# P.I of tanx and secx using Undetermined coefficients

1. Oct 24, 2012

### iVenky

I just don't get why the method of undetermined coefficients can't be applied to tan(x) and sec(x). What my book says is this-

"Since the number of terms applied by differentiating tan(x) and sec(x) is infinity".

What do they mean by that? Even the number of terms obtained by differentiating sin(x) is infinity but the method of undetermined coefficients still works for sin(x).

I would be really happy if you could clear me this doubt

Thanks a lot :)

2. Oct 24, 2012

### Vargo

Basically, the method of undetermined coefficients works on exponential, sine cosine, and polynomials and products of said functions. If you want a reason for those particular functions, then I would say that for any of those, if you differentiate them enough times you end up with a multiple of the original function (with polynomials you differentiate it enough times and you get 0 times the original function).

That's not true for tan or sec.

Regarding your book's explanation, I think the author is trying to say that successive derivatives of those functions get more and more complicated, but I'm not sure.

3. Oct 24, 2012

### iVenky

I know that with polynomials we will end up with 0 if we keep on differentiating. May be the book means the polynomials only.

4. Oct 24, 2012

### Vargo

No I don't think so. No matter how many times you differentiate sine, you will have one term in that derivative. Same with exponential or cosine. But if you differentiate tan or sec multiple times you end up with formulas that you can barely write down on a piece of paper.

However, I'm not really sure that that is the most relevant fact concerning the method of undetermined coefficients, but I think that's what the author means.

5. Oct 27, 2012

### Ocifer

Consider: sin(t)

d/dt [ sin(t) ] = cos(t)
d/dt [ cos(t) ] = - sin(t)
d/dt [ -sin(t) ] = - cos(t)
d/dt [ -cos(t) ] = sin(t)

....

It's quite easy to see that out of this process of differentiation, we only get sin(t) and cos(t) terms, ignoring coefficients. The salient fact is that if the RHS is sin(t), we get a finite number of linearly independent derivatives. We would then assume the following form for the particular solution:

A*sin(t) + B*cos(t)

By matching coefficients in the LHS and RHS of the ODE, we would be able to pin down what A,B are.

Now, consider tan(x), and take it's first 6 or 7 derivatives as an illustration.

It gets ugly (but ugliness is not the salient fact per se). What is salient is that if you continue to differentiate tan(x), you do not get a finite number of linearly independent derivatives. You can keep finding new terms in subsequent derivatives which will be linearly independent of anything in previous derivatives, and hence the derivative currently being taken will be linearly independent of the previous ones.

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It doesn't matter if there is a large number of terms in any given derivative -- what matters is that for this method we need functions which eventually eventually stop yielding linearly independent terms through differentiation. This is why sin(t), cos(t), or polynomials will work with the method of undetermined coefficients. Functions like tan(x), or t^(-n) will not work for reason discussed above.

6. Oct 28, 2012

### lurflurf

It does not mean some derivative of sine is 0 it means later derivatives can be written in terms of earlier ones (because sin''(x)+sin(x)=0)
Undetermined coefficients is much easier when we know what will produce the function
if we have
y''+y=tan(x/2)
you would need to know to guess
-x cos(x)+2 sin(x) log(cos(x/2))
or some such