Answer :
To find the Taylor polynomial [tex]\( P_9(x) \)[/tex] centered at [tex]\( x = 0 \)[/tex] for the function [tex]\( f(x) = 2x \cos(2x^2) \)[/tex], we want to express this function as a polynomial with terms up to [tex]\( x^9 \)[/tex].
### Step-by-Step Solution:
1. Understand the function: We have [tex]\( f(x) = 2x \cos(2x^2) \)[/tex]. This is a product of [tex]\( 2x \)[/tex] and the cosine function [tex]\( \cos(2x^2) \)[/tex].
2. Expand cosine using Taylor series: Recall the Taylor series for [tex]\( \cos(u) \)[/tex] around 0 is:
[tex]\[
\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \cdots
\][/tex]
In this case, substitute [tex]\( u = 2x^2 \)[/tex] to obtain the expansion:
[tex]\[
\cos(2x^2) = 1 - \frac{(2x^2)^2}{2!} + \frac{(2x^2)^4}{4!} - \frac{(2x^2)^6}{6!} + \cdots
\][/tex]
Simplify each term:
[tex]\[
\cos(2x^2) = 1 - \frac{4x^4}{2} + \cdots
\][/tex]
3. Multiply by [tex]\( 2x \)[/tex]: Multiply the expanded cosine by [tex]\( 2x \)[/tex]:
[tex]\[
2x \left(1 - 2x^4 + \cdots\right) = 2x - 4x^5 + \cdots
\][/tex]
4. Identify terms up to [tex]\( x^9 \)[/tex]: We only keep terms that contribute to the polynomial up to [tex]\( x^9 \)[/tex]:
[tex]\[
2x - 4x^5
\][/tex]
The terms involving higher powers of [tex]\( x \)[/tex] beyond [tex]\( x^5 \)[/tex] are insignificant within the degree constraint of 9.
5. Conclusion: The required Taylor polynomial [tex]\( P_9(x) \)[/tex] for the given function is:
[tex]\[
P_9(x) = 2x - 4x^5
\][/tex]
The Taylor polynomial approximates the function [tex]\( f(x) = 2x \cos(2x^2) \)[/tex] up to degree 9 when centered at [tex]\( x = 0 \)[/tex].
### Step-by-Step Solution:
1. Understand the function: We have [tex]\( f(x) = 2x \cos(2x^2) \)[/tex]. This is a product of [tex]\( 2x \)[/tex] and the cosine function [tex]\( \cos(2x^2) \)[/tex].
2. Expand cosine using Taylor series: Recall the Taylor series for [tex]\( \cos(u) \)[/tex] around 0 is:
[tex]\[
\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \cdots
\][/tex]
In this case, substitute [tex]\( u = 2x^2 \)[/tex] to obtain the expansion:
[tex]\[
\cos(2x^2) = 1 - \frac{(2x^2)^2}{2!} + \frac{(2x^2)^4}{4!} - \frac{(2x^2)^6}{6!} + \cdots
\][/tex]
Simplify each term:
[tex]\[
\cos(2x^2) = 1 - \frac{4x^4}{2} + \cdots
\][/tex]
3. Multiply by [tex]\( 2x \)[/tex]: Multiply the expanded cosine by [tex]\( 2x \)[/tex]:
[tex]\[
2x \left(1 - 2x^4 + \cdots\right) = 2x - 4x^5 + \cdots
\][/tex]
4. Identify terms up to [tex]\( x^9 \)[/tex]: We only keep terms that contribute to the polynomial up to [tex]\( x^9 \)[/tex]:
[tex]\[
2x - 4x^5
\][/tex]
The terms involving higher powers of [tex]\( x \)[/tex] beyond [tex]\( x^5 \)[/tex] are insignificant within the degree constraint of 9.
5. Conclusion: The required Taylor polynomial [tex]\( P_9(x) \)[/tex] for the given function is:
[tex]\[
P_9(x) = 2x - 4x^5
\][/tex]
The Taylor polynomial approximates the function [tex]\( f(x) = 2x \cos(2x^2) \)[/tex] up to degree 9 when centered at [tex]\( x = 0 \)[/tex].
