calculator fx-991ES integral bug found

speedbird

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Hi,

I post this here because my lecturer(DR.) thinks I'm wronge, but i'm convinced there's a problem here. so I tried to calculate the Q function(normalized) of the number 7, which is the integral of e^((-x^2)/2) from 7 to infinity. of course I can't put infinity on the calculator so therfore I put a very big number. that gave me 0. but when I integrate from 7 to lets say 8, I get 10^-12. firstly I thought it has to do with the limitation of 10^-99, but thats not the case. the thing is, that as I increase the area under the graph of the function, I get smaller numbers which is against logic. it can't be that from 7 to 8 I get 10^-12, and from 7 to 2000 I get 10^-51, and from 7 to 3000 I get 10^-84, and so on. it can't get smaller as I increase the upper limit of the integral, it doesnt make sense, the function is all positive. I decided to test this with integral solver online- guess what, the integral doesn't get smaller there, it just stays constant as I increase the upper value of the integral, which make sense because eventually the numbers get so small that it just adds 0+0+0+0.... and the integral solution number just stays the same. so what is happening with my fx-991? I failed a question on the exam because of false values which it gave me, and no one believes me, and no one can't understand my claim. what do you think? try to calculate for your self , and also try to calculate seperately in this site : https://www.integral-calculator.com/
and compare.. thank you, good day.
 
Hi,

I post this here because my lecturer(DR.) thinks I'm wronge, but i'm convinced there's a problem here. so I tried to calculate the Q function(normalized) of the number 7, which is the integral of e^((-x^2)/2) from 7 to infinity. of course I can't put infinity on the calculator so therfore I put a very big number. that gave me 0. but when I integrate from 7 to lets say 8, I get 10^-12. firstly I thought it has to do with the limitation of 10^-99, but thats not the case. the thing is, that as I increase the area under the graph of the function, I get smaller numbers which is against logic. it can't be that from 7 to 8 I get 10^-12, and from 7 to 2000 I get 10^-51, and from 7 to 3000 I get 10^-84, and so on. it can't get smaller as I increase the upper limit of the integral, it doesnt make sense, the function is all positive. I decided to test this with integral solver online- guess what, the integral doesn't get smaller there, it just stays constant as I increase the upper value of the integral, which make sense because eventually the numbers get so small that it just adds 0+0+0+0.... and the integral solution number just stays the same. so what is happening with my fx-991? I failed a question on the exam because of false values which it gave me, and no one believes me, and no one can't understand my claim. what do you think? try to calculate for your self , and also try to calculate seperately in this site : https://www.integral-calculator.com/
and compare.. thank you, good day.

Please give us something more specific to check out. State what you entered in the calculator (for one or two examples), what you got, and then the same for the website.

My guess is that the small numbers are resulting in some sort of rounding error or overflow/underflow. But it could also be that you are entering something incorrectly.

Also, what was the exam question, and what were your wrong results for that? It may be a separate issue.
 
I tried doing this on my fx-115, and got the same results (though you gave only the power of ten). You're right, it is odd. But I do think it is an issue of inaccuracy in their algorithm due to the small numbers; you'd have to ask the engineers. Ultimately, I really wouldn't expect these numbers to be valid. A math class ought to be teaching you not to trust a calculator too far ...

But I think it's important for you to tell us the exam question you got wrong, and how you solved it. I would guess that you weren't supposed to be using a calculator at all, but we'll have to see.
 
The question was to calculate the error probability of transmited symbols. everyone in the class got the same answer of Q(7), which is correct no doubt, the problem begins now when we should represent this as a number value instead of just Q(7), because we talk about symbols of a digital system which is like bilions of bits transmited, the small probability even 10^-15 have a meaning here. so everyone in the class just integrated between 7 and 100 and got 10^-24 or similar numbers, which is false of course. I'm the only one who decided to put higher upper value than 100 in the integral(to be closer to infinity) and I got 10^-84, therfore I wrote the answer as 0, because this is so small even for symbols. I didn't think that the calculator is wronge, nobody ever thinks that, it doesn't happen, you always trust it. just after I got home I suddenly realized something doesn't make sense with the area under the graph being increased, but the integral values get decreased. so now my problem is to convince this DR lecturer that there is a bug in the calculator that made everyone get the wronge answer. the answer is not 0 as I thought, and its not 10^-24 as everyone else wrote. it is actually something around 10^-11.
 
… also try to calculate seperately in this site : https://www.integral-calculator.com/ and compare …

… [the true estimate] is actually something around 10^-11.
That calculator's estimate is roughly 3.2×10-12

By the way, a number of online calculators report zero (a few of those sites can increase precision, but require additional input). Try wolframalpha or mathportal, for examples.

As everybody got the wrong answer (fx-991ES error), I assume the course handed out the calculators. If the course relies on that calculator for answers, then everybody ought to get credit.
 
