What I think makes teaching mathematics different than other teaching, is that the main work is done outside of the classroom (and outside tutoring) - solving homework problems by students themselves, with as little outside help as possible.

One of my professors said during a lecture (answering to students complaints that they don't understand): "I've never learned anything in a class - why should you?"

Alec

I meant the country, of course. In general, I think, this is more common meaning, especially in this country :)

I can't say much about educational situation in other countries because I don't have a personal experience with that, except in Russia - but that was long time ago. In Russia, laziness is a very common factor (and not necessarily negative.)

Alec

In America, in general, there is no such thing as laziness. It was founded by very energetic people, and new emigrants also are very energetic - otherwise they wouldn't even get here.

Both students and professors work very hard - the exceptions, if exist, are very rare and don't last long. So that is not a problem here.

I'll comment about motivation and willingness to share later.

Alec

In many cases, lists can be used instead of sets. If it is necessary to have a set, that can be done just using convert, for example,

G:=combinat:-permute(3,2);

        G := [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]]

convert(G,set);

           {[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]}

or shorter,

{G[]};

           {[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]}

Alec

In many cases, lists can be used instead of sets. If it is necessary to have a set, that can be done just using convert, for example,

G:=combinat:-permute(3,2);

        G := [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]]

convert(G,set);

           {[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]}

or shorter,

{G[]};

           {[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]}

Alec

Teaching math, in general, has several problems.

One is that students come unprepared - that can be partially fixed by requiring passing some tests before taking a course, but still the problem persists (except in the top 5 or 10 schools, highly selective).

Another one is that teachers are unprepared for teaching. Some schools allow Masters to teach math courses which should be totally unacceptable - like driving without a driver's license. And PhDs have only several teaching workshops during their education - they don't take such courses as "Teaching Calculus III", or "Teaching Calculus I for students with ADHD".

Not spending enough time on preparation for classes is not one of problems that I've noticed that my colleagues have. They at least read the textbook before going to class, and that takes at least 30 min. Plus, choosing the right exercises might take another 30 min.

About pointing in wrong direction - my guess would be that the person doing that doesn't know the right direction. What would be the sense in doing that on purpose?

Alec

Well, you shouldn't be sorry for the students. They are doing pretty well. Many of them (including Lisa Juan - the girl singing the song) came from specialized math schools and were prepared well for Calculus.

The question that I asked was "Do you LIKE calculus?" - the answer that they LOVE it was quite nice. Look at their faces - does that look that they are doing that for the grade? My personal grade is only 40% of their final grade - the rest comes from the midterms and the final - and it is even less actually, because it is based on the grader's (a TA) grade for quizzes and homework.

It may be unusual in other Calculus courses, but it is normal in my courses. I'm trying to make it fun (as in this site posts :), and my students perform better on the common departmental exams than students in other sections (sometimes much better.)

By the way, did you notice the Maplesoft mug in the scene with Bob? My wife is drinking coffee from it and I'd like to have another one like that for myself, but Maplesoft is that cheap that couldn't provide me with one (or with a copy of Maple 13 - which is much more expensive though.) That is one of the reasons why I don't use Maple in my classes (the other one is that Maple is unusable - but that is a different story.)

Alec

Just came home for 5 minutes, and while changing, I've noticed that I didn't answer to your question - what worries me the most (because it seemed kinda obvious). Nobody posted about that, so I've realized that it may be not that obvious as I thought. Since I don't have much time, and I'll be back only in the middle of the night, I'll answer short.

The thing that worries me the most is that series s is divergent for any positive n. In general, the series 1/(1+x) = sum (_1)^r*x*r is divergent for any x>1, and in this example, with x=((n+1)/n)^(1/3), x is obviously greater than 1 for any positive n.

The second thing that worries me is that after that substitution, one will need to do summation over r as well, and it will be also a divergent series.

Two wrongs don't make right (only 3 left turns do).

