Categories
Calculus Maths

Triple Integrals, Polar Coordinates and line integrals

Categories
Maths

summation of the powers of 4

let c = 1⁴ + 2⁴ + 3⁴ + 4⁴ + 5⁴….

thus 16c = 2⁴ + 4⁴ + 6⁴ + 8⁴…..

thus c – 32c = -31c = 1⁴ – 2⁴ + 3⁴ – 4⁴ …..

thus this can be simplified to:

-3(2² + 1²) – 7(3² + 4²) -11(5² + 6²)…

which can be equivalent to:

-3(Σ(n²)) – 4( 3² +4² + 2(5² + 6²) + 3(7² + 8²)

which can be written as:

-3(Σ(n²)) – 4( 3² +4² + 5² + 6²… + (5² + 6²…..) + (7² + 8²…..)…..)

which is equivalent to:

-3(Σ(n²)) – 4( ∞Σ(n²) – ∞Σ(n²))

which is simplified down to:

-3(Σ(n²))

thus -31c = 15/168 which is equal to 5/56 thus c = -5/1736

Categories
Maths

Pell’s Equation

Today, we shall embark on a quest to tame one of the greatest equations in history – x² – dy² = 1.

Some may view this as an mere ordinary equation but for me, it has taken hours of sweat and passion to tame this equation and launch a solution to this equation.

let us examine the equation = x² – dy² = 1

this can be simplified down to (x – y√d)(x + y√d) = 1:

Now here comes the ingenious part – squaring both the sides of the equation will not change the fundamental equation( as 1² = 1) but the following equation can be denoted as the following:

(x – y√d)²(x+y√d)²= 1

=> ((x² + dy²) -2xy√d)((x² + dy²) +2xy√d) = 1 thus we have obtained a new set of equations precisely due to the property that 1 multiplied by any number yields 1.

Categories
Theorems

Viviani’s Theorem

Suppose you pick a point in an equilateral triangle. What would the sum of the distances from that point to the sides of the triangle, where the distances are defined as length drawn from the perpendicular of 1 side length, be? The answer is the height of the triangle!

Definitions: Let’s define a side of a triangle as y Let’s define the heights of triangles APC, BPC, APB as h1, h2, h3 respectively.

Using the formula of the area of a triangle, which is 1/2 of base x height:

Thus, ½(h1 x y) + ½(h2 x y) + ½(h3 x y) = Area( whole triangle) which is the sum of the little triangles.

Therefore, ½(h1 x y) + ½(h2 x y) + ½(h3 x y) = ½(y x total height of original triangle)

Factorising out ½, we obtain (h1 x y) + (h2 x y) + (h3 x y) = y x total height of original triangle

Factorising out y, we obtain h1 + h2 + h3 = total height of original triangle!