The integral of the taylor expansion of any expression

[quietrain]

ok .lets say the expression we have is e^x

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 e^x

becos it is said that integral of e^x = e^x

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

thanks a lot!

 

[tiny-tim]

hi quietrain!

becos it is said that integral of e^x = e^x

yeees … but when we say "this is the integral of that", it's always understood that one mentally adds the words "plus a constant"

 

[quietrain]

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

 

[HallsofIvy]

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.

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

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

And, yes, integration of a power series, term by term, inside its radius of convergence (which for $e^x$, 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 $e^x$ is $e^x$" is simply wrong." What is said, or should be said, is "the integral of $e^x$ is $e^x+ C$ where C can be any number.")

 

[quietrain]

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.

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

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

And, yes, integration of a power series, term by term, inside its radius of convergence (which for $e^x$, 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 $e^x$ is $e^x$" is simply wrong." What is said, or should be said, is "the integral of $e^x$ is $e^x+ C$ where C can be any number.")

ah i see thanks a lot!