Multiplying Power Series: Help & Solutions

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To multiply the power series for e^x and arctan(x), one must apply the distributive property across the entire series, treating it similarly to the FOIL method used for binomials. The series for e^x is 1 + x + x^2/2! + x^3/3!, while arctan(x) is 0 + x - x^3/3 + x^5/5 - x^7/7. The multiplication involves combining terms from both series, specifically multiplying each term of e^x by the entire arctan series. The process can be simplified by calculating terms only up to a certain degree, such as x^5, to avoid dealing with infinite series. Ultimately, the multiplication yields a new power series that represents the product of e^x and arctan(x).
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How do I multiply power series?

Homework Statement


Find the power series:
e^x arctan(x)

Homework Equations



e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}

arctan(x) = 0 + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}

The Attempt at a Solution



So do I multiply 1 by 0, x by x and so forth? Or do I go 1 by 0, 1 by x? Or is there another way?
 
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You have to multiply 1 by the whole acrtan series, x by the whole arctan series, and so on. There might be a way to simplify it though. Wikipedia has this under "power series"

f(x)g(x) = \left(\sum_{n=0}^\infty a_n (x-c)^n\right)\left(\sum_{n=0}^\infty b_n (x-c)^n\right)

= \sum_{i=0}^\infty \sum_{j=0}^\infty a_i b_j (x-c)^{i+j}

= \sum_{n=0}^\infty \left(\sum_{i=0}^n a_i b_{n-i}\right) (x-c)^n
 
So I kind of treat it like F.O.I.L.?
 
Sort of. FOIL is the distrubutive law for (binomial)X(binomial). Here, you've got two infinitely long "polynomials". Obviously, you won't be able to write out all of the terms. :-p
 
Awesome, I think I got it, I only had to take it out to the x^5 term.
 
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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