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### Network # Finding a function from its series equivalent

D

#### Don Lancaster

Jan 1, 1970
0
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

The equivalent McLauran Series (or Taylor about zero) is found by
dividing each term by its factorial. 0.103691/1! , -0.00245142/2!...

... may be of extreme interest in finding a closed form expression
that involves trig products and possibly exponantials. The range of
interest is from 0 to 1.

The function appears continuous and monotonic with well behaved
derivatives. There is no zero offset.

The trig angle of 84.0000 degrees is also expected to play a major role
in the solution. As is the trig identity of cos(a+b) = cos(a)cos(b) -
sin(a)sin(b). As is a magic constant of 0.104528. Everything happens in

Sought after is a closed form determnistic solution that accepts the 0-1
value, the 84 degree angle, and the magic constant that evaluates to the
above series.

--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email: [email protected]

Please visit my GURU's LAIR web site at http://www.tinaja.com

G

#### Gerry Myerson

Jan 1, 1970
0
Don Lancaster said:
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

The equivalent McLauran Series (or Taylor about zero) is found by
dividing each term by its factorial. 0.103691/1! , -0.00245142/2!...

I would think you'd find the coefficients of the Taylor series by
multiplying the k-th term by k-factorial, not dividing. E.g.,
the coefficients of the power series for e^x are 1, 1, 1/2, 1/6, ...,
while the Taylor coefficients, the coefficients of (x^n)/(n!), are
1, 1, 1, 1, ....

D

#### Don Lancaster

Jan 1, 1970
0
Gerry said:
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

The equivalent McLauran Series (or Taylor about zero) is found by
dividing each term by its factorial. 0.103691/1! , -0.00245142/2!...

I would think you'd find the coefficients of the Taylor series by
multiplying the k-th term by k-factorial, not dividing. E.g.,
the coefficients of the power series for e^x are 1, 1, 1/2, 1/6, ...,
while the Taylor coefficients, the coefficients of (x^n)/(n!), are
1, 1, 1, 1, ....

Dividing plots beautifully. just checked it.
You basically have a ramp that gets VERY steep near a=0.

Also, multiplying -1517.83 by 20! might end up a tad on the largish side
of humongous. In any sane function, the higher power terms should be
quite small.

There are apparently two or three summation terms to the real answer.
One is an easily found and apparently fully determninistic linear ramp.
The second is an ever increasing "bent" variable that may or may not
need some extra help above 0.9. It may or may not have its own small
ramp component.

I suspect it does.

--
Many thanks,

Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml email: [email protected]

Please visit my GURU's LAIR web site at http://www.tinaja.com

P

#### Paul Hovnanian P.E.

Jan 1, 1970
0
Don said:
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

Question about terminology here. By the above, are you saying that
0.103691 = a1 * x^1 and
-0.00245142 = a2 * x^2
etc.

Or are you saying that 0.103691 is the coefficient of x^1, -0.00245142
is the coefficient of x^2, etc. so

f(x) = (0.103691 * x^1) + (-0.00245142 * x^2) + .....

BTW, you posted a similar, but slightly different problem a half hour
earlier with terms starting with x^0 as opposed to x^1 (above). Which is
correct?

G

#### Gerry Myerson

Jan 1, 1970
0
Don Lancaster said:
Gerry said:
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

The equivalent McLauran Series (or Taylor about zero) is found by
dividing each term by its factorial. 0.103691/1! , -0.00245142/2!...

I would think you'd find the coefficients of the Taylor series by
multiplying the k-th term by k-factorial, not dividing. E.g.,
the coefficients of the power series for e^x are 1, 1, 1/2, 1/6, ...,
while the Taylor coefficients, the coefficients of (x^n)/(n!), are
1, 1, 1, 1, ....

Dividing plots beautifully. just checked it.

Cool. But the function you're plotting isn't the one you're asking
about. The power series a_1 x + a_2 x^2 + a_3 x^3 + ... is
(obviously!) equivalent to the Taylor series
a_1 x + (2 a_2) (x^2 / 2) + (6 a_3) (x^3 / 6) + ...
and not equivalent to
a_1 x + (a_2 / 2)(x^2 / 2) + (a_3 / 6) (x^3 / 6) + ....