The second thing - while summation over r equal 0 modulo 3, obviously, could be done - because that make the summand rational and the way to find sums of rational summands is well known (well, even Maple can do that for small degrees), but the sum for all such cases could be evaluated just by replacing 1+((n+1)/n)^(1/3) with 1+((n+1)/n in the denominator - no need of using divergent series. And it is not related that much to the final sum - because the terms with r not equl to 0 modulo 3 are missing, and they couldn't be calculated that way - another techique is needed (if it is possible at all.)

The third thing I've wrote already above - that even if all that could be done, with a lot of work etc. and twice using divergent series, the final result would be the same as if one did the asymptotic of the summand for n approaching infinity from the very beginning  - that I did above - which couldn't be summated in an obvious way - at least I couldn't do that in 5 minutes that I spent on that - and I have a lot of experience in series summation - I was interested in such things about 40 - 45 years ago, and still can do them pretty well, I think.   

Alec

Just came home for 5 minutes, and while changing, I've noticed that I didn't answer to your question - what worries me the most (because it seemed kinda obvious). Nobody posted about that, so I've realized that it may be not that obvious as I thought. Since I don't have much time, and I'll be back only in the middle of the night, I'll answer short.

The thing that worries me the most is that series s is divergent for any positive n. In general, the series 1/(1+x) = sum (_1)^r*x*r is divergent for any x>1, and in this example, with x=((n+1)/n)^(1/3), x is obviously greater than 1 for any positive n.

The second thing that worries me is that after that substitution, one will need to do summation over r as well, and it will be also a divergent series.

Two wrongs don't make right (only 3 left turns do).

The second thing - while summation over r equal 0 modulo 3, obviously, could be done - because that make the summand rational and the way to find sums of rational summands is well known (well, even Maple can do that for small degrees), but the sum for all such cases could be evaluated just by replacing 1+((n+1)/n)^(1/3) with 1+((n+1)/n in the denominator - no need of using divergent series. And it is not related that much to the final sum - because the terms with r not equl to 0 modulo 3 are missing, and they couldn't be calculated that way - another techique is needed (if it is possible at all.)

The third thing I've wrote already above - that even if all that could be done, with a lot of work etc. and twice using divergent series, the final result would be the same as if one did the asymptotic of the summand for n approaching infinity from the very beginning  - that I did above - which couldn't be summated in an obvious way - at least I couldn't do that in 5 minutes that I spent on that - and I have a lot of experience in series summation - I was interested in such things about 40 - 45 years ago, and still can do them pretty well, I think.   

Alec

p:=exp(3)*(1+(x-3)+(x-3)^2/2+(x-3)^3/6);

                          /                2          3\
                          |         (x - 3)    (x - 3) |
              p := exp(3) |-2 + x + -------- + --------|
                          \            2          6    /

evalf(eval(p,x=Pi));

                             23.14034651

P:=proc(x,N) local X; eval(mtaylor(exp(X),X=3,N+1),X=x) end:

for N while not 
    evalf(evalf(P(Pi,N),26)-evalf(P(Pi,N+1),26),26)=0. 
    do od;

evalf(evalf(P(Pi,N),26),25);

                      23.14069263277926900572909

N;

                                  15

Check that,

evalf(evalf(exp(Pi),26),25);

                      23.14069263277926900572909

Alec

p:=exp(3)*(1+(x-3)+(x-3)^2/2+(x-3)^3/6);

                          /                2          3\
                          |         (x - 3)    (x - 3) |
              p := exp(3) |-2 + x + -------- + --------|
                          \            2          6    /

evalf(eval(p,x=Pi));

                             23.14034651

P:=proc(x,N) local X; eval(mtaylor(exp(X),X=3,N+1),X=x) end:

for N while not 
    evalf(evalf(P(Pi,N),26)-evalf(P(Pi,N+1),26),26)=0. 
    do od;

evalf(evalf(P(Pi,N),26),25);

                      23.14069263277926900572909

N;

                                  15

Check that,

evalf(evalf(exp(Pi),26),25);

                      23.14069263277926900572909

Alec

Also, that could be done as

eq:= sin(3*x) - cos(5*x);

                      eq := sin(3 x) - cos(5 x)

trigsubs(convert(eq,sin));

                                Pi             Pi
                  [2 cos(4 x + ----) sin(-x - ----)]
                                4              4

solve(%,AllSolutions);

