Fie \( m>n, m, n\in\mathbb{N}^* \) si \( f:\mathbb{R}\rightarrow\mathbb{R},f(x)=n2^{mx}-m2^{nx} \).
a) Demonstrati ca \( f(x)\geq f(0),\forall x \).
b) \( f \) este strict descrescatoare pe \( (-\infty,0] \) si strict crescatoare pe \( [0,\infty) \).
OLM Prahova 2009
Functie - OLM Prahova 2009
Moderators: Filip Chindea, Andrei Velicu, Radu Titiu
Functie - OLM Prahova 2009
n-ar fi rau sa fie bine 
Am sa postez solutia mea care mi se pare foarte "trasa de par" si sper sa vina altcineva cu ceva mai frumos.
a) Fixez \( x\in \mathbb{R} \) si notez \( 2^x=a \):
\( f(x)\geq f(0)\Leftrightarrow \frac{a^m-1}{m}\geq\frac{a^n-1}{n} \).
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*, g(k)=\frac{a^k-1}{k} \) si demonstrez ca e crescatoare.
\( \frac{a^k-1}{k}\leq\frac{a^{k+1}-1}{k+1}\Leftrightarrow 1+ka^{k+1}\geq(k+1)a^k \), dar din inegalitatea mediilor stim ca:
\( \frac{1+a^{k+1}+\dots+a^{k+1}}{k+1}\geq\sqrt[k+1]{a^{k(k+1)}}=a^k. \)
Deci \( g(m)\geq g(n)\Rightarrow f(x)\geq f(0) \).
b) Fixez \( x>y, x,y\in\mathbb{R} \) si notez \( 2^x=a,2^y=b \).
i) \( a>b\geq1 \)
\( f(x)>f(y)\Leftrightarrow \frac{a^m-b^m}{m}\geq\frac{a^n-b^n}{n} \).
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*,g(k)=\frac{a^k-b^k}{k} \) si demonstrez ca g e crescatoare.
\( g(k)<g(k+1)\Leftrightarrow ak\frac{a^k}{b^k}+k+1>(k+1)\frac{a^k}{b^k}+kb \) si notand \( \frac{a^k}{b^k}=t>1 \):
\( k(at+1-t-b)> t-1 \)
avem \( k\geq1 \) si
\( at+1-t-b\geq t-1\Leftrightarrow t(a-2)+2>b \) care este adevarata caci \( t(a-2)+2>a-2+2=a >b \).
Deci \( g(n)<g(m)\Rightarrow f(x)>f(y) \) deci \( f \) e strict crescatoare pe \( (0,\infty) \).
ii) \( 0<b<a\leq1 \)
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*,g(k)=\frac{a^k-b^k}{k} \) si demonstrez ca e descrescatoare.
\( g(k+1)<g(k)\Leftrightarrow k(\frac{a}{b})^ka+k+1<(k+1)(\frac{a}{b})^k+kb\Leftrightarrow(\frac{a}{b})^k>\frac{k+1-kb}{k+1-ka} \)
\( (\frac{a}{b})^k=(\frac{a}{b}-1+1)^k>k(\frac{a}{b}-1)+1>\frac{k+1-kb}{k+1-ka} \)
Ultima inegalitate e echivalenta cu \( k(\frac{a-b}{b})>\frac{k(a-b)}{k+1-ka}\Leftrightarrow k+1>ka+b \) care e adevarata caci \( 1>a>b \).
\( g(m)<g(n)\Rightarrow f(x)<f(y) \Rightarrow f \) descrescatoare pe \( (-\infty,0] \).
Am ramas surprins de cat de mult m-a chinuit problema asta cu toate ca e de locala si tot cred ca are o rezolvare mai simpla.
a) Fixez \( x\in \mathbb{R} \) si notez \( 2^x=a \):
\( f(x)\geq f(0)\Leftrightarrow \frac{a^m-1}{m}\geq\frac{a^n-1}{n} \).
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*, g(k)=\frac{a^k-1}{k} \) si demonstrez ca e crescatoare.
