$tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$?
$begingroup$
$tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $tan^{-1}x+tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $tan^{-1}x+tan^{-1}y+tan^{-1}z$ ?
begin{align}
tan^{-1}x+tan^{-1}y+tan^{-1}z&=
begin{cases}tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy < 1 \[1.5ex]
pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y>0 \[1.5ex]
-pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y<0 end{cases}\
&=
end{align}
trigonometry inverse-function
$endgroup$
add a comment |
$begingroup$
$tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $tan^{-1}x+tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $tan^{-1}x+tan^{-1}y+tan^{-1}z$ ?
begin{align}
tan^{-1}x+tan^{-1}y+tan^{-1}z&=
begin{cases}tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy < 1 \[1.5ex]
pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y>0 \[1.5ex]
-pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y<0 end{cases}\
&=
end{align}
trigonometry inverse-function
$endgroup$
$begingroup$
Well ... apart from restrictions of a similar nature that apply to the arguments of the formula for two variables ie in that case xy<1 . I seem to remember, when I once checked this out for the general case of n variables, that you get the odd symmetric polynomials of degree ≤ n in the numerator & the even symmetric [ ditto ] in the denominator ... something like that, anyway.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 18:56
$begingroup$
I'm beginning to get the gist of your question now I think - I glossed over it a bit at first ... are you saying that when you go beyond two variables you lose the property of it being valid ∀ possible values of whatever variables it imports (provided the other branches of atn be admitted) ... that even if you do admit these other branches that property ceases by reason of going from two to three variables? That would be strange if it were so, and I would not have thought it were so ... but I can see how it might actually be something we can't safely assume by default.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 7:49
add a comment |
$begingroup$
$tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $tan^{-1}x+tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $tan^{-1}x+tan^{-1}y+tan^{-1}z$ ?
begin{align}
tan^{-1}x+tan^{-1}y+tan^{-1}z&=
begin{cases}tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy < 1 \[1.5ex]
pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y>0 \[1.5ex]
-pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y<0 end{cases}\
&=
end{align}
trigonometry inverse-function
$endgroup$
$tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $tan^{-1}x+tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $tan^{-1}x+tan^{-1}y+tan^{-1}z$ ?
begin{align}
tan^{-1}x+tan^{-1}y+tan^{-1}z&=
begin{cases}tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy < 1 \[1.5ex]
pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y>0 \[1.5ex]
-pi + tan^{-1}left(dfrac{x+y}{1-xy}right)+tan^{-1}z, &xy>1,::x,y<0 end{cases}\
&=
end{align}
trigonometry inverse-function
trigonometry inverse-function
edited Dec 9 '18 at 14:43
Martin Sleziak
44.8k9118272
44.8k9118272
asked Dec 5 '18 at 18:47
ss1729ss1729
1,9211723
1,9211723
$begingroup$
Well ... apart from restrictions of a similar nature that apply to the arguments of the formula for two variables ie in that case xy<1 . I seem to remember, when I once checked this out for the general case of n variables, that you get the odd symmetric polynomials of degree ≤ n in the numerator & the even symmetric [ ditto ] in the denominator ... something like that, anyway.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 18:56
$begingroup$
I'm beginning to get the gist of your question now I think - I glossed over it a bit at first ... are you saying that when you go beyond two variables you lose the property of it being valid ∀ possible values of whatever variables it imports (provided the other branches of atn be admitted) ... that even if you do admit these other branches that property ceases by reason of going from two to three variables? That would be strange if it were so, and I would not have thought it were so ... but I can see how it might actually be something we can't safely assume by default.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 7:49
add a comment |
$begingroup$
Well ... apart from restrictions of a similar nature that apply to the arguments of the formula for two variables ie in that case xy<1 . I seem to remember, when I once checked this out for the general case of n variables, that you get the odd symmetric polynomials of degree ≤ n in the numerator & the even symmetric [ ditto ] in the denominator ... something like that, anyway.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 18:56
$begingroup$
I'm beginning to get the gist of your question now I think - I glossed over it a bit at first ... are you saying that when you go beyond two variables you lose the property of it being valid ∀ possible values of whatever variables it imports (provided the other branches of atn be admitted) ... that even if you do admit these other branches that property ceases by reason of going from two to three variables? That would be strange if it were so, and I would not have thought it were so ... but I can see how it might actually be something we can't safely assume by default.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 7:49
$begingroup$
Well ... apart from restrictions of a similar nature that apply to the arguments of the formula for two variables ie in that case xy<1 . I seem to remember, when I once checked this out for the general case of n variables, that you get the odd symmetric polynomials of degree ≤ n in the numerator & the even symmetric [ ditto ] in the denominator ... something like that, anyway.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 18:56
$begingroup$
Well ... apart from restrictions of a similar nature that apply to the arguments of the formula for two variables ie in that case xy<1 . I seem to remember, when I once checked this out for the general case of n variables, that you get the odd symmetric polynomials of degree ≤ n in the numerator & the even symmetric [ ditto ] in the denominator ... something like that, anyway.
