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- Mar 5, 2012

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Try to expand \(\displaystyle f(x+h)=e^{-(x+h)^2}\)?Use taylor series to show that the infinite series from n=0 of

$$\frac{s^n}{n!}\frac{d^n}{dq^n}(e^{-q^2})=e^{-(q+s)^2}$$

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I would prefer to get from LHS to RHSTry to expand \(\displaystyle f(x+h)=e^{-(x+h)^2}\)?

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- Mar 5, 2012

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Try to rewrite LHS as \(\displaystyle \frac{h^n}{n!}f^{(n)}(x)\)?I would prefer to get from LHS to RHS

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

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

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No... but youIf I differentiate $e^{-q^2}$ n times then I get $(-2q)^{n}e^{-q^2}$ ?

You do not actually have to differentiate.

You only need to recognize and match an nth derivative.

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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 $e^{-(q+s)^2}$No... but youwouldhave \(\displaystyle \frac{d^n}{dq^n}(e^{-q^2})\), 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.

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I find the same thing. I get: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 $e^{-s^2}$

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

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Can Opalg (or someone of similar calibre confirm this)? No disrespect Mark but Ihave it on good authority that the result is correct.I find the same thing. I get:

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

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No disrespect taken.Can Opalg (or someone of similar calibre confirm this)? No disrespect Mark but Ihave it on good authority that the result is correct.

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

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- #11

- Feb 7, 2012

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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 $n=1$.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.You do not actually have to differentiate.

You only need to recognize and match an nth derivative.

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- #14

- Mar 5, 2012

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You are supposed to sum a series of terms.I really don't know what you mean by 'match derivative'.

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 \(\displaystyle f(x+h) = e^{-(q+s)^2}\).

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I thought a taylor term was of the form $\frac{x^n}{n!}f^{n}(h)$ ?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 \(\displaystyle f(x+h) = e^{-(q+s)^2}\).

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- #16

- Mar 5, 2012

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There are many ways to write a Taylor series, distinguished by the choice of symbols (although yours would be unusual).I thought a taylor term was of the form $\frac{x^n}{n!}f^{n}(h)$ ?

What is the definition of the Taylor series you are referring to?

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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.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?

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As it happens, there is no definition on wikipedia combining $x$ and $h$ (although that is usual enough).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.

Perhaps you can write down the definition you're referring to?

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- #19

- Jan 26, 2012

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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.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.

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.

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$f(x)=f(h)+xf'(h)+\frac{x^2}{2}f''(h)+......$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, please don't try to shut down a mathematical discussion. Thanks

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- #21

- Mar 5, 2012

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This is not a proper Taylor expansion.$f(x)=f(h)+xf'(h)+\frac{x^2}{2}f''(h)+......$

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.

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Lets take $f(x)=e^x$. I will find macluarin series i.e. h=0This 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.

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

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

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- #23

- Jan 26, 2012

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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.

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- #24

- Mar 5, 2012

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You are a bit hard headed aren't you?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.

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}

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