Mixing
Representation for harmonic series $H_n$, for $n<-1$
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According to Wolfram Alpha, harmonic series $H_x$ has the following representation:
$$H_x=int_{0}^{1}frac{-1+t^x}{-1+t}dt=int_{0}^{infty}frac{1-e^{-xt}}{-1+e^t}dt,~Re(x)>-1$$
The corresponding graph for $H_x$ will then be
However, Wolfram Alpha also gives the graph for $Re(x)<1$
But either one of the two functions above is undefined for $Re(x)leq-1$. According to which function does Wolfram Alpha graph the harmonic series $H_x$, for $Re(x)<-1$? Is here such a function that satisfies both the values for $ Re(x)>-1$ and the values graphed by Wolfram Alpha when $Re(x)<-1$?
calculus sequences-and-series harmonic-numbers
asked Jan 9 at 23:56
LarryLarry
2,39131129
$endgroup$
add a comment |
$begingroup$
According to Wolfram Alpha, harmonic series $H_x$ has the following representation:
$$H_x=int_{0}^{1}frac{-1+t^x}{-1+t}dt=int_{0}^{infty}frac{1-e^{-xt}}{-1+e^t}dt,~Re(x)>-1$$
The corresponding graph for $H_x$ will then be
However, Wolfram Alpha also gives the graph for $Re(x)<1$
But either one of the two functions above is undefined for $Re(x)leq-1$. According to which function does Wolfram Alpha graph the harmonic series $H_x$, for $Re(x)<-1$? Is here such a function that satisfies both the values for $ Re(x)>-1$ and the values graphed by Wolfram Alpha when $Re(x)<-1$?
calculus sequences-and-series harmonic-numbers
asked Jan 9 at 23:56
LarryLarry
2,39131129
$endgroup$
$begingroup$
Extending is trivial, as we have $H_{x-1}=H_x-frac1x$.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:12
add a comment |
$begingroup$
According to Wolfram Alpha, harmonic series $H_x$ has the following representation:
$$H_x=int_{0}^{1}frac{-1+t^x}{-1+t}dt=int_{0}^{infty}frac{1-e^{-xt}}{-1+e^t}dt,~Re(x)>-1$$
The corresponding graph for $H_x$ will then be
However, Wolfram Alpha also gives the graph for $Re(x)<1$
But either one of the two functions above is undefined for $Re(x)leq-1$. According to which function does Wolfram Alpha graph the harmonic series $H_x$, for $Re(x)<-1$? Is here such a function that satisfies both the values for $ Re(x)>-1$ and the values graphed by Wolfram Alpha when $Re(x)<-1$?
calculus sequences-and-series harmonic-numbers
asked Jan 9 at 23:56
LarryLarry
2,39131129
$endgroup$
According to Wolfram Alpha, harmonic series $H_x$ has the following representation:
$$H_x=int_{0}^{1}frac{-1+t^x}{-1+t}dt=int_{0}^{infty}frac{1-e^{-xt}}{-1+e^t}dt,~Re(x)>-1$$
The corresponding graph for $H_x$ will then be
However, Wolfram Alpha also gives the graph for $Re(x)<1$
But either one of the two functions above is undefined for $Re(x)leq-1$. According to which function does Wolfram Alpha graph the harmonic series $H_x$, for $Re(x)<-1$? Is here such a function that satisfies both the values for $ Re(x)>-1$ and the values graphed by Wolfram Alpha when $Re(x)<-1$?
calculus sequences-and-series harmonic-numbers
calculus sequences-and-series harmonic-numbers
asked Jan 9 at 23:56
LarryLarry
2,39131129
asked Jan 9 at 23:56
LarryLarry
2,39131129
edited Jan 10 at 0:06
Larry
asked Jan 9 at 23:56
LarryLarry
2,39131129
asked Jan 9 at 23:56
LarryLarry
2,39131129
asked Jan 9 at 23:56
LarryLarry
2,39131129
2,39131129
$begingroup$
Extending is trivial, as we have $H_{x-1}=H_x-frac1x$.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:12
add a comment |
$begingroup$
Extending is trivial, as we have $H_{x-1}=H_x-frac1x$.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:12
$begingroup$
Extending is trivial, as we have $H_{x-1}=H_x-frac1x$.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:12
$begingroup$
Extending is trivial, as we have $H_{x-1}=H_x-frac1x$.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:12
add a comment |
1 Answer
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Harmonic Numbers can also be extended to all $zinmathbb{C}$, except the negative integers, as $$H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)$$
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
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$begingroup$
Harmonic Numbers can also be extended to all $zinmathbb{C}$, except the negative integers, as $$H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)$$
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
Harmonic Numbers can also be extended to all $zinmathbb{C}$, except the negative integers, as $$H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)$$
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
$endgroup$
add a comment |
$begingroup$
Harmonic Numbers can also be extended to all $zinmathbb{C}$, except the negative integers, as $$H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)$$
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
$endgroup$
Harmonic Numbers can also be extended to all $zinmathbb{C}$, except the negative integers, as $$H(z)=sum_{k=1}^inftyleft(frac1k-frac1{k+z}right)$$
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
answered Jan 10 at 0:03
robjohn♦robjohn
267k27308631
267k27308631
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Extending is trivial, as we have $H_{x-1}=H_x-frac1x$.
$endgroup$
– Simply Beautiful Art
Jan 10 at 3:12