For what values of $k$ does $(1+x)^{500+k}(1-x)^{500-k}$ exceed $10^9$?

Question

Pretty simple question, for what values of $0\leq k \leq 500$ do we have $\max\{(1+x)^{500+k}(1-x)^{500-k}|x\in[0,1]\} \geq 10^9$ ?

Some trivial observations:

The problem is equivalent to finding the smallest $k$ so that $\max\{(1+x)^{500+k}(1-x)^{500-k}|x\in[0,1]\} < 10^9$

Clearly it is equal to $(1-x^{1000-2k})(1+x)^{2k}$ so we must have $(1+x)^{2k}\geq10^9$, since $1+x\leq 2$ this implies $2k\geq 30\implies k\geq 15$. Of course, this is probably useless.

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Edit: Using the fact that maximum is reached at $\frac{k}{500}$ we can rephrase as:

$(500+k)^{500+k}(500-k)^{500-k}>500^{1000}10^9$

The following java problem solves this nicely in well under a second.

    import java.math.*;
public class eulerbla {
    public static void main(String[] args) {
        BigInteger potdiez = new BigInteger ("1000000000");
        BigInteger quinmas = new BigInteger("500");
        BigInteger quinmen = new BigInteger("500");
        BigInteger num;
        BigInteger bound = quinmen.pow(1000);
        bound= bound.multiply(potdiez);
        for(int k=0;k<=500;k++){
            num = (quinmas.pow(500+k));
            num = num.multiply(quinmen.pow(500-k));
            if(num.compareTo(bound)>=0){
                System.out.println(k);
            }
            quinmas=quinmas.add(BigInteger.ONE);
            quinmen=quinmen.subtract(BigInteger.ONE);
        }

    }

}

Output:102

Answer 1

We can compute the critical point for $(1+x)^{500+k}(1-x)^{500-k}$: $x=\frac{k}{500}$. The value at this point is $$ \left(1-\frac{k^2}{250000}\right)^{500}\left(\frac{500+k}{500-k}\right)^k $$ For $k\lt500$, this is an increasing function of $k$, the derivative of its log is $\log\left(\frac{500+k}{500-k}\right)$, and computing values shows that for $102\le k\lt500$, $(1+x)^{500+k}(1-x)^{500-k}$ will be bigger than $10^9$ for $x=\frac{k}{500}$.

For $k\lt102$, $(1+x)^{500+k}(1-x)^{500-k}$ will not exceed $10^9$ on $[0,1]$.

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We can approximate $$ \overbrace{\left(1-\frac{k^2}{250000}\right)^{500}}^{\sim e^{-\frac{k^2}{500}}}\overbrace{\left(\frac{500+k}{500-k}\right)^k\vphantom{\left(\frac{k^2}{2}\right)^5}}^{\sim e^{2\frac{k^2}{500}}} \sim e^{\frac{k^2}{500}} $$ If we solve $$ e^{\frac{k^2}{500}}=10^9 $$ we get $$ k=\sqrt{4500\log(10)}=101.792 $$

Answer 2

$\max\{(1+x)^{m+k}(1-x)^{m-k}|x\in[0,1]\} \geq 10^9 $

Set $m = 500$.

Let $a(x, k) = (1+x)^{m+k}(1-x)^{m-k} $.

$\frac{a(x, k+1)}{a(x, k)} =\frac{(1+x)^{m+k+1}(1-x)^{m-k-1}}{(1+x)^{m+k}(1-x)^{m-k}} =\frac{1+x}{1-x} \gt 1 $ so $a(x, k)$ is increasing for all $k$.

Similarly,

$\begin{array}\\ a'(x, k) &=(m+k)(1+x)^{m+k-1}(1-x)^{m-k}-(m-k)(1+x)^{m+k}(1-x)^{m-k-1}\\ &=(1+x)^{m+k-1}(1-x)^{m-k-1}((m+k)(1-x)-(m-k)(1+x))\\ &=(1+x)^{m+k-1}(1-x)^{m-k-1}((m+k)-(m-k)-x((m+k)+(m-k))\\ &=(1+x)^{m+k-1}(1-x)^{m-k-1}(2k-2mx)\\ &=0 \qquad\text{for }x = k/m\\ \end{array} $

Therefore, for fixed $k$, $a(x, k)$ is max for $x = k/m$. At this $x$,

$\begin{array}\\ b(k, m) &=a(k/m, k)\\ &=(1+k/m)^{m+k}(1-k/m)^{m-k}\\ &=\frac{(m+k)^{m+k}(m-k)^{m-k}}{m^{2m}}\\ \end{array} $

If $k = cm$, $0 < c < 1$,

$\begin{array}\\ b(cm, m) &=(1+c)^{m(1+c)}(1-c)^{m(1-c)}\\ &=\left((1+c)^{(1+c)}(1-c)^{(1-c)}\right)^m\\ \end{array} $

To find where $b(cm, m) = 10^9$, when $m = 500$, we want $g(c) =10^{9/500} \approx 1.042 $.

According to Wolfy, this is about $c=0.203$. This corresponds to $k = cm =0.203\cdot 500 \approx 101.5 $.

This agrees fairly well with your computation.