 Cuchulainn
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### exp(5) = $e^5$

That's very interesting. The algo is described in section 3.2 here, sorry in Dutch. What is noteworthy is they use Gram-Schmidt to produce an orthormal basis before LLL algo.LLL in EnglishHad a quick look at Constructive Dedekind reals looks like computing a real in an approximate interval. This looks cool. And now an upper limit How to LLL to do this?
Last edited by Cuchulainn on October 30th, 2010, 10:00 pm, edited 1 time in total. AVt
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### exp(5) = $e^5$

@wileysw:That can be done in Maple as well, though it does not have the command 'RootApproximant'.One defines a 'base' (which is one for transcendent input, not otherwise in general) bypowers and seeks for integer relations (with small numerical errors, thus depending onthe computational precision used), for which a variant of LLL is used.ApproxAlgebraicRelation:= proc(z,deg:osint, x)# z = any constant, # deg = desired degree for an approximate algebraic relation, # x = indeterminate to be used in the polynomiallocal w,u;#Digits:= Digits+3; # make your choice, Digits:= 2*Digits is finew:=Vector(deg+1, 'k -> z^(k-1)'); # = 1, z, z^2, z^3 ... w:=convert(%, list); # syntax needed for the following commandu:=IntegerRelations:-LinearDependency(evalf(w));0=sum( u[ i]*x^(i-1),i=1..deg+1); end proc;Using 14 digits (i.e. ~ hardware double precision) that replies by0 = -4491-13581*x+4778*x^2 for input z=Pi and variable x, degree=2.That says: Pi 'almost' satisfies that quadric. Within that code one can increase the accuracy - that leads to better results, but somelarger coefficients. And if one wants the 'most simple' one just increase degree, untilthe error falls below a limit (so there is a kind of trade off in combining degree andaccuracy (leading to longer expressions), what one wants to call 'most simple' ...).But that 'simple' approach is unlikely to produce those nice & curious results with veryhigh accuracy given at the Mathworld.@Cuchulainn:That 'rationalize' is degree=1, so can be done by LLL. However that is not needed at all.Essentially it (depending on computational precision) an iterated divison with remainder,until the remainder is 'killed' by the 'system epsilon' (~ precision). Silently you useit every day while coding ... (though it may differ from IEEE for other reasons).
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### exp(5) = $e^5$

QuoteThat 'rationalize' is degree=1, so can be done by LLL. However that is not needed at all.Essentially it (depending on computational precision) an iterated divison with remainder,until the remainder is 'killed' by the 'system epsilon' (~ precision). Silently you useit every day while coding ... (though it may differ from IEEE for other reasons).If I understand correctly, the iteration converges monotonically until machine accuracy cannot see the difference?It is a one-sided estimate? AVt
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### exp(5) = $e^5$

Hm ... it really is like converting a decimal input to a machine numberas the read commands do (which depend on the rounding rules, which maybe what you mean by 'one-sided'), I once went through that (do not haveit hand). Practical it is an enhanced Euclidian algorithm. It stops, assoon as all bits are used (or likewise by 'system epsilon').And a IEEE float is just a special fraction & rational number. Cuchulainn
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### exp(5) = $e^5$

Don't know if these solutions have already been given:1. Compoundingexp(x) = lim (1+x/n)^n as n -> INFINITY2. y = exp(x) solves the differential equationdy/dx = y, y(0) = 1which you solve by Euler's fd method. Maybe that's the way Euler did it at the time?? It's first-order accurate, so you can determine the accuracy based on the interview constraints (neurons work slow).
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### exp(5) = $e^5$

QuoteOriginally posted by: slopse=2.7 Ibsen Ibsen. (Ibsen was born in 1828, e = 2.718281828...) Ibsen? LOL The way I learned e = 2.718281828 in high school was that Lev Tolstoy was born in 1828. Cuchulainn
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### exp(5) = $e^5$

