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Is it possible to Hold/Defer some of outer expressions, having evaluates inner ones?

For example, I want to prepare a table of expression values, like this

Table[Defer[Sin[x]^2 + Cos[x]^2] == Sin[x]^2 + Cos[x]^2, {x, 0, 1, 0.1}] // TableForm

Currently this displays left hand sides as

Sin[x]^2 + Cos[x]^2

but can I have x-s evaluated/substituted?

UPDATE

Or just simpler example.

Suppose I write

x = 12
Defer[Sin[x]^2 + Cos[x]^2] == Sin[x]^2 + Cos[x]^2

which gives

12
Sin[x]^2 + Cos[x]^2 == 1.

but can I have

12
Sin[12]^2 + Cos[12]^2 == 1.
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    $\begingroup$ Are you working on V10? $\endgroup$
    – Kuba
    Nov 25, 2014 at 8:06
  • $\begingroup$ No, I have V9. What advantages V10 has? $\endgroup$
    – Suzan Cioc
    Nov 25, 2014 at 12:28
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    $\begingroup$ @SuzanCioc v10 has a set of functions exactly for this. Please check Inactive. $\endgroup$
    – Silvia
    Nov 25, 2014 at 12:37
  • $\begingroup$ Incredible. Why did they wait 9 versions? $\endgroup$
    – Suzan Cioc
    Nov 25, 2014 at 12:50

3 Answers 3

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Your can use Replace* inside Hold or Deffer like this:

x = 12;
test = Defer[Sin[x]^2 + Cos[x]^2] == Sin[x]^2 + Cos[x]^2

Sin[x]^2 + Cos[x]^2 == 1.

Block[{x}, test /. x -> y] /. y -> x

Sin[12.]^2 + Cos[12.]^2 == 1.

Just idea, no doubt better approach can be found.

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  • $\begingroup$ It seems to be even a bit more simple than that: Defer[Sin[x]^2 + Cos[x]^2] == Sin[x]^2 + Cos[x]^2 /.x -> 12// Simplify yields what you need. $\endgroup$ Nov 25, 2014 at 8:41
  • $\begingroup$ I assume that x has a value, so 12->12. $\endgroup$
    – Acus
    Nov 25, 2014 at 9:18
  • $\begingroup$ In my code x has no value. I do not see your point. $\endgroup$ Nov 25, 2014 at 9:26
  • $\begingroup$ Not my point. It is put in the formulation of the task: x=12; $\endgroup$
    – Acus
    Nov 25, 2014 at 9:34
  • $\begingroup$ Nice, but Alexei's solution in its present form doesn't work in a table, i.e. Table[Defer[Sin[x]^2 + Cos[x]^2] == Sin[x]^2 + Cos[x]^2 /. x -> 12 // Simplify, {x, 0, 1, 0.1}] // TableForm $\endgroup$ Nov 25, 2014 at 9:41
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Mapping Hold or HoldForm over all the elements effectively freezes the expression for easier manipulation. (This method was used in R. Maeder's Programming in Mathematica, page 137.)

x = 12.;
Replace[Map[HoldForm, Defer[Sin[x]^2 + Cos[x]^2], Infinity] ,
  HoldForm[x] -> HoldForm[12]] == Sin[x]^2 + Cos[x]^2

Sin[12]^2+Cos[12]^2==1.

And in a table:

Table[
  ReplaceAll[Map[HoldForm, Defer[Sin[x]^2 + Cos[x]^2], Infinity] ,
    HoldForm[x] -> HoldForm[Evaluate[x]]] == Sin[x]^2 + Cos[x]^2,
  {x, 0, 1, 0.1}] // TableForm

Simpler method

Table[With[{y = x}, Defer[Sin[y]^2 + Cos[y]^2] == Sin[x]^2 + Cos[x]^2],
  {x, 0, 1, 0.1}] // TableForm
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You could also use Activate/Inactivate:

TableForm @ Activate @ Table[
    Inactivate @ Defer[Sin[x]^2+Cos[x]^2] == Sin[x]^2+Cos[x]^2,
    {x, 0, 1, 0.1}
] //TeXForm

$\begin{array}{c} \sin ^2(0.)+\cos ^2(0.)=1. \\ \sin ^2(0.1)+\cos ^2(0.1)=1. \\ \sin ^2(0.2)+\cos ^2(0.2)=1. \\ \sin ^2(0.3)+\cos ^2(0.3)=1. \\ \sin ^2(0.4)+\cos ^2(0.4)=1. \\ \sin ^2(0.5)+\cos ^2(0.5)=1. \\ \sin ^2(0.6)+\cos ^2(0.6)=1. \\ \sin ^2(0.7)+\cos ^2(0.7)=1. \\ \sin ^2(0.8)+\cos ^2(0.8)=1. \\ \sin ^2(0.9)+\cos ^2(0.9)=1. \\ \sin ^2(1.)+\cos ^2(1.)=1. \\ \end{array}$

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