anonymous
  • anonymous
How do you simplify this equation into a value for y ? y = (Sqrt[2]/2) / E^((Sqrt[2]/2)^2) ​
Mathematics
chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
oh wow what is this lol
anonymous
  • anonymous
y = (Sqrt[2]/2) / E^((Sqrt[2]/2)^2) I think it's this.. ​\[y =\frac{ \frac{\sqrt{2}}{2} }{ E^{\frac{\sqrt{2}}{2}^2} ​}\]
anonymous
  • anonymous
Im supposed to be able to do this by hand... :/

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UsukiDoll
  • UsukiDoll
OMG who assigns this?
anonymous
  • anonymous
its the worst calculus course ever invented.. bill davis, horacio porta and Jerry Uhl
anonymous
  • anonymous
apparently a flipped curriculum.. where they explain nothing, and try to have the students work it out for themselves.
UsukiDoll
  • UsukiDoll
ffffffffffffffffffffffff
UsukiDoll
  • UsukiDoll
\[(\frac{\sqrt{2}}{2})^2\] I would start expanding this first.
UsukiDoll
  • UsukiDoll
how is this Calculus? More like torture.
anonymous
  • anonymous
maybe \[ (\frac{2^.5} { 2 }) (\frac{2^.5}{ 2 })\] \[ \frac{(2^.5)(2^.5)} { 4 } \]
anonymous
  • anonymous
Ive been thinking the same thing.. lol
UsukiDoll
  • UsukiDoll
ummm ... \[\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2}\]
anonymous
  • anonymous
woah.. is that true?
anonymous
  • anonymous
yes it is
UsukiDoll
  • UsukiDoll
\[\LARGE e^{\frac{1}{2}}\]
UsukiDoll
  • UsukiDoll
darn it ... I sort of forgot some of the e rules I know for \[\large e^{2+x} \rightarrow e^2e^x\]
anonymous
  • anonymous
so I think I can move the 2 in the numerator down to the denominator
UsukiDoll
  • UsukiDoll
I need to relook it up... I think it involves one of em .
anonymous
  • anonymous
yes thats true..
UsukiDoll
  • UsukiDoll
can't remember =/
myininaya
  • myininaya
hey so you are suppose to "simplify" the expression called y? and y is given as: \[y=\frac{\frac{\sqrt{2}}{2}}{e^{(\frac{\sqrt{2}}{2})^2}}\]
UsukiDoll
  • UsukiDoll
yeah and I already expanded the bottom
UsukiDoll
  • UsukiDoll
because that's an easy place to start
anonymous
  • anonymous
so I think we get y= 2^(1/2) / 2 E^(1/2)
UsukiDoll
  • UsukiDoll
where did that 2 come from? ^ at the denominator
anonymous
  • anonymous
btw is there an easy way to enter these equations? that equation tool takes me forever.
myininaya
  • myininaya
you can also write that as: \[y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}}\]
myininaya
  • myininaya
you can also chose to multiply the top and bottom by e^(1/2) to rationalize the denominator
myininaya
  • myininaya
well sorta rationalize anyways since e isn't rational :p
UsukiDoll
  • UsukiDoll
\[\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{e^\frac{1}{2}} \] I rather have this XD can't we flip the second fraction over or use one of the exponent laws for that 1/e^{something}
anonymous
  • anonymous
I cant believe these questions aren't all over open study.. I must be the only idiot in the world doing this course..
UsukiDoll
  • UsukiDoll
cough common core cough xD
myininaya
  • myininaya
yes you can do this: \[y=\frac{\sqrt{2}}{2 e^\frac{1}{2}} \cdot \frac{e^\frac{1}{2}}{e^\frac{1}{2}}\]
UsukiDoll
  • UsukiDoll
wasn't there also an ... I forgot what's it's called but if I have something like this... I can split it up...what's it called? \[\large e^{2+x} \rightarrow e^2e^x \]
myininaya
  • myininaya
that is one the law of exponents
anonymous
  • anonymous
I have this in notes... The main laws of exponents are e^(a + b) = e^a e^b e^(a - b) = e^a/e^b (e^a)^b = e^(a b) . a / E^n = a E^-n negative exponent law
UsukiDoll
  • UsukiDoll
ah... so would this work? \[\large e^{\frac{1}{2}} \rightarrow e^{1}e^{-2}\]
myininaya
  • myininaya
no
myininaya
  • myininaya
\[e^1e^{-2}=e^{1-2}=e^{-1}=\frac{1}{e}\]
UsukiDoll
  • UsukiDoll
oh xD now I remember
myininaya
  • myininaya
have you considered getting rid of the square root part on bottom as I suggested earlier ?
