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What's all this talk about TI-36 series not handling complex numbers? Sure it does.

It's possible that I just got used to handling complex numbers the TI way, so that it now seems natural, and that's why I have no endearing love for the HP33S. Plus my HP batteries just died here before it's second birthday even, while my TI is still running on the same power source from twenty years ago.

Whatever you do, bring two. $35 extra bucks is a small insurance fee against having to wash out because your calculator failed.

 
No matter what calculator do you use, please....please....please, check it against the NCEES list. I am surprised to read about calculator incidents test after test. At this stage there is no excuse for that.

 
On the Kaplan Sample Exam, they gave a 9th order polynomial characteristic equation for a control system and asked which one of four choices was not a root. When I looked at the solution in the back, there was no work shown...it just said "use your calculator to find the roots". Neither the Casio, nor the TI, will do roots of more than 2nd order polynomials (at least I couldn't find it). Anybody know of a way to "trick" the calculator into solving huge equations...or at least a shortcut to find all the roots of a large polynomial (including complex roots).

 
On the Kaplan Sample Exam, they gave a 9th order polynomial characteristic equation for a control system and asked which one of four choices was not a root. When I looked at the solution in the back, there was no work shown...it just said "use your calculator to find the roots". Neither the Casio, nor the TI, will do roots of more than 2nd order polynomials (at least I couldn't find it). Anybody know of a way to "trick" the calculator into solving huge equations...or at least a shortcut to find all the roots of a large polynomial (including complex roots).
I doubt a problem like this will appear on the real exam, but who knows. Can your calculator do polynomial division up to the level? I only used the calculator for very simple calculations on the actual exam. I remember this problem from Kaplan, and I think I either just did it by long division, or not at all.

 
I doubt a problem like this will appear on the real exam, but who knows. Can your calculator do polynomial division up to the level? I only used the calculator for very simple calculations on the actual exam. I remember this problem from Kaplan, and I think I either just did it by long division, or not at all.
I just guessed on it (correctly, I think). I was just wondering if I was missing a trick that my calc could do. I looked through the instruction book and found how to factor quadractic equations, but couldn't do any higher order equations.

 
On the Kaplan Sample Exam, they gave a 9th order polynomial characteristic equation for a control system and asked which one of four choices was not a root. When I looked at the solution in the back, there was no work shown...it just said "use your calculator to find the roots". Neither the Casio, nor the TI, will do roots of more than 2nd order polynomials (at least I couldn't find it).
I am not a EE but I am certainly pretty handy with math. :)

I would suspect that if you are given a 'higher' order equation that there should be a 'trick' that would allow you to figure out how to solve it.

Do you mind providing the equation + choices? I will take a look at it for you to see if there is anything 'simple' regarding a simplification for solution.

JR

 
I would like to see it too. I gave my book away.

In general, unless you have a real fancy calculator, I still think the easiest method has got to be to just try dividing the answers into the equation. As soon as something doesn't come out easy you can eliminate it.

Remember, this is an s^^9 problem. And assuming nothing obvious factors out, it has got to be tough. You would think if there was a simple way to do it, they would have mentioned it in the answer.

 
1.32 The characteristic equation for a given control system is

P(s) = s9 + 30 s8 + 401 s7 + 3176 s6 + 16696 s5 + 61400 s4 + 160324 s3 + 287344 s2 + 318448 s + 162240

Which of the following is not a root of this plynomial?

a. s = -2

b. s = -5 + j

c. s = -5

d. s = -3 +j2

The description in the solution says "This is the point where that expensive calculator your [sic] purchased pays for itself...Simply insert the coefficients into the machine, and the roots are returned."

 
1.32 The characteristic equation for a given control system is
P(s) = s9 + 30 s8 + 401 s7 + 3176 s6 + 16696 s5 + 61400 s4 + 160324 s3 + 287344 s2 + 318448 s + 162240

Which of the following is not a root of this plynomial?

a. s = -2

b. s = -5 + j

c. s = -5

d. s = -3 +j2

The description in the solution says "This is the point where that expensive calculator your [sic] purchased pays for itself...Simply insert the coefficients into the machine, and the roots are returned."
Six minutes? I really doubt you're going to see anything like this.

I don't think my dividing method will work fast enough, although plugging in the answers is a tried and true method for multiple choice exams. I would start with the s+2 and s+5 and try to either eliminate or include them. After that I might just guess.

Maybe there is a trick for this.

 
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This is where the structural guy who has been lurking on this thread goes "WTF is j?" I assume it's what the rest of us call 'i', eh?

either way, you have my sympathies.

 
A agree. This is well beyond the level of difficulty I would expect to see on the test. Having said that, my response to this question would be to pick B (or C) and move on to the next one. Maybe revisit if you have time at the end of the test.

 
I realize that this is above and beyond the level of difficulty of the actual test, but I was confused (as to how I was supposed to do this with my Casio) and angry (that there was no other methods on how to solve the problem in the Solution section) when I tried to understand how to solve this thing quickly.

The answer, BTW, was c.

 
Six minutes? I really doubt you're going to see anything like this.I don't think my dividing method will work fast enough, although plugging in the answers is a tried and true method for multiple choice exams. I would start with the s+2 and s+5 and try to either eliminate or include them. After that I might just guess.

