Wednesday, March 12, 2008

51+52+53+...+100 is how much more than 1+2+3+...+50? Why, 50^2 of course! Now Explain and Generalize...

Quick Updates....
Mystery Mathematician Contest ending soon...
Pi fact for today? Try explaining why the imaginary number i raised to the power of i is REAL without mentioning π somewhere! Of course, you could just ask Google to do it for you!

You'd think that the deafening silence from the 5-7-8 triangle post would discourage me - NOT! Here is an investigation for middle schoolers and up.

Typical Content Standard: Patterns, Relations, Algebra

(1) Developing strategies for comparing sums
(2) Developing algebraic generalizations
(3) A few dozen more!

Where might the first question in the title of this post be asked?
(a) SATs?
(b) End of Course Test for Algebra 2?
(c) Other standardized tests?
(d) Math contests? If so, what grade level? 7th? 8th? Higher?

If you value a question such as this, would you introduce it to middle schoolers in 6th? 7th? Prealgebra? Would you use a very different instructional approach with students in higher math courses who have reasonable algebra background? Even if you don't like this question, try it with one of your groups tomorrow and let me know what happens!

Since I have personally posed this type of question to both middle schoolers and older students, I can tell you that even strong math students often have not seen the 'compare differences of corresponding terms method'. I made up that designation but, hopefully, you can make sense of it. Do you think many high school students would attempt to find two separate sums by some method/formula (or using their calculator if allowed) they've seen?

Well, I won't give any more away, but I believe the issues of pedagogy here may transcend the problem and the math strategies:

How does one introduce this? Do you simply have this question on the white board as students enter the room and allow them to work on it individually or in small groups for 5-10 minutes? We all hear about our re-defined role as 'guides on the side' but what exactly does this look like for this activity? How do we facilitate? When do we ask leading questions? What questions would be highly effective here? I haven't even mentioned the calculator issue yet!

So many questions. So few answers...

Actually I was going to do a short video presentation of this question to demonstrate one instructional model, but, unfortunately, my dog ate my main computer which has all my files and applications. Wait - let me apologize to my pooch. He really didn't eat it or even bless it with his bodily functions. But my iBook is very sick and will need intensive care from Apple. In the meantime, I'm on a backup machine, with limited memory and lacking many of my files and applications. Excuses, excuses, excuses! Please bear with me!


Anonymous said...

Let z = i^{i}. Well, some value of that at least. Then the conjugate of z is ī raised to the ī-th power. ( I hope that Unicode character survives this post.)

But ī = -i, so the conjugate of z is (-i)^(-i).

(-i)^(-i) = 1/[(-i)^i] = [1/(-i)]^i = i^i.

So, z is its own conjugate, and is real.

mathmom said...

I will try this on my middle schoolers next week and see what they come up with. (Not the generalize part, I don't think, just the initial question.)

As you know, we've played with using Gauss' method to find sums like these. I think they would try to find the sums and subtract. I think most of them would make mistakes finding the sum from 51 to 100 because they will miscount how many terms are in that sequence.

I will be interested to see if any of them come up with the "comparing corresponding terms" method which of course makes the problem much easier!

I think a problem like this could show up at MathCounts -- maybe even in a Countdown, because it is quick to calculate once you recognize the "trick". (MathCounts is a math contest for grades 6-8.)

I don't know enough about the usual content of SATs, Algebra2 exams or other standardized tests to comment on its suitability for those types of tests.

And... I'm pretty sure Eric meant to comment on a different post -- at least I really hope so! ;-)

p.s. I'd love it if you (or your readers) would come check out and comment on my recent geometric probabilities post. Sorry for the blatant plug. ;-)

Anonymous said...


That was for the first paragraph of the article, just below the title, the Pi fact of the day.

mathmom said...

Aha, Eric, now I see!

Joshua Zucker said...

Pedagogically I think my approach to lead them to that difference would be to ask something like (to do in small groups, in maybe 15-20 minutes total -- but I work with very smart 10th graders, so for other groups it might be different):

a) What is similar and what is different about computing 1+2+3 and 101+102+103?

b) Now compare and contrast 1+2+3+4+5 and 101+102+103+104+105

c) Now compare and contrast 1+2+3+..+50 with 51+52+53+...+100.

d) Which is easier, adding 1+2+3+...+50 and then modifying that result to get the value of 51+52+53+...+100, or directly computing the value of 51+52+53+...+100? Explain why you think your choice of methods is easier.

This misses the point if it's specifically the 50^2 you're after, or if it's the recursive formula that with S(n) = sum of first n integers,
S(2n) = n^2 + S(n) or something like that,
but I think from the name you gave this method that you have the pairwise differences in mind in general, and the fact that it's 50*50 and not the product of two unequal numbers doesn't really matter for the point you're trying to make.

Dave Marain said...

