Root of a plynomial in (0,1)
$begingroup$
Define $$f_K(x)=sum_{i=K+1}^{2K} binom{2K}{i}x^{i-1}(1-x)^{2K-i}.$$ How to show that $qf_K(x)-f_K(1-x)$ has exactly one real root in $(0,1)$ for any $q > 0$ and $K geq 1$. The proof for $q=1$ goes as follows:
begin{align}
f_k(x)-f_K(1-x)&= sum_{i=K+1}^{2K} binom{2K}{i}left[x^{i-1}(1-x)^{2K-i}-(1-x)^{i-1}x^{2K-i}right]\
&= sum_{i=K+1}^{2K} binom{2K}{i} x^{2K-i}(1-x)^{2K-i}left[x^{2(i-K)-1}-(1-x)^{2(i-K)-1}right]\
&= (2x-1)sum_{i=K+1}^{2K} binom{2K}{i}x^{2K-i}(1-x)^{2K-i} sum_{r=0}^{2(i-K)-2} x^r (1-x)^{2(i-K)-2-r}
end{align}
Clearly, the above expression is $>0$ if $x > 1/2$, $<0$ if $x < 1/2$, and zero at $x=1/2$. So it has a unique root at $x=1/2$. But this argument does not generalize for other values of $q$.
real-analysis polynomials binomial-coefficients fixed-point-theorems binomial-distribution
asked Dec 6 '18 at 12:20
GogolGogol
162
$endgroup$
add a comment |
$begingroup$
Define $$f_K(x)=sum_{i=K+1}^{2K} binom{2K}{i}x^{i-1}(1-x)^{2K-i}.$$ How to show that $qf_K(x)-f_K(1-x)$ has exactly one real root in $(0,1)$ for any $q > 0$ and $K geq 1$. The proof for $q=1$ goes as follows:
begin{align}
f_k(x)-f_K(1-x)&= sum_{i=K+1}^{2K} binom{2K}{i}left[x^{i-1}(1-x)^{2K-i}-(1-x)^{i-1}x^{2K-i}right]\
&= sum_{i=K+1}^{2K} binom{2K}{i} x^{2K-i}(1-x)^{2K-i}left[x^{2(i-K)-1}-(1-x)^{2(i-K)-1}right]\
&= (2x-1)sum_{i=K+1}^{2K} binom{2K}{i}x^{2K-i}(1-x)^{2K-i} sum_{r=0}^{2(i-K)-2} x^r (1-x)^{2(i-K)-2-r}
end{align}
Clearly, the above expression is $>0$ if $x > 1/2$, $<0$ if $x < 1/2$, and zero at $x=1/2$. So it has a unique root at $x=1/2$. But this argument does not generalize for other values of $q$.
real-analysis polynomials binomial-coefficients fixed-point-theorems binomial-distribution
asked Dec 6 '18 at 12:20
GogolGogol
162
$endgroup$
add a comment |
$begingroup$
Define $$f_K(x)=sum_{i=K+1}^{2K} binom{2K}{i}x^{i-1}(1-x)^{2K-i}.$$ How to show that $qf_K(x)-f_K(1-x)$ has exactly one real root in $(0,1)$ for any $q > 0$ and $K geq 1$. The proof for $q=1$ goes as follows:
begin{align}
f_k(x)-f_K(1-x)&= sum_{i=K+1}^{2K} binom{2K}{i}left[x^{i-1}(1-x)^{2K-i}-(1-x)^{i-1}x^{2K-i}right]\
&= sum_{i=K+1}^{2K} binom{2K}{i} x^{2K-i}(1-x)^{2K-i}left[x^{2(i-K)-1}-(1-x)^{2(i-K)-1}right]\
&= (2x-1)sum_{i=K+1}^{2K} binom{2K}{i}x^{2K-i}(1-x)^{2K-i} sum_{r=0}^{2(i-K)-2} x^r (1-x)^{2(i-K)-2-r}
end{align}
Clearly, the above expression is $>0$ if $x > 1/2$, $<0$ if $x < 1/2$, and zero at $x=1/2$. So it has a unique root at $x=1/2$. But this argument does not generalize for other values of $q$.
real-analysis polynomials binomial-coefficients fixed-point-theorems binomial-distribution
asked Dec 6 '18 at 12:20
GogolGogol
162
$endgroup$
Define $$f_K(x)=sum_{i=K+1}^{2K} binom{2K}{i}x^{i-1}(1-x)^{2K-i}.$$ How to show that $qf_K(x)-f_K(1-x)$ has exactly one real root in $(0,1)$ for any $q > 0$ and $K geq 1$. The proof for $q=1$ goes as follows:
begin{align}
f_k(x)-f_K(1-x)&= sum_{i=K+1}^{2K} binom{2K}{i}left[x^{i-1}(1-x)^{2K-i}-(1-x)^{i-1}x^{2K-i}right]\
&= sum_{i=K+1}^{2K} binom{2K}{i} x^{2K-i}(1-x)^{2K-i}left[x^{2(i-K)-1}-(1-x)^{2(i-K)-1}right]\
&= (2x-1)sum_{i=K+1}^{2K} binom{2K}{i}x^{2K-i}(1-x)^{2K-i} sum_{r=0}^{2(i-K)-2} x^r (1-x)^{2(i-K)-2-r}
end{align}
Clearly, the above expression is $>0$ if $x > 1/2$, $<0$ if $x < 1/2$, and zero at $x=1/2$. So it has a unique root at $x=1/2$. But this argument does not generalize for other values of $q$.
real-analysis polynomials binomial-coefficients fixed-point-theorems binomial-distribution
real-analysis polynomials binomial-coefficients fixed-point-theorems binomial-distribution
asked Dec 6 '18 at 12:20
GogolGogol
162
asked Dec 6 '18 at 12:20
GogolGogol
162
asked Dec 6 '18 at 12:20
GogolGogol
162
asked Dec 6 '18 at 12:20
GogolGogol
162
asked Dec 6 '18 at 12:20
GogolGogol
162
162
add a comment |
add a comment |
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