# Find the MacLaurin polinomial of degree 7 for F(x)

• Nov 26th 2013, 07:36 PM
dokrbb
Find the MacLaurin polinomial of degree 7 for F(x)
The problem is as follows - Find the MacLaurin polynomial of degree 7 for $\displaystyle F(x) = \int_0^{x} sin(2t^{2})dt$

and use this polynomial to estimate the value of $\displaystyle \int_0^{0,73} sin(2x^{2})dx$

the polynomial I got is $\displaystyle \frac{2x^{3}}{3} - \frac{4x^7}{21}$ and apparently it is correct, but the estimate gives me 0.238301997 and it is not correct,

do I skip a step in solving this equation or what?

thanks for help,
dokrbb
• Nov 26th 2013, 08:55 PM
romsek
Re: Find the MacLaurin polinomial of degree 7 for F(x)
$\displaystyle \sin (x)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1} x^{2 n-1}}{(2 n-1)!}$

$\displaystyle \sin \left(2 x^2\right)=\sin \left(\left(\sqrt{2} x\right)^2\right)$

$\displaystyle \sin \left(2 x^2\right)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \left(\sqrt{2} x\right)^{2 (2 n-1)}}{(2 n-1)!}$

$\displaystyle \int \sin \left(2 x^2\right) \, dx$ = $\displaystyle \int \left(\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \left(\sqrt{2} x\right)^{2 (2 n-1)}}{(2 n-1)!}\right) \, dx$ = $\displaystyle \sum _{n=1}^{\infty } \frac{(-1)^n \left(2^{2 n-1} x^{4 n-1}\right)}{(4 n-1) (2 n-1)!}$

The first 2 terms of this are
$\displaystyle \left\{-\frac{2 x^3}{3},\frac{4 x^7}{21}\right\}$

so you missed a negative sign. Not bad!
• Nov 27th 2013, 06:00 AM
dokrbb
Re: Find the MacLaurin polinomial of degree 7 for F(x)
Quote:

Originally Posted by romsek
$\displaystyle \sin (x)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1} x^{2 n-1}}{(2 n-1)!}$

$\displaystyle \sin \left(2 x^2\right)=\sin \left(\left(\sqrt{2} x\right)^2\right)$

$\displaystyle \sin \left(2 x^2\right)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \left(\sqrt{2} x\right)^{2 (2 n-1)}}{(2 n-1)!}$

$\displaystyle \int \sin \left(2 x^2\right) \, dx$ = $\displaystyle \int \left(\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \left(\sqrt{2} x\right)^{2 (2 n-1)}}{(2 n-1)!}\right) \, dx$ = $\displaystyle \sum _{n=1}^{\infty } \frac{(-1)^n \left(2^{2 n-1} x^{4 n-1}\right)}{(4 n-1) (2 n-1)!}$

The first 2 terms of this are
$\displaystyle \left\{-\frac{2 x^3}{3},\frac{4 x^7}{21}\right\}$

so you missed a negative sign. Not bad!

thanks, but in this case, why my first part of the answer is considered correct (with this missed negative sign) and the other not(the actual result I got using this "correct" polynomial)...
• Nov 27th 2013, 09:13 AM
romsek
Re: Find the MacLaurin polinomial of degree 7 for F(x)
I'd have to see the whole problem to answer this. I'm not sure what you are saying is incorrect.
• Nov 27th 2013, 09:35 AM
dokrbb
Re: Find the MacLaurin polinomial of degree 7 for F(x)
Quote:

Originally Posted by romsek
I'd have to see the whole problem to answer this. I'm not sure what you are saying is incorrect.

The problem is as follows - Find the MacLaurin polynomial of degree 7 for $\displaystyle F(x) = \int_0^{x} sin(2t^{2})dt$

and use this polynomial to estimate the value of $\displaystyle \int_0^{0,73} sin(2x^{2})dx$

the polynomial I got is $\displaystyle \frac{2x^{3}}{3} - \frac{4x^7}{21}$ and when I use it to calculate $\displaystyle \int_0^{0,73} \frac{2x^{3}}{3} - \frac{4x^7}{21}dx$

I get something like 0.238301997

basically I need to answer to 2 questions: #1 - what is the polynomial and I have this $\displaystyle \frac{2x^{3}}{3} - \frac{4x^7}{21}$ which is correct according to the answers; and #1 - use this polynomial to estimate the value of $\displaystyle \int_0^{0,73} sin(2x^{2})dx$ which I use by calculating $\displaystyle \int_0^{0,73} \frac{2x^{3}}{3} - \frac{4x^7}{21}dx$, which gives me 0.238301997 and is not correct,

or, wait a minute, it would give me

$\displaystyle \int_0^{0,73} \frac{2x^{3}}{3} - \frac{4x^7}{21}dx =$

$\displaystyle \left( \frac{2x^{3+1}}{(3+1)3} - \frac{4x^{7+1}}{(7+1)21} \right)_0^{0,73}=$ right?
• Nov 27th 2013, 09:39 AM
romsek
Re: Find the MacLaurin polinomial of degree 7 for F(x)
no, your polynomial is the integral. Just evaluate it at the endpoints and subtract. Did you miss the step in the derivation of the series where we integrated?
• Nov 27th 2013, 09:42 AM
dokrbb
Re: Find the MacLaurin polinomial of degree 7 for F(x)
Quote:

Originally Posted by romsek
no, your polynomial is the integral. Just evaluate it at the endpoints and subtract. Did you miss the step in the derivation of the series where we integrated?

yes, you are right, can you check how much do you get, I can't belive I do a mistake when rising to a power or substracting, :)
• Nov 27th 2013, 05:12 PM
romsek
Re: Find the MacLaurin polinomial of degree 7 for F(x)
I get -0.238302 and looking at this I realized I lost a minus sign not you. It was in the step where we did the integral. The correct result is negative the result I showed.

The correct coeficients polynomial is 2/3 x3 - 4/21 x7

and so the correct answer is 0.238302.
• Nov 27th 2013, 05:25 PM
dokrbb
Re: Find the MacLaurin polinomial of degree 7 for F(x)
Quote:

Originally Posted by romsek
I get -0.238302 and looking at this I realized I lost a minus sign not you. It was in the step where we did the integral. The correct result is negative the result I showed.

The correct coeficients polynomial is 2/3 x3 - 4/21 x7

and so the correct answer is 0.238302.

funny, I got this result at first and got is an incorrect in my webwork assignment, it might be something wrong with the system since it accepts the polynomial and not this result; anyways I will show this to my prof tomorrow, but thanks, you assured me I'm not wrong ;)