Hi, any recommendations? Costco has some fancy TI's at decent prices right now (TI-84 Plus Silver, TI-89 Titanium). Thanks, Howard
The TI-89 has some excellent features built in. If it doesn't have the function native, you can always code your own. I would definitely go TI-89 over the 84 just for the pretty print. You can enter a matrix like in the following form: [1,0,0;0,1,0;0,0,1] and it will display it like: [tex] \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] [/tex] However, do not depend on a calculator for the class. It will kill you. Use it only to check your answers. The more problems you do without your calculator, the more you will get a intuitive understanding of the topic.
I second the TI-89. As long as you have the self restraint, it can be a very valuable tool. Tedious manipulations when diagonalizing a matrix can be checked upon completion, which saved me a couple of points on homework assignments. Plus, a majority of the problems your calculator can do are problems you'll be asked to show work for, i.e. solving a system or finding an inverse.
I am an "old goat" just trying to relearn some old things and learn some new things. The thing between my eyes does not work as efficiently as it used to and is occupied with making a living, raising my three boys and keeping the boss happy. Plus, I like gadgets and have not bought a calculator since getting a TI-58C 25 years ago, they have changed a bit since then! I also had an HP 11C as no self-respecting engineering student would be without one in those days. It appears that TI is now in the lead, at least for graphing calculators. Any comments on the TI Voyage 200? How big is it, is it any more useful then the 89? Howard
It's essentially the same thing as an 89, and is exactly the same as the TI-89 Titanium other than the fact that it has a qwerty key pad form factor as opposed to the standard calculator form.
Voyage isn't that big, and it is definitely nice to use a QWERTY keyboard. Go check them out in the store, and see which one you like better.
I had an hp-48g during my first linear algebra class. It was oh so exciting to get it to find a 4x4 matrix inverse for me. about once. I found it was generally faster for most of the computations we had to do to just do them by hand. I can't say it was really beneficial for learning linear algebra.
The difference between the Voyage 200 and a TI-89 is more than just the keyboard. The V200 has 2.7MB of Flash ROM and is loaded with all the APPS(Geometer's Sketchpad, Statistics pac, etc.). The granddaddy of calculators. Here's a link. Check them out. http://cgi.ebay.com/NEW-TEXAS-INSTR...ryZ50579QQssPageNameZWDVWQQrdZ1QQcmdZViewItem When/If my old TI-92 kicks the bucket, that's what I'm buying. They're worth the money. Considering how much more you get, they're not that much more expensive than the 89.
the point is that calculATORS ARE ONLY USEFUL For DOing CALCULATIONS YOU ALREADY UNDERSTAND. understanding them is the real challenge.
Any calculations small enough to plug into a hand calculator, it's easy enough to do on paper. Any larger calculations, should be left to Matlab.
Hey- Quick question I want to know if either the TI-89 Titanium or the Voyage 200 will provide step-by-step solutions to calculus problems.
No, they won't. They can do symbolic integration and differentiation as well as evaluating limits, factoring and expanding etc, but they will not do the work step by step, and I certainly hope that you do not plan on relying solely on a calculator to do all of your work for you in a calculus class because if this is the case then you will almost certainly fail.
Nope, i was actually planning on using it for my p.chem class. Which one do you think is best for pchem?
I recommend http://calculator-online.org/s/matrix/. Example the solution A*B: Given matrix A = [1 2 3] [3 2 1] [3 7 1] and B = [0 4 1] [3 3 1] [4 2 4] . Find the product A*B Consider the product A*B. The number of columns in the first factor A equal 3, number of rows in the second factor B also equal 3. The numbers coincide, therefore, the product is defined. The result is a matrix multiplication C = A*B, whose lines as much as them in the first factor, ie 3, and columns as much as them in the second factor, ie 3. So, matrix C has dimensions 3 x 3 Find: Element c1 1. In its computation involved 1-th row [1 2 3] first factor A and 1-th column [0] [3] [4] second factor B: c1 1 = (1) * (0) + (2) * (3) + (3) * (4) = 18; Element c1 2. In its computation involved 1-th row [1 2 3] first factor A and 2-th column [4] [3] [2] second factor B: c1 2 = (1) * (4) + (2) * (3) + (3) * (2) = 16; Element c1 3. In its computation involved 1-th row [1 2 3] first factor A and 3-th column [1] [1] [4] second factor B: c1 3 = (1) * (1) + (2) * (1) + (3) * (4) = 15; Element c2 1. In its computation involved 2-th row [3 2 1] first factor A and 1-th column [0] [3] [4] second factor B: c2 1 = (3) * (0) + (2) * (3) + (1) * (4) = 10; Element c2 2. In its computation involved 2-th row [3 2 1] first factor A and 2-th column [4] [3] [2] second factor B: c2 2 = (3) * (4) + (2) * (3) + (1) * (2) = 20; Element c2 3. In its computation involved 2-th row [3 2 1] first factor A and 3-th column [1] [1] [4] second factor B: c2 3 = (3) * (1) + (2) * (1) + (1) * (4) = 9; Element c3 1. In its computation involved 3-th row [3 7 1] first factor A and 1-th column [0] [3] [4] second factor B: c3 1 = (3) * (0) + (7) * (3) + (1) * (4) = 25; Element c3 2. In its computation involved 3-th row [3 7 1] first factor A and 2-th column [4] [3] [2] second factor B: c3 2 = (3) * (4) + (7) * (3) + (1) * (2) = 35; Element c3 3. In its computation involved 3-th row [3 7 1] first factor A and 3-th column [1] [1] [4] second factor B: c3 3 = (3) * (1) + (7) * (1) + (1) * (4) = 14; So, C = [18 16 15] [10 20 9] [25 35 14]
Wolfram Alpha seems to have trouble sometimes with symbolic calculations, e.g. {{cos(x),sin(x)},{-r sin(x), r cos(x)}} inv({{cos(x),sin(x)},{-r sin(x), r cos(x)}}) http://www.wolframalpha.com/
In fact, even a simple numerical calculation can give it trouble. It just got this one wrong: {{0,11},{-17,19}} * inv{{2,1},{-3,4}} I gets the inverse right, so the problem seems to lie in the actual matrix multiplication.