 Cuchulainn
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### Going off on a tangent (kind of) : find y

$x = y - \varepsilon \: sin y$ where $x = 0.8$, $\varepsilon = 0.2$
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R. van Gulik ppauper
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### Re: Going off on a tangent (kind of) : find y

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 ppauper
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### Re: Going off on a tangent (kind of) : find y

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 BigAndyD
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### Re: Going off on a tangent (kind of) : find y

To what precision? Paul
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### Re: Going off on a tangent (kind of) : find y

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. Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: Going off on a tangent (kind of) : find y

It's a sine he's caught in a vicus of reiteration and the hypotenoose is tightening! Cuchulainn
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### Re: Going off on a tangent (kind of) : find y

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.
Actually ...these are the things to do for relaxation 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.
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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik Cuchulainn
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### Re: Going off on a tangent (kind of) : find y

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.
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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik Cuchulainn
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### Re: Going off on a tangent (kind of) : find y

To what precision?
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.
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.
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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik ppauper
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Joined: November 15th, 2001, 1:29 pm

### Re: Going off on a tangent (kind of) : find y

One small question is when the eccentricity $\varepsilon > 1.$
The answer is in the stars it's straightforward to do other expansions.

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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### Re: Going off on a tangent (kind of) : find y

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.
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Approach your problem from the right end and begin with the answers. Then one day, perhaps you will find the final question..
R. van Gulik ppauper
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Joined: November 15th, 2001, 1:29 pm

### Re: Going off on a tangent (kind of) : find y

the curve looks sort of like a cubic and it goes from having 1 rea solution to 3 real solutions Paul
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### Re: Going off on a tangent (kind of) : find y

Can someone explain why this is of any interest at all?! Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: Going off on a tangent (kind of) : find y

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"). Posts: 23951
Joined: September 20th, 2002, 8:30 pm

### Re: Going off on a tangent (kind of) : find y

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