Difference between revisions of "Taylor series operator is multiplicative"

From Calculus
Jump to: navigation, search
(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...")
 
 
(2 intermediate revisions by the same user not shown)
Line 2: Line 2:
  
 
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==
Line 8: Line 13:
  
 
==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>.
Line 27: Line 34:
 
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:

Latest revision as of 22:04, 22 December 2012

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 both, and both f and g are infinitely differentiable at x_0. Then, the pointwise product of functions fg is also infinitely differentiable at x_0. Further, the Taylor series of fg at x_0 is the product of the Taylor series of f at x_0 and the Taylor series of g at x_0.

Related facts

Facts used

  1. Product rule for higher derivatives: This states that \! (fg)^{(n)} = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}g^{(k)} wherever the right side makes sense.

Proof

Given: 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 both, and both f and g are infinitely differentiable at x_0.

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

Proof: By Fact (1), we have:

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

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

We recall the expressions for the Taylor series:

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

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

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

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

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

Multiply and divide the right side by n! to get:

\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)

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

\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)

Now, using (\dagger), we get that:

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

So the product of the Taylor series for f and g is:

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

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

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

This is precisely the Taylor series for fg, and we have completed the proof.

Retrieved from "https://calculus.subwiki.org/w/index.php?title=Taylor_series_operator_is_multiplicative&oldid=2254"