Displaying 1000 terms taylor series

Hi,

I am trying to print up to 1000 terms of a Taylor series but my output either bleeds together or shows as a continuous line of numbers rather than separate fractions.

f = (x/sqrt(x^2-6*x+13));f taylor(f,x,4,500)

My output comes out as "fetch additional output" and the result is this: $\displaystyle \frac{79854682306522793194207787155926275411329758659461959227239100374575221031860704963210424694213147821527965196776756030407420591783729164672959605324317617935505546915704931}{305493636349960468205197939321361769978940274057232666389361390928129162652472045770185723510801522825687515269359046715531785342780428396973513311420091788963072442053377285222203558881953188370081650866793 And it goes on like this. Any tips or pointers? When I try 1000 terms it just gives me this error: "WARNING: 1 intermediate output message was discarded." edit retag close merge delete Comments Please provide more information: is this from the command-line, the notebook, CoCalc, SageCell? taylor(f, x, 4, 1000) works for me from the command line and the notebook. What are you doing to get LaTeX output? ( 2022-11-29 23:28:27 +0200 )edit 1 Answer Sort by » oldest newest most voted A trivial, brute-force solution to get a Latex expression of this series : # The function sage: f=(x/sqrt(x^2-6*x+13)) Its requested Taylor series sage: %time s=f.taylor(x,4,1000) CPU times: user 5.57 s, sys: 11.9 ms, total: 5.58 s Wall time: 5.29 s # Latex each ot its terms sage: %time t=list(map(lambda u:str(latex(u)), reversed(s.operands()))) CPU times: user 59.2 ms, sys: 83 µs, total: 59.3 ms Wall time: 59.3 ms # Latex of the series sage: %time p = " + ".join(t) CPU times: user 127 µs, sys: 0 ns, total: 127 µs Wall time: 133 µs  which I won't try to print since it would require about sage: ceil(len(p)/35/60) 274  274 pages typeset at 60 characters per line on 35 lines per page. Note that the result you'll get will express the development in terms of powers of$x-4$; if you need it in terms of$x$, try the same trick on se = s.expand(). This will tell you that the$x^{1000}$coefficient is about$1.3\cdot10^{-352}\$.

What are you trying to do ? Frighten the horses ?

HTH,

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