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Help with complex analysis

I need help proving that for zC z \in \mathbb{C} , given

G2,m(z):=k=0zkΓ(k+2)Γ(mk+1) G_{2,m} (z) := \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma (k+2) \Gamma (mk+1)} ,

show that
zG2,0(z)=ez1 zG_{2,0} (z) = e^{z} - 1 .


What I've done so far is
zG2,0(z)=k=0zk+1Γ(k+2)=k=0zkk!zk+1 zG_{2,0} (z) = \sum_{k=0}^{\infty} \frac{z^{k+1}}{\Gamma (k+2)} =\sum_{k=0}^{\infty} \frac{z^{k}}{k!} \frac{z}{k+1}.

But I don't know where to take it from there. I would appreciate for some guidance on what to do after this step.

Thanks
Original post by I Closed My Eyes
I need help proving that for zC z \in \mathbb{C} , given

G2,m(z):=k=0zkΓ(k+2)Γ(mk+1) G_{2,m} (z) := \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma (k+2) \Gamma (mk+1)} ,

show that
zG2,0(z)=ez1 zG_{2,0} (z) = e^{z} - 1 .


What I've done so far is
zG2,0(z)=k=0zk+1Γ(k+2)=k=0zkk!zk+1 zG_{2,0} (z) = \sum_{k=0}^{\infty} \frac{z^{k+1}}{\Gamma (k+2)} =\sum_{k=0}^{\infty} \frac{z^{k}}{k!} \frac{z}{k+1}.

But I don't know where to take it from there. I would appreciate for some guidance on what to do after this step.

Thanks

Surely this is just the series for e^z with the first term omitted (so = exp(z) - 1), so you're done?

Have to say, seems pretty advanced for year 9... :smile:
Original post by I Closed My Eyes
I need help proving that for zC z \in \mathbb{C} , given

G2,m(z):=k=0zkΓ(k+2)Γ(mk+1) G_{2,m} (z) := \sum_{k=0}^{\infty} \frac{z^{k}}{\Gamma (k+2) \Gamma (mk+1)} ,

show that
zG2,0(z)=ez1 zG_{2,0} (z) = e^{z} - 1 .


What I've done so far is
zG2,0(z)=k=0zk+1Γ(k+2)=k=0zkk!zk+1 zG_{2,0} (z) = \sum_{k=0}^{\infty} \frac{z^{k+1}}{\Gamma (k+2)} =\sum_{k=0}^{\infty} \frac{z^{k}}{k!} \frac{z}{k+1}.

But I don't know where to take it from there. I would appreciate for some guidance on what to do after this step.

Thanks


To add to the above, note that zkk!zk+1=zk+1(k+1)!\dfrac{z^k}{k!} \cdot \dfrac{z}{k+1} = \dfrac{z^{k+1}}{(k+1)!} from which it should be obvious that this is eze^z with the first term ommited
Original post by DFranklin
Surely this is just the series for e^z with the first term omitted (so = exp(z) - 1), so you're done?

Have to say, seems pretty advanced for year 9... :smile:


Original post by RDKGames
To add to the above, note that zkk!zk+1=zk+1(k+1)!\dfrac{z^k}{k!} \cdot \dfrac{z}{k+1} = \dfrac{z^{k+1}}{(k+1)!} from which it should be obvious that this is eze^z with the first term ommited


Ahhhhh thanks, I've just realised I was being quite silly and completely missed that :colondollar:

Also realised I accidentally put yr9 tag on haha
(edited 5 years ago)

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