Show that
n−1∑k=0tanh(x1nsin2(2k+14nπ))1+tanh2xtan2(2k+14nπ)=tanh(2nx)
Thank you ,This problem I take some hours,and at last I don't prove it
and This problem is from
This book have some same problem.all is not true? if not true,and we how find it or edit it somewhere?
Thank you achille hui,he told me this following maybe is true,Now How prove it?
n−1∑k=0tanhxnsin2(2k+14nπ)1+tanh2xtan2(2k+14nπ)=tanh(2nx)
Answer
Notice
cosh(2nx)=T2n(coshx) and cos(2nx)=T2n(cosx)
where T2n(z) is a
Chebyshev polynomial of the first kind. Using the 2nd relation above, it is clear the roots of T2n(x) has the
form:
±cos(2k+14nπ), for k=0,…,n−1
From this, we arrive following expansion of cosh(2nx):
cosh(2nx)=An−1∏k=0(cosh2x−cos2(2k+14nπ))
for some constant A we don't care.
Taking logarithm, differentiate w.r.t x and divide by 2n for both sides, we find:
tanh(2nx)=12nn−1∑k=02sinhxcoshxcosh2x−cos2(2k+14nπ)=1nn−1∑k=0sinhxcoshxsinh2x+sin2(2k+14nπ)=1nn−1∑k=0tanhxtanh2x+sin2(2k+14nπ)(1−tanh2x)=n−1∑k=0tanhxnsin2(2k+14nπ)1+(1sin2(2k+14nπ)−1)tanh2x=n−1∑k=0tanhxnsin2(2k+14nπ)1+tanh2xtan2(2k+14nπ)
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