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 Searching for an asymptotic to exp[0.5] sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 08/10/2014, 06:08 AM (This post was last modified: 08/16/2014, 12:25 AM by sheldonison.) There are advantages to using Tommy's 2sinh^{0.5} since there is a simple accessible closed form for the formal half iterate, unlike the exp^{0.5} which requires Kneser's Riemann mapping. In pondering an exact integral for all the derivatives asymptotic for $\text{2sinh}^{0.5}(z)$ We draw the cutpoints at the negative real axis, and using the h_n formulation, one can get approximations for the derivatives, but I ran into difficulties getting a converging solution for all of the derivatives as the magnitude of z gets arbitrarily large. The equation above is exactly equivalent to $\text{2sinh}^{0.5}(z)=2\sinh(\text{2sinh}^{0.5}(\sinh^{-1}(z/2)))\;\;$ we desire $f(z)\approx \text{2sinh}^{0.5}(z)$, a very good entire asymptotic I came up with this formulation which leads to all derivatives converging by way of the method in post#70 and post#76, by using this approximation: $\text{2sinh}^{0.5}(z)\approx \exp( \text{2sinh}^{0.5}(\sinh^{-1}(z/2)))$ $f(z)\approx \exp( \text{2sinh}^{0.5}(\sinh^{-1}(z/2)))\;\;$ where f(z) is the desired asymptotic, as per post#70/#76 There is one other step since post#70/76 used $g(z) = \ln(f(\exp(z))$, then $g(z) = \ln(f(\exp(z)) = \text{2sinh}^{0.5}(\sinh^{-1}(\exp(z)/2)))$ $g(z)= \text{2sinh}^{0.5}(z + \ln(0.5 + 0.5\sqrt{1+\frac{4}{\exp(2z)}}))\;\;$ where the sqrt term goes to 0.5 and the ln term goes to 0 as z grows exponentially, $\;g(z) \approx \text{2sinh}^{0.5}(z)$ There is also an "exact" version of g(z), which can be derived with some algebra from the equations above. However, using g_exact doesn't effect numerical results much as long $\Re(g(z))>>0$, but the sqrt term above is really important since we actually use fairly small values of z for the Taylor series numerical approximations of the nth derivatives. $g_{\text{exact}}(z)=\ln(\text{2sinh}^{0.5}(\exp(z))=g(z) + \ln(1 - \exp(2g(z)))\;\;$ g_exact can be trivially generated from g(z) $h^{-1}(z) = \frac{d}{dz}g(z)\;\;$ this is our friend h(n) which is used for the Gaussian approximation using g or g_exact(z), for the nth derivative. 2sinh(z) and exp(z) behave nearly the same for big positive numbers, but the the key difference is that exp(big negative number) is very small, which is good for the convergence of the numeric integral, and lets all of the derivatives converge, where as 2sinh(big negative number) is an even bigger negative number. This is all assuming we sum up multiple wrappings around the unit circle, until we hit the minimum magnitude, to define a converging integral for all of the derivatives as the radius gets arbitrarily large. I haven't done the numeric integrals for the derivatives of 2sinh, defined in this way, but I certainly could. This would yield another entire f(z) where f(f(z)) is an asymptotic to exp(z). I would expect it to behave very similarly in terms of convergence and error terms, with f(f(z)) eventually oscillating between larger and smaller than exp(z) an infinite number of times. The oscillations would even have nearly the same magnitude. It would also have zeros at the negative real axis, in much the same way. - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/01/2014, 10:24 PM It seems my friend ( our ? ) mick has given us ( myself and sheldon in particular ) credit for the use of " fake function theory " here is the link : http://math.stackexchange.com/questions/...for-real-x As I mentioned before z - ln(2sinh(z)/z) is asymptotic to ln(z) near the real line, but it fails to be entire ( sinh is periodic ! ). Guess we could consider the fake ln(x^2 + 1) here too. Unlike the half-exp we are considering more standard functions which might give intresting closed form results ! Since he asked for integral representations this crossed my mind : fakeln(z) = integral dz / fakesqrt(1+z^2) thereby changing