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Digamma function and Harmonic numbers

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove the following :

$\displaystyle \psi(n)= -\gamma \,+\,\sum^{n-1}_{k=1}\frac{1}{k}$​
 

chisigma

Well-known member
Feb 13, 2012
1,704
Prove the following :

$\displaystyle \psi(n)= -\gamma \,+\,\sum^{n-1}_{k=1}\frac{1}{k}$​
In...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... it has been demonstrated that, if $\displaystyle \phi(*)$ is the 'digamma function', then is ...

$\displaystyle \phi(n)= - \gamma + \sum_{k=0}^{n-1} \frac{1}{1+k}$ (1)

The little 'discrepancy' is probably justified from the fact that the definition of digamma function is a little controversial...

Kind regards

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
In...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/

... it has been demonstrated that, if $\displaystyle \phi(*)$ is the 'digamma function', then is ...

$\displaystyle \phi(n)= - \gamma + \sum_{k=0}^{n-1} \frac{1}{1+k}$ (1)

The little 'discrepancy' is probably justified from the fact that the definition of digamma function is a little controversial...

Kind regards

$\chi$ $\sigma$
Interesting , I will surely read that , thanks a lot ...
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
The little 'discrepancy' is probably justified from the fact that the definition of digamma function is a little controversial...


$\chi$ $\sigma$
Can you please elaborate on that ?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Can you please elaborate on that ?
In...

Digamma Function -- from Wolfram MathWorld

... the digamma function is defined as...

$\displaystyle \psi(x)= \frac{d}{d x} \ln \Gamma(x)$ (1)

... where...

$\displaystyle \Gamma (x)= \int_{0}^{\infty} t^{x-1}\ e^{-t}\ dt$ (2)

... that leads to write, after long efforts...

$\displaystyle \psi(n)= - \gamma + \sum_{k=1}^{n-1} \frac{1}{k}$ (3)

I adopted a slighty different approach defining the digamma function as...

$\displaystyle \phi(x) = \frac{d}{d x} \ln x!$ (4)

... where...

$\displaystyle x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt$ (5)

... and after long efforts I arrived to write...

$\displaystyle \phi(n) = - \gamma + \sum_{k=0}^{n-1} \frac{1}{k+1}$ (6)

Of course is a trivial question of definitions...

Kind regards

$\chi$ $\sigma$
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$\displaystyle \psi(x)= - \gamma + \sum_{k=1}^{n-1} \frac{1}{k}$ (3)


$\displaystyle \phi(x) = - \gamma + \sum_{k=1}^{n} \frac{1}{k}$ (6)
I assumed you meant $\psi(n)$ and $\phi(n)$ , I know this is correct for all integers but let us try to find $\psi(1)$

From (3) it is \(\displaystyle -\gamma \) but from (6) it is \(\displaystyle -\gamma+1\)

so which is correct !
 

chisigma

Well-known member
Feb 13, 2012
1,704
I have to confess one 'phobia' of my: I don't like the function $\displaystyle \Gamma (x)$ because any time I meet it I have to do terrible efforts to decide if it is $\displaystyle \Gamma(n)= (n-1)!$ or $\displaystyle \Gamma(n)= (n+1)!$... that's why I prefer the factorial function ...

$\displaystyle x!= \int_{0}^{\infty} t^{x}\ e^{-t}\ dt$ (1)

... that for $\displaystyle x=n$ of course is n!. The conclusion is that a digamma function descending from $\displaystyle \Gamma(x)$ is called $\displaystyle \psi(x)$ and the digamma function descending from $\displaystyle x!$ I called $\phi(x)$ and is...

$\displaystyle \psi(n)= \phi (n-1)$ (2)

Kind regards

$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
I have to confess one 'phobia' of my: I don't like the function $\displaystyle \Gamma (x)$ because any time I meet it I have to do terrible efforts to decide if it is $\displaystyle \Gamma(n)= (n-1)!$ or $\displaystyle \Gamma(n)= (n+1)!$... that's why I prefer the factorial function ...
I have never understood the reason for this -1 stuff with the gamma function.
Do you know where it came from?
Reading the wiki article makes it suggestive that it was a bad choice from the start.

