Tag: Cool Maths
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Linear algebra #2
I and J are the basis vectors of a coordinate system and thus in this section of the course, any vector can be denoted as a linear combination of A(I) + B(J).
The set of all possible vectors that you can reach with a linear combination of a given pair of vectors is the span of those vectors. For example, a span of 2 vectors would be a sheet cutting through the origin unless if the 2 vectors were part of a single line.
Then what is the span of a 3 dimensional vector?
1 case is that the 3rd vector lies on the span of the first two vectors thus not impacting the span of the total sum of vectors. We call the third vector, a linearly dependant vector as it can be represented as a linear combination of the other 2 vectors.
Conversely, a vector in a different direction would provide access to the whole 3d space and this vector would be known as being linearly dependent.
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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
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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.
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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!
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