the idea of a fake log to a fake sqrt such that fakesqrt(1+z^2) =/= 0 for any z. That implies ( i think ? ) ln(fakesqrt(z)) = entire ? hmmm. I wrote about fakesqrt once ... (thinking) regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/02/2014, 07:46 AM I told mick to consider fake( f ) = fake( f g )/ g. Or perhaps even fake ( f ) = fake ( f g ) / fake ( g ). That seems like a powerfull idea at first sight. So fake ( ln(x^2+1) ) = fake ( ln (x^2+1) exp(x^2) ) exp(-x^2). Naturally I wonder now about fake ( exp^[0.5](x) ) vs fake ( exp^[0.5](x) exp(x) ) exp(-x). I assume we cannot keep the property of positive derivatives for fake ( exp^[0.5](x) exp(x) ) exp(-x) , but still it seems intresting. Also crossing my mind : d/dx fake ( f ) = fake ( d/dx f ). And then there are least squares ideas. A theoretical question : Lets write fake+ for an asymptotic with positive derivatives, then does there Always exist a g(x) such that for any entire f(x) we have fake+ ( f ) = fake+ ( f g ) / g ? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/02/2014, 07:53 AM BTW fake function theory might also help to find a real periodic function with sign pattern a0 - a1 x - a2 x^2 + a3 x^3 - a4 x^4 - a5 x^5 + a6 x^6 - ... regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/03/2014, 01:04 PM (This post was last modified: 09/03/2014, 01:31 PM by sheldonison.) (09/01/2014, 10:24 PM)tommy1729 Wrote: It seems my friend ( our ? ) mick has given us ( myself and sheldon in particular ) credit for the use of " fake function theory " here is the link : http://math.stackexchange.com/questions/...for-real-x As I mentioned before z - ln(2sinh(z)/z) is asymptotic to ln(z) near the real line, but it fails to be entire ( sinh is periodic ! ). Guess we could consider the fake ln(x^2 + 1) here too. Unlike the half-exp we are considering more standard functions which might give intresting closed form results !... Thanks for the link Tommy! I've been overseas vacationing... I was able to use the basic recipe from this post to interpolate $\ln(x^2+1)\exp(x^2)$ as an even function, just as you suggested! I will post more later, either here or at Mathstack. I have also been working on an example that should allow us to make the "fake function" theory more rigorous. A very interesting asymptotic function is below. It is interesting because the corresponding $\ln(f(\exp(z)))$ function is simply $\frac{x^2}{2}$, whose derivative is x, so $h(n)=n$, so all of the Taylor series coefficients of the interpolating function have an exact closed form, as do all of the error equations. $f(x) = \exp \left$$\frac{(\ln(x))^2}{2} \right$$$ $a_n = \frac{1}{\exp(0.5n^2)\sqrt{2\pi}}\;\;$ for this function, the Gaussian approximation is also the best approximation. $f(x) \approx \sum_{n=-\infty}^{\infty} a_n x^n\;\;$ this is a converging Laurent series. We don't use the negative coefficients for the entire interpolant. The Laurent series interpolant can also be expressed in a quickly converging closed form. $\sum_{n=-\infty}^{\infty} a_n x^n\;=\;\sum_{n=-\infty}^{\infty} \exp \left$$\frac{(\ln(x)+2n\pi i)^2}{2} \right$$\;\;$ f(x) corresponds to n=0 - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/08/2014, 12:56 AM Dear Sheldon I read your answer @stackexchange. I assume your sequences are the derivatives and not the coefficients ? Otherwise it seems divergent. I intended to tell mick about the integral of (1-exp(x))/x to get a fake ln(x) or fake 1/x. --- And then I started wondering about x fake(1/x) , exp(fake(ln(x)) - x etc. --- I mentioned (1-exp(x))/x before. However apparantly that was already given a long time at mathoverflow. (as your link proves). Its easy to get a fake sqrt from a fake ln. So the main questions are these : 1) which is the best fake ln(x) ? integral (1-exp(-x)/x) - C or C_2 * +fake(ln(x) exp(x)) exp(-x) ? ( I use +fake to refer to "our" method of fake functions ; with the positive derivatives ) 2) Is there an analogue for the integral of (1 - exp(-x))/x to create a fake sqrt(x) ? ( I