$endgroup$
– AmbretteOrrisey
Dec 5 '18 at 18:56
$begingroup$
I'm beginning to get the gist of your question now I think - I glossed over it a bit at first ... are you saying that when you go beyond two variables you lose the property of it being valid ∀ possible values of whatever variables it imports (provided the other branches of atn be admitted) ... that even if you do admit these other branches that property ceases by reason of going from two to three variables? That would be strange if it were so, and I would not have thought it were so ... but I can see how it might actually be something we can't safely assume by default.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 7:49
$begingroup$
I'm beginning to get the gist of your question now I think - I glossed over it a bit at first ... are you saying that when you go beyond two variables you lose the property of it being valid ∀ possible values of whatever variables it imports (provided the other branches of atn be admitted) ... that even if you do admit these other branches that property ceases by reason of going from two to three variables? That would be strange if it were so, and I would not have thought it were so ... but I can see how it might actually be something we can't safely assume by default.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 7:49
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let
- $L(x,y,z) = tan^{-1} x + tan^{-1} y + tan^{-1} z$
- $R(x,y,z) = tan^{-1}left(frac{x+y+z - xyz}{1- xy - yz - zx}right)$
By addition formula of tangent function, we have
$$tan L(x,y,z) = tan R(x,y,z)$$
Since $tan theta$ is a periodic function with period $pi$, there is an integer valued function $N(x,y,z)$ such that $$L(x,y,z) = R(x,y,z) + N(x,y,z)pi$$
Since $tan^{-1}theta$ maps $mathbb{R}$ into $(-frac{pi}{2}, frac{pi}{2})$, we have
$$|L(x,y,z)| < frac{3pi}{2} land |R(x,y,z)| < frac{pi}{2}quadimpliesquad N(x,y,z) in { 0, pm 1 }$$
Since $tan^{-1} theta$ is a continuous function for all $theta$,
$N(x,y,z)$ will be constant over those domain where $xy+yz+zx ne 1$.
Notice
$$xy+yz+zx = 1 iff 3left(frac{x+y+z}{sqrt{3}}right)^2 - ( x^2 + y^2 + z^2 ) = 2$$
is the equation of a two sheet hyperboloid centered at origin with symmetric axis pointing along the direction $(1,1,1)$. The complement of this hyperboloid consists of $3$ connected components. One can pick a point from each of these component and figure out the value of $N(x,y,z)$ over the whole component.
