MHB Putnam Solution (Differential Calculus)

[Amad27]

Hello,

From the 2010 Putnam A2

Find all differentiable functions such that
$$f: R \implies R$$
$$f'(x) = \frac{f(x+n) - f(x)}{n}$$
For all real numbers $$x$$ and POSITIVE Integers $$n$$

Let "m" be the slope of the tangent line to the graph of f(x).

Let there be points, $$(x, y) (x1, y1)$$ such that

$$x1 > x, (x1 - x) = n$$

$$ \Delta y = f(x + n) - f(x)$$
$$y1 - y1 = \Delta y$$

$$ y - y1 = m(x - x1)$$ [equation of a line]
$$f(x + n) = m(x -x1) + f(x) $$

$$f'(x) = \frac{f(x + n) - f(x)}{x - x1} $$
$$f'(x) = \frac{m(x - x1)}{(x - x1)} $$
$$f'(x) = \frac{m}{1} = m$$
So we proved the slope is "that," now we need to prove this is a line. By definition,

$$f''(x) = \lim_{{h}\to{0}}\frac{f'(x+h) - f'(x)}{h}$$

$$f'(x) = m$$ [we proved]

$$f''(x) = \lim_{{h}\to{0}}\frac{(m-m)}{h} = \lim_{{h}\to{0}}\frac{(0)}{h}$$

The only issue is the $$\lim_{{h}\to{0}}$$ we can't direct substitute in the remaining fraction.

IS IT LEGAL TO DO THE FOLLOWING:

$$\lim_{{h}\to{0}}\frac{(0)}{h} = \lim_{{h}\to{0}}0$$

Because the 0/h takes care of it being 0, but the problem is can I do this? Without direct substitution. IF YES,

$$f''(x) = 0$$

We know the second derivative is 0, this means the first derivative doesn't change in slope, the slope f'(x) is a HORIZONTAL LINE, it is CONSTANT.

If the FIRST derivative is constant, the equation is a line. We use point slope form now.

The line we have passes through (h, k) so we get,

$$y - k = m(x - h) \implies y = mx - mh + k$$
Let $$C = k - mh$$ therefore,

$$y = mx + C$$

 

[I like Serena]

$$f''(x) = \lim_{{h}\to{0}}\frac{f'(x+h) - f'(x)}{h}$$

$$f'(x) = m$$ [we proved]

$$f''(x) = \lim_{{h}\to{0}}\frac{(m-m)}{h} = \lim_{{h}\to{0}}\frac{(0)}{h}$$

Hi Olok,

You can't just substitute $f'(x+h)=m$ here.
As it is, both numerator and denominator tend to 0, so there is still something to figure out.

You are on the right track though.
What if you evaluate your $f''(x)$ in 2 different ways with the given relationship?

Is it legal to do the following:

$$\lim_{{h}\to{0}}\frac{(0)}{h} = \lim_{{h}\to{0}}0$$

This is legal.
(No need to SHOUT.)

 

[Amad27]

Hi Olok,

You can't just substitute $f'(x+h)=m$ here.
As it is, both numerator and denominator tend to 0, so there is still something to figure out.

You are on the right track though.
What if you evaluate your $f''(x)$ in 2 different ways with the given relationship?

This is legal.
(No need to SHOUT.)

I can't evaluate $f''(x)$ is two different ways though.

$f'(x) = m$ so that is the ONLY case.

Why can't we substitution $f(x + h) = m$?

Oh wait, because it is the indeterminate form (after you do two direct subbings).

What else do you suggest. Also,

What was the second way of finding $f''(x)$?

 

[I like Serena]

I can't evaluate $f''(x)$ is two different ways though.

$f'(x) = m$ so that is the ONLY case.

Why can't we substitution $f(x + h) = m$?

Oh wait, because it is the indeterminate form (after you do two direct subbings).

What else do you suggest. Also,

What was the second way of finding $f''(x)$?

What do you get if you substitute $f'(x+h) = f(x+h+1)-f(x+h)$ respectively $f'(x) = f(x+1)-f(x)$?

 

[Amad27]

What do you get if you substitute $f'(x+h) = f(x+h+1)-f(x+h)$ respectively $f'(x) = f(x+1)-f(x)$?

I had a question instantly when I saw this. How do you say

$f'(x+h) = f(x+h+1) - f(x+h)$ In other words, how do you derive this result?

The same for
$f'(x) = f(x+1) - f(x)$, how did you derive this result here?

For the first one, is this related to what you did:?

