What is this called? A general solution?

[mesa]

If we have a function such as,

$$e=\sum_{n=0}^{\infty} f(k)$$

Where 'k' can be (almost) any real value we choose and the summation series (although unique for each value of 'k') will always be equal to 'e' exactly, what do we call this?

 

[micromass]

This makes very little sense. Either $f(k) = 0$, and then

$$\sum_{n=0}^{+\infty} f(k) = 0$$

or $f(k)\neq 0$ and

$$\sum_{n=0}^{+\infty} f(k)$$

diverges. It won't ever equal $e$.

 

[mesa]

This makes very little sense. Either $f(k) = 0$, and then

$$\sum_{n=0}^{+\infty} f(k) = 0$$

or $f(k)\neq 0$ and

$$\sum_{n=0}^{+\infty} f(k)$$

diverges. It won't ever equal $e$.

f(k) is basically to represent a part of the function in an infinite series where we can use any value (any real with the exception of one value) in place of 'k' and still get a series equal 'e'. Just to clarify this does not mean some simple function where the 'k's cancel, each series will be unique for each value of 'k'.

This would be much easier if I could just post the identity but the moderators have made it very clear that this is against guidelines as 'no personal theories are allowed' even though this is mathematics and based on proofs.

Anyway the best I have seen is on wikipedia with this family of functions:


$$e = \sum_{k=1}^\infty \frac{k^2}{2(k!)}$$
$$e = \sum_{k=1}^\infty \frac{k^3}{5(k!)}$$
$$e = \sum_{k=1}^\infty \frac{k^4}{15(k!)}$$
$$e = \sum_{k=1}^\infty \frac{k^5}{52(k!)} etc., etc.$$




Maybe this is already known? I don't know but my email is listed in my profile.

 

[micromass]

OK, so your original post was wrong and you actually mean

$$e = \sum_{k=0}^{+\infty} f(k)$$

That would make more sense.

This would be much easier if I could just post the identity but the moderators have made it very clear that this is against guidelines as 'no personal theories are allowed' even though this is mathematics and based on proofs.

I vaguely remember you saying that mathematica or wolfram alpha could verify it. In that case it's not really a new personal theory, so I'll allow it.

 

[mesa]

OK, so your original post was wrong and you actually mean

$$e = \sum_{k=0}^{+\infty} f(k)$$

That would make more sense.



I vaguely remember you saying that mathematica or wolfram alpha could verify it. In that case it's not really a new personal theory, so I'll allow it.

That was for a reciprocal gamma function, this one is not verified by mathematica but for summations up to 1000 terms has been perfect. If the moderators have no issue without a formal proof then let's take a closer look.

 

[micromass]

That was for a reciprocal gamma function, this one is not verified by mathematica but for summations up to 1000 terms has been perfect. If the moderators have no issue without a formal proof then let's take a closer look.

If you post your formula and if you ask us whether we saw this before or whether we have a reference, then I have no qualms about you posting it.

 

[mesa]

It seems it would be easier if we just do a reference request (you kick *** micromass),

$$e = \sum_{n=0}^\infty \frac{n+k}{(k+1)(n!)}$$

It seems to work for all real values (except k=-1 of course). Now this is where you guys come in and say, well that's just so and so's thing, everyone knows that one, sheesh...

 

[micromass]

It seems it would be easier if we just do a reference request (you kick *** micromass),

$$e = \sum_{n=0}^\infty \frac{n+k}{(k+1)(n!)}$$

It seems to work for all real values (except k=-1 of course). Now this is where you guys come in and say, well that's just so and so's thing, everyone knows that one, sheesh...

$$\begin{eqnarray*}<br /> \sum_{n=0}^{+\infty} \frac{n+k}{(k+1) (n!)} & = & \frac{1}{k+1}\sum_{n=0}^{+\infty} \left(\frac{n}{n!} + \frac{k}{n!}\right)\\<br /> & = & \frac{1}{k+1}\left(\sum_{n=1}^{+\infty} \frac{1}{(n-1)!} + k \sum_{n=0}^{+\infty} \frac{1}{n!}\right)\\<br /> & = & \frac{1}{k+1} (e + ke)\\<br /> & = & e<br /> \end{eqnarray*}$$

 

[mesa]

$$\begin{eqnarray*}<br /> \sum_{n=0}^{+\infty} \frac{n+k}{(k+1) (n!)} & = & \frac{1}{k+1}\sum_{n=0}^{+\infty} \left(\frac{n}{n!} + \frac{k}{n!}\right)\\<br /> & = & \frac{1}{k+1}\left(\sum_{n=0}^{+\infty} \frac{1}{(n-1)!} + k \sum_{n=0}^{+\infty} \frac{1}{n!}\right)\\<br /> & = & \frac{1}{k+1} (e + ke)\\<br /> & = & e<br /> \end{eqnarray*}$$

Damn you're good
I would imagine something so easily proved is well known?

