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These are approximate grade middles. The letter grades are only to give you an idea of how well you're doing. I don't add letter grades. I add up your scores to get the total of test points.
A | B | C | D | F |
75 | 60 | 45 | 30 | 25 |
A | B | C | D | F |
73 | 57 | 42 | 33 | ≤ 29 |
A | B | C | D | F |
100 | 75 | 52 | 37 | ≤ 32 |
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We'll examine this problem more deeply in Ch. 10.
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Corrections to the Textbook
(See below, "infinite p.o. set", for why an infinite p.o. set is relevant.)
Also, the notation Cn should be Rn.
Additions to the Textbook
Also, P(n) := P(n,n).
In particular, (n)0 := 1, since the product of no factors is defined as 1.
Each (n)k is a polynomial of degree k; that is why n can be anything that can go into a polynomial.
This is different from the book's definition in Ch. 2, but in a later chapter Brualdi will change the definition to this one.
This is correct mathematical usage. In ordinary English "unique" can mean things like "special to that person or thing". In computerese (due to illiterate computerites, I suppose) the phrase "unique users" is a corrupt way of saying "distinct users". Don't imitate this. (All computerites are not illiterate, far from it. I infer the existence of illiterate ones from the way they use English.)
By the way, I know I'm using very strong language here. I have a low opinion of people who can't distinguish one word from another. Sorry. None of you is that person, I'm sure!
We found two solutions:
The one-point line prevents any other lines from existing. So I added another requirement:
Then we found the solution
So, I asked if the problem can be solved in any other way.
Actually, I gave a stronger requirement, which you can use instead (I don't promise they give the same answers; we can work on that in Ch. 10).
I showed such a solution with 7 points and 7 lines, with 3 points on each line. But the original question was about 6 points.
This is a good discussion problem; we'll share what you come up with.
Draw a circle (it need not be perfect) and draw n points on it (not equally spaced). Draw all the chords (straight line segments) connecting the n points. Let hn = the number of regions the chords cut the disk into.
Submit your results in class for discussion. (I'm not collecting written results.)
I'm not saying you will find equality; not saying you won't.
Hint #1: How do you describe a combination of a multiset? Think "submultiset". How do you describe a submultiset? Think "repetition numbers".
Hint #2. You can do the problem directly, or you can see it as a continuation of Ch. 4, ##38–40.
Hint #3. I always say, try a small example first. What's a small example here? k = 1 and k = 2 seem like good small examples.
(Revised and corrected: the first version was overly complicated.)
Example: If the characteristic equation has roots 1 (once) and −2 (3 times), then the homogeneous general solution is
a11n + a2(−2)n + a3n(−2)n + a4n2(−2)n.
You remember 1 with multiplicity s=1 and −2 with multiplicity s=3.
Example (continued): If the non-homogeneous part is n + 3·2n − 4n·(−4)n + 7·(−4)n, then you have 1n with coefficient polynomial of degree 1, 2n with coefficient polynomial of degree 0, and (−4)n with coefficient polynomial −4n+7 of degree 1. Remember the numbers e=1 for q=1, e=0 for q=2, e=1 for q=−4.
Example (continued): We had 1 with s = 1 (Step 1) and e = 2 (Step 2), totalling 3, so we form a linear combination of n·1n=n, n2·1n=n2;
also (from Step 2) 2n; also (from Step 2) (−4)n, n·(−4)n. That will be
c1n + c2n2 + c32n + c4(−4)n + c5n·(−4)n.
You will never see such a big example in the course. It's big in order to illustrate all the possible aspects. Our problems will be much smaller.
hn = a11n + a2(−2)n + a3n(−2)n + a4n2(−2)n + Hn.
Use the initial conditions to get the values of the unknown coefficients a1, etc.