$f'(x) = [f(x + h) - f(x)]/h$
$f'(x+h) = [f(x + 2h) = f(x + h)]/h$

?? Any relation?

 

[I like Serena]

I had a question instantly when I saw this. How do you say

$f'(x+h) = f(x+h+1) - f(x+h)$ In other words, how do you derive this result?

The same for
$f'(x) = f(x+1) - f(x)$, how did you derive this result here?

You stated in your problem statement that:
$$f'(x) = \frac{f(x+n)-f(x)}{n}$$
for any $x$ and any integer $n$.In particular that means for $n=1$ that:
$$f'(x) = \frac{f(x+1)-f(x)}{1} = f(x+1)-f(x)$$

And since it holds for any $x$, it will also hold for $x+h$, meaning:
$$f'(x+h) = \frac{f((x+h)+1)-f(x+h)}{1} = f(x+h+1)-f(x+h)$$

For the first one, is this related to what you did:?

$f'(x) = [f(x + h) - f(x)]/h$
$f'(x+h) = [f(x + 2h) = f(x + h)]/h$

?? Any relation?

This comes after, together with $$\lim_{h\to 0}$$.

 

[Euge]

Hi Olok,

If you want to solve this problem, then I suggest using the condition

$\displaystyle (*) f'(x) = \frac{f(x + n) - f(x)}{n}$

(which holds for all $x\in \Bbb R$ and $n\in \Bbb N$) to show that $f'$ is 1-periodic. That is, show that $f'(x + 1) = f'(x)$ for all real $x$. After showing this, you may deduce that $f(x + 1) - f(x)$ is a constant function. On the other hand, setting $n = 1$ in $(*)$ yields $f'(x) = f(x + 1) - f(x)$ for all real $x$. Therefore $f'$ is constant. What then can you say about $f$?

 

[Amad27]

You stated in your problem statement that:
$$f'(x) = \frac{f(x+n)-f(x)}{n}$$
for any $x$ and any integer $n$.In particular that means for $n=1$ that:
$$f'(x) = \frac{f(x+1)-f(x)}{1} = f(x+1)-f(x)$$

And since it holds for any $x$, it will also hold for $x+h$, meaning:
$$f'(x+h) = \frac{f((x+h)+1)-f(x+h)}{1} = f(x+h+1)-f(x+h)$$
This comes after, together with $$\lim_{h\to 0}$$.

Hello,

This was helpful.

The Problem is this is for a special case $n=1$ we need to consider all real positive integers "n"

 

[Amad27]

Hi Olok,

If you want to solve this problem, then I suggest using the condition

$\displaystyle (*) f'(x) = \frac{f(x + n) - f(x)}{n}$

(which holds for all $x\in \Bbb R$ and $n\in \Bbb N$) to show that $f'$ is 1-periodic. That is, show that $f'(x + 1) = f'(x)$ for all real $x$. After showing this, you may deduce that $f(x + 1) - f(x)$ is a constant function. On the other hand, setting $n = 1$ in $(*)$ yields $f'(x) = f(x + 1) - f(x)$ for all real $x$. Therefore $f'$ is constant. What then can you say about $f$?

Hi there,

From the definition of periodic, I can see that if something has a period of 1 then for each one it moves it is the same shape, same place etc.. So

[a, b] it has the shape $Q$ suppose it is periodic 1. [b, c] it still has the shape $Q$ assuming $c = b + 1$

What does the period of the first derivative have anything to do with f(x+1) - f(x) being constant?

How does this prove $f(x+1) - f(x)$ is constant from this?

I might be having an issue with definitions,

What is the definition of a periodic function of period = $Q$???

Also for when you state using $n = 1$, you are simply proving this for ONLY ONE $n$ we need to prove it for all positive real integers $n$.

Thanks

 

[Euge]

Hi Olok,

I don't understand what you're saying in the second paragraph, but given $p\in \Bbb R$, a function $g : \Bbb R \to \Bbb R$ is $p$-periodic if $g(x + p) = g(x)$ for all $x$. So, again, when I say $f'$ is 1-periodic, I mean $f'(x + 1) = f'(x)$ for all $x$.

If you know that $f'$ is 1-periodic, then

$\displaystyle \frac{d}{dx}(f(x + 1) - f(x)) = f'(x + 1) - f'(x) = 0$.

Since the function $f(x + 1) - f(x)$ has derivative zero, it must be constant.

