Mixing
Proving that $frac{e^x + e^{-x}}2 le e^{x^2/2}$
$begingroup$
Prove the following inequality:
$$dfrac{e^x + e^{-x}}2 le e^{frac{x^2}{2}}$$
This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?
calculus inequality taylor-expansion exponential-function
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
$endgroup$
add a comment |
$begingroup$
Prove the following inequality:
$$dfrac{e^x + e^{-x}}2 le e^{frac{x^2}{2}}$$
This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?
calculus inequality taylor-expansion exponential-function
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
$endgroup$
$begingroup$
Why not use the full series?
$endgroup$
– Antonio Vargas
Jan 28 '14 at 18:09
add a comment |
$begingroup$
Prove the following inequality:
$$dfrac{e^x + e^{-x}}2 le e^{frac{x^2}{2}}$$
This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?
calculus inequality taylor-expansion exponential-function
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
$endgroup$
Prove the following inequality:
$$dfrac{e^x + e^{-x}}2 le e^{frac{x^2}{2}}$$
This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?
calculus inequality taylor-expansion exponential-function
calculus inequality taylor-expansion exponential-function
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
edited Jan 22 at 10:55
Martin Sleziak
44.8k10119273
44.8k10119273
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
asked Jan 28 '14 at 18:01
guynaaguynaa
656314
656314
$begingroup$
Why not use the full series?
$endgroup$
– Antonio Vargas
Jan 28 '14 at 18:09
add a comment |
$begingroup$
Why not use the full series?
$endgroup$
– Antonio Vargas
Jan 28 '14 at 18:09
$begingroup$
Why not use the full series?
$endgroup$
– Antonio Vargas
Jan 28 '14 at 18:09
$begingroup$
Why not use the full series?
$endgroup$
– Antonio Vargas
Jan 28 '14 at 18:09
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$frac{e^x+e^{-x}}{2} = frac{left(1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots right) + left(1 - x + frac{x^2}{2!} - frac{x^3}{3!} + cdots right)}{2} = frac{2 + 2 frac{x^2}{2!} + 2 frac{x^4}{4!} + cdots}{2}$$
$$ = 1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720}+ cdots$$
$$e^{frac{x^2}{2}} = 1 + frac{x^2}{2} + frac{left(frac{x^2}{2}right)^2}{2!} + frac{left(frac{x^2}{2}right)^3}{3!} + cdots = 1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + cdots$$
Basically, it comes down to comparing $displaystyle frac{1}{(2n)!}$ and $displaystyle frac{1}{2^n n!}$, because they are the coefficients for $displaystyle frac{e^x + e^{-x}}{2}$ and $e^{frac{x^2}{2}}$, respectively. We note that $$frac{1}{(2n)!} = frac{1}{1 cdot 2 cdot 3 cdots (2n-1) cdot (2n)}$$ while $$frac{1}{2^n n!} = frac{1}{2 cdot 4 cdot 6 cdots (2n-2) cdot (2n)}$$
Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$frac{e^x+e^{-x}}{2} le e^{frac{x^2}{2}}$$
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
$endgroup$
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
add a comment |
$begingroup$
Using $e^x = sum_{k=0}^infty frac{1}{k!} x^k$, we get
$$
frac{1}{2}left( e^x + e^{-x}right) = frac{1}{2} sum_{k=0}^infty frac{1}{k!} [ x^k + (-x)^k] = sum_{i=0}^infty ??? x^{2i},
$$
and
$$
e^frac{x^2}{2} = sum_{k=0}^infty ??? x^{2k}.
$$
Fill in ???, and show that each term in the first sum is smaller than the corresponding term in the latter sum.
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
$endgroup$
add a comment |
$begingroup$
Obscene overkill: since
$$ cosh(z)=prod_{ngeq 0}left(1+frac{4z^2}{(2n+1)^2pi^2}right) $$
we have
$$ logcosh(z)leq sum_{ngeq 0}frac{4z^2}{(2n+1)^2pi^2} =frac{z^2}{2}$$
and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $tanh(z)leq z$ over $[0,xinmathbb{R}^+]$.