do NOT mean exp(fake(ln(x)/2) ) Answers to these will lead to new questions. I want to remark that +fake(sqrt(x) exp(x)) exp(-x) is very likely a better approximation then exp(+fake(ln(x) exp(x)) exp(-x) / 2). I also notice we can add variables to our ideas : 1) which is the best fake ln(x) ? integral(1-exp(- A x)/x) - C or C_2 *+fake(ln(x) exp(B x)) exp(-B x) ? Which leads to more intresting questions. Anyways the most intresting is imho : 2) Is there an analogue for the integral of (1 - exp(-x))/x to create a fake sqrt(x) ? regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/08/2014, 04:15 AM (This post was last modified: 09/08/2014, 04:22 AM by sheldonison.) (09/08/2014, 12:56 AM)tommy1729 Wrote: Dear Sheldon I read your answer @stackexchange. I assume your sequences are the derivatives and not the coefficients ? Otherwise it seems divergent. I intended to tell mick about the integral of (1-exp(x))/x to get a fake ln(x) or fake 1/x. --- And then I started wondering about x fake(1/x) , exp(fake(ln(x)) - x etc. --- I mentioned (1-exp(x))/x before. However apparantly that was already given a long time at mathoverflow. (as your link proves). Its easy to get a fake sqrt from a fake ln. So the main questions are these : 1) which is the best fake ln(x) ? integral (1-exp(-x)/x) - C or C_2 * +fake(ln(x) exp(x)) exp(-x) ? ( I use +fake to refer to "our" method of fake functions ; with the positive derivatives ) 2) Is there an analogue for the integral of (1 - exp(-x))/x to create a fake sqrt(x) ? ( I do NOT mean exp(fake(ln(x)/2) ) Answers to these will lead to new questions. I want to remark that +fake(sqrt(x) exp(x)) exp(-x) is very likely a better approximation then exp(+fake(ln(x) exp(x)) exp(-x) / 2). I also notice we can add variables to our ideas : 1) which is the best fake ln(x) ? integral(1-exp(- A x)/x) - C or C_2 *+fake(ln(x) exp(B x)) exp(-B x) ? Which leads to more intresting questions. Anyways the most intresting is imho : 2) Is there an analogue for the integral of (1 - exp(-x))/x to create a fake sqrt(x) ? regards tommy1729 When I printed the numbers, they needed very very high precision for the entiref version, so the exponent isn't visible unless you scroll to the right. But it will converge, but computations are tricky with the entiref version due to precision loss as large numbers cancel each other out to generate the very small ln(x+1) assymptotic, so it is much more straightforward to work with the $f(x)=\exp(x)\ln(x+1)$ version, where entiref=f(x)exp(-x). For exp^{0.5} the method with all positive derivatives woud be much much more compact, with very few Taylor series coefficients required to represent the function, and converges pretty well. If you generate a representation instead using exp(x)exp^{0.5)(x), then it will converge faster, but requires far more terms to do so, with something like 2n terms required for f(n). Also, it will only converge on the right hand side of the complex plane, whereas the all positive derivative version converges everywhere, and even at the negative real axis, it oscillates between zero and approximately 2x exp^{0.5}(x). - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 09/08/2014, 11:03 AM I looked at sheldon's 2nd answer and noticed what partially was the motivation for one of my last questions : the plot of fake ln(x) resembles integral (1 - exp(-x) ) / x ! It was "Visually Obvious" from the psuedoperiodicity and the growth rate for Re(z) << 0. I havent considered the signs of the derivatives of integral (1 - exp(-x) / x yet. Maybe that explains alot. regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/09/2014, 04:33 AM (This post was last modified: 09/09/2014, 01:05 PM by sheldonison.) (09/08/2014, 11:03 AM)tommy1729 Wrote: I looked at sheldon's 2nd answer and noticed what partially was the motivation for one of my last questions : the plot of fake ln(x) resembles integral (1 - exp(-x) ) / x ! It was "Visually Obvious" from the psuedoperiodicity and the growth rate for Re(z) << 0. I havent