A Problem With The Integral: A Solution Armani Willis Mercy College 7.24.1.20

This is my original work, the work of A. Willis. To address recurring and annoying problems in education. All uses of my work in future works must include citation of my work.

When a person named Carl Fredrich Gauss was in what you would call middle school, the teacher issued a challenge to the class: Find the sum of 100 (E100). Apparantly, Gauss knew a little more than his other classmates, because he came to a quick and simple answer.

If Carl had rows of strings of beads counting from 1 up to 100, then he had a triangle of 101 strings of beads. Thus Ex=x(x+1)/2. E(100)=100(101)/2=5050, E(10)=10(11/2)=55 E(5)=5(6/2)=15. Do u see a problem here?

We learn that the derivative of c(x^n)=cn(x^(n-1)), Thus the continuing sum of x is the inverse: (x^(n+1))/(n+1) If n=1, then Dxdx=1, Sxdx=x(x/2). Lets try it. If x=5, Ex=5(6/2)=15, Sxdx=5(5/2)=12.5. Huh? The integral is less than the sum!

Is where the tenet for raising x or f to a power emerges, and it governs the algebraic aspect of calculus. If f=(x^n) then df=(((x+dx)^n)-(x^n))/dx So if n is 1, then df/dx is also 1. But if x=5, then is 0/0=1, no. This tenet goes according to: the indivisibility of 0 and 1.

Given that 5! = 5432 = 120 5! also equals 54321= 120 5! /= 543210=0. If x1=x, and x0=0, then dx(x1)=x-(x1)=0, dx(x0)=x-(x0)=x This is called the indivisibility of the binary numbers. it also means x/x=1, 0/x=0, x/0=0, 0/0=0.

It means that we must consider how fast dx is approaching 0. But in calculus, the rate that dx is approaching 0 is not considered, thus dx is considered to be infinitely small. The divisibility of binaries also applies to inf, since inf is a function, and must be of some measurable thing, or else we have left the realm of math.

Given that dx=0, D(x^n)=((x^n)-(x^n))/0, which is zero, but to say that it approaches zero means essentially inf, which is still 0. The algebraic derivative has no provable measure, dx must have value. df=(((x+dx)^n)-(x^n))/dx), Sdfdx=f=(x^n), The textbook asserts that y=D(x^n), not f. Calculus 1,2,3 is redundant for these reasons. It should be removed from all textbooks.

The algebraic derivation function itself is derived from the power rule. In the texts: D(x^n)=lim(dx>>0)(((x+dx)^n)-(x^n))/dx, the sums of dx leads to the hybridization n(x)^(n-1). Usage case: for(dx>0)((x+dx)^3)-(x^3)= (3x^2)dx+(dx^3)+(3dx^2)x. if dx=1, 3x^2+3x+1 is 7@x=1, 3(1)^2=3. From this point, the function cycles.

so, to get 3, dx must be 7/3. (3x^2)dx+(dx^3)+(3dx^2)x= 7+12.7+16.3=38, obviously not 7. if dx=0, zero/zero is zero, if dx= planck length, 1.6(k^-13) meters, ((4.8)(k^-13)+(1.6^3)(k^-12)+(4.8)(10^-37)) /(1.6)(k^-13) =3+((1.6^3)/k)+4.8/(k^2), or ~3.

Given that A=(For i=plank length, dx in meters, divided by plank is n, where lim(n>>na)(E(i=1>>na)((i/n)^3)(1/n)))) for f=(x^3)). for x=1, na=1/1.6*10^(-38)=6.25(10^37) lim(n>>(6.25(10^37)))((E(i^3))/(n^4)) 1/n is therefore 1, 1/n=i. lim(n>>6.25(10^37)((1/4)+(1/2n)+(1/4nn)= 1/4.

Given that x=i/n, ((b-a)/(n^2))(n(n+1)/2)= ((b-a)((1/2)+(1/2n)), or about ((b-a)/2). [0,5]:(25/2);[1,5]:((16/2),8), E(x^k)dx=(b-a)((i/n)^k)(1/n)=S(x^k)dx. As n>>6.25*10^37, the interval decreases. (n/2)(n+1)>>(n/2)(n), eccetera.

Given that S(5-0)(x^3)dx=(5^4)/4=156.25, E(n=5)(x^3) = 125+64+27+8+1=225. 225-156.25=68.75; 15-12.5=2.5, 68.75-2.5=64.25.

Given that na=6.2510^37, E(n)=n(n+1)/2, thus as n increases, E(n)>>Sndn. The increment should be n1.610^-38, and reaches 1 as n >> plank inverse. The sum of 500 is closer to the integral of 500 than the sum of 50 is to the integral of 50 by 10plank. This does not hold for infinity, because then the increment would be zero. In which case, you would be witnessing a ‘perfect’ or ‘impossible’ universe.

We have established that as n>>plank, (1/(n^(k+1)))E(i^k)>>S(x^k)dx, as n>>plank, A>>A0e^rt. Two employees, a and b, earn the same amount with interest at the end of a year. The employer want to compile the dividends of earnings in 30s and 90s. thats x1531 for An and x15*30.33 for Bn. b will make less than a exponentially.

In multiple textbooks, it is printed that y=f(x). This cannot be true because y is an axis, and so is z. thus f becomes f({x}), which contain the parallel perspectives of f. in 6 dimensions, we notice that f({x^2}) is proportional to f{[x]}, where {f(x)} contains [f{x}].

The derivative then becomes: ((((x+dx)^n)-(x^n))/dx){*x}, where D{*x} is being multiplied for each x in f. Where if for {x}, i>3, f is a supergraph. Example: for f{x=6}, Dfd{x}=E(f{x-1}df(xi)), or the partial derivative of x with respect to each x, Sfd{x}=E(f{x-1}Sfd(xi)).

The textbooks assert that y=x^n, and (dy/dx)=nx^(n-1). function f=(x^n)+(y^0)+(z^0)...eccetera, thus, (df/dx)~(nx^(n-1)) as n >> plank, (df/d[x])~f[x-1]E(n*[xi]^(n-1)), n>>plank. Please correct these formulas,observe that the sums in calculus are differential.

References

Larson Hostetler Edwards Milton Smith David M Burton Devanney C. Blanchard Hall

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