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Question Number 115534 by Hassen_Timol last updated on 26/Sep/20
Let say r^((n)) = Π_(k=0) ^(n−1) (r−k) and r^((0)) =1 With n∈N and r∈R... 1. Show that (n−1−r)^((n)) = (−1)^((n)) (r)^((n)) 2. If m≤n, show that (r^((n)) /r^((m)) )=(r−m)^((n−m)) 3. Espress r^((n+m)) as w^((n)) w′^((m)) 4. Show that (2r)^((2n)) =2^(2n) r^((n)) (r−(1/2))^((n)) Can you help me... please...
$$\mathrm{Let}\:\mathrm{say}\:{r}^{\left({n}\right)} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{k}\right)\:\mathrm{and}\:{r}^{\left(\mathrm{0}\right)} =\mathrm{1} \\ $$$$\mathrm{With}\:{n}\in\mathbb{N}\:\mathrm{and}\:{r}\in\mathbb{R}… \\ $$$$\mathrm{1}.\:\:\:\mathrm{Show}\:\mathrm{that}\:\left({n}−\mathrm{1}−{r}\right)^{\left({n}\right)} \:=\:\left(−\mathrm{1}\right)^{\left({n}\right)} \left({r}\right)^{\left({n}\right)} \\ $$$$\mathrm{2}.\:\mathrm{If}\:{m}\leqslant{n},\:\mathrm{show}\:\mathrm{that}\:\:\frac{{r}^{\left({n}\right)} }{{r}^{\left({m}\right)} }=\left({r}−{m}\right)^{\left({n}−{m}\right)} \\ $$$$\mathrm{3}.\:\mathrm{Espress}\:{r}^{\left({n}+{m}\right)} \:\mathrm{as}\:{w}^{\left({n}\right)} {w}'^{\left({m}\right)} \\ $$$$\mathrm{4}.\:\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{2}{r}\right)^{\left(\mathrm{2}{n}\right)} =\mathrm{2}^{\mathrm{2}{n}} {r}^{\left({n}\right)} \left({r}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left({n}\right)} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}…\:\mathrm{please}… \\ $$
Commented by Hassen_Timol last updated on 26/Sep/20
Could you help me... I don′t understand?
$$\mathrm{Could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}…\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}? \\ $$
Answered by aleks041103 last updated on 26/Sep/20
1. (n−1−r)^((n)) =Π_(k=0) ^(n−1) (n−1−r−k)= =Π_(k=0) ^(n−1) ((n−1−k)−r) let i=n−1−k, so if k∈[0,n−1] then i∈[n−1,0]≡[0,n−1] Π_(k=0) ^(n−1) ((n−1−k)−r)=Π_(i=0) ^(n−1) (i−r)= =Π_(i=0) ^(n−1) (r−i)(−1)=Π_(i=0) ^(n−1) (r−i)Π_(i=0) ^(n−1) (−1)= =r^((n)) (−1)^n ⇒(n−1−r)^((n)) =(−1)^n r^((n))
$$\mathrm{1}. \\ $$$$\left({n}−\mathrm{1}−{r}\right)^{\left({n}\right)} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({n}−\mathrm{1}−{r}−{k}\right)= \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\left({n}−\mathrm{1}−{k}\right)−{r}\right) \\ $$$${let}\:{i}={n}−\mathrm{1}−{k},\:{so}\:{if}\:{k}\in\left[\mathrm{0},{n}−\mathrm{1}\right]\:{then} \\ $$$${i}\in\left[{n}−\mathrm{1},\mathrm{0}\right]\equiv\left[\mathrm{0},{n}−\mathrm{1}\right] \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\left({n}−\mathrm{1}−{k}\right)−{r}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({i}−{r}\right)= \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{i}\right)\left(−\mathrm{1}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{i}\right)\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(−\mathrm{1}\right)= \\ $$$$={r}^{\left({n}\right)} \left(−\mathrm{1}\right)^{{n}} \\ $$$$\Rightarrow\left({n}−\mathrm{1}−{r}\right)^{\left({n}\right)} =\left(−\mathrm{1}\right)^{{n}} {r}^{\left({n}\right)} \\ $$
Commented by Hassen_Timol last updated on 26/Sep/20
Thank you so much, it's very nice ! thanks a lot... if you know how to do for the other questions, it would be even more nice ! But once again, thanks a lot
Commented by aleks041103 last updated on 27/Sep/20
I′ll answer 3 and 4 later
$${I}'{ll}\:{answer}\:\mathrm{3}\:{and}\:\mathrm{4}\:{later} \\ $$
Answered by aleks041103 last updated on 26/Sep/20
2. (r^((n)) /r^((m)) )=((Π_(k=0) ^(n−1) (r−k))/(Π_(k=0) ^(m−1) (r−k)))=((Π_(k=0) ^(m−1) (r−k)Π_(k=m) ^(n−1) (r−k))/(Π_(k=0) ^(m−1) (r−k))) =Π_(k=m) ^(n−1) (r−k) let k=m+i, so if k∈[m,n−1] then i∈[0,(n−m)−1] (r^((n)) /r^((m)) )=Π_(i=0) ^((n−m)−1) (r−(m+i))= =Π_(i=0) ^((n−m)−1) ((r−m)−i)=(r−m)^((n−m)) ⇒(r^((n)) /r^((m)) )=(r−m)^((n−m))
$$\mathrm{2}. \\ $$$$\:\frac{{r}^{\left({n}\right)} }{{r}^{\left({m}\right)} }=\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{k}\right)}{\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right)}=\frac{\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right)\underset{{k}={m}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{k}\right)}{\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right)} \\ $$$$=\underset{{k}={m}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{k}\right) \\ $$$${let}\:{k}={m}+{i},\:{so}\:{if}\:{k}\in\left[{m},{n}−\mathrm{1}\right]\:{then} \\ $$$${i}\in\left[\mathrm{0},\left({n}−{m}\right)−\mathrm{1}\right] \\ $$$$\frac{{r}^{\left({n}\right)} }{{r}^{\left({m}\right)} }=\underset{{i}=\mathrm{0}} {\overset{\left({n}−{m}\right)−\mathrm{1}} {\prod}}\left({r}−\left({m}+{i}\right)\right)= \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\left({n}−{m}\right)−\mathrm{1}} {\prod}}\left(\left({r}−{m}\right)−{i}\right)=\left({r}−{m}\right)^{\left({n}−{m}\right)} \\ $$$$\Rightarrow\frac{{r}^{\left({n}\right)} }{{r}^{\left({m}\right)} }=\left({r}−{m}\right)^{\left({n}−{m}\right)} \\ $$
Commented by Hassen_Timol last updated on 26/Sep/20
Thank you very much !
