Hard to define function and Conv/Dive of Integral/Series

[estro]

Homework Statement

\mbox {1. If f(x) is a positive and continuous function in [1,\infty)\ and\ \lim_{x\rightarrow \infty} f(x)=0 }

\mbox {Prove or contradict:}\\<br /> \mbox { If } \sum_{n=1}^\infty f(n) \ \mbox {is converget then} \ \int_{1}^\infty f(x)dx \ \mbox {is also convergent.}

\mbox {2. If f(x) is a positive and continuous function in [1,\infty)}

\mbox {Prove or contradict:}\\ <br /> \mbox { If } \int_{1}^\infty f(x)dx \ \mbox {is converget then} \ \sum_{n=1}^\infty f(n) \ \mbox {is also convergent.}

The Attempt at a Solution

[Edit] Counterexample for (1):

\sum_{n=1}^\infty \frac{\sin^2{(n\Pi)}}{n} \ \mbox{is convergent but } \int_{1}^\infty \frac{\sin^2{(x\Pi)}}{x}dx\ \mbox{ is divergent}[Edit] Counterexample for (2): (Still only idea)

I should somehow formally define function which is 0 most of the time but every time x is near integer n the function "draws" a triangle with area of (1/2)^n.

 

[LCKurtz]

In the thread where you posted this question previously:

https://www.physicsforums.com/showthread.php?p=2751033#post2751033

I gave you this suggestion:

How about something easier like this -- suppose an= 0 for all n so the series obviously converges. Now can you build an f(x) such that f(n) = 0 for integers n but whose area under the curve goes to infinity? Once you do that then start thinking about the other way.

Why don't you begin by addressing that.

 

[estro]

In the thread where you posted this question previously:

https://www.physicsforums.com/showthread.php?p=2751033#post2751033

I gave you this suggestion:

How about something easier like this -- suppose an= 0 for all n so the series obviously converges. Now can you build an f(x) such that f(n) = 0 for integers n but whose area under the curve goes to infinity? Once you do that then start thinking about the other way.

Why don't you begin by addressing that.

This is what I tried to do, but I have trouble making it formal.

 

[LCKurtz]

This is what I tried to do, but I have trouble making it formal.

Can't you think of a continuous function that is zero at the integers?

 

[estro]

Contradiction for 2:

\int_{1}^\infty \sin^2{(x\Pi)}dx \ \mbox{is divergent but } <br /> \sum_{1}^\infty \sin^2{(n\Pi)} \ \mbox{is convergent}

Don't know how I missed it.
Now I'll think about 1

 

[LCKurtz]

Contradiction for 2:

\int_{1}^\infty \sin{(x\Pi)}dx \ \mbox{is divergent but } <br /> \sum_{1}^\infty \sin{(n\Pi)} \ \mbox{is convergent}

Don't know how I missed it.
Now I'll think about 1

But that isn't quite a counterexample to your problem yet. [Why?]

 

[estro]

But that isn't quite a counterexample to your problem yet. [Why?]

I corrected my mistake, thanks.

 

[LCKurtz]

I corrected my mistake, thanks.

I wasn't referring to the converget vs divergent typo.

 

[estro]

I wasn't referring to the converget vs divergent typo.

Contradiction for 1:

\sum_{n=1}^\infty \frac {\sin^2{(n\Pi)}}{n} \ \mbox{is convergent but } <br /> \int_{1}^\infty \frac {\sin^2{(x\Pi)}}{x}dx \ \mbox{is divergent}

Seems right this time.

 

[estro]

Now I should look for a positive function without a limit at \infty, right?

 

[LCKurtz]

Contradiction for 1:

\sum_{n=1}^\infty \frac {\sin^2{(n\Pi)}}{n} \ \mbox{is convergent but } <br /> \int_{1}^\infty \frac {\sin^2{(x\Pi)}}{x}dx \ \mbox{is divergent}

Seems right this time.

Don't make it so difficult. Apparently you noticed that the problem requires that f(x) can't be negative, and squaring it fixes that. But why put the new denominator in there and make a big problem out of showing the integral diverges? You want to be able to calculate it easily.

Also, you still don't have a counterexample if the problem really requires f(x) to be positive. Will nonnegative do?

 

[LCKurtz]

Now I should look for a positive function without a limit at \infty, right?

You need to state more precisely what you need to do. You have to find an example of a series that does what and an integral that does what for a counterexample to 1?

[Edit] I mean a counterexample to the other one, which is 2.

 

[estro]

I made the denominator because of this requirement in (1): \lim_{x\rightarrow \infty} f(x)=0
Nonnegative will be enough, but I'll try to think about positive function.
But Ill try to give counterexample to (2).

