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I'm looking at problem #40. Before going further, this is homework, so I don't want the answer. I just want guidance, and if I'm on the wrong track, I want to be pushed in the right direction.

Single Variable Calculus - Problem # 40

So, I'm looking for L1 and L2.

Here is what I'm doing for L1

  1. Write equation
  2. Get derivative of equation
  3. Simplify
  4. Use Newton's Method to solve

My work for L1

  1. $$x^5 - (2+r)x^4 + (1+2r)x^3 - (1-r)x^2 + 2(1-r)x + r - 1 = 0$$
  2. $$ 5x^4 - 4(2+r)x^3 + 3(1+2r)x^2 - 2(1-r)x + 2(1-r) + r = 0 $$
  3. $$ 5x^4 - 8x^3 + 4rx^3 + 3x^2 + 6rx^2 - 2x - 2rx - 2r + r +2 = 0 $$

I assume after that I fill in the r value and then use Newton's Method to finish it off?

Now for L2

  1. Write equation
  2. Get derivative of equation
  3. Simplify
  4. Use Newton's Method to solve

Here is what I'm doing for L2

  1. $$ p(x) - 2rx^2 = 0 $$
  2. $$ p(x)' - 4rx $$
  3. $$ 5x^4 - 8x^3 + 4rx^3 + 3x^2 + 6rx^2 - 2x - 6rx - 2r + r +2 = 0 $$

Can someone check my work and tell me whether I'm on the correct track? If I am, do my derivatives look correct?

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the derivatives are correct. And I remember from my previous life that while calculating L1, we have a fifth-degree equation. So, most of it is right. –  Cheeku Feb 28 '13 at 17:21
    
For $L1$, you want to take the function to include the 2, that is, your $f(x)$ is the entire expression including the 2. Note, for $L2$, you have $p(x) - 2rx^2 = 0$, so you would repeat the process as above with the additional term. There may be an easier approach to $L2$. –  Amzoti Feb 28 '13 at 17:31
    
@Amzoti So for L1 I just need to keep the 2 and leave it set to 0? And for L2 I just need to add the negative sign that I missed? At Cheeku, thanks for letting me know that I'm on the right track. –  JoseBruchez Feb 28 '13 at 17:36
    
@JoseBruchez: yes, recall, you want the setup to be finding the root of $f(x) = 0$, when you setup the Newton form. –  Amzoti Feb 28 '13 at 17:37
    
@Amzoti I went back and edited my answers. If those look correct now, I can just plug those into Newton's Method and come out with my answers :) –  JoseBruchez Feb 28 '13 at 17:39

1 Answer 1

up vote 2 down vote accepted

Hint:

$(1)$ you are okay for $L1$.

$(2)$ For $L2$, you want to set:

$$w(x) = p(x) - 2rx^2 = 0$$

Using $w(x)$, repeat the same process you did for $L1$.

In other words, think of $w(x)$ as a new function of $x$.

Update

I think you have done the correct thing, but the way you are showing it is unclear. Here is what I mean, to be clear.

For L1:

$$f(x) = x^5 - (2+r)x^4 + (1+2r)x^3 - (1-r)x^2 + 2(1-r)x + r - 1$$

$$f'(x) = 2x(−1+r)+3x^2(1+2r) +4x^2(−2−r)+5x^4 + 2(1−r)$$

For L2:

$$w(x) = x^5 - (2+r)x^4 + (1+2r)x^3 - (1-r)x^2 + 2(1-r)x + r - 1 - 2rx^2$$

$$w'(x) = 2x(−1+r)+3x^2(1+2r)+4x^3(−2−r)+5x^4+2(1−r)$$

Now, you'll setup the Newton iteration using:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

using each of L1 and L2.

Update

The real root for $L_1 = 0.989989$ (there are also four imaginary roots).

The real root for $L_2 = 1.01008$ (there are also four imaginary roots).

You can compare the Newton-Rapshon results to these numbers and they should be very close.

Regards

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Sorry to bother you so long after this question, but I forgot to check my answer. Do these sound correct: L1 = 0.999702267 and L2 = 1.010078 –  JoseBruchez Mar 11 '13 at 22:21
    
What are the starting points for those values? Also, you know the exact results for each! Can you compare your $L1$ and $L2$ to the exact roots (you'll know right away)? Regards –  Amzoti Mar 11 '13 at 22:25
    
@JoseBruchez: Please see my update, I calculate the real root using a diffent method and look at how those match with yous! I did not use Newton, but looks to me like you did great! Regards –  Amzoti Mar 11 '13 at 22:42
    
Great job, Amzoti: unupvoted hero until now! ;-) –  amWhy Apr 27 '13 at 0:19
    
@amWhy: That was a nice problem to guide the students, so was fun! Thx –  Amzoti Apr 27 '13 at 0:25

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