What is this called? A general solution?

by mesa
Tags: called, solution
 HW Helper P: 2,263 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 $$\left( x \, \dfrac{d}{dx}\right) ^4\mathrm{f}(x)=\left( x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\mathrm{f}(x)\right)\right)\right)\right)$$
P: 515
 Quote by lurflurf 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?
Okay, so you are saying if we have some polynomial of 'n' in the numerator it will be equal to some value 'C' and you want to know if I can calculate 'C'? If you mean as in using some 'formula' for calculating it, no, but if given the polynomial I can figure it out.

 Quote by lurflurf How much calculus do you know?
Only Calc 1-3 (for engineers) and Differential Equations

 Quote by lurflurf Do you understand Taylor's series?
Not particularly although I have been advised to look more closely at it.
 Quote by lurflurf 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?
No, you have x followed by the differential operator so I am not sure what that means.

 Quote by lurflurf The links give general methods of expanding functions in series, they are not given in elementary calculus because they are complicated.
Okay, so these links show other methods of producing more series for a given value but they are complicated.

 Quote by lurflurf Another series is $$e=\frac{1}{1+\mathrm{erf}(1)}\sum_{n=2}^\infty \frac{1}{\Gamma(n/2)}$$
How do we get an infinite number of unique series for 'e' from that identity? Or are you saying we can use these techniques to write an infinite number of series for things like 'e'?

 Quote by lurflurf we can give series for e all day, but to what end?
'e' is a fundamental mathematical constant, anything that gives more insight or variability to it I would think is important. For example, we can use the identity I posted and with some manipulation can produce the same for ex.

 Quote by lurflurf The notation $$\left(x \, \dfrac{d}{dx}\right) ^n\mathrm{f}(x)$$ means apply n times ie $$\left( x \, \dfrac{d}{dx}\right) ^4\mathrm{f}(x)=\left( x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\mathrm{f}(x)\right)\right)\right)\right)$$
Okay, I think if you explain $$\left( x \, \dfrac{d}{dx}\right)$$ that would help. I imagine you can do better job explaining all this than the tutor :)
P: 515
 Quote by lurflurf $$\left( x \, \dfrac{d}{dx}\right) ^4\mathrm{f}(x)=\left( x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\mathrm{f}(x)\right)\right)\right)\right)$$
Wait a second...
are you just saying take the fourth derivative of 'f(x)' with a multiplication of 'x' in each step?
 HW Helper P: 2,263 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 $$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x$$
 P: 515 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?
 P: 515 $$\left( x\, \dfrac{d}{dx}\right)^n x^a=p(n) x^n$$ 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?
HW Helper
P: 2,263
sorry typo in the middle equation
that should have been
 Quote by lurflurf 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,... $$\left( x\, \dfrac{d}{dx}\right)^m x^n=n^m x^n \\ p\left( x\, \dfrac{d}{dx}\right) x^n=p(n) x^n$$ so we can find find $$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x$$
P: 515
 Quote by lurflurf sorry typo in the middle equation...
It happens, I was just about to ask about the x^a but noticed you caught that one too.

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.
 HW Helper P: 2,263 $$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x=q(x)e^x$$ 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 $$q(x)=e^{-x}p\left( x\dfrac{d}{dx}\right)e^x$$
P: 515
 Quote by lurflurf $$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x=q(x)e^x$$ 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 $$q(x)=e^{-x}p\left( x\dfrac{d}{dx}\right)e^x$$
Okay, I see where the confusion is coming in. We can use this 'technique' to generate an infinite number of solutions but we have to go through this process each time for each different polynomial of 'n' in order to generate them.

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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