Solve the equation $frac{x^7}{7}=1+sqrt[7]{10}xleft(x^2-sqrt[7]{10}right)^2$
$begingroup$
Solve in $mathbb R$ the following equation.
$$frac{x^7}{7}=1+sqrt[7]{10}xleft(x^2-sqrt[7]{10}right)^2$$
Solution
Setting $x=2times 10^{frac 1{14}}y$, the equation becomes $64y^7-112y^5+56y^3-7y=frac{7}{2sqrt{10}}$
If $|y|le 1$, then $y=cos u$ for some $u$ and equation is $cos 7u=frac{7}{2sqrt{10}}>1$, impossible.
If $y<-1$, then $y=-cosh u$ for some $u$ and equation is $cosh 7u=-frac{7}{2sqrt{10}}<0$, impossible.
If $y>1$, then $y=cosh u$ for some $u$ and equation is $cosh 7u=frac{7}{2sqrt{10}}$ and so $y=cosh(frac{cosh^{-1} frac{7}{2sqrt{10}}}7)$
It's then easy to get $cosh^{-1} frac{7}{2sqrt{10}}=lnfrac{sqrt{10}}2$ and so $y=2^{-frac 87}10^{frac 1{14}}+2^{-frac 67}10^{-frac 1{14}}$
Hence the unique root $boxed{x=sqrt[7]5+sqrt[7]2}$
My question: Could you please solve by using simpler method?
inequality hyperbolic-functions
asked Dec 18 '18 at 12:01
namnam
973
$endgroup$
add a comment |
$begingroup$
Solve in $mathbb R$ the following equation.
$$frac{x^7}{7}=1+sqrt[7]{10}xleft(x^2-sqrt[7]{10}right)^2$$
Solution
Setting $x=2times 10^{frac 1{14}}y$, the equation becomes $64y^7-112y^5+56y^3-7y=frac{7}{2sqrt{10}}$
If $|y|le 1$, then $y=cos u$ for some $u$ and equation is $cos 7u=frac{7}{2sqrt{10}}>1$, impossible.
If $y<-1$, then $y=-cosh u$ for some $u$ and equation is $cosh 7u=-frac{7}{2sqrt{10}}<0$, impossible.
If $y>1$, then $y=cosh u$ for some $u$ and equation is $cosh 7u=frac{7}{2sqrt{10}}$ and so $y=cosh(frac{cosh^{-1} frac{7}{2sqrt{10}}}7)$
It's then easy to get $cosh^{-1} frac{7}{2sqrt{10}}=lnfrac{sqrt{10}}2$ and so $y=2^{-frac 87}10^{frac 1{14}}+2^{-frac 67}10^{-frac 1{14}}$
Hence the unique root $boxed{x=sqrt[7]5+sqrt[7]2}$
My question: Could you please solve by using simpler method?
inequality hyperbolic-functions
asked Dec 18 '18 at 12:01
namnam
973
$endgroup$
3
$begingroup$
I'm really impressed at you spotting the $cos^7u$ in there.
$endgroup$
– Arthur
Dec 18 '18 at 12:13
add a comment |
$begingroup$
Solve in $mathbb R$ the following equation.
$$frac{x^7}{7}=1+sqrt[7]{10}xleft(x^2-sqrt[7]{10}right)^2$$
Solution
Setting $x=2times 10^{frac 1{14}}y$, the equation becomes $64y^7-112y^5+56y^3-7y=frac{7}{2sqrt{10}}$
If $|y|le 1$, then $y=cos u$ for some $u$ and equation is $cos 7u=frac{7}{2sqrt{10}}>1$, impossible.
If $y<-1$, then $y=-cosh u$ for some $u$ and equation is $cosh 7u=-frac{7}{2sqrt{10}}<0$, impossible.
If $y>1$, then $y=cosh u$ for some $u$ and equation is $cosh 7u=frac{7}{2sqrt{10}}$ and so $y=cosh(frac{cosh^{-1} frac{7}{2sqrt{10}}}7)$
It's then easy to get $cosh^{-1} frac{7}{2sqrt{10}}=lnfrac{sqrt{10}}2$ and so $y=2^{-frac 87}10^{frac 1{14}}+2^{-frac 67}10^{-frac 1{14}}$
Hence the unique root $boxed{x=sqrt[7]5+sqrt[7]2}$
My question: Could you please solve by using simpler method?
inequality hyperbolic-functions
asked Dec 18 '18 at 12:01
namnam
973
$endgroup$
Solve in $mathbb R$ the following equation.
$$frac{x^7}{7}=1+sqrt[7]{10}xleft(x^2-sqrt[7]{10}right)^2$$
Solution
Setting $x=2times 10^{frac 1{14}}y$, the equation becomes $64y^7-112y^5+56y^3-7y=frac{7}{2sqrt{10}}$
If $|y|le 1$, then $y=cos u$ for some $u$ and equation is $cos 7u=frac{7}{2sqrt{10}}>1$, impossible.
If $y<-1$, then $y=-cosh u$ for some $u$ and equation is $cosh 7u=-frac{7}{2sqrt{10}}<0$, impossible.
If $y>1$, then $y=cosh u$ for some $u$ and equation is $cosh 7u=frac{7}{2sqrt{10}}$ and so $y=cosh(frac{cosh^{-1} frac{7}{2sqrt{10}}}7)$
It's then easy to get $cosh^{-1} frac{7}{2sqrt{10}}=lnfrac{sqrt{10}}2$ and so $y=2^{-frac 87}10^{frac 1{14}}+2^{-frac 67}10^{-frac 1{14}}$
Hence the unique root $boxed{x=sqrt[7]5+sqrt[7]2}$
My question: Could you please solve by using simpler method?
inequality hyperbolic-functions
inequality hyperbolic-functions
asked Dec 18 '18 at 12:01
namnam
973
asked Dec 18 '18 at 12:01
namnam
973
asked Dec 18 '18 at 12:01
namnam
973
asked Dec 18 '18 at 12:01
namnam
973
asked Dec 18 '18 at 12:01
namnam
973
973
3
$begingroup$
I'm really impressed at you spotting the $cos^7u$ in there.
$endgroup$
– Arthur
Dec 18 '18 at 12:13
add a comment |
3
$begingroup$
I'm really impressed at you spotting the $cos^7u$ in there.
$endgroup$
– Arthur
Dec 18 '18 at 12:13
3
3
$begingroup$
I'm really impressed at you spotting the $cos^7u$ in there.
$endgroup$
– Arthur
Dec 18 '18 at 12:13
$begingroup$
I'm really impressed at you spotting the $cos^7u$ in there.
$endgroup$
– Arthur
Dec 18 '18 at 12:13
add a comment |
0
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$begingroup$
I'm really impressed at you spotting the $cos^7u$ in there.
$endgroup$
– Arthur
Dec 18 '18 at 12:13