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
1
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
$endgroup$
– Jack D'Aurizio
Jan 21 at 21:47
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$frac{e^x+e^{-x}}{2} = frac{left(1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots right) + left(1 - x + frac{x^2}{2!} - frac{x^3}{3!} + cdots right)}{2} = frac{2 + 2 frac{x^2}{2!} + 2 frac{x^4}{4!} + cdots}{2}$$
$$ = 1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720}+ cdots$$
$$e^{frac{x^2}{2}} = 1 + frac{x^2}{2} + frac{left(frac{x^2}{2}right)^2}{2!} + frac{left(frac{x^2}{2}right)^3}{3!} + cdots = 1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + cdots$$
Basically, it comes down to comparing $displaystyle frac{1}{(2n)!}$ and $displaystyle frac{1}{2^n n!}$, because they are the coefficients for $displaystyle frac{e^x + e^{-x}}{2}$ and $e^{frac{x^2}{2}}$, respectively. We note that $$frac{1}{(2n)!} = frac{1}{1 cdot 2 cdot 3 cdots (2n-1) cdot (2n)}$$ while $$frac{1}{2^n n!} = frac{1}{2 cdot 4 cdot 6 cdots (2n-2) cdot (2n)}$$
Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$frac{e^x+e^{-x}}{2} le e^{frac{x^2}{2}}$$
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
$endgroup$
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
add a comment |
$begingroup$
$$frac{e^x+e^{-x}}{2} = frac{left(1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots right) + left(1 - x + frac{x^2}{2!} - frac{x^3}{3!} + cdots right)}{2} = frac{2 + 2 frac{x^2}{2!} + 2 frac{x^4}{4!} + cdots}{2}$$
$$ = 1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720}+ cdots$$
$$e^{frac{x^2}{2}} = 1 + frac{x^2}{2} + frac{left(frac{x^2}{2}right)^2}{2!} + frac{left(frac{x^2}{2}right)^3}{3!} + cdots = 1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + cdots$$
Basically, it comes down to comparing $displaystyle frac{1}{(2n)!}$ and $displaystyle frac{1}{2^n n!}$, because they are the coefficients for $displaystyle frac{e^x + e^{-x}}{2}$ and $e^{frac{x^2}{2}}$, respectively. We note that $$frac{1}{(2n)!} = frac{1}{1 cdot 2 cdot 3 cdots (2n-1) cdot (2n)}$$ while $$frac{1}{2^n n!} = frac{1}{2 cdot 4 cdot 6 cdots (2n-2) cdot (2n)}$$
Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$frac{e^x+e^{-x}}{2} le e^{frac{x^2}{2}}$$
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
$endgroup$
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
add a comment |
$begingroup$
$$frac{e^x+e^{-x}}{2} = frac{left(1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots right) + left(1 - x + frac{x^2}{2!} - frac{x^3}{3!} + cdots right)}{2} = frac{2 + 2 frac{x^2}{2!} + 2 frac{x^4}{4!} + cdots}{2}$$
$$ = 1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720}+ cdots$$
$$e^{frac{x^2}{2}} = 1 + frac{x^2}{2} + frac{left(frac{x^2}{2}right)^2}{2!} + frac{left(frac{x^2}{2}right)^3}{3!} + cdots = 1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + cdots$$
Basically, it comes down to comparing $displaystyle frac{1}{(2n)!}$ and $displaystyle frac{1}{2^n n!}$, because they are the coefficients for $displaystyle frac{e^x + e^{-x}}{2}$ and $e^{frac{x^2}{2}}$, respectively. We note that $$frac{1}{(2n)!} = frac{1}{1 cdot 2 cdot 3 cdots (2n-1) cdot (2n)}$$ while $$frac{1}{2^n n!} = frac{1}{2 cdot 4 cdot 6 cdots (2n-2) cdot (2n)}$$
Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$frac{e^x+e^{-x}}{2} le e^{frac{x^2}{2}}$$
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
$endgroup$
$$frac{e^x+e^{-x}}{2} = frac{left(1 + x + frac{x^2}{2!} + frac{x^3}{3!} + cdots right) + left(1 - x + frac{x^2}{2!} - frac{x^3}{3!} + cdots right)}{2} = frac{2 + 2 frac{x^2}{2!} + 2 frac{x^4}{4!} + cdots}{2}$$
$$ = 1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720}+ cdots$$
$$e^{frac{x^2}{2}} = 1 + frac{x^2}{2} + frac{left(frac{x^2}{2}right)^2}{2!} + frac{left(frac{x^2}{2}right)^3}{3!} + cdots = 1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + cdots$$
Basically, it comes down to comparing $displaystyle frac{1}{(2n)!}$ and $displaystyle frac{1}{2^n n!}$, because they are the coefficients for $displaystyle frac{e^x + e^{-x}}{2}$ and $e^{frac{x^2}{2}}$, respectively. We note that $$frac{1}{(2n)!} = frac{1}{1 cdot 2 cdot 3 cdots (2n-1) cdot (2n)}$$ while $$frac{1}{2^n n!} = frac{1}{2 cdot 4 cdot 6 cdots (2n-2) cdot (2n)}$$
Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$frac{e^x+e^{-x}}{2} le e^{frac{x^2}{2}}$$
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
edited Jan 28 '14 at 18:38
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
answered Jan 28 '14 at 18:12
2012ssohn2012ssohn
3,59711029
3,59711029
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
add a comment |
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that.
