Understanding the solution key to a problem which shows that the integral of a sum equals a given value.
$begingroup$
Suppose that the domain of convergence of the power series
$sum_{k=0}^{infty} c_{k}x^{k}$ contains the interval $(-r, r)$.
Define $$f(x) = sum_{k=0}^{infty} c_{k}x^{k} hspace{1cm} text{
> if } |x| < r. $$
Let $[a, b] subseteq (-r, r).$ Prove that
$$int_{a}^{b} f(x) mathop{dx} = sum_{k = 0}^{infty} frac{c_{k}}{k + 1}left(b^{k + 1} - a^{k + 1}right).$$
Here's the solution I have. It might be wrong because it's not official.
Recall Theorem $5$, which states that if a sequence of integrable functions ${f_{n} : [a, b] rightarrow mathbb{R}}$ converges uniformly to the function $f : [a, b] rightarrow mathbb{R}$, then the limit function is also integrable.
So,
$$int_{a}^{b} f(x) mathop{dx} = lim_{ntoinfty} int_{a}^{b} sum_{k = 0}^{n} c_{k}x^{k} = lim_{ntoinfty} sum_{k=0}^{n}int_{a}^{b} c_{k}x^{k} mathop{dx} = lim_{ntoinfty} sum_{k=0}^{n} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right) $$
$$= sum_{k=0}^{infty} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right), $$
which is what we wanted to show. (Switching integral/summation is justified by Fubini's Theorem).
My misunderstanding comes from them citing Theorem $5$. Why is that Theorem necessary here?
real-analysis calculus limits convergence uniform-convergence
$endgroup$
add a comment |
$begingroup$
Suppose that the domain of convergence of the power series
$sum_{k=0}^{infty} c_{k}x^{k}$ contains the interval $(-r, r)$.
Define $$f(x) = sum_{k=0}^{infty} c_{k}x^{k} hspace{1cm} text{
> if } |x| < r. $$
Let $[a, b] subseteq (-r, r).$ Prove that
$$int_{a}^{b} f(x) mathop{dx} = sum_{k = 0}^{infty} frac{c_{k}}{k + 1}left(b^{k + 1} - a^{k + 1}right).$$
Here's the solution I have. It might be wrong because it's not official.
Recall Theorem $5$, which states that if a sequence of integrable functions ${f_{n} : [a, b] rightarrow mathbb{R}}$ converges uniformly to the function $f : [a, b] rightarrow mathbb{R}$, then the limit function is also integrable.
So,
$$int_{a}^{b} f(x) mathop{dx} = lim_{ntoinfty} int_{a}^{b} sum_{k = 0}^{n} c_{k}x^{k} = lim_{ntoinfty} sum_{k=0}^{n}int_{a}^{b} c_{k}x^{k} mathop{dx} = lim_{ntoinfty} sum_{k=0}^{n} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right) $$
$$= sum_{k=0}^{infty} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right), $$
which is what we wanted to show. (Switching integral/summation is justified by Fubini's Theorem).
My misunderstanding comes from them citing Theorem $5$. Why is that Theorem necessary here?
real-analysis calculus limits convergence uniform-convergence
$endgroup$
$begingroup$
How can you get the first equality?
$endgroup$
– tonychow0929
Dec 10 '18 at 6:42
$begingroup$
The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right?
$endgroup$
– joseph
Dec 10 '18 at 6:42
add a comment |
$begingroup$
Suppose that the domain of convergence of the power series
$sum_{k=0}^{infty} c_{k}x^{k}$ contains the interval $(-r, r)$.
Define $$f(x) = sum_{k=0}^{infty} c_{k}x^{k} hspace{1cm} text{
> if } |x| < r. $$
Let $[a, b] subseteq (-r, r).$ Prove that
$$int_{a}^{b} f(x) mathop{dx} = sum_{k = 0}^{infty} frac{c_{k}}{k + 1}left(b^{k + 1} - a^{k + 1}right).$$
Here's the solution I have. It might be wrong because it's not official.
Recall Theorem $5$, which states that if a sequence of integrable functions ${f_{n} : [a, b] rightarrow mathbb{R}}$ converges uniformly to the function $f : [a, b] rightarrow mathbb{R}$, then the limit function is also integrable.
So,
$$int_{a}^{b} f(x) mathop{dx} = lim_{ntoinfty} int_{a}^{b} sum_{k = 0}^{n} c_{k}x^{k} = lim_{ntoinfty} sum_{k=0}^{n}int_{a}^{b} c_{k}x^{k} mathop{dx} = lim_{ntoinfty} sum_{k=0}^{n} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right) $$
$$= sum_{k=0}^{infty} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right), $$
which is what we wanted to show. (Switching integral/summation is justified by Fubini's Theorem).
My misunderstanding comes from them citing Theorem $5$. Why is that Theorem necessary here?
real-analysis calculus limits convergence uniform-convergence
$endgroup$
Suppose that the domain of convergence of the power series
$sum_{k=0}^{infty} c_{k}x^{k}$ contains the interval $(-r, r)$.
Define $$f(x) = sum_{k=0}^{infty} c_{k}x^{k} hspace{1cm} text{
> if } |x| < r. $$
Let $[a, b] subseteq (-r, r).$ Prove that
$$int_{a}^{b} f(x) mathop{dx} = sum_{k = 0}^{infty} frac{c_{k}}{k + 1}left(b^{k + 1} - a^{k + 1}right).$$
Here's the solution I have. It might be wrong because it's not official.
