s
I think you mean P(N) is a sufficient condition for P(N+1) results in it be ing true for all N, using the standard if-then interpretation, a direct con sequence of the well-ordering principle of natural numbers which states tha t every subset contains a least element.
These discussion brings back some memories. I hope i don't have bad dreams tonight.
Following to Laplace we where tortured with Hilbert transformations.
After university i never had to solve a differential- or integral equations.
Pure Laplace is a PITA, but the Heaviside short-hand version is a beautiful engineering tool. ...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
Why go through that rigmarole:
F(N)= F(N)/F(N-1) x F(N-1)/F(N-2) x ... x F(2)/F(1) x F(1)= (0.97)^(N-1) x F(1)
Yours is also a good proof by construction, which avoids the counter-intuitive (but formally correct and mathematically ...de rigeur) proof by mathematical induction. The problem with proof by induction is that it is ok to prove formally the correctness or not of a formula, but does not help at all in dicovering the formula in the first place.
Your proof is shorter than mine and more elegant in appearance. But I think it requires more "discovery effort", as it must occur to one to form ratios and eliminate successive nominators/denominators. In my proof, the discovery step is probably easier to occur by writing all the terms and then looking at them where one could notice the same terms occuring on the left and right. In your proof, seems to me, the discovery step is more difficult to occur due to lack of visual (or other) clues, i.e. how did it occur to you to form these ratios?
An even more elegant than yours, which I think can occur (be discovered) easier is successive substitution (i.e. recursion in programming language parlance), i.e.
f(N) = k * f(N-1) = k * k * f(N-2) = k * k * f(N-3) . . . = k * k * k ... * k * f(1) = f(1) * k^(N-1)
Now, by simply arranging pebbles on a plane in a particular way, it is easy to see visually that 1 + 3 + 5 + ... (2N-1) = N^2 my point being that visual perception helps a lot in discovering such things, even if it is not concise as an algebraic formulation.
In my reply to Mr Thompson, I wanted to give him a simple way of discovering such answers easily in case he forgot or did not know the formula.
Euthymios Kappos
I still have serious hatred of Green's functions arising from that particular course. Lucky we no longer do it analytically any more.
-- Regards, Martin Brown
Our local choice was Matthews & Walker Mathematical Methods of Physics.
Always used to be popular with the turbulent flow brigade - often gave divergent power series that could only be tamed with Shank's. The pure mathematicians used to cringe at the abuse of method but they could not deny that the results it predicted matched experimental data well!
Works best on alternating series with poor convergence or divergence!
I was once very interested in extended convergence tricks and have used the rational approximation for Log(1+x) in anger several times.
Log(1+x) = x(6+x)/(6+4x) where -1/2 < x < 1
The other common one sometimes useful is
Sqrt(x) = (1+3x)/(3+x) where -1/2 < x < 2
Only really any good if you have fast hardware divide.
These tricks sometimes allow a rough polynomial approximation soluble analytically to provide a much better initial input guess for more rapid convergence of an iterative method like N-R.
-- Regards, Martin Brown
[snip]
You succeeded! Thanks again! ...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
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Not sure about that. Since the sequence is defined by F(N+1)/F(N)=0.97, i t seems pretty natural to ask what is F(N+2)/F(N), which by observation is (0.97)^2, from which you can make the short leap to F(N+k)/F(N)= (0.97)^k . Then your induction proof is the trivial F((N+1)+k)/F(N)=F((N+k)+1)/F(N
+k) x F(N+k)/F(N)=0.97 x (0.97)^k=(0.97)^(k+1), the first ratio by defi nition and the second by assumption. See, that was not so hard.e t
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,Hi Martin, (I also hated Green's functions as a student) Well I'm sure you know this, but it came as a revelation to me many years after graduating. The Green's function is just the pulse response of the system. If someone had told me that in school things would have made a lot more sense! I sometimes wish I could take some of those courses over again. It seems like I might understand them better now.
George H.
I recall the materials course where the prof explained everything using tensor calculus. Unfortunately, nobody had previously bothered to teach us tensor calculus. All the lectures were pure gibberish. I'm not at all sure how I passed that course.
Good stuff. Did you ever read Forman Acton's "Numerical Methods That Work"?
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 USA +1 845 480 2058 hobbs at electrooptical dot net http://electrooptical.net
Bell curve.
Best regards, Spehro Pefhany
-- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Yes - but along time ago. I don't possess a copy any more.
One memorable fun version of 3 term Shanks with terrible numerical stability (but you can divide it out to get something much better) is difference of geometric mean over arithmetic mean
given the partial sums a, b, c
x' = (a*c-b^2)/(a+c-2*b)
And if you feed it 1,3,7 it gives
x' = (1*7-3^2)/(1+7-2*3) = (7-9)/(8-6) = -1
Users of twos compliment arithmetic take note!
A numerically stable version that is formally equivalent is
x' = (a+2b+c)/4 - (a-c)^2/(a+c-2b)/4
Subject to typos and lapses of memory...
-- Regards, Martin Brown
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