Mixing
Prove by induction that the $n$-th derivative of $cos(x)$ is given by $cosleft(x+nfrac{pi }{2}right)$
$begingroup$
Given the function $fleft(xright)=cos(x)$. Prove by induction that for $forall nin mathbb{N}$ $f^{(n)}=cosbigl(x+ncdot frac{pi }{2}bigr)$. Being that $f^{left(1right)}$ is the first derivative and so forth.
trigonometry induction proof-explanation
edited Jan 26 at 19:12
Bernard
123k741117
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
$endgroup$
add a comment |
$begingroup$
Given the function $fleft(xright)=cos(x)$. Prove by induction that for $forall nin mathbb{N}$ $f^{(n)}=cosbigl(x+ncdot frac{pi }{2}bigr)$. Being that $f^{left(1right)}$ is the first derivative and so forth.
trigonometry induction proof-explanation
edited Jan 26 at 19:12
Bernard
123k741117
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
$endgroup$
1
$begingroup$
What do you need us to explain ? (except doing the entire proof for you )
$endgroup$
– Atmos
Jan 26 at 19:05
$begingroup$
I'm learning math all on my own. If you think it useful to do the proof for me in order for me to understand and and continue with my learning, then, by all means, do it. My resources are very limited otherwise...
$endgroup$
– Daniel Oscar
Jan 26 at 19:08
add a comment |
$begingroup$
Given the function $fleft(xright)=cos(x)$. Prove by induction that for $forall nin mathbb{N}$ $f^{(n)}=cosbigl(x+ncdot frac{pi }{2}bigr)$. Being that $f^{left(1right)}$ is the first derivative and so forth.
trigonometry induction proof-explanation
edited Jan 26 at 19:12
Bernard
123k741117
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
$endgroup$
Given the function $fleft(xright)=cos(x)$. Prove by induction that for $forall nin mathbb{N}$ $f^{(n)}=cosbigl(x+ncdot frac{pi }{2}bigr)$. Being that $f^{left(1right)}$ is the first derivative and so forth.
trigonometry induction proof-explanation
trigonometry induction proof-explanation
edited Jan 26 at 19:12
Bernard
123k741117
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
edited Jan 26 at 19:12
Bernard
123k741117
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
edited Jan 26 at 19:12
Bernard
123k741117
edited Jan 26 at 19:12
Bernard
123k741117
edited Jan 26 at 19:12
Bernard
123k741117
123k741117
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
asked Jan 26 at 19:04
Daniel OscarDaniel Oscar
35711
35711
1
$begingroup$
What do you need us to explain ? (except doing the entire proof for you )
$endgroup$
– Atmos
Jan 26 at 19:05
$begingroup$
I'm learning math all on my own. If you think it useful to do the proof for me in order for me to understand and and continue with my learning, then, by all means, do it. My resources are very limited otherwise...
$endgroup$
– Daniel Oscar
Jan 26 at 19:08
add a comment |
1
$begingroup$
What do you need us to explain ? (except doing the entire proof for you )
$endgroup$
– Atmos
Jan 26 at 19:05
$begingroup$
I'm learning math all on my own. If you think it useful to do the proof for me in order for me to understand and and continue with my learning, then, by all means, do it. My resources are very limited otherwise...
$endgroup$
– Daniel Oscar
Jan 26 at 19:08
1
1
$begingroup$
What do you need us to explain ? (except doing the entire proof for you )
$endgroup$
– Atmos
Jan 26 at 19:05
$begingroup$
What do you need us to explain ? (except doing the entire proof for you )
$endgroup$
– Atmos
Jan 26 at 19:05
$begingroup$
I'm learning math all on my own. If you think it useful to do the proof for me in order for me to understand and and continue with my learning, then, by all means, do it. My resources are very limited otherwise...
$endgroup$
– Daniel Oscar
Jan 26 at 19:08
$begingroup$
I'm learning math all on my own. If you think it useful to do the proof for me in order for me to understand and and continue with my learning, then, by all means, do it. My resources are very limited otherwise...
$endgroup$
– Daniel Oscar
Jan 26 at 19:08
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Well as requested, i will detail a little more the sketch of the proof.
Induction proof is like domino's : you need to check that the first domino falls ( true for $n=0$ here ) and then you have to check that the $n-$th domino makes the $n+1-$th fall.
Here : check the case $n=0$.
