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Find a formula for $b_m$ by evaluating both sides for $f(x)=e^{\lambda x}$ where $\lambda$ is a parameter. The formula is $\int_0^1 f(x)dx=1/2(f(0)+f(1))+\sum_{m=1}^\inf b_m(f^{(2m-1)}(1)-f^{(2m-1)}(0))$ for some unknown constants $b_m$ independent of f.

I plug in $f(x)=e^{\lambda x}$ and get that $\sum_{m=1}^\inf b_m=(2e^\lambda -2-\lambda e^\lambda-\lambda)/2\lambda e^\lambda$. I don't know how to get rid of the sum. Any thought?

J.doe
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  • Are you sure you compute well the (2m-1)th derivative of $e^{\lambda x}$ with respect to $x$ ? In your computations, one should find terms with coefficients $\lambda^{2m-1}$... In this way, the $b_n$ cannot be isolated.. – Jean Marie Feb 29 '16 at 01:08
  • I substitute x with 1first, then I get $e^\lambda$. Then take (2m-1)th derivative of it, which is still $e^\lambda$? – J.doe Feb 29 '16 at 01:16
  • no for example, the derivative of $e^{\lambda x}$ with respect to $x$ is $\lambda e^{\lambda x}$, if it is the second derivative, it is $\lambda^2e^{\lambda x} $, etc... – Jean Marie Feb 29 '16 at 01:34
  • Oh I should substitute x with 1 after I get the formula for its (2m-1)th derivative? And the answer should be $\lambda ^ {2m-1}e^\lambda$? – J.doe Feb 29 '16 at 01:43
  • And $\int_0^1 e^{\lambda x}dx=(e^\lambda -1)/\lambda $? – J.doe Feb 29 '16 at 01:46
  • That's it ! (see my answer) – Jean Marie Feb 29 '16 at 01:53
  • Hi, still I can't find a specific pattern for $b_m$. – J.doe Mar 01 '16 at 19:11

1 Answers1

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$$\dfrac{1}{\lambda}(e^{\lambda}-1)=\dfrac{1}{2}(e^{\lambda}+1)+\sum_{m=1}^{\infty}b_m\lambda^{2m-1}(e^{\lambda}-1)$$

Dividing both sides by $(e^{\lambda}-1)$, we obtain:

$$\dfrac{1}{\lambda}=\dfrac{1}{2}\dfrac{e^{\lambda}+1}{e^{\lambda}-1}+\sum_{m=1}^{\infty}b_m\lambda^{2m-1}$$

$$\dfrac{1}{\lambda}=\dfrac{1}{2}+\dfrac{1}{e^{\lambda}-1}+\sum_{m=1}^{\infty}b_m\lambda^{2m-1}$$

$$\dfrac{1}{e^{\lambda}-1}=\dfrac{1}{\lambda}-\dfrac{1}{2}-\sum_{m=1}^{\infty}b_m\lambda^{2m-1}$$

Then, multiplying by $\lambda$:

$$\dfrac{\lambda}{e^{\lambda}-1}=1-\dfrac{\lambda}{2}-\sum_{m=1}^{\infty}b_m\lambda^{2m}$$

Thus the $b_m$ are, up to the sign, the coefficients of the Taylor series of the function on the left hand side.

They are closely related to the Bernoulli numbers. see the part "generating functions" in https://en.wikipedia.org/wiki/Bernoulli_number

Jean Marie
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  • I see!! Thanks!! – J.doe Feb 29 '16 at 01:58
  • Something that I have not said is that function $f(\lambda)=\dfrac{\lambda}{e^{\lambda}-1}$ has an apparent discontinuity in $\lambda=0$, but one "forces" $f(0)=1$: this function is $C^{\infty}$ everywhere. – Jean Marie Feb 29 '16 at 07:53
  • I got a problem here: the first few terms of the Bernoulli numbers are 1 and -1/2, I can't get rid of the $\lambda /2$. And after simplifying both sides, I got $\sum_{m=1}^\infty b_m \lambda^{2m}=1/2-\lambda /2- \sum_{m=2}^\infty B_m \lambda^m/m!$, where $B_m$ stands for Bernoulli numbers. – J.doe Feb 29 '16 at 23:27
  • I tried to pull out $b_1\lambda^2$ for m=1, but it seems like that wouldn't help at all. – J.doe Feb 29 '16 at 23:28
  • In fact, the first 2 Bernoulli numbers are exceptional. But are you supposed to make the connexion with Bernoulli numbers ? – Jean Marie Feb 29 '16 at 23:48
  • I am supposed to find a formula fofr $b_m$, so I tried to connect it with Bernoulli numbers, to see if I can find any pattern. – J.doe Feb 29 '16 at 23:52