Solve Integral Problem: Find A Value

[Rochefort]

Homework Statement

1.31144371455535≈\int^{1}_{0}\sqrt{1+A^{2}\cos^{2}\left( x\right) }dx Solve for A

Homework Equations

Also if it helps 2.17363253251301≈\int^{2}_{0}\sqrt{1+\dfrac{A^{2}}{4}\cos^{2}\left( \dfrac {x}{2}\right) }dx

The Attempt at a Solution

I know that A≈1 for both equations but I don't know how to solve the equations without looking at the answers.

 

[Saitama]

Homework Statement

1.31144371455535≈\int^{1}_{0}\sqrt{1+A^{2}\cos^{2}\left( x\right) }dx Solve for A

Homework Equations

Also if it helps 2.17363253251301≈\int^{2}_{0}\sqrt{1+\dfrac{A^{2}}{4}\cos^{2}\left( \dfrac {x}{2}\right) }dx

The Attempt at a Solution

I know that A≈1 for both equations but I don't know how to solve the equations without looking at the answers.

Can you please post the exact problem statement? Are you supposed to solve the first integral using the second one or do you seek a closed form for the first one?

 

[Simon Bridge]

Have you been learning about special approximation methods recently?

 

[SammyS]

Is this problem related to your other recent threads ?

 

[Rochefort]

Have you been learning about special approximation methods recently?

No, I don't know what that is. This is a part of my own investigation and a method for solving this would help me progress.

Can you please post the exact problem statement? Are you supposed to solve the first integral using the second one or do you seek a closed form for the first one?

I don't know the exact problem statement. All I can do is produce more equations of different values (on the LHS), for values of A. If you want, I can post more equations for when A=1 and you can solve the integral with the others. Also, it can be of closed form, it doesn't matter.

Is this problem related to your other recent threads ?

No it isn't.

 

[Simon Bridge]

This is a part of my own investigation ... I don't know the exact problem statement.

This is your own investigation and you don't have an exact problem statement?

In which case, your investigation has likely become derailed.

When you do your own investigating, you must be careful to properly phrase the problems or you will arrive at nonsense.

How are you getting the numbers on the LHS?
What are you trying to achieve?

The integrals you posted have exact forms involving elliptical integrals - which basically means you have to look them up. You expressed approximate forms so you could use an approximate method, like taking a power series of the integrand; or you could use a numerical method.

You treated the same sort of integral using a numerical method (Reimman sum) in your other two posts.

 

[Rochefort]

How are you getting the numbers on the LHS?

Well I know that C=\int^{B}_{0}\sqrt {1+\dfrac {A^{2}}{B^{2}}\cos^{2}\left(\dfrac {x}{B}\right) }dx And when provided with the values of A and B I can simulate results for C. What I did was, I converted it into a Riemann sum and produced an excel spreadsheet (consisting of exactly 1048574 rows per column) to output values of C to 14 decimal places.

What are you trying to achieve?

But I was wondering if I could find the value of A when knowing the values of B and C.

What I want to produce as a final result is the graph of A versus B for a set value of C.

 

[Simon Bridge]

Presumably you mean A to be squared in that expression?

So if you fix the value of C and B, you want to get A.
That's a bit clearer.

Use a Taylor series about B/2 ... use as many orders as you need.
The integration becomes trivial.

 

[Rochefort]

Use a Taylor series about B/2 ... use as many orders as you need.
The integration becomes trivial.

I have never used a Taylor series before so I don't know what you mean exactly. Here is my attempt so far:

So from knowing the power series representation of \cos\left( x\right)\cos^{2}\left( x\right) =1+\sum^{\infty }_{n=1}\dfrac {\left(-1\right)^{n}2^{2n-1}x^{2n}}{\left( 2n\right) !} Therefore \cos^{2}\left( \dfrac{x}{B}\right) =1+\sum^{\infty }_{n=1}\dfrac {\left(-1\right)^{n}2^{2n-1}\left( \dfrac {x}{B}\right)^{2n}}{\left( 2n\right) !} So1+\left(\dfrac {A}{B}\right)^{2}\cos^{2}\left(\dfrac {x}{B}\right)=1+\left(\dfrac{A}{B}\right)^{2}+\sum^{\infty}_{n=1}\dfrac{A^{2}\left(-1\right)^{n}2^{2n-1}x^{2n}}{\left( 2n\right) !B^{2n+2}}

C=\int^{B}_{0}\sqrt{1+\left(\dfrac{A}{B}\right)^{2}+\sum^{\infty}_{n=1}\dfrac{A^{2}\left(-1\right)^{n}2^{2n-1}x^{2n}}{\left(2n\right)!B^{2n+2}}}dx Since I haven't used a Taylor series before I don't know what to do next, and at the moment, integration seems far from trivial. As I wait for a response I will try to investigate further but any input could really help clarify things.