The end result is
$$L(x,y,z) = R(x,y,z) + begin{cases}
pi, & 1 < xy+yz+zx land x+y+z > 0\
0, & 1 > xy+yz+zx\
-pi & 1 < xy+yz+zx land x+y+z < 0
end{cases}
$$
$endgroup$
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:27
|
show 4 more comments
$begingroup$
Write $a:=arctan x$ etc. so $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan a+tan b + (1-tan atan b)tan c}{1-tan atan b - (tan a+tan b)tan c}.$$If $xyne 1$, we can cancel $1-tan atan b$ to get $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan (a+b)+tan c}{1-tan (a+b)tan c}.$$If $tan (a+b)tan cne 1$ i.e. $(x+y)zne 1-xy$ i.e. $xy+yz+zx=1$, we have $$frac{x+y+z-xyz}{1-xy-yz-zx}=tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
$endgroup$
add a comment |
$begingroup$
We have that by addition formula
$$tan^{-1}x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
then
$$(tan^{-1}x+tan^{-1}y)+tan^{-1}z=tan^{-1}frac{frac{x+y}{1-xy}+z}{1-frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
- $1-xy=0$
and is certainly not well defined for
- $1-xy-yz-zx=0$
$endgroup$
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
add a comment |
$begingroup$
This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors.
Let $alpha=arctan x$, $beta=arctan y$ and $gamma=arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $alpha+beta+gamma$ is the argument
(up to an integer multiple of $2pi$) of the product
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $phi$ of a complex number $a+ib$ satisfies $tanphi=b/a$.
The claim follows from this.
Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $pi$. For example, if $x=y=z=1$ we have $arctan x=arctan y=arctan z=pi/4$ giving $3pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $arctan(-1)=-pi/4$ on the right hand side.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let
- $L(x,y,z) = tan^{-1} x + tan^{-1} y + tan^{-1} z$
- $R(x,y,z) = tan^{-1}left(frac{x+y+z - xyz}{1- xy - yz - zx}right)$
By addition formula of tangent function, we have
$$tan L(x,y,z) = tan R(x,y,z)$$
Since $tan theta$ is a periodic function with period $pi$, there is an integer valued function $N(x,y,z)$ such that $$L(x,y,z) = R(x,y,z) + N(x,y,z)pi$$
Since $tan^{-1}theta$ maps $mathbb{R}$ into $(-frac{pi}{2}, frac{pi}{2})$, we have
$$|L(x,y,z)| < frac{3pi}{2} land |R(x,y,z)| < frac{pi}{2}quadimpliesquad N(x,y,z) in { 0, pm 1 }$$
Since $tan^{-1} theta$ is a continuous function for all $theta$,
$N(x,y,z)$ will be constant over those domain where $xy+yz+zx ne 1$.
Notice
$$xy+yz+zx = 1 iff 3left(frac{x+y+z}{sqrt{3}}right)^2 - ( x^2 + y^2 + z^2 ) = 2$$
is the equation of a two sheet hyperboloid centered at origin with symmetric axis pointing along the direction $(1,1,1)$. The complement of this hyperboloid consists of $3$ connected components. One can pick a point from each of these component and figure out the value of $N(x,y,z)$ over the whole component.
The end result is
$$L(x,y,z) = R(x,y,z) + begin{cases}
pi, & 1 < xy+yz+zx land x+y+z > 0\
0, & 1 > xy+yz+zx\
-pi & 1 < xy+yz+zx land x+y+z < 0
end{cases}
$$
$endgroup$
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:27
|
show 4 more comments
$begingroup$
Let
- $L(x,y,z) = tan^{-1} x + tan^{-1} y + tan^{-1} z$
- $R(x,y,z) = tan^{-1}left(frac{x+y+z - xyz}{1- xy - yz - zx}right)$
By addition formula of tangent function, we have
$$tan L(x,y,z) = tan R(x,y,z)$$
Since $tan theta$ is a periodic function with period $pi$, there is an integer valued function $N(x,y,z)$ such that $$L(x,y,z) = R(x,y,z) + N(x,y,z)pi$$
Since $tan^{-1}theta$ maps $mathbb{R}$ into $(-frac{pi}{2}, frac{pi}{2})$, we have
$$|L(x,y,z)| < frac{3pi}{2} land |R(x,y,z)| < frac{pi}{2}quadimpliesquad N(x,y,z) in { 0, pm 1 }$$
Since $tan^{-1} theta$ is a continuous function for all $theta$,
$N(x,y,z)$ will be constant over those domain where $xy+yz+zx ne 1$.
