Difference between revisions of "009C Sample Final 2, Problem 6"
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|So, the power series of <math style="vertical-align: -5px">\sin(x^2)</math> is | |So, the power series of <math style="vertical-align: -5px">\sin(x^2)</math> is | ||
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| <math>\begin{array}{rcl} | | <math>\begin{array}{rcl} | ||
\displaystyle{\sin(x^2)} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}\\ | \displaystyle{\sin(x^2)} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}\\ | ||
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& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}.} | & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}.} | ||
\end{array}</math> | \end{array}</math> | ||
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|} | |} | ||
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|Now, to express the indefinite integral as a power series, we have | |Now, to express the indefinite integral as a power series, we have | ||
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| <math>\begin{array}{rcl} | | <math>\begin{array}{rcl} | ||
\displaystyle{\int \sin(x^2)~dx} & = & \displaystyle{\int \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\ | \displaystyle{\int \sin(x^2)~dx} & = & \displaystyle{\int \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\ | ||
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& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.} | & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.} | ||
\end{array}</math> | \end{array}</math> | ||
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|} | |} | ||
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|Hence, we have | |Hence, we have | ||
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| <math>\begin{array}{rcl} | | <math>\begin{array}{rcl} | ||
\displaystyle{\int_0^1 \sin(x^2)~dx} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n (1)^{4n+3}}{(4n+3)(2n+1)!}-\sum_{n=0}^\infty \frac{(-1)^n (0)^{4n+3}}{(4n+3)(2n+1)!}}\\ | \displaystyle{\int_0^1 \sin(x^2)~dx} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n (1)^{4n+3}}{(4n+3)(2n+1)!}-\sum_{n=0}^\infty \frac{(-1)^n (0)^{4n+3}}{(4n+3)(2n+1)!}}\\ | ||
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& = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}} | & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}} | ||
\end{array}</math> | \end{array}</math> | ||
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|} | |} | ||
Revision as of 16:09, 10 March 2017
(a) Express the indefinite integral Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin(x^2)~dx} as a power series.
(b) Express the definite integral as a number series.
| Foundations: |
|---|
| What is the power series of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin x?} |
| The power series of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin x} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.} |
Solution:
(a)
| Step 1: |
|---|
| The power series of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin x} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}.} |
| So, the power series of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(x^2)} is |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\sin(x^2)} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}.} \end{array}} |
| Step 2: |
|---|
| Now, to express the indefinite integral as a power series, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \sin(x^2)~dx} & = & \displaystyle{\int \sum_{n=0}^\infty \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\ &&\\ & = & \displaystyle{\sum_{n=0}^\infty \int \frac{(-1)^nx^{4n+2}}{(2n+1)!}~dx}\\ &&\\ & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.} \end{array}} |
(b)
| Step 1: |
|---|
| From part (a), we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}.} |
| Now, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^1 \sin(x^2)~dx=\sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}\bigg|_0^1.} |
| Step 2: |
|---|
| Hence, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int_0^1 \sin(x^2)~dx} & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n (1)^{4n+3}}{(4n+3)(2n+1)!}-\sum_{n=0}^\infty \frac{(-1)^n (0)^{4n+3}}{(4n+3)(2n+1)!}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}-0}\\ &&\\ & = & \displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}} \end{array}} |
| Final Answer: |
|---|
| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}} |
| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(4n+3)(2n+1)!}} |