The question was to calculate the error probability of transmited symbols. everyone in the class got the same answer of Q(7), which is correct no doubt, the problem begins now when we should represent this as a number value instead of just Q(7), because we talk about symbols of a digital system which is like bilions of bits transmited, the small probability even 10^-15 have a meaning here. so everyone in the class just integrated between 7 and 100 and got 10^-24 or similar numbers, which is false of course. I'm the only one who decided to put higher upper value than 100 in the integral(to be closer to infinity) and I got 10^-84, therfore I wrote the answer as 0, because this is so small even for symbols. I didn't think that the calculator is wronge, nobody ever thinks that, it doesn't happen, you always trust it. just after I got home I suddenly realized something doesn't make sense with the area under the graph being increased, but the integral values get decreased. so now my problem is to convince this DR lecturer that there is a bug in the calculator that made everyone get the wronge answer. the answer is not 0 as I thought, and its not 10^-24 as everyone else wrote. it is actually something around 10^-11.

What does DR stand for? I may be misinterpreting it. I assumed you meant the instructor has a Ph.D., but now I'm guessing otherwise. One should avoid using abbreviations that someone outside of your own context might not understand.

This topic is worth discussing with the instructor, not just for the grade, but because it is a very real issue.

First, how much precision is appropriate in your setting? Even with billions of bits (taking that literally, 10^9 or so), a probability less than 10^-11 makes no difference, right? So why would your answer of 0 be considered wrong?

Second, why would everyone but you integrate to 100? Is that something you were taught to do, or was it done in an example, or did everyone just randomly pick the same number? (At the least, you should be taught what to put in place of infinity, rather than be left to decide randomly.) And is everyone using the same kind of calculator?

Third, as I suggested before, an understanding of the limits of a calculator ought to be part of this course! If you have been taught to "always trust a calculator", you have been taught wrongly!

Looking for information about the calculator's algorithm for definite integrals, I read the manual. Read page E-34. Among other things, it says that when you use LineIO, you can enter a tolerance (though I'm not sure that helps here), and that "There may be large error in obtained integration values" (which means, "Don't trust it"). It uses the Gauss-Kronrod method of numerical integration, which I presume is an appropriate method (I'm not an expert on such things); but any algorithm has its limits. Possibly you can research the method and see whether you can explain your observations.

One way or another, your instructor (or someone else in the department) should be quite interested in how students should be taught to use the calculator for such calculations!

Now, when I integrate from 7 to 100, I get 1.3982*10^-12, not 10^-24. Am I doing something different?

Oh, yes -- there's another little issue. You said the question was about "the Q function (normalized) of the number 7, which is the integral of e^((-x^2)/2) from 7 to infinity". I looked it up, and found that, as I thought, you omitted a constant multiplier. What does "normalized" mean to you? Are you just not saying everything carefully, or are you defining it differently?
 
What does DR stand for? I may be misinterpreting it. I assumed you meant the instructor has a Ph.D., but now I'm guessing otherwise. One should avoid using abbreviations that someone outside of your own context might not understand.

This topic is worth discussing with the instructor, not just for the grade, but because it is a very real issue.

First, how much precision is appropriate in your setting? Even with billions of bits (taking that literally, 10^9 or so), a probability less than 10^-11 makes no difference, right? So why would your answer of 0 be considered wrong?
sorry I'm not a native english speaker so I forgot how to call a person who has phd, its just called a doctor in our country. by DR I meant doctor, your assumption was correct. :D

the precision appropriate in our course's setting can be less than 10^-12, actually an error probability of 10^-12 with a transmission rate of 10^9 its an error every 1000 seconds which is considered not so good. maybe 10^-15 it depends on the transmission rate. but definitely not 10^-84 as I calculated.


Second, why would everyone but you integrate to 100? Is that something you were taught to do, or was it done in an example, or did everyone just randomly pick the same number? (At the least, you should be taught what to put in place of infinity, rather than be left to decide randomly.) And is everyone using the same kind of calculator?
we never taught anything regarding the operation of a calculator, it is just considered to be obvious to operate, I study electronics engineering, we use calculator all the time, it's the end of the third year and I have never encountered such an issue.
all the other students thought that because large upper value results in 0, they should enter smaller number and just write the result, becuase everyone thought that 0 is not acceptable in this course, because of the reasons I mentioned. during the exam the examiner told us we can put any large upper value in the integral that will give us precise answer, and than told us to just put 100 if we like. he wasn't insisting, it was obvious that he wasn't aware of this issue we talk about now. but I'm the only one who thought that 0 in this case must be the answer(because I got 10^-84), even for this course's precision setting, because I tried increasing of upper values and saw the numbers getting smaller. that was the diiference between me and the others, they just put 100, wrote the result and continued without thinking. thats what they told me.


Now, when I integrate from 7 to 100, I get 1.3982*10^-12, not 10^-24. Am I doing something different?
yes, just that our final answer was Q^2, if you do (10^-12)^2 you get 10^-24. but actually the numbers are starting to get smaller after like 35-40 in the upper value, so at 100 it is already getting smaller just not enough yet to see it in the power of 10. yes you can say the others got the right answer my mistake, I know it because of their explanation, which is wronge. I found that if I want to calcaulate the closest right answer I need to find ln(10^-99), which is -227.95, and this is limit of the power of e that the calculator can handle, therefore the equation (-x^2)/2=-227.95 will give us a limit of x=21.35. so I integrate from 7 to 21.35 and this is the closest to the right answer that I can get with this calculator. after 21.35 it start to decrease first by small not visible numbers than after about 35-40 we can see the decrease and by 3000 we already get 10^-84, and so on until 0.

