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Derivative laws question numerical analysis

  • Thread starter nhrock3
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  • #1
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23ra8u0.jpg


i know the law (fg)'=f'g+fg'
but here there is epsilon
and it turns to someother sign i dont know what it says.

i cant understand the transition
?
 

Answers and Replies

  • #2
33,506
5,191
23ra8u0.jpg


i know the law (fg)'=f'g+fg'
but here there is epsilon
and it turns to someother sign i dont know what it says.

i cant understand the transition
?
There's no epsilon in what you posted. Are you talking about [itex]\prod[/itex]? Note that this is upper-case pi, which is different from the constant [itex]\pi[/itex], lower-case pi.

Upper-case pi is similar to upper-case sigma [itex]\sigma[/itex], except that it is used for products rather than sums. here's an example using the product notation.
[tex]\prod_{m = 1}^5 m = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! = 120[/tex]
 
  • #3
415
0
ok but why the derivative change epsilon to pi?
 
  • #4
33,506
5,191
What are you talking about? There is no epsilon.
 
  • #5
Cyosis
Homework Helper
1,495
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With epsilon do you mean the summation sign? If so that one is called capital sigma.

Taking the derivative of a function that is the product of two other functions yields two terms as you know, (fg)'=f'g+g'f. In our case we don't have a function consisting of the product of two functions but of a product of m functions.

[tex]
\prod_{j=0}^m f_j=f_0f_1f_2 \cdot ... \cdot f_m
[/tex]

Now taking the derivative of that product yields:

[tex]
\frac{d}{dx}\prod_{j=0}^m=\frac{d}{dx}(f_0f_1f_2 \cdot ... \cdot f_m)=f'_0f_1f_2 \cdot ... \cdot f_m+f_0f_1'f_2 \cdot ... \cdot f_m+...+f_0f_1f_2 \cdot ... \cdot f_m'
[/tex]

You can see that we get m terms that are added together and every term is the product of m terms. We can write this compactly as:

[tex]
\sum_{i=0}^m \frac{df_i}{dx}\prod_{\substack{ j=0 \\ j \neq i}}^m f_j
[/tex].
 
Last edited:

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