Mixing
Show that $lim_{hrightarrow 0}frac{e^{ih}-1}{h}=i$
$begingroup$
$hin mathbb{R}$, because we have defined the Trigonometric Functions only on $mathbb{R}$ so far.
I have a look at $e^{ih}=sum_{k=0}^{infty}frac{(ih)^k}{k!}=1+ih-frac{h^2}{2}+....$
How can one describe the nth term of the sum?
Then I look at $frac{e^{ih}-1}{h}=frac{(1-1)}{h}+i-frac{h}{2}+...=i-frac{h}{2}+....$
Again how can I describe that the nth term of the sum?
Because $frac{e^{ih}-1}{h}=sum_{k=1}^{infty}frac{frac{(ih)^k}{h}}{k!}<sum_{k=0}^{infty}frac{frac{(ih)^k}{h}}{k!}=sum_{k=1}^{infty}frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$
and $ih^{-1}$ is a complex number and the exponential-series converges absolutely for all Elements in $mathbb{C}$, I have found a convergent majorant. And I can apply the properties of Limits on $frac{e^{ih}-1}{h}forall, hin mathbb{R}$.
How can I now prove formally (i.e by chosing an explicit $delta$) that
$$forall_{epsilon>0}exists_{delta>0}forall_{hinmathbb{R}}|h-0|=|h|<deltaLongrightarrow |(frac{e^{ih}-1}{h}=i-frac{h}{2}+...)-i|<epsilon$$
I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit $delta$ ).
sequences-and-series limits
asked Jan 23 at 13:36
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
$hin mathbb{R}$, because we have defined the Trigonometric Functions only on $mathbb{R}$ so far.
I have a look at $e^{ih}=sum_{k=0}^{infty}frac{(ih)^k}{k!}=1+ih-frac{h^2}{2}+....$
How can one describe the nth term of the sum?
Then I look at $frac{e^{ih}-1}{h}=frac{(1-1)}{h}+i-frac{h}{2}+...=i-frac{h}{2}+....$
Again how can I describe that the nth term of the sum?
Because $frac{e^{ih}-1}{h}=sum_{k=1}^{infty}frac{frac{(ih)^k}{h}}{k!}<sum_{k=0}^{infty}frac{frac{(ih)^k}{h}}{k!}=sum_{k=1}^{infty}frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$
and $ih^{-1}$ is a complex number and the exponential-series converges absolutely for all Elements in $mathbb{C}$, I have found a convergent majorant. And I can apply the properties of Limits on $frac{e^{ih}-1}{h}forall, hin mathbb{R}$.
How can I now prove formally (i.e by chosing an explicit $delta$) that
$$forall_{epsilon>0}exists_{delta>0}forall_{hinmathbb{R}}|h-0|=|h|<deltaLongrightarrow |(frac{e^{ih}-1}{h}=i-frac{h}{2}+...)-i|<epsilon$$
I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit $delta$ ).
sequences-and-series limits
asked Jan 23 at 13:36
RM777RM777
38312
$endgroup$
2
$begingroup$
(1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $mathbb C$
$endgroup$
– Calvin Khor
Jan 23 at 13:42
$begingroup$
@CalvinKhor On (2), If I would put $|cdot |$ around the sums would that work?
$endgroup$
– RM777
Jan 23 at 13:47
1
$begingroup$
You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|cdot|$s everywhere)
$endgroup$
– Calvin Khor
Jan 23 at 13:49
add a comment |
$begingroup$
$hin mathbb{R}$, because we have defined the Trigonometric Functions only on $mathbb{R}$ so far.
I have a look at $e^{ih}=sum_{k=0}^{infty}frac{(ih)^k}{k!}=1+ih-frac{h^2}{2}+....$
How can one describe the nth term of the sum?
Then I look at $frac{e^{ih}-1}{h}=frac{(1-1)}{h}+i-frac{h}{2}+...=i-frac{h}{2}+....$
Again how can I describe that the nth term of the sum?