 

[Simon Bridge]

Ideally you want the entire integrand to be represented by a Taylor series, then you will end up integrating a polynomial.
You know how to do that.

Take:$$f(x)=\sqrt{1+\frac{A^2}{B^2}\cos^2\left(\frac{x}{B}\right)} \approx \sum_{k=1}^n c_k(x-a)^k : c_k=\frac{f^{(k)}(a)}{k!}$$
... a is the "center" of the series, it is common to put a=0
... n is the "order of the approximation", it is very common to stop at n=1 or n=2, you can take the sum to infinity if you want... it's just that you have to evaluate the coefficients by hand.
... $f^{(k)}(a)$ is the kth derivative of f(x) wrt x evaluated at x=a.

You need the approximation to be close to the actual function only in the region you are integrating over. You do that by choosing a and n carefully. If you pick a=0 you may need a higher value of n, I have not checked.

 

[Simon Bridge]

Aside - you continue from here:

C=\int^{B}_{0}\sqrt{1+\left(\dfrac{A}{B}\right)^{2}+\sum^{\infty}_{n=1}\dfrac{A^{2}\left(-1\right)^{n}2^{2n-1}x^{2n}}{\left(2n\right)!B^{2n+2}}}dx

... by making an approximation.
The first order approximation would stop the sum at n=1 to give:

C \approx \int^{B}_{0}\sqrt{<br /> <br /> 1+\dfrac{A^2}{B^2}-<br /> <br /> \dfrac{A^{2}}{B^{4}}x^2}dx ... which is solvable by trig substitution.

 

[Rochefort]

Aside - you continue from here:
... by making an approximation.
The first order approximation would stop the sum at n=1 to give:

C \approx \int^{B}_{0}\sqrt{<br /> <br /> 1+\dfrac{A^2}{B^2}-<br /> <br /> \dfrac{A^{2}}{B^{4}}x^2}dx ... which is solvable by trig substitution.

Thank you so much. You have been so helpful!

 

[Rochefort]

You need the approximation to be close to the actual function only in the region you are integrating over. You do that by choosing a and n carefully. If you pick a=0 you may need a higher value of n, I have not checked.

One last question. How do I find which values of "a" and "n" are the most appropriate?

 

[Simon Bridge]

Trail and error ... mostly.

Try plotting f(x), and it's approximation, on the same axes... just fudge a value for A to start with.
You only need them to match up well-enough for your purpose ... since you are doing a definite integral, you don't need to worry about whether they match up outside the region of integration.

Since you've already done a numerical method for some integrals, you can use those values to guide you.
If you really need 14dp accuracy, you may find yourself having to do as high as n=10.
It ends up depending on what you want to use the results for.

I'd intuitively pick a=B/2 and n=2 since n=1 looks like it makes the integral come out negative.

 

[Rochefort]

Trail and error ... mostly.

Try plotting f(x), and it's approximation, on the same axes... just fudge a value for A to start with.
You only need them to match up well-enough for your purpose ... since you are doing a definite integral, you don't need to worry about whether they match up outside the region of integration.

Since you've already done a numerical method for some integrals, you can use those values to guide you.
If you really need 14dp accuracy, you may find yourself having to do as high as n=10.
It ends up depending on what you want to use the results for.

I'd intuitively pick a=B/2 and n=2 since n=1 looks like it makes the integral come out negative.

I am finding it difficult to find a good approximation. I have tried different centers, "a" and different values for "A" and "B". Every time I get the summation up to the 6th order (when n=6) my computer screen freezes. Should I just assume "n" is a high value e.g. n=10? Could you suggest a way of graphing the summation without causing my computer to freeze?

 

[pasmith]

Well I know that C=\int^{B}_{0}\sqrt {1+\dfrac {A^{2}}{B^{2}}\cos^{2}\left(\dfrac {x}{B}\right) }dx

...

But I was wondering if I could find the value of A when knowing the values of B and C.

By substituting y = x/B and using \cos^2 y = 1 - \sin^2 y, you can rewrite<br /> C=\int^{B}_{0}\sqrt {1+\dfrac {A^{2}}{B^{2}}\cos^{2}\left(\dfrac {x}{B}\right) }dx as <br /> C = B \sqrt{1 + \frac{A^2}{B^2}} <br /> \int_0^1 \sqrt{1 - \frac{A^2}{A^2 + B^2}\sin^2 y}\,dy = \sqrt{A^2 + B^2} E\left(1,\frac{A}{\sqrt{A^2 + B^2}}\right) where E(\phi,k) is the incomplete elliptic integral of the second kind,<br /> E(\phi,k) = \int_0^\phi \sqrt{1 - k^2 \sin^2 y}\,dy. This suggests that it would be easier to work with k = A/\sqrt{A^2 + B^2} and B rather than A and B; you would then have <br /> C = \frac{B}{\sqrt{1 - k^2}} E(1,k), and knowing k you can recover A using
A = \frac{Bk}{\sqrt{1 - k^2}}.
You can expand \sqrt{1 - k^2 \sin^2 y} as a binomial series in k^2 \sin^2y, which can be integrated term by term (since the integral of \sin^{2n} y is known) to end up with a polynomial in k^2; but you're still going to have to solve for k numerically.

