Solving Exercise - Proving f(1 + w) > f(1) + w*f(1)

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In summary, we are asked to prove that for a strictly convex function $f(x)$ that starts at the origin and continues into the first quarter, for any $w>0$, we have $f(1+w) > f(1) + wf(1)$.
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IlanSherer
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Hello :)

Help please about the following exercise :

The function f(x) comes up and is an convex which gets out from (0,0) and continues into the first quarter.
Prove that for all w > 0, takes place f(1 + w) > f(1) + w*f(1).

I have been trying for two days to solve it :\

Thanks a lot!
 
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  • #2
IlanSherer said:
Hello :)

Help please about the following exercise :

The function f(x) comes up and is an convex which gets out from (0,0) and continues into the first quarter.
Prove that for all w > 0, takes place f(1 + w) > f(1) + w*f(1).

I have been trying for two days to solve it :\

Thanks a lot!

Hi IlanSherer! Welcome to MHB! (Smile)

A function is defined to be convex if the line segment between any two points on the graph of the function lies above or on the graph.

Let's pick the points $(0,0)$ and $(1,f(1))$.
Then any point on the corresponding line segment is above the graph, and any point on the line to the right, is below the graph.
And that line is given by $y=f(1) x$.
Yes?

It looks like this:
\begin{tikzpicture}[xscale=3]
\def\x{3};
\draw[<->] (0,0) -- (3.3,0) node
{$x$};
\draw[<->] (0,0) -- (0,9.5) node[above] {$y$};
\draw[help lines] (1,-0.1) node[below] {1} -- (1,1) -- (-0.03,1) node
{f(1)};
\draw[help lines] (\x,\x) -- (-0.03,\x) node
{f(1)+wf(1)};
\draw[help lines] (\x,-0.1) node[below] {1+w} -- (\x,{\x^2}) -- (-0.03,{\x^2}) node
{f(1+w)};
\draw[domain=-0:3.1, variable=\x, red, ultra thick] plot ({\x}, {(\x)^2}) node
{y=f(x)};
\draw[blue, ultra thick] (0,0) -- (3.2,3.2) node
{y=f(1)x};
\draw[blue, ultra thick] (0,0) -- (1,0) -- node
{f(1)} (1,1) -- node[above] {w} (\x,1) -- node
{wf(1)} (\x,\x);
\end{tikzpicture}

So for $w \ge 0$, we have:
$$f(1+w) \ge f(1)(1+w) = f(1) + wf(1)$$

And if the function is strictly convex, we have for $w>0$:
$$f(1+w) > f(1) + wf(1)$$​
 

FAQ: Solving Exercise - Proving f(1 + w) > f(1) + w*f(1)

What is the purpose of solving the exercise "Proving f(1 + w) > f(1) + w*f(1)"?

The purpose of this exercise is to demonstrate the use of algebraic manipulation and mathematical reasoning to prove a statement or inequality.

What is the significance of f(1 + w) and f(1) in this exercise?

f(1 + w) and f(1) represent two different values of a function f, where f(1 + w) is the value at 1 + w and f(1) is the value at 1. These values are used to prove the inequality f(1 + w) > f(1) + w*f(1).

What does the term "w*f(1)" represent in the inequality?

The term "w*f(1)" represents the product of the variable w and the value of the function f at 1. This term is included in the inequality to show the effect of w on the overall value of the function.

How can algebraic manipulation be used to prove this inequality?

Algebraic manipulation involves rearranging and simplifying equations using mathematical rules and properties. In this exercise, we can manipulate the expression f(1 + w) > f(1) + w*f(1) by expanding the parentheses and combining like terms on both sides of the inequality.

Are there any restrictions on the values of w and f(1) in order for the inequality to hold true?

Yes, there are certain restrictions on the values of w and f(1) for the inequality to hold true. For example, w must be a positive real number and f(1) must be a positive or negative real number. The specific restrictions may vary depending on the function f being used.

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