Nice Eric! I knew if anyone could do this without π, you could! Here's one approach with π:
We can write i as
i = e^(iπ/2).
Then i^i = (e^(iπ/2))^i
= ((e^(π/2))^i)^i
= (e^(π/2))^(-1), which is real!
I probably messed up some parentheses here. This is not nearly as neat as your argument but students can make some sense of this as soon as they understand that e^(ix) = cos(x) + isin(x). Help me out here, Eric -- in either explanation, isn't there some issue of a principal value for i^i?

Ok, I think I missed the main theme of this post! Joshua, this post had several key issues embedded:
(1) Develop the idea of pairwise differences or as I like to explain it: To find the difference of the sums, we can instead find the sum of the differences - an important point IMO.
(2) Using your notation for S(n):
Stating and demonstrating that
[S(2n)-S(n)] - S(n) =
S(2n) - 2S(n) = n^2.
For example:
S(6) - 2S(3) = 21-12 = 9 = 3^2.
This symbolic form of course is not intended for middle schoolers nor would it be trivial for advanced students at first.
(3) Yes, my focus here was on consecutive sums of consecutive positive integers as opposed to 1,2,3 and 101,102,103 which nicely demonstrates pairwise differences but doesn't lead to the same algebraic formulation.
(4) Most importantly, the role of the instructor in developing these ideas. Developing an effective sequence of questions here and knowing when to intervene is a subtle art that can be learned from experience and watching models of teachers who have refined this art. I don't believe enough of this kind of training takes place in the preservice education of young math teachers. This is one of the reasons I started this blog. Joshua, your comments will help these teachers...

Anonymous said...


It all depends on which value of the logarithm you take. Since e^{2πi} = 1, e^{2nπi} = 1 for any integer n. This means that the logarithm is a 'multi-valued function' on the complex plane. Imagine taking the standard logarithm near z = 1, and extending it by continuity along the unit circle. When you get back to 1, your value of logarithm is 2πi.

The Riemann surface of log z, the topological space of pairs (z, log z) for any value of the logarithm, is sort of a helicoid; you can buy models in the pasta aisle of your supermarket.

So, the values of log i are πi/2 + 2nπi. Then,

i^i = e^{i log i} = e^{-π/2 -2nπ}.

The principal value would have n = 0, but that would be a convention, and would be of much less significance than the standard principal value of the square root.

Dave Marain said...

I seem to recall studying the analytic continuation of log(z) in complex analysis and its multi-valued nature (it got more complicated with sheaves, etc., but that's way too vague in my mind). Again, thanks for clarifying this for me.

Anonymous said...

By looking sharp one can see that the sum of consecutive integer
numbers between a and b can be computed by multiplying the amount
of numbers that we want to add (= b-a) by the mean of the numbers
(= a+(b-a)/2):

a+...+b = (b-a+1) * (a + (b-a)/2)

With this formula it's easy to compare sums.


a=1 , b=50 : (50-1+1)*(1+(50-1)/2) = 50*(1+(49)/2) = 1275
a=51, b=100: (100-51+1)*(51+(100-51)/2) = 50*(51+49/2)= 3775

Test: 3775-1275 = 2500 = 50*50

The above example is a special case where the sums of
a+...+b and b+...+c are compared and where b is c/2.

Lets see how the general formula for this special case

(c-b+1)*(b+(c-b)/2) - (b-a+1)*(a+(b-a)/2)
= (b/2+b+1)*(b+(b/2 -b)/2 - (b-a+1)*(a+(b-a)/2)
= ...
= ((c-a+1)/2) *( (b+(c-b)/2) - (a+(b-a)/2) )

which is simply the difference of the means of the
two sums multiplied by the total number of elements
divided by two.

The formulas seem very complex but
if one simply remembers to multiply the mean of the numbers by the amount of the numbers .. you can
derive your own formula very quickly.

Anonymous said...

Sheets, Dave, not sheaves. Sheaves are something different that also appears in algebraic geometry and Riemann surfaces.

Joshua Zucker said...

Going back to the very first comment:
(-i)^(-i) = 1/[(-i)^i] = [1/(-i)]^i = i^i.

I'm not sure I buy this. I don't think it's true that 1/(z^k) = (1/z)^k in general for complex numbers.

For instance:
-i = 1/i = 1/((-1)^(1/2)) =? (1/-1)^(1/2) = (-1)^(1/2) = i

So -i = i?

I think we have to deal with the multivaluedness of these exponent things. Maybe this is more or less directly related to the multivalued ln, because
i^i = e^(ln(i^i)) = e^(i * ln(i)) = ... with the obvious ensuing ambiguities.

Anyway, what I'm really asking is not about the multivaluedness but about whether the proof that the conjugate of i^i = i^i is a valid proof or not.

mathmom said...

Hi Dave,

I did try this out, and wrote about it on my blog.

I was pretty much right on in my predictions -- they all used the Gaussian summing method and subtracted (with much inaccuracy), though I did get 2 of 8 students who came up with a better way when asked to find a second way.