Recall Theorem $5$, which states that if a sequence of integrable functions ${f_{n} : [a, b] rightarrow mathbb{R}}$ converges uniformly to the function $f : [a, b] rightarrow mathbb{R}$, then the limit function is also integrable.
So,
$$int_{a}^{b} f(x) mathop{dx} = lim_{ntoinfty} int_{a}^{b} sum_{k = 0}^{n} c_{k}x^{k} = lim_{ntoinfty} sum_{k=0}^{n}int_{a}^{b} c_{k}x^{k} mathop{dx} = lim_{ntoinfty} sum_{k=0}^{n} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right) $$
$$= sum_{k=0}^{infty} left(frac{c_{k}}{k + 1}right)left(b^{k + 1} - a^{k + 1}right), $$
which is what we wanted to show. (Switching integral/summation is justified by Fubini's Theorem).
My misunderstanding comes from them citing Theorem $5$. Why is that Theorem necessary here?
real-analysis calculus limits convergence uniform-convergence
real-analysis calculus limits convergence uniform-convergence
edited Dec 10 '18 at 12:36
J.G.
27.2k22843
27.2k22843
asked Dec 10 '18 at 6:39
josephjoseph
496111
496111
$begingroup$
How can you get the first equality?
$endgroup$
– tonychow0929
Dec 10 '18 at 6:42
$begingroup$
The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right?
$endgroup$
– joseph
Dec 10 '18 at 6:42
add a comment |
$begingroup$
How can you get the first equality?
$endgroup$
– tonychow0929
Dec 10 '18 at 6:42
$begingroup$
The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right?
$endgroup$
– joseph
Dec 10 '18 at 6:42
$begingroup$
How can you get the first equality?
$endgroup$
– tonychow0929
Dec 10 '18 at 6:42
$begingroup$
How can you get the first equality?
$endgroup$
– tonychow0929
Dec 10 '18 at 6:42
$begingroup$
The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right?
$endgroup$
– joseph
Dec 10 '18 at 6:42
$begingroup$
The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right?
$endgroup$
– joseph
Dec 10 '18 at 6:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Notice that $forall x in (-r,r),$
$$f(x) = sum_{k = 0}^{infty} c_kx^k = lim_{n to infty} sum_{k = 0}^{n} c_kx^k.$$
Hence
$$int_{a}^{b} f(x) dx = int_{a}^{b} lim_{n to infty} sum_{k = 0}^{n}c_kx^k dx.$$
Now apply Theorem $5$ to pull out the limit.
$endgroup$
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
add a comment |
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
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oldest
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$begingroup$
Notice that $forall x in (-r,r),$
$$f(x) = sum_{k = 0}^{infty} c_kx^k = lim_{n to infty} sum_{k = 0}^{n} c_kx^k.$$
Hence
$$int_{a}^{b} f(x) dx = int_{a}^{b} lim_{n to infty} sum_{k = 0}^{n}c_kx^k dx.$$
Now apply Theorem $5$ to pull out the limit.
$endgroup$
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
add a comment |
$begingroup$
Notice that $forall x in (-r,r),$
$$f(x) = sum_{k = 0}^{infty} c_kx^k = lim_{n to infty} sum_{k = 0}^{n} c_kx^k.$$
Hence
$$int_{a}^{b} f(x) dx = int_{a}^{b} lim_{n to infty} sum_{k = 0}^{n}c_kx^k dx.$$
Now apply Theorem $5$ to pull out the limit.
$endgroup$
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
add a comment |
$begingroup$
Notice that $forall x in (-r,r),$
$$f(x) = sum_{k = 0}^{infty} c_kx^k = lim_{n to infty} sum_{k = 0}^{n} c_kx^k.$$
Hence
$$int_{a}^{b} f(x) dx = int_{a}^{b} lim_{n to infty} sum_{k = 0}^{n}c_kx^k dx.$$
Now apply Theorem $5$ to pull out the limit.
$endgroup$
Notice that $forall x in (-r,r),$
$$f(x) = sum_{k = 0}^{infty} c_kx^k = lim_{n to infty} sum_{k = 0}^{n} c_kx^k.$$
Hence
$$int_{a}^{b} f(x) dx = int_{a}^{b} lim_{n to infty} sum_{k = 0}^{n}c_kx^k dx.$$
Now apply Theorem $5$ to pull out the limit.
answered Dec 10 '18 at 6:46
tonychow0929tonychow0929
29825
29825
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
add a comment |
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
Can't you just switch the sum b/c of Fubini's Theorem? Write the sum without the limit first. Then switch it out.
$endgroup$
– joseph
Dec 10 '18 at 6:48
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
$begingroup$
You use Fubini to switch the integral/summation signs (the second equality). You need Theorem 5 to switch integral/limit signs.
$endgroup$
– tonychow0929
Dec 10 '18 at 6:51
add a comment |
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$begingroup$
How can you get the first equality?
$endgroup$
– tonychow0929
Dec 10 '18 at 6:42
$begingroup$
The first equality just plugs in the definition of $f(x)$, which is defined in the problem statement, right?
$endgroup$
– joseph
Dec 10 '18 at 6:42