Proving $sin(2x) = 2sin(x)cos(x)$ using Cauchy product
$begingroup$
I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}
trigonometry cauchy-product
$endgroup$
add a comment |
$begingroup$
I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}
trigonometry cauchy-product
$endgroup$
3
$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55
$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58
add a comment |
$begingroup$
I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}
trigonometry cauchy-product
$endgroup$
I've stumbled upon this question. I can't understand why all even terms of the Cauchy product are $0$, since we add only positive numbers:
begin{equation*}
c_n = begin{cases} sumlimits_{k=0}^{m} frac{(-1)^{k}(-1)^{m-k}}{(2k+1)!(2m-2k)!} & n = 2m + 1 \ hspace{32 pt} 0 & n = 2m end{cases}
end{equation*}
trigonometry cauchy-product
trigonometry cauchy-product
asked Dec 17 '18 at 14:53
DreikäsehochDreikäsehoch
3048
3048
3
$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55
$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58
add a comment |
3
$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55
$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58
3
3
$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55
$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55
$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58
$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$$c_n=sum_{k=0}^n a_kb_{n-k}$$
where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then
i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;
ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.
Hence $c_n=sum_{k=0}^n 0=0$.
By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.
$endgroup$
add a comment |
$begingroup$
Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044043%2fproving-sin2x-2-sinx-cosx-using-cauchy-product%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$c_n=sum_{k=0}^n a_kb_{n-k}$$
where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then
i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;
ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.
Hence $c_n=sum_{k=0}^n 0=0$.
By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.
$endgroup$
add a comment |
$begingroup$
Note that
$$c_n=sum_{k=0}^n a_kb_{n-k}$$
where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then
i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;
ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.
Hence $c_n=sum_{k=0}^n 0=0$.
By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.
$endgroup$
add a comment |
$begingroup$
Note that
$$c_n=sum_{k=0}^n a_kb_{n-k}$$
where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then
i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;
ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.
Hence $c_n=sum_{k=0}^n 0=0$.
By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.
$endgroup$
Note that
$$c_n=sum_{k=0}^n a_kb_{n-k}$$
where $a_k$ is the coefficient of $x^k$ of the series of $sin(x)$ and $b_{n-k}$ is the coefficient of $x^{n-k}$ of the series of $cos(x)$. Now if $n$ is even then
i) if $k$ is odd then $n-k$ is odd and $b_{n-k}=0$ which implies that $a_kb_{n-k}=0$;
ii) if $k$ is even then $a_{k}=0$ which implies that $a_kb_{n-k}=0$.
Hence $c_n=sum_{k=0}^n 0=0$.
By using the same argument we can show that the product of an "even" series with and an "odd" series is an "odd" series.
edited Dec 17 '18 at 15:05
answered Dec 17 '18 at 15:00
Robert ZRobert Z
97.4k1066137
97.4k1066137
add a comment |
add a comment |
$begingroup$
Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$
$endgroup$
add a comment |
$begingroup$
Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$
$endgroup$
add a comment |
$begingroup$
Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$
$endgroup$
Since$$sin x=0+x+0times x^2-frac1{3!}x^3+cdots$$and$$cos x=1+0times x-frac1{2!}x^2+0times x^3+cdots$$when you compute a term of even order of the Cauchy product of these series, you get a sum with zeros everywhere; for instance, the coeffiecient of $x^2$ is$$0timesleft(-frac1{2!}right)+1times0+0times1=0.$$
answered Dec 17 '18 at 15:03
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044043%2fproving-sin2x-2-sinx-cosx-using-cauchy-product%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
All even terms of the Cauchy product are 0 because $sin(2x)$ is an odd function.
$endgroup$
– Robert Z
Dec 17 '18 at 14:55
$begingroup$
@RobertZ but how does this result from the Cauchy product?
$endgroup$
– Dreikäsehoch
Dec 17 '18 at 14:58