Notice
$$xy+yz+zx = 1 iff 3left(frac{x+y+z}{sqrt{3}}right)^2 - ( x^2 + y^2 + z^2 ) = 2$$
is the equation of a two sheet hyperboloid centered at origin with symmetric axis pointing along the direction $(1,1,1)$. The complement of this hyperboloid consists of $3$ connected components. One can pick a point from each of these component and figure out the value of $N(x,y,z)$ over the whole component.
The end result is
$$L(x,y,z) = R(x,y,z) + begin{cases}
pi, & 1 < xy+yz+zx land x+y+z > 0\
0, & 1 > xy+yz+zx\
-pi & 1 < xy+yz+zx land x+y+z < 0
end{cases}
$$
$endgroup$
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
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– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
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– AmbretteOrrisey
Dec 6 '18 at 8:27
|
show 4 more comments
$begingroup$
Let
- $L(x,y,z) = tan^{-1} x + tan^{-1} y + tan^{-1} z$
- $R(x,y,z) = tan^{-1}left(frac{x+y+z - xyz}{1- xy - yz - zx}right)$
By addition formula of tangent function, we have
$$tan L(x,y,z) = tan R(x,y,z)$$
Since $tan theta$ is a periodic function with period $pi$, there is an integer valued function $N(x,y,z)$ such that $$L(x,y,z) = R(x,y,z) + N(x,y,z)pi$$
Since $tan^{-1}theta$ maps $mathbb{R}$ into $(-frac{pi}{2}, frac{pi}{2})$, we have
$$|L(x,y,z)| < frac{3pi}{2} land |R(x,y,z)| < frac{pi}{2}quadimpliesquad N(x,y,z) in { 0, pm 1 }$$
Since $tan^{-1} theta$ is a continuous function for all $theta$,
$N(x,y,z)$ will be constant over those domain where $xy+yz+zx ne 1$.
Notice
$$xy+yz+zx = 1 iff 3left(frac{x+y+z}{sqrt{3}}right)^2 - ( x^2 + y^2 + z^2 ) = 2$$
is the equation of a two sheet hyperboloid centered at origin with symmetric axis pointing along the direction $(1,1,1)$. The complement of this hyperboloid consists of $3$ connected components. One can pick a point from each of these component and figure out the value of $N(x,y,z)$ over the whole component.
The end result is
$$L(x,y,z) = R(x,y,z) + begin{cases}
pi, & 1 < xy+yz+zx land x+y+z > 0\
0, & 1 > xy+yz+zx\
-pi & 1 < xy+yz+zx land x+y+z < 0
end{cases}
$$
$endgroup$
Let
- $L(x,y,z) = tan^{-1} x + tan^{-1} y + tan^{-1} z$
- $R(x,y,z) = tan^{-1}left(frac{x+y+z - xyz}{1- xy - yz - zx}right)$
By addition formula of tangent function, we have
$$tan L(x,y,z) = tan R(x,y,z)$$
Since $tan theta$ is a periodic function with period $pi$, there is an integer valued function $N(x,y,z)$ such that $$L(x,y,z) = R(x,y,z) + N(x,y,z)pi$$
Since $tan^{-1}theta$ maps $mathbb{R}$ into $(-frac{pi}{2}, frac{pi}{2})$, we have
$$|L(x,y,z)| < frac{3pi}{2} land |R(x,y,z)| < frac{pi}{2}quadimpliesquad N(x,y,z) in { 0, pm 1 }$$
Since $tan^{-1} theta$ is a continuous function for all $theta$,
$N(x,y,z)$ will be constant over those domain where $xy+yz+zx ne 1$.
