I was watching this tutorial and at timestamp 3:03 he did this:
This blows my mind because I didn't expect that to actually work, how the heck does that work ???? For example, if I have vector (2, 3, 4), that means:
$$ {(2, 3, 4)}^{(8, 8, 8)} $$
How does that work mathematically?
The Shader Graph documentation for Power Node says:
The following example code represents one possible outcome of this node.
void Unity_Power_float4(float4 A, float4 B, out float4 Out) { Out = pow(A, B); }
And if we then look up the documentation for the pow() function using float4 it says (emphasis mine):
pow(float4, float4)
Returns the componentwise result of raising x to the power y.
Thus the comment by @PepeOjeda is correct and with respect to Shader Graph, you have: $$(2,3,4)^{(8,8,8)} = (2^8,3^8,4^8)$$
From a float4 is a vector perspective this might feel like misleading overload. But from a SIMD (single instruction, multiple data) perspective, the operation makes sense. In that context, the float4 can be seen as parallel data packaged together for
parallel processing, hence the componentwise math operation. As @DMGregory points out, this convention is common in in shader languages and many operators are treated this way. For example, if you search the Math class docs for "returns the componentwise", you'll see over a dozen functions operate as described.
float4 aren't always thought of as vectors in the mathematical sense, but as parallel streams of data that all need to go through the same series of operations, so we might as well do them four at a time in parallel. pow(a, b) then doesn't mean "raise vector a to the power of vector b" but "apply the power operation to these four separate pairs of bases and exponents, in parallel". The same happens even for basic multiplication.
\$\endgroup\$
(2⁸, 3⁸,4⁸), since that is how shaders tend to handle scalar math on vectors. I can't really verify this, though. \$\endgroup\$