Because $frac{e^{ih}-1}{h}=sum_{k=1}^{infty}frac{frac{(ih)^k}{h}}{k!}<sum_{k=0}^{infty}frac{frac{(ih)^k}{h}}{k!}=sum_{k=1}^{infty}frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$
and $ih^{-1}$ is a complex number and the exponential-series converges absolutely for all Elements in $mathbb{C}$, I have found a convergent majorant. And I can apply the properties of Limits on $frac{e^{ih}-1}{h}forall, hin mathbb{R}$.
How can I now prove formally (i.e by chosing an explicit $delta$) that
$$forall_{epsilon>0}exists_{delta>0}forall_{hinmathbb{R}}|h-0|=|h|<deltaLongrightarrow |(frac{e^{ih}-1}{h}=i-frac{h}{2}+...)-i|<epsilon$$
I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit $delta$ ).
sequences-and-series limits
asked Jan 23 at 13:36
RM777RM777
38312
$endgroup$
$hin mathbb{R}$, because we have defined the Trigonometric Functions only on $mathbb{R}$ so far.
I have a look at $e^{ih}=sum_{k=0}^{infty}frac{(ih)^k}{k!}=1+ih-frac{h^2}{2}+....$
How can one describe the nth term of the sum?
Then I look at $frac{e^{ih}-1}{h}=frac{(1-1)}{h}+i-frac{h}{2}+...=i-frac{h}{2}+....$
Again how can I describe that the nth term of the sum?
Because $frac{e^{ih}-1}{h}=sum_{k=1}^{infty}frac{frac{(ih)^k}{h}}{k!}<sum_{k=0}^{infty}frac{frac{(ih)^k}{h}}{k!}=sum_{k=1}^{infty}frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$
and $ih^{-1}$ is a complex number and the exponential-series converges absolutely for all Elements in $mathbb{C}$, I have found a convergent majorant. And I can apply the properties of Limits on $frac{e^{ih}-1}{h}forall, hin mathbb{R}$.
How can I now prove formally (i.e by chosing an explicit $delta$) that
$$forall_{epsilon>0}exists_{delta>0}forall_{hinmathbb{R}}|h-0|=|h|<deltaLongrightarrow |(frac{e^{ih}-1}{h}=i-frac{h}{2}+...)-i|<epsilon$$
I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit $delta$ ).
sequences-and-series limits
sequences-and-series limits
asked Jan 23 at 13:36
RM777RM777
38312
asked Jan 23 at 13:36
RM777RM777
38312
asked Jan 23 at 13:36
RM777RM777
38312
asked Jan 23 at 13:36
RM777RM777
38312
asked Jan 23 at 13:36
RM777RM777
38312
38312
2
$begingroup$
(1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $mathbb C$
$endgroup$
– Calvin Khor
Jan 23 at 13:42
$begingroup$
@CalvinKhor On (2), If I would put $|cdot |$ around the sums would that work?
$endgroup$
– RM777
Jan 23 at 13:47
1
$begingroup$
You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|cdot|$s everywhere)
$endgroup$
– Calvin Khor
Jan 23 at 13:49
add a comment |
2
$begingroup$
(1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $mathbb C$
$endgroup$
– Calvin Khor
Jan 23 at 13:42
$begingroup$
@CalvinKhor On (2), If I would put $|cdot |$ around the sums would that work?
$endgroup$
– RM777
Jan 23 at 13:47
1
$begingroup$
You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|cdot|$s everywhere)
$endgroup$
– Calvin Khor
Jan 23 at 13:49
2
2
$begingroup$
(1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $mathbb C$
$endgroup$
– Calvin Khor
Jan 23 at 13:42
$begingroup$
(1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $mathbb C$
$endgroup$
– Calvin Khor
Jan 23 at 13:42
$begingroup$
@CalvinKhor On (2), If I would put $|cdot |$ around the sums would that work?
