Let $$h(t) = \frac{\sinh(t)}{t}.$$ Let $$ f_i(t) = h\left(\frac{t}{2^i} \right)^{2^i}, $$ where $i\ge 0$ is an integer.
Is there anyway to get a series expansion of $f_i(t)$ without replacing $i$ with a fixed integer?
If I understand your question correctly, you can use taylor:
sage: var('i,t')
(i, t)
sage: h(t) = sinh(t)/t
sage: fi(t) = h(t/2^i)^(2^i)
sage: fi(t).taylor(t, 0, 10)
1/359251200*(385*2^(4*i) - 1540*2^(3*i) + 2684*2^(2*i) - 2288*2^i + 768)*t^10/2^(9*i) + 1/5443200*(175*2^(3*i) - 420*2^(2*i) + 404*2^i - 144)*t^8/2^(7*i) + 1/45360*(35*2^(2*i) - 42*2^i + 16)*t^6/2^(5*i) + 1/360*(5*2^i - 2)*t^4/2^(3*i) + 1/6*t^2/2^i + 1
Asked: 2017-06-06 10:38:46 +0200
Seen: 734 times
Last updated: Jun 06 '17
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