Hey guys,

I'm trying to prove that f(x)=x^4 is continuous on R, using the e-delta argument, but I can't find a way to break up x^4-a^4 effectively in the forms of x-a. ANybody have any idea here?

I appreciate any hints and help!

Thanks a lot!

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- Sep 6th 2012, 07:58 PMmfxuusCOntinuity
Hey guys,

I'm trying to prove that f(x)=x^4 is continuous on R, using the e-delta argument, but I can't find a way to break up x^4-a^4 effectively in the forms of x-a. ANybody have any idea here?

I appreciate any hints and help!

Thanks a lot! - Sep 6th 2012, 08:08 PMSworDRe: COntinuity
$\displaystyle x^4 - a^4 = (x-a)(a^3+xa^2+x^2a+x^3)$

That help? If thats insufficient, maybe prove using similar identities by induction that if x^n is continuous, then x^(n+1) is too? Sorry if this doesn't help, I don't know much about the kind of stuff your trying to do :p - Sep 6th 2012, 09:43 PMjohnsomeoneRe: COntinuity
Fix a>0.

Pick any epsilon > 0.

Choose some positive delta < min { epsilon/(15a^3), a }.

Note that x^4 - a^4 = ( x - a )( x^3 + ax^2 + (a^2)x + a^3 )

When |x-a|<delta, have a-delta < x < a+delta, so, since 0< delta < a, get 0 < x < 2a.

Then | x^3 + ax^2 + (a^2)x + a^3 | = x^3 + ax^2 + (a^2)x + a^3 < (2a)^3 + a(2a)^2 + (a^2)(2a) + a^3 = 15 a^3.

Thus |x-a|<delta implies |x^4 - a^4| = | x - a | * | x^3 + ax^2 + (a^2)x + a^3 | < | x - a | * ( 15 a^3 ) < (delta)( 15 a^3 ) < (epsilon/(15a^3))( 15 a^3 ) = epsilon

Thus 0<|x-a|<delta implies |x^4 - a^4| < epsilon.

Thus lim x->a of x^4 exists and equals a^4.

Thus the function f defined by f(x)=x^4 is continuous at x=a for all a>0.

Since f is even, f is also continuous for x < 0 ( lim x->(-a) f(x) = lim t->a f(-t) = lim t->a f(t) = f(a) = f(-a) ).

Showing f is continuous at 0 is very simple (use delta = the fourth root of epsilon).

Thus f is continuous on all of R. - Sep 6th 2012, 10:22 PMDevenoRe: COntinuity
there is no reason to put any restriction on a.

what we CAN do is restrict how far away from a we let x be. for example, we can require $\displaystyle \delta \leq 1$.

then:

|x - a| < 1

but |x| - |a| ≤ |x - a| < 1, so:

|x| < 1 + |a|.

therefore:

$\displaystyle |x^3 + ax^2 + a^2x + a^3| \leq |x^3| + |ax^2| + |a^2x| + |a^3|$

$\displaystyle = |x|^3 + |a||x|^2 + |a|^2|x| + |a|^3 < (1 + |a|)^3 + |a|(1 + |a|)^2 + |a|^2(1 + |a|) + |a|^3$

$\displaystyle = (1 + 3|a| + 3|a|^2 + |a|^3) + |a|(1 + 2|a| + |a|^2) + |a|^2 + |a|^3 + |a|^3$

$\displaystyle = 1 + 4|a| + 6|a|^3 + 4|a|^3$. for convenience let us call this number L.

then, letting $\displaystyle \delta = \min\{\frac{\epsilon}{L},1\}$, we have:

$\displaystyle |x^4 - a^4| = |x - a||x^3 + ax^2 + a^2x + a^3| < \left(\frac{\epsilon}{L}\right)(L) = \epsilon$.

a similar (but much more awkward to write down) proof shows that:

$\displaystyle f(x) = x^n$ is continuous at any real number a, for any natural number n.