I'm fluent in the fluxions 'Cause I antidifferentiate Leibniz didn't find the fundamentals My code to this day, postdate. by parts is your great foe Product rule can't stand the test of time. Newton: Oh really? Leibniz sank so low I use dots because I'm prime Integ. Physics is stupid can’t you see I guess the Apple doesn’t fall far from the tree Just like you, your math is bad dy/dx is the only thing that’s rad. You took my work and twisted it I came up first, try carbon dating Your just another old misfit Only Newton has a 5-star rating Leibniz: Oh no Leibniz is on top Why the hell would you put a dot Have you ever even thought Without the product rule today calculus wouldn’t be hot. Newton: Your notation is derivative You are way too acquisitive Can't even differentiate Or keep your bad notation straight Leibniz: You didn’t publish your works Until 1693 What in the world is a Fluxion? You’re such a phony Newton: Oh yeah Newton for the win Because Leibniz is a sin To your screen integrals pin Let's toss dx in the bin His notation's weak as tin And his arguments are thin My proof's stolen from a djinn I have no equal twin. Last edited by The_Imaginarium (J19:54:31) Substitute this into our equation: (e^iz - e^-iz)/2i = 2 Since x^-1 is just the multiplicative inverse of x^1, (e^iz - (1/e^iz))/2i = 2 multiply by 2i e^iz - 1/e^iz = 4i Multiply by e^iz (e^iz)^2 - 1 = 4i(e^iz) Form a quadratic equation (e^iz)^2 - 4i(e^iz) - 1 = 0 Solve the quadratic equation: e^iz = 4i ± sqrt(4i^2 +4)/2 simplify: = 4i ± sqrt(-12)/2 remove 2i from the square root = 4i ± 2i(sqrt(3))/2 divide each term by 2 = 2i ± i(sqrt(3)) factor i = i(2 ± sqrt(3)) the quadratic equation is solved! Now we know that e^iz = i(2 ± sqrt(3)) take the natural log to eliminate e iz = ln(i(2 ± sqrt(3)) Use the addition property of logarithms iz = ln(i) + ln(2 ± sqrt(3)) To solve the natural logarithm of i, we know that a complex number in exponential form is represented as z = re^ix Where r is the magnitude of the number and x is the argument Substitute z for i i = re^iz add in the corresponding values (these can be found by graphing i) i = 1e^i(pi/2) take the natural logarithm ln(i) = ln(1) + ln(e^i(pi/2)) ln(i) = i(pi/2) incorporate this back into the equation iz =i(pi/2) + ln(2 ± sqrt(3)) multiply by the multiplicative inverse of i z = pi/2 + 1/i(ln(2 ± sqrt(3))) No one likes i in the denominator, so multiply by i/i z = pi/2 - ln(2 ± sqrt(3)) Add in 2pi(n), because sine is periodic z = pi/2 - ln(2i ± sqrt(3)) + 2pi(n) where n is a positive integer There, we have one of the answers in a different way And sine is simply opposite e^iz = cos(z) - isin(z) Algebraic elimination: e^iz = cos(z) + isin(z) - e^-iz = cos(z) - isin(z) _ e^iz - e^-iz = 2isin(z) divide 2i to isolate sine (e^iz - e^-iz)/2i = sin(z) Great, we have the complex definition of sine. I also began with Euler's formula e^iz = cos(z) + isin(z) change z to negative e^-iz = cos(-z) + isin(-z) since cosine is symmetrical, whether or not z is negative does not matter. Raihan142857 wrote:snipI solved it slightly differently. z doesn't necessarily have to be real, so you can use that to find the inverse sine of basically any number You can manipulate the equation a bit e^(i * (-z)) = cos (-z) + i sin(-z) = cos (z) - i sin(z) cos z - i sin(z) - cos z + i sin(z) -2i sin(z) e^(-iz) - e ^ (iz) = -2i sin(z) (e ^ (-iz) - e ^ (iz)) / (-2i) = sin(z) So if sin(z) = 2, then (e ^ (-iz) - e ^ (iz)) / (-2i) = 2 e ^ (-iz) - e ^ (iz) = -4i If you substitute x = e ^ iz then 1/x - x = -4i 1 - x ^ 2 = -4ix -x ^ 2 + 4ix + 1 = 0 Using the quadratic formula, x = (-4i +- sqrt(-16 - 4(-1))) / -2 = 2i +- sqrt(-12) / -2 = 2i +- sqrt(12) i / -2 = 0.268i e ^ iz = 0.268i if a is the real part of z and b is the imaginary part then e ^ i(a + bi) = 0.268i e ^ (-b + ai) = 0.268i e^(-b) e^ai = 0.268i e ^ (-b) (cos (a) + i sin(a)) = 0.268i It's pretty obvious that e ^ -b is |0.268i| and a is its argument. Here's how I'd solve sin(x) = 2 Euler's formula states that e^(iz) = cos z + i sin z. Wolfram is able to solve problems involving complex numbers. The_Imaginarium wrote:-snip- That makes sense, the first two tools are used for solving algebra problems, while Wolfram is used for solving general problems.
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