UsukiDoll
  • UsukiDoll
e and the other exponent laws are different \[n^{a-b} = \frac{n^a}{n^b}\]
myininaya
  • myininaya
n can be e
UsukiDoll
  • UsukiDoll
yeah \[\frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}) = \frac{2}{4} \rightarrow \frac{1}{2} \]
UsukiDoll
  • UsukiDoll
oh whoops yeah for something else not the exponent law thing
anonymous
  • anonymous
maybe log.. can be applied?
myininaya
  • myininaya
remember this: \[y=\frac{\sqrt{2}}{2 e^\frac{1}{2}} \cdot \frac{e^\frac{1}{2}}{e^\frac{1}{2}} \\ y= \frac{\sqrt{2} e^\frac{1}{2}}{2 e}=\frac{\sqrt{2 e}}{2e}\]
myininaya
  • myininaya
anyways that is the form I would leave it in but you may think the square thing looks better on bottom and if you think that you can multiply top and bottom by sqrt(2e) and simplify from there but this form above is the one I would leave it as
myininaya
  • myininaya
square root thing*
anonymous
  • anonymous
My calculator says this will end up being 1/Sqrt[2E] and I would get a value of 0.428882
myininaya
  • myininaya
ok if you want that other result as I said you can put the square root thing in the bottom by multiplying top and bottom by sqrt(2e) and simplifying from there
anonymous
  • anonymous
are we on track?
myininaya
  • myininaya
\[y=\frac{\sqrt{2e}}{2e} \\ \text{ this is my preferred form } \text{ multiply } \sqrt{2e} \text{ on bottom and \top and simplify from } \\ \text{ there if you want } y=\frac{1}{\sqrt{2e}}\]
dan815
  • dan815
sounds like a fun course :)
anonymous
  • anonymous
lol, dan you would own this course.. :)
anonymous
  • anonymous
is there a trick to turning this into a value that I can set into a chart.. I'm supposed to plot points with this, by hand.
myininaya
  • myininaya
example: \[\frac{\sqrt{u}}{u} \\ \frac{\sqrt{u}}{u} \cdot \frac{\sqrt{u}}{\sqrt{u}} =\frac{u}{u \sqrt{u}}=\frac{1}{\sqrt{u}}\]
anonymous
  • anonymous
ah I like that perspective.. thnx. myininaya, I will add that to the list..
UsukiDoll
  • UsukiDoll
oh I see.. after we got \[e^\frac{1}{2}\] we changed it in radical form and rationalize!
dan815
  • dan815
|dw:1434788368933:dw|
myininaya
  • myininaya
@hughfuve I still don't understand why you wanted to write it in this particular form
UsukiDoll
  • UsukiDoll
\[\large y=\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{e}} \] \[\frac{\sqrt{2}}{2\sqrt{e}}\] \[\frac{\sqrt{2}}{2\sqrt{e}} \cdot \frac{\sqrt{e}}{\sqrt{e}}\] \[\frac{\sqrt{2e}}{2e}\]
dan815
  • dan815
u wanna graph it? it would be a horizontal line
myininaya
  • myininaya
And is there a bigger question? You say you want to graph this? this is y=constant where we know the constant is between 0 and 1 since we had \[\frac{\sqrt{2e}}{2e} \text{ where } 0<\sqrt{2e}<2e \text{ dividing both sides by } 2e \text{ gives } \\ 0<\frac{\sqrt{2e}}{2e}<1\]
anonymous
  • anonymous
Oh Im just trying to get it to a place, where I can calculate a number , I need to arrive at 0.428882
anonymous
  • anonymous
and do it by hand... there are no calculators allowed.
UsukiDoll
  • UsukiDoll
O_O!
dan815
  • dan815
u can use those series to approximate e
anonymous
  • anonymous
well, that's just a single point.. on a graph, but it's a critical point.. it's the maxima
dan815
  • dan815
writing out the answer as y=1/sqrt(2e), is the nicest way imo
anonymous
  • anonymous
this equation is the 1st derivative[0]
myininaya
  • myininaya
Can you give the whole question?
anonymous
  • anonymous
okay... 1 sec.. typing it in..
anonymous
  • anonymous
Examine the derivative of f[x] = x E^(-x^2) to give a reasonably good hand sketch of the curve y = f[x] on the axes below. What are the maximum and minimum values f[x] can have ?
UsukiDoll
  • UsukiDoll
first derivative will have something like y' = ....
