Mixing
Prove that $f(n)ge 2^{n-1}$
This is from a Brazilian math contest for college students (OBMU):
Let $f: (0,+infty) to (0,+infty)$ be a infinitely differentiable function such that
- For all positive integer $k$ and positive real $x$, $f^{(k)}(x)> 0$ (where $f^{(k)}$ is the kth derivative).
- For all positive integer $m$, $f(m)$ is a positive integer.
Prove that $f(n)ge 2^{n-1}$ for all positive integer $n$.
Attempt
By the mean value theorem, we have
$$f(2)-f(1) = f'(c_1),space c_1 in (1,2)$$
$$f(3)-f(2) = f'(c_2),space c_2 in (2,3)$$
$$vdots $$
$$f(n)-f(n-1) = f'(c_{n-1}),space c_2 in (n,n-1) $$
Then, $f'(c_k)$ is positive integer for all $k in {1,2,cdots, n-1}$. Besides, $f'$ is strictly increasing. Thus, $f'(c_k) ge k$. Adding all the inequalities, we get
$$f(n) ge sum_{k=1}^{n-1}k + f(1) = frac{n(n-1)}{2} + f(1) $$
real-analysis inequality contest-math
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
add a comment |
This is from a Brazilian math contest for college students (OBMU):
Let $f: (0,+infty) to (0,+infty)$ be a infinitely differentiable function such that
- For all positive integer $k$ and positive real $x$, $f^{(k)}(x)> 0$ (where $f^{(k)}$ is the kth derivative).
- For all positive integer $m$, $f(m)$ is a positive integer.
Prove that $f(n)ge 2^{n-1}$ for all positive integer $n$.
Attempt
By the mean value theorem, we have
$$f(2)-f(1) = f'(c_1),space c_1 in (1,2)$$
$$f(3)-f(2) = f'(c_2),space c_2 in (2,3)$$
$$vdots $$
$$f(n)-f(n-1) = f'(c_{n-1}),space c_2 in (n,n-1) $$
Then, $f'(c_k)$ is positive integer for all $k in {1,2,cdots, n-1}$. Besides, $f'$ is strictly increasing. Thus, $f'(c_k) ge k$. Adding all the inequalities, we get
$$f(n) ge sum_{k=1}^{n-1}k + f(1) = frac{n(n-1)}{2} + f(1) $$
real-analysis inequality contest-math
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
Hint: Induct on $n$.
– user10354138
Nov 21 '18 at 15:36
@user10354138 could you be more specific?
– Rafael Deiga
Nov 21 '18 at 17:07
add a comment |
This is from a Brazilian math contest for college students (OBMU):
Let $f: (0,+infty) to (0,+infty)$ be a infinitely differentiable function such that
- For all positive integer $k$ and positive real $x$, $f^{(k)}(x)> 0$ (where $f^{(k)}$ is the kth derivative).
- For all positive integer $m$, $f(m)$ is a positive integer.
Prove that $f(n)ge 2^{n-1}$ for all positive integer $n$.
Attempt
By the mean value theorem, we have
$$f(2)-f(1) = f'(c_1),space c_1 in (1,2)$$
$$f(3)-f(2) = f'(c_2),space c_2 in (2,3)$$
$$vdots $$
$$f(n)-f(n-1) = f'(c_{n-1}),space c_2 in (n,n-1) $$
Then, $f'(c_k)$ is positive integer for all $k in {1,2,cdots, n-1}$. Besides, $f'$ is strictly increasing. Thus, $f'(c_k) ge k$. Adding all the inequalities, we get
$$f(n) ge sum_{k=1}^{n-1}k + f(1) = frac{n(n-1)}{2} + f(1) $$
real-analysis inequality contest-math
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
This is from a Brazilian math contest for college students (OBMU):
Let $f: (0,+infty) to (0,+infty)$ be a infinitely differentiable function such that
- For all positive integer $k$ and positive real $x$, $f^{(k)}(x)> 0$ (where $f^{(k)}$ is the kth derivative).
- For all positive integer $m$, $f(m)$ is a positive integer.
Prove that $f(n)ge 2^{n-1}$ for all positive integer $n$.
Attempt
By the mean value theorem, we have
$$f(2)-f(1) = f'(c_1),space c_1 in (1,2)$$
$$f(3)-f(2) = f'(c_2),space c_2 in (2,3)$$
$$vdots $$
$$f(n)-f(n-1) = f'(c_{n-1}),space c_2 in (n,n-1) $$
Then, $f'(c_k)$ is positive integer for all $k in {1,2,cdots, n-1}$. Besides, $f'$ is strictly increasing. Thus, $f'(c_k) ge k$. Adding all the inequalities, we get
$$f(n) ge sum_{k=1}^{n-1}k + f(1) = frac{n(n-1)}{2} + f(1) $$
real-analysis inequality contest-math
real-analysis inequality contest-math
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
asked Nov 21 '18 at 2:39
Rafael Deiga
662311
662311
Hint: Induct on $n$.
