Welcome to our community

Be a part of something great, join today!

Taylor series

Jameson

Administrator
Staff member
Jan 26, 2012
4,055
I just realized that ILS gave a solution using a definition of a Taylor Series I hadn't seen before. The logic and result are clearly correct but I thought about trying it from the definition I am used to.

This might be wrong so I am hoping someone can check it and comment if necessary. The end result is correct but there is one spot where I am unsure, so read with scrutiny and skepticism. :)
---------------------------------
We are given \(\displaystyle \frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})\) and want to try to match it to this form: \(\displaystyle \sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} (x-a)^n\)

There is one term that they share which represents the nth derivative at $a$. These must match up so $f(a)=e^{-q^2}$. There are two things we need to know however, $f(x)$ and $a$. There are perhaps various combinations to try but the one that comes to mind first is $f(x)=e^{-x^2}$ and $a=q$.

Looking at $s^n$ it seems that this should match up with $(x-a)^n$ so we find that $x-a=s$ and since $a=q$ that is equivalent to $x-q=s$. Finally that gives us that $x=s+q$. If we plug that into $f(x)=e^{-x^2}$ then we have our result.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,322
I just realized that ILS gave a solution using a definition of a Taylor Series I hadn't seen before. The logic and result are clearly correct but I thought about trying it from the definition I am used to.

This might be wrong so I am hoping someone can check it and comment if necessary. The end result is correct but there is one spot where I am unsure, so read with scrutiny and skepticism.
---------------------------------
We are given \(\displaystyle \frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})\) and want to try to match it to this form: \(\displaystyle \sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} (x-a)^n\)

There is one term that they share which represents the nth derivative at $a$. These must match up so $f(a)=e^{-q^2}$. There are two things we need to know however, $f(x)$ and $a$. There are perhaps various combinations to try but the one that comes to mind first is $f(x)=e^{-x^2}$ and $a=q$.

Looking at $s^n$ it seems that this should match up with $(x-a)^n$ so we find that $x-a=s$ and since $a=q$ that is equivalent to $x-q=s$. Finally that gives us that $x=s+q$. If we plug that into $f(x)=e^{-x^2}$ then we have our result.
Looks good. :)
What are you unsure about?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,055
Looks good. :)
What are you unsure about?
I wasn't sure about $f(a)=e^{-q^2}$, although it is logical and fits. Pretty much anything that involves making conclusions about functions in this way makes me nervous. Not being able to rigorously prove a claim I'm using makes me feel like I don't have a right to use the claim, even if it's correct.

Anyway, glad to read that my solution works. I felt like my hints were not very effective and I don't really know where Fermat is feeling stuck so in this case a full solution seemed best.

@Fermat - what do you think about the way I did it? Does it make more sense when using the definition you are used to?