Use of method of undetermined coefficients

[vineethbs]

Use of "method of undetermined coefficients"

Hi all,

Suppose I have a equation

f(z+1) - f(z) = z^{1/2} , \forall z \geq 0eq (1)

then is it possible to solve this equation by the method of undetermined coefficients ?

It is usually seen in textbooks that the forcing function is taken to be sinusoidal or polynomial or exponential when the method of undetermined coefficients is used. Why is this so ? What kind of properties must the forcing function satisfy so that this method can be used ?

In the above eq (1), if suppose I assume that f(z) is say
c_{1} z^{1/2} + c_{2} + c_{3} z^{-1/2} + \cdots
and then substitute in eq (1),
then can I do a Taylor series expansion for an arbitrarily large z ?
for eg :
c_{1} (z + 1)^{1/2} = c_{1} z^{1/2} * (1 + \frac{1}{z})^{1/2}

and then expand out the 1 + .. term in a Taylor series expansion ?

Does this lead to a series solution ?Thanks in advance for spending your time on this

 

[HallsofIvy]

Hi all,

Suppose I have a equation

f(z+1) - f(z) = z^{1/2} , \forall z \geq 0eq (1)

then is it possible to solve this equation by the method of undetermined coefficients ?

It is usually seen in textbooks that the forcing function is taken to be sinusoidal or polynomial or exponential when the method of undetermined coefficients is used. Why is this so ? What kind of properties must the forcing function satisfy so that this method can be used ?

Essentially because those are the kind of solution you would expect for a linear difference (or differential) equation with constant coefficients because a linear operator with constant coefficients will map those kinds of functions into themselves. If the "forcing function" (more physics terminology- Bah!) is not of that kind you cannot use the "method of undetermined coefficients" because you cannot expect the correct function to be a square root. You will need to use something like "variation of parameters".

In the above eq (1), if suppose I assume that f(z) is say
c_{1} z^{1/2} + c_{2} + c_{3} z^{-1/2} + \cdots
and then substitute in eq (1),
then can I do a Taylor series expansion for an arbitrarily large z ?
for eg :
c_{1} (z + 1)^{1/2} = c_{1} z^{1/2} * (1 + \frac{1}{z})^{1/2}

and then expand out the 1 + .. term in a Taylor series expansion ?

Does this lead to a series solution ?


Thanks in advance for spending your time on this