# A few questions.Watch

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#1
(1) I know that the integral of 1/(2x-1) dx = (1/2)ln(2x-1).
Why can it not also be done as INT 1/2(x-.5) = (1/2)INT(1/x-.5) = (1/2)ln(x-0.5), which gives a completely different answer.

(2) Concerning the equation coshu = x
e^u + e^-u = 2x
e^2u + 1 = 2xe^u
e^2u - 2xe^u + 1 = 0

e^u = 2x(+-)rt[4x^2 - 4] / 2
e^u = x(+-)rt[x^2-1] (x>=1)

u = ln[x+-rt(x^2 - 1)]. (x>=1)
arcoshx = ln[x+-rt(x^2-1)]

However, my P5 book neglects the negative root.
It then later says in an answer to a question that coshx=2 has two roots, but arcosh2 has only one root. If they were to include the negative root (as above however) then arcosh 2 would have two roots. Surely coshx=2 and x=arcosh2 should be equivelant?
If they were to include the negative root illustrated in the formula

0
14 years ago
#2
(Original post by Gaz031)
(1) I know that the integral of 1/(2x-1) dx = (1/2)ln(2x-1).
Why can it not also be done as INT 1/2(x-.5) = (1/2)INT(1/x-.5) = (1/2)ln(x-0.5), which gives a completely different answer.
Your two answers may be different - but they are only a constant apart - you forgot to put in the constant of integration.

(Original post by Gaz031)
[(2) Concerning the equation coshu = x
e^u + e^-u = 2x
e^2u + 1 = 2xe^u
e^2u - 2xe^u + 1 = 0

e^u = 2x(+-)rt[4x^2 - 4] / 2
e^u = x(+-)rt[x^2-1] (x>=1)

u = ln[x+-rt(x^2 - 1)]. (x>=1)
arcoshx = ln[x+-rt(x^2-1)]

However, my P5 book neglects the negative root.
It then later says in an answer to a question that coshx=2 has two roots, but arcosh2 has only one root. If they were to include the negative root (as above however) then arcosh 2 would have two roots. Surely coshx=2 and x=arcosh2 should be equivelant?
If they were to include the negative root illustrated in the formula

If x>1 then arccoshx has two possibilities. (Think about the graph.) By ignoring the - and taking the + you are choosing the positive value of arccoshx.

The thing is, you want arccosh to be a function, just like you want arcsin to be - even though sinx = 1/2 has infinitely many solutions we choose

arcsin(1/2) = pi/6.
0
14 years ago
#3
(Original post by Gaz031)
(1) I know that the integral of 1/(2x-1) dx = (1/2)ln(2x-1).
Why can it not also be done as INT 1/2(x-.5) = (1/2)INT(1/x-.5) = (1/2)ln(x-0.5), which gives a completely different answer.

(2) Concerning the equation coshu = x
e^u + e^-u = 2x
e^2u + 1 = 2xe^u
e^2u - 2xe^u + 1 = 0

e^u = 2x(+-)rt[4x^2 - 4] / 2
e^u = x(+-)rt[x^2-1] (x>=1)

u = ln[x+-rt(x^2 - 1)]. (x>=1)
arcoshx = ln[x+-rt(x^2-1)]

However, my P5 book neglects the negative root.
It then later says in an answer to a question that coshx=2 has two roots, but arcosh2 has only one root. If they were to include the negative root (as above however) then arcosh 2 would have two roots. Surely coshx=2 and x=arcosh2 should be equivelant?
If they were to include the negative root illustrated in the formula

1) I think u forgot the constant, let's put the limits in it, I hope they give u the same answer.
2) Actually, the book forgot the negative root. However the function arcoshx just gives you one solution. Because ... I forgot what it's called. But remember arcsinx or arccosx ... they just take one value .... but you can get others by plus 2kpi.
What I mean is If y = f(x) so y just have only one value although x has many values. Here, u = arcoshx, so u just have one root. But the question ask x = coshu -> u can have more than 1 for each value of x.
0
#4
(1) Thanks. I now realise they are simply an addition of ln2 apart.

(2) Arcsin(0.5) = u has many solutions that are simply gained by adding suitable multiples though. It isn't restricted to just one like they are restricting arcosh. I'm still not happy with this. They're missing a root to make it easier to write out the function? Surely that's sheer laziness. Lets not draw parallels with the trigonetric functions as this result is actually derived using algebra. They don't miss out the negative root for the general solution to quadratic equations! Sorry for being picky but i like to have everything nailed down if i'm going to study this at university.
So is there any really good reason why the negative root is rejected?
0
14 years ago
#5
(Original post by Gaz031)
(2) Arcsin(0.5) = u has many solutions that are simply gained by adding suitable multiples though. It isn't restricted to just one like they are restricting arcosh. I'm still not happy with this. They're missing a root to make it easier to write out the function? Surely that's sheer laziness. Lets not draw parallels with the trigonetric functions as this result is actually derived using algebra. They don't miss out the negative root for the general solution to quadratic equations! Sorry for being picky but i like to have everything nailed down if i'm going to study this at university.
So is there any really good reason why the negative root is rejected?
No it doesn't - arcsin(1/2) = pi/3 is the only value. Functions are allowed to only take one value, this is part of the definition of what it is to be a function. In the same way rt(2) means the positive root of 2 - it doesn't denote two numbers and that decision has been made to make rt(x) into a function for positive x.
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