You have
$$
cosleft(x+0.frac{pi}{2}right)=cosleft(xright)=f^{left(0right)}left(xright)
$$
So the given relation is true. Now suppose that it is true until a certain rank $n$, then you need to consider $f^{left(n+1right)}(x)$. You have
$$
f^{left(n+1right)}(x)=left(f^{left(nright)}left(xright)right)'=left(cosleft(x+nfrac{pi}{2}right)right)'=-sinleft(x+nfrac{pi}{2}right)=cosleft(x+frac{pi}{2}+nfrac{pi}{2}right)
$$
The last equality comes from $displaystyle cosleft(x+frac{pi}{2}right)=-sinleft(xright)$ ( check a trigonometric circle to see this ).
Hence we've shown
$$
f^{left(n+1right)}(x)=cosleft(x+left(n+1right)frac{pi}{2}right)
$$
The relation is hence TRUE at rank $n+1$.
By induction the relation is true for ALL $n$.
answered Jan 26 at 19:16
AtmosAtmos
4,830420
$endgroup$
add a comment |
$begingroup$
Here is a proof that doesn't use a recurrence.
I give it because it can be rather appealing under the condition to have some practice with complex numbers.
Let us start from the famous De Moivre formula :
$$cos(x)+i sin(x)=e^{ix} tag{1}$$
$n$ times differentiation of both sides gives :
$$cos^{(n)}(x)+i sin^{(n)}(x)=i^n e^{ix}tag{2}$$
The RHS of (2) can be transformed
$$cos^{(n)}(x)+i sin^{(n)}(x)=e^{itfrac{pi}{2}n}e^{ix}=e^{i(x+tfrac{pi}{2}n)}tag{3}$$
Using anew De Moivre formula, we get :
$$cos^{(n)}(x)+i sin^{(n)}(x)=cos(x+tfrac{pi}{2}n)+isin(x+tfrac{pi}{2}n)tag{4}$$
Taking the real part of both sides terminates the proof.
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
$endgroup$
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
add a comment |
$begingroup$
Hint:
You just have to know that the first derivative, $f'(x)=-sin x$, can be written as
$$f'(x)=sinbigl(x+tfracpi 2bigr)$$
from a well-known trigonometry formula, then an easy induction will complete the proof.
answered Jan 26 at 19:16
BernardBernard
123k741117
$endgroup$
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%2f3088624%2fprove-by-induction-that-the-n-th-derivative-of-cosx-is-given-by-cos-lef%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well as requested, i will detail a little more the sketch of the proof.
Induction proof is like domino's : you need to check that the first domino falls ( true for $n=0$ here ) and then you have to check that the $n-$th domino makes the $n+1-$th fall.
Here : check the case $n=0$.
You have
$$
cosleft(x+0.frac{pi}{2}right)=cosleft(xright)=f^{left(0right)}left(xright)
$$
So the given relation is true. Now suppose that it is true until a certain rank $n$, then you need to consider $f^{left(n+1right)}(x)$. You have
$$
f^{left(n+1right)}(x)=left(f^{left(nright)}left(xright)right)'=left(cosleft(x+nfrac{pi}{2}right)right)'=-sinleft(x+nfrac{pi}{2}right)=cosleft(x+frac{pi}{2}+nfrac{pi}{2}right)
$$
The last equality comes from $displaystyle cosleft(x+frac{pi}{2}right)=-sinleft(xright)$ ( check a trigonometric circle to see this ).
Hence we've shown
$$
f^{left(n+1right)}(x)=cosleft(x+left(n+1right)frac{pi}{2}right)
$$
The relation is hence TRUE at rank $n+1$.
By induction the relation is true for ALL $n$.
answered Jan 26 at 19:16
AtmosAtmos
4,830420
$endgroup$
add a comment |
$begingroup$
Well as requested, i will detail a little more the sketch of the proof.
Induction proof is like domino's : you need to check that the first domino falls ( true for $n=0$ here ) and then you have to check that the $n-$th domino makes the $n+1-$th fall.
Here : check the case $n=0$.
You have
$$
cosleft(x+0.frac{pi}{2}right)=cosleft(xright)=f^{left(0right)}left(xright)
$$
So the given relation is true. Now suppose that it is true until a certain rank $n$, then you need to consider $f^{left(n+1right)}(x)$. You have
$$
f^{left(n+1right)}(x)=left(f^{left(nright)}left(xright)right)'=left(cosleft(x+nfrac{pi}{2}right)right)'=-sinleft(x+nfrac{pi}{2}right)=cosleft(x+frac{pi}{2}+nfrac{pi}{2}right)
$$
The last equality comes from $displaystyle cosleft(x+frac{pi}{2}right)=-sinleft(xright)$ ( check a trigonometric circle to see this ).
