- #1
nhrock3
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why we do tailor around 1 and not 0
?
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Why Tailoring Around 1 Makes More Sense
- Thread starter nhrock3
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Homework Help Overview
The discussion revolves around the use of Taylor series in the context of logarithmic functions, particularly focusing on the implications of tailoring around the point 1 instead of 0. Participants explore the significance of this choice in relation to loss of significance when dealing with values of x and y that are close to each other.
Discussion Character
- Exploratory, Assumption checking, Conceptual clarification
Approaches and Questions Raised
- Participants question why Taylor series are preferred around 1 rather than 0, and how this choice affects accuracy in calculations involving logarithmic differences. There is also inquiry into the nature of the loss of significance problem when x and y are nearly equal, and how the Taylor series might address this issue.
Discussion Status
Some participants have provided insights into the limitations of numerical precision and how approximating logarithmic expressions can mitigate issues arising from small differences in values. However, there is no explicit consensus on the effectiveness of the Taylor series in resolving the loss of significance problem.
Contextual Notes
Participants are working under the constraints of a homework problem that involves logarithmic functions and numerical precision, with specific attention to the behavior of these functions when inputs are close in value.
why we do tailor around 1 and not 0
?
- #2
tiny-tim
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Hi nhrock3! 
You can "Taylor" around any value where the function is sufficiently continuous and differentiable.
But it isn't at z = 0, so you can't "Taylor" there!
(and I expect they've chosen z = 1 because it's easiest to calculate the derivatives there!
)
You can "Taylor" around any value where the function is sufficiently continuous and differentiable.
But it isn't at z = 0, so you can't "Taylor" there!
(and I expect they've chosen z = 1 because it's easiest to calculate the derivatives there!
)- #3
nhrock3
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so when x=y we have a problem
how transforming it into a taylor series solves this accuracy problem?
how transforming it into a taylor series solves this accuracy problem?
- #4
Mark44
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Why do you think there is a problem when x = y? If x = y, log(x/y) = log(1) = 0.
- #5
nhrock3
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the question states how to get over loss of significance problem in here
f=ln x -ln y
x,y>0
and the solution says that we have a problem when x and y are close to each other
how transforming it into a taylor series solves this accuracy problem?
f=ln x -ln y
x,y>0
and the solution says that we have a problem when x and y are close to each other
how transforming it into a taylor series solves this accuracy problem?
- #6
Mark44
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I think that the idea here is that if x is close to y, then x/y is close to 1. You could have a situation where the actual value of x/y was 1.000000000001253, but because of limitations in computing precision, you might lose the part after the zeros.
By using a Taylor series to approximate log(x) - log(y) = log(z) [itex]\approx[/itex] z - 1, the part after the zeros above is now significant.
By using a Taylor series to approximate log(x) - log(y) = log(z) [itex]\approx[/itex] z - 1, the part after the zeros above is now significant.
- #7
nhrock3
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so what if " x/y is close to 1"
"limitations in computing precision"
how does this limitation get solves by this method?
"limitations in computing precision"
how does this limitation get solves by this method?
- #8
Mark44
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Try it with a calculator. Enter .00000000000125, and then add 1. Many calculators will display an answer of 1 since they aren't able to maintain enough digits of precision to display the actual answer.
So if you have two numbers that are very different in relative size, (e.g. 1 vs. .00000000000125), adding them causes the loss of digits. If you can strip off the 1, though, there's no problem in storing or computing with the part to the right of all the zeros. In computers, floating point numbers are stored in a way that is similar to scientific notation. Instead of being stored as .00000000000125, it would be stored something like 1.25 X 10-12. Not exactly like that, since the base is 2, not 10, and there are some other differences.
number like 1.00000000000125, where one part is very large
So if you have two numbers that are very different in relative size, (e.g. 1 vs. .00000000000125), adding them causes the loss of digits. If you can strip off the 1, though, there's no problem in storing or computing with the part to the right of all the zeros. In computers, floating point numbers are stored in a way that is similar to scientific notation. Instead of being stored as .00000000000125, it would be stored something like 1.25 X 10-12. Not exactly like that, since the base is 2, not 10, and there are some other differences.
number like 1.00000000000125, where one part is very large
- #9
nhrock3
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the question states how to get over loss of significance problem in here
f=ln x -ln y?
f=ln x -ln y=ln z
the tailor series for ln z is:
[tex]\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(Z-1)}{k}[/tex]
but how this expression equals z-1 ?
f=ln x -ln y?
f=ln x -ln y=ln z
the tailor series for ln z is:
[tex]\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(Z-1)}{k}[/tex]
but how this expression equals z-1 ?
- #10
Mark44
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The summation doesn't "equal" z - 1. It's "approximately equal" to z - 1. What they have done is discard all terms of degree 2 or larger in the series.
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