# 3rd degree exponential polynomial

1. Mar 30, 2008

### W3bbo

[SOLVED] 3rd degree exponential polynomial

1. The problem statement, all variables and given/known data

Derived from the original question:

"Reduce to find x"

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

3. 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.

2. Mar 30, 2008

### cristo

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

3. Mar 30, 2008

### 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

4. Mar 30, 2008

### W3bbo

No, that's my problem. :)

5. Mar 30, 2008

### cristo

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

6. Mar 30, 2008

### rocomath

It really doesn't, lol.

7. Mar 31, 2008

### W3bbo

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:

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

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

8. Mar 31, 2008

### cristo

Staff Emeritus
$$e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \Rightarrow 2e^{3x}- e^{2x} - 2e^x=-1$$, 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!

9. Mar 31, 2008

### W3bbo

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:

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

Last edited: Mar 31, 2008
10. Mar 31, 2008

### rocomath

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

11. Apr 1, 2008

### Kurret

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

12. Apr 1, 2008

### W3bbo

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