calling math/chem geniuses
January 25, 2011 8:51 AM Subscribe
When doing mixed-operation arithmetic keeping significant figures, do you round after each operation- according to the rules for that particular step? Or do you keep everything in long-form calculator numbers, then round at the final answer?
This is for an intro Chemistry class, and while the book makes clear the different procedures for addition vs multiplication, it doesn't do any example problems featuring both- yet has problems like that in the homework problems. Best I can tell, you're supposed to round to the SF after each operation, but then there was this problem:
(0.0267 + .0019)(4.626)
---------------------------------
28.7794
=
(.0286)(4.626)
--------------------
28.7794
=
.1323036
-------------
28.7794
=.004597163 = 4.6 x10e-3 (use the .0019 for SF rounding?)
The answer listed in the back is 4.48 x10e-3.
This was the first problem in the series, and I don't think I realized you're supposed to round after each operation (if you actually are).
Keeping SF after the addition and multiplication gets me .1323/28.7794 - which is .00459704 or 4.6x10e-3 (4 SF).
It's driving me crazy! Could it be a mistake or just something I missed? I've been doing this problem over and over, admittedly math is not my strongest suit.
This is for an intro Chemistry class, and while the book makes clear the different procedures for addition vs multiplication, it doesn't do any example problems featuring both- yet has problems like that in the homework problems. Best I can tell, you're supposed to round to the SF after each operation, but then there was this problem:
(0.0267 + .0019)(4.626)
---------------------------------
28.7794
=
(.0286)(4.626)
--------------------
28.7794
=
.1323036
-------------
28.7794
=.004597163 = 4.6 x10e-3 (use the .0019 for SF rounding?)
The answer listed in the back is 4.48 x10e-3.
This was the first problem in the series, and I don't think I realized you're supposed to round after each operation (if you actually are).
Keeping SF after the addition and multiplication gets me .1323/28.7794 - which is .00459704 or 4.6x10e-3 (4 SF).
It's driving me crazy! Could it be a mistake or just something I missed? I've been doing this problem over and over, admittedly math is not my strongest suit.
Response by poster: or wait, would 4 SF in the final answer be 4.597x10e-3?
Either way, not 4.48.
posted by tremspeed at 9:00 AM on January 25, 2011
Either way, not 4.48.
posted by tremspeed at 9:00 AM on January 25, 2011
You need to maintain the correct number of significant figures at each step. So when you perform the multiplication in Step 2, you need to round to 3 significant figures (0.0286 has 3 significant figures, 4.626 has 4, so the rule for multiplication that the answer cannot contain any more than the least amount). After Step 3 when you divide, you wound need to round to 3 significant figures again.
posted by muddgirl at 9:02 AM on January 25, 2011
posted by muddgirl at 9:02 AM on January 25, 2011
But I still don't get their "right answer". Honestly, the back of the book is often wrong.
posted by muddgirl at 9:03 AM on January 25, 2011
posted by muddgirl at 9:03 AM on January 25, 2011
(the answer I get from what you wrote is 4.59 x 10e-3)
posted by muddgirl at 9:08 AM on January 25, 2011
posted by muddgirl at 9:08 AM on January 25, 2011
I once worked for a textbook publisher. We issued errata pages periodically with corrections to problems in a given book. Call your book's publisher, after giving Excel a shot at doing the problem for you.
posted by SMPA at 9:10 AM on January 25, 2011
posted by SMPA at 9:10 AM on January 25, 2011
Yeah, there is an error either in the problem or the answer. It should be 4.60E-3, I believe.
I was taught that first you do the calculation with all the digits. Do not round, do not truncate, just do the calculation. If you do this, say with a calculator, you get 0.004597163.
Then you start examining the problem. As muddgirl said, the 0.0286 with 3 sig figs wins out. So you can keep 3 sig figs in the final answer: 0.0460 or 4.60E-3.
posted by Fortran at 9:11 AM on January 25, 2011
I was taught that first you do the calculation with all the digits. Do not round, do not truncate, just do the calculation. If you do this, say with a calculator, you get 0.004597163.
Then you start examining the problem. As muddgirl said, the 0.0286 with 3 sig figs wins out. So you can keep 3 sig figs in the final answer: 0.0460 or 4.60E-3.
posted by Fortran at 9:11 AM on January 25, 2011
You need to maintain the correct number of significant figures at each step. So when you perform the multiplication in Step 2, you need to round to 3 significant figures (0.0286 has 3 significant figures, 4.626 has 4, so the rule for multiplication that the answer cannot contain any more than the least amount).
Why would you do this? It introduces unnecessary rounding errors.
posted by mr_roboto at 9:13 AM on January 25, 2011
Why would you do this? It introduces unnecessary rounding errors.
posted by mr_roboto at 9:13 AM on January 25, 2011
Whoops. That is 0.00460. I got the answer right in scientific notation, just not in decimal.