considered the signs of the derivatives of integral (1 - exp(-x) / x yet. Maybe that explains alot. regards tommy1729 I reread your post. So, to clarify, you are interested in comparing the fake function for $\ln(x) = \exp(-z)\text{fake}(\exp(z)\ln(z))$ with the integral (which I don't yet understand), integral (1 - exp(-x) ) / x. So far, I have only answered the Op's question, and generated the "fake" function for $\ln(z+1) \approx \exp(-z)\text{fake}(\exp(z)\ln(z+1))$ but doing so for $\ln(x)$ or $\sqrt{x}$ or $\exp^{0.5}(x)$ or $\frac{1}{x}$ would presumably be straightforward enough using the equation I posted in my 2nd answer; copied below. $a_n =\lim_{r\to\infty} \int_{-\pi}^{\pi} \frac{1}{2\pi}e^{-n(r+ix)}(\exp(e^{r+ix})\ln(1+e^{r+ix}))\; \mathrm{d}x\;\;$ fake function for exp(z)ln(z+1), replace ln(1+e^{}) with other positive valued function $\text{fake}(z) = \exp(-z) \, \sum_{n=0}^{\infty} a_n x^n\;\;$ the fake(z) zeros are probably always near the imaginary axis $\approx 2\pi i$ apart. Also, I posted the first 18 zeros of the asymptotic function for $\exp(x)\ln(x+1)$ at MSE, along with an approximation. It would be interesting if this equation is related to the integral solution posted on Mathoverflow, but I don't understand that solution (yet). - Sheldon sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 09/09/2014, 06:26 PM (This post was last modified: 09/10/2014, 03:56 PM by sheldonison.) (09/09/2014, 04:33 AM)sheldonison Wrote: (09/08/2014, 11:03 AM)tommy1729 Wrote: I looked at sheldon's 2nd answer and noticed what partially was the motivation for one of my last questions : the plot of fake ln(x) resembles integral (1 - exp(-x) ) / x ! It was "Visually Obvious" from the psuedoperiodicity and the growth rate for Re(z) << 0. I havent considered the signs of the derivatives of integral (1 - exp(-x) / x yet. Maybe that explains alot. regards tommy1729... So far, I have only answered the Op's question, and generated the "fake" function for $\ln(z+1) \approx \exp(-z)\text{fakelnx}(\exp(z)\ln(z+1))$ but doing so for $\ln(x)$ would presumably be straightforward enough Hey Tommy, Actually, its much more elegant to generate the fake function for ln(x) (than ln(x+1)), and, as you correctly guessed, it is exactly the same as the integral of (1 - exp(-x) ) / x solution! Of course, I can't prove it (yet), but numerically, its exact to the limit that I can generate the fake function Taylor series. The "fakelnx" solution equation simplifies for ln(x) to: $a_n =\lim_{r\to\infty} \int_{-\pi}^{\pi} \frac{1}{2\pi}e^{-n(r+ix)}\exp(e^{r+ix})(r+ix)\; \mathrm{d}x\;\;$ for large n, the integral at r=ln(n) works extremely well $\text{fakelnx}(z) = \exp(-z) \, \sum_{n=0}^{\infty} a_n x^n\;\;$ The integral solution is a simple entire Taylor series, though it does require twice as many terms to get good convergence. Do you have any links to the integral of integral of (1 - exp(-x) ) / x solution to the asymptotic approximation of ln(x) in the complex plane, maybe showing where the zeros are, and how to derive the a0 coefficient? I generated the Taylor series and the zeros and the complex plane graph for the fake function, which is visually the same as before. The approximation is the zeros of f(x) are approximately the zeros of $\exp(x)\ln(x)+x^{-1}-x^{-2}$ zeros of the asymptotic Taylor series of f(x), where $\ln(x)\approx \exp(-x)f(x)\;$ these zeros are exactly 1 greater than the zeros of the $\exp(x)\ln(x+1)$ asymptotic I posted on mathstack, in answering Mick's question. Code:zero[1]= 0.6763550778654 zero[2]= -3.014193883364 + 6.790134224598*I zero[3]= -3.782847137525 + 13.31987455364*I zero[4]= -4.248507475714 + 19.72207503441*I zero[5]= -4.583409659397 + 26.07792021174*I zero[6]= -4.844835249373 + 32.41104823228*I zero[7]= -5.059115679636 + 38.73113851434*I zero[8]= -5.240572105211 + 45.04296385204*I zero[9]= -5.397864071944 + 51.34917648054*I zero[10]= -5.536627830306 + 57.65137956884*I zero[11]= -5.660737008441 + 63.95060483952*I zero[12]= -5.772967940649 + 70.24754933585*I zero[13]= -5.875378096927 + 76.54270265230*I zero[14]= -5.969534334368 + 82.83641978994*I zero[15]= -6.056657240617 + 89.12896507913*I zero[16]= -6.137716050140 + 95.42053981217*I zero[17]= -6.213493139944 + 101.7113002628*I zero[18]= -6.284629098684 + 108.0013698029*I Taylor series for f(x), where $\ln(x)\approx \exp(-x)f(x)\;\;$ The error term for x=100 is 3.7E-46 An approximation for large n is $a_n\approx \ln(n+0.5)/n!