Signed area graph problem

11252016, 07:55 PM
Post: #1




Signed area graph problem
Possible graphing app problem:
In home, integral from 0 to 1 of (sqrt(tan(X)+1)  sqrt(sin(X)+1))/x^3 correctly produced 0.226409969741. In the graphing app, signed area from 0 to 1 correctly fills in the area of interest, but produces the hourglass symbol...after several minutes of the hourglass, I gave up and had to do a pinhole reset to exit out of the problem. 

11252016, 08:46 PM
Post: #2




RE: Signed area graph problem
Out of morbid curiosity, I then tried using 1 E12 as lower limit in graphing app as was necessary on my CASIO 9860gii due to the function being undefined at X=0. Still showed endless hourglass requiring pinhole reset.


11252016, 09:26 PM
Post: #3




RE: Signed area graph problem
Finally, tried solve d(function)/dx in home, and produced the endless hourglass necessitating pinhole reset.


11252016, 10:32 PM
Post: #4




RE: Signed area graph problem
(11252016 07:55 PM)lrdheat Wrote: Possible graphing app problem: The Signed Area feature of Function simply passes the work off to other parts of the system. What expression are you entering in Home? When I tried ∫(F1(X),X,0,1) in Home (with F1(X) defined as (sqrt(TAN(X)+1)sqrt(SIN(X)+1))/X^3) my emulator behaved just as the Signed Area feature did. A significant amount of computation time (~14s) was required before the approximate answer came up. I thought perhaps it was locked up, given the time it was taking and the nature of the posting. With either way of finding the integral, pressing ON/C aborted the calculations. Which version of the software are you using? Are you using an emulator or a physical calculator? 

11262016, 12:07 AM
Post: #5




RE: Signed area graph problem
Physical calculator, latest version (10637). Home approximation takes about 1 minute 14 seconds. While it may eventually produce an answer in the graphing app, I gave up after perhaps 4 minutes.
By comparison, WP 34S and CASIO 9860gii nearly instantaneous. 

11262016, 01:29 AM
Post: #6




RE: Signed area graph problem
(11262016 12:07 AM)lrdheat Wrote: Physical calculator, latest version (10637). Home approximation takes about 1 minute 14 seconds. … From "approximation", I'm guessing you're not entering ∫(F1(X),X,0,1) in Home (with F1(X) defined appropriately) — is that the case? The Signed Area feature uses a similar code path as ∫() to compute the signed area. 

11262016, 01:44 AM
(This post was last modified: 11262016 01:45 AM by Mark Hardman.)
Post: #7




RE: Signed area graph problem
(11262016 12:07 AM)lrdheat Wrote: Physical calculator, latest version (10637). Home approximation takes about 1 minute 14 seconds. While it may eventually produce an answer in the graphing app, I gave up after perhaps 4 minutes. Using the function's plot on my physical Prime, I get a result of 0.226409969741 for the Signed Area in about 1'15". Mark Hardman Ceci n'est pas une signature. 

11262016, 02:35 AM
Post: #8




RE: Signed area graph problem
I think I have found an oddity when evaluating the function...as we approach zero from above, the function approaches .25 (which the Prime correctly produces when I take the limit). When substituting small numbers into the function, it does approach .25, but smaller than .001, such as .0001, Prime comes up with 0.
Perhaps this is the problem. I am in standard display as opposed to fix option. Is there some other setting that is producing this behavior? 

11262016, 02:52 AM
Post: #9




RE: Signed area graph problem
More oddity...while function should approac .25 as "X" becomes small, Prime shows
X .0003 F(X) .3703+ .0004 .15625 .0005 .24 Why? 

11262016, 03:10 AM
Post: #10




RE: Signed area graph problem
Guess it's due to numerator approaching zero due to convergence of tan,sin approaching zero...beyond the 12 or 15 digit resolution of the machine.
Why does the Home algorithm succeed, likewise the WP 34S and CASIO 9860gii? My Prime does not succeed in the graphing app signed area 01 for the function... 

