Airy Equation

[mario99]

A few days ago I worked hard to find the Airy Equation's solution. I found two solutions using Power Series but I don't know how to use them for approximation. $$Ai(1) \approx 0.13529$$ and $$Bi(1) \approx 1.20742$$

Any help would be appreciated!

 

[topsquark]

A few days ago I worked hard to find the Airy Equation's solution. I found two solutions using Power Series but I don't know how to use them for approximation. $$Ai(1) \approx 0.13529$$ and $$Bi(1) \approx 1.20742$$

Any help would be appreciated!

Could you possibly be any more vague? What is your question? If it refers to another thread, then you need to be posting it in the appropriate thread. If it's a new question then you need to post the whole problem, as well as any work you've done on it.

-Dan

 

[mario99]

Thank you topsquark for helping me. My question is how to use my Airy infinite series function to have an approximation of 0.13529 or 1.20742 when x = 1?

 

[topsquark]

Thank you topsquark for helping me. My question is how to use my Airy infinite series function to have an approximation of 0.13529 or 1.20742 when x = 1?

And, again, what is your Airy infinite series function?

-Dan

 

[mario99]

It is the same functions in the link except my functions have the variable x, y1(x) and y2(x). Do I have to write them again in here?

Sosmath.com

www.sosmath.com

 

[blamocur]

Do I have to write them again in here?

You don't if you expect someone else to spend all the necessary time collecting the information together. But the chances of someone doing that are rather small. You are more likely to get help if you follow @topsquark's suggestions in post #2, i.e. either post in the relevant thread, or make you post self-contained.

 

[topsquark]

The problem is that I am still unsure as to what your question is. I know what the Airy functions are. What I don't know are things like

A few days ago I worked hard to find the Airy Equation's solution. I found two solutions using Power Series but I don't know how to use them for approximation.

and gave two numbers, without any explanation of what you are trying to do. Now you are saying

It is the same functions in the link except my functions have the variable x, y1(x) and y2(x). Do I have to write them again in here?

Please state your full question, showing any work that you have done on it. Otherwise we can't understand what you are asking or what you already know, two pieces of information that are absolutely critical if we are to be of any help to you.

-Dan

 

[mario99]

Thank you blamocur for helping me.

$$y(x) = y_1(x) + y_2(x)$$
$$y_1(x) = 1 + \sum_{n=1}^{\infty}\frac{x^{3n}}{[2 \cdot 3][5 \cdot 6] .... [ (3n - 1) \cdot (3n) ]}$$
$$y_2(x) = x + \sum_{n=1}^{\infty}\frac{x^{3n+1}}{[3 \cdot 4][6 \cdot 7] .... [ (3n) \cdot (3n + 1) ]}$$
How to use this infinite series to calculate the function up to 5 decimal places when x = 1 and n goes from n = 1 to n = 100?

$$y_{n=100}(1) = \ ???$$
How to use this infinite series function to get a correct solution up to five decimal places?
$$y(1) = 0.13529$$or
$$y(1) = 1.20742$$
Thank you very much Mario for spending 40 minutes writing the latex code! ?

 

[blamocur]

y(1)=0.13529or
y(1)=1.20742y(1) = 1.20742y(1)=1.20742

Do $y(1)$ mean $y_1(1)$ in one place and $y_2(1)$ in another? If that is the case then your numerical values are different from what I get. In particular both values must be larger than 1.

As for how to compute the actual values, you can probably do it with a simple calculator. In my quick and dirty scripts the precision of 5 decimal digits is achieved after only 4 iterations.

P.S. Spending 40 minutes writing LaTeX code is not a useless exercise. It will be taking you much less time pretty soon. Personally, I find myself useng LaTeX code instead of paper and pencil when trying to work out formulas for myself.

 

[mario99]

Do $y(1)$ mean $y_1(1)$ in one place and $y_2(1)$ in another? If that is the case then your numerical values are different from what I get. In particular both values must be larger than 1.

