- #1

nhrock3

- 415

- 0

why we do tailor around 1 and not 0

?

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter nhrock3
- Start date

- #1

nhrock3

- 415

- 0

why we do tailor around 1 and not 0

?

Last edited:

- #2

tiny-tim

Science Advisor

Homework Helper

- 25,838

- 255

You can "Taylor" around

But it

(and I expect they've chosen z = 1 because it's easiest to calculate the derivatives there! )

- #3

nhrock3

- 415

- 0

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

Mentor

- 36,332

- 8,293

Why do you think there is a problem when x = y? If x = y, log(x/y) = log(1) = 0.

- #5

nhrock3

- 415

- 0

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

Mentor

- 36,332

- 8,293

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

- 415

- 0

"limitations in computing precision"

how does this limitation get solves by this method?

- #8

Mark44

Mentor

- 36,332

- 8,293

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

number like 1.00000000000125, where one part is very large

- #9

nhrock3

- 415

- 0

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

Mentor

- 36,332

- 8,293

Share:

- Replies
- 4

- Views
- 600

- Last Post

- Replies
- 4

- Views
- 485

- Replies
- 10

- Views
- 326

- Last Post

- Replies
- 3

- Views
- 831

- Replies
- 4

- Views
- 319

- Replies
- 6

- Views
- 977

- Replies
- 1

- Views
- 367

- Replies
- 3

- Views
- 1K

- Replies
- 4

- Views
- 373

- Replies
- 3

- Views
- 482