$endgroup$
– RM777
Jan 23 at 13:47
$begingroup$
@CalvinKhor On (2), If I would put $|cdot |$ around the sums would that work?
$endgroup$
– RM777
Jan 23 at 13:47
1
1
$begingroup$
You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|cdot|$s everywhere)
$endgroup$
– Calvin Khor
Jan 23 at 13:49
$begingroup$
You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|cdot|$s everywhere)
$endgroup$
– Calvin Khor
Jan 23 at 13:49
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:
Theorem. Suppose the continuous functions $f_n=f_n(h)$ taking values in $mathbb C$ are such that $sum_{n=0}^infty f_n(h) := lim_{Ntoinfty} sum_{n=0}^N f_n(h)$ converges absolutely uniformly on some interval $hin [-a,a]$ (i.e. $sum_{n=0}^infty |f_n|_{infty} < infty $). Then
$$ lim_{hto 0} sum_{n=0}^infty f_n(h) = sum_{n=0}^infty f_n(0)$$
Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $N$th partial sums $F_N(h) := sum_{n=0}^N f_n(h)$ are continuous, and they converge pointwise on $[-a,a]$ to $F(h):=sum_{n=0}^infty f_n(h)$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $[-a,a]$. Hence, $F$ is continuous at all points in $[-a,a]$, and in particular at $0$. The result follows.
Application: $$frac{e^{ih}-1}{h}\= frac{(sum_{n=0}^infty (ih)^n/n! )- 1} {h} \= frac{sum_{n=1}^infty (ih)^n/n! } {h} \= sum_{n=1}^infty frac{i(ih)^{n-1}}{n!} \= sum_{m=0}^infty frac{i(ih)^{m}}{(m+1)!} $$
After you verify we can apply the theorem, this yields $lim_{hto 0 } frac{e^{ih}-1}{h} = i+0+0+dots = i$.
PS make sure you realise that all infinite sums are defined in terms of limits in $mathbb C$ (e.g. $a_n to a inmathbb C$ iff $|a_n - a| to 0$)
PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
1
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
|
show 1 more comment
$begingroup$
Hint: Use Euler's formula and split the limit into well known trigonometric limits.
answered Jan 23 at 13:51
JamJam
5,00821431
$endgroup$
add a comment |
$begingroup$
Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.
To reiterate the question for clarity we have $hinmathbb{R} $ and the symbol $e^{ih} $ is defined by the series $sum_{n=0}^{infty} (ih) ^n/n! $. Let's assume $|h|< 1$ and observe that
begin{align*}
left|frac {e^{ih} - 1}{h}-iright|&=left|frac{i^2h}{2!}+frac{i^3h^2}{3!}+dotsright|\
&leqfrac{|h|}{2!}+frac{|h|^2}{3!}+dots\
&leqfrac{|h|} {2}+frac{|h|^2}{2^2}+frac{|h|^3}{2^3}+dots\
&=frac{|h|}{2-|h|}text{ (sum of infinite GP)} \
&<|h|
end{align*}
Thus given any $epsilon>0$ if we choose $delta=min(1,epsilon) $ then we have $$left|frac{e^{ih} - 1}{h}-iright|<epsilon $$ whenever $0<|h|<delta$. Therefore by definition of limit we have $$lim_{hto 0}dfrac{e^{ih}-1}{h}=i$$
Hardy uses the above limit to prove the formula $$e^{ix} =cos x+isin x ,forall xinmathbb {R} $$ The idea is to show that the function $f(x) =e^{ix} $ is differentiable with derivative $f'(x) =if(x) $ and then consider the function $$g(x) =(cos x-isin x) f(x) $$ It is easily proved that $g'(x) =0$ and hence $g$ is constant. Thus $g(x) =g(0)=1$ and $f(x) =cos x+isin x$.