\( \frac{a^k-1}{k}\leq\frac{a^{k+1}-1}{k+1}\Leftrightarrow 1+ka^{k+1}\geq(k+1)a^k \), dar din inegalitatea mediilor stim ca:
\( \frac{1+a^{k+1}+\dots+a^{k+1}}{k+1}\geq\sqrt[k+1]{a^{k(k+1)}}=a^k. \)
Deci \( g(m)\geq g(n)\Rightarrow f(x)\geq f(0) \).
b) Fixez \( x>y, x,y\in\mathbb{R} \) si notez \( 2^x=a,2^y=b \).
i) \( a>b\geq1 \)
\( f(x)>f(y)\Leftrightarrow \frac{a^m-b^m}{m}\geq\frac{a^n-b^n}{n} \).
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*,g(k)=\frac{a^k-b^k}{k} \) si demonstrez ca g e crescatoare.
\( g(k)<g(k+1)\Leftrightarrow ak\frac{a^k}{b^k}+k+1>(k+1)\frac{a^k}{b^k}+kb \) si notand \( \frac{a^k}{b^k}=t>1 \):
\( k(at+1-t-b)> t-1 \)
avem \( k\geq1 \) si
\( at+1-t-b\geq t-1\Leftrightarrow t(a-2)+2>b \) care este adevarata caci \( t(a-2)+2>a-2+2=a >b \).
Deci \( g(n)<g(m)\Rightarrow f(x)>f(y) \) deci \( f \) e strict crescatoare pe \( (0,\infty) \).
ii) \( 0<b<a\leq1 \)
Construiesc \( g:\mathbb{N}^*\rightarrow\mathbb{N}^*,g(k)=\frac{a^k-b^k}{k} \) si demonstrez ca e descrescatoare.
\( g(k+1)<g(k)\Leftrightarrow k(\frac{a}{b})^ka+k+1<(k+1)(\frac{a}{b})^k+kb\Leftrightarrow(\frac{a}{b})^k>\frac{k+1-kb}{k+1-ka} \)
\( (\frac{a}{b})^k=(\frac{a}{b}-1+1)^k>k(\frac{a}{b}-1)+1>\frac{k+1-kb}{k+1-ka} \)
Ultima inegalitate e echivalenta cu \( k(\frac{a-b}{b})>\frac{k(a-b)}{k+1-ka}\Leftrightarrow k+1>ka+b \) care e adevarata caci \( 1>a>b \).
\( g(m)<g(n)\Rightarrow f(x)<f(y) \Rightarrow f \) descrescatoare pe \( (-\infty,0] \).
Am ramas surprins de cat de mult m-a chinuit problema asta cu toate ca e de locala si tot cred ca are o rezolvare mai simpla.
n-ar fi rau sa fie bine
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
\( \odot\ \ \ \{m,n\}\subset \mathbb{N}^* \) , \( m\ >\ n\ ,\ 0\ <\ a\ \ne\ 1 \) \( \Longrightarrow\ \) \( \frac {a^m-1}{m}\ \>\ \frac {a^n-1}{n}\ (*)\ . \)
Metoda 1. Se arata usor ca \( (*)\ \Longleftrightarrow\ f(a)\equiv na^m-ma^n+m-n\ >\ 0\ . \) Vom aplica schema lui Horner pentru radacina \( a:=1 \) (dubla !).