 

[mesa]

So what do we call a solution of this type?

And I would imagine something so easily proved is well know?

 

[mesa]

-bump-

I appreciate you all waiting for my return before posting :)

So here is the final form:

$$e = \sum_{n=0}^\infty \frac{mn+k}{(k+m)(n!)}$$

So this series will create a unique series expansion for 'e' for any input for 'k' or 'm' (real or imaginary) with the exception of course where:

$$k+m=0$$

And people say work in series summations is complete, take that classical analysis!

 

[HomogenousCow]

-bump-

I appreciate you all waiting for my return before posting :)

So here is the final form:

$$e = \sum_{n=0}^\infty \frac{mn+k}{(k+m)(n!)}$$

So this series will create a unique series expansion for 'e' for any input for 'k' or 'm' (real or imaginary) with the exception of course where:

$$k+m=0$$

And people say work in series summations is complete, take that classical analysis!

You can manipulate that expression like micromass did with the other one and see that it isn't anything new.

 

[mesa]

You can manipulate that expression like micromass did with the other one and see that it isn't anything new.

Micromass laid out a beautifully simple proof and it can easily be used to show the latter is also true ('k' and 'm') by replacing the '1' with an 'm' so this function is easy to prove as well.

An interesting thing about proofs, I had one of my Calculus Professors review another identity I wrote and he was eventually able to write a simple proof for it as well and he quickly declared it therefore must be known. The research Professors at my University took a different attitude towards the identity and have instructed me to "not waste my time with him".

I am new to this so don't know the difference between what is known or not. Can you perhaps show a reference to the identity? This general series can produce an infinite number of unique series expansions for 'e' so it would seem this should be commonly known but I have been unable to turn it up.

 

[micromass]

You're right, your identity isn't commonly known and it is probably even new. But that doesn't make it very useful. So finding a new identity is easy, but you should also really explain why people would be interested in such an identity. I don't think your identity doesn't offer anything new.

 

[lurflurf]

Generalize further you have
$$\left(C_1+C_2 x\dfrac{d}{dx}\right)\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty \left(C_1+C_2 x\dfrac{d}{dx}\right) a_n x^n=\sum_{n=0}^\infty \left(C_1+C_2 n\right) a_n x^n$$
or further still
$$p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty p\left( x\dfrac{d}{dx}\right) a_n x^n=\sum_{n=0}^\infty p\left( n\right) a_n x^n$$
for a polynomial p
which can be proved by using uniform convergence to justify term by term differentiation

 

[mesa]

You're right, your identity isn't commonly known and it is probably even new. But that doesn't make it very useful. So finding a new identity is easy,

Are you referring to something along the lines of what lurflurf posted?

I've hardly brought Calculus into series yet (outside of writing series for the Euler Mascheroni constant and 0th degree polygamma general solutions I haven't used it at all).

What other series do we have for creating an infinite number of unique series expansions for 'e'? I am up to a whopping '2' (including this one :P), math is so broad today it can be difficult to find your way around...

...but you should also really explain why people would be interested in such an identity. I don't think your identity doesn't offer anything new.

I would love to! Although it seems PF is not the place for such discussions.

 

[lurflurf]

For usefulness I would say the formula is more useful the other way
That is I cannot think (though there are surely some) of an example where
$$\sum_{n=0}^\infty a_n x^n=\left[ p\left( x\dfrac{d}{dx}\right)\right]^{-1} \sum_{n=0}^\infty p\left( n\right) a_n x^n$$
would be useful, however
$$\sum_{n=0}^\infty p\left( n\right) a_n x^n=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty a_n x^n$$
is a very useful way to find
$$\sum_{n=0}^\infty p\left( n\right) a_n x^n$$
when we know
$$\sum_{n=0}^\infty a_n x^n$$
These come up all the times in many areas such as probability where besides examples involving e^x we have many involving 1/(1-x) like
$$\sum_{n=0}^\infty n^2 x^n$$
and sums like
$$\sum_{n=0}^N n^3=\lim_{x\rightarrow 1} \sum_{n=0}^N n^3 x^n$$
That you can now easily calculate