You're last paragraph indicates that you're misunderstanding the Putnam question. You have to determine all differentiable functions $f : \Bbb R \to \Bbb R$ which satisfy

$\displaystyle f'(x) = \frac{f(x + n) - f(x)}{n}$

for all positive integers $n$ and real numbers $x$. So you start with a differentiable $f : \Bbb R \to \Bbb R$ which satisfies the above condition and use that condition to find what $f$ is. It will turn out that $f$ will be a linear function. Next, you must show that all linear functions satisfy the above condition. What you're suggesting is proving the condition you're assuming, which is circular reasoning.

 

[Amad27]

Hi Olok,

I don't understand what you're saying in the second paragraph, but given $p\in \Bbb R$, a function $g : \Bbb R \to \Bbb R$ is $p$-periodic if $g(x + p) = g(x)$ for all $x$. So, again, when I say $f'$ is 1-periodic, I mean $f'(x + 1) = f'(x)$ for all $x$.

If you know that $f'$ is 1-periodic, then

$\displaystyle \frac{d}{dx}(f(x + 1) - f(x)) = f'(x + 1) - f'(x) = 0$.

Since the function $f(x + 1) - f(x)$ has derivative zero, it must be constant.

You're last paragraph indicates that you're misunderstanding the Putnam question. You have to determine all differentiable functions $f : \Bbb R \to \Bbb R$ which satisfy

$\displaystyle f'(x) = \frac{f(x + n) - f(x)}{n}$

for all positive integers $n$ and real numbers $x$. So you start with a differentiable $f : \Bbb R \to \Bbb R$ which satisfies the above condition and use that condition to find what $f$ is. It will turn out that $f$ will be a linear function. Next, you must show that all linear functions satisfy the above condition. What you're suggesting is proving the condition you're assuming, which is circular reasoning.

Alright, that is helpful.

$f'(x) = \frac{f(x+n) - f(x)}{n}$

Therefore,
$f'(x+1) = \frac{f(x+n+1) - f(x+1)}{n}$

But the question is, how do you prove this is 1-periodic?

We must prove $f'(x + 1) = f'(x) = \frac{f(x+n+1) - f(x+1)}{n} = \frac{f(x+n) - f(x)}{n}$

I'm stuck again.

 

[Euge]

Fix $x\in\Bbb R$. For all positive integers $n$,

$\displaystyle f'(x + 1) = \frac{f(x + n + 1) - f(x + 1)}{n} = \frac{f(x + n + 1) - f(x) + f(x) - f(x + 1)}{n}$

$\displaystyle = \frac{f(x + n + 1) - f(x)}{n} + \frac{f(x) - f(x + 1)}{n}$

$\displaystyle = \frac{f(x + n + 1) - f(x)}{n + 1} \frac{n + 1}{n} + \frac{f(x) - f(x + 1)}{n}$

$\displaystyle = f'(x)\frac{n + 1}{n} + \frac{f(x) - f(x + 1)}{n}$.

Taking the limit as $n \to \infty$ yields $f'(x + 1) = f'(x)$. Since $x$ was arbitrary, $f'$ is 1-periodic.

 

[Amad27]

Fix $x\in\Bbb R$. For all positive integers $n$,

$\displaystyle f'(x + 1) = \frac{f(x + n + 1) - f(x + 1)}{n} = \frac{f(x + n + 1) - f(x) + f(x) - f(x + 1)}{n}$

$\displaystyle = \frac{f(x + n + 1) - f(x)}{n} + \frac{f(x) - f(x + 1)}{n}$

$\displaystyle = \frac{f(x + n + 1) - f(x)}{n + 1} \frac{n + 1}{n} + \frac{f(x) - f(x + 1)}{n}$

$\displaystyle = f'(x)\frac{n + 1}{n} + \frac{f(x) - f(x + 1)}{n}$.

Taking the limit as $n \to \infty$ yields $f'(x + 1) = f'(x)$. Since $x$ was arbitrary, $f'$ is 1-periodic.

Before I continue with this method.

I would like to put out there, this is LARGELY intuitive.

This just wouldn't strike an average person, I had no idea how you came up with this.

What made you think of this idea? Especially the
$\frac{n+1}{n}$ How did you think of this?

 

[I like Serena]

Before I continue with this method.

I would like to put out there, this is LARGELY intuitive.

This just wouldn't strike an average person, I had no idea how you came up with this.

What made you think of this idea? Especially the
$\frac{n+1}{n}$ How did you think of this?

As an alternative, you might go back to your own:
$$f''(x) = \lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}$$
and substitute $f'(x+h)=f(x+h+1)-f(x+h)$ respectively $f'(x)=f(x+1)-f(x)$.
These substitutions are the simplest possible applications of what is given in your problem statement, so it makes sense to explore them.

