E to the -y plus trial and error math?

In summary, a co-worker was trying to solve a compound interest bond equation and was told to use a trial and error approach. However, using a numerical method like Newton's method is actually less work. Another method is to substitute u for e^(-0.5y) and solve the resulting quartic equation. A complex conjugate pair and two real roots were obtained, but the negative solution can be rejected if looking for a real solution. However, this negative solution cannot be rejected if allowing for complex solutions. Each of the four solutions to the polynomial generates an infinite family of complex solutions to the original equation.
  • #1
PhanthomJay
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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??
 
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  • #2
If you substitute u for [itex] e^{- \frac {1}{2} y} [/itex] you get a quartic equation, for which 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.
 
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  • #3
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.
 
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  • #4
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!
 
  • #5
D H said:
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.
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##.
 
  • #6
D H said:
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##.
Yes, I realized that afterwards..thanks for the correction!
 
  • #7
willem2 said:
If you substitute u for [itex] e^{- \frac {1}{2} y} [/itex] you get a quartic equation, for which 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.

I attempted this method and received 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
 
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  • #8
Quagz said:
I attempted this method and received 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.
 
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  • #9
The answer given in the text example was in fact y = .0677
 

1. What is "E to the -y plus trial and error math"?

"E to the -y plus trial and error math" refers to a mathematical approach where the value of a variable is iteratively adjusted until a desired outcome is achieved. The term "E to the -y" is often used in this context to represent the error or difference between the desired outcome and the current value of the variable. This method is commonly used in optimization problems where an exact solution cannot be easily obtained.

2. How does "E to the -y plus trial and error math" work?

The process of "E to the -y plus trial and error math" involves repeatedly adjusting the value of a variable based on the error or difference between the desired outcome and the current value. This is typically done using a predetermined algorithm or by manually making adjustments until the desired outcome is achieved. The process continues until the error is minimized or eliminated.

3. What types of problems can be solved using "E to the -y plus trial and error math"?

"E to the -y plus trial and error math" can be used to solve a variety of problems, including optimization problems, root-finding problems, and equation solving. It is often used in situations where an exact solution cannot be easily determined or when an approximate solution is sufficient.

4. What are the advantages of using "E to the -y plus trial and error math"?

One of the main advantages of "E to the -y plus trial and error math" is its versatility. It can be applied to a wide range of problems and can often provide a solution even when other methods fail. Additionally, it does not require advanced mathematical knowledge and can be easily implemented using a computer or by hand.

5. Are there any limitations to using "E to the -y plus trial and error math"?

While "E to the -y plus trial and error math" can be a useful approach in many situations, it also has some limitations. It can be time-consuming and may require multiple iterations to obtain a satisfactory solution. Additionally, it may not always provide the most accurate or precise solution, as it relies on approximation rather than exact calculation.

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