Need help with complicated algebra for graduate engineering course

  • Thread starter blue mango
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  • #1
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Homework Statement


w=(PI*(P0-PL)*R^4*epsilon^3*row)/6*mu*L)*(1-1/2*epsilon)

show that this can be obtained using
w=(PI*(P0-PL)*R^4*row)/8*mu*L)*((1-kappa^4)-((1-kappa^2)^2/ln(1/kappa))
by setting kappa equal to 1-epsilon and expanding the expression for w in powers of epsilon. this requires using the Taylor series ln(1-epsilon)=-epsilon-1/2*epsilon^2-1/3*epsilon^3-...


The Attempt at a Solution



so far I have plugged in 1-epsilon for all the kappas, expanded everything and canceled what I could and have ended up with this:
w=(PI*(P0-PL)*R^4*row)/8*mu*L)*epsilon*((-3*x^6+8*x^5-8*x^4+40*x^2-96*x+96)/(3*x^3+4*x^2+6*x+12))

I found an online polynomial long division calculator but I still end up with a jumbled mess. I was wondering if anyone had any ideas or did I make a mistake in the middle somewhere and that is why things aren't turning out nicely. Thanks so much.
 

Answers and Replies

  • #2
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It's would be nice if you use LaTex.
You might check your answer by using Mathematica. It's really useful program.
 
  • #3
marcusl
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What is x? Do you mean epsilon?
Apply the binomial expansion to the denominator so everything is in the numerator.
And, yes, Latex would be a big improvement :eek:)
 
  • #4
vela
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Expand it piece by piece to keep things manageable. What are the expansions about [itex]\epsilon=0[/itex] of

[tex]\begin{align*}
1-\kappa^4 &= 1-(1-\epsilon)^4 \\
(1-\kappa^2)^2 &= [1-(1-\epsilon)^2]^2 \\
\log(1/\kappa) &= \log(1/(1-\epsilon))
\end{align*}
[/tex]

The first two are just polynomials, so you can multiply them out.
 
  • #5
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Expand it piece by piece to keep things manageable. What are the expansions about [itex]\epsilon=0[/itex] of

[tex]\begin{align*}
1-\kappa^4 &= 1-(1-\epsilon)^4 \\
(1-\kappa^2)^2 &= [1-(1-\epsilon)^2]^2 \\
\log(1/\kappa) &= \log(1/(1-\epsilon))
\end{align*}
[/tex]

The first two are just polynomials, so you can multiply them out.


I did this and that is how I ended up with the big jumbled mess of a polynomial that I have now. And yes x does mean epsilon.
 
  • #6
vela
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So show us what you got before you did any simplifications and perhaps tell us what you tried next. This problem is just essentially doing algebra, so if you need advice on how to do it, you need to show us your actual work. Right now, all we can tell you is you made an algebra mistake somewhere.
 
  • #7
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Here is what I started with:
[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21w%3D%5Cpi%28P_0-P_L%29R%5E4%5Cvarrho/8%5Cmu%20L%29%5Cleft%5B%20%281-%281-%5Cvarepsilon%29%5E4%29-%28%281-%281-%5Cvarepsilon%29%29%5E2%29/%28ln%281/%281-%5Cvarepsilon%29%29%5Cright%5D.gif [Broken]

Here is what I have now:
<img alt="w=\pi(P_0-P_L)R^4\varrho/8\mu L)\varepsilon((-3\varepsilon^6+8\varepsilon^5-8\varepsilon^4+40\varepsilon^2-96\varepsilon+96)/(3\varepsilon^3+4\varepsilon^2+6\varepsilon+12))" src=[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21w%3D%5Cpi%28P_0-P_L%29R%5E4%5Cvarrho/8%5Cmu%20L%29%5Cvarepsilon%28%28-3%5Cvarepsilon%5E6%2B8%5Cvarepsilon%5E5-8%5Cvarepsilon%5E4%2B40%5Cvarepsilon%5E2-96%5Cvarepsilon%2B96%29/%283%5Cvarepsilon%5E3%2B4%5Cvarepsilon%5E2%2B6%5Cvarepsilon%2B12%29%29.gif [Broken][/URL] align=center border=0><img alt="w=\pi(P_0-P_L)R^4\varrho/8\mu L)\varepsilon((-3\varepsilon^6+8\varepsilon^5-8\varepsilon^4+40\varepsilon^2-96\varepsilon+96)/(3\varepsilon^3+4\varepsilon^2+6\varepsilon+12))" src=[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21w%3D%5Cpi%28P_0-P_L%29R%5E4%5Cvarrho/8%5Cmu%20L%29%5Cvarepsilon%28%28-3%5Cvarepsilon%5E6%2B8%5Cvarepsilon%5E5-8%5Cvarepsilon%5E4%2B40%5Cvarepsilon%5E2-96%5Cvarepsilon%2B96%29/%283%5Cvarepsilon%5E3%2B4%5Cvarepsilon%5E2%2B6%5Cvarepsilon%2B12%29%29.gif [Broken][/URL] align=center border=0><img alt="w=\pi(P_0-P_L)R^4\varrho/8\mu L)\varepsilon((-3\varepsilon^6+8\varepsilon^5-8\varepsilon^4+40\varepsilon^2-96\varepsilon+96)/(3\varepsilon^3+4\varepsilon^2+6\varepsilon+12))" src=[PLAIN]http://www.texify.com/img/%5CLARGE%5C%21w%3D%5Cpi%28P_0-P_L%29R%5E4%5Cvarrho/8%5Cmu%20L%29%5Cvarepsilon%28%28-3%5Cvarepsilon%5E6%2B8%5Cvarepsilon%5E5-8%5Cvarepsilon%5E4%2B40%5Cvarepsilon%5E2-96%5Cvarepsilon%2B96%29/%283%5Cvarepsilon%5E3%2B4%5Cvarepsilon%5E2%2B6%5Cvarepsilon%2B12%29%29.gif [Broken][/URL] align=center border=0>
 
Last edited by a moderator:
  • #8
vela
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Enclose LaTeX code in between [ tex] and [ /tex] tags (without spaces) to get it to render. There's a bug in the forum where the preview shows the wrong image. Just reload the page to get the images update.

Again, I'll note that showing what you started with and then the final mess without showing any intermediate steps isn't helpful at all.
 

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