Notice
$$xy+yz+zx = 1 iff 3left(frac{x+y+z}{sqrt{3}}right)^2 - ( x^2 + y^2 + z^2 ) = 2$$
is the equation of a two sheet hyperboloid centered at origin with symmetric axis pointing along the direction $(1,1,1)$. The complement of this hyperboloid consists of $3$ connected components. One can pick a point from each of these component and figure out the value of $N(x,y,z)$ over the whole component.
The end result is
$$L(x,y,z) = R(x,y,z) + begin{cases}
pi, & 1 < xy+yz+zx land x+y+z > 0\
0, & 1 > xy+yz+zx\
-pi & 1 < xy+yz+zx land x+y+z < 0
end{cases}
$$
answered Dec 5 '18 at 19:54
achille huiachille hui
95.9k5132258
95.9k5132258
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:27
|
show 4 more comments
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:27
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
thanx, i have difficulty in understanding the derivation, though i am working on it. Mean time I have checked for the case where $x=sqrt{3},y=1,z=frac{1}{sqrt{3}}$. We have $xy+yz+zx=3.30>1$ and $x+y+z>0$. LHS=$frac{pi}{3}+frac{pi}{4}+frac{pi}{6}=frac{3pi}{4}$. RHS=$pi+tan^{-1}(-1)=pi-frac{pi}{4}=frac{3pi}{4}$=LHS. THus, verified !
$endgroup$
– ss1729
Dec 5 '18 at 21:09
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
I think I see. Obviously if x>0, y>0, & xy >1, in the expression with the single arctan & rational function as argument, the arctan is going to need to be taken to be the next 'occurence' of it - the 'branch' (if I've used that term strictly correctly) that goes from π/2 to 3π/2. If you allow that, the values of x & y are unrestricted ... is that so - have I understood it aright? I've tended to neglect this kind of thing a bit, as I've found that in situations where this addition formula arises, the values of x & y tend to be such as to obviate the need to bring 'other branches' into it.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 6:57
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
Are you saying that you can have any value of x y & z, ie any point in xyz space? That the sheets of the hyperboloid are the surfaces that when (x,y,z) crosses it changes value, being -1 inside the negative 'lobe', +1 inside the positive lobe, & 0 between the two lobes; but that no restriction is imposed on what (x,y,z) might actually be? That for three variables three branches of atn are sufficient to take care of all (x,y,z)? If (x,y,z)→(∞,∞,∞) then the extreme upper edge of the top branch; & if (-∞,-∞,-∞) the the extreme lower edge of the bottom branch?
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:16
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
@AmbretteOrrisey Yes, aside from the restriction that $xy+yz+zx ne 1$, there is no other restriction on $(x,y,z)$. If you look at the function $tan^{-1}theta$, it is well behaved for all finite $theta$. The only place the $N(x,y,z)$ can change value is when $(x,y,z)$ cross the hyperboloid.
$endgroup$
– achille hui
Dec 6 '18 at 8:23
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:27
$begingroup$
I can see something else: every time you introduce a new variable, the numerator & the denominator alternate between having the higher degree polynomial - so as all the variables go of to ∞ the atn alternates between going to 0 & π/2, and likewise for -∞; so the range of atn increases by half the range of a branch above & below each time.