Oh, and that final zero is significant, of course, so you can't just say 4.6E-3 as you've lost a significant digit.
posted by Fortran at 9:13 AM on January 25, 2011
Oh, and that final zero is significant, of course, so you can't just say 4.6E-3 as you've lost a significant digit.
posted by Fortran at 9:13 AM on January 25, 2011
Yeah, I'm probably mis-remembering. Most internet sources say that you round off at the end. My apologies - I agree with 4.60 x 10e-3
posted by muddgirl at 9:24 AM on January 25, 2011
posted by muddgirl at 9:24 AM on January 25, 2011
Every time you round you make your answer less exact. As a general rule, you should round last. The only exception to this is if the instructions specifically tell you not to wait to round. But I've only seen that where the point of the problem was to learn about significant figures and not about the math.
posted by theichibun at 9:24 AM on January 25, 2011
posted by theichibun at 9:24 AM on January 25, 2011
Response by poster: ok, and i'm glad I'm not just crazy. thanks!
but if you are looking at an SF for the solution based on the original numbers, why not the .0019 - 2 significant figures?
posted by tremspeed at 9:31 AM on January 25, 2011
but if you are looking at an SF for the solution based on the original numbers, why not the .0019 - 2 significant figures?
posted by tremspeed at 9:31 AM on January 25, 2011
Best answer: "When using numbers in calculations, it's a good idea to keep one digit beyond the significant digits. Once the final answer is calculated, it may be expressed with the correct number of significant digits." (via)
Good practice is to maintain an extra significant digit throughout your calculations, and round to the appropriate number of significant digits at the end of the process. However, you need to be conscious of the significant digits that would result from each step. Assuming the final answer should have 2 significant digits in this case is incorrect, because 0.0019 is used in an addition step, not a multiplication step. The rule for addition is that your sum should have as many decimal places as the addend with the least number of decimal places.
Here is what each individual step would look like. Think of each step as a totally separate problem at this point. I will use the answer from the previous step, carrying through one extra digit than there are significant figures. I have put the appropriate significant figures from each individual step in italics (for now, ignoring any limitations on sig figs from previous steps, carrying through extra digits, etc.)
0.0267+ 0.0019 = 0.02860
0.02860*4.626 = 0.13230
0.13230/28.7794 = 0.0045970
Now look at the answers from each step, and look for the answer that most limits your significant figures. As it turns out, the first step has an answer with three significant figures, and the others have more. Therefore, your final answer should have three significant figures.
tl;dr Round at the end, but pay attention to how many sig figs each individual step would give you. Use the smallest of those.
posted by pemberkins at 9:34 AM on January 25, 2011
Good practice is to maintain an extra significant digit throughout your calculations, and round to the appropriate number of significant digits at the end of the process. However, you need to be conscious of the significant digits that would result from each step. Assuming the final answer should have 2 significant digits in this case is incorrect, because 0.0019 is used in an addition step, not a multiplication step. The rule for addition is that your sum should have as many decimal places as the addend with the least number of decimal places.
Here is what each individual step would look like. Think of each step as a totally separate problem at this point. I will use the answer from the previous step, carrying through one extra digit than there are significant figures. I have put the appropriate significant figures from each individual step in italics (for now, ignoring any limitations on sig figs from previous steps, carrying through extra digits, etc.)
0.0267+ 0.0019 = 0.02860
0.02860*4.626 = 0.13230
0.13230/28.7794 = 0.0045970
Now look at the answers from each step, and look for the answer that most limits your significant figures. As it turns out, the first step has an answer with three significant figures, and the others have more. Therefore, your final answer should have three significant figures.
tl;dr Round at the end, but pay attention to how many sig figs each individual step would give you. Use the smallest of those.
posted by pemberkins at 9:34 AM on January 25, 2011
Sorry, I fail at formatting. The last step should read
0.13230/28.7794 = 0.00459704
But, the basic idea above is correct. Just pay attention to which numbers are used for addition and which are used for multiplication when you are determining your final sig figs.
I still don't know why the book says 4.48, unless I've done some arithmetic wrong (possible!).
posted by pemberkins at 9:37 AM on January 25, 2011
0.13230/28.7794 = 0.00459704
But, the basic idea above is correct. Just pay attention to which numbers are used for addition and which are used for multiplication when you are determining your final sig figs.
I still don't know why the book says 4.48, unless I've done some arithmetic wrong (possible!).
posted by pemberkins at 9:37 AM on January 25, 2011
Response by poster: great. thanks all. I'll have to try peberkins method on the rest of the HW problems, but I feel alot better knowing that it's likely the 'right answer' was not.
posted by tremspeed at 9:47 AM on January 25, 2011
posted by tremspeed at 9:47 AM on January 25, 2011
I still don't know why the book says 4.48, unless I've done some arithmetic wrong (possible!).
