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A Geometric Way to Visualise sin(x + y) and cos(x + y)

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The angle–addition identities for sine and cosine often appear as algebraic formulas, but they can also be understood by combining two right triangles in a simple geometric construction. The calculations for the side lengths follow directly from the definitions of sine and cosine. sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y − sin x sin y Start with a right triangle of angle y and hypotenuse 1. From basic trigonometry, its horizontal and vertical sides are: cos y and sin y. Next, attach a second right triangle with angle x . Its hypotenuse is the side of length cos y from the first triangle, so its adjacent and opposite sides become: adjacent = cos x · cos y opposite = sin x · cos y Likewise, if the first triangle's vertical side sin y is used as a hypotenuse in a similar way, it contributes: adjacent = cos x · sin y opposite = sin x · sin y When the horizontal components are combined, they give the expression for cos(x + y...
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The Method of Differences — A Clean Proof of the Sum of Cubes

The Method of Differences — A Clean Proof of the Sum of Cubes The method of differences is a remarkably elegant tool for evaluating finite sums. When each term of a series can be written in the form f(r+1) − f(r) , the sum “collapses” — all interior terms cancel, leaving only a boundary expression. This behaviour is called a telescoping sum . 1) Telescoping Sums Assume the general term u r can be written as: u r = f(r+1) − f(r). Then the finite sum from r = 1 to r = n becomes: Σ u r = Σ ( f(r+1) − f(r) ). To see what happens, write out a few terms: u₁ = f(2) − f(1) u₂ = f(3) − f(2) u₃ = f(4) − f(3) ⋮ uₙ = f(n+1) − f(n) When these are added, everything cancels except the first and last pieces: Σ u r = f(n+1) − f(1). This is the essence of the method: interior structure disappears, leaving just the difference between the final and initial states. 2) A Classic Application — The Sum of Cubes We will use this technique to prove the well-known ...
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2×2 Orthogonal Matrix Mastery — A Generalised Construction

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2×2 Orthogonal Matrix Mastery — A Generalised Construction Orthogonal matrices in two dimensions reveal one of the cleanest structures in linear algebra. A 2×2 matrix is orthogonal when its columns (and rows) satisfy two conditions: They are perpendicular (their dot product is zero); They have unit length (their magnitude is one). This article presents a clear generalisation: any pair of perpendicular vectors with equal magnitude can be normalised to form an orthogonal matrix. 1. Begin with two perpendicular vectors Let the first vector be: (a, b) A perpendicular vector can be chosen as: (−b, a) Their dot products confirm orthogonality: (a, b) · (−b, a) = −ab + ab = 0 (a, −b) · (b, a) = ab − ab = 0 2. Compute their shared magnitude Both vectors have the same length: |(a, b)| = √(a² + b²) We can therefore normalise each one by dividing by √(a² + b²). 3. Form the matrix using the normalised vectors Place the two normalised vectors...
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