\( \overline{\underline{\left\|\ \begin{array}{cccccccccccccc}
* & \ \ \ \ a^m\ \ \ \ & \ \ \ \ a^{m-1}\ \ \ \ & \ \ \ a^{m-2}\ \ \ \ & \ldots & \ \ \ \ a^{n+1}\ \ \ \ & \ \ \ \ a^n\ \ \ \ & \ \ \ \ a^{n-1}\ \ \ \ & \ \ \ \ a^{n-2}\ \ \ \ & \ldots & \ \ \ \ a^2\ \ \ \ & \ \ \ \ a^1\ \ \ \ & \ \ \ \ a^0\\\\\\\\\\
* & n & 0 & 0 & \ldots & 0 & -m & 0 & 0 & \ldots & 0 & 0 & m-n\\\\\\\\\\
= & === & === & === & === & === & === & === & === & === & === & === & ===\\\\\\\\\\
1 & n & n & n & \ldots & n & n-m & n-m & n-m & \ldots & n-m & n-m & \underline{\overline{\left|\ 0\ \right|}}\\\\\\\\\\
1 & n & 2n & 3n & \ldots & (m-n)n & (m-n)(n-1) & (m-n)(n-2) & (m-n)(n-3) & \ldots & m-n & \underline{\overline{\left|\ 0\ \right|}} & *\end{array}\ \ \right\|}} \)
In concluzie, \( f(a)=(a-1)^2\ \cdot\ \left[\ n\ \cdot\ \sum_{k=1}^{m-n}\ ka^{m-k-1}\ +\ (m-n)\ \cdot\ \sum_{k=0}^{n-2}\ (k+1)a^k\ \right]\ >\ 0\ . \)
Metoda 2. Se arata usor ca \( (*)\ \Longleftrightarrow\ (a-1)E\ >\ 0 \) , unde \( E\equiv na^n\sum_{k=0}^{m-n-1}a^k-(m-n)\sum_{k=0}^{n-1}a^k\ . \)
Cazul 1. \( \overline{\underline{\left\|\ 0\ <\ a\ <\ 1\ \right\|}} \) Pentru \( k\in\mathbb N^* \) avem \( a^k\ <\ 1 \) si \( E\ <\ na^n\sum_{k=0}^{m-n-1}1-(m-n)\sum_{k=0}^{n-1}a^k= \)
\( n(m-n)a^n-(m-n)\sum_{k=0}^{n-1}a^k=(m-n)\left(na^n-\sum_{k=0}^{n-1}a^k\right)=(m-n)\left(\sum_{k=0}^{n-1}a^n-\sum_{k=0}^{n-1}a^k\right)= \)
\( (m-n)\sum_{k=0}^{n-1}a^k\left(a^{n-k}-1\right)\ <\ 0\ \Longrightarrow\ E\ <\ 0\ \Longrightarrow\ (a-1)\cdot E\ >\ 0\ \Longleftrightarrow\ (*)\ . \)
Cazul 2. \( \overline{\underline{\left\|\ a\ >\ 1\ \right\|}} \) Pentru \( k\in\mathbb N^* \) avem \( a^k\ >\ 1 \) si \( E\ >\ na^n\sum_{k=0}^{m-n-1}1-(m-n)\sum_{k=0}^{n-1}a^k= \)
\( n(m-n)a^n-(m-n)\sum_{k=0}^{n-1}a^k=(m-n)\left(na^n-\sum_{k=0}^{n-1}a^k\right)=(m-n)\left(\sum_{k=0}^{n-1}a^n-\sum_{k=0}^{n-1}a^k\right)= \)
\( (m-n)\sum_{k=0}^{n-1}a^k\left(a^{n-k}-1\right)\ >\ 0\ \Longrightarrow\ E\ >\ 0\ \Longrightarrow\ (a-1)\cdot E\ >\ 0\ \Longleftrightarrow\ (*)\ . \)
\( \odot\ \ m\in\mathbb N^*\ \Longrightarrow\ \ \ \) \( \underline{\overline{\left\|\ \begin{array}{ccccc}
0<b<a<1 & \ \Longrightarrow\ & \frac {a^{m+1}-b^{m+1}}{m+1} & \ <\ & \frac {a^{m}-b^{m}}{m}\\\\\\\\\\
1<b<a & \ \Longrightarrow\ & \frac {a^{m+1}-b^{m+1}}{m+1} & \ >\ & \frac {a^{m}-b^{m}}{m}\end{array}\ \right\|}} \)
Notam \( E_m=m\left(a^{m+1}-b^{m+1}\right)-(m+1)\left(a^m-b^m\right)\ ,\ n\in\mathbb N^* \).
Se observa ca \( E_{m+1}-E_m=(m+1)\left[a^m(a-1)^2-b^m(b-1)^2\right]\ . \)
Cazul 1. \( 1\ <\ b\ <\ a\ \Longrightarrow\ \left|\ \begin{array}{ccc}
a^m & > & b^m\\\\\\\\\\
(a-1)^2 & > & (b-1)^2\end{array}\right|\ \Longrightarrow\ E_{m+1}\ >\ E_m\ (\forall )\ m\in\mathbb N^*\ \Longrightarrow \)
\( (\forall )\ m\in\mathbb N^* \) , \( E_m\ >\ E_1=(a-b)(a+b-2)\ >\ 0\ \Longrightarrow\ E_m\ >\ 0\ \Longrightarrow\ \frac {a^{m+1}-b^{m+1}}{m+1}\ >\ \frac {a^{m}-b^{m}}{m}\ . \)
Cazul 2. \( 0\ <\ b\ <\ a\ <\ 1\ . \) In acest caz nu mai putem proceda ca la cazul precedent.