Maybe there is a trick for this.
I know I'm going to feel like an idiot when I see a response to this question, but how does the "plugging in" method work? I think I understand that by plugging in the "real" answers, roots will make P(s) = 0. Where that breaks down in my mind is working with the complex answers.

 
I know I'm going to feel like an idiot when I see a response to this question, but how does the "plugging in" method work? I think I understand that by plugging in the "real" answers, roots will make P(s) = 0. Where that breaks down in my mind is working with the complex answers.
No, at least I don't think you're an idiot.

Like I said, I would try the real roots first. Once I got to the complex roots I would probably just guess. Unless you had a calculator that could do that sort of thing.

You would have a 50-50 chance at that point, assuming you didn't find the clinker first.

At least that's what i would do on the exam.

another possiblity is you come back at the end of the exam if you have time, convert one of the complex numbers to polar form, and try to plug it in that way. Will that help? I don't know. Probably better than all those j^8s. Then again, maybe I'm messing this all up. I would just guess between the two.

Maybe somebody else will come up with something better.

 
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I agree with the other posters - I wouldn't expect a problem of that level of difficulty on the exam.

However, I would like to point out some relevant principles:

1. The Fundamental Theorem of Algebra tells us that a polynomial of n degree has n roots, some of which can be multiple roots. This means that we know that this polynomial of degree NINE has NINE roots.

2. Descartes' Rule of Signs tells us:

  • The number of positive roots is either equal to the number of variations in sign or less than that by an even number; and
  • The number of negative roots is either equal to the number of variations in sign or less than that by an even number.
In other words - look for how many times the sign changes. Then apply the rules above.

For the case of your example,

P(s) = zero sign changes; therefore, there aren't any POSITIVE roots

P(-S) = -s9 + 30 s8 - 401 s7 + 3176 s6 - 16696 s5 + 61400 s4 - 160324 s3 + 287344 s2 - 318448 s + 162240

Note that there are nine sign changes, meaning that there are nine NEGATIVE roots.

3. If a polynomial has REAL coefficients, then either all of the roots are real or there are an even number of non-real complex roots in conjugate pairs.

Since we are looking for the answer that DOESN'T fit - that means we know we can eliminate answers b. and d. because one of the real numbers doesn't fit. Subsequent analysis (Plug-N-Chug) shows that -2 fits, so the answer for which one not fitting is -5 --> Answer c. is correct.

The important take away message is to understand the algebraic principles that allow you to narrow your choices.

I hope this helps. Let me know if you have any questions.

JR

 
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I agree with the other posters - I wouldn't expect a problem of that level of difficulty on the exam.
However, I would like to point out some relevant principles:

1. The Fundamental Theorem of Algebra tells us that a polynomial of n degree has n roots, some of which can be multiple roots. This means that we know that this polynomial of degree NINE has NINE roots.

2. Descartes' Rule of Signs tells us:

  • The number of positive roots is either equal to the number of variations in sign or less than that by an even number; and
  • The number of negative roots is either equal to the number of variations in sign or less than that by an even number.
In other words - look for how many times the sign changes. Then apply the rules above.

For the case of your example,

P(s) = zero sign changes; therefore, there aren't any POSITIVE roots

P(-S) = -s9 + 30 s8 - 401 s7 + 3176 s6 - 16696 s5 + 61400 s4 - 160324 s3 + 287344 s2 - 318448 s + 162240

Note that there are nine sign changes, meaning that there are nine NEGATIVE roots.

3. If a polynomial has REAL coefficients, then either all of the roots are real or there are an even number of non-real complex roots in conjugate pairs.

Since we are looking for the answer that DOESN'T fit - that means we know we can eliminate answers b. and d. because one of the real numbers doesn't fit. Subsequent analysis (Plug-N-Chug) shows that -2 fits, so the answer for which one not fitting is -5 --> Answer c. is correct.

The important take away message is to understand the algebraic principles that allow you to narrow your choices.

I hope this helps. Let me know if you have any questions.

JR
Look at JR bustin out the Descartes after 5 on Friday! :w00t:

:appl: :appl: :appl: :appl: :appl:

 
Very good!

Of course, very few people, except JR are going to have this stuff memorized. So, in the real world, if you just plugged in the real numbers you would have found the -5 without having to look it up.

 
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I realize that this is above and beyond the level of difficulty of the actual test, but I was confused (as to how I was supposed to do this with my Casio) and angry (that there was no other methods on how to solve the problem in the Solution section) when I tried to understand how to solve this thing quickly.
If it makes you feel any better, I know of no way to solve this monster using the Casio calculators. I could do it on my TI-85, but not on one of the approved Casios. I'm not saying it can't be done, I just don't know how if it can. But it's a moot point; you shouldn't need to solve anything nearly this complicated on the exam.

 
Look at JR bustin out the Descartes after 5 on Friday! :w00t: :appl: :appl: :appl: :appl: :appl:
Thanks ... like I said ... don't be hatin' on my analysis !!! :p

I wish I was bustin' more than algebra and my ass to get ready for a family/friend party this weekend on a Friday evening!

Very good!Of course, very few people, except JR are going to have this stuff memorized. So, in the real world, if you just plugged in the real numbers you would have found the -5 without having to look it up.
:D

If it makes you feel any better, I know of no way to solve this monster using the Casio calculators. I could do it on my TI-85, but not on one of the approved Casios.
Did you say solution ??

B)

JR

 

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