- #1
vineethbs
- 8
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Use of "method of undetermined coefficients"
Hi all,
Suppose I have a equation
[tex]f(z+1) - f(z) = z^{1/2} , \forall z \geq 0 [/tex]eq (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
[tex]c_{1} z^{1/2} + c_{2} + c_{3} z^{-1/2} + \cdots[/tex]
and then substitute in eq (1),
then can I do a Taylor series expansion for an arbitrarily large z ?
for eg :
[tex]c_{1} (z + 1)^{1/2} = c_{1} z^{1/2} * (1 + \frac{1}{z})^{1/2}[/tex]
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
Hi all,
Suppose I have a equation
[tex]f(z+1) - f(z) = z^{1/2} , \forall z \geq 0 [/tex]eq (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
[tex]c_{1} z^{1/2} + c_{2} + c_{3} z^{-1/2} + \cdots[/tex]
and then substitute in eq (1),
then can I do a Taylor series expansion for an arbitrarily large z ?
for eg :
[tex]c_{1} (z + 1)^{1/2} = c_{1} z^{1/2} * (1 + \frac{1}{z})^{1/2}[/tex]
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
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