Here is some light reading
http://www.ams.org/journals/tran/1928-030-01/S0002-9947-1928-1501425-4/S0002-9947-1928-1501425-4.pdf
http://www.emis.de/journals/HOA/IJMMS/Volume13_4/643718.pdf
http://www.rowan.edu/open/depts/math/osler/Taylor%27s%20series%20Generalized%20.pdf
http://mathworld.wolfram.com/BuermannsTheorem.html
http://mathworld.wolfram.com/DarbouxsFormula.html
http://mathworld.wolfram.com/TeixeirasTheorem.html

 

[mesa]

For usefulness I would say the formula is more useful the other way
That is I cannot think (though there are surely some) of an example where
$$\sum_{n=0}^\infty a_n x^n=\left[ p\left( x\dfrac{d}{dx}\right)\right]^{-1} \sum_{n=0}^\infty p\left( n\right) a_n x^n$$
would be useful, however
$$\sum_{n=0}^\infty p\left( n\right) a_n x^n=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty a_n x^n$$
is a very useful way to find
$$\sum_{n=0}^\infty p\left( n\right) a_n x^n$$
when we know
$$\sum_{n=0}^\infty a_n x^n$$
These come up all the times in many areas such as probability where besides examples involving e^x we have many involving 1/(1-x) like
$$\sum_{n=0}^\infty n^2 x^n$$
and sums like
$$\sum_{n=0}^N n^3=\lim_{x\rightarrow 1} \sum_{n=0}^N n^3 x^n$$
That you can now easily calculate

Here is some light reading
http://www.ams.org/journals/tran/1928-030-01/S0002-9947-1928-1501425-4/S0002-9947-1928-1501425-4.pdf
http://www.emis.de/journals/HOA/IJMMS/Volume13_4/643718.pdf
http://www.rowan.edu/open/depts/math/osler/Taylor%27s%20series%20Generalized%20.pdf
http://mathworld.wolfram.com/BuermannsTheorem.html
http://mathworld.wolfram.com/DarbouxsFormula.html
http://mathworld.wolfram.com/TeixeirasTheorem.html

I bet there is a lot of good information in your post however I am having trouble figuring out what you are trying to show. After looking at it for awhile yesterday I eventually threw in the towel and went to the advanced math tutoring lab and spoke with one of our grad students (specializes in algebraic number theory) but he wasn't sure what you were getting at either.

Unfortunately if the graduate Mathematics major is confused then this lowly Engineering undergrad doesn't stand a chance. I bet your post is chock full of useful information so I will attempt the tutoring lab again today, if all else fails then I will go knocking on some doors.

So in the interim, are you aware of any other series summations 'general solutions' capable of producing an infinite number of unique series summations for 'e'?

 

[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?
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
$$\left( x \, \dfrac{d}{dx}\right) ^4\mathrm{f}(x)=\left( x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\mathrm{f}(x)\right)\right)\right)\right)$$

 

[mesa]

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.

How much calculus do you know?

Only Calc 1-3 (for engineers) and Differential Equations

Do you understand Taylor's series?

Not particularly although I have been advised to look more closely at it.

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.

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.

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'?

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 e^x.

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 :)

 

[mesa]

$$\left( x \, \dfrac{d}{dx}\right) ^4\mathrm{f}(x)=\left( x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\left(x \, \dfrac{d}{dx}\mathrm{f}(x)\right)\right)\right)\right)$$

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

 

[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,...
$$\left( x\, \dfrac{d}{dx}\right)^m x^n=n^m x^n \\
\left( x\, \dfrac{d}{dx}\right)^n x^a=p(n) x^n$$

so we can find find
$$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x$$

 

[mesa]

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?

 

[mesa]

$$\left( x\, \dfrac{d}{dx}\right)^n x^a=p(n) x^n$$

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?

 

[lurflurf]

sorry typo in the middle equation
that should have been

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
$$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x$$

 

[mesa]

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.

 

[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
$$q(x)=e^{-x}p\left( x\dfrac{d}{dx}\right)e^x$$

 

[mesa]

$$\sum_{n=0}^\infty p\left( n\right) \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)\sum_{n=0}^\infty \frac{x^n}{n!}=p\left( x\dfrac{d}{dx}\right)e^x=q(x)e^x$$
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?