Mixing
Connection between powers of $i$ and derivatives of $sin$ and $cos$?
$begingroup$
There are plenty of connections between trigonometry and the complex plane. One particular comparison seems striking to me, and that’s the four-step cycle inherent in exponentiating $i$ to various powers and taking derivatives of $sin$. Namely:
$$i^0=1, i^1=i, i^2=-1, i^3=-i, dots$$
$$f(x)=sin x,f’(x)=cos x,f’’(x)=-sin x,f’’’(x)=-cos x,dots$$
I’m aware of the connection between $e^{itheta}=isintheta+costheta$ and $frac{dleft(e^{it}right)}{dt}=ie^{it}$, so there’s already a connection between trigonometry, derivatives, and the unit circle in the imaginary plane. Is there an extension of this, or something else entirely, that connects the four-step cycle of differentiating $sin$ and that of exponentiating $i$?
complex-analysis derivatives trigonometry exponentiation
asked Jan 15 at 2:01
DonielFDonielF
505515
$endgroup$
add a comment |
$begingroup$
There are plenty of connections between trigonometry and the complex plane. One particular comparison seems striking to me, and that’s the four-step cycle inherent in exponentiating $i$ to various powers and taking derivatives of $sin$. Namely:
$$i^0=1, i^1=i, i^2=-1, i^3=-i, dots$$
$$f(x)=sin x,f’(x)=cos x,f’’(x)=-sin x,f’’’(x)=-cos x,dots$$
I’m aware of the connection between $e^{itheta}=isintheta+costheta$ and $frac{dleft(e^{it}right)}{dt}=ie^{it}$, so there’s already a connection between trigonometry, derivatives, and the unit circle in the imaginary plane. Is there an extension of this, or something else entirely, that connects the four-step cycle of differentiating $sin$ and that of exponentiating $i$?
complex-analysis derivatives trigonometry exponentiation
asked Jan 15 at 2:01
DonielFDonielF
505515
$endgroup$
2
$begingroup$
The equations you wrote are basically the relevant equations. Have you differentiated $e^{ix}$ 4 times?
$endgroup$
– Mark S.
Jan 15 at 2:05
$begingroup$
@MarkS. ...May I smack myself now for not seeing this?
$endgroup$
– DonielF
Jan 15 at 2:06
add a comment |
$begingroup$
There are plenty of connections between trigonometry and the complex plane. One particular comparison seems striking to me, and that’s the four-step cycle inherent in exponentiating $i$ to various powers and taking derivatives of $sin$. Namely:
$$i^0=1, i^1=i, i^2=-1, i^3=-i, dots$$
$$f(x)=sin x,f’(x)=cos x,f’’(x)=-sin x,f’’’(x)=-cos x,dots$$
I’m aware of the connection between $e^{itheta}=isintheta+costheta$ and $frac{dleft(e^{it}right)}{dt}=ie^{it}$, so there’s already a connection between trigonometry, derivatives, and the unit circle in the imaginary plane. Is there an extension of this, or something else entirely, that connects the four-step cycle of differentiating $sin$ and that of exponentiating $i$?
complex-analysis derivatives trigonometry exponentiation
asked Jan 15 at 2:01
DonielFDonielF
505515
$endgroup$
There are plenty of connections between trigonometry and the complex plane. One particular comparison seems striking to me, and that’s the four-step cycle inherent in exponentiating $i$ to various powers and taking derivatives of $sin$. Namely:
$$i^0=1, i^1=i, i^2=-1, i^3=-i, dots$$
$$f(x)=sin x,f’(x)=cos x,f’’(x)=-sin x,f’’’(x)=-cos x,dots$$
I’m aware of the connection between $e^{itheta}=isintheta+costheta$ and $frac{dleft(e^{it}right)}{dt}=ie^{it}$, so there’s already a connection between trigonometry, derivatives, and the unit circle in the imaginary plane. Is there an extension of this, or something else entirely, that connects the four-step cycle of differentiating $sin$ and that of exponentiating $i$?
complex-analysis derivatives trigonometry exponentiation
complex-analysis derivatives trigonometry exponentiation
asked Jan 15 at 2:01
DonielFDonielF
505515
asked Jan 15 at 2:01
DonielFDonielF
505515
asked Jan 15 at 2:01
DonielFDonielF
505515
asked Jan 15 at 2:01
DonielFDonielF
505515
asked Jan 15 at 2:01
DonielFDonielF
505515
505515
2
$begingroup$
The equations you wrote are basically the relevant equations. Have you differentiated $e^{ix}$ 4 times?
$endgroup$
– Mark S.
Jan 15 at 2:05
$begingroup$
@MarkS. ...May I smack myself now for not seeing this?
