3rd degree exponential polynomial

[W3bbo]

[SOLVED] 3rd degree exponential polynomial

Homework Statement

Derived from the original question:

"Reduce to find x"

$$2e^{3x} - e^{2x} - 2e^x = 1$$

The Attempt at a Solution

Inspection fails, since the answer is transcendental. I have the answer from my CAS, but I can't figure it out myself.

 

[cristo]

If you make the substitution u=e^x, then the equation reduces to $2u^3-u^2-2u-1=0$. Can you solve this?

 

[happyg1]

I haven't worked the thing out , but you might try letting $$u=e^x$$ and try that.
What was the original question?
I'll play with it more here after we eat.
CC

 

[W3bbo]

If you make the substitution u=e^x, then the equation reduces to $2u^3-u^2-2u-1=0$. Can you solve this?

No, that's my problem. :)

 

[cristo]

No, that's my problem. :)

Are you sure that's the right equation? It doesn't have "nice" solutions. What is the original question?

 

[rocomath]

Are you sure that's the right equation? It doesn't have "nice" solutions. What is the original question?

It really doesn't, lol.

 

[W3bbo]

Are you sure that's the right equation? It doesn't have "nice" solutions. What is the original question?

The original question was "Find x, using the definitions of the hyperbolics in terms of exponentials"

...from this expression:

$$2 = {\mathop{\rm Cosech}\nolimits} \left( x \right) - 2{\mathop{\rm Coth}\nolimits} \left( x \right)$$

Here's my working so far, reducing down to the final polynomial:

$$\displaylines{<br /> {\mathop{\rm Cosech}\nolimits} \left( x \right) = {2 \over {e^x - e^{ - x} }} \cr <br /> {\mathop{\rm Coth}\nolimits} \left( x \right) = {{{\mathop{\rm Cosh}\nolimits} \left( x \right)} \over {{\mathop{\rm Sinh}\nolimits} \left( x \right)}} = {{{\textstyle{1 \over 2}}\left( {e^x + e^{ - x} } \right)} \over {{\textstyle{1 \over 2}}\left( {e^x - e^{ - x} } \right)}} = {{e^{2x} + 1} \over {e^{2x} - 1}} \cr}$$

$$\displaylines{<br /> 2 = {\mathop{\rm Cosech}\nolimits} \left( x \right) - 2{\mathop{\rm Coth}\nolimits} \left( x \right) \cr <br /> = {2 \over {e^x - e^{ - x} }} - 2\left( {{{e^{2x} + 1} \over {e^{2x} - 1}}} \right) \cr <br /> 1 = {1 \over {e^x - e^{ - x} }} - {{e^{2x} + 1} \over {e^{2x} - 1}} \cr <br /> = {{e^{2x} - 1} \over {\left( {e^x - e^{ - x} } \right)\left( {e^{2x} - 1} \right)}} - \left( {{{e^{2x} + 1} \over {e^{2x} - 1}} \times {{e^x - e^{ - x} } \over {e^x - e^{ - x} }}} \right) \cr <br /> = {{e^{2x} - 1} \over {e^{3x} - 2e^x + e^{ - x} }} - {{e^{3x} - e^{ - x} } \over {e^{3x} - 2e^x + e^{ - x} }} \cr <br /> = {{e^{2x} - 1 - e^{3x} + e^{ - x} } \over {e^{3x} - 2e^x + e^{ - x} }} \cr <br /> e^{3x} - 2e^x + e^{ - x} = e^{2x} - 1 - e^{3x} + e^{ - x} \cr <br /> e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \cr <br /> 2e^{3x} - e^{2x} - 2e^x = 1 \cr}$$

 

[cristo]

Your mistake's in your last line!
$$e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \Rightarrow 2e^{3x}- e^{2x} - 2e^x=-1$$, so then the equation you want to solve, letting u=e^x, is $2u^3-u^2-2u+1=0$, which does have nice solutions!

 

[W3bbo]

Your mistake's in your last line!
$$e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \Rightarrow 2e^{3x}- e^{2x} - 2e^x=-1$$, so then the equation you want to solve, letting u=e^x, is $2u^3-u^2-2u+1=0$, which does have nice solutions!

Indeed, that solved it. Thank you.

I have one more question to ask:

I need to show that:

$${\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\mathop{\rm Ln}\nolimits} \left( {1 + \sqrt 2 } \right)$$

Here's my working so far, I don't know where to go from the bottom line:

$$\displaylines{<br /> {\mathop{\rm Artanh}\nolimits} \left( x \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{x + 1} \over {x - 1}}} \right) \cr <br /> {\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right) = {\textstyle{1 \over {\sqrt 2 }}} \cr <br /> {\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{1 + {\textstyle{1 \over {\sqrt 2 }}}} \over {1 - {\textstyle{1 \over {\sqrt 2 }}}}}} \right) \cr <br /> = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{{{\sqrt 2 + 1} \over {\sqrt 2 }}} \over {{{\sqrt 2 - 1} \over {\sqrt 2 }}}}} \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{\left( {\sqrt 2 + 1} \right)\sqrt 2 } \over {\left( {\sqrt 2 - 1} \right)\sqrt 2 }}} \right) \cr <br /> = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {{{\sqrt 2 + 1} \over {\sqrt 2 - 1}}} \right) \cr <br /> {{\sqrt 2 + 1} \over {\sqrt 2 - 1}} = {{\sqrt 2 + 1} \over {\sqrt 2 - 1}} \times {{1 + \sqrt 2 } \over {1 + \sqrt 2 }} \cr <br /> = {{3 + 2\sqrt 2 } \over 1} = 3 + 2\sqrt 2 \cr <br /> {\mathop{\rm Artanh}\nolimits} \left( {{\mathop{\rm Sin}\nolimits} \left( {{\textstyle{\pi \over 4}}} \right)} \right) = {\textstyle{1 \over 2}}{\mathop{\rm Ln}\nolimits} \left( {3 + 2\sqrt 2 } \right) \cr <br /> = {\mathop{\rm Ln}\nolimits} \left( {\sqrt {3 + 2\sqrt 2 } } \right) \cr <br /> = \cr} <br />$$

 

[rocomath]

The original question was "Find x, using the definitions of the hyperbolics in terms of exponentials"

WOW, way to leave that out. Next time you ask for help, make sure you post the whole question.

 

[Kurret]

for your second question, you only need to show that
$$\sqrt {3 + 2\sqrt 2 } =1+\sqrt2$$
and indeed:
$$\sqrt {3 + 2\sqrt 2 }=\sqrt {1 + 2\sqrt 2 + 2 }=\sqrt {1^2 + 2\sqrt2 + (\sqrt2)^2 }=\sqrt {(1+\sqrt2)^2}=1+\sqrt2$$

 

[W3bbo]

for your second question, you only need to show that
$$\sqrt {3 + 2\sqrt 2 } =1+\sqrt2$$
and indeed:
$$\sqrt {3 + 2\sqrt 2 }=\sqrt {1 + 2\sqrt 2 + 2 }=\sqrt {1^2 + 2\sqrt2 + (\sqrt2)^2 }=\sqrt {(1+\sqrt2)^2}=1+\sqrt2$$

Whoa, that's voodoo. How did you figure that out?