If X = A^9 , then relative error in X is how many times the relative error in A?
Answers
I assume you mean f = x^3. Then the relative error in f is 3 times the relative error in X. Proof: take log of both sides: ln F = 3 ln X. Now differentiate both sides: delta f/f = 3 delta X/X.
Sorry for the mix up between lower and upper cases!
Answer:
y=x
n
⇒
dx
dy
=nx
n−1
Approximate error in y is dy=(
dx
dy
)Δx
=nx
n−1
Δx
Relative error in y is
y
dy
=
x
n
Δx
Approximate error in x is dx=(
dy
dx
)Δy
=
nx
n−1
1
Δy
Relative error in x is
x
dx
=
nx
n
1
Δy
Required ratio =
nx
n
1
Δy
x
n
Δx
=n
2
x
n−1
Δy
Δx
=
1
n
So, the ratio of relative errors in y and x is n:1.
Explanation:
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