On wiki I can see that there is also a pi function introduced by Gauss, that does behave like the factorial function.
Shouldn't that one be preferred over this gamma function?
$$\Pi(n) = \Gamma(n+1) = n!$$
Wiki only states the $\Gamma$ is dominant in literature.
 

chisigma

Well-known member
Feb 13, 2012
1,704
I assumed you meant $\psi(n)$ and $\phi(n)$ , I know this is correct for all integers but let us try to find $\psi(1)$

From (3) it is \(\displaystyle -\gamma \) but from (6) it is \(\displaystyle -\gamma+1\)

so which is correct !
Your question is very useful because permits me to remark an important topic... from my definition of digammma function descends that for x = n is...

$\displaystyle \phi(n) = - \gamma + \sum_{k=0}^{n-1} \frac{1}{k+1}$ (1)

... so that...

$\displaystyle \psi(1) = \phi(0) = - \gamma$ (2)

... which is correct. From the 'standard definition' reported on 'MonsterWolfram' descends that for x=n is...

$\displaystyle \psi(n)= - \gamma + \sum_{k=1}^{n-1} \frac{1}{k}$ (3)

... so that...

$\displaystyle \psi(1) = - \gamma + \sum_{k=1}^{0} \frac{1}{k} = \infty$ (4)

Gulp! (Tmi)...


Kind regards


$\chi$ $\sigma$
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Your question is very useful because permits me to remark an important topic... from my definition of digammma function descends that for x = n is...

$\displaystyle \phi(n) = - \gamma + \sum_{k=0}^{n-1} \frac{1}{k+1}$ (1)

... so that...

$\displaystyle \psi(1) = \phi(0) = 1 - \gamma$ (2)

... which is correct. From the 'standard definition' reported on 'MonsterWolfram' descends that for x=n is...

$\displaystyle \psi(n)= - \gamma + \sum_{k=1}^{n-1} \frac{1}{k}$ (3)

... so that...

$\displaystyle \psi(1) = - \gamma + \sum_{k=1}^{0} \frac{1}{k} = \infty$ (4)

Gulp! (Tmi)...


Kind regards


$\chi$ $\sigma$
Wait!
Wolfram says $\psi(1)=-\gamma + H_0 = -\gamma$.
See here.
A sum that has an upper bound below the lower bound is an empty sum.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
According to the Weierstrass representation of the gamma function we can get :

\(\displaystyle \psi(x) = -\gamma-\frac{1}{x}+\sum^{\infty}_{n=1}\frac{x}{(n+x)}\)

Now we can put x=1 so we have :

\(\displaystyle \psi(1) = -\gamma-1+\sum^{\infty}_{n=1}\frac{1}{n(n+1)}\)


We know that : \(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n(n+1)}=1\)

so \(\displaystyle \psi(1)=-\gamma\)

I described some digamma values here post #19
 

chisigma

Well-known member
Feb 13, 2012
1,704
Wait!
Wolfram says $\psi(1)=-\gamma + H_0 = -\gamma$.
See here.
A sum that has an upper bound below the lower bound is an empty sum.
'MonsterWolfram' sometime seems a little in contradiction with himself... in...

Harmonic Number -- from Wolfram MathWorld

...the following definition of the $\displaystyle H_{n}$ is reported...

$\displaystyle H_{n}= \sum_{k=1}^{n} \frac{1}{k}$ (1)

... and few lines after the $\displaystyle H_{n}$ are defined as the solution of the difference equation...

$\displaystyle H_{n} = H_{n-1} + \frac{1}{n},\ H_{1}=1$ (2)

... so that the element $\displaystyle H_{0}$ is in any case undefined...

Avoiding any type of useless controversial I symply say that the function $\displaystyle \psi(*)$ is, in my opinion, badly defined and leads sometimes to difficulties so that I prefer to use the function $\displaystyle \phi(*)$ that leads to 'secure results'...

My opinion of course...

Kind regards

$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Ah well, more specifically it says:
Based on their definition, harmonic numbers satisfy the obvious recurrence equation
(3)

with
.

In other words, this is a consequence of the definition - not the definition itself.

The article only says about the definition
A harmonic number is a number of the form
arising from truncation of the harmonic series. A harmonic number can be expressed analytically as


Then in the article it is extended to more domains.

As I see it, the article doesn't say anything about $H_0$ and it doesn't specifically make it undefined.
It just leaves it sort of hanging.
So the article is a bit sloppy with the definition... and the wiki article is too.