Question

HELP PLEASE

Answer 1

I dont understand why I got this wrong.

Answer 2

Here is the picture, showing my work.

Answer 3

@518nad @zepdrix @mathmate @SolomonZelman

Answer 4

yes, this is correct!

Answer 5

But my webassign isn't taking my answer. Am i inputting it incorrectly?

Answer 6

I really don't know why the website isn't taking it, but
\(\color{black}{\displaystyle \cos(x)=1+\frac{1}{2}x^2-\frac{1}{24}x^4+\frac{1}{720}x^6 }\)
is correct.

Answer 7

Hmm... Mkay, Ill keep looking. Haha
Thanks for verifying my answer though! :)

Answer 8

maybe you entered \(\color{black}{\displaystyle \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^kx^{2k}}{(2k)!} }\) ?

Answer 9

If that is what you entered, then you surely and of course got it wrong (because this is not 6 terms, it's all the terms)

Answer 10

Well it's asking for each term. T_1, T_2, T_3, etc.
So I put down each term

Answer 11

Up to 6th degree, yes you are good.

Answer 12

Oh, T's are Taylor polynomials not terms.

Answer 13

Well, if you got it incorrectly for entering correct terms, that is the only option I see.

Answer 14

So, enter tailor polynomial (TP) of degree 0, then TP of degree 1, then TP of degree 2, and so on till (and incl) 6th degree TP.

Answer 15

Im confused. Sorry, what do you mean by taylor polynomial? D:

Answer 16

What is the Taylor Polynomial for cos(x) of degree 0?

Answer 17

Enter that for T0.

Answer 18

Then, the Taylor Polynomial for cos(x) of degree 1.
Enter this for T1.

Answer 19

And for each \(0\le i \le 6\), enter \(i\)th degree Taylor Polynomial, into the corresponding \(T_i\) cell.

Answer 20

Taylor Polynomial?
So like...
T_0 = 1
T_1 = 0
T_2 = -1/2x^2

.... etc?

Answer 21

Cos that's what I had entered

Answer 22

What is the Taylor Polynomial of 0th degree for cos(x)?

Answer 23

I dont know... D:

Answer 24

Well, Taylor Polynomial of degree 0, would just be
\(\color{black}{\displaystyle \cos(x)=1=\sum_{k=0}^{0}\frac{(-1)^k\cdot x^{2k}}{(2k)!} }\)

Answer 25

This is for \(T_0\).

Answer 26

Where did you get that summation from?

Answer 27

That is what you have obtained earlier.
(Also, I've seen a few times before)

Answer 28

Agree with this result for \(T_0\)?

Answer 29

Yes

Answer 30

Now, any \(n\)th degree polynomial is going to be:
\(\color{black}{\displaystyle \cos(x)=\sum_{k=0}^{\color{red}{n}}\frac{(-1)^k\cdot x^{2k}}{(2k)!} }\)

Answer 31

Haha no worries. You meant k I am assuming?

Answer 32

Well, neth degree polynomial approximation ...
\(\color{black}{\displaystyle \cos(x)=1 }\) is the 0th degree approximation.
you don't have any non-zero \(x^1\) terms in the expansion,
so, (anybody correct me if I am wrong, but) for 1st degree
polynomial you would still have \(\color{black}{\displaystyle \cos(x)=1 }\).

Answer 33

You found then that the next term that follows is: 1, 0, \(\color{red}{(x^2/2)}\),
so your quadratic (2nd degree) polynomial would be \(\color{black}{\displaystyle \cos(x)=1+\frac{1}{x^2} }\).

Answer 34

and so forth, just keep including the terms that you found.

Answer 35

For example (a pretty abstract one, but)
\(\color{black}{\displaystyle f(x)=a_1+a_2x^2+a_3x^3+a_5x^5+a_8x^8 }\),

then,
\(\color{black}{\displaystyle T_0=a_1 }\)
\(\color{black}{\displaystyle T_1=a_1 }\) (no non-zero linear terms)
\(\color{black}{\displaystyle T_2=a_1+a_2x^2 }\)
\(\color{black}{\displaystyle T_3=a_1+a_2x^2+a_3x^3 }\)
\(\color{black}{\displaystyle T_4=a_1+a_2x^2+a_3x^3 }\) (no \(x^4\) terms, so it is the same as \(T_3\))
\(\color{black}{\displaystyle T_5=a_1+a_2x^2+a_3x^3+a_5x^5 }\)
\(\color{black}{\displaystyle T_6=a_1+a_2x^2+a_3x^3+a_5x^5 }\) (no \(x^6\) terms, so it is the same as \(T_5\))
\(\color{black}{\displaystyle T_7=a_1+a_2x^2+a_3x^3+a_5x^5 }\) (no \(x^7\) terms, so it is the same as \(T_7\))
\(\color{black}{\displaystyle T_8=a_1+a_2x^2+a_3x^3+a_5x^5+a_8x^8 }\)

Answer 36

This way you have to "include" the terms in your situation ...
(Except that you have a different case)

Answer 37

Oh, I understand what you are saying. I got it right. Thanks!

Answer 38

yw