About the sum function.

Example: calculate the sum of 1/(x^2+x+1) from x=1 to x=10. The result appeared immediately after 1 second. Now, change x=10 to x=12. It take considerable time to solve this problem. I have no solution.

Indeed, summing from

x=1 to

x=12 takes 5 secs. Not a lot of time, but considerably larger. It seems that ClassPad "gets confused" if you add small terms (those corresponding to

x=11 and

x=12), to the sum of the first 10 terms (which is a considerably larger number). I

*suppose* that this is the case, because

sum(1/(x^2+x+1) for

x=11 to

x=12 is computed immediately. In general, adding very small number to e.g., unity, may lead to unexpected results. This is an unavoidable problem in computer arithmetics. For example, most computers give 1+8*10^-7=1, or 1+3*10^-16=1, depending on the precision of the machine and the programming language used.

The obvious workaround is to avoid adding very small numbers to large numbers, or at least to reduce such additions to the minimum. In this case, you get the correct result immediately if you add the sum from

x=1 to

x=10 to the sum from

x=11 to

x=12. I'm afraid that there is no better solution.

About solving trigonometric equations: it takes much more time than the first OS (1.0) in solving any kind of trig-eq.

I have upgraded the OS immediately after buying my ClassPad, so I cannot tell everything about that. I think that you should post this in the "Bugs" topic, adding examples of computation times, if you can.

About integral: calculate the area of circle x^2+y^2=R^2:

S = 2*integral(sqrt(R^2-x^2),x,-R,R). In OS 1.24, it cannot solve (real mode). In OS 2.0, it can solve this problem in real mode.

However, if you calculate the area as formula:

S=4*integral(sqrt(R^2-x^2),x,0,R), then in real mode, it cannot solve.

Well, although ClassPad's integration has many problems, I don't think that there is a bug here. First of all, let me say this: there are

*much better* ways to compute this area, given that it has a circular symmetry (see below for this). That being said, if you

*insist* to compute the area the way you do, here is what happens: ClassPad does not know anything about

R, hence the problem. For example, if

R is imaginary, then the integral is also "Non-Real"; you can overpass this problem by switching to "real" mode. But, still, the result cannot be obtained without any further knowledge about

R. To understand what happens here, calculate the corresponding indefinite integral (in "real" mode). You will see that the answer contains the term

x/|R|, which is equal to

x/R, if

R is positive, or to

-x/R, if

R is negative, or even indeterminate, if

R=0. So, ClassPad cannot "decide" which is the case. You can "help" ClassPad to decide as follows: first compute the indefinite integral then do:

(ans|{x=R,R>0})-(ans|{x=0})This way, you can compute

integral(sqrt(R^2-x^2),x,0,R) easily. You tell ClassPad that

R is positive, so there is no confusion. This gives the correct answer,

pi R^2/4. As you can see, there is no bug here. In fact, if ClassPad was able to give the expected answer without any knowledge about

R, that would be a bug.

However,

*you should avoid computing areas that have circular symmetry that way*. There are two much more convenient ways to do this. You may probably know that, but i'll post the alternative solutions anyway:

(1) Compute the area in polar coordinates by using

(1/2)*integral(r(phi)^2,phi,phi1,phi2),

where

phi is the polar angle,

r(phi) is your function transformed in polar coordinates, and

phi1,

phi2 are the minimum and maximum angle, respectively. In your case,

4*integral(sqrt(R^2-x^2),x,0,R) is transformed to

2*integral(R,phi,0,pi/2),

and the result is obtained in no time, no matter what the mode is ("real" or "complex").

(2) Compute the area by the corresponding double integral, also in polar coordinates. In your case, you have to compute

4*integral(integral(r,r,0,R),phi,0,pi/2)This is computed in no time as well.

As you can see, computer algebra cannot be used to solve every problem "blindly". Usually, you should transform the problem at hand to the most simple form, then use a computer to do the rest.