gorica
  • gorica
Vector space
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
the anticipation is killing me!!
gorica
  • gorica
Let \[V=\{f:(0,2) \rightarrow R| f\ is\ continuous\ and\ \lim_{x \rightarrow o^+}f(x)=\lim_{x \rightarrow 2^-}f(x)=0 \}\]Show that \[(V,+, \cdot, R)\]is vector space. I know I have to show thatfor all x,y in V, x+y is in V and that for all x in V and all α in R, αx is in V.
amistre64
  • amistre64
hmm, so something of a visual like this maybe? |dw:1377792310168:dw|

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gorica
  • gorica
|dw:1377792442827:dw|or like this? I don't think it's much about how does them look like.
amistre64
  • amistre64
ive got abstract algebra group stuff cluttering my head at the moment ...
amistre64
  • amistre64
right, f is some general function with the stated properties what do you recall about the requirments to define a vector space? and subspace?
amistre64
  • amistre64
identity, inverse, and zero is what im thinking of, but those relate to groups ...
gorica
  • gorica
What I wrote above is actually for subspace. I think I could solve this if somebody could tell me definition for \[\lim_{x \rightarrow a^+} f(x)=0\] and \[\lim_{x \rightarrow a^-}f(x)=0\]
amistre64
  • amistre64
limit from the left and right
amistre64
  • amistre64
|dw:1377792859241:dw|
amistre64
  • amistre64
in other words: as x approaches 2 from the left, the value approaches 0 as x approaches 0 from the right, the value approaches 0
amistre64
  • amistre64
it could just as well be the function: x(x-2)
gorica
  • gorica
That is clear to me, but I need some definition as \[(\forall \epsilon > 0)(\exists \delta>0(\forall x_1,x_2 \in [0,1])(|x_1-x_2|<\delta \rightarrow |f(x_1)-f(x_2)|<\epsilon)\] is for continuity of f
gorica
  • gorica
I mean, I need it in this form so I can use it
anonymous
  • anonymous
This link has some info on one-sided limit definitions near the bottom of the page: http://tutorial.math.lamar.edu/Classes/CalcI/DefnOfLimit.aspx
amistre64
  • amistre64
Pauls site? yeah, thats always a good bet
anonymous
  • anonymous
By the way @gorica, do you have to show it's a vector space or a subspace? Because what you mentioned has to do with showing a space is a subspace.
gorica
  • gorica
I have to show that it is vector space.
terenzreignz
  • terenzreignz
Do we actually have to go through the nitty-gritty definition of a limit? Why not just use \[\Large \lim_{x\rightarrow a^+}f(x) = L \quad \implies \quad \lim_{x\rightarrow a^+}cf(x) = cL\]
terenzreignz
  • terenzreignz
And of course, the other bit, \[\Large \lim_{x\rightarrow a^+}f(x) = L \quad \& \quad \lim_{x\rightarrow a^+}g(x) = M\]\[\Large \implies \lim_{x\rightarrow a^+}[f(x) +g(x)]= L+M\]
gorica
  • gorica
That's enough? If V is a subset of some vector space I can show only those two. But I don't know if V is a subset of some vector space. Do you maybe know? Or I have to show all properties of vector space?
terenzreignz
  • terenzreignz
Let's start with basics (unless you've already done these) Suppose f and g are two elements of V. Then show that f+g is also in V... is this already done?
gorica
  • gorica
If it is enough to write only what you wrote, it is done. I thought to try to show it by definition.
terenzreignz
  • terenzreignz
Overkill, don't you think? I mean... does your linear algebra instructor actually want to use the Real-Analysis limit? With the epsilon and delta? D: ...sadistic
gorica
  • gorica
Not linear algebra... Metric and normed spaces :D
terenzreignz
  • terenzreignz
Oh... crud. Then... enjoy the \(\varepsilon\) and \(\delta\) then XD So... have you proved this? \[\Large \lim_{x\rightarrow a^+}f(x) = L \quad \implies \quad \lim_{x\rightarrow a^+}cf(x) = cL\]
gorica
  • gorica
Thanks for encouraging me :D I haven't.