$endgroup$
– GinKin
Feb 26 '14 at 18:16
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
$begingroup$
Then you can either write it as $$1 + frac{x^2}{2} + frac{x^4}{24} + frac{x^6}{720} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + frac{x^2}{2} + frac{x^4}{8} + frac{x^6}{48} + O(x^8)$$ or as $$sum_{n=0}^{infty} frac{x^{2n}}{2^n n!}$$
$endgroup$
– 2012ssohn
Feb 27 '14 at 0:53
add a comment |
$begingroup$
Using $e^x = sum_{k=0}^infty frac{1}{k!} x^k$, we get
$$
frac{1}{2}left( e^x + e^{-x}right) = frac{1}{2} sum_{k=0}^infty frac{1}{k!} [ x^k + (-x)^k] = sum_{i=0}^infty ??? x^{2i},
$$
and
$$
e^frac{x^2}{2} = sum_{k=0}^infty ??? x^{2k}.
$$
Fill in ???, and show that each term in the first sum is smaller than the corresponding term in the latter sum.
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
$endgroup$
add a comment |
$begingroup$
Using $e^x = sum_{k=0}^infty frac{1}{k!} x^k$, we get
$$
frac{1}{2}left( e^x + e^{-x}right) = frac{1}{2} sum_{k=0}^infty frac{1}{k!} [ x^k + (-x)^k] = sum_{i=0}^infty ??? x^{2i},
$$
and
$$
e^frac{x^2}{2} = sum_{k=0}^infty ??? x^{2k}.
$$
Fill in ???, and show that each term in the first sum is smaller than the corresponding term in the latter sum.
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
$endgroup$
add a comment |
$begingroup$
Using $e^x = sum_{k=0}^infty frac{1}{k!} x^k$, we get
$$
frac{1}{2}left( e^x + e^{-x}right) = frac{1}{2} sum_{k=0}^infty frac{1}{k!} [ x^k + (-x)^k] = sum_{i=0}^infty ??? x^{2i},
$$
and
$$
e^frac{x^2}{2} = sum_{k=0}^infty ??? x^{2k}.
$$
Fill in ???, and show that each term in the first sum is smaller than the corresponding term in the latter sum.
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
$endgroup$
Using $e^x = sum_{k=0}^infty frac{1}{k!} x^k$, we get
$$
frac{1}{2}left( e^x + e^{-x}right) = frac{1}{2} sum_{k=0}^infty frac{1}{k!} [ x^k + (-x)^k] = sum_{i=0}^infty ??? x^{2i},
$$
and
$$
e^frac{x^2}{2} = sum_{k=0}^infty ??? x^{2k}.
$$
Fill in ???, and show that each term in the first sum is smaller than the corresponding term in the latter sum.
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
answered Jan 28 '14 at 18:10
JiKJiK
4,8841232
4,8841232
add a comment |
add a comment |
$begingroup$
Obscene overkill: since
$$ cosh(z)=prod_{ngeq 0}left(1+frac{4z^2}{(2n+1)^2pi^2}right) $$
we have
$$ logcosh(z)leq sum_{ngeq 0}frac{4z^2}{(2n+1)^2pi^2} =frac{z^2}{2}$$
and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $tanh(z)leq z$ over $[0,xinmathbb{R}^+]$.
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
1
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
$endgroup$
– Jack D'Aurizio
Jan 21 at 21:47
add a comment |
$begingroup$
Obscene overkill: since
$$ cosh(z)=prod_{ngeq 0}left(1+frac{4z^2}{(2n+1)^2pi^2}right) $$
we have
$$ logcosh(z)leq sum_{ngeq 0}frac{4z^2}{(2n+1)^2pi^2} =frac{z^2}{2}$$
and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $tanh(z)leq z$ over $[0,xinmathbb{R}^+]$.
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
1
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
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– Jack D'Aurizio
Jan 21 at 21:47
add a comment |
$begingroup$
Obscene overkill: since
$$ cosh(z)=prod_{ngeq 0}left(1+frac{4z^2}{(2n+1)^2pi^2}right) $$
we have
$$ logcosh(z)leq sum_{ngeq 0}frac{4z^2}{(2n+1)^2pi^2} =frac{z^2}{2}$$
and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $tanh(z)leq z$ over $[0,xinmathbb{R}^+]$.
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
$endgroup$
Obscene overkill: since
$$ cosh(z)=prod_{ngeq 0}left(1+frac{4z^2}{(2n+1)^2pi^2}right) $$
we have
$$ logcosh(z)leq sum_{ngeq 0}frac{4z^2}{(2n+1)^2pi^2} =frac{z^2}{2}$$
and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $tanh(z)leq z$ over $[0,xinmathbb{R}^+]$.
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
answered Jan 21 at 21:31
Jack D'AurizioJack D'Aurizio
291k33284666
291k33284666
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
1
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
$endgroup$
– Jack D'Aurizio
Jan 21 at 21:47
add a comment |
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
1
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
$endgroup$
– Jack D'Aurizio
Jan 21 at 21:47
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
$begingroup$
Mind providing an elementary proof of the first identity?
$endgroup$
– Dwagg
Jan 21 at 21:43
1
1
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
$endgroup$
– Jack D'Aurizio
Jan 21 at 21:47
$begingroup$
@Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function.
$endgroup$
– Jack D'Aurizio
Jan 21 at 21:47
add a comment |
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$begingroup$
Why not use the full series?
$endgroup$
– Antonio Vargas
Jan 28 '14 at 18:09