Mixing
![Functional equations and Determination of Function [on hold]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
A Better Approximation at Lower Input Values
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I'm doing some research on an algorithm and to give an upper bound I wanted to simplify the term, call it $T$, below.
beginequation
T(i)=sum_k=1^i-1secBig(big(frac34big)^k-1big(45^circbig)Big),
textrm for igeq 2endequation
I've been able to come up with the approximation $T leq i$, and this approximation is good enough after the first few values of $i$, but the lower the value of $i$, the more accurate I need $T$ to be. Is there a more accurate approximation at lower values of $i$ for this?
summation algorithms approximation
asked Jul 27 at 14:55
RoryHector
9012
 |Â
show 2 more comments
up vote
0
down vote
favorite
I'm doing some research on an algorithm and to give an upper bound I wanted to simplify the term, call it $T$, below.
beginequation
T(i)=sum_k=1^i-1secBig(big(frac34big)^k-1big(45^circbig)Big),
textrm for igeq 2endequation
I've been able to come up with the approximation $T leq i$, and this approximation is good enough after the first few values of $i$, but the lower the value of $i$, the more accurate I need $T$ to be. Is there a more accurate approximation at lower values of $i$ for this?
summation algorithms approximation
asked Jul 27 at 14:55
RoryHector
9012
Additional term as i increases tend to 1. Does that help? What lower values of i do you want? The minimum is 2.
– herb steinberg
Jul 27 at 15:00
@herbsteinberg If I had to give an example then I would say $T leq i$ is a good approximation for $i > 10$, but the lower the value of $i$, the more important that $T(i)$ be accurate
– RoryHector
Jul 27 at 15:08
Noe that sec > 1 for each k in you series. My only suggestion is try using the exponential representation $sec(x)=frac2e^ix+e^-ix$, although it might be clumsy.
– herb steinberg
Jul 27 at 15:26
The statement "$Tle i$" is a little confusing, since T is a function of i.
– herb steinberg
Jul 27 at 15:29
It's just saying that the output of the function, $T$, is less than the input of the function, $i$.
– RoryHector
Jul 27 at 16:11
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm doing some research on an algorithm and to give an upper bound I wanted to simplify the term, call it $T$, below.
beginequation
T(i)=sum_k=1^i-1secBig(big(frac34big)^k-1big(45^circbig)Big),
textrm for igeq 2endequation
I've been able to come up with the approximation $T leq i$, and this approximation is good enough after the first few values of $i$, but the lower the value of $i$, the more accurate I need $T$ to be. Is there a more accurate approximation at lower values of $i$ for this?
summation algorithms approximation
asked Jul 27 at 14:55
RoryHector
9012
I'm doing some research on an algorithm and to give an upper bound I wanted to simplify the term, call it $T$, below.
beginequation
T(i)=sum_k=1^i-1secBig(big(frac34big)^k-1big(45^circbig)Big),
textrm for igeq 2endequation
I've been able to come up with the approximation $T leq i$, and this approximation is good enough after the first few values of $i$, but the lower the value of $i$, the more accurate I need $T$ to be. Is there a more accurate approximation at lower values of $i$ for this?
summation algorithms approximation
asked Jul 27 at 14:55
RoryHector
9012
edited Jul 27 at 16:42
asked Jul 27 at 14:55
RoryHector
9012
asked Jul 27 at 14:55
RoryHector
9012
asked Jul 27 at 14:55
RoryHector
9012
9012
Additional term as i increases tend to 1. Does that help? What lower values of i do you want? The minimum is 2.
– herb steinberg
Jul 27 at 15:00
@herbsteinberg If I had to give an example then I would say $T leq i$ is a good approximation for $i > 10$, but the lower the value of $i$, the more important that $T(i)$ be accurate
– RoryHector
Jul 27 at 15:08
Noe that sec > 1 for each k in you series. My only suggestion is try using the exponential representation $sec(x)=frac2e^ix+e^-ix$, although it might be clumsy.
– herb steinberg
Jul 27 at 15:26
The statement "$Tle i$" is a little confusing, since T is a function of i.
– herb steinberg
Jul 27 at 15:29
It's just saying that the output of the function, $T$, is less than the input of the function, $i$.
– RoryHector
Jul 27 at 16:11
 |Â
show 2 more comments
Additional term as i increases tend to 1. Does that help? What lower values of i do you want? The minimum is 2.
– herb steinberg
Jul 27 at 15:00
@herbsteinberg If I had to give an example then I would say $T leq i$ is a good approximation for $i > 10$, but the lower the value of $i$, the more important that $T(i)$ be accurate
– RoryHector
Jul 27 at 15:08
Noe that sec > 1 for each k in you series. My only suggestion is try using the exponential representation $sec(x)=frac2e^ix+e^-ix$, although it might be clumsy.