terenzreignz
  • terenzreignz
Well... okay, let's do it...
terenzreignz
  • terenzreignz
Assume that \[\Large \lim_{x\rightarrow a^+}f(x) = L\]
terenzreignz
  • terenzreignz
Then, for all \(\large \varepsilon>0\) there exists a \(\large \delta>0\) such that whenever\[\Large x -a < \delta\] then \[\Large \left|f(x) - L\right|<\Large \frac{\varepsilon}{c}\] Catch me so far?
gorica
  • gorica
I think I do :D You skipped few steps but I got it :D
terenzreignz
  • terenzreignz
Actually it's wrong, c could be negative... this is more accurate: \[\Large \left|f(x) - L\right| < \frac{\varepsilon}{|c|}\]
terenzreignz
  • terenzreignz
So that we could multiply both sides by |c| giving \[\Large |c|\left|f(x) - L\right| < \varepsilon\]\[\Large \left|cf(x) - cL\right| < \varepsilon\]
terenzreignz
  • terenzreignz
Thus proving that \[\Large \lim_{x\rightarrow a^+}cf(x) = cL\]
terenzreignz
  • terenzreignz
Got it? ^_^
gorica
  • gorica
got it :)
terenzreignz
  • terenzreignz
Now, let's do the other one, the sum... Suppose we have \[\Large \lim_{x\rightarrow a^+}f(x) = L\]and\[\Large \lim_{x\rightarrow a^+}g(x) = M\]
terenzreignz
  • terenzreignz
Then we begin the proof...
terenzreignz
  • terenzreignz
Let \(\large \varepsilon >0\) Then there exist \(\large \delta_1,\delta_2>0\) such that \[\Large x-a < \delta_1 \implies \left|f(x)-L\right|<\varepsilon\]\[\Large x-a < \delta_2 \implies \left|g(x) -M\right|<\varepsilon\] Catch me so far?
gorica
  • gorica
yes, go on
terenzreignz
  • terenzreignz
Wait, neater this way...\[\Large x-a < \delta_1 \implies \left|f(x)-L\right|<\frac{\varepsilon}2\]\[\Large x-a < \delta_2 \implies \left|g(x) -M\right|<\frac{\varepsilon}2\]
terenzreignz
  • terenzreignz
After all, \(\Large \frac{\varepsilon}2\) is just as arbitrary.
terenzreignz
  • terenzreignz
Anyway, choose \(\large \delta = \min\left\{\delta_1,\delta_2\right\} \)
terenzreignz
  • terenzreignz
So that whenever \(\Large x-a < \delta\) then both \[\Large \left|f(x) -L\right|<\frac \varepsilon 2\] and \[\Large \left|g(x)-M\right|<\frac \varepsilon 2\]hold
terenzreignz
  • terenzreignz
Now, we can simply add these two inequalities... \[\Large \left|f(x)-L\right|+\left|g(x)-M\right| < \varepsilon\]
terenzreignz
  • terenzreignz
Do you see where this is going?
terenzreignz
  • terenzreignz
TRIANGLE INEQUALITY. |a + b| \(\le\) |a| + |b|
gorica
  • gorica
Yes, it's ok, I've got it when I saw it. \[|f(x)+g(x)-(L+M)|<\epsilon\]
terenzreignz
  • terenzreignz
That's good ^_^ And THAT... proves it :D
terenzreignz
  • terenzreignz
So you now have pretty much proved the only two limit-properties that you need to prove that V is vector space... unless you also want to prove that the sum of two continuous functions is continuous? :3
gorica
  • gorica
No, I think it would be enough to write it by words :D But tell me, is it enough to show only what you showed now or I have to show all properties of a vector space? Or you know that V is a subset of some vector space?
terenzreignz
  • terenzreignz
That really depends on your instructor... -.-
amistre64
  • amistre64
nice job .. i personally never know how far to take a proof. When do we stop trying to reinvent the wheel after all?
terenzreignz
  • terenzreignz
^_^
gorica
  • gorica
ok, I hope she will be happy when she sees a lot of epsilons and deltas so she won't ask for more :D Thank you very much :)

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