Perturbation approximation of the period of a pendulum

[limofunder]

Homework Statement

Find the perturbation approximation of the following in terms of powers of θ₀.
T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{d\theta}{\sqrt{cos\theta - cos\theta_0}}
It is helpful to first perform the change of variable u = θ/θ_0 in the integral

Homework Equations

so the relevant equations are the taylor expansion of θ₀, which would be
θ₀=θ₀₀+εθ₀₁+ε² θ₀₂/2

The Attempt at a Solution

so then making the change of variable suggested, we have
T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{\theta_0 du}{\sqrt{cos(u\theta_0) - cos\theta_0}}
is it correct to say that the perturbation approximation is then
\frac{\theta_0_0 + \epsilon \theta_0_1+<br /> \frac{\epsilon^2 \theta_0_2}{2}...}{\sqrt{cos(u\theta_0_0)-\epsilon sin(u\theta_0_1)-\epsilon^2cos(u\theta_0_2)-cost\theta_0_0)-\epsilon sin(theta_0_1)-\epsilon^2cos(\theta_0_2)...-cos\theta_0}}=0

or am i doing something wrong?

 

[gabbagabbahey]

Homework Statement

Find the perturbation approximation of the following in terms of powers of θ₀.
T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{d\theta}{\sqrt{cos\theta - cos\theta_0}}
It is helpful to first perform the change of variable u = θ/θ₀ in the integral

Homework Equations

so the relevant equations are the taylor expansion of θ₀, which would be
θ₀=θ₀₀+εθ₀₁+ε² θ₀₂/2

Huh?!

What are \theta_{0_0}, \theta_{0_1} and \theta_{0_2} supposed to represent?

The Attempt at a Solution

so then making the change of variable suggested, we have
T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{\theta_0 du}{\sqrt{cos(u\theta_0) - cos\theta_0}}
is it correct to say that the perturbation approximation is then
\frac{\theta_0_0 + \epsilon \theta_0_1+<br /> \frac{\epsilon^2 \theta_0_2}{2}...}{\sqrt{cos(u\theta_0_0)-\epsilon sin(u\theta_0_1)-\epsilon^2cos(u\theta_0_2)-cost\theta_0_0)-\epsilon sin(theta_0_1)-\epsilon^2cos(\theta_0_2)...-cos\theta_0}}=0

or am i doing something wrong?

What you have written seems nonsensical to me.

The way I would interpret the question is that you are supposed to Taylor expand the integrand about the point \theta=\theta_0, then integrate your Taylor Series term by term in order to end up with a series of the form:

\sum_{n=0}^\infty a_n \theta_0^n

that represents your integral.

 

[limofunder]

i erroneously thought that the
θ₀₀+εθ₀₁+ε2 θ₀₂/2
terms represented the expansion about θ₀.

so then following the help you gave, the expansion of the integrand would be
following Taylor series:
f(0)+f'(0)+f''(0)... where we only go the second order since the original problem asks us to go to O(θ²) (I forgot to list this in the original question).
we get
\frac{\theta_0}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{2 (cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)}{2 (cosu\theta_0-cos\theta_0)^{3/2}}*\theta_0
+ \frac{2 (cosu\theta_0-cos\theta_0)^{3/2}[(sin\theta_0-usinu\theta_0)-\theta_0(cos\theta_0-u^2cosu\theta_0)]-6[(cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)]((sin\theta_0-usinu\theta_0)(cosu\theta_0-cos\theta_0)^{1/2}}{4(cosu\theta_0-cos\theta_0)^3}
of course, last term would also be multiplied by θ₀²/2.when substituting in 0 for fⁿ, however, everything becomes zero, so what did i do wrong?

 

[gabbagabbahey]

i erroneously thought that the
θ₀₀+εθ₀₁+ε2 θ₀₂/2
terms represented the expansion about θ₀.

so then following the help you gave, the expansion of the integrand would be
following Taylor series:
f(0)+f'(0)+f''(0)... where we only go the second order since the original problem asks us to go to O(θ²) (I forgot to list this in the original question).
we get
\frac{\theta_0}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{2 (cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)}{2 (cosu\theta_0-cos\theta_0)^{3/2}}*\theta_0
+ \frac{2 (cosu\theta_0-cos\theta_0)^{3/2}[(sin\theta_0-usinu\theta_0)-\theta_0(cos\theta_0-u^2cosu\theta_0)]-6[(cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)]((sin\theta_0-usinu\theta_0)(cosu\theta_0-cos\theta_0)^{1/2}}{4(cosu\theta_0-cos\theta_0)^3}
of course, last term would also be multiplied by θ₀²/2.when substituting in 0 for fⁿ, however, everything becomes zero, so what did i do wrong?

Since you know 0\leq u\leq 1 for your entire integration interval, try expanding f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}} about u=0 instead.

 

[limofunder]

So the expansion of
<br /> f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}}
about u=0 gives:
f(u)+θ₀f'(u)+θ₀²/2
yielding
\frac{1}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{u\sin(u\theta_0)}{2 (\cos(u\theta_0)-\cos(\theta_0)^{3/2}}+\frac{\sin(u\theta_0)+u^2\cos(u\theta_0))(\cos(u\theta_0)-\cos(\theta_0))+\frac{3}{2}u^2\sin^2(u\theta_0)}{4(\cos(u\theta_0)-\cos(\theta_0)^{3/2}}
thus about u=0, everything but the first term becomes 0, giving:
\frac{1}{\sqrt{1-\cos(\theta_0)}}}
how can we solve about bigOH(θ₀²)?

 

[gabbagabbahey]

So the expansion of
<br /> f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}}
about u=0 gives:
f(u)+θ₀f'(u)+θ₀²/2

No, the expansion about u=0 is:

f(u)=f(0)+f&#039;(0)u+\frac{f&#039;&#039;(0)}{2!}u^2+\frac{f&#039;&#039;&#039;(0)}{3!}u^3+\ldots

And since f(u) is a function of u, the differentiation is with respect to u, not \theta_0.

 

[limofunder]

I think I finally understand it, then we use the fact that u=θ/θ₀, and rewrite the expansion in terms of θ₀?