$endgroup$
– DonielF
Jan 15 at 2:06
add a comment |
2
$begingroup$
The equations you wrote are basically the relevant equations. Have you differentiated $e^{ix}$ 4 times?
$endgroup$
– Mark S.
Jan 15 at 2:05
$begingroup$
@MarkS. ...May I smack myself now for not seeing this?
$endgroup$
– DonielF
Jan 15 at 2:06
2
2
$begingroup$
The equations you wrote are basically the relevant equations. Have you differentiated $e^{ix}$ 4 times?
$endgroup$
– Mark S.
Jan 15 at 2:05
$begingroup$
The equations you wrote are basically the relevant equations. Have you differentiated $e^{ix}$ 4 times?
$endgroup$
– Mark S.
Jan 15 at 2:05
$begingroup$
@MarkS. ...May I smack myself now for not seeing this?
$endgroup$
– DonielF
Jan 15 at 2:06
$begingroup$
@MarkS. ...May I smack myself now for not seeing this?
$endgroup$
– DonielF
Jan 15 at 2:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
H/t Mark S for pointing out the obvious.
Taking the equation $e^{ix}=isin x+cos x$ and differentiating it four times yields the following:
$$e^{ix}=isin x+cos x\ie^{ix}=icos x-sin x\-e^{ix}=-isin x-cos x\-ie^{ix}=-icos x+sin x\dots$$
The derivative of the left side rotates through the powers of $i$ and the derivative of the right side rotates through the derivatives of $sin$ and $cos$.
Excuse me while I go repeatedly smack myself in the face.
answered Jan 15 at 2:43
DonielFDonielF
505515
$endgroup$
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%2f3073998%2fconnection-between-powers-of-i-and-derivatives-of-sin-and-cos%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
H/t Mark S for pointing out the obvious.
Taking the equation $e^{ix}=isin x+cos x$ and differentiating it four times yields the following:
$$e^{ix}=isin x+cos x\ie^{ix}=icos x-sin x\-e^{ix}=-isin x-cos x\-ie^{ix}=-icos x+sin x\dots$$
The derivative of the left side rotates through the powers of $i$ and the derivative of the right side rotates through the derivatives of $sin$ and $cos$.
Excuse me while I go repeatedly smack myself in the face.
answered Jan 15 at 2:43
DonielFDonielF
505515
$endgroup$
add a comment |
$begingroup$
H/t Mark S for pointing out the obvious.
Taking the equation $e^{ix}=isin x+cos x$ and differentiating it four times yields the following:
$$e^{ix}=isin x+cos x\ie^{ix}=icos x-sin x\-e^{ix}=-isin x-cos x\-ie^{ix}=-icos x+sin x\dots$$
The derivative of the left side rotates through the powers of $i$ and the derivative of the right side rotates through the derivatives of $sin$ and $cos$.
Excuse me while I go repeatedly smack myself in the face.
answered Jan 15 at 2:43
DonielFDonielF
505515
$endgroup$
add a comment |
$begingroup$
H/t Mark S for pointing out the obvious.
Taking the equation $e^{ix}=isin x+cos x$ and differentiating it four times yields the following:
$$e^{ix}=isin x+cos x\ie^{ix}=icos x-sin x\-e^{ix}=-isin x-cos x\-ie^{ix}=-icos x+sin x\dots$$
The derivative of the left side rotates through the powers of $i$ and the derivative of the right side rotates through the derivatives of $sin$ and $cos$.
Excuse me while I go repeatedly smack myself in the face.
answered Jan 15 at 2:43
DonielFDonielF
505515
$endgroup$
H/t Mark S for pointing out the obvious.
Taking the equation $e^{ix}=isin x+cos x$ and differentiating it four times yields the following:
$$e^{ix}=isin x+cos x\ie^{ix}=icos x-sin x\-e^{ix}=-isin x-cos x\-ie^{ix}=-icos x+sin x\dots$$
The derivative of the left side rotates through the powers of $i$ and the derivative of the right side rotates through the derivatives of $sin$ and $cos$.
Excuse me while I go repeatedly smack myself in the face.
answered Jan 15 at 2:43
DonielFDonielF
505515
answered Jan 15 at 2:43
DonielFDonielF
505515
answered Jan 15 at 2:43
DonielFDonielF
505515
answered Jan 15 at 2:43
DonielFDonielF
505515
505515
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%2f3073998%2fconnection-between-powers-of-i-and-derivatives-of-sin-and-cos%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
2
$begingroup$
The equations you wrote are basically the relevant equations. Have you differentiated $e^{ix}$ 4 times?
$endgroup$
– Mark S.
Jan 15 at 2:05
$begingroup$
@MarkS. ...May I smack myself now for not seeing this?
$endgroup$
– DonielF
Jan 15 at 2:06