Dear all,
can someone help me and say me if he agrees with what i did?
Here is my problem:
Let $\mathbf{Y} \in \mathbb{R}^n$, $p,p_0,m \in \mathbb{N}$ such that $p-p_0 \geq 2$, $\Gamma(\cdot)$ the gamma function and $H$ a projection matrix onto a linear subspace $M \subset \mathbb{R}^n$.
I have to show that the following series converges (it is not necessary true): [math]\sum_{k=0}^\infty \left(\frac{||H\mathbf{Y}||}{2||\mathbf{Y}||}\right)^{2k} \frac{e^{-\frac{m}{2}-2k}(2k)^{2k}(m+2k)^{\frac{m}{2}+2k}}{k!||\mathbf{Y}||^{m}\Gamma\left(k + \frac{p-p_0}{2}\right)}[/math].
To do so, i used the root test:
[math]R =\lim_{k\rightarrow \infty} \left[\left(\frac{||H\mathbf{Y}||}{2||\mathbf{Y}||}\right)^{2k} \frac{e^{-\frac{m}{2}-2k}(2k)^{k}(m+2k)^{\frac{m}{2}+k}}{k!||\mathbf{Y}||^{m}\Gamma\left(k + \frac{p-p_0}{2}\right)}\right]^{\frac{1}{k}}[/math][math]= \frac{||H\mathbf{Y}||^2e^{-2}}{4||\mathbf{Y}||^2} \lim_{k\rightarrow \infty} 2k(m+2k)\frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{(k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}}||\mathbf{Y}||^{\frac{m}{2k}}}[/math][math]= \frac{ ||H\mathbf{Y}||^2e^{-2} }{ 4||\mathbf{Y}||^2 } \lim_{k\rightarrow \infty} \frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{||\mathbf{Y}||^{\frac{m}{2k}}} \frac{ 2k(m+2k) }{ (k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} }[/math]
Then, I used [math]\lim_{k\rightarrow \infty} \frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{||\mathbf{Y}||^{\frac{m}{2k}}} = 1[/math]
With the Stirling formula $\Gamma(z + b) \sim \sqrt{2\pi}e^{-z}z^{z+b-1/2}$ for $b\in \mathbb{R}$, i have:
[math]\lim_{k \rightarrow \infty} (k!)^{1/k} = e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln(\Gamma(k+1))}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\sqrt{2\pi}e^{-k-1}(k+1)^{k+1/2}\right)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{\ln\left(\sqrt{2\pi}\right)}{k} - \frac{k+1}{k} + \frac{k+1/2}{k}\ln(k+1)}[/math] [math]= \lim_{k\rightarrow\infty} (k+1)[/math] and
[math]\lim_{k\rightarrow\infty}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} = e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\Gamma\left(k+\frac{p-p_0}{2}\right)\right)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\sqrt{2\pi}e^{-k}(k)^{k+\frac{p-p_0-1}{2}}\right)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{\ln\left(\sqrt{2\pi}\right)}{k} - \frac{k}{k} + \frac{k}{k}\ln(k)+ \frac{p-p_0-1}{2k}\ln(k)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\ln(k)\left(\frac{\ln\left(\sqrt{2\pi}\right)}{k\ln(k)} - \frac{1}{\ln(k)} + 1 + \frac{p-p_0-1}{2k}\right)}[/math] [math]= \lim_{k\rightarrow\infty} k[/math].
Then we have:
[math]\lim_{k\rightarrow\infty} \frac{ 2k(m+2k) }{ (k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} }[/math][math]= \lim_{k\rightarrow\infty} \frac{ 2k(m+2k) }{ (k+1)k} = 4[/math].
Finally, we then have:
[math]R = \frac{ ||H\mathbf{Y}||^2e^{-2} }{ 4||\mathbf{Y}||^2 } \cdot 1 \cdot 4 = \frac{ ||H\mathbf{Y}||^2e^{-2} }{ ||\mathbf{Y}||^2 }[/math]
because [math]\frac{||H\mathbf{Y}||}{||\mathbf{Y}||} = \frac{||\mathbf{Y}|| - ||(I_n-H)\mathbf{Y}||}{||\mathbf{Y}||} \leq 1[/math] we have $R \leq e^{-2} < 1$.
can someone help me and say me if he agrees with what i did?
