Question

Show that the map Q[X] → Q[X], sum^{n}{i=0}(a_nX^i) → sum^{n}{i=0}(a_n(2X + 3)^i) , is an automorphism of Q[X],

Answer #1

Let ϕ:Q[x]→Q[x] be given by ϕ(f(x))=f(2x+3). First note that if φ:Q→Q is given by φ(x)=2x+3, we can re-write ϕ as ϕ(f)=f∘φ.

We want to see if **ϕ is an automorphism**.

Firstly, we check that ϕ is well defined. Clearly φ is a
polynomial, and the composition of polynomials is a polynomial, so
**ϕ is well defined**.

Now we want to check that **it preserves the
structure**. Let f,g∈Q[x]. Then

ϕ((f+g)(x))=(f+g)(2x+3)=f(2x+3)+g(2x+3)=ϕ(f(x))+ϕ(g(x))

Similarly,

ϕ((fg)(x))=(fg)(2x+3)=f(2x+3)g(2x+3)=ϕ(f(x))ϕ(g(x))

Finally it suffices to show that **ϕ is
bijective**.

First we show **ϕ is injective**.

Let f,g∈Q[x] and suppose ϕ(f)=ϕ(g).

Then f∘φ=g∘φ,

so f=f∘φ∘φ^{−1}=g∘φ∘φ^{−1}=g (as we can easily
see that φ is invertible.)

Similarly, we can show that **ϕ is
surjective**.

Thus, **ϕ is an automorphism.**

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A population of values has a normal distribution with
μ=58.6μ=58.6 and σ=59.9σ=59.9. You intend to draw a random sample
of size n=11n=11.
Find the probability that a single randomly selected value is
greater than 26.1.
P(X > 26.1)...

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