Welcome to our community

Be a part of something great, join today!

Taylor series

Fermat

Active member
Nov 3, 2013
188
Use taylor series to show that the infinite series from n=0 of

$$\frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})=e^{-(q+s)^2}$$
 
Last edited by a moderator:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Use taylor series to show that the infinite series from n=0 of

$$\frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})=e^{-(q+s)^2}$$
Try to expand \(\displaystyle f(x+h)=e^{-(x+h)^2}\)?
 

Fermat

Active member
Nov 3, 2013
188

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888

Fermat

Active member
Nov 3, 2013
188
Try to rewrite LHS as \(\displaystyle \frac{h^n}{n!}f^{(n)}(x)\)?
If I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?

Then, $\frac{(-2sq)^n}{n!}=e^{-2sq}$ Still missin a factor of $e^{-s^2}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
If I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?
No... but you would have \(\displaystyle \frac{d^n}{dq^n}(e^{-q^2})\), which I believe is what you are looking for...
You do not actually have to differentiate.
You only need to recognize and match an nth derivative.
 

Fermat

Active member
Nov 3, 2013
188
No... but you would have \(\displaystyle \frac{d^n}{dq^n}(e^{-q^2})\), which I believe is what you are looking for...
You do not actually have to differentiate.
You only need to recognize and match an nth derivative.
No, \(\displaystyle \frac{d^n}{dq^n}(e^{-q^2})\) is part of the LHS which is where I am starting at and then trying to get $e^{-(q+s)^2}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?

Then, $\frac{(-2sq)^n}{n!}=e^{-2sq}$ Still missin a factor of $e^{-s^2}$
I find the same thing. I get:

\(\displaystyle \sum_{n=0}^{\infty}\left[\frac{s^n}{n!}\cdot\frac{d^n}{dq^n}\left(e^{-q^2} \right) \right]=e^{-q^2-2qs}\)
 

Fermat

Active member
Nov 3, 2013
188
I find the same thing. I get:

\(\displaystyle \sum_{n=0}^{\infty}\left[\frac{s^n}{n!}\cdot\frac{d^n}{dq^n}\left(e^{-q^2} \right) \right]=e^{-q^2-2qs}\)
Can Opalg (or someone of similar calibre confirm this)? No disrespect Mark but Ihave it on good authority that the result is correct.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Can Opalg (or someone of similar calibre confirm this)? No disrespect Mark but Ihave it on good authority that the result is correct.
No disrespect taken. :D

In fact, we have both made an error in not applying the product rule for the $n$th derivative on the left side.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
If I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?
If I differentiate $e^{-q^2}$ once then I get $-2qe^{-q^2}$. If I differentiate it a second time (using the product rule) then I get $(-2+4q^2)e^{-q^2}$. The third derivative then comes out as $8qe^{-q^2} -2q(-2+4q^2)e^{-q^2} = (12q - 8q^3)e^{-q^2}.$ It doesn't look like $(-2q)^{n}e^{-q^2}$ except when $n=1$.

You do not actually have to differentiate.
You only need to recognize and match an nth derivative.
Take this advice seriously! I think that ILS has the right approach.
 

Fermat

Active member
Nov 3, 2013
188
Its not a 'show that' question so I cannot start the RHS (the answer). That's why I need to start with the LHS. I didn't make that clear- sorry.
 

Fermat

Active member
Nov 3, 2013
188
I really don't know what you mean by 'match derivative'.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
I really don't know what you mean by 'match derivative'.
You are supposed to sum a series of terms.
The trick to get it summed, is to match it with a Taylor series.
If you can find a function $f$, such that each of your terms matches the corresponding Taylor term, you can write the infine sequence as an expression with that $f$.

Each Taylor term is of the form \(\displaystyle \frac{h^n}{n!}f^{(n)}(x)\), which belongs to the expansion of \(\displaystyle f(x+h)\).
So you need to find an $f$, such that \(\displaystyle f^{(n)}(x)\) matches part of your term.
As it is, you happen to have an nth derivative (which would be the reason to try to match the series with a Taylor expansion in the first place).
So let's try to match that nth derivative with \(\displaystyle f^{(n)}(x)\).
Your nth derivative is \(\displaystyle \frac{d^n}{dq^n}(e^{-q^2})\).
So let's try \(\displaystyle f(x) = e^{-x^2}\).

Consequently, $s$ will have to match $h$.
And from there you have a complete match.
\begin{cases}
h &=& s \\
x &=& q \\
f(x) &=& e^{-x^2}
\end{cases}
It follows that the series sums to \(\displaystyle f(x+h) = e^{-(q+s)^2}\).
 

Fermat

Active member
Nov 3, 2013
188
You are supposed to sum a series of terms.
The trick to get it summed, is to match it with a Taylor series.
If you can find a function $f$, such that each of your terms matches the corresponding Taylor term, you can write the infine sequence as an expression with that $f$.