$ notice that a0=-EulerPhi.... Code:{fakelnx ~= exp(x)ln(x)        -0.57721566490153286060651209008240243104215933593992 +x^ 1*  0.42278433509846713939348790991759756895784066406008 +x^ 2*  0.46139216754923356969674395495879878447892033203004 +x^ 3*  0.20935294473863341212113687387515515038186233289890 +x^ 4*  0.062754902851325019696950885135455454262132249891392 +x^ 5*  0.014217647236931670606056843693757757519093116644945 +x^ 6*  0.0026010893543034265824909554304411077346636675889723 +x^ 7*  0.00039992886467373214216990973269340087819458289819786 +x^ 8*  0.000053091306496914930469651414999373522472735560687431 +x^ 9*  0.0000062052264910348355038785718801969012819990325122033 +x^10*  0.00000064807996832746944104041450724367601885246057044432 +x^11*  0.000000061193825155719196346860601420828885238864788984523 +x^12*  0.0000000052734584045421671870651342044124190391765942791647 +x^13*  0.00000000041800375714695256408146599816261296296183472186151 +x^14*  3.0676750196048734913951264010092504902004812301510 E-11 +x^15*  2.0960977712820477707694251472447541297613548216991 E-12 +x^16*  1.3399328403787010148398795138474559812733276238735 E-13 +x^17*  8.0473377230715071910109939392797598907980150072089 E-15 +x^18*  4.5575165515318719197961986571001851250184494711409 E-16 +x^19*  2.4419594231569027457714518260362673329003457163469 E-17 +x^20*  1.2415312996950121971781158661103143683232247056530 E-18 +x^21*  6.0052582893257307543283973356372371365643437116337 E-20 +x^22*  2.7701028196591984032708574136646644636352129965898 E-21 +x^23*  1.2212106614632631494121854491216924290726114882837 E-22 +x^24*  5.1555334882259347204278024257071006128577447887905 E-24 +x^25*  2.0880011964279117817559003831596589939642935331579 E-25 +x^26*  8.1261429194621529090640377639159175389943080330102 E-27 +x^27*  3.0436962289259660631660351601039469159158981129458 E-28 +x^28*  1.0987482576059515865241269919203814411407055597183 E-29 +x^29*  3.8277869672151700119161577810252405452867105558248 E-31 +x^30*  1.2884956145011096894515623320939005339760596313537 E-32 +x^31*  4.1956673061827253837027808998413659853977660722504 E-34 +x^32*  1.3230222542910227561339742527403906609614580021133 E-35 +x^33*  4.0440563330660445648926094976778450972008364935595 E-37 +x^34*  1.1993905613003694656955658397091397527362477928820 E-38 +x^35*  3.4544804407400038762504447316858932217789985937355 E-40 +x^36*  9.6704517872301951114371135786099107617811821656656 E-42 +x^37*  2.6332719970328988715110651429505782573566576970054 E-43 +x^38*  6.9799779712718465546370325439885222772809906938190 E-45 +x^39*  1.8023083766248671765676856777913109113410290185636 E-46 +x^40*  4.5364113775403775876548026941017712014457115955210 E-48 +x^41*  1.1137328014175125933567546129869861759845314537825 E-49 +x^42*  2.6686909717828667444479935338379898294931402896110 E-51 +x^43*  6.2447513499076481310115732329654147057038279739689 E-53 +x^44*  1.4278113133454478974855760221290914072671436197014 E-54 +x^45*  3.1914910315391793365975572167819747254191359216660 E-56 +x^46*  6.9775308381625959301427270564390623481298207420352 E-58 +x^47*  1.4928078985749365678746849007271857212954266040863 E-59 +x^48*  3.1267986971785162607060674915230911476461510402713 E-61 +x^49*  6.4147723352279225531453584591845593166236564138976 E-63 +x^50*  1.2895303658788508267631349532278490303245075237949 E-64 +x^51*  2.5411320108221725479286061393374532447727295564903 E-66 +x^52*  4.9106346348668154497164547717815377655007541397561 E-68 +x^53*  9.3094850686649472206253875793277373051016813465776 E-70 +x^54*  1.7320008191374067477204080720504599283120867287661 