UsukiDoll
  • UsukiDoll
oh this is another question...
anonymous
  • anonymous
this is where this equation came from... I worked out the derivative.. and then put it in as x for f[x]
myininaya
  • myininaya
so when you set f'=0 you got 1-2x^2=0 right?
anonymous
  • anonymous
maybe I was completely on the wrong track though.
anonymous
  • anonymous
argh.. I ballsed up there hey?
anonymous
  • anonymous
I had this...
anonymous
  • anonymous
f'[x] = -2x^2 E^(-x^2) + E^(-x^2)
myininaya
  • myininaya
right and you can factor the exp(-x^2) out
anonymous
  • anonymous
0 = E^(-x^2) -2x^2 E^(-x^2)
anonymous
  • anonymous
head slap!
myininaya
  • myininaya
\[f'=e^{-x^2}(1-2x^2) \\ e^{-x^2} \neq 0 \text{ so we only have \to solve } 1-2x^2=0\]
anonymous
  • anonymous
I see...
anonymous
  • anonymous
looking at the equation now and it's jumping out at me.
myininaya
  • myininaya
but looking at \[f(x)=xe^{-x^2} \ \\ f(\frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}e^{-(\frac{1}{\sqrt{2}})^2}=\frac{1}{\sqrt{2}}e^{-\frac{1}{2}}=\frac{1}{\sqrt{2} e^\frac{1}{2}}=\frac{1}{\sqrt{2e}} \\ \text{ you already got this \above } \\ \text{ now you can find the other value } f(\frac{-1}{\sqrt{2}})\]
myininaya
  • myininaya
but you still need to determine if they are max,min,neither
myininaya
  • myininaya
e^(-x^2) is positive for all x (1-2x^2) is a parabola; it is what will tell us if the derivative is positive or negative
anonymous
  • anonymous
2nd derivative? or is there a better way
myininaya
  • myininaya
this is a rough graph of y=1-2x^2 |dw:1434789467495:dw| we are just using this graph to determine when f' is positive or negative you know where f' is: \[f'(x)=e^{-x^2}(1-2x^2)\]
UsukiDoll
  • UsukiDoll
2nd derivative is concavity.
myininaya
  • myininaya
looking at this graph can you tell me for what intervals y is positive and what intervals y is negative
anonymous
  • anonymous
somewhere between +/- 1/2 and 1/1
anonymous
  • anonymous
.5 .. .7 ... 1 ?
anonymous
  • anonymous
sqrt 2 = about 1.4 ?
UsukiDoll
  • UsukiDoll
1/1 = 1 though
myininaya
  • myininaya
I'm trouble understanding what you are saying ... umm... in the picture I drew of the parabola y=1-2x^2 y is positive when the graph is above the x-axis it is positive on (-1/sqrt(2),1/sqrt(2)) y is negative when the graph is below the x-axis it is negative on (-inf,-1/sqrt(2)) U (1/sqrt(2),inf)
anonymous
  • anonymous
thats why I was thinking ... 1/1 ... 1/sqrt(2) ... 1/2 somewhere in the middle.
anonymous
  • anonymous
and came up with about .7
UsukiDoll
  • UsukiDoll
sqrt(2) = 1.41 1... 1/1.41 = 0.70 1/2 = 0.5
anonymous
  • anonymous
or do we only need to know if positive or negative?
anonymous
  • anonymous
on the 2nd derivative.. I just learned that if f[x] is a possible extrema and f''[x] =0 then it is not maxima or minima, but if f''[x] = negative then f[x] is a maxima, and if f''[x] is positive then f[x] is a minima.. I think its a Fermat law ...
anonymous
  • anonymous
perl suggested it earlier on a problem.
myininaya
  • myininaya
this means: \[f'(x)=e^{-x^2}(1-2x^2) \\ f' \text{ is positive on } (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}} ) \text{ which means } f \text{ is increasing there } \\ f' \text{ is negative on } (-\infty, \frac{-1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \text{ which means } f \\ \text{ is decreasing there }\]
anonymous
  • anonymous
so we have our parabola orientation. sweet.
myininaya
  • myininaya
anyways so lets make a little sign chart here for the derivative thing: if it switches from increasing to decreasing at a critical number then at that critical number we have a max if it switches from decreasing to increasing at a critical number then at that critical number we have a min |dw:1434790458729:dw|
myininaya
  • myininaya
so using this can you determine which of f(-1/sqrt(2)) or f(1/sqrt(2)) will give us a max and which will give us a min?
myininaya
  • myininaya
remember the f'=- means f decreasing and f'=+ means f increasing
anonymous
  • anonymous
yes min in the -x max in the +x
myininaya
  • myininaya
right so the min value is f(-1/sqrt(2)) and the max value is f(1/sqrt(2))
anonymous
  • anonymous
/ \ 0 \ /
myininaya
  • myininaya
so we know our graph looks something like this between -1/sqrt(2) and 1/sqrt(2) |dw:1434790667090:dw|
myininaya
  • myininaya
now the other thing we know our function is decreasing on the other intervals ...so we know we have something like this for outside that interval: |dw:1434790765867:dw| I didn't finish the tail ends because we don't know the end behavior of the graph yet
anonymous
  • anonymous
and those points will be at.. y = f[ +/- 1/Sqrt[2] ]
myininaya
  • myininaya
I ask you to consider the following limits for end behavior : \[\lim_{x \rightarrow \infty}f(x) \text{ and } \lim_{x \rightarrow -\infty}f(x)\]
anonymous
  • anonymous
oh, actually..they gave me the points for that.. so we can skip that..
