Taylor series representation as an infinite sum

Question

I want to see the Taylor series representation for arbitrary functions, e.g. $\sin$. With the Series[] command, I can only see the first $n$ terms. Is there the possibility to show the infinite sum representation?

Thanks for the help!

Basheer Algohi · Accepted Answer · 2015-05-15T21:29:45.433

9

Borrowing from Chip Hurst:

f = Inactive[Sum][
  Assuming[n > 0, SeriesCoefficient[Sin[x], {x, 0, n}]] x^n, {n, 
   0, Infinity}]

enter image description here

ExpToTrig[Activate[f]] // Simplify

(*Sin[x]*)

edited May 15 '15 at 21:29

answered May 15 '15 at 21:07

Basheer Algohi

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Your Assuming[] does nothing unless there is a Simplify[] or FullSimplify[] within. – J. M.'s missing motivation May 15 '15 at 22:01
5

@J. M. Actually there is a difference (in V10.1): http://i.stack.imgur.com/ZwYxv.png. -- Algohi, one can also use FullSimplify instead of the Simplify @ ExpToTrig[..]. – Michael E2 May 16 '15 at 02:14
@Michael, version 10 is looking weirder to me with each passing day… :o – J. M.'s missing motivation May 16 '15 at 03:16
@MichaelE2 I did not know that FullSimplify can do that. I checked other examples and FullSimplify gives same results as ExpToTrig. Thanks – Basheer Algohi May 16 '15 at 04:13

score 7 · Answer 2 · answered May 15 '15 at 19:57

7

Try this

FullSimplify[SeriesCoefficient[Sin[x], {x, 0, n}], n >= 0 && n ∈ Integers]

((-1)^(1 + n) Sin[n π/2])/n!

answered May 15 '15 at 19:57

Greg Hurst

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Bit then I am expressing Sin with another Sin. For Sin(x) I know the Row. I am trying to get some result like: Sum[((-1)^k*x^(2*k + 1))/(2*k + 1)!, {k, 0, Infinity}] . – McLawrence May 15 '15 at 20:15
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@McLawrence Yes you can post code in comments as well: enclose it in single backticks like this: ` ` – MarcoB May 15 '15 at 20:18

Taylor series representation as an infinite sum

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