Failing ε-N proof

Taylor series ?

>>8676753
the equation in the picture is fucked up
it should be |(m - sqrt(m^2 + 1))/(2m + 2sqrt(m^2 + 1))| < epsilon. Obviously, (-1)*(m+sqrt(m^2 +1))*(m-sqrt(m^2 +1))/(2(m+sqrt(m^2 +1))^2) = 1/(2(m+sqrt(m^2 +1))^2) < epsilon. Clearly, 1/(2(m+sqrt(m^2 +1))^2) < 1/(2(2m)^2), so, choose m such that 1/(2(2m)^2) < epsilon, 8m^2>1/(epsilon), m>1/sqrt(8*epsilon). Proof: Take m>1/sqrt(8*epsilon), then 8m^2>1/(epsilon), 1/(2(2m)^2) < epsilon, 1/(2(m+sqrt(m^2+1))^2) < 1/(2(2m)^2)<epsilon, |(m-sqrt(m^2+1))*(m+sqrt(m^2+1))/(2(m+sqrt(m^2+1))^2)|<epsilon,
|(m-sqrt(m^2 +1))/(2(m+sqrt(m^2 +1)))|<epsilon
|(2m - (m + sqrt(m^2 +1)))/(2/(m+sqrt(m^2 +1)))|<epsilon
|(m/(m+sqrt(m^2+1))) - 1/2| < epsilon.

>>8676758
You mean developing the taylor series? That would rely on derivatives and we can only use that the reals are a complete totally ordered field and the definition of convergence of a sequence.

I think the main issue I'm having is that it has +1 on the radical, because if that was -1 I could square and add +1 to both sides. I guess I'll try to work out the problem with -1 under the radical and see exactly how and why that would change the problem and try to adapt my strategy.

If anyone has pointers, they're still more than welcome.

>>8676812
protip: correct your arithmetic

>>8676810
>the equation in the picture is fucked up
It is? What step is wrong? I don't see how you get to
[eqn]
\left|\frac{n - \sqrt{n^2 + 1}}{2n + 2\sqrt{n^2 + 1}}\right| < \varepsilon
[/eqn]

>>8676814
Ugh the square. Thanks a bunch, I should be able to find it now.

>>8676830
Ignore this. There isn't a square missing as anywhere.

>>8676814
I don't see the arithmetic error. Can you point it out please?

>>8676828
|(m(m^2 + 1 - m^2)/(sqrt(m^2+1) +m) - 1/2| < epsilon. Simplifying, we get |m/(sqrt(m^2 +1) +m) - 1/2|<epsilon. Now, using arithmetic, we can simplify this to |2m/(2(sqrt(m^2 +1) +m))- (sqrt(m^2 +1) +m)/(2(sqrt(m^2 +1) +m))|<epsilon. Which becomes |(m - sqrt(m^2 +1))/(2(sqrt(m^2+1)+m))|<epsilon. This is where the blackboard differs

>>8676814
>>8676841
Wow I reviewed this several times and never caught forgetting the multiplication by 2 in the numerator.

Thanks a lot for pointing it out.

>>8676812
I don't know what you are talking about. I mean using Taylor expansion of a function, like so :

[math] \sqrt{1 + \frac{1}{n2}} = \frac{1}{2n2} + o(\frac{1}{n2})[/math]

thus [[eqn] n(\sqrt{n2 + 1} - n) = n2 \left( \sqrt{1 + \frac{1}{n2}} - 1\right) = n2 \left( \frac{1}{2n2} + o(\frac{1}{n2})\right) = \frac{1}{2} + o(1) [/eqn]

where o(1) is something that tends towards 0

>>8676753
Multiply by
Sqrt (m^2 + 1) / sqrt (m^2 +1)

Then you get
abs ( 1 / ( 2 + 1/m))

Is easy from here because m is increasing so 1/m is decreasing.

Therefore, you can use 3*epsilon and the rest is easy from there.

>>8677011
Nvm. You can still do this but you have to make an argument for

m/sqrt (m^2 + 1)

Which is increasing so it won't work.

Try using quotient theorem?

Or if you want to do it without Taylor expansion, factorising was still the way to go:

[math] n(\sqrt{n^2 + 1} - n) = n^2 ( \sqrt{ 1 + \frac{1}{n^2} } - 1) = n^2 \frac{1 + \frac{1}{n^2} -1}{\sqrt{ 1 + \frac{1}{n^2} } + 1} = \frac{1}{\sqrt{ 1 + \frac{1}{n^2} } + 1} \rightarrow_{n \rightarrow \infty} \frac{1}{2} [/math]

>>8676753

I think you'll find your problem in the fact that you decided to work under the 'real' numbers

>>8677040
Correct but doesn't use epsilon-delta definitions.

File: 1336861186638.jpg (25KB, 500x364px) Image search: [Google]

25KB, 500x364px

>>8677742
It's implied in the last step.

>>8677742
For 0<e<1/2 suppose N=(1-2e)/sqrt{8e}, then
N^2+1=(1-2e)^2/8/e+1
=(1+2e)^2/8/e
sqrt{N^2+1}=(1+2e)/sqrt{8e}
sqrt{N^2+1}-N=4e/sqrt{8e}
N(sqrt{N^2+1}-N)=(1-2e)4e/8/e=(1-2e)/2=1/2-e
so N(sqrt{N^2+1}-N)-1/2=-e
Maybe show that this also holds for inequalities?