Fermat said:Use taylor series to show that the infinite series from n=0 of
$$\frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})=e^{-(q+s)^2}$$
I like Serena said:Try to expand \(\displaystyle f(x+h)=e^{-(x+h)^2}\)?
Fermat said:I would prefer to get from LHS to RHS
I like Serena said:Try to rewrite LHS as \(\displaystyle \frac{h^n}{n!}f^{(n)}(x)\)?
Fermat said:If I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?
I like Serena said: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 said: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 $e^{-s^2}$
MarkFL said: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 said:Can Opalg (or someone of similar calibre confirm this)? No disrespect Mark but Ihave it on good authority that the result is correct.
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 $n=1$.Fermat said:If I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?
Take this advice seriously! I think that ILS has the right approach.I like Serena said:You do not actually have to differentiate.
You only need to recognize and match an nth derivative.
Fermat said:I really don't know what you mean by 'match derivative'.
I like Serena said: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 said:I thought a taylor term was of the form $\frac{x^n}{n!}f^{n}(h)$ ?
I like Serena said: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 said: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.
Fermat said: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.
I like Serena said: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?
Fermat said:$f(x)=f(h)+xf'(h)+\frac{x^2}{2}f''(h)+...$
I like Serena said: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 said: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 said: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.
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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.
I like Serena said:Looks good. :)
What are you unsure about?
The Taylor series expansion of $e^{-(q+s)^2}$ is an infinite sum of terms that approximates the function $e^{-(q+s)^2}$ at a given point. It is given by the formula:
$e^{-(q+s)^2} = \sum_{n=0}^{\infty} \frac{(-1)^n (q+s)^{2n}}{n!}$
The Taylor series expansion of $e^{-(q+s)^2}$ is useful because it allows us to approximate the value of $e^{-(q+s)^2}$ at any point by using a finite number of terms. This can be helpful in situations where it is difficult or impossible to evaluate the function directly, or when we need a more precise approximation than what a calculator can provide.
The interval of convergence for the Taylor series expansion of $e^{-(q+s)^2}$ is the set of all values of $(q+s)$ for which the series converges. In this case, the interval of convergence is $(-\infty, \infty)$, meaning that the series will converge for all real values of $(q+s)$.
The Taylor series expansion of $e^{-(q+s)^2}$ is closely related to the Gaussian function, also known as the bell curve. In fact, the Gaussian function can be defined as $e^{-(x^2)}$. Therefore, the Taylor series expansion of $e^{-(q+s)^2}$ can be thought of as a generalization of the Gaussian function to a function of two variables, $(q+s)$.
Technically, yes. However, since the Taylor series expansion is an infinite sum, it would require an infinite number of terms to calculate the exact value at any point. In practice, we can use a finite number of terms to get a close approximation, but the more terms we use, the more accurate our approximation will be.