Ai(1) = 0.13529 and Bi(1) = 1.20742 are Airy function's solutions taken from Wolfram. At least one of them must be the Airy function I derived by Power Series. I don't know if y(1) = 0.13529 and y(1) = 1.20742 means y1(x) and y2(x) respectively. Maybe, Ai(x) = y1(x) + y2(x) and Bi(x) another Airy functions that its answers can be derived from Ai(x). This function comes with two constants a0 and a1. Maybe because I got rid of them, the function has stretched or compressed somehow to give wrong results.

$$y_1(x) = a_0 + \sum_{n=1}^{\infty}a_0\frac{x^{3n}}{[2 \cdot 3][5 \cdot 6] .... [ (3n - 1) \cdot (3n) ]}$$
$$y_2(x) = a_1x + \sum_{n=1}^{\infty}a_1\frac{x^{3n+1}}{[3 \cdot 4][6 \cdot 7] .... [ (3n) \cdot (3n + 1) ]}$$

As for how to compute the actual values, you can probably do it with a simple calculator. In my quick and dirty scripts the precision of 5 decimal digits is achieved after only 4 iterations.

What do you mean by 4 iterations? Does it mean from n = 1 to n = 4? If it does mean this, did you use my infinite series to do the 4 iterations? Can I use Wolfram calculator to do the 4 iterations at once? If yes, how to do that?

I tried to use Wolfram to sum the function index by index by hand (frustrating) from n = 1 to n = 6 and never got close to the answers of Ai(1) and Bi(1). I tried to use y1(1) alone, then y2(1) alone, then combined them but always different answers from Ai(1) and Bi(1). If this infinite series is a correct solution to Airy's equation, there must be a way to use it to approximate the function to the required precision.

P.S. Spending 40 minutes writing LaTeX code is not a useless exercise. It will be taking you much less time pretty soon. Personally, I find myself useng LaTeX code instead of paper and pencil when trying to work out formulas for myself.

?

 

[topsquark]

Thank you blamocur for helping me.

$$y(x) = y_1(x) + y_2(x)$$
$$y_1(x) = 1 + \sum_{n=1}^{\infty}\frac{x^{3n}}{[2 \cdot 3][5 \cdot 6] .... [ (3n - 1) \cdot (3n) ]}$$
$$y_2(x) = x + \sum_{n=1}^{\infty}\frac{x^{3n+1}}{[3 \cdot 4][6 \cdot 7] .... [ (3n) \cdot (3n + 1) ]}$$
How to use this infinite series to calculate the function up to 5 decimal places when x = 1 and n goes from n = 1 to n = 100?

$$y_{n=100}(1) = \ ???$$
How to use this infinite series function to get a correct solution up to five decimal places?
$$y(1) = 0.13529$$or
$$y(1) = 1.20742$$
Thank you very much Mario for spending 40 minutes writing the latex code! ?

Thank you.

Neither of these y(x) formulas are Ai(x) nor Bi(x). The A and B functions are approximations to (orthogonal) solutions to Airy's equation near a given value of x. Could you tell us what value of x your power series is based near? If I had to guess, I'd say it would be for x near 0 (which is what you want if you are trying to find a solution at x = 1.)

You should be able to plug this into W|A or MathLab or something.

-Dan

 

[mario99]

Thank you.

Neither of these y(x) formulas are Ai(x) nor Bi(x). The A and B functions are approximations to (orthogonal) solutions to Airy's equation near a given value of x. Could you tell us what value of x your power series is based near? If I had to guess, I'd say it would be for x near 0 (which is what you want if you are trying to find a solution at x = 1.)

You should be able to plug this into W|A or MathLab or something.

-Dan

Thank you for clarifying that Ai(x) and Bi(x) are slightly different than my equation. You are right that this infinite series was solved based near zero. Nevertheless its radius of convergence is infinite. The only benefit of finding solutions near zero on this particular infinite series (based near zero) is that the function will converge faster which means we need less indices or iterations (if that is the same meaning) to compute the function.