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
$begingroup$
Show that $lim_{hrightarrow 0}frac{e^{ih}-1}{h}=i$:
As we know: $ cos theta + i sin theta = e^{i theta}$
So, $$lim limits_{hrightarrow 0} [frac { cos h + i sin h - 1 } {h} ] = lim limits_{h rightarrow 0} [ frac {1 - frac{h^2}{2!} + phi_1 (h^4) + i( h - frac{h^3}{3!}... phi_2(h^5) )-1} {h} ] = lim limits_{h rightarrow0} frac { i(1 - higher space powersspace ofspace h) }{1} = i$$
Alternatively,
$$e^{ih} = 1 + ih + frac{-h^2}{2!}...$$ will give the same result, just take terms of $i$ to one side and none $i$ ones to other, then take $i$ common and cancel 1 from numerator, then take $h$ common and cancel it from denominator, you'll get the desired result!
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
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%2f3084479%2fshow-that-lim-h-rightarrow-0-fraceih-1h-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:
Theorem. Suppose the continuous functions $f_n=f_n(h)$ taking values in $mathbb C$ are such that $sum_{n=0}^infty f_n(h) := lim_{Ntoinfty} sum_{n=0}^N f_n(h)$ converges absolutely uniformly on some interval $hin [-a,a]$ (i.e. $sum_{n=0}^infty |f_n|_{infty} < infty $). Then
$$ lim_{hto 0} sum_{n=0}^infty f_n(h) = sum_{n=0}^infty f_n(0)$$
Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $N$th partial sums $F_N(h) := sum_{n=0}^N f_n(h)$ are continuous, and they converge pointwise on $[-a,a]$ to $F(h):=sum_{n=0}^infty f_n(h)$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $[-a,a]$. Hence, $F$ is continuous at all points in $[-a,a]$, and in particular at $0$. The result follows.
Application: $$frac{e^{ih}-1}{h}\= frac{(sum_{n=0}^infty (ih)^n/n! )- 1} {h} \= frac{sum_{n=1}^infty (ih)^n/n! } {h} \= sum_{n=1}^infty frac{i(ih)^{n-1}}{n!} \= sum_{m=0}^infty frac{i(ih)^{m}}{(m+1)!} $$
After you verify we can apply the theorem, this yields $lim_{hto 0 } frac{e^{ih}-1}{h} = i+0+0+dots = i$.
PS make sure you realise that all infinite sums are defined in terms of limits in $mathbb C$ (e.g. $a_n to a inmathbb C$ iff $|a_n - a| to 0$)
PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
1
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
|
show 1 more comment
$begingroup$
Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:
Theorem. Suppose the continuous functions $f_n=f_n(h)$ taking values in $mathbb C$ are such that $sum_{n=0}^infty f_n(h) := lim_{Ntoinfty} sum_{n=0}^N f_n(h)$ converges absolutely uniformly on some interval $hin [-a,a]$ (i.e. $sum_{n=0}^infty |f_n|_{infty} < infty $). Then
$$ lim_{hto 0} sum_{n=0}^infty f_n(h) = sum_{n=0}^infty f_n(0)$$
Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $N$th partial sums $F_N(h) := sum_{n=0}^N f_n(h)$ are continuous, and they converge pointwise on $[-a,a]$ to $F(h):=sum_{n=0}^infty f_n(h)$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $[-a,a]$. Hence, $F$ is continuous at all points in $[-a,a]$, and in particular at $0$. The result follows.
Application: $$frac{e^{ih}-1}{h}\= frac{(sum_{n=0}^infty (ih)^n/n! )- 1} {h} \= frac{sum_{n=1}^infty (ih)^n/n! } {h} \= sum_{n=1}^infty frac{i(ih)^{n-1}}{n!} \= sum_{m=0}^infty frac{i(ih)^{m}}{(m+1)!} $$
After you verify we can apply the theorem, this yields $lim_{hto 0 } frac{e^{ih}-1}{h} = i+0+0+dots = i$.
