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Topic: Loss of trig memory... (Read 1333 times) |
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jpk2009
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Loss of trig memory...
« on: May 2nd, 2010, 12:29pm » |
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You are rushing to class to take your final exam and you slip, fall, and hit your head on the floor. This injury causes you to completely lose your memory about the derivatives and anti-derivatives of all the the trig functions as well as their power series expansions.
You panic when you see the integration problem on the exam
int exp(-x) cos(x) dx
but you ace the exam.
What did you do?
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jpk2009
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Re: Loss of trig memory...
« Reply #2 on: May 2nd, 2010, 4:06pm » |
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That's it. It takes a little work to simply the product after integration but it works. I just learned this approach this week and so it excited me enough to share it someway.
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Obob
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Re: Loss of trig memory...
« Reply #3 on: May 2nd, 2010, 4:12pm » |
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In some sense, if you remember the differentiation formula for the complex exponential, you remember them for sin and cos as well:
exp(ix) = cos(x) + i sin(x)
-sin(x) + i cos(x) = i(cos(x) + i sin(x))= i exp(i x) = d/dx (exp(ix)) = d/dx(cos(x) + i sin(x)) = d/dx (cos(x)) + i d/dx (sin(x)),
from which it follows that d/dx (cos(x)) = -sin(x) and d/dx (sin(x)) = cos(x).
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jpk2009
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Re: Loss of trig memory...
« Reply #4 on: May 2nd, 2010, 4:18pm » |
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Cool. But that does require me to remember something about the derivatives of the trig functions. I am not suppose to know that but thanks for pointing this out.
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Obob
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Re: Loss of trig memory...
« Reply #5 on: May 2nd, 2010, 4:21pm » |
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All I'm saying is that if you remember cos(x) = (exp(ix)+exp(-ix))/2 (which is one-half of the Euler identity exp(ix) = cos(x) + i sin(x)) and you remember the derivative of exp(x), then you essentially know the derivatives of the trig functions.
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jpk2009
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Re: Loss of trig memory...
« Reply #6 on: May 2nd, 2010, 6:39pm » |
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Oh yeah you're right. Thanks.
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Michael Dagg
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Re: Loss of trig memory...
« Reply #7 on: Aug 9th, 2010, 10:56am » |
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You didn't say what class it is so the answers will vary.
I think the best answer is to reconstruct the lost memory
since you need only know a few facts to compute the limits of
(cos(x+h)-cos(x))/h and sin(x+h)-sin(x))/h, both as h->0,
and then you can get the antiderivatives using the FTC.
Perhaps a better way to have put your question across would
be "how does one solve the problem without using integration by
parts or Stieltjes integration."
An engineering student or student with a little complex analysis
background may calculate
\int cos(x) dx
= Re[\int e^{ix} dx]
= Re[1/i e^{ix}] + C
= Re[-ie^{ix}] + C
= Re[-i(cos(x) + i sin(x))] + C
= Re[-i cos(x) + sin(x)] + C
= Re[sin(x) - i cos(x)] + C
= sin(x) + C
and then do a similar thing for \int sin(x) dx (neither
of which require differentiation or antiderviation of
cosine or sine. From these two and by the FTC you'll have
enough restored memory to use integration by parts.
Better yet, notice that e^{-x} cos(x) = e^{-x} Re[e^{ix}]
and so
\int e^{-x} cos(x) dx
= Re[\int e^{-x} e^{ix} dx]
= Re[\int e^{(-1+i)x} dx]
= Re[1/(-1+i) e^{(-1+i)x)}] + C
= Re[1/(-1+i) (1-i)/(1-i) e^{(-1+i)x)}] + C
= Re[-1/2(1+i) e^{(-1+i)x}] + C
= Re[-1/2(1+i) e^{-x} e^{ix}] + C
= Re[-1/2(1+i) e^{-x}(cos(x) + i sin(x))] + C
= Re[-1/2 e^{-x} (1+i)(cos(x) + i sin(x))] + C
= Re[-1/2 e^{-x}(cos(x) + i sin(x) + i cos(x) + i^2 sin(x))] + C
= Re[-1/2 e^{-x}(cos(x) - sin(x)) + e^{-x} (i sin(x) + i cos(x))] + C
= -1/2 e^{-x}(cos(x) - sin(x)) + C,
= -1/2 e^{-x}cos(x) + 1/2 e^{-x} sin(x) + C ,
thus requiring no knowledge of the dervrivative, antiderivative
among the cosine, sine functions.
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| « Last Edit: Aug 9th, 2010, 10:58am by Michael Dagg » |
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Regards,
Michael Dagg
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