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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