Taylor series operator is multiplicative - Revision history

Vipul at 22:04, 22 December 2012

2012-12-22T22:04:51Z

← Older revision		Revision as of 22:04, 22 December 2012
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	==Proof==		==Proof==

			<center>{{#widget:YouTube\|id=dyClhbYakPI}}</center>

	'''Given''': <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of both, and both <math>f</math> and <math>g</math> are infinitely differentiable at <math>x_0</math>.		'''Given''': <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of both, and both <math>f</math> and <math>g</math> are infinitely differentiable at <math>x_0</math>.

Vipul: /* Proof */

2012-07-04T16:07:13Z

Proof

← Older revision		Revision as of 16:07, 4 July 2012
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	We now want to multiply these Taylor series. To do this, we need to determine the coefficient of <math>(x - x_0)^n</math> in the product.		We now want to multiply these Taylor series. To do this, we need to determine the coefficient of <math>(x - x_0)^n</math> in the product.

	<math>(x - x_0)^n</math> can arise in the product by picking a factor <math>(x - x_0)^k</math> from the Taylor series of <math>f</math> and a factor <math>(x - x_0)^{n-k}</math> from the Taylor series of <math>g</math>. The coefficient arising from such a product is <math>\frac{f^{(k)}(x_0)g^{(n-k)}(x_0)}{k!(n - k)!}</math>. The overall coefficient is thus:		<math>(x - x_0)^n</math> can arise in the product by picking a factor <math>(x - x_0)^k</math> from the Taylor series of <math>g</math> and a factor <math>(x - x_0)^{n-k}</math> from the Taylor series of <math>f</math>. The coefficient arising from such a product is <math>\frac{f^{(n-k)}(x_0)g^{(k)}(x_0)}{(n - k)!k!}</math>. The overall coefficient is thus:

	<math>\mbox{Coefficient of } (x - x_0)^n = \sum_{k=0}^n \frac{f^{(k)}(x_0)g^{(n-k)}(x_0)}{k!(n - k)!}</math>		<math>\mbox{Coefficient of } (x - x_0)^n = \sum_{k=0}^n \frac{f^{(n-k)}(x_0)g^{(k)}(x_0)}{(n - k)!k!}</math>

	Multiply and divide the right side by <matH>n!</math> to get:		Multiply and divide the right side by <matH>n!</math> to get:

	<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n- k)!} f^{(k)}(x_0)g^{(n-k)}(x_0)</math>		<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n- k)!} f^{(n-k)}(x_0)g^{(k)}(x_0)</math>

	We have that <math>\frac{n!}{k!(n- k)!} = \binom{n}{k}</math>, and we get:		We have that <math>\frac{n!}{k!(n- k)!} = \binom{n}{k}</math>, and we get:

	<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} f^{(k)}(x_0)g^{(n-k)}(x_0)</math>		<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x_0)g^{(k)}(x_0)</math>

	Now, using <math>(\dagger)</math>, we get that:		Now, using <math>(\dagger)</math>, we get that:

Vipul at 15:52, 30 June 2012

2012-06-30T15:52:21Z

← Older revision		Revision as of 15:52, 30 June 2012
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	Suppose <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of both, and both <math>f</math> and <math>g</math> are infinitely differentiable at <math>x_0</math>. Then, the [[pointwise product of functions]] <math>fg</math> is also infinitely differentiable at <math>x_0</math>. Further, the [[fact about::Taylor series]] of <math>fg</math> at <math>x_0</math> is the product of the Taylor series of <math>f</math> at <math>x_0</math> and the Taylor series of <math>g</math> at <math>x_0</math>.		Suppose <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of both, and both <math>f</math> and <math>g</math> are infinitely differentiable at <math>x_0</math>. Then, the [[pointwise product of functions]] <math>fg</math> is also infinitely differentiable at <math>x_0</math>. Further, the [[fact about::Taylor series]] of <math>fg</math> at <math>x_0</math> is the product of the Taylor series of <math>f</math> at <math>x_0</math> and the Taylor series of <math>g</math> at <math>x_0</math>.

			==Related facts==

			* [[Taylor series operator is linear]]
			* [[Taylor series operator commutes with differentiation]]

	==Facts used==		==Facts used==

Vipul: Created page with "==Statement== Suppose $f$ and $g$ are functions defined on subsets of the reals such that $x_0$ is a point in the interior of the domain of b..."

