Maclaurin polynomial problem

[mikhail_ccp]

I have following problem:
Find a fourth-degree polynomial which has the same first three terms with the approximation of f(x) = (1 - x)ln(1 - x) at x = 0.

There is my solution. My math professor told me that this solution is incorrect. Where did I do a mistake? It seems like my solution is correct.

 

[topsquark]

Hmmm... [math]\dfrac{d}{dx} \dfrac{1}{x - 1} = \dfrac{d}{dx} (x - 1)^{-1}[/math]. What is the sign on that again?

-Dan

 

[Steven G]

Please please do not write f⁽²⁾(0) = [math]\dfrac{1}{1-x}=1[/math]
Maybe f²(0) = 1 but f²(0) [math]\neq \dfrac{1}{1-x}\ and \ \dfrac {1}{1-x} \neq 1[/math]. Therefore we can not conclude that f²(0) = 1

Of course I have the same complaint for all the equations you wrote.

Please write math correctly, especially at your level. Thanks

 

[mikhail_ccp]

Please please do not write f⁽²⁾(0) = [math]\dfrac{1}{1-x}=1[/math]
Maybe f²(0) = 1 but f²(0) [math]\neq \dfrac{1}{1-x}\ and \ \dfrac {1}{1-x} \neq 1[/math]. Therefore we can not conclude that f²(0) = 1

Of course I have the same complaint for all the equations you wrote.

Please write math correctly, especially at your level. Thanks

Thank you for you answer.
But the problem is not in the solution. The professor did not see my solution. He saw only my result.
The question is - my answer correct or not?

 

[mikhail_ccp]

Hmmm... [math]\dfrac{d}{dx} \dfrac{1}{x - 1} = \dfrac{d}{dx} (x - 1)^{-1}[/math]. What is the sign on that again?

-Dan

Why did you take derivative of (x - 1)? The expression is (1-x)ln(1-x).

 

[topsquark]

Why did you take derivative of (x - 1)? The expression is (1-x)ln(1-x).

Your expression for [math]f^{(2)}(x)[/math] is correct. But when you took the derivative of it to get [math]f^{(3)}(x)[/math] you were off by a minus sign. I was trying to tell you to double check that calculation.

-Dan

 

[mikhail_ccp]

Your expression for [math]f^{(2)}(x)[/math] is correct. But when you took the derivative of it to get [math]f^{(3)}(x)[/math] you were off by a minus sign. I was trying to tell you to double check that calculation.

-Dan

There is my solution to find third derivative. Could you check, am I right or not?

 

[topsquark]

(Ahem.) I screwed it up. Yes, you are correct.

-Dan

 

[Steven G]

Thank you for you answer.
But the problem is not in the solution. The professor did not see my solution. He saw only my result.
The question is - my answer correct or not?

(Ahem.) I screwed it up. Yes, you are correct.

-Dan

Sorry, but it is corner time.

 

[Steven G]

Thank you for you answer.
But the problem is not in the solution. The professor did not see my solution. He saw only my result.
The question is - my answer correct or not?

It doesn't matter if you submitted the work to your professor or not. It is not valid. All equal signs must be equal. This is not debatable.

As far as the final answer you got I do not see anything wrong.

 

[mikhail_ccp]

It doesn't matter if you submitted the work to your professor or not. It is not valid. All equal signs must be equal. This is not debatable.

As far as the final answer you got I do not see anything wrong.

Thank you. Of course, I agree with you.

 

[mikhail_ccp]

(Ahem.) I screwed it up. Yes, you are correct.

-Dan

Thank you, Dan.

 

[Cubist]

Find a fourth-degree polynomial which has the same first three terms with the approximation of f(x) = (1 - x)ln(1 - x) at x = 0.

I find the wording of the part I've made bold (in the quote) very confusing. Do they mean that P(x) can be different to the Maclaurin series that you've found, so long as the first three terms match?

 

[mikhail_ccp]

This is the problem. I cannot understand where I did a mistake. My professor told me that the result is incorrect, but he cannot explain why.
He simply sent me the answer, he insists, which is correct: -x + x^2/2 + x^3/6 - x^4/3. He did not explain how he received this answer.
I cannot understand what he needs. Maybe someone has the experience with such problems in this forum, but I did not face with such problems in textbook and in the Internet.

 

[Steven G]

We all think that your answer is correct. In fact it is the correct Maclaurin polynomial.

I was thinking exactly what Cubist was thinking. Do they mean that P(x) can be different to the Maclaurin series that you've found, so long as the first three terms match? That is some strange wording!

 

[Steven G]

There are two cases. Case 1, your professor made a careless mistake and you are correct. Case 2, there is a way of doing this problem with that strange wording to get your professor's answer that no one here (yet) understands. So I would not worry about this at all.

If your professor is correct I am sure that someone will come by and explain this now that the answer is given.