Let r = xi +yj +zk and r = r. Find each of the following. (Give your answers in terms of r.) a) gradient * r b) gradient * (r r) c) gradient2r3
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The gradient of a function w=f(x,y,z) is the vector function: displaymath62 ... The level curves are the ellipses 4x^2+y^2=c. ... Converting this to a unit vector, we have <2,1> /sqrt ...
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By symmetry with the rest of the series, I have to suppose the first term is actually 1/(2·4) so the question is to find (we take a common 1212 factor out of all terms)
12∑∞m=0∏mn=02n+12n+412∑m=0∞∏n=0m2n+12n+4
We can stick the inner part into Wolfram Alpha, just to see if it is reasonable to calculate, and we get that
∏mn=02n+12n+4=1π√Γ(m+32)Γ(m+3)∏n=0m2n+12n+4=1πΓ(m+32)Γ(m+3)
Once we know the answer, it is relatively easy to prove it, take out a common factor of 2 from each internal fraction, and multiply everything out:
∏mn=02(n+12)2(n+2)=∏mn=0n+12n+2=∏mn=0n+12(m+2)!∏n=0m2(n+12)2(n+2)=∏n=0mn+12n+2=∏n=0mn+12(m+2)!
The product at the numerator is exactly Γ(m+32)Γ(12)=1π√Γ(m+32)Γ(m+32)Γ(12)=1πΓ(m+32), by the multiplicative property of ΓΓ.
So now we just have to find (again, take out the common 1π√1π factor):
12π√∑∞m=0Γ(m+32)Γ(m+3)12π∑m=0∞Γ(m+32)Γ(m+3)
If you plug ∑∞m=0Γ(m+32)Γ(m+3)∑m=0∞Γ(m+32)Γ(m+3) back to Alpha, the answer is π−−√π , but in this case I haven’t been able to find out the actual operations to get to this result.
☛ One possible way is to use duplication formula to get Γ(m+32)=π√Γ(2m+2)22n+1Γ(m+1)Γ(m+32)=πΓ(2m+2)22n+1Γ(m+1) so the fraction becomes (as Γ(n+1)=n!Γ(n+1)=n!):
π−−√∑∞m=0(2m+1)!22m+1m!(m+2)!π∑m=0∞(2m+1)!22m+1m!(m+2)!
In any case the raw result is
12∑∞m=0∏mn=02n+12n+412∑m=0∞∏n=0m2n+12n+4
We can stick the inner part into Wolfram Alpha, just to see if it is reasonable to calculate, and we get that
∏mn=02n+12n+4=1π√Γ(m+32)Γ(m+3)∏n=0m2n+12n+4=1πΓ(m+32)Γ(m+3)
Once we know the answer, it is relatively easy to prove it, take out a common factor of 2 from each internal fraction, and multiply everything out:
∏mn=02(n+12)2(n+2)=∏mn=0n+12n+2=∏mn=0n+12(m+2)!∏n=0m2(n+12)2(n+2)=∏n=0mn+12n+2=∏n=0mn+12(m+2)!
The product at the numerator is exactly Γ(m+32)Γ(12)=1π√Γ(m+32)Γ(m+32)Γ(12)=1πΓ(m+32), by the multiplicative property of ΓΓ.
So now we just have to find (again, take out the common 1π√1π factor):
12π√∑∞m=0Γ(m+32)Γ(m+3)12π∑m=0∞Γ(m+32)Γ(m+3)
If you plug ∑∞m=0Γ(m+32)Γ(m+3)∑m=0∞Γ(m+32)Γ(m+3) back to Alpha, the answer is π−−√π , but in this case I haven’t been able to find out the actual operations to get to this result.
☛ One possible way is to use duplication formula to get Γ(m+32)=π√Γ(2m+2)22n+1Γ(m+1)Γ(m+32)=πΓ(2m+2)22n+1Γ(m+1) so the fraction becomes (as Γ(n+1)=n!Γ(n+1)=n!):
π−−√∑∞m=0(2m+1)!22m+1m!(m+2)!π∑m=0∞(2m+1)!22m+1m!(m+2)!
In any case the raw result is
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