# Roots of a function

Hello,

Given the following function

```
f(x)=1/(x+1)^(1/3)
```

I would like to assign the 3 roots of this function to three different functions; say f1(x),f2(x),f3(x). How can I do that with sage?

Roots of a function

Hello,

Given the following function

```
f(x)=1/(x+1)^(1/3)
```

I would like to assign the 3 roots of this function to three different functions; say f1(x),f2(x),f3(x). How can I do that with sage?

2

**Assuming** that, as suggested by `tmonteil`

, you mean a function per cube root branch, the following one-liner :

```
sage: (f1,f2,f3)=tuple(((1/u)*1/(x+1)^3).function(x) for u in (x^3==1).roots(multiplicities=False))
```

does what you mean :

```
sage: [f1, f2, f3]
[x |--> 1/((x + 1)^3*(1/2*I*sqrt(3) - 1/2)),
x |--> 1/((x + 1)^3*(-1/2*I*sqrt(3) - 1/2)),
x |--> (x + 1)^(-3)]
```

But remember that the expression `1/(1+x)^(1/3)`

will evaluate to *some* cube root of `x+1`

, with no specification of which one :

```
sage: ((x+1)^(1/3)).subs(x=7)
2
sage: ((x+1)^(1/3)).subs(x=-9)
2*(-1)^(1/3)
sage: ((x+1)^(1/3)).subs(x=-9).n()
1.00000000000000 + 1.73205080756888*I
```

*Caveat emptor*... HTH,

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Asked: ** 2020-06-08 21:51:11 +0200 **

Seen: **269 times**

Last updated: **Jun 08 '20**

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What do you mean by

root? Do you meanbranchof the cube root ?