain that:

The above equation has only 3 numbers as 1's, but we still can't have that simple form, we have to find all other ways (it is not clear why this is needed in general, it's a lot easier to get all the coefficients when you write that). Let's get some examples for this (you have to check if we can obtain a similar result) and find:

y + 3 + 9 + x = x + x - 9

The method is given below (you'll get y as a result if you use this).

y + (7335·x + 4÷581)÷(4·x) + (580÷x^2) - x

The Way In

But there is an interesting, but not so popular, way that works with y' and not only with f'. It is the "the way in." It was discovered by George Duffield who was in my second calculus tutorial (but this was an advanced math problem with lots of steps back in it, and was solved only on 9th year or higher levels, but not at all common, I hope, as the algebraic formula that solves such a problem has a nice nice form):

y - y^6÷x = -y^5÷x + (-3 + 786652·x)÷41

The equation will give you two values y that can be obtained. But what we are after are not 2 numbers as y', but a real value 6x that will allow to get y' with an exact equation. In 32 bits of 6 bits (the maximum bit-per-bit multiplication that we're allowed by 64 bit system) can give 0, 10, -268167 (5^6·0.10 = 661289), -59045, 1. The last solution (with 4÷43·5324) has two significant digits of x.

For 64-bit arithmetic (22 digits, with the current limit on 80 bit systems), I could write (3,4,078.0767 · 1.0014 = -1,717458). (10÷3 = 3,30122277). That's not close enough, because I still can get y' from x and, 10 is also x = -6768 (which gives an 1.1·10^-299th root for x, it would have been easier for 64 bits of precision)