PHY302 BF Quiz 60 points - Open tutorial - 60 minutes
Note: This quiz consists of 4 short problems followed by one longer problem with two parts. Be
sure to attempt all of the required problems. You must show your work to receive credit! Partial credit
will be given. If you cannot do a calculation but know what calculation you should be doing, describe
it in words.
1. (15 points) Find an expression for J5/2(x) using properties of the Bessel function, given
J1/2(x) = r2
πx sin x,
J3/2(x) = r2
πx 1
xsin x−cos x.
Solution:
We will choose property BF 35:
Jm−1(x) + Jm+1(x) = 2m
xJm(x)
set m=3
2⇒
J1/2(x) + J5/2(x) = 3
xJ3/2(x)
J5/2(x) = 3
xJ3/2(x)−J1/2(x)
J5/2(x) = r2
πx 3
x2sin x−3
xcos x−sin x
J5/2(x) = r2
πx 3
x2−1sin x−3
xcos x.
There are several other valid derivations, using e.g. BF 38 or BF 36, but they require a bit more
algebra.
2. (5 points) Compute
Γ(101)
98! .
Solution:
We use Γ(n)=(n−1)!:
Γ(101)
98! =100!
98! = 100 ∗99 = 9900.
3. (10 points) Compute
∞
X
m=−∞
3mJm(x).
Solution:
This follows from the Bessel generating function:
F(x, h) =
∞
X
m=−∞
hmJm(x) = exp x
2h−1
h
set h= 3 ⇒
∞
X
m=−∞
3mJm(x) = exp x
23−1
3= exp 4x
3.
4. (5 bonus points) Compute
∞
X
m=−∞
Jm(x).
Solution:
This also follows from the Bessel generating function:
F(x, h) =
∞
X
m=−∞
hmJm(x) = exp x
2h−1
h
set h= 1 ⇒
∞
X
m=−∞
Jm(x) = exp x
21−1
1= 1.
There are other ways to prove this result, but this one’s nice and compact.
5. (30 points) Consider the differential equation
x3d2y
dx2+1
xy= 0.
(a) Find the change of variables required to transform this equation into Bessel’s equation.
Solution:
From Handy Hint # 13, we have
x2d2y
dx2+ (1 −2α)xdy
dx +hα2+β2(C2x2β−m2)iy= 0,for y(x) = xαf(Cxβ),
where fis a Bessel function of order m.
To recover our equation, we first note that we have no dy
dx terms, so we must have
1−2α= 0 α=1
2.
Next, we need to match the coefficient of the d2y
dx2terms. We divide our starting equation
by x, obtaining
x2d2y
dx2+1
x2y= 0.
2
Now we can match the ycoefficient directly. We find
1
x2=1
4−β2m2+β2C2x2β.
where we have already plugged in α= 1/2. Matching the power on the xterm requires
β=−1. We then have to match the constant terms, finding m=±1/2β=±1/2. Lastly,
we set the coefficients of the 1/x2terms to match, and find C2= 1/β2= 1, so C= 1.
Overall, we have the change of variables
y(x) = √xf (1/x), m =±1/2.
(b) Write its general solution. Note: the general solution should contain two linearly indepen-
dent solutions. Solution:
The general solution is easy to write once we have the Bessel form. We find
y=A√xJ1/21
x+B√xY1/21
x
Because m= 1/2 is not an integer, it’s also ok to use J−1/2as the second linearly indepen-
dent solution.
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