Maclaurin Series for ##\int_{0}^{x} \cos{t^2} \cdot dt##

• Potatochip911
In summary: So the series is 1 + \frac{1}{2!} x^2 + \frac{1}{4!} x^4 + \cdots, and evaluating at x gives the sum above.In summary, to find the Maclaurin series of the integral of cos(t2), we first write the Maclaurin series for cos(t2), then integrate that series term-by-term. Finally, we evaluate the resulting series at x and at 0 to get the Maclaurin series for the integral. Alternatively, we can use the formula for finding the Maclaurin series of an integral and the Maclaurin series for cos(t2) to get the same result.
Potatochip911

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.

Potatochip911 said:

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.

Potatochip911
Mark44 said:
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.

Potatochip911 said:
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
$$f(0) + f'(0) x + \frac{1}{2!} f''(0) x^2 + \cdots,$$
whose terms are easy to get, since
$$\begin{array}{rcl} f(0) &=& \int_0^0 \cos (t^2) \, dt = 0\\ f'(x) &=& \cos (x^2) \; \Rightarrow f'(0) = \cos^2(0) = 1 \\ f''(x) &=& \frac{d}{dx} \cos(x^2) = -2 x \sin(x^2) \; \Rightarrow f''(0) = 0 \\ &\vdots& \end{array}$$
In fact, ##f'(0), f''(0), \ldots## are just the coefficients in the Maclauren expansion of ##\cos(x^2)##, multiplied by the appropriate factorial.

Potatochip911

What is a Maclaurin Series?

A Maclaurin Series is an infinite series used to represent a function as a sum of power functions. It is named after Scottish mathematician Colin Maclaurin.

How do you find the Maclaurin Series for a function?

The Maclaurin Series can be found by taking the derivatives of the function at a specific point and plugging them into the general Maclaurin Series formula. The formula is: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

Why is the Maclaurin Series useful for integration?

The Maclaurin Series allows us to approximate the value of a function at a specific point, which can be useful for integration. By using the Maclaurin Series, we can convert an integral into a series, which can then be easily evaluated using known methods.

What is the Maclaurin Series for ##\int_{0}^{x} \cos{t^2} \cdot dt##?

The Maclaurin Series for ##\int_{0}^{x} \cos{t^2} \cdot dt## is: ##1 - \frac{1}{2}x^2 + \frac{1}{12}x^4 - \frac{1}{720}x^6 + \frac{1}{30240}x^8 - ...##

Can the Maclaurin Series be used to find an exact value for ##\int_{0}^{x} \cos{t^2} \cdot dt##?

No, the Maclaurin Series is an infinite series and can only be used to approximate the value of the integral. To find an exact value, the integral must be evaluated using other methods such as integration by parts or substitution.

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