我建立一個homecooked功能來恢復的象征衍生表達和呼叫的衍生功能。
Pkg.add("Symbolics")
using Symbolics
function derivative(exp,x)
@variables x
Dx=Differential(x)
?e = exp
return (expand_derivatives(Dx(?e))), first(substitute.(expand_derivatives(Dx(?e)), (Dict(x => ξ),)))
end
# 結果：：導數
derivative(-x^2 0.1*sin(x) 2*sin(x)^2,x)
# 結果：（錯誤）
ERROR: MethodError: no method matching ^(::StepRangeLen{Float64, Base.TwicePrecis
ion{Float64}, Base.TwicePrecision{Float64}}, ::Int64)
Closest candidates are:
^(::Union{AbstractChar, AbstractString}, ::Integer) at strings/basic.jl:718
^(::LinearAlgebra.Symmetric{var"#s814", S} where {var"#s814"<:Real, S<:(Abstrac
tMatrix{var"#s814"} where var"#s814"<:var"#s814")}, ::Integer) at /build/julia/sr
c/julia-1.6.3/usr/share/julia/stdlib/v1.6/LinearAlgebra/src/symmetric.jl:868
^(::LinearAlgebra.Symmetric{var"#s814", S} where {var"#s814"<:Complex, S<:(Abst
ractMatrix{var"#s814"} where var"#s814"<:var"#s814")}, ::Integer) at /build/julia
/src/julia-1.6.3/usr/share/julia/stdlib/v1.6/LinearAlgebra/src/symmetric.jl:869
...
Stacktrace:
[1] macro expansion
@ ./none:0 [inlined]
[2] literal_pow(f::typeof(^), x::StepRangeLen{Float64, Base.TwicePrecision{Float
64}, Base.TwicePrecision{Float64}}, #unused#::Val{2})
@ Base ./none:0
[3] top-level scope
@ REPL[191]:1
以互動方式呼叫我希望該函式執行的操作（預期行為）：
@variables x
Dx=Differential(x)
?e = -x^2 0.1*sin(x) 2*sin(x)^2
(expand_derivatives(Dx(?e)))
# 結果：：0.1cos(x) 4cos(x)*sin(x) - 2x
d?(ξ) = first(substitute.(expand_derivatives(Dx(?e)), (Dict(x => ξ),)))
# 結果：：dφ
d?(1)
# 結果：：-0.12737491576182247
uj5u.com熱心網友回復：
您可以將exp變成一個匿名函式，x作為變數接收，如下所示：
julia> function derivative(exp, ξ)
@variables x
Dx=Differential(x)
?e = exp(x) # <-- change here
return (expand_derivatives(Dx(?e))), first(substitute.(expand_derivatives(Dx(?e)), (Dict(x => ξ),)))
end
julia> derivative(x->-x^2 0.1*sin(x) 2*sin(x)^2, 1)
(0.1cos(x) 4cos(x)*sin(x) - (2x), -0.12737491576182247)
如果您愿意，也可以使用do 語法，這相當于傳遞x -> ....
julia> derivative(1) do x
-x^2 0.1*sin(x) 2*sin(x)^2
end
(0.1cos(x) 4cos(x)*sin(x) - (2x), -0.12737491576182247)
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