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P5 hyperbolic functions
Hi
I know the reason for arsinhx=ln(x+√(x²+1)) and not ±
but why can arcoshx not =ln(x±√(x²-1)), surely it could be ± and does not have to be +
I know the reason for arsinhx=ln(x+√(x²+1)) and not ±
but why can arcoshx not =ln(x±√(x²-1)), surely it could be ± and does not have to be +
Scroll to see replies
lgs98jonee
Hi
I know the reason for arsinhx=ln(x+√(x²+1)) and not ±
but why can arcoshx not =ln(x±√(x²-1)), surely it could be ± and does not have to be +
I know the reason for arsinhx=ln(x+√(x²+1)) and not ±
but why can arcoshx not =ln(x±√(x²-1)), surely it could be ± and does not have to be +
I asked the same question when I met it! Still have the pencil notes in the margin of my text book!
Aitch
[Sits patiently and waits for someone to provide an answer...]
I also wondered that.
I think ln[x+(x^2-1)^0.5] is just regarded as the principial root of coshy=x and that you can easily do +-ln[x+(x^2-1)^0.5] to find both roots as cosh is an even function.
It's similar to how sinx=0.5 has an infinite number of roots, but x=pi/6 is considered to be the principial one.
I'd be interested to hear it if someone had a proper explanation though.
I think ln[x+(x^2-1)^0.5] is just regarded as the principial root of coshy=x and that you can easily do +-ln[x+(x^2-1)^0.5] to find both roots as cosh is an even function.
It's similar to how sinx=0.5 has an infinite number of roots, but x=pi/6 is considered to be the principial one.
I'd be interested to hear it if someone had a proper explanation though.
Gaz031
I also wondered that.
I think ln[x+(x^2-1)^0.5] is just regarded as the principial root of coshy=x and that you can easily do +-ln[x+(x^2-1)^0.5] to find both roots as cosh is an even function.
It's similar to how sinx=0.5 has an infinite number of roots, but x=pi/6 is considered to be the principial one.
I'd be interested to hear it if someone had a proper explanation though.
I think ln[x+(x^2-1)^0.5] is just regarded as the principial root of coshy=x and that you can easily do +-ln[x+(x^2-1)^0.5] to find both roots as cosh is an even function.
It's similar to how sinx=0.5 has an infinite number of roots, but x=pi/6 is considered to be the principial one.
I'd be interested to hear it if someone had a proper explanation though.
It doesnt look like there is one but the textbook claims that it can be shown
lgs98jonee
It doesnt look like there is one but the textbook claims that it can be shown
It is true, but its never written like that. Instead it is written
arcosh x = ± ln(x+√(x²-1))
Since 1/(x-√(x²-1)) = x+√(x²-1)
Makes sence because its easy to see that cosh is symmetrical about x=0.
Just use the relationship between sinh and cosh. If you know the reason for arcsinh having the positive root, it sorta carries over to arccosh.
* arsinhx=ln(x+√(x²+1))
consider
coshy = x
sinhy = √(x² -1)
this implies that
arcsinh[ √(x² -1) ] = y = arccoshx
Hence in (*) setting x to √(x² -1)
arcsinh[ √(x² -1) ] = ln(√(x²-1) + x) = arccoshx
* arsinhx=ln(x+√(x²+1))
consider
coshy = x
sinhy = √(x² -1)
this implies that
arcsinh[ √(x² -1) ] = y = arccoshx
Hence in (*) setting x to √(x² -1)
arcsinh[ √(x² -1) ] = ln(√(x²-1) + x) = arccoshx
JamesF
It is true, but its never written like that. Instead it is written
arcosh x = ± ln(x+√(x²-1))
Since 1/(x-√(x²-1)) = x+√(x²-1)
Makes sence because its easy to see that cosh is symmetrical about x=0.
arcosh x = ± ln(x+√(x²-1))
Since 1/(x-√(x²-1)) = x+√(x²-1)
Makes sence because its easy to see that cosh is symmetrical about x=0.
Hmm perhaps not, I can't say I really know how it's written conventionally in the real world of maths but I know I've been asked to justify arccosh having the positive root in P5 before and that's how I did it.
I think it has to do with the fact that coshx isn't 1-1. However, if we were to consider x>=0, then we can get a 1-1 mapping. In fact, since cosh0=1, we can directly see that the domain of its inverse function is x>=1 and its range is y>=0 (that is if we consider coshx for x>=0).
