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General => General Technical Chat => Topic started by: J4e8a16n on July 11, 2013, 09:45:58 pm
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Hi,
I checked an integration 'error?' in my book . See the red ! .
Unfortunately the book is in french. Anyhow . . .
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Yeah, should be -1/3 ln |5-I|. Somebody missed a factor of three...
In my experience you don't usually make it far into a textbook with a bunch of math before you find an arithmetic error. Maybe my school just picks shitty books, though...
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Hmm? Let's work it backwards:
f(I) = -1/3 ln(15 - 3I)
Let u = 15 - 3I, then df/dI = (df/du)(du/dI), where
f(u) = -1/3 ln u
So:
df/du = (-1/3)(1/u)
du/dI = -3
df/dI = (-1/3)(1/u)(-3) = 1/u = 1 / (15 - 3I)
Which is what the book has.
So what did I get wrong?
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Actually, you're right, thank you!
The catch here is that we came to different answers through different methods, but both are correct because of the constant that is being ignored.
ln(5 - i) = ln(15 - 3i) - ln(3)
And ln(3) is of course a constant.
(I used a lowercase i for legibility)
Kind of ashamed I missed that, I used to grade math papers... :-[
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Why not that way?
integration sign ((1/ (15-3i) ) dI) =
ln|(1/ (15-3i) | =
-ln|(15-3i) | =
-|15-3i| = e^(t-C)
and be stuck with no 1/3 :-/
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Why not that way?
integration sign ((1/ (15-3i) ) dI) =
ln|(1/ (15-3i) | =
-ln|(15-3i) | =
-|15-3i| = e^(t-C)
and be stuck with no 1/3 :-/
Because your working is wrong at the line indicated in red. The antiderivative of 1/x is ln x, and you are missing a factor of (-1/3):
http://www.wolframalpha.com/input/?i=integrate+1%2F%2815-3i%29+di (http://www.wolframalpha.com/input/?i=integrate+1%2F%2815-3i%29+di)
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substitute u = 15-3 i
du/di = 0-3
and du = -3 di:
du/-3 = di
1/-3 *du = di
ok
-1/3 log(15+3i)) constant = t- C
-1/3 log(15+3i)) = t- (C +constant)
log(15+3i)) =-3( t- C)
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Why Wolfgang say?
Substitute back for u = 15-3 i: = -1/3 log(-3 (-5+i))+constant
Which is equivalent for restricted i values to:
Answer: |
| = -1/3 log(5-i)+constant
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substitute u = 15-3 i
du/di = 0-3
and du = -3 di:
du/-3 = di
1/-3 *du = di
ok
-1/3 log(15+3i)) constant = t- C
-1/3 log(15+3i)) = t- (C +constant)
log(15+3i)) =-3( t- C)
Be careful with your signs. In several places there you have changed a "-" into a "+". If you do that in a real problem you will lose marks in a test, or build something that doesn't work the way you expect...
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For a long time, almost every paper and book I read had errors in the equations, probably so that those who don't understand the maths can't make use of them.