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Maxima and Minima (vector calculus)


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I don't think she was doing any more than eyeballing the graph. It's maybe easiest to see at x=0, y seems to be about 1 at its max, and so too when y=0 x is about 1.

Depends how much experience you have in R. Try following this example. Wolfram alpha might be better if easy is all you want.

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https://www.wolframalpha.com/input?i=plot+(x^2%2By^2)exp((-x^2-y^2))

The intuitive idea you can get from highlighting x=1 and y=1 curves:

image.png.46ee63783b1c7206a96b8f61c7e98afe.png

And another analytic tool is using polar coordinates,

\[ \left(x^{2}+y^{2}\right)\exp\left(-x^{2}-y^{2}\right)=r^{2}\exp\left(-r^{2}\right) \]

and then do a Taylor-series expansion around r=1 in a small positive parameter \( \varepsilon \),

\[ r=1+\varepsilon \]

\[ \varepsilon>0 \]

\[ r^{2}\exp\left(-r^{2}\right)=\left(1+\varepsilon\right)^{2}\exp\left(-\left(1+\varepsilon\right)^{2}\right)=\left(1+2\varepsilon+\varepsilon^{2}\right)\exp\left(-1-2\varepsilon-\varepsilon^{2}\right)= \]

\[ =e^{-1}\left(1+2\varepsilon+\varepsilon^{2}\right)\exp\left(-2\varepsilon-\varepsilon^{2}\right)= \]

\[ =e^{-1}\left(1+2\varepsilon+\textrm{order}\left(\varepsilon^{2}\right)\right)\left(1-2\varepsilon+\textrm{order}\left(\varepsilon^{2}\right)\right)= \]

\[ =e^{-1}\left(1-4\varepsilon^{2}\right)+\textrm{order}\left(\varepsilon^{2}\right) \]

And, as you see, the first-order correction is negative no matter what orientation with respect to the origin (just where the bump is) you place yourself.

Same goes for 1-epsilon, of course.

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The easiest way to visualise this since the circle is symmetric is to consider the simpler function

f(p) = pe-p,  where p = (x2 + y2)

any plane section will have this form.

When p = 0.5 f(p) is approximately 0.3

When p =1, f(p) is approximately 0.4

When p = 2 f(p) is approximately 0.2

and f(p) is continuous so there is a max between p = (x2 + y2) = 0.3 and p = (x2 + y2) = 2

 

plot8.jpg.b2f8e08de001e4bdbd974540e0937c91.jpg

 

Edited by studiot
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Posted (edited)
4 hours ago, joigus said:

https://www.wolframalpha.com/input?i=plot+(x^2%2By^2)exp((-x^2-y^2))

The intuitive idea you can get from highlighting x=1 and y=1 curves:

image.png.46ee63783b1c7206a96b8f61c7e98afe.png

And another analytic tool is using polar coordinates,

 

(x2+y2)exp(x2y2)=r2exp(r2)

 

and then do a Taylor-series expansion around r=1 in a small positive parameter ε ,

 

r=1+ε

 

 

ε>0

 

 

r2exp(r2)=(1+ε)2exp((1+ε)2)=(1+2ε+ε2)exp(12εε2)=

 

 

=e1(1+2ε+ε2)exp(2εε2)=

 

 

=e1(1+2ε+order(ε2))(12ε+order(ε2))=

 

 

=e1(14ε2)+order(ε2)

 

And, as you see, the first-order correction is negative no matter what orientation with respect to the origin (just where the bump is) you place yourself.

Same goes for 1-epsilon, of course.   

So, we will get the same first order correction for [math]1-\epsilon[/math]

Isn't it?

 

6 hours ago, Prometheus said:

I don't think she was doing any more than eyeballing the graph. It's maybe easiest to see at x=0, y seems to be about 1 at its max, and so too when y=0 x is about 1.

Depends how much experience you have in R. Try following this example. Wolfram alpha might be better if easy is all you want.

In which R package De Jong function is available ?

Edited by Dhamnekar Win,odd
removed reply from the quote
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49 minutes ago, Dhamnekar Win,odd said:

Taylor series expansion of the function (r2er2;r;1;2)=e10.73576r2   From this expansion, what is the conclusion we can draw out?   

 

I understood your all calculations but didn't  understand your secondlast sentence 'the first order correction is - ve" .

 

Would you explain your answer a little bit more?

Sorry. I meant the next-to-zeroth-order correction. If you Taylor-series expand you immediately see it's a maximum, because there0s no 1st-order correction and the 2nd-order correction is negative. That's what I mean, and sorry for the confusion.

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On 4/13/2022 at 3:43 PM, Dhamnekar Win,odd said:

In which R package De Jong function is available ?

You don't need a package if you can specify the function yourself, which you did in the OP. Check out the link i embedded above - you just need to respecify the function.

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1 hour ago, Prometheus said:

You don't need a package if you can specify the function yourself, which you did in the OP. Check out the link i embedded above - you just need to respecify the function.

> fdejong <- function(x,y){return(x^2 +y^2)* exp(-(x^2+y^2))}
> x<- seq(-1,1 , length=2)
> y <- x
> z <- outer (x,y, fdejong)
> require(lattice)
Loading required package: lattice
> wireframe(z, drape=T, col.regions=rainbow(100))

 

image.png.f4e153e2da3b2a23961f88652c09717e.png

 

What is this? I don't get it.😕🤔😧

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On 4/15/2022 at 1:29 PM, Dhamnekar Win,odd said:

What is this? I don't get it.😕🤔😧

Learning to code is learning to troubleshoot. I suggest you go through each line of code separately and see if they're doing what you think they should do. I don't use r, but i can see 2 places this snippet of code fails.

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