So either you're not doing what you say you're doing, or else
the function you're plotting has nothing to do with the function
you say interests you.

C

#### Chip Eastham

Jan 1, 1970
0
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

The equivalent McLauran Series (or Taylor about zero) is found by
dividing each term by its factorial. 0.103691/1! , -0.00245142/2!...

... may be of extreme interest in finding a closed form expression
that involves trig products and possibly exponantials. The range of
interest is from 0 to 1.

The function appears continuous and monotonic with well behaved
derivatives. There is no zero offset.

The trig angle of 84.0000 degrees is also expected to play a major role
in the solution. As is the trig identity of cos(a+b) = cos(a)cos(b) -
sin(a)sin(b). As is a magic constant of 0.104528. Everything happens in

Sought after is a closed form determnistic solution that accepts the 0-1
value, the 84 degree angle, and the magic constant that evaluates to the
above series.

Is the function known to be periodic?

A Fourier series expansion, if so, would be more
illuminating than a power series.

regards, chip

J

#### Jon

Jan 1, 1970
0
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

Search for the article "Computing the Generating Function of a Series Given its First Few Terms" by Bergeron and Plouffe. I dont have the URL, but it is a free download. It doesn't provide a bullet proof method, but gives some suggestions.

W

#### whit3rd

Jan 1, 1970
0
It is usually a fairly simple matter to get from a trig or exponantial
function of one sort or another to its underlying series form.

But how can you get from a known accurate series expression to a
nonobvious and crucially esoteric equivalent function?

Specifically, the "raw" power series

[-1517.83 5094.6 821.18 -29457.7 61718.9 -61268.8 30448.6 -4770.84
-269.684 -2892.14 3300.63 -1460.88 213.578 78.8959 -49.2164 12.3083
-1.74731 0.149743 -0.00245142 0.103691]

where 0.103691 is the x^1 term, -0.00245142 is x^2 etc...

This is the art of curve fitting. You need one additional piece of
information, the error (tolerance) of each of the measurements,
to do it properly.

Since you have specified this over the range 0 to 1, the classical
approach would be to use a set of functions that are orthogonal
over this range. Since you already have a polynomial,
a kind of Legendre polynomial is your starting point (I think the
Legendre polynomials are orthogonal over (-1, +1), so there's
some scaling and shifting involved).

Your goal is, from that starting point, to find a simpler expression
that holds the same data (i.e. reproduces the same function to
the given tolerance). Sine/Cosine, polynomial, Bessel, there
are lots of sets of orthogonal functions to choose from, and
one of the sets might have all-but-one-term-vanishes character.

If there is any OTHER information about the boundary conditions,
(if it's periodic, use sines; if it doesn't blow up at infinity,
don't use
polynomials) that can help, you need to consider that now.
Finally, there's a theorem, the Wiener-Hopf theorem, that
states that a set of fit functions is optimal if its autocorrelation
is the same as that of the data. That means that data with
big ripples doesn't fit well with sinewaves with little tiny ripples,
but you can find that kind of thing out the hard way, too.

For orthogonal functions, the fit procedure is simple: there's a
unique
right set of coefficients, and a procedure to find it (inner product
of your data and the function). For other kinds of functions,
like wavelets and fractals and such, it gets ... challenging.

Your starting point has 20 coefficients, so anything that expresses
the data with 19 or less, and hits within the measurement tolerance,
is some kind of success.

Finally, I should note that the 'measurement tolerance' defines a kind
of weight, a statistical weight function, that cannot be ignored.
The orthogonal polynomials (Legendre polynomials) with
constant weight are very different from the orthogonal polynomials
(Chebyshev polynomials) with (x(1-x)) weight, and the weight
is part of that 'inner product' step as well.

Usually, one starts with a graph of the function, makes a guess as
to its form, then looks at a graph of the difference-from-the-guess.
If you get lucky with a guess, and it was a natural measurement
you started with, you've just created a scientific theory. Kudos!

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