Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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**imcute****Member**- Registered: 2022-09-28
- Posts: 176

i^2=-1

j^2=1

so k^2=0

e^kx=cosk x+ksink x

e^kx=1+kx+0x/2+0k^2/6...=1+kx

cosk x=1

sink x=x

so tank x=x asink x=x Acosk x=1 acosk x=anything atank x=x

e^ix is on the x^2+y^2=1

e^jx is on the x^2-y^2=1

so e^kx is on the x^2+0y^2=1 or x=+-1

e^ix x is half the area of a sector

e^jx x is half the area of a hyperbola sector thingy

then e^kx x is half the area of a triangle(does work with cosk and sink)

conclusion:the zero factor crumples stuff up a lot

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