QuoteYou can get e^5~148 with sub-1% accuracy without pen and paper if you remember that ln(2)~0.693. Try it.Here's another way and you only have to remember that exp(0) = 11. Compute e == exp(1) using the fact that the exponential satisfies du/dx = u, u(0) = 1. Solve using Euler's method U(n+1) = (1+h)*U(n). Compute e ~ U(N) = (1+h)^N.Perhaps surprisingly, since this work on logarithms had come so close to recognising the number e, when e is first "discovered" it is not through the notion of logarithm at all but rather through a study of compound interest. In 1683 Jacob Bernoulli looked at the problem of compound interest and, in examining continuous compound interest, he tried to find the limit of (1 + 1/n)n as n tends to infinity. He used the binomial theorem to show that the limit had to lie between 2 and 3 so we could consider this to be the first approximation found to e. Also if we accept this as a definition of e, it is the first time that a number was defined by a limiting process. He certainly did not recognise any connection between his work and that on logarithms. 2. Now e^5 = (e^2) * (e^2) * e.
Last edited by Cuchulainn on January 29th, 2011, 11:00 pm, edited 1 time in total. ExSan
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### exp(5) = $e^5$

QuoteOriginally posted by: CuchulainnHere's another way and you only have to remember that exp(0) = 11. Compute e == exp(1) using the fact that the exponential satisfies du/dx = u, u(0) = 1. Solve using Euler's method U(n+1) = (1+h)*U(n). Compute e ~ U(N) = (1+h)^N. Thanks to Cuchulainn for the hints ! Numerical Ordinary Differential EquationsFrom the Downloadable File extract program ExSan_EULER_ODE_EXP to your desktop and execute. (see updated post)
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### exp(5) = $e^5$

A more accurate approx is to use the Cayley transform (aka Crank Nicolson)du/dx = u(U(n+1) - U(n))/h = (U(n+1) + U(n))/2or U(n+1) = alpha*U(n)alpha = (1 +h/2)/(1 - h/2)
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### exp(5) = $e^5$

QuoteOriginally posted by: CuchulainnA more accurate approx is to use the Cayley transform (aka Crank Nicolson)du/dx = u(U(n+1) - U(n))/h = (U(n+1) + U(n))/2or U(n+1) = alpha*U(n)alpha = (1 +h/2)/(1 - h/2)neat
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### exp(5) = $e^5$

In general, it's better to solve for du/dx = -u, compute CN and then invert the answer! For the rest, it's the same as before.
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### exp(5) = $e^5$

IMPROVED EULER'S METHOD -Heun's Formula - Midpoint methodRegular Euler Method: exp( 5 ) ~ 129.1299 error = -12.99 % step = .1Improved Euler Method: exp( 5 ) ~ 147.2699 error = --0.77 % step = .1From the Downloadable File extract program ExSan_IMP_EULER_ODE_EXP to your desktop and execute. (see updated post)
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### exp(5) = $e^5$

QuoteOriginally posted by: CuchulainnIn general, it's better to solve for du/dx = -u, compute CN and then invert the answer! For the rest, it's the same as before.For systemsdU/dt = QU + rWe wish to compute a bounded solution as t -> infinity and this is determined by Q (See Varga 1962) and when the eigenvalues are < 0 then limit U(t) = Inverse(Q)*r as t -> infinity @exsanWhat is the order of accuracy h^n for your two schemes (1 and 2?) The convergence of CT to exp(-z) http://en.wikipedia.org/wiki/Pad%C3%A9_table
Last edited by Cuchulainn on February 7th, 2011, 11:00 pm, edited 1 time in total. ExSan
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### exp(5) = $e^5$

Runge - Kutta 4 OrderRK4 exp( 5 ) ~ 31.71365 error = -78.63 % step = 0.1RK4 exp( 5 ) ~ 28.37667 error = - -80.88 % step = 0.01Basic EULER Method is the most stable, so far, for ode du/dx = u , other methods are stable only for certain values of step (h)From the Downloadable File extract program ExSan_RK_EXP to your desktop and execute.
Attachments ExSan_V.4.01.Y_RungeKutta_exp(x).zip
Last edited by ExSan on February 8th, 2011, 11:00 pm, edited 1 time in total. Cuchulainn
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### exp(5) = $e^5$

One scheme that might be useful are Exponential integrators esp. page 8.Have not tried this but I once did exponential fitting for this DE.
Last edited by Cuchulainn on March 4th, 2011, 11:00 pm, edited 1 time in total.  