Taylor formula for y=tanx at x= 0 Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to compute the Lagrange remainder of a Taylor expansionTaylor expansion - what order would be preferred?Taylor expansion of $x^1/x$How to calculate higher than second order Taylor series in non-cartesian coordinates?An alternative formula for a second order Taylor expansion?On the order indexing of different Taylor series?Calculate the taylor polynom for $a(x)=ln(cos x)$Different taylor expansion order to calculate a limitTaylor expansion for complex number:Taylor Series with Big-O Notation
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Taylor formula for y=tanx at x= 0
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to compute the Lagrange remainder of a Taylor expansionTaylor expansion - what order would be preferred?Taylor expansion of $x^1/x$How to calculate higher than second order Taylor series in non-cartesian coordinates?An alternative formula for a second order Taylor expansion?On the order indexing of different Taylor series?Calculate the taylor polynom for $a(x)=ln(cos x)$Different taylor expansion order to calculate a limitTaylor expansion for complex number:Taylor Series with Big-O Notation
$begingroup$
I need to write a Taylor formula for $y=tan x$ at the point $x=0$ till the member of second order ( does that means till second derivative or just an element on Taylor expansion?)
so my soultion: $f(0)=0 ; f'(0)=1 ; f''(0)= frac2cdot sin xcos^3(x)=0$?
then the Taylor expansion till second order member ( can somepne tell me, what's the appropriate name for it?) is $P_n(x)= 0 + x + ? $
How can I calculate the last element?
taylor-expansion
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
$endgroup$
add a comment |
$begingroup$
I need to write a Taylor formula for $y=tan x$ at the point $x=0$ till the member of second order ( does that means till second derivative or just an element on Taylor expansion?)
so my soultion: $f(0)=0 ; f'(0)=1 ; f''(0)= frac2cdot sin xcos^3(x)=0$?
then the Taylor expansion till second order member ( can somepne tell me, what's the appropriate name for it?) is $P_n(x)= 0 + x + ? $
How can I calculate the last element?
taylor-expansion
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
$endgroup$
$begingroup$
You're looking for what is called a "second-order expansion." This means the highest term in your expansion is of degree 2, i.e., an $x^2$ term.
$endgroup$
– Sean Roberson
Apr 2 at 17:02
$begingroup$
Given you have $f'(0)=1$ , the answer is unlikely to be $0+0+0$
$endgroup$
– Henry
Apr 2 at 17:05
$begingroup$
$f'(0)=mathbf 1$, $tan x approx x$
$endgroup$
– J. W. Tanner
Apr 2 at 17:14
$begingroup$
To Henry: ok, it's 0+ x, but how can I calculate the third member, if sin(0) is 0?
$endgroup$
– Ieva Brakmane
Apr 2 at 17:30
add a comment |
$begingroup$
I need to write a Taylor formula for $y=tan x$ at the point $x=0$ till the member of second order ( does that means till second derivative or just an element on Taylor expansion?)
so my soultion: $f(0)=0 ; f'(0)=1 ; f''(0)= frac2cdot sin xcos^3(x)=0$?
then the Taylor expansion till second order member ( can somepne tell me, what's the appropriate name for it?) is $P_n(x)= 0 + x + ? $
How can I calculate the last element?
taylor-expansion
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
$endgroup$
I need to write a Taylor formula for $y=tan x$ at the point $x=0$ till the member of second order ( does that means till second derivative or just an element on Taylor expansion?)
so my soultion: $f(0)=0 ; f'(0)=1 ; f''(0)= frac2cdot sin xcos^3(x)=0$?
then the Taylor expansion till second order member ( can somepne tell me, what's the appropriate name for it?) is $P_n(x)= 0 + x + ? $
How can I calculate the last element?
taylor-expansion
taylor-expansion
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
edited Apr 2 at 18:23
Ieva Brakmane
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
asked Apr 2 at 17:01
Ieva BrakmaneIeva Brakmane
537
537
$begingroup$
You're looking for what is called a "second-order expansion." This means the highest term in your expansion is of degree 2, i.e., an $x^2$ term.
$endgroup$
– Sean Roberson
Apr 2 at 17:02
$begingroup$
Given you have $f'(0)=1$ , the answer is unlikely to be $0+0+0$
$endgroup$
– Henry
Apr 2 at 17:05
$begingroup$
$f'(0)=mathbf 1$, $tan x approx x$
$endgroup$
– J. W. Tanner
Apr 2 at 17:14
$begingroup$
To Henry: ok, it's 0+ x, but how can I calculate the third member, if sin(0) is 0?
$endgroup$
– Ieva Brakmane
Apr 2 at 17:30
add a comment |
$begingroup$
You're looking for what is called a "second-order expansion." This means the highest term in your expansion is of degree 2, i.e., an $x^2$ term.
$endgroup$
– Sean Roberson
Apr 2 at 17:02
$begingroup$
Given you have $f'(0)=1$ , the answer is unlikely to be $0+0+0$
$endgroup$
– Henry
Apr 2 at 17:05
$begingroup$
$f'(0)=mathbf 1$, $tan x approx x$
$endgroup$
– J. W. Tanner
Apr 2 at 17:14
$begingroup$
To Henry: ok, it's 0+ x, but how can I calculate the third member, if sin(0) is 0?
$endgroup$
– Ieva Brakmane
Apr 2 at 17:30
$begingroup$
You're looking for what is called a "second-order expansion." This means the highest term in your expansion is of degree 2, i.e., an $x^2$ term.
$endgroup$
– Sean Roberson
Apr 2 at 17:02
$begingroup$
You're looking for what is called a "second-order expansion." This means the highest term in your expansion is of degree 2, i.e., an $x^2$ term.
