Author Topic: Real Analysis (Read 15390 times) Tweet Share

QuantumJG · « **on:** March 01, 2010, 08:03:32 pm »

Ok I have hit a rather embarrassing but confusing question.

What is x²?

Actually according to wikipedia,

$x^{2} = \sum _{n = 0} ^{x - 1} (2n + 1)$

QuantumJG · « **Reply #1 on:** March 01, 2010, 08:12:06 pm »

Ok is it $x \times x$ ?

QuantumJG · « **Reply #2 on:** March 01, 2010, 09:04:26 pm »

Another question asks me to define e^x

$e^{x} = 1 + \dfrac{x}{1!} + \dfrac{x^{2} }{2!} + \dfrac{x^{3} }{3!} + ... + \dfrac{x^{n} }{n!} , where n \in \mathbb{Z} ^{+}$

$\therefore e^{x} = \sum _{n = 0} ^{ \infty } \dfrac{x^{n} }{ n! }$

$where n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1 = 1 \times 2 \times ... \times (n-2) \times (n-1) \times n$

$\therefore n! = \prod _{k=1} ^{n} k , where k \in \mathbb{Z} \le n$

Is this definition ok for e^x?

TrueTears · « **Reply #3 on:** March 01, 2010, 09:10:19 pm »

well the way you have defined e^x is centered at 0, which is a Maclaurin series, it could be centered around other numbers. You could also add that the radius of convergence is infinity, so the series converges for all x.

Maybe you could also use taylor's inequality and prove that e^x is indeed equal to this specific taylor expansion. Which will prove that your definition is correct.

QuantumJG · « **Reply #4 on:** March 01, 2010, 09:14:49 pm »

We haven't looked at that stuff yet!

TrueTears · « **Reply #5 on:** March 01, 2010, 09:15:06 pm »

Oh, it's fun as stuff, omg you will love it when you get to it !!

it's kinda like epsilon - delta proofs, i love it!!

but if you are not up to it, do u know WHY e^x can be expanded like that? I think that is important to know, maybe you can try prove it

QuantumJG · « **Reply #6 on:** March 01, 2010, 09:39:28 pm »

Quote from: TrueTears on March 01, 2010, 09:15:06 pm

Oh, it's fun as stuff, omg you will love it when you get to it !!

it's kinda like epsilon - delta proofs, i love it!!

but if you are not up to it, do u know WHY e^x can be expanded like that? I think that is important to know, maybe you can try prove it

Our professor didn't discuss why, but I remember in physics our professor was saying it's from the Taylor series, is that true? Then again he didn't really explain what a Taylor series is.

TrueTears · « **Reply #7 on:** March 01, 2010, 09:43:07 pm »

if f has a power series expansion: $f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^n$ then the maclaruin series is just a special case of it: namely, centered at 0. (a = 0)

but that's not the slick part, the coolest part is the relationship between the coefficients of the power series f(x) and the derivatives of f(x), thats the part i love the most!!

humph · « **Reply #8 on:** March 01, 2010, 11:12:21 pm »

Wait until you get to Laurent series in complex analysis and its relation to residues and contour integrals. That's pretty cool

/0 · « **Reply #9 on:** March 02, 2010, 01:19:11 am »

Quote from: QuantumJG on March 01, 2010, 08:12:06 pm

Ok is it $x \times x$ ?

Isn't it just a matter of definition?

QuantumJG · « **Reply #10 on:** March 02, 2010, 03:01:19 pm »

Quote from: TrueTears on March 01, 2010, 09:43:07 pm

if f has a power series expansion: $f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ then the maclaruin series is just a special case of it: namely, centered at 0. (a = 0)

but that's not the slick part, the coolest part is the relationship between the coefficients of the power series f(x) and the derivatives of f(x), thats the part i love the most!!

So what this is impliying that if,

$f(x) = e^{x}$

$then, f(a=0) = e^{0} = 1$

So

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ becomes $f(x) = \sum_{n=0}^{\infty} \frac{ x^{n} }{n!}$

TrueTears · « **Reply #11 on:** March 02, 2010, 09:33:42 pm »

Quote from: QuantumJG on March 02, 2010, 03:01:19 pm

Quote from: TrueTears on March 01, 2010, 09:43:07 pm
if f has a power series expansion: $f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ then the maclaruin series is just a special case of it: namely, centered at 0. (a = 0)

but that's not the slick part, the coolest part is the relationship between the coefficients of the power series f(x) and the derivatives of f(x), thats the part i love the most!!

So what this is impliying that if,

$f(x) = e^{x}$

$then, f(a=0) = e^{0} = 1$

So

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^{n}$ becomes $f(x) = \sum_{n=0}^{\infty} \frac{ x^{n} }{n!}$

don't you mean when f(x) is centered around 0? not f(a=0)?

it's not implying anything... just that the expansion is "centered" around 0 so that the power series expansion provides a good approximation around 0 xD

QuantumJG · « **Reply #12 on:** March 06, 2010, 09:12:53 pm »

Prove that:

$tan ^{-1} x = sin ^{-1} ( \dfrac{x}{ \sqrt{ 1 + x^{2} } } )$

What would be a good way to do this proof?

TrueTears · « **Reply #13 on:** March 06, 2010, 09:28:11 pm »

$\tan\left(\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\right)$

$\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$ Right angle triangle, opposite is $x$ , hypotenuse is $\sqrt{1+x^2}$

Thus tan of this angle is $x$ .

Thus $\tan\left(\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)\right) = x$

QuantumJG · « **Reply #14 on:** March 11, 2010, 01:54:43 pm »

What is the difference between $\mathbb{Q} [x]$ and $\mathbb{Q} [[x]]$ ?

Also when you are asked:

write sin(x) as an element of $\mathbb{Q} [[x]]$ is that just:

sin(x) = $x - \dfrac{x ^{3} }{3!} + \dfrac{x ^{5} }{5!} - \dfrac{x ^{7} }{7!}$

Also what is: $\mathbb{Q} (x)$ & $\mathbb{Q} ((x))$ ?

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