# Any method a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. f(x)=x^2cos x^2

Question
Series
Any method
a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.
$$f(x)=x^2\cos x^2$$

2020-11-23
Here the objective is to find the Taylor’s series of $$f(x)=x^2\cos x^2$$ around a=0
Taylor’s series of f(x) around x=a is
$$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...+\frac{f^n(a)}{n!}(x-a)^n+R(z)$$
Let $$f(x)=\cos x,$$ and a=0
$$f(x)=\cos x\to f(0)=\cos0=1$$
$$f'(x)=-\sin x\to f'(0)=-\sin0=0$$
$$f''(x)=-\cos x\to f''(0)=-\cos0=-1$$
$$f'''(x)=\sin x\to f'''(0)=\sin0=0$$
$$f^4(x)=\cos x\to f^4(0)=\cos0=1$$
Therefore the required Taylor’s series of $$\cos(x)$$ is
$$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\frac{f^4(a)}{4!}(x-a)^4+...$$
$$\cos x=1+\frac{0}{1!}(x-0)+\frac{(-1)}{2!}(x-0)^2+\frac{0}{3!}(x-0)^3+\frac{1}{4!}(x-0)^4+\frac{0}{5!}(x-0)^5+\frac{(-1)}{6!}(x-0)^6...$$
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$
$$\cos x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k}$$
To find the Taylor's series of $$\cos x^2$$, replace x by $$x^2$$ in Taylor's series of $$\cos x$$, so
$$\cos x=1-\frac{(x^2)^2}{2!}+\frac{(x^4)^2}{4!}-\frac{(x^6)^2}{6!}+...$$
$$\cos x=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$$
To find the Taylor's series of $$x^2\cos x^2$$, replace x by $$x^2$$ in Taylor's series of $$\cos x^2$$, so
$$x^2\cos x=x(1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...)$$
$$x^2\cos x=x^2-\frac{x^6}{2!}+\frac{x^{10}}{4!}-\frac{x^{14}}{6!}+...$$
$$x^2\cos x=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{4k+2}$$

### Relevant Questions

Taylor series and interval of convergence
a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
c. Determine the interval of convergence of the series.
$$f(x)=\log_3(x+1),a=0$$
Taylor series
a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.
b. Write the power series using summation notation.
$$f(x)=2^x,a=1$$
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$\displaystyle{f{{\left({x}\right)}}}={\text{cosh}{{\left({2}{x}-{2}\right)}}},{a}={1}$$
Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
$$f(x)=\tan^{-1}4x,a=0$$
Binomial series
a. Find the first four nonzero terms of the binomial series centered at 0 for the given function.
b. Use the first four terms of the series to approximate the given quantity.
$$f(x)=(1+x)^{\frac{2}{3}}$$, approximate $$(1.02)^{\frac{2}{3}}$$
Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{\sqrt{{{x}}}}}}$$ with $$\displaystyle{a}={4}$$, approximate $$\displaystyle{\frac{{{1}}}{{\sqrt{{{3}}}}}}$$
Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when $$Z=1.3$$ and $$H=0.05$$;
Assume that you do not have vales of the area beyond $$z=1.2$$ in the table; i.e. you may need to use the extrapolation.
Check your calculated value and compare with the values in the table $$[for\ z=1.3\ and\ H=0.05]$$.
Calculate your percentage of error in the estimation.
How do I solve this problem using extrapolation?
$$\begin{array}{|c|c|}\hline Z+H & Prob. & Extrapolation \\ \hline 1.20000 & 0.38490 & Differences \\ \hline 1.21000 & 0.38690 & 0.00200 \\ \hline 1.22000 & 0.38880 & 0.00190 \\ \hline 1.23000 & 0.39070 & 0.00190 \\ \hline 1.24000 & 0.39250 & 0.00180 \\ \hline 1.25000 & 0.39440 & 0.00190 \\ \hline 1.26000 & 0.39620 & 0.00180 \\ \hline 1.27000 & 0.39800 & 0.00180 \\ \hline 1.28000 & 0.39970 & 0.00170 \\ \hline 1.29000 & 0.40150 & 0.00180 \\ \hline 1.30000 & 0.40320 & 0.00170 \\ \hline 1.31000 & 0.40490 & 0.00170 \\ \hline 1.32000 & 0.40660 & 0.00170 \\ \hline 1.33000 & 0.40830 & 0.00170 \\ \hline 1.34000 & 0.41010 & 0.00180 \\ \hline 1.35000 & 0.41190 & 0.00180 \\ \hline \end{array}$$

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
i.If $$a_{n} and f(n) satisfy the requirements of the integral test, then \sum_{n=1}^{\infty} a_n = \int_1^{\infty} f(x)dx$$.
ii. The series $$\sum_{n=1}^{\infty} \frac{1}{n^p} converges if p > 1 and diverges if p \leq 1$$.
$$f(x)=x+e^{-x}, a=0$$