Finding a function, given derivative

[Maybe_Memorie]

Homework Statement

Prove that there is a function s, defined on all of R, such that

s(0) = 0 and s′(x) = (1 + x^4)^(−1/2)

Show that s is bounded.

Homework Equations

Integration by parts

The Attempt at a Solution

Right, I tried using integration by parts. That didn't get me anywhere.

A hint to get me started would be very helpful. Thanks :)

 

[Mark44]

Homework Statement

Prove that there is a function s, defined on all of R, such that

s(0) = 0 and s′(x) = (1 + x^4)^(−1/2)

Show that s is bounded.

Homework Equations

Integration by parts

The Attempt at a Solution

Right, I tried using integration by parts. That didn't get me anywhere.

A hint to get me started would be very helpful. Thanks :)

I think that you are misreading this problem. It doesn't ask you to find s(x) - it asks you to show that s(x) is bounded.

 

[SVXX]

@Mark : What other way can he use to prove that 's' exists besides integration? It is given that after proving that s(x) exists, to show that it is bounded.
The integral isn't an elementary one though...can't be done using Riemann integrals.

 

[Mark44]

This problem looks to me like it might be an application of the Fundamental Theorem of Calculus.
\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}

From this it's easy to see that s(0) = 0 and s'(x) = 1/sqrt(1 + x^4).

I haven't worked it through, but this is what I'd start with.

 

[SammyS]

Find a function f(x) such that f(x) > s'(x) for all real x and such that: \int_0^\infty f(x)\,dx converges. Then s(x) is bounded above.

This works because s'(x) > 0 for all real x.

Show that s(x) is an odd function to get a lower bound.

Hint to help find such a function, f :

\frac{1}{x^2}>\frac{1}{\sqrt{1+x^4}}\quad\text{and}\quad\int_{a}^\infty\frac{1}{x^2}\,dx=\frac{1}{a}\ \ \text{ for }\ \ a>0\,.

A piecewise function will do the trick.

 

[Maybe_Memorie]

This problem looks to me like it might be an application of the Fundamental Theorem of Calculus.
\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}

From this it's easy to see that s(0) = 0 and s'(x) = 1/sqrt(1 + x^4).

I haven't worked it through, but this is what I'd start with.

I'm kind of lost as to what to do next

 

[SammyS]

Define: r'(x)=min(1, 1/x²), and r(0)=0.

Show that r'(x) ≥ s'(x) for all real x, in fact, r'(x) > s'(x) except for x = 0 at which they're equal.

r(x) > s(x) for x > 0.

Use integration to find r(x).

Both r'(x) & s'(x) are even functions which take on only positive values.

Both r(x) & s(x) are odd functions and monotonic increasing. (Why?)

Toss in a few other details, & you're done.

 

[Maybe_Memorie]

I found s(x) and showed that it was unique.

I'm not sure how to show that it is bounded though

 

[SammyS]

I found s(x) and showed that it was unique.

I'm not sure how to show that it is bounded though

What did you get for s(x)?

s'(x) > 0, so that says that s(x) is monotonic increasing.

If you can determine the limits, lim_x→±∞ s(x), that should give you the bounds.

If you use the function r'(x) from my previous post, that will also give you bounds, just not the least upper bound or greatest lower bound.

 

[judowrestler1]

well the integral ends up being
ln((x^2)+sqrt(1+(x^4)))
so at 0
ln((0^2)+sqrt(1+(0^4)))
ln(sqrt(1))
ln(1)
0

 

[Maybe_Memorie]

I noticed the question never actually said "Find s(x)". All I had to do was show it exists, is defined on all of R, is unique, and is bounded.

\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}

s(0) clearly equals 0 and s(x) is clearly defined on all of R.

Suppose f(x) also has the desired properties.
Then f'(x) = s'(x)
So f(x) = s(x) + C , C is a constant.

s(0) = 0 = f(0) therefore C = 0.
Then f(x) = s(x) and the function is unique.


Now how do I show it is bounded?

 

[HallsofIvy]

well the integral ends up being
ln((x^2)+sqrt(1+(x^4)))
so at 0
ln((0^2)+sqrt(1+(0^4)))
ln(sqrt(1))
ln(1)
0

?
If F is defined by
F(x)= \int_0^x f(t)dt
for any integrable function f, then F(0)= 0.

 

[hunt_mat]

You know the derivative, my first impulse would be to find out the values of the function s'(x) at some various points, say 0,\pm\infty.

 

[Maybe_Memorie]

You know the derivative, my first impulse would be to find out the values of the function s'(x) at some various points, say 0,\pm\infty.

x = 0, s'(x) = 1
x = plus or minus infinity, s'(x) = 0

 

[SammyS]

well the integral ends up being
ln((x^2)+sqrt(1+(x^4)))
so at 0
ln((0^2)+sqrt(1+(0^4)))
ln(sqrt(1))
ln(1)
0

Not true.

d[ln((x^2)+sqrt(1+(x^4)))]/dx = 2x/√(1+x⁴) ≠ s'(x)

 

[SammyS]

I noticed the question never actually said "Find s(x)". All I had to do was show it exists, is defined on all of R, is unique, and is bounded.

\text{Let} s(x) = \int_0^x \frac{dt}{\sqrt{1+t^4}}

s(0) clearly equals 0 and s(x) is clearly defined on all of R.

Suppose f(x) also has the desired properties.
Then f'(x) = s'(x)
So f(x) = s(x) + C , C is a constant.

s(0) = 0 = f(0) therefore C = 0.
Then f(x) = s(x) and the function is unique.Now how do I show it is bounded?

Use the comparison test with s'(x) and the r'(x) I suggested.

Another way to describe r'(x) is:

r&#039;(x)=\left\{\begin{array}{cc}1,&amp;\mbox{ if }<br /> -1\leq x\leq 1\\ \displaystyle \frac{1}{x^2}\ \ , &amp; \mbox{ otherwise } \end{array}\right.

Compare the graphs of r'(x) and s'(x) . This may help you see how finding r(x) can help show that s(x) is bounded.

 

[SammyS]

Here is a link to a plot by WolframAlpha showing the two functions: r'(x) and s'(x), where r'(x) is defined as:

r&#039;(x)=\left\{\begin{array}{cc}1,&amp;\mbox{ if }<br /> -1\leq x\leq 1\\ \\ \displaystyle \frac{1}{x^2}\ \ , &amp; \mbox{ otherwise } \end{array}\right.

The function, r(x), where r(0) = 0, is easily found by integration:

r(x) = \int_0^x r&#039;(t)\,dt

r(x)=\left\{\begin{array}{cc}x,&amp;\mbox{ if }<br /> -1\leq x\leq 1\\ \\ \displaystyle 2-\frac{1}{x}\ \ , &amp; \mbox{ if }<br /> x&gt;1 \\ \\ \displaystyle -2-\frac{1}{x}\ \ , &amp; \mbox{ if }<br /> x&lt;-1 \end{array}\right.

.