# Math Help - Show that the function is a decreasing function

1. ## Show that the function is a decreasing function

The problem is:

Show that the function $f(K) = \sum\limits_{i = 1}^K \binom{K}{i} {x^{K - i}{y^i}}$ is a monotonically decreasing function when $x+y \le 0.5$, $x \ge 0$ and $y \ge 0$, $K$ is a positive integer number.

A proof is as follows (but I cannot understand):

Consider the function $f(t)$, with $t \in {R^ + }$. The first order derivative of $f(t)$ with respect to $t$ is:

$f{(t)^'} = \ln (x + y){x^t}\{ {[{x^{ - 1}}{(x + y)^t} - \ln (x)\ln (x + y)]^{ - 1}}\}$

Because ${[{x^{ - 1}}{(x + y)^t} - \ln (x)\ln (x + y)]^{ - 1}} \ge 0$ when $x+y \le 0.5$, so $f{(t)^'} \le 0$, and finally, we obtain the function $f(t)$ is monotonically decreasing function. The function $f(K)$ is a special case of $f(t)$, and we have the proof for the problem.

The above proof is very short, and I cannot understand that. Can everyone give me some ideas?. Thanks in advance.

2. Where exactly do you not understand the proof? What step doesn't make sense?

+ First, how the first order derivative of $f(t)$ is obtained?. I have implemented in Mathematica but the result cannot be simplified to that in the proof.

According to my knowlegde, when $t \in R^+$, the function $f(t)$ is generalized from $f(K)$ and can be written as:

$f(t) = \int_1^t {\binom{t}{u}} {x^{t - u}}{y^u}du$. Is this right?. And if right, how to derive the $f{(t)^'}$.

+ Second, the way to show that ${[{x^{ - 1}}{(x + y)^t} - \ln (x)\ln (x + y)]^{ - 1}} \ge 0$ when $x+y \le 0.5$.

Thanks Ackbeet very much.

4. + First, if K is an integer, the expression simplifies greatly, using the binomial theorem. I think you're meant to work with that. See where that leads you. I will wait on your second question until you've worked out the first.

5. If K is an integer, the function $f(K)$ can be expressed as:

$f(K) = {(x + y)^K} - {x^K}$. So the first order derivative of $f(K)$ with respect to $K$ is:

$f(K)' = {(x + y)^K}\ln (x + y) - {x^K}\ln (x)$

And then, the problem turns into: prove that the $f(K)' \le 0$ when $x+y \le 0.5$, $x \ge 0$, $y \ge 0$ and $K \in N^+$.

And how this problem can be solved?. Thank you.

6. Originally Posted by lptuyen
The problem is:

Show that the function $f(K) = \sum\limits_{i = 1}^K \binom{K}{i} {x^{K - i}{y^i}}$ is a monotonically decreasing function when $x+y \le 0.5$, $x \ge 0$ and $y \ge 0$, $K$ is a positive integer number.

A proof is as follows (but I cannot understand):

Consider the function $f(t)$, with $t \in {R^ + }$. The first order derivative of $f(t)$ with respect to $t$ is:

$f{(t)^'} = \ln (x + y){x^t}\{ {[{x^{ - 1}}{(x + y)^t} - \ln (x)\ln (x + y)]^{ - 1}}\}$

Because ${[{x^{ - 1}}{(x + y)^t} - \ln (x)\ln (x + y)]^{ - 1}} \ge 0$ when $x+y \le 0.5$, so $f{(t)^'} \le 0$, and finally, we obtain the function $f(t)$ is monotonically decreasing function. The function $f(K)$ is a special case of $f(t)$, and we have the proof for the problem.

The above proof is very short, and I cannot understand that. Can everyone give me some ideas?. Thanks in advance.
The 'proof' You have shown is really complicate and pratically impossible to undestand... very confortable and undestable is to consider that is...

$f(k)= (x+y)^{k} -x^{k}$ (1)

... and from (1) You derive...

$\Delta(k)= f(k+1)-f(k)= (x+k)^{k}\ (x+y-1) + x^{k} (1-x)$ (2)

The first term of (2) is <0, the second >0 and because is $x^{k}\ (1-x) < (x+y)^{k}\ (1-x-y)$ for all k is $\Delta(k)<0$ ...

Kind regards

$\chi$ $\sigma$

7. To chisigma:

$x^{k}\ (1-x) < (x+y)^{k}\ (1-x-y)$ for all $k$ ...
I've checked. This is true only when $(x+y) \le 0.5$, $\forall K \in N^+$ . And how to show this in calculus?. This turns into the key point of my problem. Could you give me some derivation steps. Thank chisigma.

8. Originally Posted by lptuyen
To chisigma:

I've checked. This is true only when $(x+y) \le 0.5$, $\forall K \in N^+$ . And how to show this in calculus?. This turns into the key point of my problem. Could you give me some derivation steps. Thank chisigma.
All right!... You have to consider the expression...

$f(y)= (x+y)^{k} (1-x-y)$ (1)

Of course for y=0 is...

$f(0)= x^{k}\ (1-x)$ (2)

... and the inequality...

$x^{k}\ (1-x) < (x+y)^{k}\ (1-x-y)$ (3)

... is reduced to an identity. The (3) is true if the second term is increasing with y so that we can compute...

$f'(y)= (x+y)^{k-1}\ \{1-(1+\frac{1}{k})\ (x+y)\}$ (4)

... and it is easy to see that is $f'(y)>0$ for $k \ge 1$ only if $x+y<\frac{1}{2}$ ...

Kind regards

$\chi$ $\sigma$

9. Oh, I see. Thank chisigma so much!. Hope to see you again.