         {x = 1/16 Pi + 1/4 Pi _Z1~}, {x = -1/4 Pi + Pi _Z2~}

about(_Z1,_Z2);

Originally _Z1, renamed _Z1~:
  is assumed to be: integer

Originally _Z2, renamed _Z2~:
  is assumed to be: integer

Alec

Also, that could be done as

eq:= sin(3*x) - cos(5*x);

                      eq := sin(3 x) - cos(5 x)

trigsubs(convert(eq,sin));

                                Pi             Pi
                  [2 cos(4 x + ----) sin(-x - ----)]
                                4              4

solve(%,AllSolutions);

         {x = 1/16 Pi + 1/4 Pi _Z1~}, {x = -1/4 Pi + Pi _Z2~}

about(_Z1,_Z2);

Originally _Z1, renamed _Z1~:
  is assumed to be: integer

Originally _Z2, renamed _Z2~:
  is assumed to be: integer

Alec

Sorry, Alejandro. I looked briefly at what you did and it seemed completely wrong. I reread it again and it started to make some sense. The problem with this approach is that even if the formulas for given r can be found, the result, after summation over r, still will look like C+sum(c(k)*Zeta(k),k=2..infinity). In some cases that could be evaluated though (but, probably, not in Maple).

I didn't spend much time on doing this series (first, because I don't have much free time, and second, because it didn't seem interesting, especially without knowing the roots of the problem and how it could be related to something interesting.)

Now, I tried the following - using f1 from your post,

asympt(f1,n,10);

   1      1      1       65       223       227       2275
  ---- - ---- + ---- - ------ + ------- - ------- + --------
     2      3      4        5         6         7          8
  2 n    3 n    4 n    324 n    1296 n    1458 n    15552 n

             177419        1
         - ---------- + O(---)
                    9      10
           1259712 n      n

After summation over n, that looks close to

sum((-1)^k*Zeta(k)/k,k=2..infinity);

                        infinity
                         -----       k
                          \      (-1)  Zeta(k)
                           )     -------------
                          /            k
                         -----
                         k = 2

evalf(%);

                             0.5772156649

evalf(gamma);

                             0.5772156649

which is close to the answer, and if the rest could be evaluated in a similar manner, which could be, perhaps, done, since the first terms look more or less computable,

1/5-65/324;

                                  -1
                                 ----
                                 1620

1/6-223/1296;

                                  -7
                                 ----
                                 1296

1/7-227/1458;

                                -131
                                -----
                                10206

But still it seems as if we could just get another infinite series.

Alec

Sorry, Alejandro. I looked briefly at what you did and it seemed completely wrong. I reread it again and it started to make some sense. The problem with this approach is that even if the formulas for given r can be found, the result, after summation over r, still will look like C+sum(c(k)*Zeta(k),k=2..infinity). In some cases that could be evaluated though (but, probably, not in Maple).

I didn't spend much time on doing this series (first, because I don't have much free time, and second, because it didn't seem interesting, especially without knowing the roots of the problem and how it could be related to something interesting.)

Now, I tried the following - using f1 from your post,

asympt(f1,n,10);

   1      1      1       65       223       227       2275
  ---- - ---- + ---- - ------ + ------- - ------- + --------
     2      3      4        5         6         7          8
  2 n    3 n    4 n    324 n    1296 n    1458 n    15552 n

             177419        1
         - ---------- + O(---)
                    9      10
           1259712 n      n

After summation over n, that looks close to

sum((-1)^k*Zeta(k)/k,k=2..infinity);

                        infinity
                         -----       k
                          \      (-1)  Zeta(k)
                           )     -------------
                          /            k
                         -----
                         k = 2

evalf(%);

                             0.5772156649

evalf(gamma);

                             0.5772156649

which is close to the answer, and if the rest could be evaluated in a similar manner, which could be, perhaps, done, since the first terms look more or less computable,

1/5-65/324;

                                  -1
                                 ----
                                 1620

1/6-223/1296;

                                  -7
                                 ----
                                 1296

1/7-227/1458;

                                -131
                                -----
                                10206

But still it seems as if we could just get another infinite series.

Alec