Maclaurin Series

  • #1

Homework Statement


Find the Maclaurin series of ##\int_{0}^{x} \cos{t^2} \cdot dt ##

Homework Equations


3. The Attempt at a Solution [/B]
I normally have some idea how to go about solving these but for this one I just can't figure out where to start. I tried doing it with ##\int_{0}^{x} \cos{t}\cdot dt ## and then after evaluating that series replacing ##x## with ##x^2## but that didn't work at all. I checked what the integral of the original function was on wolfram alpha and it looked quite complicated so I'm assuming I'm supposed to calculate the integral of a simpler function then replace the x but I can't figure out what the simpler function is.
 

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  • #2
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Homework Statement


Find the Maclaurin series of ##\int_{0}^{x} \cos{t^2} \cdot dt ##

Homework Equations


3. The Attempt at a Solution [/B]
I normally have some idea how to go about solving these but for this one I just can't figure out where to start. I tried doing it with ##\int_{0}^{x} \cos{t}\cdot dt ## and then after evaluating that series replacing ##x## with ##x^2## but that didn't work at all. I checked what the integral of the original function was on wolfram alpha and it looked quite complicated so I'm assuming I'm supposed to calculate the integral of a simpler function then replace the x but I can't figure out what the simpler function is.
Write the Maclaurin series for cos(t2), then integrate that series term-by-term to get the Maclaurin series of the integral you posted. Finally, evaluate your series at x and at 0, just like you would normally do with a definite integral.
 
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  • #3
Write the Maclaurin series for cos(t2), then integrate that series term-by-term to get the Maclaurin series of the integral you posted. Finally, evaluate your series at x and at 0, just like you would normally do with a definite integral.
Okay so $$\cos{t}=1-\frac{1}{2!}t^2+\frac{1}{4!}t^4-\frac{1}{6!}t^6+...\\
\cos{t^2}=1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+... \\
\int_{0}^{x} \cos{t^2}\cdot dt=\int_{0}^{x} (1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+...)dt \\
\int_{0}^{x} \cos{t^2} \cdot dt=x-\frac{1}{5*2!}x^5+\frac{1}{9*4!}x^9-\frac{1}{12*6!}x^{13}+... \\
\int_{0}^{x} \cos{t^2} \cdot dt=\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{(4n+1)!(2n)!} $$
Thanks this ended up giving the correct answer.
 
  • #4
Ray Vickson
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Okay so $$\cos{t}=1-\frac{1}{2!}t^2+\frac{1}{4!}t^4-\frac{1}{6!}t^6+...\\
\cos{t^2}=1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+... \\
\int_{0}^{x} \cos{t^2}\cdot dt=\int_{0}^{x} (1-\frac{1}{2!}t^4+\frac{1}{4!}t^8-\frac{1}{6!}t^{12}+...)dt \\
\int_{0}^{x} \cos{t^2} \cdot dt=x-\frac{1}{5*2!}x^5+\frac{1}{9*4!}x^9-\frac{1}{12*6!}x^{13}+... \\
\int_{0}^{x} \cos{t^2} \cdot dt=\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+1}}{(4n+1)!(2n)!} $$
Thanks this ended up giving the correct answer.

Alternatively: if ##f(x) = \int_0^x \cos (t^2) \, dt## the Maclauren series is
[tex] f(0) + f'(0) x + \frac{1}{2!} f''(0) x^2 + \cdots,[/tex]
whose terms are easy to get, since
[tex] \begin{array}{rcl}
f(0) &=& \int_0^0 \cos (t^2) \, dt = 0\\
f'(x) &=& \cos (x^2) \; \Rightarrow f'(0) = \cos^2(0) = 1 \\
f''(x) &=& \frac{d}{dx} \cos(x^2) = -2 x \sin(x^2) \; \Rightarrow f''(0) = 0 \\
&\vdots&
\end{array}
[/tex]
In fact, ##f'(0), f''(0), \ldots## are just the coefficients in the Maclauren expansion of ##\cos(x^2)##, multiplied by the appropriate factorial.
 
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