MHB Proving Convexity & Estimating Function $f(x)=(1+x)^n$ for $n\in \mathbb{N}$

[mathmari]

Hey!

I want to show that the function $f(x)=(1+x)^n, x\geq -1$ is for $n\in \mathbb{N}$ convex.

So that the function is convex it must hold $f''(x)>0$.

The second derivative is $f''(x)=n(n-1)(1+x)^{n-2}$.
It holds that $n>0$ and $n-1\geq 0$. We also have that $x\geq -1$. Therefore, we get that $f''(x)\geq 0$ for $n\in \mathbb{N}$, or not? (Wondering)

Is there a problem if we have $\geq 0$ instead of $>0$ ? (Wondering)

Or do we have to use the bernoulli inequality or an other way? (Wondering) Then I want to determine the equation of the tangent at $x_0=0$.

The equation of the tangent at $x_0=0$ is $t(x)=f'(0)x+f(0)$.

We have that $f'(x)=n(1+x)^{n-1}$. Therefore we get $$t(x)=nx+1$$ right? (Wondering) Then I want to say what estimate we get for the function $f(x)$ from the above information. What exactly do we have to say here? What is meant by "estimate" ? (Wondering)

 

[caffeinemachine]

Hey!

I want to show that the function $f(x)=(1+x)^n, x\geq -1$ is for $n\in \mathbb{N}$ convex.

So that the function is convex it must hold $f''(x)>0$.

The second derivative is $f''(x)=n(n-1)(1+x)^{n-2}$.
It holds that $n>0$ and $n-1\geq 0$. We also have that $x\geq -1$. Therefore, we get that $f''(x)\geq 0$ for $n\in \mathbb{N}$, or not? (Wondering)

Is there a problem if we have $\geq 0$ instead of $>0$ ? (Wondering)

Or do we have to use the bernoulli inequality or an other way? (Wondering) Then I want to determine the equation of the tangent at $x_0=0$.

The equation of the tangent at $x_0=0$ is $t(x)=f'(0)x+f(0)$.

We have that $f'(x)=n(1+x)^{n-1}$. Therefore we get $$t(x)=nx+1$$ right? (Wondering) Then I want to say what estimate we get for the function $f(x)$ from the above information. What exactly do we have to say here? What is meant by "estimate" ? (Wondering)

Showing that $f(x)=(1+x)^n$ is convex for $x\geq -1$ is as good as showing that $g(x)=x^n$ is convex for $x\geq 0$. Now is it easier?

 

[mathmari]

Showing that $f(x)=(1+x)^n$ is convex for $x\geq -1$ is as good as showing that $g(x)=x^n$ is convex for $x\geq 0$. Now is it easier?

Using the method with the second derivative we have the following:

$$f''(x)=n(n-1)(1+x)^{n-2}\geq 0 \text{ for } x\geq -1 \text{ and } n\in \mathbb{N}$$ right? (Wondering)

Having that $f''(x)\geq 0$, we get that $f$ is convex (but not strictly convex), or not? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Is there a problem if we have $\geq 0$ instead of $>0$ ? (Wondering)

It means it's convex but not strictly convex.

We have that $f'(x)=n(1+x)^{n-1}$. Therefore we get $$t(x)=nx+1$$ right? (Wondering)

Then I want to say what estimate we get for the function $f(x)$ from the above information. What exactly do we have to say here? What is meant by "estimate" ? (Wondering)

The function $t(x)=1+nx$ is already an 'estimate' for $f(x)$.
That is, it's not the real function, but it approximates it up to the first degree.
We can get higher order Taylor expansions to get better 'estimates', such as:
$$f(x) \approx 1 + nx + \frac 1{2!}f''(0)x^2 = 1 + nx + \frac 12 n(n-1)x^2$$
This one may be intended, since we already have the information to calculate $f''(0)$. (Thinking)

 

[mathmari]

The function $t(x)=1+nx$ is already an 'estimate' for $f(x)$.
That is, it's not the real function, but it approximates it up to the first degree.
We can get higher order Taylor expansions to get better 'estimates', such as:
$$f(x) \approx 1 + nx + \frac 1{2!}f''(0)x^2 = 1 + nx + \frac 12 n(n-1)x^2$$
This one may be intended, since we already have the information to calculate $f''(0)$. (Thinking)

Ah ok. For the estimate do we use also the information that $f$ is convex? (Wondering)

 

[I like Serena]

Ah ok. For the estimate do we use also the information that $f$ is convex? (Wondering)

Not really.

If we want we could use the information that $f''(x)\ge 0$, which is what it means for a twice differentiable function to be convex.
To do so we can write the Taylor expansion as:
$$f(x) = 1 + nx + \frac 1{2!} f''(\xi) x^2$$
where $0\le \xi \le x$ (consequence of the mean value theorem).

Since $f''(\xi) \ge 0$ it follows that:
$$f(x) \ge 1 + nx$$
(Thinking)

 

[mathmari]

Not really.

If we want we could use the information that $f''(x)\ge 0$, which is what it means for a twice differentiable function to be convex.
To do so we can write the Taylor expansion as:
$$f(x) = 1 + nx + \frac 1{2!} f''(\xi) x^2$$
where $0\le \xi \le x$ (consequence of the mean value theorem).

Since $f''(\xi) \ge 0$ it follows that:
$$f(x) \ge 1 + nx$$
(Thinking)

I see! Thank you very much! (Happy)