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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

let y = ae^x + be^(-x)

show that for any values a and b where a ≠ 0 and b ≠ 0 there exists two values A and B such that y equals either A sinh (x + B) or A cosh (x + B)

*Last edited by mikau (2006-10-27 05:04:52)*

A logarithm is just a misspelled algorithm.

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**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

I'm pretty sure that if you allow these numbers to be complex you can actually always pick A and B so that this expression equals Asinh(x+B). It looks as though the problem you're trying to solve requires you to keep your constants real. Then depending on the values of a and b you may only be able to get one of the sinh and the cosh expressions to work without having lograthim of a negative number involved.

I recommend this approach: use the exponential expansions for your hyperbolic functions:

[align=center]

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You want to show that there are some values you can pick for A and B such that y=y1 or y=y2. Start with y1 and pick A in terms of a and B such that the positive exponential in x has the correct coefficient, and then see if you can figure out what B needs to be so that the negative exponential in x has the correct coefficient. Your final answer will only be real depending on the relative signs of a and b. You will then need to show that your y2 expression takes care of the cases of a and b that give imaginary results for your y1 answer.

Does that make sense? Let me know if you need more detailed advice and I'll try to get back to it sometime in the next couple of days.

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