The book says 4.48 because the grad student assigned to proof/answer this chapter was distracted or bored or some such. I imagine in a paper somewhere there is a 4.60 that should be 4.48...
posted by Fortran at 9:47 AM on January 25, 2011
The book says 4.48 because the grad student assigned to proof/answer this chapter was distracted or bored or some such. I imagine in a paper somewhere there is a 4.60 that should be 4.48...
posted by Fortran at 9:47 AM on January 25, 2011
This looks like a pretty good summary of rounding rules for chemistry. In my biotech QC experience, people just simplify it to "keep all digits in play, then round at the end, and 5 gets rounded up."
posted by Quietgal at 9:48 AM on January 25, 2011
posted by Quietgal at 9:48 AM on January 25, 2011
Good practice is to maintain an extra significant digit throughout your calculations, and round to the appropriate number of significant digits at the end of the process. However, you need to be conscious of the significant digits that would result from each step.
You need to understand your limits here. These rules work fine under normal circumstances, only a few calculation steps, but can bite you if you are doing long computation. Optimization, non-linear fits, finite element analysis, quantum chemical energy minimization calculations can easily produce garbage if you follow the rules above.
Every cycle through one of these iterative calculation reduces the significance of your result because numerical digitization errors magnify through each iteration. Errors arise because the base-2 arithmetic that computers use does not map well onto the base-10 arithmetic we normally use. This is a form of Quantization Error, and typically affects the last stored digit, the least significant digit of a digital number. With successive calculation, these errors magnify and can, over the course of several million calculations, overwhelm the result. This is particularly true if you use a short numerical form, a real/float/word (8 bit) representation for example, instead of a double word (16 bit). Even doubles can have issues with large problems.
As a result, I'm always cautious, perhaps excessively so with intermediate rounding. Even with something like a digital calculator or Excel*, you want to keep as much of the intermediate representations as you can, as you are fighting not one source of uncertainty but two: the (implied) uncertainty in the terminal digit of your inputs as well as the digitization error in the terminal value of the binary representation of your number. If these two overlap, or are even with a couple of orders of magnitude of each other, even a small number of calculations can cause problems.
So, my rules:
1. Keep as much of the intermediate representations as you can.
2. Use a digital form that's appropriate, at least two digits longer than your least significant digit. For more than 6 significant digits, you must always use a double word representation, for example. You can run into issues with hand-held calculators here.
3. If doing iterative calculations or calculations where small differences between large numbers are important, ANOVA for example, you need to be even more careful, often requiring double doubles or other big-number representations*.
*Excel does small differences between big number notably poorly and in a non-standard way. Never completely trust the stat functions in Excel.
posted by bonehead at 10:20 AM on January 25, 2011 [1 favorite]
You need to understand your limits here. These rules work fine under normal circumstances, only a few calculation steps, but can bite you if you are doing long computation. Optimization, non-linear fits, finite element analysis, quantum chemical energy minimization calculations can easily produce garbage if you follow the rules above.
Every cycle through one of these iterative calculation reduces the significance of your result because numerical digitization errors magnify through each iteration. Errors arise because the base-2 arithmetic that computers use does not map well onto the base-10 arithmetic we normally use. This is a form of Quantization Error, and typically affects the last stored digit, the least significant digit of a digital number. With successive calculation, these errors magnify and can, over the course of several million calculations, overwhelm the result. This is particularly true if you use a short numerical form, a real/float/word (8 bit) representation for example, instead of a double word (16 bit). Even doubles can have issues with large problems.
As a result, I'm always cautious, perhaps excessively so with intermediate rounding. Even with something like a digital calculator or Excel*, you want to keep as much of the intermediate representations as you can, as you are fighting not one source of uncertainty but two: the (implied) uncertainty in the terminal digit of your inputs as well as the digitization error in the terminal value of the binary representation of your number. If these two overlap, or are even with a couple of orders of magnitude of each other, even a small number of calculations can cause problems.
So, my rules:
1. Keep as much of the intermediate representations as you can.
2. Use a digital form that's appropriate, at least two digits longer than your least significant digit. For more than 6 significant digits, you must always use a double word representation, for example. You can run into issues with hand-held calculators here.
3. If doing iterative calculations or calculations where small differences between large numbers are important, ANOVA for example, you need to be even more careful, often requiring double doubles or other big-number representations*.
*Excel does small differences between big number notably poorly and in a non-standard way. Never completely trust the stat functions in Excel.
posted by bonehead at 10:20 AM on January 25, 2011 [1 favorite]
You need to understand your limits here. These rules work fine under normal circumstances, only a few calculation steps, but can bite you if you are doing long computation. Optimization, non-linear fits, finite element analysis, quantum chemical energy minimization calculations can easily produce garbage if you follow the rules above.
You are, of course, correct. I ought to have said good practice for the purposes of Gen Chem homework (where I feel it is both tedious and unnecessary to maintain many extraneous digits, particularly if you are asked to demonstrate any of the math by hand). Longer and more complex computations will require more care.
posted by pemberkins at 11:04 AM on January 25, 2011
You are, of course, correct. I ought to have said good practice for the purposes of Gen Chem homework (where I feel it is both tedious and unnecessary to maintain many extraneous digits, particularly if you are asked to demonstrate any of the math by hand). Longer and more complex computations will require more care.
posted by pemberkins at 11:04 AM on January 25, 2011
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posted by madcaptenor at 9:00 AM on January 25, 2011 [1 favorite]