Question

] Show that if n is a natural number and α, β are real numbers with β > 0 then there exists a real function f with derivatives of all orders such that: (i) |f^(k)(x)| ≤ β for k ∈ {0, 1, ..., n − 1} and x ∈ (−∞, ∞); (ii) f^(k)(0) = 0 for k ∈ {0, 1, ..., n − 1}; (iii) f ⁽ⁿ⁾(0) = α.

Answer 1

Given

n is natural number

$\alpha \space ,\beta$ are real numbers

$\beta$ > 0

(i) Yes derivative exists as seen below

The function is defined

As follows

|f^(k)(x)| ≤ β

k ∈ {0, 1, ..., n − 1} and x ∈ (−∞, ∞); 

As we know

$\beta$ > 0

|f^(k)(x)| ∈ (0, ∞);

As the function is always positive , as it is under mod

And the function is less than $\beta$

(ii)

Yes derivative exists as seen below

The function is defined

As follows

f^(k)(0) = 0

for k ∈ {0, 1, ..., n − 1}

As we can see than , k^th order derivative of the function is 0

$\implies$ that the function is defined and derivative exists , nth order function is also defined

It can also be used to find the roots and k+1 order to find monotonocity of the function on graph

(iii)

Yes derivative exists as seen below

The function is defined

As follows

f ⁽ⁿ⁾(0) = α.

$\implies$

(n+1 )th order derivative is zero

$\implies$

At nth order derivative $\alpha$ is the root of the function

$\implies$

f($\alpha$ ) = 0