It would thus seem necessary to use a root-finding algorithm to solve <br /> \frac{B}{\sqrt{1 - k^2}} E(1,k) - C = 0<br /> for k and to integrate numerically (or, if possible, use a library function to evaluate E(1,k)) at each step of that algorithm.

 

[Rochefort]

Also I don't think you can solve for "A"

 

[Rochefort]

@pasmith:

That is way beyond my understanding (at the moment.) Could you provide more steps/explanation to your method. Also if n=10 for C=\int^{B}_{0}\sqrt{1+\left(\dfrac{A}{B}\right)^{2}+\sum^{n}_{k=1} \dfrac{A^{2}\left(-1\right)^{k}2^{2k-1}x^{2k}}{\left(2k\right)!B^{2k+2}}}dx could you show me how to solve for "A" using trigonometric substitution.

 

[Simon Bridge]

The bottom line is that you have set yourself a non-trivial task requiring some fairly advanced maths.
You'll have to experiment a bit to learn how to do it. You can look up "elliptical functions" online.

To get your computer to handle the tougher calculations, you need to use a program other than a spreadsheet.
Do the computation in stages.

You'll have to decide how good an answer you want and how badly you want it.
It looks to me like you are trying to solve a problem the hard way.

 

[Rochefort]

Sorry, accidentally posted via my phone, my question is bellow.

 

[Rochefort]

Aside - you continue from here:
... by making an approximation.
The first order approximation would stop the sum at n=1 to give:

C \approx \int^{B}_{0}\sqrt{<br /> <br /> 1+\dfrac{A^2}{B^2}-<br /> <br /> \dfrac{A^{2}}{B^{4}}x^2}dx ... which is solvable by trig substitution.

By solvable I assume you mean you can make "A" the subject. If you can prove this for n=1 I can finish my investigation (without using elliptical integrals). So please could you prove this?

 

[dirk_mec1]

Stopping the sum at n=1 is a very bad estimation you know that right?

 

[Rochefort]

Stopping the sum at n=1 is a very bad estimation you know that right?

Yes, if you can make "A" the subject please do!

 

[Simon Bridge]

By solvable I assume you mean you can make "A" the subject. If you can prove this for n=1 I can finish my investigation (without using elliptical integrals). So please could you prove this?

No - I mean the integral is solvable.

Do you not know how to do trig substitutions?
It is not my habit to do a lot of maths for people without getting paid for it.
Look at it in form:
$$\int_0^c \sqrt{a^2-b^2x^2}\;dx$$ then you can substitute $x=\frac{a}{b}\sin\theta$

To solve the result for A, you will probably have to make a further approximation - like use "small angles". You should be able to solve for A if you used a Taylor expansion for the entire integrand.

Inability to do Taylor series and trig substitutions would suggest you are biting off more than you can chew and you'll be better off backtracking and discussing the actual problem you are trying to use these calculations to solve.
There is probably a better way.

 

[Rochefort]

You can expand \sqrt{1 - k^2 \sin^2 y} as a binomial series in k^2 \sin^2y, which can be integrated term by term (since the integral of \sin^{2n} y is known) to end up with a polynomial in k^2; but you're still going to have to solve for k numerically.

I have followed your method up to this point. I am stuck on what you have told me to do here.

 

[sankalpmittal]

I have followed your method up to this point. I am stuck on what you have told me to do here.

I guess that you can express cos square x term in cos2x. What you get ?

 

[Ray Vickson]

I have followed your method up to this point. I am stuck on what you have told me to do here.

Rather than trying series, etc., why not use a purely numerical method? To solve
F(A) = C, \: \text{ where } \:F(A) \equiv \int_0^B \sqrt{ \left( 1 + \frac{A^2}{B^2} \cos^2(x/B) \right)} \, dx
you can approximate the integral via Simpson's rule, for example, so that for any given value of A you can get an numerical approximation as accurate as you want by including enough terms. You can also get a numerical (approximate) value of $dF(A)/dA$ by numerical integration. That allows you to improve the successive values of $A$ via Newton's method, for example. Since you could be use Simpson's rule with fixed numbers of intervals on $x$, you could do everything in a spreadsheet if you really need to do that.