Then see where this takes you...

 

[Euge]

Hi Olok,

Since the condition that $f$ satisfies involves a sequence of functions, I know that I have to take the limit of a sequence at some point. Since

$\displaystyle f'(x) = \frac{f(x + n) - f(x)}{n}$

for all $n\in \Bbb N$ and all $x\in \Bbb R$, in particular,

$\displaystyle f'(x) = \frac{f(x + n + 1) - f(x)}{n + 1}$

for all $n\in N$ and $x\in \Bbb R$. So in the expression

$\displaystyle \frac{f(x + n + 1) - f(x)}{n}$,

I multiply by $\frac{n + 1}{n+ 1}$ to get

$\displaystyle \frac{f(x + n + 1) - f(x)}{n + 1}\frac{n + 1}{n} = f'(x) \frac{n + 1}{n} \to f'(x)$ as $n\to \infty$.

 

[Amad27]

As an alternative, you might go back to your own:
$$f''(x) = \lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}$$
and substitute $f'(x+h)=f(x+h+1)-f(x+h)$ respectively $f'(x)=f(x+1)-f(x)$.
These substitutions are the simplest possible applications of what is given in your problem statement, so it makes sense to explore them.

Then see where this takes you...

Hello,

I am still confused, how is

$f'(x + h) = f(x + h + 1) - f(x + h)$?

From what we know,

$f'(x) = [f(x+h) - f(x)]/h$

Oh so you are considering the interval where $h = \Delta x$ = 1?

Just a question, how could the relation between the period(1) and $\Delta x$ be called?
By Period(1) I mean a period of 1.

I mean how is

Period(1) = $\Delta x$?

 

[I like Serena]

Hello,

I am still confused, how is

$f'(x + h) = f(x + h + 1) - f(x + h)$?

That is because $f'(x)=f(x+1)-f(x)$ holds for any $x$.
That includes that it holds for $(x+h)$ instead of $x$.
See post http://mathhelpboards.com/calculus-10/putnam-solution-differential-calculus-11789-post55388.html#post55388.

From what we know,

$f'(x) = [f(x+h) - f(x)]/h$

This is not (necessarily) true.
It should be:
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$

Oh so you are considering the interval where $h = \Delta x$ = 1?

Just a question, how could the relation between the period(1) and $\Delta x$ be called?
By Period(1) I mean a period of 1.

I mean how is

Period(1) = $\Delta x$?

Err... no. I don't know what you mean here.
Well... we can pick $\Delta x=1$ if we want to...

 

[Amad27]

That is because $f'(x)=f(x+1)-f(x)$ holds for any $x$.
That includes that it holds for $(x+h)$ instead of $x$.
See post http://mathhelpboards.com/calculus-10/putnam-solution-differential-calculus-11789-post55388.html#post55388.
This is not (necessarily) true.
It should be:
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$

Err... no. I don't know what you mean here.
Well... we can pick $\Delta x=1$ if we want to...

But we need to prove it is true for ALL positive integers $n$. We can't just choose $\Delta x=1$

Wait I think I know what you are trying to do.

If you want to prove period-1 then you choose $\Delta x=1$
If you want to prove period-2 then you choose $\Delta x=2$
etc...?

 

[I like Serena]

But we need to prove it is true for ALL positive integers $n$. We can't just choose $\Delta x=1$

Wait I think I know what you are trying to do.

If you want to prove period-1 then you choose $\Delta x=1$
If you want to prove period-2 then you choose $\Delta x=2$
etc...?

Erm... I'm not trying to do anything with periodicity...

 

[Amad27]

Err... no. I don't know what you mean here.
Well... we can pick $\Delta x=1$ if we want to...

That is my question, WHY can you pick delta(x) as YOUR choice?

It says prove it for all positive integers $n = \Delta x$ not just for $\Delta x = 1$

What is the logic behind this letting $n = 1$?

 

[I like Serena]

That is my question, WHY can you pick delta(x) as YOUR choice?

It says prove it for all positive integers $n = \Delta x$ not just for $\Delta x = 1$

What is the logic behind this letting $n = 1$?

We begin with $n=1$ because it is the simplest case.
After we have found all solutions, we have to verify if those solutions also satisfy other $n$
... and they do.

 

[Amad27]

We begin with $n=1$ because it is the simplest case.
After we have found all solutions, we have to verify if those solutions also satisfy other $n$
... and they do.

Hello,

Let's try this approach, let n = 1 so

$f'(x) = f(x+1) - f(x)$

Jeez, this is tough, I just can't seem to find a next lead. =(