Find maximum and minimum of $\sum\limits_{k=0}^n a^kb^{n-k}$ for $a, b\ge 0$ with $a^2 + b^2 = 1$

Question

Let $n$ be a positive integer and $a,b$ non negative real numbers such that $a^2+b^2=1$. Find the maximum and the minimum of $$\sum_{k=0}^n a^kb^{n-k}.$$ I can prove that up to $n=5$ the minimum is attained for $(a,b)=(1,0)$, but for $n=6$, is changes to $a=b$. Is it true that this is the point of minimum for all $n\geq 6$? What about the maximum and what happens if we have $a,b,c$ with sum of squares $1$?

Answer 1

Let $$\sum_{k=0}^na^kb^{n-k}\geq(n+1)\left(\sqrt{\frac{a^2+b^2}{2}}\right)^n$$ be true.

Thus, since it should be true for $a\rightarrow+\infty$ (we can do it because the last inequality is homogeneous), we obtain: $$\frac{n+1}{2^{\frac{n}{2}}}\leq1,$$ which gives $n\geq5.31972...$, id est, $n\geq6.$

Now, we'll prove that for any $n\geq6$ we have:$$\min_{a^2+b^2=1,a\geq0,b\geq0}\sum_{k=0}^na^kb^{n-k}=\frac{n+1}{2^{\frac{n}{2}}}.$$

Indeed, we need to prove that $$\sum_{k=0}^na^kb^{n-k}\geq\frac{n+1}{2^{\frac{n}{2}}}\left(a^2+b^2\right)^{\frac{n}{2}}$$ or $f(x)\geq0,$ where $$f(x)=\ln\sum_{k=0}^nx^k-\ln\frac{n+1}{2^{\frac{n}{2}}}-\frac{n}{2}\ln(x^2+1)$$ for $x\geq0.$

But $$f'(x)=\frac{\sum\limits_{k=0}^nkx^{k-1}}{\sum\limits_{k=0}^nx^k}-\frac{nx}{x^2+1}=\frac{\sum\limits_{k=0}^n\left(kx^{k+1}+kx^{k-1}-nx^{k+1}\right)}{(x^2+1)\sum\limits_{k=0}^nx^k}=$$ $$=\frac{-x^n+\sum\limits_{k=0}^{n-2}(2k+2-n)x^{k+1}+1}{(x^2+1)\sum\limits_{k=0}^nx^k}.$$

We see that a polynomial $$-x^n+\sum\limits_{k=0}^{n-2}(2k+2-n)x^{k+1}+1$$ has three sign changes of coefficients,

which by the Descartes' rule of signs https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs

says that our polynomial has three positive roots maximum.

But easy to see that $f'\left(1-\frac{1}{n}\right)<0$, $f'\left(1+\frac{1}{n}\right)>0$ and since $f'(0)>0$ and $f'(n)<0,$ we obtain that

there are $0<u<1$ and $v>1$, for which $f'(u)=f'(v)=0$.

Also, easy to see that $f'(1)=0,$ $f(0)>0$ (because for $n\geq6$ we have $\frac{n+1}{2^{\frac{n}{2}}}<1$)

and $\lim\limits_{x\rightarrow+\infty}f(x)>0,$ which says that $f$ gets a minimal value for $x=1$, which ends a proof.

To find a maximal value it's not so easy.

We'll find a maximal value for $n=6$.

Let $a=bx,$ where $x>1$ if $x=1$, we obtain a value of $$\frac{\sum\limits_{k=0}^6a^kb^{n-k}}{\left(\sqrt{a^2+b^2}\right)^6}=\frac{7}{8}$$ and we'll see that we can get a greater value.

