Hyperbolic functions problem
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If $p^2sinh x+q^2cosh x = r^2$ has at least one root, how do I show that $r^4 > p^4-q^4?$
hyperbolic-functions
edited Nov 20 at 15:36
KM101
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asked Nov 20 at 15:30
Rik
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up vote
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down vote
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If $p^2sinh x+q^2cosh x = r^2$ has at least one root, how do I show that $r^4 > p^4-q^4?$
hyperbolic-functions
edited Nov 20 at 15:36
KM101
3,466417
asked Nov 20 at 15:30
Rik
183
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $p^2sinh x+q^2cosh x = r^2$ has at least one root, how do I show that $r^4 > p^4-q^4?$
hyperbolic-functions
edited Nov 20 at 15:36
KM101
3,466417
asked Nov 20 at 15:30
Rik
183
If $p^2sinh x+q^2cosh x = r^2$ has at least one root, how do I show that $r^4 > p^4-q^4?$
hyperbolic-functions
hyperbolic-functions
edited Nov 20 at 15:36
KM101
3,466417
asked Nov 20 at 15:30
Rik
183
edited Nov 20 at 15:36
KM101
3,466417
asked Nov 20 at 15:30
Rik
183
edited Nov 20 at 15:36
KM101
3,466417
edited Nov 20 at 15:36
KM101
3,466417
edited Nov 20 at 15:36
KM101
3,466417
3,466417
asked Nov 20 at 15:30
Rik
183
asked Nov 20 at 15:30
Rik
183
asked Nov 20 at 15:30
Rik
183
183
add a comment |
add a comment |
1 Answer
1
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0
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Your equation is equivalent to:
$$(p^2+q^2)e^x+(p^2-q^2)e^{-x}=2r^2$$
or:
$$(p^2+q^2)e^{2x}-2r^2e^x+(p^2-q^2)=0tag{1}$$
This is a quadratic equation with respect to $t=e^x$. If (1) has at least one root, it means that there has to be one positive solution of (1). And it is possible if:
$$D=(2r^2)^2-4(p^2+q^2)(p^2-q^2)ge 0$$
...which leads directly to:
$$r^4ge p^4-r^4$$
answered Nov 20 at 15:47
Oldboy
6,0081628
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your equation is equivalent to:
$$(p^2+q^2)e^x+(p^2-q^2)e^{-x}=2r^2$$
or:
$$(p^2+q^2)e^{2x}-2r^2e^x+(p^2-q^2)=0tag{1}$$
This is a quadratic equation with respect to $t=e^x$. If (1) has at least one root, it means that there has to be one positive solution of (1). And it is possible if:
$$D=(2r^2)^2-4(p^2+q^2)(p^2-q^2)ge 0$$
...which leads directly to:
$$r^4ge p^4-r^4$$
answered Nov 20 at 15:47
Oldboy
6,0081628
add a comment |
up vote
0
down vote
Your equation is equivalent to:
$$(p^2+q^2)e^x+(p^2-q^2)e^{-x}=2r^2$$
or:
$$(p^2+q^2)e^{2x}-2r^2e^x+(p^2-q^2)=0tag{1}$$
This is a quadratic equation with respect to $t=e^x$. If (1) has at least one root, it means that there has to be one positive solution of (1). And it is possible if:
$$D=(2r^2)^2-4(p^2+q^2)(p^2-q^2)ge 0$$
...which leads directly to:
$$r^4ge p^4-r^4$$
answered Nov 20 at 15:47
Oldboy
6,0081628
add a comment |
up vote
0
down vote
up vote
0
down vote
Your equation is equivalent to:
$$(p^2+q^2)e^x+(p^2-q^2)e^{-x}=2r^2$$
or:
$$(p^2+q^2)e^{2x}-2r^2e^x+(p^2-q^2)=0tag{1}$$
This is a quadratic equation with respect to $t=e^x$. If (1) has at least one root, it means that there has to be one positive solution of (1). And it is possible if:
$$D=(2r^2)^2-4(p^2+q^2)(p^2-q^2)ge 0$$
...which leads directly to:
$$r^4ge p^4-r^4$$
answered Nov 20 at 15:47
Oldboy
6,0081628
Your equation is equivalent to:
$$(p^2+q^2)e^x+(p^2-q^2)e^{-x}=2r^2$$
or:
$$(p^2+q^2)e^{2x}-2r^2e^x+(p^2-q^2)=0tag{1}$$
This is a quadratic equation with respect to $t=e^x$. If (1) has at least one root, it means that there has to be one positive solution of (1). And it is possible if:
$$D=(2r^2)^2-4(p^2+q^2)(p^2-q^2)ge 0$$
...which leads directly to:
$$r^4ge p^4-r^4$$
answered Nov 20 at 15:47
Oldboy
6,0081628
answered Nov 20 at 15:47
Oldboy
6,0081628
answered Nov 20 at 15:47
Oldboy
6,0081628
answered Nov 20 at 15:47
Oldboy
6,0081628
6,0081628
add a comment |
add a comment |
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