Let's obtain the logarithmic form of arcoshx:
x = coshu = 0.5(e^u + e^(-u))
=> e^(2u) - x.e^(u) + 1 = 0
Using the quadratic formula:
e^u = x ± sqrt(x² - 1)
u = ln[x ± sqrt(x²-1)] = arcoshx
The tricky part is to spot the following:
[x - sqrt(x²-1)][x + sqrt(x²-1)] = x² - (x²-1) = 1
Now:
x - sqrt(x²-1) = 1/[x + sqrt(x²-1)] = [x + sqrt(x²-1)]^(-1)
Thus we can re-write arcoshx as:
arcoshx = ln{[x + sqrt(x²-1)]^(±1)} = ± ln[x + sqrt(x²-1)]
But we know that arcoshx>=0 by definition, so we disregard the - sign and end up with:
arcosh = ln[x + sqrt(x²-1)]
Let's obtain the logarithmic form of arcoshx:
x = coshu = 0.5(e^u + e^(-u))
=> e^(2u) - x.e^(u) + 1 = 0
Using the quadratic formula:
e^u = x ± sqrt(x² - 1)
u = ln[x ± sqrt(x²-1)] = arcoshx
The tricky part is to spot the following:
[x - sqrt(x²-1)][x + sqrt(x²-1)] = x² - (x²-1) = 1
Now:
x - sqrt(x²-1) = 1/[x + sqrt(x²-1)] = [x + sqrt(x²-1)]^(-1)
Thus we can re-write arcoshx as:
arcoshx = ln{[x + sqrt(x²-1)]^(±1)} = ± ln[x + sqrt(x²-1)]
But we know that arcoshx>=0 by definition, so we disregard the - sign and end up with:
arcosh = ln[x + sqrt(x²-1)]
Spenceman_
Hmm perhaps not, I can't say I really know how it's written conventionally in the real world of maths but I know I've been asked to justify arccosh having the positive root in P5 before and that's how I did it.
Because its a function and thus cannot be 1 to many, so you take the positive solution?
Gaz031
Why do we know that?
From my first paragraph:
In fact, since cosh0=1, we can directly see that the domain of its inverse function is x>=1 and its range is y>=0 (that is if we consider coshx for x>=0).
To make it clearer:
f(x) = coshx, x R, x>=0 (which means range>=1)
f-1(x) = arcoshx, x R, x>=1 (range = domain of f(x), i.e. >=0)
dvs
From my first paragraph:
In fact, since cosh0=1, we can directly see that the domain of its inverse function is x>=1 and its range is y>=0 (that is if we consider coshx for x>=0).
To make it clearer:
f(x) = coshx, x R, x>=0 (which means range>=1)
f-1(x) = arcoshx, x R, x>=1
In fact, since cosh0=1, we can directly see that the domain of its inverse function is x>=1 and its range is y>=0 (that is if we consider coshx for x>=0).
To make it clearer:
f(x) = coshx, x R, x>=0 (which means range>=1)
f-1(x) = arcoshx, x R, x>=1
Why are we restricting ourselves to considering coshx only for x>=0, surely if x isn't defined as positive then we have to define arcoshx as +-ln[x+rt(x^2-1)]?
Gaz031
Why do we know that? That isn't explaining anything. You've derived +-ln[x+rt(x^2-1)] which we all know then explained the '-' away by 'definition'?
Because its multivalued, so at the start you need to define which value you are going to take, otherwise you dont have a function anymore. Exactly the same as taking the positive square root, and hence the need for the plus-minus sign.
JamesF
Because its multivalued, so at the start you need to define which value you are going to take, otherwise you dont have a function anymore. Exactly the same as taking the positive square root, and hence the need for the plus-minus sign.
Hrm..
If we need to take only one value then why don't we instead define it as ln[x-rt(x^2-1)] (alternatively writable as -ln[x+rt(x^2-1)])?
Coshx=y and x=arcoshy don't represent the same relationship then? [In the first x can be positive or negative, in the second x is only positive]
yeh that's nice, getting the +/- outside.
Gaz031
Why are we restricting ourselves to considering coshx only for x>=0, surely if x isn't defined as positive then we have to define arcoshx as +-ln[x+rt(x^2-1)]?
Because coshx isn't 1-1 (it's multivalued), so its inverse doesn't exist for all the values coshx is defined for. Hence we must define a set of values for a domain on which coshx is 1-1. A set of values that works is [0, inf) (which is the domain of f(x)=coshx), and its image-set is [1, inf) (which is the range of f(x)=coshx).
So the inverse function has domain [1, inf) and range [0, inf), which means it's always positive for all x it's defined for.
Gaz031
Hrm..
If we need to take only one value then why don't we instead define it as ln[x-rt(x^2-1)] (alternatively writable as -ln[x+rt(x^2-1)])?
Coshx=y and x=arcoshy don't represent the same relationship then? [In the first x can be positive or negative, in the second x is only positive]
If we need to take only one value then why don't we instead define it as ln[x-rt(x^2-1)] (alternatively writable as -ln[x+rt(x^2-1)])?
Coshx=y and x=arcoshy don't represent the same relationship then? [In the first x can be positive or negative, in the second x is only positive]
You can if you like, you just have to pick one or the other at the start otherwise you wont have a 1-1 map.
Gaz031
Why is it so bad for an inverse to be a 1 to 2 mapping?
Sorry to be so persistent, even though I know things are 'defined' against me. I guess this just isn't one of those things I'm going to be like.
Sorry to be so persistent, even though I know things are 'defined' against me. I guess this just isn't one of those things I'm going to be like.
An inverse function can't exist if the function isn't one-one.
Think of it this way..
Suppose you have a function f:X->Y that is 1-1. This means that for every Y there is a unique X, and so we can define a function g:Y->X, which is defined as the inverse function of f:X->Y. If f:X->Y didn't have a 1-1 mapping, then we simply can't obtain a unique X for every Y, and hence its inverse doesn't exist.
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