– herb steinberg
Jul 27 at 15:26
The statement "$Tle i$" is a little confusing, since T is a function of i.
– herb steinberg
Jul 27 at 15:29
It's just saying that the output of the function, $T$, is less than the input of the function, $i$.
– RoryHector
Jul 27 at 16:11
Additional term as i increases tend to 1. Does that help? What lower values of i do you want? The minimum is 2.
– herb steinberg
Jul 27 at 15:00
Additional term as i increases tend to 1. Does that help? What lower values of i do you want? The minimum is 2.
– herb steinberg
Jul 27 at 15:00
@herbsteinberg If I had to give an example then I would say $T leq i$ is a good approximation for $i > 10$, but the lower the value of $i$, the more important that $T(i)$ be accurate
– RoryHector
Jul 27 at 15:08
@herbsteinberg If I had to give an example then I would say $T leq i$ is a good approximation for $i > 10$, but the lower the value of $i$, the more important that $T(i)$ be accurate
– RoryHector
Jul 27 at 15:08
Noe that sec > 1 for each k in you series. My only suggestion is try using the exponential representation $sec(x)=frac2e^ix+e^-ix$, although it might be clumsy.
– herb steinberg
Jul 27 at 15:26
Noe that sec > 1 for each k in you series. My only suggestion is try using the exponential representation $sec(x)=frac2e^ix+e^-ix$, although it might be clumsy.
– herb steinberg
Jul 27 at 15:26
The statement "$Tle i$" is a little confusing, since T is a function of i.
– herb steinberg
Jul 27 at 15:29
The statement "$Tle i$" is a little confusing, since T is a function of i.
– herb steinberg
Jul 27 at 15:29
It's just saying that the output of the function, $T$, is less than the input of the function, $i$.
– RoryHector
Jul 27 at 16:11
It's just saying that the output of the function, $T$, is less than the input of the function, $i$.
– RoryHector
Jul 27 at 16:11
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
Using that $sec x<1+x^2/2$ you can get
$$
T(i)<i-1+fracpi^2126Bigl(9-16Bigl(frac916Bigr)^iBigr)=i-0.709042-1.25328Bigl(frac916Bigr)^i.
$$
This can be improved using more terms in the Taylor series of $sec x$ around $x=0$.
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Using that $sec x<1+x^2/2$ you can get
$$
T(i)<i-1+fracpi^2126Bigl(9-16Bigl(frac916Bigr)^iBigr)=i-0.709042-1.25328Bigl(frac916Bigr)^i.
$$
This can be improved using more terms in the Taylor series of $sec x$ around $x=0$.
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
add a comment |Â
up vote
0
down vote
Using that $sec x<1+x^2/2$ you can get
$$
T(i)<i-1+fracpi^2126Bigl(9-16Bigl(frac916Bigr)^iBigr)=i-0.709042-1.25328Bigl(frac916Bigr)^i.
$$
This can be improved using more terms in the Taylor series of $sec x$ around $x=0$.
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using that $sec x<1+x^2/2$ you can get
$$
T(i)<i-1+fracpi^2126Bigl(9-16Bigl(frac916Bigr)^iBigr)=i-0.709042-1.25328Bigl(frac916Bigr)^i.
$$
This can be improved using more terms in the Taylor series of $sec x$ around $x=0$.
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
Using that $sec x<1+x^2/2$ you can get
$$
T(i)<i-1+fracpi^2126Bigl(9-16Bigl(frac916Bigr)^iBigr)=i-0.709042-1.25328Bigl(frac916Bigr)^i.
$$
This can be improved using more terms in the Taylor series of $sec x$ around $x=0$.
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
answered Jul 27 at 18:13
Julián Aguirre
64.5k23894
64.5k23894
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864460%2fa-better-approximation-at-lower-input-values%23new-answer', 'question_page');
);
Post as a guest
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
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
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
Additional term as i increases tend to 1. Does that help? What lower values of i do you want? The minimum is 2.
– herb steinberg
Jul 27 at 15:00
@herbsteinberg If I had to give an example then I would say $T leq i$ is a good approximation for $i > 10$, but the lower the value of $i$, the more important that $T(i)$ be accurate
– RoryHector
Jul 27 at 15:08
Noe that sec > 1 for each k in you series. My only suggestion is try using the exponential representation $sec(x)=frac2e^ix+e^-ix$, although it might be clumsy.
– herb steinberg
Jul 27 at 15:26
The statement "$Tle i$" is a little confusing, since T is a function of i.
– herb steinberg
Jul 27 at 15:29
It's just saying that the output of the function, $T$, is less than the input of the function, $i$.
– RoryHector
Jul 27 at 16:11