PS make sure you realise that all infinite sums are defined in terms of limits in $mathbb C$ (e.g. $a_n to a inmathbb C$ iff $|a_n - a| to 0$)
PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
1
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
|
show 1 more comment
$begingroup$
Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:
Theorem. Suppose the continuous functions $f_n=f_n(h)$ taking values in $mathbb C$ are such that $sum_{n=0}^infty f_n(h) := lim_{Ntoinfty} sum_{n=0}^N f_n(h)$ converges absolutely uniformly on some interval $hin [-a,a]$ (i.e. $sum_{n=0}^infty |f_n|_{infty} < infty $). Then
$$ lim_{hto 0} sum_{n=0}^infty f_n(h) = sum_{n=0}^infty f_n(0)$$
Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $N$th partial sums $F_N(h) := sum_{n=0}^N f_n(h)$ are continuous, and they converge pointwise on $[-a,a]$ to $F(h):=sum_{n=0}^infty f_n(h)$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $[-a,a]$. Hence, $F$ is continuous at all points in $[-a,a]$, and in particular at $0$. The result follows.
Application: $$frac{e^{ih}-1}{h}\= frac{(sum_{n=0}^infty (ih)^n/n! )- 1} {h} \= frac{sum_{n=1}^infty (ih)^n/n! } {h} \= sum_{n=1}^infty frac{i(ih)^{n-1}}{n!} \= sum_{m=0}^infty frac{i(ih)^{m}}{(m+1)!} $$
After you verify we can apply the theorem, this yields $lim_{hto 0 } frac{e^{ih}-1}{h} = i+0+0+dots = i$.
PS make sure you realise that all infinite sums are defined in terms of limits in $mathbb C$ (e.g. $a_n to a inmathbb C$ iff $|a_n - a| to 0$)
PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
$endgroup$
Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:
Theorem. Suppose the continuous functions $f_n=f_n(h)$ taking values in $mathbb C$ are such that $sum_{n=0}^infty f_n(h) := lim_{Ntoinfty} sum_{n=0}^N f_n(h)$ converges absolutely uniformly on some interval $hin [-a,a]$ (i.e. $sum_{n=0}^infty |f_n|_{infty} < infty $). Then
$$ lim_{hto 0} sum_{n=0}^infty f_n(h) = sum_{n=0}^infty f_n(0)$$
Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $N$th partial sums $F_N(h) := sum_{n=0}^N f_n(h)$ are continuous, and they converge pointwise on $[-a,a]$ to $F(h):=sum_{n=0}^infty f_n(h)$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $[-a,a]$. Hence, $F$ is continuous at all points in $[-a,a]$, and in particular at $0$. The result follows.
Application: $$frac{e^{ih}-1}{h}\= frac{(sum_{n=0}^infty (ih)^n/n! )- 1} {h} \= frac{sum_{n=1}^infty (ih)^n/n! } {h} \= sum_{n=1}^infty frac{i(ih)^{n-1}}{n!} \= sum_{m=0}^infty frac{i(ih)^{m}}{(m+1)!} $$
After you verify we can apply the theorem, this yields $lim_{hto 0 } frac{e^{ih}-1}{h} = i+0+0+dots = i$.
PS make sure you realise that all infinite sums are defined in terms of limits in $mathbb C$ (e.g. $a_n to a inmathbb C$ iff $|a_n - a| to 0$)
PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
edited Jan 23 at 22:14
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
answered Jan 23 at 14:11
Calvin KhorCalvin Khor
12.4k21439
12.4k21439
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
1
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
|
show 1 more comment
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
1
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
Nice answer, Calvin. I was wondering if you had a source or proof of the theorem?
$endgroup$
– Jam
Jan 23 at 14:15
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
@Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help?
