Expansion for $cos$ at $ntheta = C + fracC^324h^2 + frac3C^5640h^4 + o(h^4)$ as a polynomial in $h$

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We have $C>0, h>0$ fixed constants. And $cos theta = 1-frach^22C^2$, for $theta in [0,pi]$. Using Taylor series expansion,

$$theta = Ch + fracC^324h^3 + frac3C^5640h^5 + o(h^5).$$
Now taking $n=1/h$, we have
$$ntheta = C + fracC^324h^2 + frac3C^5640h^4 + o(h^4).$$

Then, we get

$$cos(ntheta) = cos C - fracC^324 sin Ch^2 - frac27sin C + 5C cos C5760 C^5h^4 + o(h^4).$$

How do we get this last expansion for $cos (ntheta)$? I would greatly appreciate any help.

calculus real-analysis taylor-expansion

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edited Aug 6 at 11:49

asked Aug 6 at 7:51

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We have $C>0, h>0$ fixed constants. And $cos theta = 1-frach^22C^2$, for $theta in [0,pi]$. Using Taylor series expansion,

$$theta = Ch + fracC^324h^3 + frac3C^5640h^5 + o(h^5).$$
Now taking $n=1/h$, we have
$$ntheta = C + fracC^324h^2 + frac3C^5640h^4 + o(h^4).$$

Then, we get

$$cos(ntheta) = cos C - fracC^324 sin Ch^2 - frac27sin C + 5C cos C5760 C^5h^4 + o(h^4).$$

How do we get this last expansion for $cos (ntheta)$? I would greatly appreciate any help.

calculus real-analysis taylor-expansion

share|cite|improve this question

edited Aug 6 at 11:49

asked Aug 6 at 7:51

takecare

2,25811431

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

We have $C>0, h>0$ fixed constants. And $cos theta = 1-frach^22C^2$, for $theta in [0,pi]$. Using Taylor series expansion,

$$theta = Ch + fracC^324h^3 + frac3C^5640h^5 + o(h^5).$$
Now taking $n=1/h$, we have
$$ntheta = C + fracC^324h^2 + frac3C^5640h^4 + o(h^4).$$

Then, we get

$$cos(ntheta) = cos C - fracC^324 sin Ch^2 - frac27sin C + 5C cos C5760 C^5h^4 + o(h^4).$$

How do we get this last expansion for $cos (ntheta)$? I would greatly appreciate any help.

calculus real-analysis taylor-expansion

share|cite|improve this question

edited Aug 6 at 11:49

asked Aug 6 at 7:51

takecare

2,25811431

We have $C>0, h>0$ fixed constants. And $cos theta = 1-frach^22C^2$, for $theta in [0,pi]$. Using Taylor series expansion,

$$theta = Ch + fracC^324h^3 + frac3C^5640h^5 + o(h^5).$$
Now taking $n=1/h$, we have
$$ntheta = C + fracC^324h^2 + frac3C^5640h^4 + o(h^4).$$

Then, we get

$$cos(ntheta) = cos C - fracC^324 sin Ch^2 - frac27sin C + 5C cos C5760 C^5h^4 + o(h^4).$$

How do we get this last expansion for $cos (ntheta)$? I would greatly appreciate any help.

calculus real-analysis taylor-expansion

share|cite|improve this question

edited Aug 6 at 11:49

asked Aug 6 at 7:51

takecare

2,25811431

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 11:49

edited Aug 6 at 11:49

edited Aug 6 at 11:49

asked Aug 6 at 7:51

takecare

2,25811431

asked Aug 6 at 7:51

takecare

2,25811431

asked Aug 6 at 7:51

takecare

2,25811431

2,25811431

add a comment | 

add a comment | 

1 Answer
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We are given that
$$ ntheta = C + D ,quad D := fracC^324h^2 + frac3C^5640h^4 + O(h^6). $$
Using the addition formula $ cos(x+y) = cos x cos y - sin x sin y $
we get the equation
$$ cos ntheta = cos C cos D - sin C sin D . $$
Now $ cos D = 1 - fracC^61152h^4 + O(h)^6 $ and
$ sin D = fracC^324h^2 + frac3C^5640h^4 + O(h)^6. $
Substituting we get
$$ cos ntheta = cos C Big( 1 - fracC^61152h^4 + O(h)^6 Big)
- sin C Big(fracC^324h^2 + frac3C^5640h^4 + O(h)^6 Big) $$
and your result follows.

