proving definite integral value by induction
$begingroup$
I have been trying to prove the following by induction but all my efforts have failed. the original problem is to prove that
$int_{-1}^{1}(1-x^{2})^kdx = frac{2^{2k+1}(k!)^{2}}{(2k+1)!}$
Where $k$ is a positive integer.
after expanding the function using the binomial theorem and taking the definite integral I got this
$int_{-1}^{1}(1-x^{2})^kdx = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
Now i need to show that
$ frac{2^{2k+1}(k!)^{2}}{(2k+1)!} = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
I took the base case $k=1$ and it works for $1$
Any help will be much appreciated.
Thank you in advance.
definite-integrals induction
asked Jan 14 at 11:37
user451552user451552
875
$endgroup$
add a comment |
$begingroup$
I have been trying to prove the following by induction but all my efforts have failed. the original problem is to prove that
$int_{-1}^{1}(1-x^{2})^kdx = frac{2^{2k+1}(k!)^{2}}{(2k+1)!}$
Where $k$ is a positive integer.
after expanding the function using the binomial theorem and taking the definite integral I got this
$int_{-1}^{1}(1-x^{2})^kdx = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
Now i need to show that
$ frac{2^{2k+1}(k!)^{2}}{(2k+1)!} = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
I took the base case $k=1$ and it works for $1$
Any help will be much appreciated.
Thank you in advance.
definite-integrals induction
asked Jan 14 at 11:37
user451552user451552
875
$endgroup$
$begingroup$
Should that be $$sum_{i=0}^kfrac{binom{k}{i}2(-1)^{k-i}}{2(k-i)+1},$$ perhaps? It doesn't really make sense as written.
$endgroup$
– Cameron Buie
Jan 14 at 11:46
$begingroup$
Oh! Yes that's what it means sorry for that misconception
$endgroup$
– user451552
Jan 14 at 11:56
add a comment |
$begingroup$
I have been trying to prove the following by induction but all my efforts have failed. the original problem is to prove that
$int_{-1}^{1}(1-x^{2})^kdx = frac{2^{2k+1}(k!)^{2}}{(2k+1)!}$
Where $k$ is a positive integer.
after expanding the function using the binomial theorem and taking the definite integral I got this
$int_{-1}^{1}(1-x^{2})^kdx = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
Now i need to show that
$ frac{2^{2k+1}(k!)^{2}}{(2k+1)!} = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
I took the base case $k=1$ and it works for $1$
Any help will be much appreciated.
Thank you in advance.
definite-integrals induction
asked Jan 14 at 11:37
user451552user451552
875
$endgroup$
I have been trying to prove the following by induction but all my efforts have failed. the original problem is to prove that
$int_{-1}^{1}(1-x^{2})^kdx = frac{2^{2k+1}(k!)^{2}}{(2k+1)!}$
Where $k$ is a positive integer.
after expanding the function using the binomial theorem and taking the definite integral I got this
$int_{-1}^{1}(1-x^{2})^kdx = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
Now i need to show that
$ frac{2^{2k+1}(k!)^{2}}{(2k+1)!} = sum_{i=0}^{k}frac {binom{k}{i}2(-1)^{k-i}}{2 (k-i)+1}$
I took the base case $k=1$ and it works for $1$
Any help will be much appreciated.
Thank you in advance.
definite-integrals induction
definite-integrals induction
asked Jan 14 at 11:37
user451552user451552
875
asked Jan 14 at 11:37
user451552user451552
875
edited Jan 14 at 12:02
user451552
asked Jan 14 at 11:37
user451552user451552
875
asked Jan 14 at 11:37
user451552user451552
875
asked Jan 14 at 11:37
user451552user451552
875
875
$begingroup$
Should that be $$sum_{i=0}^kfrac{binom{k}{i}2(-1)^{k-i}}{2(k-i)+1},$$ perhaps? It doesn't really make sense as written.
$endgroup$
– Cameron Buie
Jan 14 at 11:46
$begingroup$
Oh! Yes that's what it means sorry for that misconception
$endgroup$
– user451552
Jan 14 at 11:56
add a comment |
$begingroup$
Should that be $$sum_{i=0}^kfrac{binom{k}{i}2(-1)^{k-i}}{2(k-i)+1},$$ perhaps? It doesn't really make sense as written.
$endgroup$
– Cameron Buie
Jan 14 at 11:46
$begingroup$
Oh! Yes that's what it means sorry for that misconception
$endgroup$
– user451552
Jan 14 at 11:56
$begingroup$
Should that be $$sum_{i=0}^kfrac{binom{k}{i}2(-1)^{k-i}}{2(k-i)+1},$$ perhaps? It doesn't really make sense as written.
$endgroup$
– Cameron Buie
Jan 14 at 11:46
$begingroup$
Should that be $$sum_{i=0}^kfrac{binom{k}{i}2(-1)^{k-i}}{2(k-i)+1},$$ perhaps? It doesn't really make sense as written.
$endgroup$
– Cameron Buie
Jan 14 at 11:46
$begingroup$
Oh! Yes that's what it means sorry for that misconception
$endgroup$
– user451552
Jan 14 at 11:56
$begingroup$
Oh! Yes that's what it means sorry for that misconception
$endgroup$
– user451552
Jan 14 at 11:56
add a comment |
1 Answer
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$begingroup$
I suggest to suppose the thesis valid for $n-1$ and then calculate
$$int_{-1}^1(1-x^2)(1-x^2)^{n-1}dx$$
and integrate by parts using the inductive hypotesis.
answered Jan 14 at 11:45
GianniGianni
387
$endgroup$
add a comment |
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$begingroup$
I suggest to suppose the thesis valid for $n-1$ and then calculate
$$int_{-1}^1(1-x^2)(1-x^2)^{n-1}dx$$
and integrate by parts using the inductive hypotesis.
answered Jan 14 at 11:45
GianniGianni
387
$endgroup$
add a comment |
$begingroup$
I suggest to suppose the thesis valid for $n-1$ and then calculate
$$int_{-1}^1(1-x^2)(1-x^2)^{n-1}dx$$
and integrate by parts using the inductive hypotesis.
answered Jan 14 at 11:45
GianniGianni
387
$endgroup$
add a comment |
$begingroup$
I suggest to suppose the thesis valid for $n-1$ and then calculate
$$int_{-1}^1(1-x^2)(1-x^2)^{n-1}dx$$
and integrate by parts using the inductive hypotesis.
answered Jan 14 at 11:45
GianniGianni
387
$endgroup$
I suggest to suppose the thesis valid for $n-1$ and then calculate
$$int_{-1}^1(1-x^2)(1-x^2)^{n-1}dx$$
and integrate by parts using the inductive hypotesis.
answered Jan 14 at 11:45
GianniGianni
387
answered Jan 14 at 11:45
GianniGianni
387
answered Jan 14 at 11:45
GianniGianni
387
answered Jan 14 at 11:45
GianniGianni
387
387
add a comment |
add a comment |
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$begingroup$
Should that be $$sum_{i=0}^kfrac{binom{k}{i}2(-1)^{k-i}}{2(k-i)+1},$$ perhaps? It doesn't really make sense as written.
$endgroup$
– Cameron Buie
Jan 14 at 11:46
$begingroup$
Oh! Yes that's what it means sorry for that misconception
$endgroup$
– user451552
Jan 14 at 11:56