A co worker has the following equation: [itex]3e^{-.5y} +3e^{-y} + 3e^{-1.5y} + 103e^{-2y} =98.39[/itex] Solve for y. Some sort of compound interest bond equation I am told, or something like that. He has been told that to solve for y, one must use a trial and error approach. True??
If you substitute u for [itex] e^{- \frac {1}{2} y} [/itex] you get a quartic equation, for wich an exact solution exists. Type the equation in WolframAlpha to get a meaningless jumble of really large numbes and lots of square and cube root signs. It's less work to solve the quartic with a numerical method like newton's method than to use the formula for the quartic equation.
There is an exact solution. Let ##x=\exp(y/2)##. Then your equation is equivalent to ##103x^4+3x^3_3x^2_3x-98.39=0##. This is a quartic equation, so it is solvable, exactly. Then solve for y. Simple! Not so simple. Solving cubics is a bear of a problem. Solving quartics? That's a megafauna bear of a problem. Solving this numerically is non-trivial. Newton's method doesn't work very well on this problem. You need to use something else such as [strike]the secant method[/strike] the midpoint method. Edit: The secant method doesn't work very well here either because f(x) is almost flat between -1 and +1.
willem2 and DH...thanks! This forum is loaded with some very brilliant minds. He couldn't find an answer anywhere on-line or from his college finance professors, so I told him not to worry, I would get an answer through the best site on the web. Thanks again!
Massive typos there. I should have said Let ##x=\exp(-y/2)##. Then your equation is equivalent to ##103x^4+3x^3+3x^2+3x-98.39=0##.
I attempted this method and recieved reasonable answers of a complex conjugate pair and two real roots. However you obviously cannot ln a negative number which reduces the outcome to three possibilities. Once divided buy the -0.5, ending up with y=0.0677, 0.008-∏i and -13.807+∏i
You can reject the negative solution to ##3(x+x^2+x^3)+103x^4=98.39## (and also the two complex solutions) if you are looking for a real solution to the original equation. You cannot reject that negative solution if you allow complex solutions to the original equation. In fact, each of the four solutions to the polynomial generates an infinite family of complex solutions to the original equation. If some complex valued ##y## is a solution to that original equation, then so is ##y+4k\pi## for any integer k.