Oh, yes -- there's another little issue. You said the question was about "the Q function (normalized) of the number 7, which is the integral of e^((-x^2)/2) from 7 to infinity". I looked it up, and found that, as I thought, you omitted a constant multiplier. What does "normalized" mean to you? Are you just not saying everything carefully, or are you defining it differently?
there is a constant multiplier of 1/(square root of 2 pai), by "normalized" I mean the standard deviation is 1 and the average is 0(in the normal distribution of probability).
I omitted the multiplier on purpose because it is not a factor in the integral, the issue still exists with or without it


thank you about manual comment , I will look it up also
 
That calculator's estimate is roughly 3.2×10-12

By the way, a number of online calculators report zero (a few of those sites can increase precision, but require additional input). Try wolframalpha or mathportal, for examples.

As everybody got the wrong answer (fx-991ES error), I assume the course handed out the calculators. If the course relies on that calculator for answers, then everybody ought to get credit.

yes as I replyed to peterson I made a little mistake the final answer was Q^2, so everybody actually got an answer which is close the right one, by mistake..
 
sorry I'm not a native english speaker so I forgot how to call a person who has phd, its just called a doctor in our country. by DR I meant doctor, your assumption was correct. :D

Okay; the second time you said "DR lecturer", and I thought that meant "lecturer in a course about DR", whatever that might be. In America we abbreviate "doctor" as "Dr.", so you were not far off. (In England they don't use the period.)

the precision appropriate in our course's setting can be less than 10^-12, actually an error probability of 10^-12 with a transmission rate of 10^9 its an error every 1000 seconds which is considered not so good. maybe 10^-15 it depends on the transmission rate. but definitely not 10^-84 as I calculated.

we never taught anything regarding the operation of a calculator, it is just considered to be obvious to operate, I study electronics engineering, we use calculator all the time, it's the end of the third year and I have never encountered such an issue.

One reason it surprises me that you would not be taught to be careful about accepting a calculator's results is that your course appears to emphasize avoiding errors (in a different sense)! I would expect precision to be something you would be very careful about, so you would be aware of rounding errors.

all the other students thought that because large upper value results in 0, they should enter smaller number and just write the result, becuase everyone thought that 0 is not acceptable in this course, because of the reasons I mentioned. during the exam the examiner told us we can put any large upper value in the integral that will give us precise answer, and than told us to just put 100 if we like. he wasn't insisting, it was obvious that he wasn't aware of this issue we talk about now. but I'm the only one who thought that 0 in this case must be the answer(because I got 10^-84), even for this course's precision setting, because I tried increasing of upper values and saw the numbers getting smaller. that was the diiference between me and the others, they just put 100, wrote the result and continued without thinking. thats what they told me.

That is unfortunately imprecise; but it fits my expectation that everyone must have been told something about 100. So you were really the only one who took him at his word, and "put any large upper value"! You should get extra credit for thinking, and so discovering this issue.

yes, just that our final answer was Q^2, if you do (10^-12)^2 you get 10^-24. but actually the numbers are starting to get smaller after like 35-40 in the upper value, so at 100 it is already getting smaller just not enough yet to see it in the power of 10. yes you can say the others got the right answer my mistake, I know it because of their explanation, which is wronge. I found that if I want to calcaulate the closest right answer I need to find ln(10^-99), which is -227.95, and this is limit of the power of e that the calculator can handle, therefore the equation (-x^2)/2=-227.95 will give us a limit of x=21.35. so I integrate from 7 to 21.35 and this is the closest to the right answer that I can get with this calculator. after 21.35 it start to decrease first by small not visible numbers than after about 35-40 we can see the decrease and by 3000 we already get 10^-84, and so on until 0.

I see you have been doing some further experimentation and thinking, which is great. I can't confirm that your conclusion about how to determine the calculator's limit is necessarily correct, but it makes some sense. It might be of interest to write to Casio and ask them if you are right; their designers would have more to say about it (I hope).

there is a constant multiplier of 1/(square root of 2 pai), by "normalized" I mean the standard deviation is 1 and the average is 0(in the normal distribution of probability).
I omitted the multiplier on purpose because it is not a factor in the integral, the issue still exists with or without it

Good -- this is what I meant when I said you might be "not saying everything carefully", but just saying what you felt was necessary.

thank you about manual comment , I will look it up also

Again, I hope you share all this good thinking with your instructor, and perhaps even have a good conversation about it in class.
 
It might be of interest to write to Casio and ask them if you are right; their designers would have more to say about it (I hope).
.
I already tried contacting them but they told me they don't deal with those calculators because it's not american.. I told them it doesn't matter which model, and now waiting for a reply..
 
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