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answered Aug 6 at 13:43

Somos

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

We are given that
$$ ntheta = C + D ,quad D := fracC^324h^2 + frac3C^5640h^4 + O(h^6). $$
Using the addition formula $ cos(x+y) = cos x cos y - sin x sin y $
we get the equation
$$ cos ntheta = cos C cos D - sin C sin D . $$
Now $ cos D = 1 - fracC^61152h^4 + O(h)^6 $ and
$ sin D = fracC^324h^2 + frac3C^5640h^4 + O(h)^6. $
Substituting we get
$$ cos ntheta = cos C Big( 1 - fracC^61152h^4 + O(h)^6 Big)
- sin C Big(fracC^324h^2 + frac3C^5640h^4 + O(h)^6 Big) $$
and your result follows.

share|cite|improve this answer

answered Aug 6 at 13:43

Somos

11.7k11033

add a comment | 

up vote
1
down vote



accepted

We are given that
$$ ntheta = C + D ,quad D := fracC^324h^2 + frac3C^5640h^4 + O(h^6). $$
Using the addition formula $ cos(x+y) = cos x cos y - sin x sin y $
we get the equation
$$ cos ntheta = cos C cos D - sin C sin D . $$
Now $ cos D = 1 - fracC^61152h^4 + O(h)^6 $ and
$ sin D = fracC^324h^2 + frac3C^5640h^4 + O(h)^6. $
Substituting we get
$$ cos ntheta = cos C Big( 1 - fracC^61152h^4 + O(h)^6 Big)
- sin C Big(fracC^324h^2 + frac3C^5640h^4 + O(h)^6 Big) $$
and your result follows.

share|cite|improve this answer

answered Aug 6 at 13:43

Somos

11.7k11033

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

We are given that
$$ ntheta = C + D ,quad D := fracC^324h^2 + frac3C^5640h^4 + O(h^6). $$
Using the addition formula $ cos(x+y) = cos x cos y - sin x sin y $
we get the equation
$$ cos ntheta = cos C cos D - sin C sin D . $$
Now $ cos D = 1 - fracC^61152h^4 + O(h)^6 $ and
$ sin D = fracC^324h^2 + frac3C^5640h^4 + O(h)^6. $
Substituting we get
$$ cos ntheta = cos C Big( 1 - fracC^61152h^4 + O(h)^6 Big)
- sin C Big(fracC^324h^2 + frac3C^5640h^4 + O(h)^6 Big) $$
and your result follows.

share|cite|improve this answer

answered Aug 6 at 13:43

Somos

11.7k11033

We are given that
$$ ntheta = C + D ,quad D := fracC^324h^2 + frac3C^5640h^4 + O(h^6). $$
Using the addition formula $ cos(x+y) = cos x cos y - sin x sin y $
we get the equation
$$ cos ntheta = cos C cos D - sin C sin D . $$
Now $ cos D = 1 - fracC^61152h^4 + O(h)^6 $ and
$ sin D = fracC^324h^2 + frac3C^5640h^4 + O(h)^6. $
Substituting we get
$$ cos ntheta = cos C Big( 1 - fracC^61152h^4 + O(h)^6 Big)
- sin C Big(fracC^324h^2 + frac3C^5640h^4 + O(h)^6 Big) $$
and your result follows.

share|cite|improve this answer

answered Aug 6 at 13:43

Somos

11.7k11033

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 13:43

Somos

11.7k11033

answered Aug 6 at 13:43

Somos

11.7k11033

answered Aug 6 at 13:43

Somos

11.7k11033

11.7k11033

add a comment | 

add a comment | 

 

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