Hence we've shown
$$
f^{left(n+1right)}(x)=cosleft(x+left(n+1right)frac{pi}{2}right)
$$
The relation is hence TRUE at rank $n+1$.
By induction the relation is true for ALL $n$.
answered Jan 26 at 19:16
AtmosAtmos
4,830420
$endgroup$
add a comment |
$begingroup$
Well as requested, i will detail a little more the sketch of the proof.
Induction proof is like domino's : you need to check that the first domino falls ( true for $n=0$ here ) and then you have to check that the $n-$th domino makes the $n+1-$th fall.
Here : check the case $n=0$.
You have
$$
cosleft(x+0.frac{pi}{2}right)=cosleft(xright)=f^{left(0right)}left(xright)
$$
So the given relation is true. Now suppose that it is true until a certain rank $n$, then you need to consider $f^{left(n+1right)}(x)$. You have
$$
f^{left(n+1right)}(x)=left(f^{left(nright)}left(xright)right)'=left(cosleft(x+nfrac{pi}{2}right)right)'=-sinleft(x+nfrac{pi}{2}right)=cosleft(x+frac{pi}{2}+nfrac{pi}{2}right)
$$
The last equality comes from $displaystyle cosleft(x+frac{pi}{2}right)=-sinleft(xright)$ ( check a trigonometric circle to see this ).
Hence we've shown
$$
f^{left(n+1right)}(x)=cosleft(x+left(n+1right)frac{pi}{2}right)
$$
The relation is hence TRUE at rank $n+1$.
By induction the relation is true for ALL $n$.
answered Jan 26 at 19:16
AtmosAtmos
4,830420
$endgroup$
Well as requested, i will detail a little more the sketch of the proof.
Induction proof is like domino's : you need to check that the first domino falls ( true for $n=0$ here ) and then you have to check that the $n-$th domino makes the $n+1-$th fall.
Here : check the case $n=0$.
You have
$$
cosleft(x+0.frac{pi}{2}right)=cosleft(xright)=f^{left(0right)}left(xright)
$$
So the given relation is true. Now suppose that it is true until a certain rank $n$, then you need to consider $f^{left(n+1right)}(x)$. You have
$$
f^{left(n+1right)}(x)=left(f^{left(nright)}left(xright)right)'=left(cosleft(x+nfrac{pi}{2}right)right)'=-sinleft(x+nfrac{pi}{2}right)=cosleft(x+frac{pi}{2}+nfrac{pi}{2}right)
$$
The last equality comes from $displaystyle cosleft(x+frac{pi}{2}right)=-sinleft(xright)$ ( check a trigonometric circle to see this ).
Hence we've shown
$$
f^{left(n+1right)}(x)=cosleft(x+left(n+1right)frac{pi}{2}right)
$$
The relation is hence TRUE at rank $n+1$.
By induction the relation is true for ALL $n$.
answered Jan 26 at 19:16
AtmosAtmos
4,830420
answered Jan 26 at 19:16
AtmosAtmos
4,830420
answered Jan 26 at 19:16
AtmosAtmos
4,830420
answered Jan 26 at 19:16
AtmosAtmos
4,830420
4,830420
add a comment |
add a comment |
$begingroup$
Here is a proof that doesn't use a recurrence.
I give it because it can be rather appealing under the condition to have some practice with complex numbers.
Let us start from the famous De Moivre formula :
$$cos(x)+i sin(x)=e^{ix} tag{1}$$
$n$ times differentiation of both sides gives :
$$cos^{(n)}(x)+i sin^{(n)}(x)=i^n e^{ix}tag{2}$$
The RHS of (2) can be transformed
$$cos^{(n)}(x)+i sin^{(n)}(x)=e^{itfrac{pi}{2}n}e^{ix}=e^{i(x+tfrac{pi}{2}n)}tag{3}$$
Using anew De Moivre formula, we get :
$$cos^{(n)}(x)+i sin^{(n)}(x)=cos(x+tfrac{pi}{2}n)+isin(x+tfrac{pi}{2}n)tag{4}$$
Taking the real part of both sides terminates the proof.
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
$endgroup$
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
add a comment |
$begingroup$
Here is a proof that doesn't use a recurrence.
I give it because it can be rather appealing under the condition to have some practice with complex numbers.