– user10354138
Nov 21 '18 at 15:36
@user10354138 could you be more specific?
– Rafael Deiga
Nov 21 '18 at 17:07
add a comment |
Hint: Induct on $n$.
– user10354138
Nov 21 '18 at 15:36
@user10354138 could you be more specific?
– Rafael Deiga
Nov 21 '18 at 17:07
Hint: Induct on $n$.
– user10354138
Nov 21 '18 at 15:36
Hint: Induct on $n$.
– user10354138
Nov 21 '18 at 15:36
@user10354138 could you be more specific?
– Rafael Deiga
Nov 21 '18 at 17:07
@user10354138 could you be more specific?
– Rafael Deiga
Nov 21 '18 at 17:07
add a comment |
1 Answer
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We have
$$f(n+1) = f(n) + int_n^{n+1} f'(t) , d t$$
Since $f(n+1) > f(n)$, because $f'(x) >0$, and $f(n+1)$ and $f(n)$ are integers, we see that $int_n^{n+1} f'(t) , d t$ is an integer. Define now
$$h_1(x) := int_x^{x+1} f'(t) , dt.$$
Note that $h'_1(x) = f'(x+1)-f'(x)$. Also note that $h_1(x)$ is integer-valued on $mathbb{N}$ and has the property that $h^{(i)}_1(x) >0$, i.e. $h_1$ has the same properties as $f$.
We can proceed in this manner to obtain functions $h_n$ with the properties (1) and (2). (For example, apply the same argument on $h_1$ instead of $f$ in order to get $h_2$.) This construction satisfies $$h_i(n+1) = h_i(n) + h_{i+1}(n).$$
We prove by induction that $h_i(j) ge 2^{j-1}$ for all $i in mathbb{N}_0$, whereby we define $h_0 =f$, simultaneously.
- If $j=1$ then $h_i(1)$ is an integer (for all $i in mathbb{N}_0$) and thus the lower bound is satisfied.
- Assume that the statement was already proven for all $ 1 le j le n$ and $i in mathbb{N}_0$. Then we have
$$h_i(n+1) = h_i(n) + h_{i+1}(n) ge 2^{n-1} + 2^{n-1} = 2^{n}.$$
As a special case, we get $f(n) = h_0(n) ge 2^{n-1}$.
answered Nov 22 '18 at 9:39
p4sch
4,770217
add a comment |
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We have
$$f(n+1) = f(n) + int_n^{n+1} f'(t) , d t$$
Since $f(n+1) > f(n)$, because $f'(x) >0$, and $f(n+1)$ and $f(n)$ are integers, we see that $int_n^{n+1} f'(t) , d t$ is an integer. Define now
$$h_1(x) := int_x^{x+1} f'(t) , dt.$$
Note that $h'_1(x) = f'(x+1)-f'(x)$. Also note that $h_1(x)$ is integer-valued on $mathbb{N}$ and has the property that $h^{(i)}_1(x) >0$, i.e. $h_1$ has the same properties as $f$.
We can proceed in this manner to obtain functions $h_n$ with the properties (1) and (2). (For example, apply the same argument on $h_1$ instead of $f$ in order to get $h_2$.) This construction satisfies $$h_i(n+1) = h_i(n) + h_{i+1}(n).$$
We prove by induction that $h_i(j) ge 2^{j-1}$ for all $i in mathbb{N}_0$, whereby we define $h_0 =f$, simultaneously.
- If $j=1$ then $h_i(1)$ is an integer (for all $i in mathbb{N}_0$) and thus the lower bound is satisfied.
- Assume that the statement was already proven for all $ 1 le j le n$ and $i in mathbb{N}_0$. Then we have
$$h_i(n+1) = h_i(n) + h_{i+1}(n) ge 2^{n-1} + 2^{n-1} = 2^{n}.$$
As a special case, we get $f(n) = h_0(n) ge 2^{n-1}$.
answered Nov 22 '18 at 9:39
p4sch
4,770217
add a comment |
We have
$$f(n+1) = f(n) + int_n^{n+1} f'(t) , d t$$
Since $f(n+1) > f(n)$, because $f'(x) >0$, and $f(n+1)$ and $f(n)$ are integers, we see that $int_n^{n+1} f'(t) , d t$ is an integer. Define now
$$h_1(x) := int_x^{x+1} f'(t) , dt.$$
Note that $h'_1(x) = f'(x+1)-f'(x)$. Also note that $h_1(x)$ is integer-valued on $mathbb{N}$ and has the property that $h^{(i)}_1(x) >0$, i.e. $h_1$ has the same properties as $f$.
