Newton Raphson Method - Confused

tcawley29

Hi guys,
I have been trying to revise newton raphson for my maths exam in college and I'm having a problem.
I can calculate the values shown in x1, x2, and x3 fine.
However I get confused on x4.
On my calculator for x4 under column f(x) I get 3.7521856 E -10.
The picture here shows the answer to be 2.33 E -10.
Please tell me where I'm going wrong.
Thanks,
Thomas

entangled

Hey,
As far as I can tell, you're right.
Possibly a typo in the book or something like that.

tcawley29

entangled wrote: »

Hey,
As far as I can tell, you're right.
Possibly a typo in the book or something like that.

Thats what I thought too. But then I looked at another example and my final answer is always much bigger than the lecturer's final answer.
Thanks though

tcawley29

Here is another example of where it all goes wrong in the final step.
The page says the answer is 3.2918 E -6.
I get 1.7909648 E -6.

Ciaran

tcawley29 wrote: »

Here is another example of where it all goes wrong in the final step.
The page says the answer is 3.2918 E -6.
I get 1.7909648 E -6.

Again the book's wrong. Is it just these two or are there more errors in the book?

tcawley29

Ciaran wrote: »

Again the book's wrong. Is it just these two or are there more errors in the book?

Couldn't tell ya. The lecturer just comes into class and gives us handouts like this once in a while. Its quite possible that he made these questions up himself (Usually does).

On the bright side, thanks everyone for letting me know I was doing it right. Nice easy question for the exam next week I'm guessing

MathsManiac

The scanned sheets are not wrong.

The issue is related to rounding errors. If you retain the full available precision in the calculations, by using the memory button on your calculator, then you will get the answers on the sheet, in both cases. If you start each calulation afresh, retaining only the level of precision displayed in the table, then your numbers will start to deviate gradually from those shown, which have been calculated to a greater level of precision than they are displayed. (They look to me as though they have been pasted from an Excel sheet, by the way, and I have replicated them in Excel and got the same answers.)

For example, the last number in the first row of the first sheet shows 0.089744. However, this is really 0.089743589743589743589... (a recurring decimal). These small errors gradually become detectable as the iterative process continues.

You'll still get the same value for the root, of course.

tcawley29

MathsManiac wrote: »

The scanned sheets are not wrong.

The issue is related to rounding errors. If you retain the full available precision in the calculations, by using the memory button on your calculator, then you will get the answers on the sheet, in both cases. If you start each calulation afresh, retaining only the level of precision displayed in the table, then your numbers will start to deviate gradually from those shown, which have been calculated to a greater level of precision than they are displayed. (They look to me as though they have been pasted from an Excel sheet, by the way, and I have replicated them in Excel and got the same answers.)

For example, the last number in the first row of the first sheet shows 0.089744. However, this is really 0.089743589743589743589... (a recurring decimal). These small errors gradually become detectable as the iterative process continues.

You'll still get the same value for the root, of course.

Just tried there and you're 100% correct. Didn't get the exact same answer but I was a hell of a lot closer to it than what I got before.
Thanks MathsManiac.
Now I have to learn integration with area/volume before wednesday and I'm golden

Ciaran

MathsManiac wrote: »

The scanned sheets are not wrong.

The issue is related to rounding errors. If you retain the full available precision in the calculations, by using the memory button on your calculator, then you will get the answers on the sheet, in both cases. If you start each calulation afresh, retaining only the level of precision displayed in the table, then your numbers will start to deviate gradually from those shown, which have been calculated to a greater level of precision than they are displayed. (They look to me as though they have been pasted from an Excel sheet, by the way, and I have replicated them in Excel and got the same answers.)

For example, the last number in the first row of the first sheet shows 0.089744. However, this is really 0.089743589743589743589... (a recurring decimal). These small errors gradually become detectable as the iterative process continues.

You'll still get the same value for the root, of course.

Ah, very nice. I wondered if there was something systematic going on as it was only the value of the function for the last iteration that was different in both cases; the derivative was the same.