11262016, 03:55 AM
Post: #11




RE: Signed area graph problem
(11262016 03:10 AM)lrdheat Wrote: Guess it's due to numerator approaching zero due to convergence of tan,sin approaching zero...beyond the 12 or 15 digit resolution of the machine. What, precisely, are you entering in Home? 

11262016, 04:43 AM
Post: #12




RE: Signed area graph problem
Hi...
I put the function that I quoted in my original post in home...it works in home, but not in the graphing app. How do I define a function as opposed to simply entering the function into home to the enter into the function app...it sounds as if you had a different experience in the graphing app when doing so (although why that would be the case is another question, and why CASIO and WP 34S, while suffering from same difficulty with the function for numbers close to zero, still come up with a good answer for the integral from 0 to 1). How does the Prime come up with a good answer for the integral in home, and why does it take so long compared to the other calculators that I referred to? Puzzled, but interested! 

11262016, 04:49 PM
(This post was last modified: 11262016 04:49 PM by Mark Hardman.)
Post: #13




RE: Signed area graph problem
(11262016 02:52 AM)lrdheat Wrote: More oddity...while function should approac .25 as "X" becomes small, Prime shows Because the function fluctuates wildly near the discontinuity at X=0. While it does approach 0.25 as it approaches the discontinuity, it does so in an illbehaved manner. Again, I have no problem at all getting this function to integrate on my physical Prime using the graphing app. Yes, it takes more than a minutebut not much more. Maybe you could try resetting the Function app and the Home settings using ShiftClear. Mark Hardman Ceci n'est pas une signature. 

11262016, 07:35 PM
Post: #14




RE: Signed area graph problem
Does the function behave this way in reality, or is it an artifact of tan X, sin X converging for very low values, producing delta tan X, sin X below 12 digit resolution of the calculator, or below higher resolution of more sophisticated math software?


11262016, 07:49 PM
Post: #15




RE: Signed area graph problem
(11262016 07:35 PM)lrdheat Wrote: Does the function behave this way in reality, or is it an artifact of tan X, sin X converging for very low values, producing delta tan X, sin X below 12 digit resolution of the calculator, or below higher resolution of more sophisticated math software? The function behaves this way in reality. https://www.wolframalpha.com/input/?i=pl...4+and+1E4 Ceci n'est pas une signature. 

11262016, 08:34 PM
Post: #16




RE: Signed area graph problem
Hi Mark,
I'm still wondering if such a graph is an artifact of the numerator becoming smaller than can be handled smoothly with a finite calculation resolution inherent in any mathematical software... Without going into high level math, is there an intuitive explanation that would explain why the function should not approach it's limit of .25 smoothly? I did let the Prime have several minutes to calculate signed area in the graphing app, and it did come up with a good answer. Any idea as to why the WP 34S or CASIO 9860gii would be close to immediate in finding an answer vs what the Prime does in home and graphing? Thanks for your thoughts! 

11272016, 06:09 PM
Post: #17




RE: Signed area graph problem
From an intuitive point of view, how can this function be approaching a specific limit f(X), X=0 of 1/4 if the function is not approaching that specific limit? (I.e., function becomes increasingly erratic as we get arbitrarily close to X?)
This is the motivation of my question, and my guess as to why the function appears to behave erratically when X approaches zero. 

11272016, 07:06 PM
Post: #18




RE: Signed area graph problem
(11272016 06:09 PM)lrdheat Wrote: From an intuitive point of view, how can this function be approaching a specific limit f(X), X=0 of 1/4 if the function is not approaching that specific limit? (I.e., function becomes increasingly erratic as we get arbitrarily close to X?) In the range we are interested in, the solution to √(TAN(X)+1) = √(SIN(X)+1) is X=0. Both functions are well behaved in that interval. I concede that there is no reason for the original function to behave erratically other than a lack of precision as X approaches 0. Sorry to have taken you down the wrong trail. Mark Hardman Ceci n'est pas une signature. 

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