The question is now how to compute my Airy function to 5 decimal places? y(1) = ? (I don't have an actual solution to compare with!)

 

[blamocur]

What do you mean by 4 iterations? Does it mean from n = 1 to n = 4? If it does mean this, did you use my infinite series to do the 4 iterations? Can I use Wolfram calculator to do the 4 iterations at once? If yes, how to do that?

I was computing the series from your post #8 for $x=1$. They converge quite fast, but the answers I get are $y_1(1) \approx 1.17230$ and $y_2(1) \approx 1.08534$, and it required only $1 \leq n \leq 4$. Computing for larger $n$ did not change the result, which means the rest of the sum did not add more than 0.00001.

 

[mario99]

I was computing the series from your post #8 for $x=1$. They converge quite fast, but the answers I get are $y_1(1) \approx 1.17230$ and $y_2(1) \approx 1.08534$, and it required only $1 \leq n \leq 4$. Computing for larger $n$ did not change the result, which means the rest of the sum did not add more than 0.00001.

My solution agrees with yours.

y(1) = y1(1) + y2(1) = 2.25764

Why does this Airy solution give different results from other Airy solutions? I thought that different methods of solving the same differential equation will lead to the same approximation!

 

[topsquark]

My solution agrees with yours.

y(1) = y1(1) + y2(1) = 2.25764

Why does this Airy solution give different results from other Airy solutions? I thought that different methods of solving the same differential equation will lead to the same approximation!

Someone else is going to have to chime in with an "official" answer, but I suspect the difference is that yours is a power series solution and the Ai and Bi are asymptotic solutions.

-Dan

 

[mario99]

Thank you topsquark and blamocur for digging with me so that I could sum my Airy deeply.

Someone else is going to have to chime in with an "official" answer, but I suspect the difference is that yours is a power series solution and the Ai and Bi are asymptotic solutions.

-Dan

I will be waiting for this charming someone to fill up the remaining gaps officially.


Dr.Peterson said it before, everyday you learn something new.

 

[blamocur]

I will be waiting for this charming someone to fill up the remaining gaps officially.

Instead of waiting I'd concentrate on stating the problem in a clearer way

 

[mario99]

Instead of waiting I'd concentrate on stating the problem in a clearer way

I don't agree with you on this. Three years ago, I used to state problems in a clear systematic way. Guess what happened? I got banned by Dan. Go back to all my posts in that website and you will see no one single vagueness.

Dr.Peterson once said to a member (not me) here in this website, I will not spoon feed you. It was clear that it was intended to show some work from the OP of that post. I would rather use this sentence to you Seniors (with no offense). I will not spoon feed you by stating the problem in a clearer way because you as a Senior needs to think the hard way to understand and help vague students. I see it more exciting in this way for both sides the Seniors and the Juniors. Forgive me if my words in this reply seem offensive.

 

[topsquark]

I don't agree with you on this. Three years ago, I used to state problems in a clear systematic way. Guess what happened? I got banned by Dan. Go back to all my posts in that website and you will see no one single vagueness.

Dr.Peterson once said to a member (not me) here in this website, I will not spoon feed you. It was clear that it was intended to show some work from the OP of that post. I would rather use this sentence to you Seniors (with no offense). I will not spoon feed you by stating the problem in a clearer way because you as a Senior needs to think the hard way to understand and help vague students. I see it more exciting in this way for both sides the Seniors and the Juniors. Forgive me if my words in this reply seem offensive.

You didn't get banned for stating a problem in a clear and concise manner. You got banned because you were rude, didn't follow forum guidelines, and didn't want to do your own work. Pretty much the same reason you've been banned everywhere else.

If you don't want to "spoon feed" us seniors then you won't get the help you need. We understand the material, you don't. It's your call.

-Dan

 

[khansaheb]

I will not spoon feed you by stating the problem in a clearer way because you as a Senior needs to think the hard way to understand and help vague students.

When you are begging for help - you cannot choose the way the help is given to you.

If you hire a paid-tutor, s/he will spoon-feed you according to your taste.