Power series representation of a function

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To find a power series representation for f(x) = ∫₀ˣ sin(t³) dt, the power series for sin(t³) can be used. The correct approach involves integrating the series term-by-term, leading to f(x) = ∫₀ˣ ∑ₙ₌₀^∞ [(-1)ⁿ (t³)^(2n+1)] / (2n+1)! dt. This results in a new series where the terms are adjusted for integration, specifically changing the exponent and dividing by (n+1). Care must be taken with the exponents during integration to avoid mistakes. The final representation will be a power series in terms of x.
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Homework Statement



Find a power series representation for f(x) using termwise integration, where f(x) = \int_{0}^{x} sin(t^3) dt.


Homework Equations





The Attempt at a Solution



I've never done this before, but apparently, if I have a power series representation for sin(t^3), I can use the fact that f(x) = \int_{0}^{x} f(t) dt = \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}. Is that right?

So my approach was to use the power series representation of sine, replacing sin x with sin t^3 - this gave me sin(t^3) \approx \sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+1}}{(2n+1)!}. Now do I just change this to the sum
\sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+3}}{(2n+3)! (n+1)}? It's not at all clear to me that this is what I'm supposed to do...
 
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quasar_4 said:

Homework Statement



Find a power series representation for f(x) using termwise integration, where f(x) = \int_{0}^{x} sin(t^3) dt.


Homework Equations





The Attempt at a Solution



I've never done this before, but apparently, if I have a power series representation for sin(t^3), I can use the fact that f(x) = \int_{0}^{x} f(t) dt = \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}. Is that right?
I think you might confuse yourself by having two different definitions for f.
f(x) = \int_{0}^{x} sin(t^3) dt = \int_{0}^{x}\sum_{n=0}^{\infty} \frac{(-1)^n (t^3)^{2n+1}}{(2n+1)!} dt
Now you're set up to integrate term-by-term to get an antiderivative, which you can evaluate at x and at 0.

quasar_4 said:
So my approach was to use the power series representation of sine, replacing sin x with sin t^3 - this gave me sin(t^3) \approx \sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+1}}{(2n+1)!}. Now do I just change this to the sum
\sum_{n=0}^{\infty} \frac{(-1)^n (x^3)^{2n+3}}{(2n+3)! (n+1)}? It's not at all clear to me that this is what I'm supposed to do...

I think you made a mistake with your exponent. (t^3)^(2n+1) = t^(6n + 3). When you integrate that, you get something different from what you show.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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