49t+245e^(-t/5)-545=0, by some other means than a calculator or wolfram?

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49t+245e^(-t/5)-545=0, by some other means than a calculator or wolfram?

Mathematics
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if x is very small than 1, then e^x can be expanded as 1+x+x^2/2! and higher order terms can be neglected let's assume -t/5 << 1 then e^(-t/5)=1-t/5+t^2/50 substitute this in the equation 49t+245(1-t/5+t^2/50)-545=0 49t+245-49t+4.9t^2-545=0 combining like terms 4.9t^2=300 \[t=\sqrt{\frac{300}{4.9}}\] \[t=7.8246\]
cool man, series expansion, thanks
  • phi
but t/5 is not << 1

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phi we don't know about that, so I just assumed it
it would be if the premise is that t is small to begin with, in that case t/5 would be smaller
but it is that assumption
thanks unless you have a belligerent way to do it phi
  • phi
did you plug your answer back into the original equation?
phi you're right it won't work, we'll have to include higher order terms
not yet, I just wanted to get an idea of where to go before I worked on it. It seems reasonable if I do an longer expansion
I just didn't think of doing the expansion in the first place
rather couldn't
I've got to roll, but thanks for the help guys
Yeah daomowon , now you include the cubic power, find t and substitute back, if it doesn't then include higher power
Do you mean no computers and no wolfram? A problem like this without calculators would take a while!
  • phi
It's not clear what the constraints are on how to do this problem. But generally, you would solve it using numerical techniques. such as http://en.wikipedia.org/wiki/Newton's_method see attached pdf for result
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I would approach this using numerical methods, like Newton's method. We should note that the middle (e^(-t/5) term is insignificant when t>0, which gives the first approximation as x=545/49=11.1. It converge to10.5116 in a few more iterations. When t<0, both terms come into play, but e^(-t/5) has more influence. It should do the same towards -6.
yeah, I figure if I can get at these types of things without computers and stuff, I can get at wider selection of material better in the long run. thanks for all the insight though, I couldn't make the leap I needed in the beginning, but this all helps a whole lot

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