$endgroup$
– Sean Roberson
Apr 2 at 17:02
$begingroup$
Given you have $f'(0)=1$ , the answer is unlikely to be $0+0+0$
$endgroup$
– Henry
Apr 2 at 17:05
$begingroup$
Given you have $f'(0)=1$ , the answer is unlikely to be $0+0+0$
$endgroup$
– Henry
Apr 2 at 17:05
$begingroup$
$f'(0)=mathbf 1$, $tan x approx x$
$endgroup$
– J. W. Tanner
Apr 2 at 17:14
$begingroup$
$f'(0)=mathbf 1$, $tan x approx x$
$endgroup$
– J. W. Tanner
Apr 2 at 17:14
$begingroup$
To Henry: ok, it's 0+ x, but how can I calculate the third member, if sin(0) is 0?
$endgroup$
– Ieva Brakmane
Apr 2 at 17:30
$begingroup$
To Henry: ok, it's 0+ x, but how can I calculate the third member, if sin(0) is 0?
$endgroup$
– Ieva Brakmane
Apr 2 at 17:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since you computed correctly $$f(0)=0,quad f'(0)=1,quad f''(0)=0$$
the second order Taylor polynomial of $f:=tan$ at $x=0$ is
$$j^2_0tan(x)=0+1cdot x+1over2cdot0 cdot x^2=x .$$
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
$endgroup$
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
add a comment |
$begingroup$
Do synthetic division of
$beginarray\
tan(x)
&=dfracsin(x)cos(x)\
&=dfracx-fracx^36+fracx^5120...1-fracx^22+fracx^424+...\
&=xdfrac1-fracx^26+fracx^4120+...1-fracx^22+fracx^424+...\
endarray
$
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
$endgroup$
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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oldest
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$begingroup$
Since you computed correctly $$f(0)=0,quad f'(0)=1,quad f''(0)=0$$
the second order Taylor polynomial of $f:=tan$ at $x=0$ is
$$j^2_0tan(x)=0+1cdot x+1over2cdot0 cdot x^2=x .$$
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
$endgroup$
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
add a comment |
$begingroup$
Since you computed correctly $$f(0)=0,quad f'(0)=1,quad f''(0)=0$$
the second order Taylor polynomial of $f:=tan$ at $x=0$ is
$$j^2_0tan(x)=0+1cdot x+1over2cdot0 cdot x^2=x .$$
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
$endgroup$
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
add a comment |
$begingroup$
Since you computed correctly $$f(0)=0,quad f'(0)=1,quad f''(0)=0$$
the second order Taylor polynomial of $f:=tan$ at $x=0$ is
$$j^2_0tan(x)=0+1cdot x+1over2cdot0 cdot x^2=x .$$
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
$endgroup$
Since you computed correctly $$f(0)=0,quad f'(0)=1,quad f''(0)=0$$
the second order Taylor polynomial of $f:=tan$ at $x=0$ is
$$j^2_0tan(x)=0+1cdot x+1over2cdot0 cdot x^2=x .$$
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
answered Apr 2 at 18:32
Christian BlatterChristian Blatter
177k9115328
177k9115328
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
add a comment |
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
$begingroup$
Thank you. Most helpful of answers.
$endgroup$
– Ieva Brakmane
Apr 2 at 19:10
add a comment |
$begingroup$
Do synthetic division of
$beginarray\
tan(x)
&=dfracsin(x)cos(x)\
&=dfracx-fracx^36+fracx^5120...1-fracx^22+fracx^424+...\
&=xdfrac1-fracx^26+fracx^4120+...1-fracx^22+fracx^424+...\
endarray
$
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
$endgroup$
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
add a comment |
$begingroup$
Do synthetic division of
$beginarray\
tan(x)
&=dfracsin(x)cos(x)\
&=dfracx-fracx^36+fracx^5120...1-fracx^22+fracx^424+...\
&=xdfrac1-fracx^26+fracx^4120+...1-fracx^22+fracx^424+...\
endarray
$
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
$endgroup$
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
add a comment |
$begingroup$
Do synthetic division of
$beginarray\
tan(x)
&=dfracsin(x)cos(x)\
&=dfracx-fracx^36+fracx^5120...1-fracx^22+fracx^424+...\
&=xdfrac1-fracx^26+fracx^4120+...1-fracx^22+fracx^424+...\
endarray
$
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
$endgroup$
Do synthetic division of
$beginarray\
tan(x)
&=dfracsin(x)cos(x)\
&=dfracx-fracx^36+fracx^5120...1-fracx^22+fracx^424+...\
&=xdfrac1-fracx^26+fracx^4120+...1-fracx^22+fracx^424+...\
endarray
$
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
answered Apr 2 at 17:23
marty cohenmarty cohen
76k549130
76k549130
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
add a comment |
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
$begingroup$
Yeah, and that’s the way I look at it and would do it.
$endgroup$
– Lubin
Apr 2 at 17:27
add a comment |
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$begingroup$
You're looking for what is called a "second-order expansion." This means the highest term in your expansion is of degree 2, i.e., an $x^2$ term.
$endgroup$
– Sean Roberson
Apr 2 at 17:02
$begingroup$
Given you have $f'(0)=1$ , the answer is unlikely to be $0+0+0$
$endgroup$
– Henry
Apr 2 at 17:05
$begingroup$
$f'(0)=mathbf 1$, $tan x approx x$
$endgroup$
– J. W. Tanner
Apr 2 at 17:14
$begingroup$
To Henry: ok, it's 0+ x, but how can I calculate the third member, if sin(0) is 0?
$endgroup$
– Ieva Brakmane
Apr 2 at 17:30