The integral of the taylor expansion of any expression

  • Thread starter quietrain
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Main Question or Discussion Point

ok .lets say the expression we have is ex

the taylor expansion becomes 1+x+x2/2+...

integrating becomes x+x2/2+x3/6+...+c

so how do we know that c = 1? for it to become back to ex

becos it is said that integral of ex = ex

do we just let x be 0 to find c = 1? does it work for all expressions?

thanks a lot!
 

Answers and Replies

  • #2
tiny-tim
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hi quietrain! :smile:
becos it is said that integral of ex = ex
yeees … but when we say "this is the integral of that", it's always understood that one mentally adds the words "plus a constant" :wink:
 
  • #3
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hi tim, so how do we know that the constant is 1?

is it just by letting x = 0?

does it work for all expressions? because i was told it won't work for sinx
 
  • #4
HallsofIvy
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We don't! I have no idea where you got the impression that the constant of integration had to be 1- it doesn't unless you are given some additional information about the integral. Again, the (indefinite) integral or "anti-derivative" of any function is determined only up to an additive constant.

[tex]\int e^x dx= e^x+ C[/tex].
In terms of the Taylor series
[tex]\int (1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot)dx= C+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot[/tex]
where C can be any number.

Since you know that [itex]e^x= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot[/itex] for that to "come back to [itex]e^x[/itex]", obviously, C would have to be 1 but, again, [itex]\int e^x dx= e^x+ C[/itex], not just [itex]e^x[/itex].

And, yes, integration of a power series, term by term, inside its radius of convergence (which for [itex]e^x[/itex], sin(x), cos(x), etc. is "all x") always works.

(I would disagree with tiny time that "one mentally adds the words "plus a constant". One should add it explicitely! "It is said that the integral of [itex]e^x[/itex] is [itex]e^x[/itex]" is simply wrong." What is said, or should be said, is "the integral of [itex]e^x[/itex] is [itex]e^x+ C[/itex] where C can be any number.")
 
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  • #5
654
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We don't! I have no idea where you got the impression that the constant of integration had to be 1- it doesn't unless you are given some additional information about the integral. Again, the (indefinite) integral or "anti-derivative" of any function is determined only up to an additive constant.

[tex]\int e^x dx= e^x+ C[/tex].
In terms of the Taylor series
[tex]\int (1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot)dx= C+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot[/tex]
where C can be any number.

Since you know that [itex]e^x= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot[/itex] for that to "come back to [itex]e^x[/itex]", obviously, C would have to be 1 but, again, [itex]\int e^x dx= e^x+ C[/itex], not just [itex]e^x[/itex].

And, yes, integration of a power series, term by term, inside its radius of convergence (which for [itex]e^x[/itex], sin(x), cos(x), etc. is "all x") always works.

(I would disagree with tiny time that "one mentally adds the words "plus a constant". One should add it explicitely! "It is said that the integral of [itex]e^x[/itex] is [itex]e^x[/itex]" is simply wrong." What is said, or should be said, is "the integral of [itex]e^x[/itex] is [itex]e^x+ C[/itex] where C can be any number.")
ah i see thanks a lot!
 

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