Let us start from the famous De Moivre formula :
$$cos(x)+i sin(x)=e^{ix} tag{1}$$
$n$ times differentiation of both sides gives :
$$cos^{(n)}(x)+i sin^{(n)}(x)=i^n e^{ix}tag{2}$$
The RHS of (2) can be transformed
$$cos^{(n)}(x)+i sin^{(n)}(x)=e^{itfrac{pi}{2}n}e^{ix}=e^{i(x+tfrac{pi}{2}n)}tag{3}$$
Using anew De Moivre formula, we get :
$$cos^{(n)}(x)+i sin^{(n)}(x)=cos(x+tfrac{pi}{2}n)+isin(x+tfrac{pi}{2}n)tag{4}$$
Taking the real part of both sides terminates the proof.
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
$endgroup$
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
add a comment |
$begingroup$
Here is a proof that doesn't use a recurrence.
I give it because it can be rather appealing under the condition to have some practice with complex numbers.
Let us start from the famous De Moivre formula :
$$cos(x)+i sin(x)=e^{ix} tag{1}$$
$n$ times differentiation of both sides gives :
$$cos^{(n)}(x)+i sin^{(n)}(x)=i^n e^{ix}tag{2}$$
The RHS of (2) can be transformed
$$cos^{(n)}(x)+i sin^{(n)}(x)=e^{itfrac{pi}{2}n}e^{ix}=e^{i(x+tfrac{pi}{2}n)}tag{3}$$
Using anew De Moivre formula, we get :
$$cos^{(n)}(x)+i sin^{(n)}(x)=cos(x+tfrac{pi}{2}n)+isin(x+tfrac{pi}{2}n)tag{4}$$
Taking the real part of both sides terminates the proof.
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
$endgroup$
Here is a proof that doesn't use a recurrence.
I give it because it can be rather appealing under the condition to have some practice with complex numbers.
Let us start from the famous De Moivre formula :
$$cos(x)+i sin(x)=e^{ix} tag{1}$$
$n$ times differentiation of both sides gives :
$$cos^{(n)}(x)+i sin^{(n)}(x)=i^n e^{ix}tag{2}$$
The RHS of (2) can be transformed
$$cos^{(n)}(x)+i sin^{(n)}(x)=e^{itfrac{pi}{2}n}e^{ix}=e^{i(x+tfrac{pi}{2}n)}tag{3}$$
Using anew De Moivre formula, we get :
$$cos^{(n)}(x)+i sin^{(n)}(x)=cos(x+tfrac{pi}{2}n)+isin(x+tfrac{pi}{2}n)tag{4}$$
Taking the real part of both sides terminates the proof.
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
answered Jan 26 at 22:12
Jean MarieJean Marie
30.9k42155
30.9k42155
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
add a comment |
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
$begingroup$
This is the sort of thing that made me love mathematics!
$endgroup$
– Daniel Oscar
Jan 27 at 14:25
add a comment |
$begingroup$
Hint:
You just have to know that the first derivative, $f'(x)=-sin x$, can be written as
$$f'(x)=sinbigl(x+tfracpi 2bigr)$$
from a well-known trigonometry formula, then an easy induction will complete the proof.
answered Jan 26 at 19:16
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
You just have to know that the first derivative, $f'(x)=-sin x$, can be written as
$$f'(x)=sinbigl(x+tfracpi 2bigr)$$
from a well-known trigonometry formula, then an easy induction will complete the proof.
answered Jan 26 at 19:16
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
You just have to know that the first derivative, $f'(x)=-sin x$, can be written as
$$f'(x)=sinbigl(x+tfracpi 2bigr)$$
from a well-known trigonometry formula, then an easy induction will complete the proof.
answered Jan 26 at 19:16
BernardBernard
123k741117
$endgroup$
Hint:
You just have to know that the first derivative, $f'(x)=-sin x$, can be written as
$$f'(x)=sinbigl(x+tfracpi 2bigr)$$
from a well-known trigonometry formula, then an easy induction will complete the proof.
answered Jan 26 at 19:16
BernardBernard
123k741117
answered Jan 26 at 19:16
BernardBernard
123k741117
answered Jan 26 at 19:16
BernardBernard
123k741117
answered Jan 26 at 19:16
BernardBernard
123k741117
123k741117
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%2f3088624%2fprove-by-induction-that-the-n-th-derivative-of-cosx-is-given-by-cos-lef%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
1
$begingroup$
What do you need us to explain ? (except doing the entire proof for you )
$endgroup$
– Atmos
Jan 26 at 19:05
$begingroup$
I'm learning math all on my own. If you think it useful to do the proof for me in order for me to understand and and continue with my learning, then, by all means, do it. My resources are very limited otherwise...
$endgroup$
– Daniel Oscar
Jan 26 at 19:08