Polynomial problem for homework

[juantheron]

If $$\mathbf{p(x) = 2009-2008x^{100}+2007x^{99}-2006x^{98}+.....+1909x}$$
Then Calculate $$\mathbf{p(2008)}$$

 

[stevenb]

If $$\mathbf{p(x) = 2009-2008x^{100}+2007x^{99}-2006x^{98}+.....+1909x}$$
Then Calculate $$\mathbf{p(2008)}$$

Homework Statement

Homework Equations

The Attempt at a Solution

Are you trying to do it exactly via algebra/arithmetic, or approximately with a computer program?

You can do this out in a straightforward way exactly if you use the summation (Sigma) notation. I've attached a scan of a sketched out way to approach the solution. Note that I did this very quickly and so I expect some mistakes there. However, you should be able to pick up on the method.

 

[juantheron]

Are you trying to do it exactly via algebra/arithmetic, or approximately with a computer program?

You can do this out in a straightforward way exactly if you use the summation (Sigma) notation. I've attached a scan of a sketched out way to approach the solution. Note that I did this very quickly and so I expect some mistakes there. However, you should be able to pick up on the method.

thanks stevenb

 

[stevenb]

thanks stevenb

You are welcome.

I hate to leave an answer that I know is wrong, even if my intent was to give the method more than the correct answer. Tonight I was bored, so thought I would take the time to work this out correctly, just for the record, in case anyone comes across this thread in the future.

I attached the correct way to work it out exactly. I know this is correct because I took step #1 and step #10 and fed them into Maxima to make sure they agree. Of course, the answer is hundreds of digits long once expanded out, but both answers agree to the last digit.

Note that one could work out a general formula in terms of x rather that the one value of x=2008, by following the exact same procedure.

 

[stevenb]

Note that one could work out a general formula in terms of x rather that the one value of x=2008, by following the exact same procedure.

And, making certain this horse has truly been beaten to death, I might as well post that too, since I worked it out in another fit of boredom. I should have done it that way the first time.

This works out to the following, which was verified with Maxima which expands it back to the original formula 2009-2008x^100+2007x^99 ... 1909x

$${{2009+5927x+3917x^2-2009x^{101}-2008x^{102}}\over{(x+1)^2}}$$

 

[Ray Vickson]

You are welcome.

I hate to leave an answer that I know is wrong, even if my intent was to give the method more than the correct answer. Tonight I was bored, so thought I would take the time to work this out correctly, just for the record, in case anyone comes across this thread in the future.

I attached the correct way to work it out exactly. I know this is correct because I took step #1 and step #10 and fed them into Maxima to make sure they agree. Of course, the answer is hundreds of digits long once expanded out, but both answers agree to the last digit.

Note that one could work out a general formula in terms of x rather that the one value of x=2008, by following the exact same procedure.

Just as a matter of interest: how do you post an attachment to this forum? I see no tabs or buttons or menu items that look like "attach" commands.

RGV

 

[berkeman]

Are you trying to do it exactly via algebra/arithmetic, or approximately with a computer program?

You can do this out in a straightforward way exactly if you use the summation (Sigma) notation. I've attached a scan of a sketched out way to approach the solution. Note that I did this very quickly and so I expect some mistakes there. However, you should be able to pick up on the method.

Please do not do the student's work for them. That is against the rules. Especially when the student showed no effort at all in working out the problem.

 

[berkeman]

Just as a matter of interest: how do you post an attachment to this forum? I see no tabs or buttons or menu items that look like "attach" commands.

RGV

When you are in the Advanced Reply (or New Topic) window, look next to the little smiley face pulldown menu for the paper clip "Attachments" pulldown menu. Clicking on that should get you to the Attachments dialog box.


EDIT -- since this thread is now locked, PM me if you have further questions about attachments.