# Tag Info

9

Try ((# - Sqrt[#])/(# + Sqrt[#]) == (#^2 - #)/36) & /@ {1, 4, 8} {True, True, False} or (x |-> ((x - Sqrt[x])/(x + Sqrt[x]) == (x^2 - x)/36)) /@ {1, 4, 8} {True, True, False} ( |-> is Function, version 12.3!)

7

f = Expand @* SubtractSides; f[x^2 + (x + 82)^2 == (x + 100)^2] -3276 - 36 x + x^2 == 0 TraditionalForm @ % PolynomialForm[%%, TraditionalOrder -> True]

7

Perhaps Map is what you are looking for Map[(((x - Sqrt[x])/(x + Sqrt[x]) == (x^2 - x)/36)) /. x -> # &, {1,4, 8}] (*{True, True, False}*)

4

You can also use ReplaceAll, you just have to tell it to do multiple replacements for the different values of x: ((x - Sqrt[x])/(x + Sqrt[x]) == (x^2 - x)/36) /. {{x -> 1}, {x -> 4}, {x -> 8}} (* {True, False, True} *)

3

This is a great example to use the ComplexityFunctionoption to Simplify and the undocumented PolynomialForm. expr = x^2 + (x + 82)^2 == (x + 100)^2; Simplify[expr, ComplexityFunction -> (If[MatchQ[#1, _ == 0], 0, 1] &)] PolynomialForm[%, TraditionalOrder -> True] -3276 - 36 x + x^2 == 0 x^2 - 36 x - 3276 == 0

2

There are many ways of doing this, but I'd typically do it this way expr = x^2 + (x + 82)^2 == (x + 100)^2; # - Last[expr] & /@ % (* x^2 + (82 + x)^2 - (100 + x)^2 == 0 *)

1

\$Version (* "12.3.0 for Mac OS X x86 (64-bit) (May 10, 2021)" *) Clear["Global`*"] c[t_] = (2 - Sqrt[t + 3])/(1 - t); Re "to see its behavior at infinity" Asymptotic was introduced in version 12.1 c2[t_] = Asymptotic[c[t], t -> Infinity] (* 1/Sqrt[t] *) Or using Series c3[t_] = Series[c[t], {t, Infinity, 1}] // Normal (...

1

As an alternative you could use Fold Diffn1[f_, x_Symbol, n_Integer?Positive] := Fold[D[#1, {x, #2}] &, f, Range[n, 1, -1]] Diffn1[f[x], x, 3] (* Derivative[6][f][x] *) Or it can be simplified as Diffn2[f_, x_Symbol, n_Integer?Positive] := D[f, {x, n (n + 1)/2}] These are equivalent And @@ Table[Diffn1[f[x], x, n] == Diffn2[f[x], x, n], {n, 1, 20}]...

1

(* get our operators like this: *) n=3; operators = Array[Function[{y}, D[y, {x, #}]] &, n] (** returns: {Function[{y}, D[y, {x, 1}]], Function[{y}, D[y, {x, 2}]], Function[{y}, D[y, {x, 3}]]} **) (* Apply composition to put them together *) composed = Composition@@operators; (* try it out: *) composed[x^6] (* 720 *) (* verify it matches this ...

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