$endgroup$
– Calvin Khor
Jan 23 at 14:16
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
It does. Thank you
$endgroup$
– Jam
Jan 23 at 14:17
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
$begingroup$
@Jam you're welcome
$endgroup$
– Calvin Khor
Jan 23 at 14:21
1
1
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
$begingroup$
@RM777 I have added the proof to the answer, as in the comment to Jam
$endgroup$
– Calvin Khor
Jan 23 at 22:11
|
show 1 more comment
$begingroup$
Hint: Use Euler's formula and split the limit into well known trigonometric limits.
answered Jan 23 at 13:51
JamJam
5,00821431
$endgroup$
add a comment |
$begingroup$
Hint: Use Euler's formula and split the limit into well known trigonometric limits.
answered Jan 23 at 13:51
JamJam
5,00821431
$endgroup$
add a comment |
$begingroup$
Hint: Use Euler's formula and split the limit into well known trigonometric limits.
answered Jan 23 at 13:51
JamJam
5,00821431
$endgroup$
Hint: Use Euler's formula and split the limit into well known trigonometric limits.
answered Jan 23 at 13:51
JamJam
5,00821431
answered Jan 23 at 13:51
JamJam
5,00821431
answered Jan 23 at 13:51
JamJam
5,00821431
answered Jan 23 at 13:51
JamJam
5,00821431
5,00821431
add a comment |
add a comment |
$begingroup$
Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.
To reiterate the question for clarity we have $hinmathbb{R} $ and the symbol $e^{ih} $ is defined by the series $sum_{n=0}^{infty} (ih) ^n/n! $. Let's assume $|h|< 1$ and observe that
begin{align*}
left|frac {e^{ih} - 1}{h}-iright|&=left|frac{i^2h}{2!}+frac{i^3h^2}{3!}+dotsright|\
&leqfrac{|h|}{2!}+frac{|h|^2}{3!}+dots\
&leqfrac{|h|} {2}+frac{|h|^2}{2^2}+frac{|h|^3}{2^3}+dots\
&=frac{|h|}{2-|h|}text{ (sum of infinite GP)} \
&<|h|
end{align*}
Thus given any $epsilon>0$ if we choose $delta=min(1,epsilon) $ then we have $$left|frac{e^{ih} - 1}{h}-iright|<epsilon $$ whenever $0<|h|<delta$. Therefore by definition of limit we have $$lim_{hto 0}dfrac{e^{ih}-1}{h}=i$$
Hardy uses the above limit to prove the formula $$e^{ix} =cos x+isin x ,forall xinmathbb {R} $$ The idea is to show that the function $f(x) =e^{ix} $ is differentiable with derivative $f'(x) =if(x) $ and then consider the function $$g(x) =(cos x-isin x) f(x) $$ It is easily proved that $g'(x) =0$ and hence $g$ is constant. Thus $g(x) =g(0)=1$ and $f(x) =cos x+isin x$.
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
$begingroup$
Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.
To reiterate the question for clarity we have $hinmathbb{R} $ and the symbol $e^{ih} $ is defined by the series $sum_{n=0}^{infty} (ih) ^n/n! $. Let's assume $|h|< 1$ and observe that
begin{align*}
left|frac {e^{ih} - 1}{h}-iright|&=left|frac{i^2h}{2!}+frac{i^3h^2}{3!}+dotsright|\
&leqfrac{|h|}{2!}+frac{|h|^2}{3!}+dots\
&leqfrac{|h|} {2}+frac{|h|^2}{2^2}+frac{|h|^3}{2^3}+dots\
&=frac{|h|}{2-|h|}text{ (sum of infinite GP)} \
&<|h|
end{align*}
Thus given any $epsilon>0$ if we choose $delta=min(1,epsilon) $ then we have $$left|frac{e^{ih} - 1}{h}-iright|<epsilon $$ whenever $0<|h|<delta$. Therefore by definition of limit we have $$lim_{hto 0}dfrac{e^{ih}-1}{h}=i$$
Hardy uses the above limit to prove the formula $$e^{ix} =cos x+isin x ,forall xinmathbb {R} $$ The idea is to show that the function $f(x) =e^{ix} $ is differentiable with derivative $f'(x) =if(x) $ and then consider the function $$g(x) =(cos x-isin x) f(x) $$ It is easily proved that $g'(x) =0$ and hence $g$ is constant. Thus $g(x) =g(0)=1$ and $f(x) =cos x+isin x$.
