3rd degree exponential polynomial

In summary, the conversation discusses finding the solution to the equation 2e^{3x} - e^{2x} - 2e^x = 1 and determining the original question of the problem. The conversation also includes a question about showing a specific equality and the solution is given using algebraic manipulation.
  • #1
W3bbo
31
0
[SOLVED] 3rd degree exponential polynomial

Homework Statement



Derived from the original question:

"Reduce to find x"

[tex]2e^{3x} - e^{2x} - 2e^x = 1[/tex]

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.
 
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  • #2
If you make the substitution u=e^x, then the equation reduces to [itex]2u^3-u^2-2u-1=0[/itex]. Can you solve this?
 
  • #3
I haven't worked the thing out , but you might try letting [tex]u=e^x[/tex] and try that.
What was the original question?
I'll play with it more here after we eat.
CC
 
  • #4
cristo said:
If you make the substitution u=e^x, then the equation reduces to [itex]2u^3-u^2-2u-1=0[/itex]. Can you solve this?

No, that's my problem. :)
 
  • #5
W3bbo said:
No, that's my problem. :)

Are you sure that's the right equation? It doesn't have "nice" solutions. What is the original question?
 
  • #6
cristo said:
Are you sure that's the right equation? It doesn't have "nice" solutions. What is the original question?
It really doesn't, lol.
 
  • #7
cristo said:
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:

[tex]2 = {\mathop{\rm Cosech}\nolimits} \left( x \right) - 2{\mathop{\rm Coth}\nolimits} \left( x \right)[/tex]

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

[tex]\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}[/tex]

[tex]\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}[/tex]
 
  • #8
Your mistake's in your last line!
[tex]e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \Rightarrow 2e^{3x}- e^{2x} - 2e^x=-1 [/tex], so then the equation you want to solve, letting u=e^x, is [itex]2u^3-u^2-2u+1=0[/itex], which does have nice solutions!
 
  • #9
cristo said:
Your mistake's in your last line!
[tex]e^{3x} - 2e^x = e^{2x} - 1 - e^{3x} \Rightarrow 2e^{3x}- e^{2x} - 2e^x=-1 [/tex], so then the equation you want to solve, letting u=e^x, is [itex]2u^3-u^2-2u+1=0[/itex], which does have nice solutions!

Indeed, that solved it. Thank you.

I have one more question to ask:

I need to show that:

[tex]{\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)[/tex]

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

[tex]$\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} $

[/tex]
 
Last edited:
  • #10
W3bbo said:
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.
 
  • #11
for your second question, you only need to show that
[tex]\sqrt {3 + 2\sqrt 2 } =1+\sqrt2
[/tex]
and indeed:
[tex]\sqrt {3 + 2\sqrt 2 }=\sqrt {1 + 2\sqrt 2 + 2 }=\sqrt {1^2 + 2\sqrt2 + (\sqrt2)^2 }=\sqrt {(1+\sqrt2)^2}=1+\sqrt2[/tex]
 
  • #12
Kurret said:
for your second question, you only need to show that
[tex]\sqrt {3 + 2\sqrt 2 } =1+\sqrt2
[/tex]
and indeed:
[tex]\sqrt {3 + 2\sqrt 2 }=\sqrt {1 + 2\sqrt 2 + 2 }=\sqrt {1^2 + 2\sqrt2 + (\sqrt2)^2 }=\sqrt {(1+\sqrt2)^2}=1+\sqrt2[/tex]

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

1. What is a 3rd degree exponential polynomial?

A 3rd degree exponential polynomial is a mathematical function of the form f(x) = ax³ + bx² + cx + d, where a, b, c, and d are constants and x is the variable. The term "exponential" refers to the presence of the variable x in an exponent, such as x³. This type of polynomial is also known as a cubic polynomial.

2. How is a 3rd degree exponential polynomial graphed?

A 3rd degree exponential polynomial can be graphed by plotting points on a coordinate plane and connecting them with a smooth curve. The number of points needed to accurately graph the polynomial depends on the complexity of its coefficients and the desired level of precision.

3. What are the key features of a 3rd degree exponential polynomial?

The key features of a 3rd degree exponential polynomial include its degree, leading coefficient, and roots. The degree of the polynomial is the highest exponent in the equation, which in this case is 3. The leading coefficient is the coefficient of the highest degree term, and the roots are the values of x that make the polynomial equal to zero.

4. How are 3rd degree exponential polynomials used in real life?

3rd degree exponential polynomials can be used to model real-life situations such as population growth, financial investments, and chemical reactions. They can also be used to approximate other types of functions, such as trigonometric functions, and to solve equations that involve multiple variables.

5. What is the difference between a 3rd degree exponential polynomial and a 3rd degree polynomial?

A 3rd degree exponential polynomial includes terms with the variable x raised to an exponent, while a 3rd degree polynomial may not. Additionally, a 3rd degree polynomial may include terms with variables raised to different exponents, while a 3rd degree exponential polynomial only includes terms with the variable raised to the same exponent (in this case, 3).

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