The output of this integral would be confusing to a student:
The answer is quite simply "6", as returned by a 48,49,50. Perhaps labeling these two results could help lower the confusion.
Three ways to work around this are: use "0." and "4." as limits of integration and get the answer "6.", use "0" and "4" and disable "Exact" in the CAS settings and get "6.", or perform the calculation in Home and get "6.".
In my opinion this is confusing as the three models prior to the Prime were able to give the correct exact answer "6" without selecting "approx" in the CAS settings (using "stock" flag settings set on the 48).
Are we always required to use decimal values in the CAS?
[HP Prime] "Error while checking exact value with approximate value, returning both!"
|
|
« Next Oldest | Next Newest »
|
▼
12-05-2013, 10:13 AM
▼
12-05-2013, 11:06 AM
The issue is that there really is not great support for the nthroot form due to personal preferences of the CAS author. Use ^1/3 and it works fine. TW ▼
12-05-2013, 11:57 AM
In fact, the CAS's nthroot errors are somewhat predictable:
▼
12-05-2013, 12:40 PM
The conventions for nth root and 1/n power in xcas are explained in pages 64 and 65 of They are different from those in Mathematica(TM). ▼
12-05-2013, 01:47 PM
Thank you for the reference; I'm looking into it :)
In my opinion, the software author's method of interpreting powers and nth roots "taints" the CAS and propigates errors throughout the system. Or I'm just not understanding something here?
Edited: 5 Dec 2013, 1:58 p.m. ▼
12-05-2013, 02:21 PM
Actually, both limits produce the correct result in XCAS in Linux: Edited: 5 Dec 2013, 2:22 p.m. ▼
12-05-2013, 04:09 PM
It looks like the xcas in the Prime is not up to the official xcas. Besides the Prime cas seems to have changed between updates. Here is what I obtain with an older emulator (the one I have running in Linux under wine)
12-05-2013, 05:31 PM
Chris, In your first example you are taking the limit as b goes to 0 of (1 + b). This result is 1. You are then taking that result to the power of 1/b. This gives the correct result of 1 (1^n=1).
You need to add an extra pair of parenthesis starting at the limit and closing at the end of the expression. This gives the correct result of e. limit(1+b,b,0)^(1/b) = 1
Edited: 5 Dec 2013, 5:41 p.m. ▼
12-05-2013, 05:58 PM
Yes. You are correct ! Obviously PEBKAC in this case. ▼
12-05-2013, 06:02 PM
Well, I think this needs work on our end to clarify what is happening. It is an easy mistake to make since there are times it can not be clear to what the function is being applied. I've flagged another mark on that item to float it up higher in the queue. TW |