Tag: sum

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
Maths

summation of the squares

we all know Ramanujan’s summation of the positive integers which is equal to – 1/12

Here’s a proof for all non mathematicians:

Let the sum 1 + 2+ 3 + 4 + 5 + 6…. = c

thus 4c = 4 + 8 + 12 + 16 +20….

therefore -3c = 1 -2 + 3 – 4 + 5 -6 … (1)

which is equal to ( 1 -1 + 1 – 1 +1…)². (2)

then what is the sum of 1 + 1 – 1 + 1 – 1….?

let a equal to the previous sum thus -a = -1 +1 -1+1… thus subtracting 1 equation from another yields 2a = 1 thus a = 1/2. substituting this into (2), we obtain -3c = 1/4 thus c = -1/12

what is the sum of odd numbers or Σ(2n+1)?

analysing equation (1), we can rearrange this as (1 + 3 + 5 + 7…) – 2( 1 + 2 + 3 + 4 + 5…) = Σ(2n+1) – 2(-1/12) which is equivalent to 1/4 thus by the distributive property, we obtain Σ(2n+1) = 1/12. (3)

thus what is the Σ(n²)?

  1. lets denote this sum as c which is 1 + 4 + 9 + 16 +25 +36 + 49 + 64 + 81 + 100..
  2. thus 8c = 8 + 32 + 72+ 128
  3. hence -7c = 1 – 4 + 9 – 16 +25 – 36….. = (1-4) + (9-16) + (25-36)….
  4. which is equal to -3 -7 – 11 – 15 = – ( 3 + 7 + 11 + 15 + 19…..) = -( Σ(2n+1) – (1 + 5 +9 +13 + 17….) (4)
  5. let 1 + 5 + 9 + 13 +17… be denoted as s
  6. by inspection, we can deduce that 3+7 + 11 + 15… = (1+5)/2 + (5+9)/2 ….. = 1/2 + 5 + 9 + 13 + 17… = s – 1/2
  7. therefore substituting this into (4), we yield the following equation s – 1/2 = 1/12 -s thus s = 7/24 thus -7c = -( -5/24) = 5/24 thus c = -5/168

What is the Σ(2n+1)²?

from (4), we know that 5/24 = 1 -4 + 9 -16 + 25… = 1 + 9 + 25 + 49… – 4( 1 +4 + 9 +16 + 25…) = Σ(2n+1) – 4(-5/168) = Σ(2n+1) + 5/42 therefore Σ(2n+1) = 5/24 – 5/42 = 15/168 or 5/56

I will endeavour to send more derivations of this technique