anonymous
  • anonymous
its on a plot... two points near + / - (2,0)
myininaya
  • myininaya
and I didn't mean to draw a 1 at those points by the way let me redo that
myininaya
  • myininaya
|dw:1434790888537:dw|
myininaya
  • myininaya
|dw:1434790920439:dw| you still need to finish the tail ends of the graphs where are they decreasing from do you think
myininaya
  • myininaya
decreasing from and to *
anonymous
  • anonymous
nice
anonymous
  • anonymous
if you wanted now to work out the square root of 2E by hand.. is there a trick? other than trial and error and squeezing ?
myininaya
  • myininaya
you can use a calculator ... or...you can say this: \[2
anonymous
  • anonymous
nice
myininaya
  • myininaya
and remember sqrt(2e)/(2e) is the same number as 1/sqrt(2e) we just wrote it differently
anonymous
  • anonymous
gotcha
myininaya
  • myininaya
summary: \[f(x)=xe^{-x^2} \\ f'(x)=e^{-x^2}+x(-2x)e^{-x^2} \\ f'(x)=e^{-x^2}(1-2x^2) \\ \text{ we know } e^{-x^2} >0 \text{ for all } x \\ 1-2x^2=0 \text{ when } x=\pm \frac{1}{\sqrt{2}} \text{ or if you prefer \to write as } x=\pm \frac{\sqrt{2}}{2} \\ \text now f(\pm \frac{\sqrt{2}}{2})=\pm \frac{1}{\sqrt{2e}} \text{ or if you prefer } \pm \frac{\sqrt{2e}}{2e} \\ \text{ also looking at } f' \\ \text{ we have } f'>0 \text{ when } x \in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}) \text{ so } f \text{ is increasing on } (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) \\ \text{ we have } f'<0 \text{ when } x \in (-\infty,\frac{-1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \\ \text{ so } f \text{ is decreasing on } (-\infty,\frac{-1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}},\infty) \\ \text{ so this tells us our max } f(\frac{\sqrt{2}}{2})=\frac{1}{\sqrt{2e}} \text{ and our min is } f(-\frac{\sqrt{2}}{2})=-\frac{1}{\sqrt{2e}} \] so from this information we can roughly sketch part of the graph: |dw:1434791775024:dw| the only left for you is to work out the end behavior the graph
anonymous
  • anonymous
oh thats amazing help.. you out did yourself here.. I got those end behaviours.. this equation is decaying to 0 because of that exponential denominator f[x] = x/ E^(x^2)
myininaya
  • myininaya
you probably could use the fact your function is odd to justify the end behavior of your graph as: as we approach pos/neg infinity the graph is getting closer to y=0
myininaya
  • myininaya
|dw:1434792046243:dw| yep so you would end up with something like this
anonymous
  • anonymous
gotcha.. phew! exactly what I was thinking..
anonymous
  • anonymous
man, I wish every problem wasn't like this. I got 11 more on this assignment.. just like this one.. lol.
anonymous
  • anonymous
but this will get me a long way to solving the rest..
myininaya
  • myininaya
good luck on the rest
myininaya
  • myininaya
myininaya is probably going to have to go now unless you have anymore questions about this particular question?
anonymous
  • anonymous
thank you thank you again.. I hope the rest of your night are fun problems.
anonymous
  • anonymous
Im good
myininaya
  • myininaya
I actually wasted my whole. I should have been in bed like hours ago. It is 4:25 AM here.
myininaya
  • myininaya
my whole night*
anonymous
  • anonymous
lol.. oh wow.. well I appreciate your time very very much, you're a life saver.
myininaya
  • myininaya
I wouldn't say wasted. I had fun. Played some video games and did some math problems. Fun fun.
myininaya
  • myininaya
anyways goodnight(sort of )
anonymous
  • anonymous
here's a sneak peak at my next challenge.. please dont answer ... To Plot this by hand ... f[x] = (x^2 - 2 Log[x])/2
anonymous
  • anonymous
sigh.. going to be a long night. lol.
myininaya
  • myininaya
http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtI.aspx This site maybe helpful too. Not sure if you ever heard of it.
UsukiDoll
  • UsukiDoll
at least @hughfuve got his moneys worth and Qualified Help unlike the last time I came on one of his QH questions.

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