$endgroup$
– AmbretteOrrisey
Dec 6 '18 at 8:27
|
show 4 more comments
$begingroup$
Write $a:=arctan x$ etc. so $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan a+tan b + (1-tan atan b)tan c}{1-tan atan b - (tan a+tan b)tan c}.$$If $xyne 1$, we can cancel $1-tan atan b$ to get $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan (a+b)+tan c}{1-tan (a+b)tan c}.$$If $tan (a+b)tan cne 1$ i.e. $(x+y)zne 1-xy$ i.e. $xy+yz+zx=1$, we have $$frac{x+y+z-xyz}{1-xy-yz-zx}=tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
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add a comment |
$begingroup$
Write $a:=arctan x$ etc. so $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan a+tan b + (1-tan atan b)tan c}{1-tan atan b - (tan a+tan b)tan c}.$$If $xyne 1$, we can cancel $1-tan atan b$ to get $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan (a+b)+tan c}{1-tan (a+b)tan c}.$$If $tan (a+b)tan cne 1$ i.e. $(x+y)zne 1-xy$ i.e. $xy+yz+zx=1$, we have $$frac{x+y+z-xyz}{1-xy-yz-zx}=tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
$endgroup$
add a comment |
$begingroup$
Write $a:=arctan x$ etc. so $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan a+tan b + (1-tan atan b)tan c}{1-tan atan b - (tan a+tan b)tan c}.$$If $xyne 1$, we can cancel $1-tan atan b$ to get $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan (a+b)+tan c}{1-tan (a+b)tan c}.$$If $tan (a+b)tan cne 1$ i.e. $(x+y)zne 1-xy$ i.e. $xy+yz+zx=1$, we have $$frac{x+y+z-xyz}{1-xy-yz-zx}=tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
$endgroup$
Write $a:=arctan x$ etc. so $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan a+tan b + (1-tan atan b)tan c}{1-tan atan b - (tan a+tan b)tan c}.$$If $xyne 1$, we can cancel $1-tan atan b$ to get $$frac{x+y+z-xyz}{1-xy-yz-zx}=frac{tan (a+b)+tan c}{1-tan (a+b)tan c}.$$If $tan (a+b)tan cne 1$ i.e. $(x+y)zne 1-xy$ i.e. $xy+yz+zx=1$, we have $$frac{x+y+z-xyz}{1-xy-yz-zx}=tan (a+b+c).$$We want to get $a+b+c$ from that, which isn't as simple as taking arctangents; it has the subtleties claimed in your question (albeit not the title).
answered Dec 5 '18 at 18:56
J.G.J.G.
25.6k22539
25.6k22539
add a comment |
add a comment |
$begingroup$
We have that by addition formula
$$tan^{-1}x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
then
$$(tan^{-1}x+tan^{-1}y)+tan^{-1}z=tan^{-1}frac{frac{x+y}{1-xy}+z}{1-frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
- $1-xy=0$
and is certainly not well defined for
- $1-xy-yz-zx=0$
$endgroup$
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
add a comment |
$begingroup$
We have that by addition formula
$$tan^{-1}x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
then
$$(tan^{-1}x+tan^{-1}y)+tan^{-1}z=tan^{-1}frac{frac{x+y}{1-xy}+z}{1-frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
- $1-xy=0$
and is certainly not well defined for
- $1-xy-yz-zx=0$
$endgroup$
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
add a comment |
$begingroup$
We have that by addition formula
$$tan^{-1}x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
then
$$(tan^{-1}x+tan^{-1}y)+tan^{-1}z=tan^{-1}frac{frac{x+y}{1-xy}+z}{1-frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
- $1-xy=0$
and is certainly not well defined for
- $1-xy-yz-zx=0$
$endgroup$
We have that by addition formula
$$tan^{-1}x+tan^{-1}y=tan^{-1}frac{x+y}{1-xy}$$
then
$$(tan^{-1}x+tan^{-1}y)+tan^{-1}z=tan^{-1}frac{frac{x+y}{1-xy}+z}{1-frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
- $1-xy=0$
and is certainly not well defined for
- $1-xy-yz-zx=0$
edited Dec 5 '18 at 20:06
answered Dec 5 '18 at 18:50
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
add a comment |
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
so u saying the above expression is true for all $x,y,z$ except $xy=1,yz=1,zx=1$ and $xy+yz+zx=1$.
$endgroup$
– ss1729
Dec 5 '18 at 20:03
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 Maybe I could be more precise, what I'm claiming is that how we have obtained the equality those point could be problematic and then we need to check those cases directly. For exmaple the last one always is requested. I didn't chek the others.
