$I_1=int_0^1 frac 1 1+frac 1 sqrt x dx$How to prove that $lim limits_nrightarrow infty fracF_n+1F_n=fracsqrt5+12$Fibonacci nth termArctangents, Fibonacci numbers, and the golden ratioMean and Variance of Fibonacci NumbersEvaluating: $I_1 = intsin^-1 left(sqrtfracxx+a;right) dx$$ I = int_0^1 ((y')^2-y^2)dx $ and $I_1= int_0^1 (y' + y tan x)^2dx$Closed form of series involving Fibonacci numbersLimit involving fibonacci series with an integralCalculation of integral $int_0^afracsqrt xsqrt x + sqrta-xdx$Powers of the golden ratio
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$I_1=int_0^1 frac 1 1+frac 1 sqrt x dx$
How to prove that $lim limits_nrightarrow infty fracF_n+1F_n=fracsqrt5+12$Fibonacci nth termArctangents, Fibonacci numbers, and the golden ratioMean and Variance of Fibonacci NumbersEvaluating: $I_1 = intsin^-1 left(sqrtfracxx+a;right) dx$$ I = int_0^1 ((y')^2-y^2)dx $ and $I_1= int_0^1 (y' + y tan x)^2dx$Closed form of series involving Fibonacci numbersLimit involving fibonacci series with an integralCalculation of integral $int_0^afracsqrt xsqrt x + sqrta-xdx$Powers of the golden ratio
$begingroup$
If we let $I_2= int_0^1 frac 1 1+frac 1 1+sqrt x dx$ and so on, we’re tasked with solving $I_n$. We can show that $I_2 = int frac 1+sqrt x 1-sqrt x + 1 = int frac 1+sqrt x 2+sqrt x $
And that $I_3=int frac 2+sqrt x3+2sqrt x$, and that in general, using $F_n$ to denote the nth Fibonacci number,
$$I_n=int^1_0 fracF_n+F_n-1sqrt xF_n+1+F_n sqrt xdx$$
However, at this point I’m uncertain how to proceed; we obviously have some ratio of Fibonacci’s which looks like the golden ratio, but I’m unsure what substitutions to make to have the golden ratio more clearly emerge as a useful quantity.
calculus integration sequences-and-series fibonacci-numbers
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
$endgroup$
add a comment |
$begingroup$
If we let $I_2= int_0^1 frac 1 1+frac 1 1+sqrt x dx$ and so on, we’re tasked with solving $I_n$. We can show that $I_2 = int frac 1+sqrt x 1-sqrt x + 1 = int frac 1+sqrt x 2+sqrt x $
And that $I_3=int frac 2+sqrt x3+2sqrt x$, and that in general, using $F_n$ to denote the nth Fibonacci number,
$$I_n=int^1_0 fracF_n+F_n-1sqrt xF_n+1+F_n sqrt xdx$$
However, at this point I’m uncertain how to proceed; we obviously have some ratio of Fibonacci’s which looks like the golden ratio, but I’m unsure what substitutions to make to have the golden ratio more clearly emerge as a useful quantity.
calculus integration sequences-and-series fibonacci-numbers
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
$endgroup$
3
$begingroup$
With the substitution $y= sqrtx$, you should be able to compute explicitely the integral... no ?
$endgroup$
– TheSilverDoe
Mar 14 at 15:56
1
$begingroup$
Golden ratio is not going to appear in person, unless you take the limit at $ntoinfty$.
$endgroup$
– Ivan Neretin
Mar 14 at 15:57
add a comment |
$begingroup$
If we let $I_2= int_0^1 frac 1 1+frac 1 1+sqrt x dx$ and so on, we’re tasked with solving $I_n$. We can show that $I_2 = int frac 1+sqrt x 1-sqrt x + 1 = int frac 1+sqrt x 2+sqrt x $
And that $I_3=int frac 2+sqrt x3+2sqrt x$, and that in general, using $F_n$ to denote the nth Fibonacci number,
$$I_n=int^1_0 fracF_n+F_n-1sqrt xF_n+1+F_n sqrt xdx$$
However, at this point I’m uncertain how to proceed; we obviously have some ratio of Fibonacci’s which looks like the golden ratio, but I’m unsure what substitutions to make to have the golden ratio more clearly emerge as a useful quantity.