2012-06-30T15:51:23Z

Created page with "==Statement== Suppose <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of b..."

New page

==Statement==

Suppose <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of both, and both <math>f</math> and <math>g</math> are infinitely differentiable at <math>x_0</math>. Then, the [[pointwise product of functions]] <math>fg</math> is also infinitely differentiable at <math>x_0</math>. Further, the [[fact about::Taylor series]] of <math>fg</math> at <math>x_0</math> is the product of the Taylor series of <math>f</math> at <math>x_0</math> and the Taylor series of <math>g</math> at <math>x_0</math>.

==Facts used==

# [[uses::Product rule for higher derivatives]]: This states that <math>\! (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)}</math> [[concept of equality conditional to existence of one side|wherever the right side makes sense]].

==Proof==

'''Given''': <math>f</math> and <math>g</math> are functions defined on subsets of the reals such that <math>x_0</math> is a point in the interior of the domain of both, and both <math>f</math> and <math>g</math> are infinitely differentiable at <math>x_0</math>.

'''To prove''': The pointwise product <math>fg</math> is infinitely differentiable at <math>x_0</math> and its Taylor series at <math>x_0</math> is the product of the Taylor series of <math>f</math> at <math>x_0</math> and the Taylor series of <math>g</math> at <math>x_0</math>.

'''Proof''': By Fact (1), we have:

<math>\! (fg)^{(n)}(x_0) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x_0)g^{(k)}(x_0) \qquad (\dagger)</math>

with equality holding wherever the right side makes sense. Since the right side always makes sense, so does the left side, so <math>fg</math> is infinitely differentiable.

We recall the expressions for the Taylor series:

<math>\mbox{Taylor series for } f \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k</math>

<math>\mbox{Taylor series for } g \mbox{ at } x_0 = \sum_{k=0}^\infty \frac{g^{(k)}(x_0)}{k!}(x - x_0)^k</math>

We now want to multiply these Taylor series. To do this, we need to determine the coefficient of <math>(x - x_0)^n</math> in the product.

<math>(x - x_0)^n</math> can arise in the product by picking a factor <math>(x - x_0)^k</math> from the Taylor series of <math>f</math> and a factor <math>(x - x_0)^{n-k}</math> from the Taylor series of <math>g</math>. The coefficient arising from such a product is <math>\frac{f^{(k)}(x_0)g^{(n-k)}(x_0)}{k!(n - k)!}</math>. The overall coefficient is thus:

<math>\mbox{Coefficient of } (x - x_0)^n = \sum_{k=0}^n \frac{f^{(k)}(x_0)g^{(n-k)}(x_0)}{k!(n - k)!}</math>

Multiply and divide the right side by <matH>n!</math> to get:

<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n- k)!} f^{(k)}(x_0)g^{(n-k)}(x_0)</math>

We have that <math>\frac{n!}{k!(n- k)!} = \binom{n}{k}</math>, and we get:

<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} f^{(k)}(x_0)g^{(n-k)}(x_0)</math>

Now, using <math>(\dagger)</math>, we get that:

<math>\mbox{Coefficient of } (x - x_0)^n = \frac{1}{n!} (fg)^{(n)}(x_0)</math>

So the product of the Taylor series for <math>f</math> and <math>g</math> is:

<math>\sum_{n=0}^\infty \frac{(fg)^{(n)}(x_0)}{n!}(x - x_0)^n</math>

Since <math>n</math> is a dummy variable, it can be replaced by the dummy variable <math>k</math>, giving:

<math>\sum_{k=0}^\infty \frac{(fg)^{(k)}(x_0)}{k!}(x - x_0)^k</math>

This is precisely the Taylor series for <math>fg</math>, and we have completed the proof.

Taylor series operator is multiplicative - Revision history

Vipul at 22:04, 22 December 2012

Vipul: /* Proof */

Vipul at 15:52, 30 June 2012

Vipul: Created page with "==Statement== Suppose f and g are functions defined on subsets of the reals such that x_0 is a point in the interior of the domain of b..."

Vipul: Created page with "==Statement== Suppose $f$ and $g$ are functions defined on subsets of the reals such that $x_0$ is a point in the interior of the domain of b..."