$endgroup$
– gimusi
Dec 5 '18 at 20:05
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
$begingroup$
@ss1729 I see know the the secon one was equivalent to the first + third! I've updated that.
$endgroup$
– gimusi
Dec 5 '18 at 20:07
add a comment |
$begingroup$
This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors.
Let $alpha=arctan x$, $beta=arctan y$ and $gamma=arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $alpha+beta+gamma$ is the argument
(up to an integer multiple of $2pi$) of the product
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $phi$ of a complex number $a+ib$ satisfies $tanphi=b/a$.
The claim follows from this.
Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $pi$. For example, if $x=y=z=1$ we have $arctan x=arctan y=arctan z=pi/4$ giving $3pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $arctan(-1)=-pi/4$ on the right hand side.
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$begingroup$
This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors.
Let $alpha=arctan x$, $beta=arctan y$ and $gamma=arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $alpha+beta+gamma$ is the argument
(up to an integer multiple of $2pi$) of the product
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $phi$ of a complex number $a+ib$ satisfies $tanphi=b/a$.
The claim follows from this.
Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $pi$. For example, if $x=y=z=1$ we have $arctan x=arctan y=arctan z=pi/4$ giving $3pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $arctan(-1)=-pi/4$ on the right hand side.
$endgroup$
add a comment |
$begingroup$
This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors.
Let $alpha=arctan x$, $beta=arctan y$ and $gamma=arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $alpha+beta+gamma$ is the argument
(up to an integer multiple of $2pi$) of the product
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $phi$ of a complex number $a+ib$ satisfies $tanphi=b/a$.
The claim follows from this.
Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $pi$. For example, if $x=y=z=1$ we have $arctan x=arctan y=arctan z=pi/4$ giving $3pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $arctan(-1)=-pi/4$ on the right hand side.
$endgroup$
This follows from the fact the the argument of a product of complex numbers is the sum of the arguments of the factors.
Let $alpha=arctan x$, $beta=arctan y$ and $gamma=arctan z$. These are the arguments of the complex numbers $z_1=1+ix$, $z_2=1+iy$ and $z_3=1+iz$ respectively.
In light of the above fact we see that $alpha+beta+gamma$ is the argument
(up to an integer multiple of $2pi$) of the product
$$
z_1z_2z_3=(1+ix)(1+iy)(1+iz)=(1-xy-yz-zx)+i(x+y+z-xyz).
$$
But the argument $phi$ of a complex number $a+ib$ satisfies $tanphi=b/a$.
The claim follows from this.
Just be mindful of the lingering uncertainty in the value of the inverse tangent up to an integer multiple of $pi$. For example, if $x=y=z=1$ we have $arctan x=arctan y=arctan z=pi/4$ giving $3pi/4$ on the left hand side. But, $x+y+z-xyz=2$, $1-xy-yz-zx=-2$, so we have $arctan(-1)=-pi/4$ on the right hand side.
edited Dec 6 '18 at 10:03
answered Dec 6 '18 at 9:56
Jyrki LahtonenJyrki Lahtonen
109k13169372
109k13169372
add a comment |
add a comment |
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Well ... apart from restrictions of a similar nature that apply to the arguments of the formula for two variables ie in that case xy<1 . I seem to remember, when I once checked this out for the general case of n variables, that you get the odd symmetric polynomials of degree ≤ n in the numerator & the even symmetric [ ditto ] in the denominator ... something like that, anyway.
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– AmbretteOrrisey
Dec 5 '18 at 18:56
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I'm beginning to get the gist of your question now I think - I glossed over it a bit at first ... are you saying that when you go beyond two variables you lose the property of it being valid ∀ possible values of whatever variables it imports (provided the other branches of atn be admitted) ... that even if you do admit these other branches that property ceases by reason of going from two to three variables? That would be strange if it were so, and I would not have thought it were so ... but I can see how it might actually be something we can't safely assume by default.
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– AmbretteOrrisey
Dec 6 '18 at 7:49