You could even use the EXCEL Solver tool to solve the equation; the cell containing $A$ would be your variable cell and the cell containing $C$ would be your target cell; you ask the Solver to set the cell containing your (Simpson's rule) approximation to $F(A)$ equal to the target cell. The Solver options page allows you to try different solution methods and to set tolerances, etc. Often, the best approach is to start with a not-too-strict tolerance, then tighten the tolerance and re-solve, starting from the previous solution.

 

[Rochefort]

Using Numerical Methods

Rather than trying series, etc., why not use a purely numerical method?

I agree, using a numerical method would be easier. Here is my method:

Using

$\frac{B}{\sqrt{1 - k^2}} E(1,k) - C = 0$

I can find $k$ as a root

when knowing the values of B and C

For example, when $B=2$ and $C=2.17363253251301$ (I know $A=1$ using my Riemann sum) $$\frac{2E(1,k)}{\sqrt{1 - k^2}}- 2.17363253251301 = \dfrac {2\int ^{1}_{0}\sqrt {1-k^{2}\sin ^{2}y} dy}{\sqrt {1-k^{2}}}-2.17363253251301=0$$ When I type this into Wolfram|Alpha I get what is in the "Attached Thumbnails."

Using

$A = \frac{Bk}{\sqrt{1 - k^2}}$

$$A ≈ \frac{0.447215806740405B}{\sqrt{1 - 0.447215806740405^2}}=\frac{0.447215806740405\cdot 2}{\sqrt{1 - 0.447215806740405^2}}≈1.00000618061640...$$ Which is accurate enough.

Now my question is, how could I - when knowing the values of $B$ and $C$- find $k$ using numerical methods? I want to be able to use this method in excel i.e. to be able to use multiple columns to determine $k$."

 

[Ray Vickson]

I agree, using a numerical method would be easier. Here is my method:

Using I can find $k$ as a root For example, when $B=2$ and $C=2.17363253251301$ (I know $A=1$ using my Riemann sum) $$\frac{2E(1,k)}{\sqrt{1 - k^2}}- 2.17363253251301 = \dfrac {2\int ^{1}_{0}\sqrt {1-k^{2}\sin ^{2}y} dy}{\sqrt {1-k^{2}}}-2.17363253251301=0$$ When I type this into Wolfram|Alpha I get what is in the "Attached Thumbnails."

Using $$A ≈ \frac{0.447215806740405B}{\sqrt{1 - 0.447215806740405^2}}=\frac{0.447215806740405\cdot 2}{\sqrt{1 - 0.447215806740405^2}}≈1.00000618061640...$$ Which is accurate enough.

Now my question is, how could I - when knowing the values of $B$ and $C$- find $k$ using numerical methods? I want to be able to use this method in excel i.e. to be able to use multiple columns to determine $k$."

I meant something different by "numerical methods"; I meant that (for given numerical value of A) that the integral would be evaluated numerically, rather than analytically. So, if one used, say Simpson's rule with f fixed number of sub-intervals---the same for all values of A---you could bypass the elliptic functions altogether and do everything in more "elementary" software. You could even use EXCEL to evaluate (approximately) the integral and the built-in Solver tool to obtain the solution of the equation.

Of course, applying Elliptic functions and then solving the resulting equation numerically is OK too, provided that you have access to the required tools such as Maple or Mathematica.

 

[Rochefort]

for given numerical value of A.

I am trying to find a method that gives me values of $A$ given values of $B$ and $C$.

 

[Simon Bridge]

I am trying to find a method that gives me values of $A$ given values of $B$ and $C$.

You have been supplied with two - but you responded by asking me to work out the integral for you as well.

You have to be prepared to put some more of the work in yourself.
If you will not take advise, we cannot help you.

 

[Rochefort]

You have been supplied with two - but you responded by asking me to work out the integral for you as well.

I wanted you to prove that I could make $A$ the subject for when $n=1$ since I thought that you suggested that it was possible when using trig substitution and it didn't seem so. But you later specified

You should be able to solve for A if you used a Taylor expansion for the entire integrand.

Which I gave up on since I found it difficult to find my "order of approximation."

You can expand \sqrt{1 - k^2 \sin^2 y} as a binomial series $$...$$
$\frac{B}{\sqrt{1 - k^2}} E(1,k) - C = 0$

I think that this method is easier if I can solve for $k$ but I am also finding it difficult to do so, yet I haven't given up on this method.

You have to be prepared to put some more of the work in yourself.
If you will not take advise, we cannot help you.

So I am taking advice. I just ask questions because, for me, it takes longer to solve a specific problem by myself than when I simply ask someone with more insight and experience with more advanced mathematics to guide me with hints on how to solve such a problem. So therefore you can help me, but you may chose not to for reasons such as: it is not your habit to do a lot of maths for people without getting paid for it, which is perfectly acceptable. Regardless, I will continue to put the work into this problem until I resolve it.