Here is my problem:
Let $\mathbf{Y} \in \mathbb{R}^n$, $p,p_0,m \in \mathbb{N}$ such that $p-p_0 \geq 2$, $\Gamma(\cdot)$ the gamma function and $H$ a projection matrix onto a linear subspace $M \subset \mathbb{R}^n$.
I have to show that the following series converges (it is not necessary true): [math]\sum_{k=0}^\infty \left(\frac{||H\mathbf{Y}||}{2||\mathbf{Y}||}\right)^{2k} \frac{e^{-\frac{m}{2}-2k}(2k)^{2k}(m+2k)^{\frac{m}{2}+2k}}{k!||\mathbf{Y}||^{m}\Gamma\left(k + \frac{p-p_0}{2}\right)}[/math].
To do so, i used the root test:
[math]R =\lim_{k\rightarrow \infty} \left[\left(\frac{||H\mathbf{Y}||}{2||\mathbf{Y}||}\right)^{2k} \frac{e^{-\frac{m}{2}-2k}(2k)^{k}(m+2k)^{\frac{m}{2}+k}}{k!||\mathbf{Y}||^{m}\Gamma\left(k + \frac{p-p_0}{2}\right)}\right]^{\frac{1}{k}}[/math][math]= \frac{||H\mathbf{Y}||^2e^{-2}}{4||\mathbf{Y}||^2} \lim_{k\rightarrow \infty} 2k(m+2k)\frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{(k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}}||\mathbf{Y}||^{\frac{m}{2k}}}[/math][math]= \frac{ ||H\mathbf{Y}||^2e^{-2} }{ 4||\mathbf{Y}||^2 } \lim_{k\rightarrow \infty} \frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{||\mathbf{Y}||^{\frac{m}{2k}}} \frac{ 2k(m+2k) }{ (k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} }[/math]
Then, I used [math]\lim_{k\rightarrow \infty} \frac{e^{\frac{-m}{2k}}(m+2k)^{\frac{m}{2k}}}{||\mathbf{Y}||^{\frac{m}{2k}}} = 1[/math]
With the Stirling formula $\Gamma(z + b) \sim \sqrt{2\pi}e^{-z}z^{z+b-1/2}$ for $b\in \mathbb{R}$, i have:
[math]\lim_{k \rightarrow \infty} (k!)^{1/k} = e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln(\Gamma(k+1))}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\sqrt{2\pi}e^{-k-1}(k+1)^{k+1/2}\right)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{\ln\left(\sqrt{2\pi}\right)}{k} - \frac{k+1}{k} + \frac{k+1/2}{k}\ln(k+1)}[/math] [math]= \lim_{k\rightarrow\infty} (k+1)[/math] and
[math]\lim_{k\rightarrow\infty}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} = e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\Gamma\left(k+\frac{p-p_0}{2}\right)\right)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{1}{k}\ln\left(\sqrt{2\pi}e^{-k}(k)^{k+\frac{p-p_0-1}{2}}\right)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\frac{\ln\left(\sqrt{2\pi}\right)}{k} - \frac{k}{k} + \frac{k}{k}\ln(k)+ \frac{p-p_0-1}{2k}\ln(k)}[/math] [math]= e^{\lim_{k\rightarrow\infty}\ln(k)\left(\frac{\ln\left(\sqrt{2\pi}\right)}{k\ln(k)} - \frac{1}{\ln(k)} + 1 + \frac{p-p_0-1}{2k}\right)}[/math] [math]= \lim_{k\rightarrow\infty} k[/math].
Then we have:
[math]\lim_{k\rightarrow\infty} \frac{ 2k(m+2k) }{ (k!)^{\frac{1}{k}}\Gamma\left(k+\frac{p-p_0}{2}\right)^{\frac{1}{k}} }[/math][math]= \lim_{k\rightarrow\infty} \frac{ 2k(m+2k) }{ (k+1)k} = 4[/math].
Finally, we then have:
[math]R = \frac{ ||H\mathbf{Y}||^2e^{-2} }{ 4||\mathbf{Y}||^2 } \cdot 1 \cdot 4 = \frac{ ||H\mathbf{Y}||^2e^{-2} }{ ||\mathbf{Y}||^2 }[/math]
because [math]\frac{||H\mathbf{Y}||}{||\mathbf{Y}||} = \frac{||\mathbf{Y}|| - ||(I_n-H)\mathbf{Y}||}{||\mathbf{Y}||} \leq 1[/math] we have $R \leq e^{-2} < 1$.