 

[LCKurtz]

I made the denominator because of this requirement in (1): \lim_{x\rightarrow \infty} f(x)=0
Nonnegative will be enough, but I'll try to think about positive function.
But Ill try to give counterexample to the second one first.

Woops, I didn't see the \lim_{x\rightarrow \infty} f(x)=0 thing. Has that always been there? Anyway, what you now have seems to work for the nonnegative case, assuming you are actually going to show the integral diverges (not difficult).

 

[estro]

You need to state more precisely what you need to do. You have to find an example of a series that does what and an integral that does what for a counterexample to 1?

[Edit] I mean a counterexample to the other one, which is 2.

\mbox {Now I need to prove or contradict: If f(x) is a nonnegative and continuous function in [1,\infty)}

\mbox {Prove or contradict:}\\ <br /> \mbox { If } \int_{1}^\infty f(x)dx \ \mbox {is converget then} \ \sum_{n=1}^\infty f(n) \ \mbox {is also convergent.}

I think I have one...
[Edit] No, I don't have one. Now I'm trying to think about something that will fail the n'th term test.

 

[LCKurtz]

\mbox {Now I need to prove or contradict: If f(x) is a nonnegative and continuous function in [1,\infty)}

\mbox {Prove or contradict:}\\ <br /> \mbox { If } \int_{1}^\infty f(x)dx \ \mbox {is converget then} \ \sum_{n=1}^\infty f(n) \ \mbox {is also convergent.}

I think I have one...
[Edit] No, I don't have one. Now I'm trying to think about something that will fail the n'th term test.

Think about a_n=1. That will give a divergent series by failing the nth term test. Now build an f(x) ... This is the real meat of this problem.

 

[estro]

Think about a_n=1. That will give a divergent series by failing the nth term test. Now build an f(x) ... This is the real meat of this problem.

I'm trying to think about a function which is 0 almost always but every time x is near integer the function builds very very thin triangle (triangle height is 1)...

 

[LCKurtz]

I'm trying to think about a function which is 0 almost always but every time x is near integer the function builds very very thin triangle (triangle height is 1)...

Good, you're onto it. Make it so the areas of the triangles converge when you add them all up. I think you can take it from here.

 

[estro]

Good, you're onto it. Make it so the areas of the triangles converge when you add them all up. I think you can take it from here.

I actually have bad sense for making function from my mind formal, can you give me a tip?

trying something like this \ \sin^2{(e^x)}

 

[LCKurtz]

I'm trying to think about a function which is 0 almost always but every time x is near integer the function builds very very thin triangle (triangle height is 1)...

I actually have bad sense for making function from my mind formal, can you give me a tip?

trying something like this \ \sin^2{(e^x)}

Don't make it so complicated. Use the triangles you mentioned above. Fix it so the triangle at integer n has area a_n where ∑a_n is a convergent series of your choice.

 

[estro]

Don't make it so complicated. Use the triangles you mentioned above. Fix it so the triangle at integer n has area a_n where ∑a_n is a convergent series of your choice.

Now, it's obvious that my function should build triangles near x = n, while the triangle area at n should be (1/2)^n. But I have no idea how to define such functions.

 

[LCKurtz]

Now, it's obvious that my function should build triangles near x = n, while the triangle area at n should be (1/2)^n. But I have no idea how to define such functions.

You can't figure out how wide the base of a triangle of height 1 needs to be so that its area is (1/2)ⁿ? Or how to write the equations of the straight lines forming its sides? You are at least in calculus, right? I don't think anyone here is going to do that for you.

 

[estro]

You can't figure out how wide the base of a triangle of height 1 needs to be so that its area is (1/2)ⁿ? Or how to write the equations of the straight lines forming its sides? You are at least in calculus, right? I don't think anyone here is going to do that for you.

This is what I thought about:

f(x) = \left\{<br /> \begin{array}{c l}<br /> 1 &amp; x \in n\\<br /> g(x) &amp; x \in [n-(1/2)^n,n)\\<br /> h(x) &amp; x \in (n,n+(1/2)^n]\\<br /> 0 &amp; \mbox{ rest cases}<br /> \end{array}<br /> \right.<br />

Should I include something like this?

h(x)=\frac{x-n+(1/2)^n}{(1/2)^n}.Can I instead do it this way?
g(x) is a straight line connecting (x,1) with (x+(1/2)^x,1).Actually I've just started my academic quest after 3 years brake in IDF, so I'm little slow and rusty. (I'm now self study Calc 1 and 2, enrolled in Open University to gather some credit and momentum before the real thing).

 

[Dick]

Define f(x)=x for x in [0,1], f(x)=2-x for x in [1,2] and f(x)=0 otherwise. Now f(x) is a 'bump' function. You can define other functions like shifts f_n(x)=f(x-n), scalings f_a(x)=f(a*x) and multiples f_b(x)=b*f(x). What do they look like? Now combine those operations to make the triangles you want and sum them.

