Sum of squares of average change is less than

[caffeinemachine]

Hello MHB,

I have no good ideas on how to go about solving the following:

Let $f:[a,b]\to\mathbb R$ and $g:[a,b]\to\mathbb R$ be real values functions both of which are differentiable in $(a,b)$. Show that there is an $x\in(a,b)$ such that $$\left(\frac{f(b)-f(a)}{b-a}\right)^2+\left(\frac{g(b)-g(a)}{b-a}\right)^2\leq (f'(x))^2+(g'(x))^2$$

Please help.

 

[Opalg]

Re: Sum of squares of average change is less than ...

Hello MHB,

I have no good ideas on how to go about solving the following:

Let $f:[a,b]\to\mathbb R$ and $g:[a,b]\to\mathbb R$ be real values functions both of which are differentiable in $(a,b)$. Show that there is an $x\in(a,b)$ such that $$\left(\frac{f(b)-f(a)}{b-a}\right)^2+\left(\frac{g(b)-g(a)}{b-a}\right)^2\leq (f'(x))^2+(g'(x))^2$$

Please help.

Here's an idea to get you started. Define a path in 2-dimensional space by the function $h:[a,b] \to \mathbb{R}^2$ given by $h(t) = \bigl(f(t),g(t)\bigr)\ (a\leqslant t\leqslant b).$ Then the distance between the endpoints is less than (or equal to) the length of the path.

 

[caffeinemachine]

Re: Sum of squares of average change is less than ...

Here's an idea to get you started. Define a path in 2-dimensional space by the function $h:[a,b] \to \mathbb{R}^2$ given by $h(t) = \bigl(f(t),g(t)\bigr)\ (a\leqslant t\leqslant b).$ Then the distance between the endpoints is less than (or equal to) the length of the path.

Here's an idea to get you started. Define a path in 2-dimensional space by the function $h:[a,b] \to \mathbb{R}^2$ given by $h(t) = \bigl(f(t),g(t)\bigr)\ (a\leqslant t\leqslant b).$ Then the distance between the endpoints is less than (or equal to) the length of the path.

So the above says that:

$\sqrt{(f(b)-f(a))^2+(g(b)-g(a))^2}\leq \displaystyle\int_a^b\left[\sqrt{(f'(x))^2+(g'(x))^2}\right]dx$. Now we use the fact that if $h:[a,b]\to\mathbb R$ is a continuous function then $\int_a^bh(x)dx\leq h(x_0)(b-a)$ for some $x_0\in (a,b)$.

Is this correct?

 

[Opalg]

Re: Sum of squares of average change is less than ...

So the above says that:

$\sqrt{(f(b)-f(a))^2+(g(b)-g(a))^2}\leq \displaystyle\int_a^b\left[\sqrt{(f'(x))^2+(g'(x))^2}\right]dx$. Now we use the fact that if $h:[a,b]\to\mathbb R$ is a continuous function then $\int_a^bh(x)dx\leq h(x_0)(b-a)$ for some $x_0\in (a,b)$.

Is this correct?

That's exactly what I was thinking of. (Nod)

 

[blue_raver22]

I would approach this problem by first breaking it down into smaller, more manageable parts. I would start by looking at the left side of the inequality and trying to understand what it represents. The sum of squares of average change can be interpreted as the average rate of change of both functions over the interval [a,b], squared and added together. This can also be thought of as the total change in both functions over the interval, squared and added together.

Next, I would look at the right side of the inequality and try to understand what it represents. The right side is the sum of squares of the derivatives of both functions evaluated at some point x in the interval (a,b). This can be interpreted as the instantaneous rate of change of both functions at a specific point x.

Based on this interpretation, it makes intuitive sense that the total change over an interval would be less than or equal to the instantaneous change at a specific point. This is because the instantaneous change at a point is the limit of the average change over smaller and smaller intervals. Therefore, it is reasonable to expect that there would be at least one point x in the interval (a,b) where the instantaneous change is larger than or equal to the average change over the entire interval.

To prove this mathematically, I would use the Mean Value Theorem, which states that for a differentiable function over an interval, there exists at least one point where the derivative is equal to the average rate of change over that interval. This point can be used to show that the right side of the inequality is greater than or equal to the left side, thus proving the statement.

In conclusion, as a scientist, I would approach this problem by first understanding the concepts behind the inequality and then using mathematical tools such as the Mean Value Theorem to prove it. This approach allows for a deeper understanding of the problem and its solution, rather than simply relying on a formula or algorithm.