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
add a comment |
$begingroup$
Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.
To reiterate the question for clarity we have $hinmathbb{R} $ and the symbol $e^{ih} $ is defined by the series $sum_{n=0}^{infty} (ih) ^n/n! $. Let's assume $|h|< 1$ and observe that
begin{align*}
left|frac {e^{ih} - 1}{h}-iright|&=left|frac{i^2h}{2!}+frac{i^3h^2}{3!}+dotsright|\
&leqfrac{|h|}{2!}+frac{|h|^2}{3!}+dots\
&leqfrac{|h|} {2}+frac{|h|^2}{2^2}+frac{|h|^3}{2^3}+dots\
&=frac{|h|}{2-|h|}text{ (sum of infinite GP)} \
&<|h|
end{align*}
Thus given any $epsilon>0$ if we choose $delta=min(1,epsilon) $ then we have $$left|frac{e^{ih} - 1}{h}-iright|<epsilon $$ whenever $0<|h|<delta$. Therefore by definition of limit we have $$lim_{hto 0}dfrac{e^{ih}-1}{h}=i$$
Hardy uses the above limit to prove the formula $$e^{ix} =cos x+isin x ,forall xinmathbb {R} $$ The idea is to show that the function $f(x) =e^{ix} $ is differentiable with derivative $f'(x) =if(x) $ and then consider the function $$g(x) =(cos x-isin x) f(x) $$ It is easily proved that $g'(x) =0$ and hence $g$ is constant. Thus $g(x) =g(0)=1$ and $f(x) =cos x+isin x$.
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
$endgroup$
Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.
To reiterate the question for clarity we have $hinmathbb{R} $ and the symbol $e^{ih} $ is defined by the series $sum_{n=0}^{infty} (ih) ^n/n! $. Let's assume $|h|< 1$ and observe that
begin{align*}
left|frac {e^{ih} - 1}{h}-iright|&=left|frac{i^2h}{2!}+frac{i^3h^2}{3!}+dotsright|\
&leqfrac{|h|}{2!}+frac{|h|^2}{3!}+dots\
&leqfrac{|h|} {2}+frac{|h|^2}{2^2}+frac{|h|^3}{2^3}+dots\
&=frac{|h|}{2-|h|}text{ (sum of infinite GP)} \
&<|h|
end{align*}
Thus given any $epsilon>0$ if we choose $delta=min(1,epsilon) $ then we have $$left|frac{e^{ih} - 1}{h}-iright|<epsilon $$ whenever $0<|h|<delta$. Therefore by definition of limit we have $$lim_{hto 0}dfrac{e^{ih}-1}{h}=i$$
Hardy uses the above limit to prove the formula $$e^{ix} =cos x+isin x ,forall xinmathbb {R} $$ The idea is to show that the function $f(x) =e^{ix} $ is differentiable with derivative $f'(x) =if(x) $ and then consider the function $$g(x) =(cos x-isin x) f(x) $$ It is easily proved that $g'(x) =0$ and hence $g$ is constant. Thus $g(x) =g(0)=1$ and $f(x) =cos x+isin x$.
