gaussian elimination method
- Thread starter PG1995
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The problem is that you aren't treating y the same way that you are treating x. You have three equations and you are trying to find the pair (x,y) that satisfy all of the equations simultaneously.
So you have
y = -6x + 1
y = -2x + 2
When you add these, you get:
2y = -8x + 3
or
y = -4x + 1.5
So you have
y = -6x + 1
y = -2x + 2
When you add these, you get:
2y = -8x + 3
or
y = -4x + 1.5
Thank you, WBahn.
I had thought I had found an answer to Q2 but that wasn't completely true. Anyway, at least now I have general idea what it's all about.
I was reading this Wikipedia article which indirectly provides answer to Q2. I'm only stuck at understanding what exactly four-digit arithmetic means. Could you please help me with this? Thank you.
Reference(s):
http://en.wikipedia.org/wiki/Numerical_stability
http://en.wikipedia.org/wiki/Significant_figures
Regards
PG
I had thought I had found an answer to Q2 but that wasn't completely true. Anyway, at least now I have general idea what it's all about.
I was reading this Wikipedia article which indirectly provides answer to Q2. I'm only stuck at understanding what exactly four-digit arithmetic means. Could you please help me with this? Thank you.
Reference(s):
http://en.wikipedia.org/wiki/Numerical_stability
http://en.wikipedia.org/wiki/Significant_figures
Regards
PG
Attachments
They are just using four digits as an example.
The point is that when you are working with a finite precision (number of significant figures), situations can arise in which you lose all of your sig figs due to round off errors.
Consider subtracting x=234.6497 from y=234.4503. A common rule of thumb is to report values to 3 sig figs, and so most people carry 4 sig figs when working the problem. If you applied that here, you would If you have x=234.6 and y=234.5 and would get (x-y)=0.1 and report that as 0.1000. But if you had carried 7 sig figs in the work, you would have gotten (x-y)=0.194 and reported that as 0.1940. That's nearly a 50% error in the result.
While computers carry more sig figs (an IEEE double precision floating point value has about 15 sig figs), we tend to use them to crunch a lot more numbers an accumulated round off errors can rob us of the precision we think we are working with pretty easily. And anytime you are adding/subtracting a large number and a small number, you tend to lose information. Sometimes that information is insignificant, but sometimes all of the important information is contained in that small difference.
The point is that when you are working with a finite precision (number of significant figures), situations can arise in which you lose all of your sig figs due to round off errors.
Consider subtracting x=234.6497 from y=234.4503. A common rule of thumb is to report values to 3 sig figs, and so most people carry 4 sig figs when working the problem. If you applied that here, you would If you have x=234.6 and y=234.5 and would get (x-y)=0.1 and report that as 0.1000. But if you had carried 7 sig figs in the work, you would have gotten (x-y)=0.194 and reported that as 0.1940. That's nearly a 50% error in the result.
While computers carry more sig figs (an IEEE double precision floating point value has about 15 sig figs), we tend to use them to crunch a lot more numbers an accumulated round off errors can rob us of the precision we think we are working with pretty easily. And anytime you are adding/subtracting a large number and a small number, you tend to lose information. Sometimes that information is insignificant, but sometimes all of the important information is contained in that small difference.
Thanks a lot, WBahn.
Though I still have some queries to make, your last post was very helpful. Please have a look on the attachment and kindly help me with the query there.
I understand that when several calculations are required to reach a final solution, then a round-off error from each stage in calculation process gets propagated to the next stage which increases the overall round-off error, and final round-off error at the end of a calculation is called an accumulated error. Could you please provide me some simple example to see what really happens as you have effectively shown me in your last post that how reducing number of significant figures can cause error? Thank you.
Regards
PG
Though I still have some queries to make, your last post was very helpful. Please have a look on the attachment and kindly help me with the query there.
I understand that when several calculations are required to reach a final solution, then a round-off error from each stage in calculation process gets propagated to the next stage which increases the overall round-off error, and final round-off error at the end of a calculation is called an accumulated error. Could you please provide me some simple example to see what really happens as you have effectively shown me in your last post that how reducing number of significant figures can cause error? Thank you.
Regards
PG
Attachments
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