We need to find $\max\limits_{x>1}f(x),$ where $$f(x)=\ln\frac{\sum\limits_{k=0}^6x^k}{(x^2+1)^3}=\ln(x^7-1)-\ln(x-1)-3\ln(x^2+1).$$ Now, $$f'(x)=-\frac{(x^2-1)(x^4-4x^3-x^2-4x+1)}{(x^2+1)\sum\limits_{k=0}^6x^k},$$ which says that a maximal value occurs for a greatest root of the equation $x^4-4x^3-x^2-4x+1=0,$ which after using substitution $x^2+1=2ax$ gives $$x_{max}=\frac{2+\sqrt7+\frac{\sqrt[4]7}{\sqrt2}(1+\sqrt7)}{2},$$ which gives a maximal value: $$\frac{7(2\sqrt7-1)}{27}\approx1.1126...$$ For $n=7$ I got the following maximal value $$\frac{2\sqrt{2(229+63\sqrt{14})}}{25\sqrt5}\approx1.09073...$$

Answer 2

Not a full answer

By $x=\frac ab \ge 0$ we have

$$\sum_{k=0}^n a^kb^{n-k}=\frac{x^{n+1}-1}{(x^2+1)^\frac n2(x-1)}=f_n(x)$$

and for

• $n=0 \implies f_0(x)=1$
• $n=1 \implies f_1(x)=\frac{x^{2}-1}{(x^2+1)^\frac 12(x-1)}$ with max $\sqrt 2$ at $x=1$, min $1$ at $x=0$
• $n=2 \implies f_2(x)=\frac{x^{3}-1}{(x^2+1)(x-1)}$ with max $\frac32$ at $x=1$, min $1$ at $x=0$
• $n=3 \implies f_3(x)=\frac{x^{4}-1}{(x^2+1)^\frac32(x-1)}$ with max $\sqrt 2$ at $x=1$, min $1$ at $x=0$
• $n=4 \implies f_4(x)=\frac{x^{5}-1}{(x^2+1)^2(x-1)}$ with max $\frac54$ at $x=1$, min $1$ at $x=0$
• $n=5 \implies f_5(x)=\frac{x^{6}-1}{(x^2+1)^\frac52(x-1)}$ with max $\approx 1.1513$ at $x\approx 0.31949$ and $x\approx 3.1300$, min $1$ at $x=0$
• $n=6 \implies f_6(x)=\frac{x^{7}-1}{(x^2+1)^3(x-1)}$ with max $\frac 7{27}(2\sqrt 7 -1)$ at $x=1+\frac{\sqrt 7}2\pm \frac12\sqrt{7+4\sqrt 7}$, min $\frac 78$ at $x= 1$
• $n=7 \implies f_7(x)=\frac{x^{8}-1}{(x^2+1)^\frac72(x-1)}$ with max $\approx 1.0907$ at $x\approx 0.17980$ and $x\approx 5.5619$, min $\frac{\sqrt 2}2$ at $x= 1$
• $\ldots$
• $n=50 \implies f_{51}(x)=\frac{x^{51}-1}{(x^2+1)^{25}(x-1)}$ with max $\approx 1$, min $\approx 1.52\cdot 10^{-6}$

From this empirical evaluation it seems that the maximum is $\frac 32$ attained at $x=1$ for the case $n=2$ while the minimum seems descresing at zero with $n$.

Answer 3

Remarks: List some closed form expression of the maximum, for $n = 6, 7, \cdots, 11$. For $n=9, 11$, the expression is complicated which is not given.

$n = 6$: We have $$f_{\max} = \frac{14\sqrt 7 - 7}{27}.$$

$n = 7$: We have $$f_{\max} = \frac{2}{125}\sqrt{2290 + 630\sqrt{14}}.$$

$n = 8$: We have $$f_{\max} = \frac{9\sqrt{33}}{8}\sin \left(\frac{\pi}{6} + \frac13\arccos\frac{59\sqrt{33}}{363}\right) - \frac{45}{16}.$$ (Note: It is a real root of $256U^3 + 2160U^2 - 1944U - 729 = 0$. Also, we have a radical expression.)

$n = 9$: $f_{\max}$ is a real root of $387420489U^8 - 1628614600U^6 - 3632250000U^4 + 5643600000U^2 + 20000000 = 0$.

$n = 10$: We have $$f_{\max} = \frac{70279}{6250} - \frac{10142}{3125}\sqrt{11} + \frac{11}{6250}\sqrt{85635517-25787632\sqrt{11}}.$$ (Note: It is a real root of $3125\,{U}^{4}-140558\,{U}^{3}-11253\,{U}^{2}+151734\,U+14641 = 0$.)

$n = 11$: $f_{\max}$ is a real root of $94143178827\,{U}^{8}-124801907328\,{U}^{6}+20769573376\,{U}^{4}+ 2266398720\,{U}^{2}+1769472 = 0$.