User:Egm6341.s10.team3.ynahn81/HW5-6

= (6) Find coefficients $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ from Bernoulli numbers = Ref: Lecture notes [[media:Egm6341.s10.mtg28.djvu|p.28-2]]

Problem Statement
Using Bernoulli numbers obtained by the Taylor series expansion of $$\displaystyle x\,coth(x)$$,


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x\,coth(x) = \sum_{r=0}^{\infty}\underbrace{\frac{B_{2r}}{(2r)!}}_{-\bar{d_{2r}}}x^{2r} $$  
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i) Verify $$\displaystyle \bar{d_{2}}, \; \bar{d_{4}}, \; \bar{d_{6}}$$ by comparing the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$. ii) Compute $$\displaystyle \bar{d_{8}}, \; \bar{d_{10}}$$

Solution
i)

Let's explicitly express first several terms of the Taylor series expansion of $$\displaystyle x\,coth(x)$$,


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x\,coth(x) = B_{0} + \frac{B_2}{2!}x^{2} + \frac{B_4}{4!}x^{4} + \frac{B_6}{6!}x^{6} + \frac{B_8}{8!}x^{8} + \frac{B_{10}}{10!}x^{10} + \cdot\cdot\cdot. $$ 
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Note that $$\displaystyle B_{0}=1, \, B_{2}=\frac{1}{6}, \, B_{4}=-\frac{1}{30}, \, B_{6}=\frac{1}{42}$$.

Then, based on the following relation between $$\displaystyle \bar{d_{2r}} $$ and Bernoulli numbers,


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\frac{B_{2r}}{(2r)!} = -\bar{d_{2r}}, $$ 
 * $$\displaystyle
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we can obtain $$\displaystyle \bar{d_{2}}, \,\bar{d_{4}}, \,\bar{d_{6}}$$ as follows.


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$$\displaystyle \bar{d_{2}}=-\frac{B_{2}}{2!}=-\frac{1}{2}\times\frac{1}{6}=-\frac{1}{12} $$ 
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$$\displaystyle \bar{d_{4}}=-\frac{B_{4}}{4!}=\frac{1}{24}\times\frac{1}{30}=\frac{1}{720} $$ 
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$$\displaystyle \bar{d_{6}}=-\frac{B_{6}}{6!}=-\frac{1}{720}\times\frac{1}{42}=-\frac{1}{30240} $$ 
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Above results are exactly same with the results from $$\displaystyle \bar{d_{2r}} = \frac{P_{2r}(1)}{2^{2r}}$$ as we obtained at problem 4 in HW5.

ii)

Since $$\displaystyle B_{8}=-\frac{1}{30}, \, B_{10}=\frac{5}{66}$$,


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$$\displaystyle \bar{d_{8}}=-\frac{B_{8}}{8!}=\frac{1}{40320}\times\frac{1}{30}=\frac{1}{1209600} $$ 
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$$\displaystyle \bar{d_{10}}=-\frac{B_{10}}{10!}=-\frac{1}{3628800}\times\frac{5}{66}=-\frac{1}{47900160} $$ 
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