- Cuchulainn
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[$]x = y - \varepsilon \: sin y[$] where [$]x = 0.8[$], [$]\varepsilon = 0.2[$]

http://www.datasimfinancial.com

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

maple gives y=0.9643338877

you can solve it by iteration

[$]y_{n+1}=0.8+0.2 \sin y_n[$]

when I tried that, it converges fairly quickly

you can solve it by iteration

[$]y_{n+1}=0.8+0.2 \sin y_n[$]

when I tried that, it converges fairly quickly

if I write it as [$]1-\epsilon=y-\epsilon \sin y[$] (why? it feels right, that's y) and expand y as a series in [$]\epsilon[$], the first 4 terms gives 0.9644........

[$]y=y_0+\epsilon y_1 +\epsilon^2 y_2 +\cdots[$]

y0=1

y1= sin 1 -1

etc

[$]y=y_0+\epsilon y_1 +\epsilon^2 y_2 +\cdots[$]

y0=1

y1= sin 1 -1

etc

And where does it come from? Possible clues are Cuch's obsessions: Joyce, Russians, numerics, exp(5),...Iceland? But most likely the title of the thread. To get [$]y[$] appearing as a linear and a trig term suggests something involving both triangles and circles...And why have [$]x[$] and [$]\epsilon[$] and not just the numbers? These questions are so much more interesting than 0.964... I'm afraid.

- Traden4Alpha
**Posts:**23951**Joined:**

It's a sine he's caught in a vicus of reiteration and the hypotenoose is tightening!

- Cuchulainn
**Posts:**61589**Joined:****Location:**Amsterdam-
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Actually ...these are the things to do for relaxationAnd where does it come from? Possible clues are Cuch's obsessions: Joyce, Russians, numerics, exp(5),...Iceland? But most likely the title of the thread. To get [$]y[$] appearing as a linear and a trig term suggests something involving both triangles and circles...And why have [$]x[$] and [$]\epsilon[$] and not just the numbers? These questions are so much more interesting than 0.964... I'm afraid.

I have deliberately scoped the problem to work with numbers for the moment, to focus the mind.

ppauper gets an A+ for solving the (initial) problem. We can build on his solution (I see about 7 other solutions offhand). One small question is when the eccentricity [$]\varepsilon > 1.[$]

The answer is in the stars

As ppauper says: It is developing skill to answer the question that is posed.

Last edited by Cuchulainn on May 17th, 2017, 4:41 pm, edited 4 times in total.

http://www.datasimfinancial.com

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

- Cuchulainn
**Posts:**61589**Joined:****Location:**Amsterdam-
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It's a sine he's caught in a vicus of reiteration and the hypotenoose is tightening!

You are one the right track. Get it working then right then optimised. It is a vicus problem, indeed.

Last edited by Cuchulainn on May 17th, 2017, 4:35 pm, edited 2 times in total.

http://www.datasimfinancial.com

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

- Cuchulainn
**Posts:**61589**Joined:****Location:**Amsterdam-
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ppauper has two solutions 1) Using Banach fixed point theorem (it is a contraction for [$]\varepsilon < 1[$]) and convergence is linear. It can be improved to second order by Aitken.To what precision?

2) I don't know the answer in the asymptotic expansion case. But I am sure it is a piece of cake for many Wilmotters.

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

it's straightforward to do other expansions.One small question is when the eccentricity [$]\varepsilon > 1.[$]

The answer is in the stars

if you look at [$]x=y-\epsilon \sin y[$] with [$]x=1-\epsilon[$]

you can expand about [$]\epsilon=1[$], say [$]\epsilon=1+\mu[$]

and for large [$]\epsilon[$], then [$]1/\epsilon[$] is small and expand in that

- Cuchulainn
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I get the same answers as ppauper's findings. Fixed point needs ~ 6 iterations while Aitken does it in 3 for a tolerance of 1.0e-9.

When the eccentricity is 1.5 fixed point needs 36 iteration and Aitken in 6.

The fixed point iteration hangs when [$]\varepsilon = 2[$]. Aitken and Bessel's original solution give 2.29645562918859.

When the eccentricity is 1.5 fixed point needs 36 iteration and Aitken in 6.

The fixed point iteration hangs when [$]\varepsilon = 2[$]. Aitken and Bessel's original solution give 2.29645562918859.

http://www.datasim.nl

Every Time We Teach a Child Something, We Keep Him from Inventing It Himself

Jean Piaget

about eps=2.131615 something happens

the curve looks sort of like a cubic and it goes from having 1 rea solution to 3 real solutions

the curve looks sort of like a cubic and it goes from having 1 rea solution to 3 real solutions

Can someone explain why this is of any interest at all?!

- Traden4Alpha
**Posts:**23951**Joined:**

Note that a Russian engineer would say the solution does not exist. The "y" inside the sine must have units of an angle but sin(y) will be a unit-less ratio of distances that cannot be subtracted from an angle (the lead "y").

- Traden4Alpha
**Posts:**23951**Joined:**

Banach has appealch?Can someone explain why this is of any interest at all?!

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