Se observa ca \( m(1-a)+(1-b)>0 \) , adica \( m+1>ma+b\ \Longrightarrow\ \frac {a^{m+1}-b^{m+1}}{m+1}<\frac {a^{m+1}-b^{m+1}}{ma+b} \) .
Insa \( \ \frac {a^{m+1}-b^{m+1}}{ma+b}<\frac {a^m-b^m}{m}\ \Longleftrightarrow\ \left(a^m-b^m\right)-mb^{m-1}(a-b)\ >\ 0\ \Longleftrightarrow\ \)
\( a^{m-1}+a^{m-2}b+\ldots +ab^{m-2}+b^{m-1}-mb^{m-1}\ >\ 0\ \Longleftrightarrow \)
\( \left(a^{m-1}-b^{m-1}\right)+b\left(a^{m-2}-b^{m-2}\right)+\ldots +b^{m-2}\left(a-b\right)\ >\ 0 \) , ceea ce este adevarat.
Metoda 1. Se arata usor ca \( (*)\ \Longleftrightarrow\ f(a)\equiv na^m-ma^n+m-n\ >\ 0\ . \) Vom aplica schema lui Horner pentru radacina \( a:=1 \) (dubla !).
\( \overline{\underline{\left\|\ \begin{array}{cccccccccccccc}
* & \ \ \ \ a^m\ \ \ \ & \ \ \ \ a^{m-1}\ \ \ \ & \ \ \ a^{m-2}\ \ \ \ & \ldots & \ \ \ \ a^{n+1}\ \ \ \ & \ \ \ \ a^n\ \ \ \ & \ \ \ \ a^{n-1}\ \ \ \ & \ \ \ \ a^{n-2}\ \ \ \ & \ldots & \ \ \ \ a^2\ \ \ \ & \ \ \ \ a^1\ \ \ \ & \ \ \ \ a^0\\\\\\\\\\
* & n & 0 & 0 & \ldots & 0 & -m & 0 & 0 & \ldots & 0 & 0 & m-n\\\\\\\\\\
= & === & === & === & === & === & === & === & === & === & === & === & ===\\\\\\\\\\
1 & n & n & n & \ldots & n & n-m & n-m & n-m & \ldots & n-m & n-m & \underline{\overline{\left|\ 0\ \right|}}\\\\\\\\\\
1 & n & 2n & 3n & \ldots & (m-n)n & (m-n)(n-1) & (m-n)(n-2) & (m-n)(n-3) & \ldots & m-n & \underline{\overline{\left|\ 0\ \right|}} & *\end{array}\ \ \right\|}} \)
In concluzie, \( f(a)=(a-1)^2\ \cdot\ \left[\ n\ \cdot\ \sum_{k=1}^{m-n}\ ka^{m-k-1}\ +\ (m-n)\ \cdot\ \sum_{k=0}^{n-2}\ (k+1)a^k\ \right]\ >\ 0\ . \)
Metoda 2. Se arata usor ca \( (*)\ \Longleftrightarrow\ (a-1)E\ >\ 0 \) , unde \( E\equiv na^n\sum_{k=0}^{m-n-1}a^k-(m-n)\sum_{k=0}^{n-1}a^k\ . \)
Cazul 1. \( \overline{\underline{\left\|\ 0\ <\ a\ <\ 1\ \right\|}} \) Pentru \( k\in\mathbb N^* \) avem \( a^k\ <\ 1 \) si \( E\ <\ na^n\sum_{k=0}^{m-n-1}1-(m-n)\sum_{k=0}^{n-1}a^k= \)
\( n(m-n)a^n-(m-n)\sum_{k=0}^{n-1}a^k=(m-n)\left(na^n-\sum_{k=0}^{n-1}a^k\right)=(m-n)\left(\sum_{k=0}^{n-1}a^n-\sum_{k=0}^{n-1}a^k\right)= \)