 

[estro]

Thank you both for the guidance.

I think I got it:

<br /> f(x) = \left\{<br /> \begin{array}{c l}<br /> \frac{x-n}{(1/2)^n}, &amp; \mbox{ if } |n-x|\leq(1/2)^n \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 0, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br /> <br />

[Edit] No, actually I've made mistake, working to fix it...

 

[LCKurtz]

Think about the triangle based at n, its base is

\left[ n - \frac 1 {2^n},n + \frac 1 {2^n}\right]

and its vertex is at (n,1). Just write the equations of the two straight lines forming its sides.

 

[estro]

<br /> <br /> f(x) = \left\{<br /> \begin{array}{c l}<br /> \frac{x-n+(1/2)^n}{(1/2)^n}, &amp; \mbox{ if } |n-x|\leq(1/2)^n \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 0, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br /> <br />

Thank you, I appreciate your help!

 

[LCKurtz]

<br /> <br /> f(x) = \left\{<br /> \begin{array}{c l}<br /> \frac{x-n+(1/2)^n}{(1/2)^n}, &amp; \mbox{ if } |n-x|\leq(1/2)^n \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 0, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br /> <br />

Thank you, I appreciate your help!

Still not quite right. The non-zero part of your function has positive slope (the coefficient of x) but part of your triangle must have negative slope. Do the two sides separately.

 

[LCKurtz]

This is what I thought about:

f(x) = \left\{<br /> \begin{array}{c l}<br /> 1 &amp; x \in n\\<br /> g(x) &amp; x \in [n-(1/2)^n,n)\\<br /> h(x) &amp; x \in (n,n+(1/2)^n]\\<br /> 0 &amp; \mbox{ rest cases}<br /> \end{array}<br /> \right.<br />

Yes, that is a good way to express your answer. I would call it f_n(x) once you get the g(x) and h(x) right. Then your final answer is:

f(x) = ∑ f_n(x).

 

[estro]

<br /> <br /> f(x) = \left\{<br /> \begin{array}{c l}<br /> \frac{x-n+(1/2)^n}{(1/2)^n}, &amp; \mbox{if } x \in [n-(1/2)^n, n] \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> \frac{n-x-(1/2)^n}{(1/2)^n}, &amp; \mbox{if } x \in (n, n+(1/2)^n] \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 0, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br /> <br />

 

[LCKurtz]

When x = n your second one gives -1, so it's still wrong. And if you simplify your first one by multiplying numerator and denominator by 2ⁿyou can write it as

1 + 2ⁿ(x-n)

You should be able to write the second piece similarly. Keeping it simple is always a good idea.

 

[estro]

When x = n your second one gives -1, so it's still wrong. And if you simplify your first one by multiplying numerator and denominator by 2ⁿyou can write it as

1 + 2ⁿ(x-n)

You should be able to write the second piece similarly. Keeping it simple is always a good idea.

But the second one deals with x\in (n,n+(1/2)^n] or I missed something?

 

[LCKurtz]

But the second one deals with x\in (n,n+(1/2)^n] or I missed something?

Sure, the second one applies to that interval but your formula for it is wrong. You should get 1 when x = n in both of them.

 

[estro]

<br /> <br /> f(x) = \left\{<br /> \begin{array}{c l}<br /> 2^n(x-n)-1, &amp; \mbox{if } x \in [n-(1/2)^n, n] \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 2^n(n-x)+1, &amp; \mbox{if } x \in (n, n+(1/2)^n] \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 0, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br /> <br />

 

[LCKurtz]

Haven't we gone over this before:

1. Your function must have a negative slope on the right side of the interval.

2. When you put x = n in either formula, you have to get 1.

If your formula doesn't satisfy both of those, it is wrong. Think about it. I have to leave now for today.

 

[estro]

I believe it was fatigue blunder (embarrassing).

<br /> <br /> f(x) = \left\{<br /> \begin{array}{c l}<br /> 2^n(x-n)+1, &amp; \mbox{if } x \in [n-(1/2)^n, n] \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 2^n(n-x)+1, &amp; \mbox{if } x \in (n, n+(1/2)^n] \mbox{ while } n=min\{ |\lfloor x \rfloor - x|, |\lfloor x+1 \rfloor - x| \}\\<br /> \\<br /> 0, &amp; \mbox{otherwise}<br /> \end{array}<br /> \right.<br />

Many times I have simple idea for a solution fairly fast, but when I try to apply it and actually write equations my intuition is letting me down and I get lost.
This was a good example for it, without your help most likely I would spent a week(it actually happens to me occasionally) crunching this problem even though having the idea.