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What is this called? A general solution? 
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#19
Apr814, 04:19 PM

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P: 2,264

Could you explain the parts that are confusing?
I have questions too. Back in #3 do you know how the numbers in the family can be calculated? Do you see how $$\sum_{n=0} p(n)/n!=C \, e$$ For some constant c and how to find C? How much calculus do you know? Do you understand Taylor's series? Do you understand the generalization of your formula I GAVE IN #15 hand how it gives the above formula and a way to find C? The links give general methods of expanding functions in series, they are not given in elementary calculus because they are complicated. Another series is $$e=\frac{1}{1+\mathrm{erf}(1)}\sum_{n=2}^\infty \frac{1}{\Gamma(n/2)}$$ we can give series for e all day, but to what end? The notation $$\left(x \, \dfrac{d}{dx}\right) ^n\mathrm{f}(x)$$ means apply n times ie $$\left( x \, \dfrac{d}{dx}\right) ^4\mathrm{f}(x)=\left( x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\mathrm{f}(x)\right)\right)\right)\right)$$ 


#20
Apr814, 07:42 PM

P: 553




#21
Apr814, 08:01 PM

P: 553

are you just saying take the fourth derivative of 'f(x)' with a multiplication of 'x' in each step? 


#22
Apr914, 03:42 AM

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P: 2,264

Sorry about the confusion. Yes I am talking about taking the derivative then multiplying by x possibly multiple times. That is called homogeneous differentiation because it does not change the degree of nonconstant polynomials. For example if n=1,2,3,...
$$\left( x\, \dfrac{d}{dx}\right)^m x^n=n^m x^n \\ \left( x\, \dfrac{d}{dx}\right)^n x^a=p(n) x^n$$ so we can find find $$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x$$ 


#23
Apr1014, 09:23 AM

P: 553

I understand this part,
$$\left( x\, \dfrac{d}{dx}\right)^m x^n=n^m x^n$$ as we can see for whatever value of the exponent of 'x' we get a multiplier of that to to the 'm' power on the right hand side and of course the x^n is preserved but I am unsure what you are doing here, $$\left( x\, \dfrac{d}{dx}\right)^n x^a=p(n) x^n$$ Why isn't the right side x^a? Is p(n)=a^n or is it something else in order to get x^n on the right hand side? 


#24
Apr1014, 09:39 AM

P: 553

$$\left( x\, \dfrac{d}{dx}\right)^n x^a=p(n) x^n$$
The only way I can see to make this work is if $$p(n)=\frac{a^n*x^a}{x^n}$$ but that seems redundant so am I missing something? 


#25
Apr1014, 11:31 AM

HW Helper
P: 2,264

sorry typo in the middle equation
that should have been 


#26
Apr1014, 12:48 PM

P: 553

Okay so you have this next, $$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x$$ I see how this can be a useful tool for calculating for some polynomial of 'n' without having to take the infinite series which is wonderful but how do we get an infinite number of unique series for 'e' or 'e^x' with this technique? I am going to look back over your previous posts now that I have this new information. 


#27
Apr1014, 06:12 PM

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P: 2,264

$$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x=q(x)e^x$$
for some polynomial q(x) that can be found from p then $$e^x=\frac{1}{q(x)}\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}$$ there are an infinite number of polynomials and each gives us a series for e^x or e if we let x=1 there are some better ways to find q given p but if p is not too high in degree we can use basic calculus $$q(x)=e^{x}p\left( x\dfrac{d}{dx}\right)e^x$$ 


#28
Apr1114, 12:06 PM

P: 553

I would like to post my e^x solution (if the moderators have no objection), I didn't use Calculus for any of the derivations and my general solutions are substantially easier to use. In the meantime lurflurf, this is a wonderful method! I will be sure to use it in the future. I am curious, where did it come from (as in who figured it out)? You posted several links earlier, I would imagine they contain several other strategies, would you mind picking one and we review it like we just did here? 


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