E-71 +x^55*  3.1634128462587723403395241011643887317355583461067 E-73 +x^56*  5.6740670925058372131902540824043736094208642710853 E-75 +x^57*  9.9977930694762194846582425729533808769160304100329 E-77 +x^58*  1.7310924308536906046868980013353193814008830857469 E-78 +x^59*  2.9462764640942144757766343127462958590848324312570 E-80 +x^60*  4.9304904483845128374647134189799241414168541873582 E-82 +x^61*  8.1150684720533352796443669019602589412242975959324 E-84 +x^62*  1.3140072248285872935738443991756488960169177564096 E-85 +x^63*  2.0937318816588576190548470868569796570892351623049 E-87 +x^64*  3.2837701779251999087544223660466681359531516106385 E-89 +x^65*  5.0706074505672195579606100954869165403321429950741 E-91 +x^66*  7.7105729621622077873599101447226642146050406632429 E-93 +x^67*  1.1549241677675466249115772771205673568929741784449 E-94 +x^68*  1.7043476000648078825576088011256477299848855419037 E-96 +x^69*  2.4785382153355181142456349095181055097381201721554 E-98 +x^70*  3.5526949372480863188649463736826916909206153008007 E-100 +x^71*  5.0203563701334727144493939245675258370560353143692 E-102 +x^72*  6.9953986729220286705202294337362466359365648123984 E-104 +x^73*  9.6133828215490126811877251911240979265377813165857 E-106 +x^74*  1.3031910289910555096467725707115006055650061583240 E-107 +x^75*  1.7429624018520381177865219755515130838010329675454 E-109 +x^76*  2.3003500654527736291739536225264614802191025323224 E-111 +x^77*  2.9964128829325443326861647274649569589040207571395 E-113 +x^78*  3.8528762378249015228297200937610483334683261690396 E-115 +x^79*  4.8912078360079331167384967468993573540157848553962 E-117 +x^80*  6.1314753459808002074757652090764529053387058610005 E-119 +x^81*  7.5910188553896589198254614231271632732503148617811 E-121 +x^82*  9.2829943312081258653178716695729078376848711536329 E-123 +x^83*  1.1214866876330017067715237090532928812497782757604 E-124 +x^84*  1.3386952029851194214521788933697680384334504041334 E-126 +x^85*  1.5791117029921800906334375373586368217339140584430 E-128 +x^86*  1.8409759452832727603177383660269076889285324983831 E-130 +x^87*  2.1215176144112587986738668725960014270213332854535 E-132 +x^88*  2.4169419937195637104358582916013620975707654328002 E-134 +x^89*  2.7224715496431876894301253674263470340599943575343 E-136 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+x^279*  9.4038556368308995397568913793379947091035625133076 E-563 +x^280*  3.3606492343665157329697229608898756515437035031841 E-565 +x^281*  1.1967156657039872496806340384815551596429394434664 E-567 +x^282*  4.2463406814885223111667505102375130743499390920123 E-570 +x^283*  1.5014132019840775339042565143810507402574337843156 E-572 +x^284*  5.2899625245005917799920451377492618316178715347366 E-575 +x^285*  1.8572797469458407989057767546916334157379318177279 E-577 +x^286*  6.4980009176000772430924985714948958044962090377158 E-580 +x^287*  2.2655061720295298733579683215468706433711662893603 E-582 +x^288*  7.8711655714334002805698052630685889213059378931640 E-585 +x^289*  2.7252503713886831916892869041842233261363727385871 E-587 +x^290*  9.4031320795565700348527537599976981415263498968553 E-590 +x^291*  3.2332747214058454654056553776814625407003183140089 E-592 +x^292*  1.1079540657533670750213709293353852119014479585542 E-594 +x^293*  3.7836859640197928030255035689763898809807362270184 E-597 +x^294*  1.2877384366771383455442173002611824335164100683808 E-599 +x^295*  4.3678177857035932803561463137741736885158507046764 E-602 +x^296*  1.4764904522846115813849812490633608203940578183013 E-604 +x^297*  4.9742890139828740800335481562783343835900102686241 E-607 +x^298*  1.6702079985532166301461419574819243928674364798095 E-609 +x^299*  5.5892582156573519333552413618290111316747972794917 E-612 +x^300*  1.8641751918061524199262751313929480833397412117845 E-614; } Complex plane plot from -40, to +100, real, and -10 to +100 imag, grids every 10 units     - Sheldon « Next Oldest | Next Newest »

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