We can proceed in this manner to obtain functions $h_n$ with the properties (1) and (2). (For example, apply the same argument on $h_1$ instead of $f$ in order to get $h_2$.) This construction satisfies $$h_i(n+1) = h_i(n) + h_{i+1}(n).$$
We prove by induction that $h_i(j) ge 2^{j-1}$ for all $i in mathbb{N}_0$, whereby we define $h_0 =f$, simultaneously.
- If $j=1$ then $h_i(1)$ is an integer (for all $i in mathbb{N}_0$) and thus the lower bound is satisfied.
- Assume that the statement was already proven for all $ 1 le j le n$ and $i in mathbb{N}_0$. Then we have
$$h_i(n+1) = h_i(n) + h_{i+1}(n) ge 2^{n-1} + 2^{n-1} = 2^{n}.$$
As a special case, we get $f(n) = h_0(n) ge 2^{n-1}$.
answered Nov 22 '18 at 9:39
p4sch
4,770217
add a comment |
We have
$$f(n+1) = f(n) + int_n^{n+1} f'(t) , d t$$
Since $f(n+1) > f(n)$, because $f'(x) >0$, and $f(n+1)$ and $f(n)$ are integers, we see that $int_n^{n+1} f'(t) , d t$ is an integer. Define now
$$h_1(x) := int_x^{x+1} f'(t) , dt.$$
Note that $h'_1(x) = f'(x+1)-f'(x)$. Also note that $h_1(x)$ is integer-valued on $mathbb{N}$ and has the property that $h^{(i)}_1(x) >0$, i.e. $h_1$ has the same properties as $f$.
We can proceed in this manner to obtain functions $h_n$ with the properties (1) and (2). (For example, apply the same argument on $h_1$ instead of $f$ in order to get $h_2$.) This construction satisfies $$h_i(n+1) = h_i(n) + h_{i+1}(n).$$
We prove by induction that $h_i(j) ge 2^{j-1}$ for all $i in mathbb{N}_0$, whereby we define $h_0 =f$, simultaneously.
- If $j=1$ then $h_i(1)$ is an integer (for all $i in mathbb{N}_0$) and thus the lower bound is satisfied.
- Assume that the statement was already proven for all $ 1 le j le n$ and $i in mathbb{N}_0$. Then we have
$$h_i(n+1) = h_i(n) + h_{i+1}(n) ge 2^{n-1} + 2^{n-1} = 2^{n}.$$
As a special case, we get $f(n) = h_0(n) ge 2^{n-1}$.
answered Nov 22 '18 at 9:39
p4sch
4,770217
We have
$$f(n+1) = f(n) + int_n^{n+1} f'(t) , d t$$
Since $f(n+1) > f(n)$, because $f'(x) >0$, and $f(n+1)$ and $f(n)$ are integers, we see that $int_n^{n+1} f'(t) , d t$ is an integer. Define now
$$h_1(x) := int_x^{x+1} f'(t) , dt.$$
Note that $h'_1(x) = f'(x+1)-f'(x)$. Also note that $h_1(x)$ is integer-valued on $mathbb{N}$ and has the property that $h^{(i)}_1(x) >0$, i.e. $h_1$ has the same properties as $f$.
We can proceed in this manner to obtain functions $h_n$ with the properties (1) and (2). (For example, apply the same argument on $h_1$ instead of $f$ in order to get $h_2$.) This construction satisfies $$h_i(n+1) = h_i(n) + h_{i+1}(n).$$
We prove by induction that $h_i(j) ge 2^{j-1}$ for all $i in mathbb{N}_0$, whereby we define $h_0 =f$, simultaneously.
- If $j=1$ then $h_i(1)$ is an integer (for all $i in mathbb{N}_0$) and thus the lower bound is satisfied.
- Assume that the statement was already proven for all $ 1 le j le n$ and $i in mathbb{N}_0$. Then we have
$$h_i(n+1) = h_i(n) + h_{i+1}(n) ge 2^{n-1} + 2^{n-1} = 2^{n}.$$
As a special case, we get $f(n) = h_0(n) ge 2^{n-1}$.
answered Nov 22 '18 at 9:39
p4sch
4,770217
edited Nov 22 '18 at 9:58
answered Nov 22 '18 at 9:39
p4sch
4,770217
answered Nov 22 '18 at 9:39
p4sch
4,770217
answered Nov 22 '18 at 9:39
p4sch
4,770217
4,770217
add a comment |
add a comment |
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Hint: Induct on $n$.
– user10354138
Nov 21 '18 at 15:36
@user10354138 could you be more specific?
– Rafael Deiga
Nov 21 '18 at 17:07