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
edited Jan 24 at 3:16
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
answered Jan 24 at 2:31
Paramanand SinghParamanand Singh
50.7k557168
50.7k557168
add a comment |
add a comment |
$begingroup$
Show that $lim_{hrightarrow 0}frac{e^{ih}-1}{h}=i$:
As we know: $ cos theta + i sin theta = e^{i theta}$
So, $$lim limits_{hrightarrow 0} [frac { cos h + i sin h - 1 } {h} ] = lim limits_{h rightarrow 0} [ frac {1 - frac{h^2}{2!} + phi_1 (h^4) + i( h - frac{h^3}{3!}... phi_2(h^5) )-1} {h} ] = lim limits_{h rightarrow0} frac { i(1 - higher space powersspace ofspace h) }{1} = i$$
Alternatively,
$$e^{ih} = 1 + ih + frac{-h^2}{2!}...$$ will give the same result, just take terms of $i$ to one side and none $i$ ones to other, then take $i$ common and cancel 1 from numerator, then take $h$ common and cancel it from denominator, you'll get the desired result!
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
$begingroup$
Show that $lim_{hrightarrow 0}frac{e^{ih}-1}{h}=i$:
As we know: $ cos theta + i sin theta = e^{i theta}$
So, $$lim limits_{hrightarrow 0} [frac { cos h + i sin h - 1 } {h} ] = lim limits_{h rightarrow 0} [ frac {1 - frac{h^2}{2!} + phi_1 (h^4) + i( h - frac{h^3}{3!}... phi_2(h^5) )-1} {h} ] = lim limits_{h rightarrow0} frac { i(1 - higher space powersspace ofspace h) }{1} = i$$
Alternatively,
$$e^{ih} = 1 + ih + frac{-h^2}{2!}...$$ will give the same result, just take terms of $i$ to one side and none $i$ ones to other, then take $i$ common and cancel 1 from numerator, then take $h$ common and cancel it from denominator, you'll get the desired result!
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
add a comment |
$begingroup$
Show that $lim_{hrightarrow 0}frac{e^{ih}-1}{h}=i$:
As we know: $ cos theta + i sin theta = e^{i theta}$
So, $$lim limits_{hrightarrow 0} [frac { cos h + i sin h - 1 } {h} ] = lim limits_{h rightarrow 0} [ frac {1 - frac{h^2}{2!} + phi_1 (h^4) + i( h - frac{h^3}{3!}... phi_2(h^5) )-1} {h} ] = lim limits_{h rightarrow0} frac { i(1 - higher space powersspace ofspace h) }{1} = i$$
Alternatively,
$$e^{ih} = 1 + ih + frac{-h^2}{2!}...$$ will give the same result, just take terms of $i$ to one side and none $i$ ones to other, then take $i$ common and cancel 1 from numerator, then take $h$ common and cancel it from denominator, you'll get the desired result!
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
$endgroup$
Show that $lim_{hrightarrow 0}frac{e^{ih}-1}{h}=i$:
As we know: $ cos theta + i sin theta = e^{i theta}$
So, $$lim limits_{hrightarrow 0} [frac { cos h + i sin h - 1 } {h} ] = lim limits_{h rightarrow 0} [ frac {1 - frac{h^2}{2!} + phi_1 (h^4) + i( h - frac{h^3}{3!}... phi_2(h^5) )-1} {h} ] = lim limits_{h rightarrow0} frac { i(1 - higher space powersspace ofspace h) }{1} = i$$
Alternatively,
$$e^{ih} = 1 + ih + frac{-h^2}{2!}...$$ will give the same result, just take terms of $i$ to one side and none $i$ ones to other, then take $i$ common and cancel 1 from numerator, then take $h$ common and cancel it from denominator, you'll get the desired result!
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
edited Jan 27 at 12:54
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
answered Jan 23 at 15:15
Abhas Kumar SinhaAbhas Kumar Sinha
304115
304115
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%2f3084479%2fshow-that-lim-h-rightarrow-0-fraceih-1h-i%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
$begingroup$
(1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $mathbb C$
$endgroup$
– Calvin Khor
Jan 23 at 13:42
$begingroup$
@CalvinKhor On (2), If I would put $|cdot |$ around the sums would that work?
$endgroup$
– RM777
Jan 23 at 13:47
1
$begingroup$
You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|cdot|$s everywhere)
$endgroup$
– Calvin Khor
Jan 23 at 13:49