Evaluate with numbers from list

Question

This code works:

x=2
y/.Solve[y==x^2]
Out[138]= 2
Out[139]= {4}

So above I put number in variable x and getting the result of corresponding y

Now I want to give list with several values in list in variable x and get evaluated values of corresponding y's. How can I do that?

I tried this but it does not works:

x={1,2,3,4}
y/.Solve[y==x^2]
Out[140]= {1,2,3,4}
Out[141]= y

Expected output is:

{1,4,9,16}

Solve needs the variable you're solving for in the second argument, and you want to solve for the array - Array[y, 4]/.First[Solve[Array[y, 4] == x^2, Array[y, 4]]]. But obviously you can just do {1,2,3,4}^2 — flinty, Commented Jul 13, 2020 at 0:16
@flinty This does not works: x = {1, 2, 3, 4} y /. Solve[Array[y, 4] == x^2, Array[y, 4]]. How can I fix it? — vasili111, Commented Jul 13, 2020 at 0:22
What doesn't work? It works perfectly fine for me - you need a semicolon after your first list. — flinty, Commented Jul 13, 2020 at 0:25

Blueka · Accepted Answer · 2020-07-13 16:34:06Z

2

Maybe you can use the following codes:

x = {1, 2, 3, 4};Solve[y == z^2] /. z -> x

The final result is：

{{y -> {1, 4, 9, 16}}}

answered Jul 13, 2020 at 16:34

Blueka

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TumbiSapichu · Accepted Answer · 2020-07-13 00:32:55Z

1

Why not bypass Solve, and apply the square directly:

x={1, 2, 3, 4};

y = x^2

{1, 4, 9, 16}

More complicated operations:

y= somethingComplicated[#]&/@x

answered Jul 13, 2020 at 0:12

TumbiSapichu

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$\begingroup$ Thank you. I need a way to use x={1, 2, 3, 4} in y/.Solve[y==x^2] . $\endgroup$
– vasili111
Commented Jul 13, 2020 at 0:23
1

$\begingroup$ If bypassing Solve, you only need x = {1, 2, 3, 4}; y = x^2 since Power is Listable $\endgroup$
– Bob Hanlon
Commented Jul 13, 2020 at 0:30

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2 Answers 2