Can someone help with this demonstration?

[marscf]

this is written in portuguese but basically I can't solve the first one i) where they ask to show that f(x) (given above) it's the same as the summation given bellow.

Please if someone could help!

 

[Deleted member 4993]

View attachment 19885
this is written in portuguese but basically I can't solve the first one i) where they ask to show that f(x) (given above) it's the same as the summation given bellow.

Please if someone could help!

Do you know how to expand \(\displaystyle e^{-(x-1)^2}\) using McLaurin series?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

 

[marscf]

No I dont know, I didnt post any of my resolution because I really didnt come up to anything. Basically im stuck in the begining

 

[pka]

First even if the PO were in perfect English, I am not sure what is expected.
I noticed right off that \(f(x)=\dfrac{1}{e^{(x-1)^2}}\)
Let use \(\exp((x-1)^2)=e^{(x-1)^2}\) for economy of notation.
So \(\exp (x) =\displaystyle \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} \) Thus \(\exp (-(x-1)^2) =\displaystyle \sum\limits_{n = 0}^\infty {\dfrac{{{(-1)^n}}}{{n!}}}(x-1)^{2n} \)

 

[marscf]

First even if the PO were in perfect English, I am not sure what is expected.
I noticed right off that \(f(x)=\dfrac{1}{e^{(x-1)^2}}\)
Let use \(\exp((x-1)^2)=e^{(x-1)^2}\) for economy of notation.
So \(\exp (x) =\displaystyle \sum\limits_{n = 0}^\infty {\dfrac{{{x^n}}}{{n!}}} \) Thus \(\exp (-(x-1)^2) =\displaystyle \sum\limits_{n = 0}^\infty {\dfrac{{{(-1)^n}}}{{n!}}}(x-1)^{2n} \)

That helped! It would be even more useful if you could tell me what you based yourself on to conclude the last part of the problem (this: \(\exp (-(x-1)^2) =\displaystyle \sum\limits_{n = 0}^\infty {\dfrac{{{(-1)^n}}}{{n!}}}(x-1)^{2n} \) )

Thanks again!

 

[Dr.Peterson]

You haven't yet told us what you have learned that you can use. I think you said you haven't learn about Maclaurin series. What series do you know?

That's what pka used; see here: https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions

 

[marscf]

You haven't yet told us what you have learned that you can use. I think you said you haven't learn about Maclaurin series. What series do you know?

That's what pka used; see here: https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions

The truth is that theoretically I have learned about Maclaurin series, but I can't apply or identify them when solving problems.

 

[Dr.Peterson]

The truth is that theoretically I have learned about Maclaurin series, but I can't apply or identify them when solving problems.

In this case, you may be expected to have memorized the simplest cases in the list in my reference, e.g. e^x, ln(1+x), 1/(1+x), sin(x), cos(x).

Or, they may expect you to derive the series for e^x, using the formula for Maclaurin series, as this is the easiest possible case.

In any case, once you know that series, pka in post #4 showed how to use it to answer your problem. The key is to recognize that the exponent [MATH]x^2 - 2x + 1[/MATH] is a perfect square.