calculus integration sequences-and-series fibonacci-numbers
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
$endgroup$
If we let $I_2= int_0^1 frac 1 1+frac 1 1+sqrt x dx$ and so on, we’re tasked with solving $I_n$. We can show that $I_2 = int frac 1+sqrt x 1-sqrt x + 1 = int frac 1+sqrt x 2+sqrt x $
And that $I_3=int frac 2+sqrt x3+2sqrt x$, and that in general, using $F_n$ to denote the nth Fibonacci number,
$$I_n=int^1_0 fracF_n+F_n-1sqrt xF_n+1+F_n sqrt xdx$$
However, at this point I’m uncertain how to proceed; we obviously have some ratio of Fibonacci’s which looks like the golden ratio, but I’m unsure what substitutions to make to have the golden ratio more clearly emerge as a useful quantity.
calculus integration sequences-and-series fibonacci-numbers
calculus integration sequences-and-series fibonacci-numbers
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
asked Mar 14 at 15:53
Thor KamphefnerThor Kamphefner
836
836
3
$begingroup$
With the substitution $y= sqrtx$, you should be able to compute explicitely the integral... no ?
$endgroup$
– TheSilverDoe
Mar 14 at 15:56
1
$begingroup$
Golden ratio is not going to appear in person, unless you take the limit at $ntoinfty$.
$endgroup$
– Ivan Neretin
Mar 14 at 15:57
add a comment |
3
$begingroup$
With the substitution $y= sqrtx$, you should be able to compute explicitely the integral... no ?
$endgroup$
– TheSilverDoe
Mar 14 at 15:56
1
$begingroup$
Golden ratio is not going to appear in person, unless you take the limit at $ntoinfty$.
$endgroup$
– Ivan Neretin
Mar 14 at 15:57
3
3
$begingroup$
With the substitution $y= sqrtx$, you should be able to compute explicitely the integral... no ?
$endgroup$
– TheSilverDoe
Mar 14 at 15:56
$begingroup$
With the substitution $y= sqrtx$, you should be able to compute explicitely the integral... no ?
$endgroup$
– TheSilverDoe
Mar 14 at 15:56
1
1
$begingroup$
Golden ratio is not going to appear in person, unless you take the limit at $ntoinfty$.
$endgroup$
– Ivan Neretin
Mar 14 at 15:57
$begingroup$
Golden ratio is not going to appear in person, unless you take the limit at $ntoinfty$.
$endgroup$
– Ivan Neretin
Mar 14 at 15:57
add a comment |
1 Answer
1
active
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votes
$begingroup$
We define $$j=j(a,b)=int_0^1fraca+sqrt xb+sqrt xdx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=int_0^1fraca_1+a_2sqrt xa_3+a_4sqrt xdx=fraca_2a_4jleft(fraca_1a_2,fraca_3a_4right)$$
So then we see that
$$beginalign
j&=int_0^1fraca-b+b+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+int_0^1fracb+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+1
endalign$$
Then we set $x=u^2Rightarrow dx=2udu$:
$$beginalign
int_0^1fracdxb+sqrt x&=2int_0^1fracub+udu\
&=2int_0^1frac-b+b+ub+udu\
&=2-2bint_0^1fracdub+udu\
&=2-2bln|u+b|]_0^1\
&=2+2blnleft|fracbb+1right|
endalign$$
So $$j(a,b)=1+2(a-b)left(1+blnleft|fracbb+1right|right)$$
And your integral is given by $$I_n=j^*(F_n,F_n-1,F_n+1,F_n)=fracF_n-1F_njleft(fracF_nF_n-1,fracF_n+1F_nright)$$
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
We define $$j=j(a,b)=int_0^1fraca+sqrt xb+sqrt xdx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=int_0^1fraca_1+a_2sqrt xa_3+a_4sqrt xdx=fraca_2a_4jleft(fraca_1a_2,fraca_3a_4right)$$
So then we see that
$$beginalign
j&=int_0^1fraca-b+b+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+int_0^1fracb+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+1
endalign$$
Then we set $x=u^2Rightarrow dx=2udu$:
$$beginalign
int_0^1fracdxb+sqrt x&=2int_0^1fracub+udu\
&=2int_0^1frac-b+b+ub+udu\
&=2-2bint_0^1fracdub+udu\
&=2-2bln|u+b|]_0^1\
&=2+2blnleft|fracbb+1right|
endalign$$
So $$j(a,b)=1+2(a-b)left(1+blnleft|fracbb+1right|right)$$
And your integral is given by $$I_n=j^*(F_n,F_n-1,F_n+1,F_n)=fracF_n-1F_njleft(fracF_nF_n-1,fracF_n+1F_nright)$$
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
$endgroup$
add a comment |
$begingroup$
We define $$j=j(a,b)=int_0^1fraca+sqrt xb+sqrt xdx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=int_0^1fraca_1+a_2sqrt xa_3+a_4sqrt xdx=fraca_2a_4jleft(fraca_1a_2,fraca_3a_4right)$$
So then we see that
$$beginalign
j&=int_0^1fraca-b+b+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+int_0^1fracb+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+1
endalign$$
Then we set $x=u^2Rightarrow dx=2udu$:
$$beginalign
int_0^1fracdxb+sqrt x&=2int_0^1fracub+udu\
&=2int_0^1frac-b+b+ub+udu\
&=2-2bint_0^1fracdub+udu\
&=2-2bln|u+b|]_0^1\
&=2+2blnleft|fracbb+1right|
endalign$$
So $$j(a,b)=1+2(a-b)left(1+blnleft|fracbb+1right|right)$$
And your integral is given by $$I_n=j^*(F_n,F_n-1,F_n+1,F_n)=fracF_n-1F_njleft(fracF_nF_n-1,fracF_n+1F_nright)$$
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
$endgroup$
add a comment |
$begingroup$
We define $$j=j(a,b)=int_0^1fraca+sqrt xb+sqrt xdx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=int_0^1fraca_1+a_2sqrt xa_3+a_4sqrt xdx=fraca_2a_4jleft(fraca_1a_2,fraca_3a_4right)$$
So then we see that
$$beginalign
j&=int_0^1fraca-b+b+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+int_0^1fracb+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+1
endalign$$
Then we set $x=u^2Rightarrow dx=2udu$:
$$beginalign
int_0^1fracdxb+sqrt x&=2int_0^1fracub+udu\
&=2int_0^1frac-b+b+ub+udu\
&=2-2bint_0^1fracdub+udu\
&=2-2bln|u+b|]_0^1\
&=2+2blnleft|fracbb+1right|
endalign$$
So $$j(a,b)=1+2(a-b)left(1+blnleft|fracbb+1right|right)$$
And your integral is given by $$I_n=j^*(F_n,F_n-1,F_n+1,F_n)=fracF_n-1F_njleft(fracF_nF_n-1,fracF_n+1F_nright)$$
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
$endgroup$
We define $$j=j(a,b)=int_0^1fraca+sqrt xb+sqrt xdx$$
So we see that
$$j^*(a_1,a_2,a_3,a_4)=int_0^1fraca_1+a_2sqrt xa_3+a_4sqrt xdx=fraca_2a_4jleft(fraca_1a_2,fraca_3a_4right)$$
So then we see that
$$beginalign
j&=int_0^1fraca-b+b+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+int_0^1fracb+sqrt xb+sqrt xdx\&=(a-b)int_0^1fracdxb+sqrt x+1
endalign$$
Then we set $x=u^2Rightarrow dx=2udu$:
$$beginalign
int_0^1fracdxb+sqrt x&=2int_0^1fracub+udu\
&=2int_0^1frac-b+b+ub+udu\
&=2-2bint_0^1fracdub+udu\
&=2-2bln|u+b|]_0^1\
&=2+2blnleft|fracbb+1right|
endalign$$
So $$j(a,b)=1+2(a-b)left(1+blnleft|fracbb+1right|right)$$
And your integral is given by $$I_n=j^*(F_n,F_n-1,F_n+1,F_n)=fracF_n-1F_njleft(fracF_nF_n-1,fracF_n+1F_nright)$$
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
edited Mar 16 at 18:54
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
answered Mar 14 at 18:12
clathratusclathratus
5,1651338
5,1651338
add a comment |
add a comment |
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meYh8 jK4rih,cp r YeuJa7Z4S,4v7uQrk1LH,hqPr7emVpu29ccAhRRi3ACuY
3
$begingroup$
With the substitution $y= sqrtx$, you should be able to compute explicitely the integral... no ?
$endgroup$
– TheSilverDoe
Mar 14 at 15:56
1
$begingroup$
Golden ratio is not going to appear in person, unless you take the limit at $ntoinfty$.
$endgroup$
– Ivan Neretin
Mar 14 at 15:57