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How can a series be equal to this integral?

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opus

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1. The problem statement, all variables and given/known data
Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##

2. Relevant equations


3. The attempt at a solution
So I didn't really know what I was doing but I did end up with the correct solution.
What I did was to find a Taylor Series for the integrand, this turned out to be ##\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}##

Now I have ##\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx##
After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have ##\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx##
This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after ##\sum## in brackets?

Eventually, I end up evaluating and I get ##\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}##

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.
 

opus

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Also, apologies for the hideous notation. I don't really know what goes where, and this type of problem seems very "choppy" for me right now. So everything is pretty far from nice to look at :cry:
 

opus

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Also again....
Here's an attachment of my work in case there were specific things that need to be addressed.
 

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Delta2

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...
Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.
I don't understand why you think that a series can never equal an integral. A series in the case it converges, it is equal to a real number. A definite integral also is equal to a real number... Why cant we have that two real numbers are equal???
 

opus

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I don't understand why you think that a series can never equal an integral. A series in the case it converges, it is equal to a real number. A definite integral also is equal to a real number... Why cant we have that two real numbers are equal???
Here’s what I mean that goes along with the lat part of my first post that is part of my confusion. Not the “bargraph” like piece represents a series.
 

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Delta2

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Ok I think I understand now what you trying to say. Still it isn't quite true for two reasons:
1) The interval of integration can be anything when we take our integral , it is not necessarily have to be from 0 to +infinite. And indeed in our case the interval of integration is from 0 to 1.
2) I can define a step-wise function such that ##f(x)=a_n , x\in[n,n+1)## and the integral of that function from 0 to +infinite ##\int_0^{+\infty}f(x)dx =\sum a_n##would be equal to the series .
 
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Don't overthink this.
Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##
Evaluate the integral (it's an easy substitution) and you get a number for which you can find a Taylor series. Voila!
 

vela

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Eventually, I end up evaluating and I get
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}$$
Use the fact that ##6n+3 = 3(2n+1)##. You should be able to identify the resulting series.
 

opus

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Ok I think I understand now what you trying to say. Still it isn't quite true for two reasons:
1) The interval of integration can be anything when we take our integral , it is not necessarily have to be from 0 to +infinite. And indeed in our case the interval of integration is from 0 to 1.
2) I can define a step-wise function such that ##f(x)=a_n , x\in[n,n+1)## and the integral of that function from 0 to +infinite ##\int_0^{+\infty}f(x)dx =\sum a_n##would be equal to the series .
Ok yes that answered my question, thank you.
 

opus

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Don't overthink this.

Evaluate the integral (it's an easy substitution) and you get a number for which you can find a Taylor series. Voila!
So let me ask this. Say the integral has bounds [0,2]. When finding a Taylor series, is the center of the Taylor series relevant? I want to say it definitely would be, but I don't have a good argument for this.

Use the fact that ##6n+3 = 3(2n+1)##. You should be able to identify the resulting series.
Thanks!
 

Orodruin

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The series you have is not the Riemann sum of the integral. It is just a number of different contributions to it, integration is linear.

Compare with how you would integrate 1+x ...
 

Delta2

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From what I understood, his point was that: how can a series be equal to the integral of any function from 0 to infinity, since the area under a random curve that extends from 0 to infinity, cannot equal to a series because a series is the area that is just a sum of bars and it just approximates the area under that random curve, that is it approximates our integral (in best case scenario). I believe I answered this point at post #6.
 
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So let me ask this. Say the integral has bounds [0,2]. When finding a Taylor series, is the center of the Taylor series relevant? I want to say it definitely would be, but I don't have a good argument for this.
I don't think the center is relevant here. For example, the Maclaurin series for ##e^x## is ##1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dots##, so it's straightforward to find, say, ##e^1##.

From what I understood, his point was that: how can a series be equal to the integral of any function from 0 to infinity
Not relevant here. The integral in question is over the interval [0, 1].
Delta2 said:
, since the area under a random curve that extends from 0 to infinity, cannot equal to a series because a series is the area that is just a sum of bars and it just approximates the area under that random curve, that is it approximates our integral (in best case scenario). I believe I answered this point at post #6.
 

Ray Vickson

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1. The problem statement, all variables and given/known data
Determine the series that is equal to the integral ##\int_0^1 x^2\cos(x^3)dx##

2. Relevant equations


3. The attempt at a solution
So I didn't really know what I was doing but I did end up with the correct solution.
What I did was to find a Taylor Series for the integrand, this turned out to be ##\sum_{n=0}^\infty (-1)n\frac{x^{6n+2}}{(2n)!}##

Now I have ##\int_0^1\sum_{n=0}^\infty (-1)^n\frac{x^{6n+2}}{(2n)!}dx##
After some research, I found that for power series (for which a Taylor series is) you can put the integral inside the sum (if this is incorrect, please let me know).

Now I have ##\sum_{n=0}^\infty (-1)^n\frac{1}{(2n)!}\int_0^1 x^{6n+2}dx##
This looks like a mess, but not sure how I should have grouped this the proper way. That is, should I have put everything after ##\sum## in brackets?

Eventually, I end up evaluating and I get ##\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!(6n+3)}##

Now normally, a series would never equal an integral to my understanding because series are for positive integers. But this is a power series, which involves x. So is it true that this is why the series can equal the integral in this case? Because x can take on ANY real positive values now?

I guess thinking about a series like a "bar graph" kind of thing isn't helping me much in trying to understand this.
If your problem was to evaluate the indefinite integral ##\int x^2 \cos(x^3) \, dx## then, indeed, you would get an infinite series involving ##x##. However, you have a definite integral in which ##x = 1##, so that is why you don't see an ##x## in your answer. No rocket science involved here!
 
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If your problem was to evaluate the indefinite integral ##\int x^2 \cos(x^3) \, dx## then, indeed, you would get an infinite series involving ##x##.
This is true, but somewhat misleading. Finding an antiderivative is very straightforward, and results in a simple function. As an additional step, you can rewrite that function as an infinite series.
Ray Vickson said:
However, you have a definite integral in which ##x = 1##, so that is why you don't see an ##x## in your answer. No rocket science involved here!
 

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