2
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*For further study of series, they can be put into a different form.

Normal[Series[1/(1 - x), {x, 0, 10}]]

This serie 1/(1-x) = 1 + x + ... ,must be first integrated and then multiplied by -1 for both sides of the equation.

I was thinking on the Map function, but how to do that ?

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  • $\begingroup$ Map is the right function to use. Which part are you having difficulty? $\endgroup$
    – xzczd
    Feb 28, 2022 at 10:36
  • $\begingroup$ Its mapping about a equation, but i think i have to split the mapping for the LHS and RHS of the equation Normal[Series[1/(1 - x), {x, 0, 10}] $\endgroup$
    – janhardo
    Feb 28, 2022 at 10:56
  • $\begingroup$ You don't need any splitting. Please read the document of Map carefully, especially the Generalizations & Extensions section. If you still don't get it, execute f/@(aaa==bbb) and observe. $\endgroup$
    – xzczd
    Feb 28, 2022 at 11:10
  • $\begingroup$ Thanks, I will do that and see what happens $\endgroup$
    – janhardo
    Feb 28, 2022 at 11:19
  • 1
    $\begingroup$ @janhardo Could you explain why you haven't accepted any answers to your questions? $\endgroup$
    – Artes
    Mar 6, 2022 at 0:35

3 Answers 3

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eqn = 1/(1 - x) == Normal@Series[1/(1 - x), {x, 0, 10}]
func = expr |-> -Integrate[expr, x];
func /@ eqn

Alternatively:

Map[func][eqn]

To be more compact:

-Integrate[#, x] & /@ (1/(1 - x) == Normal@Series[1/(1 - x), {x, 0, 10}])
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  • $\begingroup$ Never could make it to this..., This expression -Integrate[#1, x] & as pure function is standing for a indefinite integral operator ? Next step to do is multiplying by -1 with Map $\endgroup$
    – janhardo
    Feb 28, 2022 at 12:24
  • $\begingroup$ Wait.. this minus sign in -Integrate[#1, x] is -1 multiplying to Integrate , so the equation is now intergated and multiplied. Yes, the post is crossing.. $\endgroup$
    – janhardo
    Feb 28, 2022 at 12:58
  • $\begingroup$ @janhardo Yes, if you want to first integrate and then multiply by -1, just define another pure function in a similar manner as I've done above. $\endgroup$
    – xzczd
    Feb 28, 2022 at 13:01
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    $\begingroup$ Unfortunately, I am not yet strong in coming up with pure functions. Need to get more familiar with it. $\endgroup$
    – janhardo
    Feb 28, 2022 at 13:09
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    $\begingroup$ @janhardo the suggestion that was made that you should be accepting an answer by clicking on the checkmark is good practice in general for the site. This site is meant to be useful for other users that might face similar difficulties and it is easier for them to know which is the preferred approach to a particular question. In the answer by xzczd you can find THREE (3) approach that answer your question. I am just mentioning this as a general comment. It's a (+1) to be honest. $\endgroup$
    – user49048
    Feb 28, 2022 at 20:22
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eqn1 = 1/(1 - x) == Series[1/(1 - x), {x, 0, 10}]

enter image description here

As an alternative to Map you can use ApplySides

eqn2 = ApplySides[-Integrate[#, x] &, eqn1]

enter image description here

Series[eqn2[[1]], {x, 0, 11}] == eqn2[[2]]

(* True *)
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  • $\begingroup$ is useful ,thanks $\endgroup$
    – janhardo
    Feb 28, 2022 at 20:22
1
$\begingroup$

Try this

Integrate[(-1)*Normal[Series[1/(1 - x), {x, 0, 10}]], x]

(*  -x - x^2/2 - x^3/3 - x^4/4 - x^5/5 - x^6/6 - x^7/7 - x^8/8 - x^9/9 - \
x^10/10 - x^11/11  *)

Have fun!

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1
  • $\begingroup$ @ Alexei Boulbitch , yes that is the familiar integrating for the RHS $\endgroup$
    – janhardo
    Feb 28, 2022 at 12:27

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