\( (m-n)\sum_{k=0}^{n-1}a^k\left(a^{n-k}-1\right)\ <\ 0\ \Longrightarrow\ E\ <\ 0\ \Longrightarrow\ (a-1)\cdot E\ >\ 0\ \Longleftrightarrow\ (*)\ . \)
Cazul 2. \( \overline{\underline{\left\|\ a\ >\ 1\ \right\|}} \) Pentru \( k\in\mathbb N^* \) avem \( a^k\ >\ 1 \) si \( E\ >\ na^n\sum_{k=0}^{m-n-1}1-(m-n)\sum_{k=0}^{n-1}a^k= \)
\( n(m-n)a^n-(m-n)\sum_{k=0}^{n-1}a^k=(m-n)\left(na^n-\sum_{k=0}^{n-1}a^k\right)=(m-n)\left(\sum_{k=0}^{n-1}a^n-\sum_{k=0}^{n-1}a^k\right)= \)
\( (m-n)\sum_{k=0}^{n-1}a^k\left(a^{n-k}-1\right)\ >\ 0\ \Longrightarrow\ E\ >\ 0\ \Longrightarrow\ (a-1)\cdot E\ >\ 0\ \Longleftrightarrow\ (*)\ . \)
\( \odot\ \ m\in\mathbb N^*\ \Longrightarrow\ \ \ \) \( \underline{\overline{\left\|\ \begin{array}{ccccc}
0<b<a<1 & \ \Longrightarrow\ & \frac {a^{m+1}-b^{m+1}}{m+1} & \ <\ & \frac {a^{m}-b^{m}}{m}\\\\\\\\\\
1<b<a & \ \Longrightarrow\ & \frac {a^{m+1}-b^{m+1}}{m+1} & \ >\ & \frac {a^{m}-b^{m}}{m}\end{array}\ \right\|}} \)
Notam \( E_m=m\left(a^{m+1}-b^{m+1}\right)-(m+1)\left(a^m-b^m\right)\ ,\ n\in\mathbb N^* \).
Se observa ca \( E_{m+1}-E_m=(m+1)\left[a^m(a-1)^2-b^m(b-1)^2\right]\ . \)
Cazul 1. \( 1\ <\ b\ <\ a\ \Longrightarrow\ \left|\ \begin{array}{ccc}
a^m & > & b^m\\\\\\\\\\
(a-1)^2 & > & (b-1)^2\end{array}\right|\ \Longrightarrow\ E_{m+1}\ >\ E_m\ (\forall )\ m\in\mathbb N^*\ \Longrightarrow \)
\( (\forall )\ m\in\mathbb N^* \) , \( E_m\ >\ E_1=(a-b)(a+b-2)\ >\ 0\ \Longrightarrow\ E_m\ >\ 0\ \Longrightarrow\ \frac {a^{m+1}-b^{m+1}}{m+1}\ >\ \frac {a^{m}-b^{m}}{m}\ . \)
Cazul 2. \( 0\ <\ b\ <\ a\ <\ 1\ . \) In acest caz nu mai putem proceda ca la cazul precedent.
Se observa ca \( m(1-a)+(1-b)>0 \) , adica \( m+1>ma+b\ \Longrightarrow\ \frac {a^{m+1}-b^{m+1}}{m+1}<\frac {a^{m+1}-b^{m+1}}{ma+b} \) .
Insa \( \ \frac {a^{m+1}-b^{m+1}}{ma+b}<\frac {a^m-b^m}{m}\ \Longleftrightarrow\ \left(a^m-b^m\right)-mb^{m-1}(a-b)\ >\ 0\ \Longleftrightarrow\ \)
\( a^{m-1}+a^{m-2}b+\ldots +ab^{m-2}+b^{m-1}-mb^{m-1}\ >\ 0\ \Longleftrightarrow \)
\( \left(a^{m-1}-b^{m-1}\right)+b\left(a^{m-2}-b^{m-2}\right)+\ldots +b^{m-2}\left(a-b\right)\ >\ 0 \) , ceea ce este adevarat.