Each Taylor term is of the form \(\displaystyle \frac{h^n}{n!}f^{(n)}(x)\), which belongs to the expansion of \(\displaystyle f(x+h)\).
So you need to find an $f$, such that \(\displaystyle f^{(n)}(x)\) matches part of your term.
As it is, you happen to have an nth derivative (which would be the reason to try to match the series with a Taylor expansion in the first place).
So let's try to match that nth derivative with \(\displaystyle f^{(n)}(x)\).
Your nth derivative is \(\displaystyle \frac{d^n}{dq^n}(e^{-q^2})\).
So let's try \(\displaystyle f(x) = e^{-x^2}\).

Consequently, $s$ will have to match $h$.
And from there you have a complete match.
\begin{cases}
h &=& s \\
x &=& q \\
f(x) &=& e^{-x^2}
\end{cases}
It follows that the series sums to \(\displaystyle f(x+h) = e^{-(q+s)^2}\).
I thought a taylor term was of the form $\frac{x^n}{n!}f^{n}(h)$ ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
I thought a taylor term was of the form $\frac{x^n}{n!}f^{n}(h)$ ?
There are many ways to write a Taylor series, distinguished by the choice of symbols (although yours would be unusual).
What is the definition of the Taylor series you are referring to?
 

Fermat

Active member
Nov 3, 2013
188
There are many ways to write a Taylor series, distinguished by the choice of symbols (although yours would be unusual).
What is the definition of the Taylor series you are referring to?
It is the definition on wikipedia. This is not solved by a simple change of symbols because in my definition the derivative is w.r.t x and then evaluated at h.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
It is the definition on wikipedia. This is not solved by a simple change of symbols because in my definition the derivative is w.r.t x and then evaluated at h.
As it happens, there is no definition on wikipedia combining $x$ and $h$ (although that is usual enough).
Perhaps you can write down the definition you're referring to?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
It is the definition on wikipedia. This is not solved by a simple change of symbols because in my definition the derivative is w.r.t x and then evaluated at h.
You aren't given $f(x)$ in this problem. You have $s,q$ as your variables. I like Serena was just writing those in terms of the standard variables $x,h$ (or $x,a$ as I am used to) for Taylor Series to make the connection clear.

The main insight needed to solve this problem is definitely recognizing that this is a Taylor Series in disguise. Then you need to decide what the function, $f$, is and where the series is centered around. It all fits together nicely and ILS has nicely posted all the pieces already.
 

Fermat

Active member
Nov 3, 2013
188
As it happens, there is no definition on wikipedia combining $x$ and $h$ (although that is usual enough).
Perhaps you can write down the definition you're referring to?
$f(x)=f(h)+xf'(h)+\frac{x^2}{2}f''(h)+......$

Jameson, please don't try to shut down a mathematical discussion. Thanks
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
$f(x)=f(h)+xf'(h)+\frac{x^2}{2}f''(h)+......$
This is not a proper Taylor expansion.
Perhaps you can check the wiki page?
When we have a Taylor expansion that you are comfortable with (and that is correct), we can match the symbols.
 

Fermat

Active member
Nov 3, 2013
188
This is not a proper Taylor expansion.
Perhaps you can check the wiki page?
When we have a Taylor expansion that you are comfortable with (and that is correct), we can match the symbols.
Lets take $f(x)=e^x$. I will find macluarin series i.e. h=0

$f^{n}(0)=1$ for all n. So by my definition we have

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$ which is correct.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
I have no idea what you mean. I was commenting on what ILS wrote. The choice of what letters you use in the Taylor Series formula isn't important as long it's clear what it means. You have a function of some variable, not necessary $x$, that is centered at some point.

Please make an effort be more respectful in your interaction with others at MHB. If you didn't find my post helpful then please let me know in a constructive way. You have a lot of people here working with you and trying to help, so let's try to be on the same team. If this isn't clear or you want to discuss it further we can talk privately.

I'll let ILS and others continue from here.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Lets take $f(x)=e^x$. I will find macluarin series i.e. h=0

$f^{n}(0)=1$ for all n. So by my definition we have

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$ which is correct.
You are a bit hard headed aren't you?
I take it you did not check the wiki page?

The 2 most common forms of the Taylor expansion are:
\begin{array}{}
f(x)&=&f(a)+\frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + ... \\
f(x+h)&=&f(x)+\frac{f'(x)}{1!}h + \frac{f''(x)}{2!}h^2 + ...
\end{array}
 

Fermat

Active member
Nov 3, 2013
188
Thank you for your patience. I still do not see the second form (which I have never seen)on the wikipedia page, but I will take your word for it.

Jameson, your post seemed a not too subtle way of saying 'you've got an answer, now please go away'.