Answered by aleks041103 last updated on 27/Sep/20
3. r^((n+m)) =Π_(k=0) ^(n+m−1) (r−k)= =Π_(k=0) ^(n−1) (r−k)Π_(k=n) ^(n+m−1) (r−k)= =r^((n)) Π_(k=n) ^(n+m−1) (r−k) let k=n+i so Π_(k=n) ^(n+m−1) (r−k)=Π_(i=0) ^(m−1) (r−n−i)=(r−n)^((m)) ⇒r^((n+m)) =r^((n)) (r−n)^((m)) By analogy r^((n+m)) =r^((m)) (r−m)^((n)) So r^((n+m)) =(r−m)^((n)) r^((m)) =r^((n)) (r−n)^((m))
$$\mathrm{3}. \\ $$$${r}^{\left({n}+{m}\right)} =\underset{{k}=\mathrm{0}} {\overset{{n}+{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right)= \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{k}\right)\underset{{k}={n}} {\overset{{n}+{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right)= \\ $$$$={r}^{\left({n}\right)} \underset{{k}={n}} {\overset{{n}+{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right) \\ $$$${let}\:{k}={n}+{i}\:{so} \\ $$$$\underset{{k}={n}} {\overset{{n}+{m}−\mathrm{1}} {\prod}}\left({r}−{k}\right)=\underset{{i}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\prod}}\left({r}−{n}−{i}\right)=\left({r}−{n}\right)^{\left({m}\right)} \\ $$$$\Rightarrow{r}^{\left({n}+{m}\right)} ={r}^{\left({n}\right)} \left({r}−{n}\right)^{\left({m}\right)} \\ $$$${By}\:{analogy} \\ $$$${r}^{\left({n}+{m}\right)} ={r}^{\left({m}\right)} \left({r}−{m}\right)^{\left({n}\right)} \\ $$$${So} \\ $$$${r}^{\left({n}+{m}\right)} =\left({r}−{m}\right)^{\left({n}\right)} {r}^{\left({m}\right)} ={r}^{\left({n}\right)} \left({r}−{n}\right)^{\left({m}\right)} \\ $$
Answered by aleks041103 last updated on 27/Sep/20
4. (2r)^((2n)) =Π_(k=0) ^(2n−1) (2r−k)=Π_(k=0) ^(2n−1) (r−(k/2))2= =2^(2n) Π_(k=0) ^(2n−1) (r−(k/2)) Π_(k=0) ^(2n−1) (r−(k/2))= =Π_(k=0;even k) ^(2n−1) (r−(k/2))Π_(k=0;odd k) ^(2n−1) (r−(k/2))= =Π_(k=0) ^(n−1) (r−((2k)/2))Π_(k=0) ^(n−1) (r−((2k+1)/2))= =Π_(k=0) ^(n−1) (r−k)Π_(k=0) ^(n−1) (r−(1/2)−k) =r^((n)) (r−(1/2))^((n)) ⇒(2r)^((2n)) =2^(2n) r^((n)) (r−(1/2))^((n))
$$\mathrm{4}. \\ $$$$\left(\mathrm{2}{r}\right)^{\left(\mathrm{2}{n}\right)} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{r}−{k}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\prod}}\left({r}−\frac{{k}}{\mathrm{2}}\right)\mathrm{2}= \\ $$$$=\mathrm{2}^{\mathrm{2}{n}} \underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\prod}}\left({r}−\frac{{k}}{\mathrm{2}}\right) \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\prod}}\left({r}−\frac{{k}}{\mathrm{2}}\right)= \\ $$$$=\underset{{k}=\mathrm{0};{even}\:{k}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\prod}}\left({r}−\frac{{k}}{\mathrm{2}}\right)\underset{{k}=\mathrm{0};{odd}\:{k}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\prod}}\left({r}−\frac{{k}}{\mathrm{2}}\right)= \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−\frac{\mathrm{2}{k}}{\mathrm{2}}\right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−{k}\right)\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({r}−\frac{\mathrm{1}}{\mathrm{2}}−{k}\right) \\ $$$$={r}^{\left({n}\right)} \left({r}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left({n}\right)} \\ $$$$\Rightarrow\left(\mathrm{2}{r}\right)^{\left(\mathrm{2}{n}\right)} =\mathrm{2}^{\mathrm{2}{n}} {r}^{\left({n}\right)} \left({r}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left({n}\right)} \\ $$$$ \\ $$

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