Understanding the solution key to a problem which shows that the integral of a sum equals a given value.












1












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 12:36









J.G.

31.6k23149




31.6k23149










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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
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1 Answer
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active

oldest

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1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 6:46









tonychow0929tonychow0929

28825




28825












  • $begingroup$
    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












  • $begingroup$
    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




















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